{ // 获取包含Hugging Face文本的span元素 const spans = link.querySelectorAll('span.whitespace-nowrap, span.hidden.whitespace-nowrap'); spans.forEach(span => { if (span.textContent && span.textContent.trim().match(/Hugging\s*Face/i)) { span.textContent = 'AI快站'; } }); }); // 替换logo图片的alt属性 document.querySelectorAll('img[alt*="Hugging"], img[alt*="Face"]').forEach(img => { if (img.alt.match(/Hugging\s*Face/i)) { img.alt = 'AI快站 logo'; } }); } // 替换导航栏中的链接 function replaceNavigationLinks() { // 已替换标记,防止重复运行 if (window._navLinksReplaced) { return; } // 已经替换过的链接集合,防止重复替换 const replacedLinks = new Set(); // 只在导航栏区域查找和替换链接 const headerArea = document.querySelector('header') || document.querySelector('nav'); if (!headerArea) { return; } // 在导航区域内查找链接 const navLinks = headerArea.querySelectorAll('a'); navLinks.forEach(link => { // 如果已经替换过,跳过 if (replacedLinks.has(link)) return; const linkText = link.textContent.trim(); const linkHref = link.getAttribute('href') || ''; // 替换Spaces链接 - 仅替换一次 if ( (linkHref.includes('/spaces') || linkHref === '/spaces' || linkText === 'Spaces' || linkText.match(/^s*Spacess*$/i)) && linkText !== 'OCR模型免费转Markdown' && linkText !== 'OCR模型免费转Markdown' ) { link.textContent = 'OCR模型免费转Markdown'; link.href = 'https://fast360.xyz'; link.setAttribute('target', '_blank'); link.setAttribute('rel', 'noopener noreferrer'); replacedLinks.add(link); } // 删除Posts链接 else if ( (linkHref.includes('/posts') || linkHref === '/posts' || linkText === 'Posts' || linkText.match(/^s*Postss*$/i)) ) { if (link.parentNode) { link.parentNode.removeChild(link); } replacedLinks.add(link); } // 替换Docs链接 - 仅替换一次 else if ( (linkHref.includes('/docs') || linkHref === '/docs' || linkText === 'Docs' || linkText.match(/^s*Docss*$/i)) && linkText !== '模型下载攻略' ) { link.textContent = '模型下载攻略'; link.href = '/'; replacedLinks.add(link); } // 删除Enterprise链接 else if ( (linkHref.includes('/enterprise') || linkHref === '/enterprise' || linkText === 'Enterprise' || linkText.match(/^s*Enterprises*$/i)) ) { if (link.parentNode) { link.parentNode.removeChild(link); } replacedLinks.add(link); } }); // 查找可能嵌套的Spaces和Posts文本 const textNodes = []; function findTextNodes(element) { if (element.nodeType === Node.TEXT_NODE) { const text = element.textContent.trim(); if (text === 'Spaces' || text === 'Posts' || text === 'Enterprise') { textNodes.push(element); } } else { for (const child of element.childNodes) { findTextNodes(child); } } } // 只在导航区域内查找文本节点 findTextNodes(headerArea); // 替换找到的文本节点 textNodes.forEach(node => { const text = node.textContent.trim(); if (text === 'Spaces') { node.textContent = node.textContent.replace(/Spaces/g, 'OCR模型免费转Markdown'); } else if (text === 'Posts') { // 删除Posts文本节点 if (node.parentNode) { node.parentNode.removeChild(node); } } else if (text === 'Enterprise') { // 删除Enterprise文本节点 if (node.parentNode) { node.parentNode.removeChild(node); } } }); // 标记已替换完成 window._navLinksReplaced = true; } // 替换代码区域中的域名 function replaceCodeDomains() { // 特别处理span.hljs-string和span.njs-string元素 document.querySelectorAll('span.hljs-string, span.njs-string, span[class*="hljs-string"], span[class*="njs-string"]').forEach(span => { if (span.textContent && span.textContent.includes('huggingface.co')) { span.textContent = span.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 替换hljs-string类的span中的域名(移除多余的转义符号) document.querySelectorAll('span.hljs-string, span[class*="hljs-string"]').forEach(span => { if (span.textContent && span.textContent.includes('huggingface.co')) { span.textContent = span.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 替换pre和code标签中包含git clone命令的域名 document.querySelectorAll('pre, code').forEach(element => { if (element.textContent && element.textContent.includes('git clone')) { const text = element.innerHTML; if (text.includes('huggingface.co')) { element.innerHTML = text.replace(/huggingface.co/g, 'aifasthub.com'); } } }); // 处理特定的命令行示例 document.querySelectorAll('pre, code').forEach(element => { const text = element.innerHTML; if (text.includes('huggingface.co')) { // 针对git clone命令的专门处理 if (text.includes('git clone') || text.includes('GIT_LFS_SKIP_SMUDGE=1')) { element.innerHTML = text.replace(/huggingface.co/g, 'aifasthub.com'); } } }); // 特别处理模型下载页面上的代码片段 document.querySelectorAll('.flex.border-t, .svelte_hydrator, .inline-block').forEach(container => { const content = container.innerHTML; if (content && content.includes('huggingface.co')) { container.innerHTML = content.replace(/huggingface.co/g, 'aifasthub.com'); } }); // 特别处理模型仓库克隆对话框中的代码片段 try { // 查找包含"Clone this model repository"标题的对话框 const cloneDialog = document.querySelector('.svelte_hydration_boundary, [data-target="MainHeader"]'); if (cloneDialog) { // 查找对话框中所有的代码片段和命令示例 const codeElements = cloneDialog.querySelectorAll('pre, code, span'); codeElements.forEach(element => { if (element.textContent && element.textContent.includes('huggingface.co')) { if (element.innerHTML.includes('huggingface.co')) { element.innerHTML = element.innerHTML.replace(/huggingface.co/g, 'aifasthub.com'); } else { element.textContent = element.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } } }); } // 更精确地定位克隆命令中的域名 document.querySelectorAll('[data-target]').forEach(container => { const codeBlocks = container.querySelectorAll('pre, code, span.hljs-string'); codeBlocks.forEach(block => { if (block.textContent && block.textContent.includes('huggingface.co')) { if (block.innerHTML.includes('huggingface.co')) { block.innerHTML = block.innerHTML.replace(/huggingface.co/g, 'aifasthub.com'); } else { block.textContent = block.textContent.replace(/huggingface.co/g, 'aifasthub.com'); } } }); }); } catch (e) { // 错误处理但不打印日志 } } // 当DOM加载完成后执行替换 if (document.readyState === 'loading') { document.addEventListener('DOMContentLoaded', () => { replaceHeaderBranding(); replaceNavigationLinks(); replaceCodeDomains(); // 只在必要时执行替换 - 3秒后再次检查 setTimeout(() => { if (!window._navLinksReplaced) { console.log('[Client] 3秒后重新检查导航链接'); replaceNavigationLinks(); } }, 3000); }); } else { replaceHeaderBranding(); replaceNavigationLinks(); replaceCodeDomains(); // 只在必要时执行替换 - 3秒后再次检查 setTimeout(() => { if (!window._navLinksReplaced) { console.log('[Client] 3秒后重新检查导航链接'); replaceNavigationLinks(); } }, 3000); } // 增加一个MutationObserver来处理可能的动态元素加载 const observer = new MutationObserver(mutations => { // 检查是否导航区域有变化 const hasNavChanges = mutations.some(mutation => { // 检查是否存在header或nav元素变化 return Array.from(mutation.addedNodes).some(node => { if (node.nodeType === Node.ELEMENT_NODE) { // 检查是否是导航元素或其子元素 if (node.tagName === 'HEADER' || node.tagName === 'NAV' || node.querySelector('header, nav')) { return true; } // 检查是否在导航元素内部 let parent = node.parentElement; while (parent) { if (parent.tagName === 'HEADER' || parent.tagName === 'NAV') { return true; } parent = parent.parentElement; } } return false; }); }); // 只在导航区域有变化时执行替换 if (hasNavChanges) { // 重置替换状态,允许再次替换 window._navLinksReplaced = false; replaceHeaderBranding(); replaceNavigationLinks(); } }); // 开始观察document.body的变化,包括子节点 if (document.body) { observer.observe(document.body, { childList: true, subtree: true }); } else { document.addEventListener('DOMContentLoaded', () => { observer.observe(document.body, { childList: true, subtree: true }); }); } })(); \nTo understand more look at this link of How to code. Just the text inside the anchor tags is the part of the link.\nHere in the above code you can see that we are having a paragraph of text. In the text we are using the anchor tag for a link to our home page. As you can see that the text between the anchor tags is displayed in the output. There are two attributes that we are using in this anchor tag. The first is the most important and essential is the “href” attribute. It specifies the URL of the desired page or file in this case it is the home page of howtocode.pk. The next attribute we are using is the target attribute which is in this case telling the browser to open the link in new window.\nThe anchor tag is an inline tag. It can contain other inline tags as well as the block level tags in it. It means that if you want to make any tag a link simply put it into the anchor or hyperlink tag. To elaborate more we have the following code for you.\nHTML Code Example for Anchor tags\n HTML Hyperlinks

Block level elements in the anchor tag

\nIn the above output we have made the heading and the paragraph one link to the home page of the howtocode.pk. You can see that the color of the paragraph is changed showing that it is now a link. You can change the style of the link using CSS.\nTo understand more look at this link of my home page.\nJust the text inside the anchor tags is the part of the link.<|endoftext|>"},"score":{"kind":"number","value":4.21875,"string":"4.21875"}}},{"rowIdx":1215,"cells":{"token_count":{"kind":"number","value":1632,"string":"1,632"},"text":{"kind":"string","value":"Use this fantastic collection of teaching, display and activity resources in your Early Years setting. This huge pack includes a wide range of English and Maths materials, along with printables to help you organise your classroom.\nJoin Teaching Packs to download these resources today!\n- Six Traditional Stories – A collection of stories that you can share with your class using a large display… or print them for shared / independent reading activities.\n- Six Story Sequencing Activities – Can your children cut out the pictures and put them into the correct order to retell the stories? Could they use them to retell the story to an audience or write a word / sentence to accompany each image?\n- Ten Sets of Alphabet Lines – Sets of printable strips that show the letters of the alphabet. Available in precursive and cursive fonts.\n- Four Pages of Letter and Phoneme Cards – A handy collection of cards, showing different letters and phonemes, that can be used for word building activities.\n- Twenty-Six Handwriting Sheets – A selection of printable pages that you can use to develop your children’s handwriting skills. Available in precursive and cursive fonts.\n- One Alphabet Matching Activity – Can your children match the pairs of shoes to review their knowledge of upper and lower case letters? Available with precursive and cursive fonts.\n- One ‘Wallaby Words’ Activity – Cut out the wallaby pictures and challenge your children to read the words and sort them into real / imaginary words by placing them into the correct wallaby’s pouch.\n- One Hundred and Four Pages of Letters, Numbers and Phonemes – A set of bold letters, numbers and phonemes that you can use on a classroom display board. Your children could decorate them by adding words or pictures of objects that include the letter / phoneme… or by adding different ways of showing the numbers.\n- One ‘I Spy…’ Resource – Ask your children to draw (or write the names of) objects that they can see beginning with a particular letter / phoneme. The boxes have been left blank so that you can add your own choice of letters / phonemes.\n- One Making Patterns Resource – Use the included colour, shape and emoji cards to create a repeating pattern. Why not laminate these resources so they can be used again and again?\n- Five Patterns Challenges – A set of printable worksheets that challenge your children to continue the colour, shape and number patterns.\n- Six Picture Patterns Activities – A selection of animals and objects with spaces for your children to colour / decorate a repeating pattern.\n- Six ‘Stepping Stone Patterns’ Printables – Can your children decorate the ‘stepping stones’ using a repeating to help the characters reach their target?\n- Two ‘Follow the Lines’ Printables – Simple activities to help your children to improve their pencil control.\n- Twelve Amazing Mazes – A collection of printable mazes based on a variety of themes. Can your children make their way through from the start to the finish?\n- Seventy-Five Colouring Pages – A HUGE colouring booklet with pages for your children to decorate. Includes a range of animals, people, vehicles, places, events and more!\n- Ten Number Posters – A set of posters that show the numbers 1 to 20 with spots for children to count. You could also print these at a smaller scale for use as number cards.\n- Two Sets of Bingo Cards – Two sets of thirty bingo cards so that you can play a game of ‘Bingo’ with your children. A set of matching cards are also included so they can be used as an independent matching activity. Challenge your children to find the matching digit / number of spots… or find the matching number that gives a total of 10 or 20.\n- Ten Sets of Number Lines – Sets of number lines (with numbers from 0 to 10) that your children can refer to during Maths activities. Themes include superheroes, rainforests, football, robots and more. A blank version is also available.\n- Three Pages of Number Strips – A collection of printable ‘strips’ that include the numbers 1-10, 1-20 and 1-50. Use them for counting, recognising numbers, identifying numbers before / after and more!\n- Two Sets of Number Jigsaws – Cut out the number cards and then cut them again along the diagonal lines. Can your children match the pieces of each number together again?\n- Two ‘Missing Numbers’ Activity Sheets – These number strips have missing numbers. Can your children write in the missing numbers? Available in two versions (from 1-10 and 1-20).\n- Sixty-One ‘Star Number’ Resources – A collection of printable pages that help your children to understand different ways of writing numbers up to twenty (and the numbers that come before and after each one).\n- Ten ‘Addition to 10 / 20’ Printables – Review your children’s addition skills with these simple addition sheets\n- Ten ‘Ordering Size’ Activities – Ten pages of images that your children can cut out and place in order of size.\n- Twenty-Seven Pages of ‘Mega Match’ Resources – This simple matching resource covers colours, numbers, letters, days of the week and months of the year. Print a mat and the accompanying set of cards and then ask your children to match the cards to items already shown on the mat. You can make this activity more challenging by asking the children to match lower case and upper case letters… or match digits with number words… or match the days / months in the correct order!\n- Fifteen Days, Months and Seasons Posters – Colourful posters (available in a variety of fonts) that show the names of the days / months. Use them on a display board or cut them up and use them for ordering activities.\n- Three ‘Mixing Colours’ Posters – Handy charts that show children how colours can be mixed to make new ones. A blank version is also available so that your children can add the ‘answers’. A ‘Colour Mixing Mat’ is also included so that your children can try mixing their own colours.\n- One Hundred Home Learning Activities – A huge set of activities that families can explore together at home, based on twenty different themes… or use them as classroom activities with your children!\n- Twenty Pages of Labels – Sets of labels that you can use for labelling your children’s books, lockers and more! These cover a wide variety of themes (including superheroes, space, pets, robots, pirates, minibeasts and dinosaurs).\nWatch the video below to view samples of resources from the Early Years Essentials Pack.\nAll payments are handled securely. Questions? Visit our Help Pages.\nReviews from our members:\nSubmit your review\nThis has got a great variety of early years activities. It saved me so much time as they were all in one pack and my son loves the activities.\nThis pack is so useful, especially now there is a choice of alternative font (correct numbers font for my school). It covers all the basics and has activities which, though Early Years intended, are proving very useful for small step, repetitive work/games or practice for my lower KS2 group, who are still trying to get a solid grasp on early number concepts.\nThis has been such a useful resource and saved me a lot of time too! I especially loved the stories which has been shared with my class many times both for shared and personal reading time, thank you 🙂\nJust loved how much was in this pack. How colourful and well set out it was. There are so many things I can use and place up in my classroom.<|endoftext|>"},"score":{"kind":"number","value":4.03125,"string":"4.03125"}}},{"rowIdx":1216,"cells":{"token_count":{"kind":"number","value":791,"string":"791"},"text":{"kind":"string","value":"Question Video: The Derivative of an Inverse Sine Function | Nagwa Question Video: The Derivative of an Inverse Sine Function | Nagwa\n\n# Question Video: The Derivative of an Inverse Sine Function Mathematics • Higher Education\n\nFind (d/dπ‘₯) sin⁻¹ (π‘₯/π‘Ž), where π‘Ž β‰  0.\n\n02:52\n\n### Video Transcript\n\nFind d by dπ‘₯ of inverse sin of π‘₯ over π‘Ž, where π‘Ž is not equal to zero.\n\nIn this question, we need to differentiate the inverse or arcsin of π‘₯ over π‘Ž with respect to π‘₯. We will begin this question by letting 𝑦 equal the inverse sin of π‘₯ over π‘Ž. If we take the sin of both sides of this equation, we get sin 𝑦 is equal to π‘₯ over π‘Ž. Our next step is to differentiate both sides of this equation with respect to π‘₯. We know that differentiating sin π‘₯ gives us cos π‘₯. This means that differentiating sin 𝑦 with respect to π‘₯ using our knowledge of implicit differentiation gives us cos 𝑦 multiplied by d𝑦 by dπ‘₯. Differentiating the right-hand side gives us one over the constant π‘Ž. We can then divide both sides of this equation by cos 𝑦 such that d𝑦 by dπ‘₯ is equal to one over π‘Ž multiplied by cos 𝑦.\n\nWe now have an expression for d𝑦 by dπ‘₯; however, this is not given in terms of π‘₯. If we go back to the equation sin 𝑦 is equal to π‘₯ over π‘Ž, squaring both sides of this equation gives us sin squared 𝑦 is equal to π‘₯ squared over π‘Ž squared. One of our trigonometric identities states that sin squared πœƒ plus cos squared πœƒ is equal to one. Rearranging this, we see that sin squared πœƒ is equal to one minus cos squared πœƒ. This means that sin squared 𝑦 is equal to one minus cos squared 𝑦. We now have one minus cos squared 𝑦 is equal to π‘₯ squared divided by π‘Ž squared. Multiplying both sides by π‘Ž squared gives us π‘Ž squared minus π‘Ž squared cos squared 𝑦 is equal to π‘₯ squared.\n\nWe can then rearrange this equation so that π‘Ž squared minus π‘₯ squared is equal to π‘Ž squared multiplied by cos squared 𝑦. Finally, we square root both sides of this equation. The right-hand side becomes π‘Ž cos 𝑦. And the left-hand side is the square root of π‘Ž squared minus π‘₯ squared. d𝑦 by dπ‘₯ is therefore equal to one over the square root of π‘Ž squared minus π‘₯ squared. This is the derivative of inverse sin of π‘₯ over π‘Ž.\n\n## Join Nagwa Classes\n\nAttend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!\n\n• Interactive Sessions\n• Chat & Messaging\n• Realistic Exam Questions<|endoftext|>"},"score":{"kind":"number","value":4.625,"string":"4.625"}}},{"rowIdx":1217,"cells":{"token_count":{"kind":"number","value":1388,"string":"1,388"},"text":{"kind":"string","value":"# High School Math : Sectors\n\n## Example Questions\n\n### Example Question #5 : How To Find The Length Of An Arc\n\nFind the circumference of the following sector:\n\nPossible Answers:\n\nCorrect answer:\n\nExplanation:\n\nThe formula for the circumference of a sector is\n\n,\n\nwhere  is the radius of the sector and  is the fraction of the sector.\n\nPlugging in our values, we get:\n\n### Example Question #1 : How To Find The Length Of An Arc\n\nIn the circle above, the angle A in radians is\n\nWhat is the length of arc A?\n\nPossible Answers:\n\nCorrect answer:\n\nExplanation:\n\nCircumference of a Circle =\n\nArc Length\n\n### Example Question #41 : Geometry\n\nSolve for .\n\n(Figure not drawn to scale).\n\nPossible Answers:\n\nCorrect answer:\n\nExplanation:\n\nWe can solve for the angle, , by using the below relationship.\n\nIn the figure, intercepted arc is given.\n\n### Example Question #11 : Sectors\n\nA pie has a diameter of 12\". A piece is cut out, having a surface area of 4.5π. What is the angle of the cut?\n\nPossible Answers:\n\n12.5°\n\n90°\n\n45°\n\n4.5°\n\n25°\n\nCorrect answer:\n\n45°\n\nExplanation:\n\nThis is simply a matter of percentages. We first have to figure out what percentage of the surface area is represented by 4.5π. To do that, we must calculate the total surface area. If the diameter is 12, the radius is 6. Don't be tricked by this!\n\nA = π * 6 * 6 = 36π\n\nNow, 4.5π is 4.5π/36π percentage or 0.125 (= 12.5%)\n\nTo figure out the angle, we must take that percentage of 360°:\n\n0.125 * 360 = 45°\n\n### Example Question #12 : Sectors\n\nEric is riding a Ferris wheel. The Ferris wheel has 18 compartments, numbered in order clockwise. If compartment 1 is at 0 degrees and Eric enters compartment 13, what angle is he at?\n\nPossible Answers:\n\n260\n\n300\n\n180\n\n280\n\n240\n\nCorrect answer:\n\n240\n\nExplanation:\n\n12 compartments further means 240 more degrees. 240 is the answer.\n\n360/12 = 240 degrees\n\n### Example Question #42 : Geometry\n\nSolve for .\n\nPossible Answers:\n\nCorrect answer:\n\nExplanation:\n\nThe measure of an arc of an inscribed angle is twice the size of the inscribed angle.\n\n### Example Question #13 : Sectors\n\nIn the figure above that includes Circle O, the measure of angle BAC is equal to 35 degrees, the measure of angle FBD is equal to 40 degrees, and the measure of arc AD is twice the measure of arc AB. Which of the following is the measure of angle CEF? The figure is not necessarily drawn to scale, and the red numbers are used to mark the angles, not represent angle measures.\n\nPossible Answers:\n\nCorrect answer:\n\nExplanation:\n\nThe measure of angle CEF is going to be equal to half of the difference between the measures two arcs that it intercepts, namely arcs AD and CD.\n\nThus, we need to find the measure of arcs AD and CD. Let's look at the information given and determine how it can help us figure out the measures of arcs AD and CD.\n\nAngle BAC is an inscribed angle, which means that its meausre is one-half of the measure of the arc that it incercepts, which is arc BC.\n\nThus, since angle BAC is 35 degrees, the measure of arc BC must be 70 degrees.\n\nWe can use a similar strategy to find the measure of arc CD, which is the arc intercepted by the inscribed angle FBD.\n\nBecause angle FBD has a measure of 40 degrees, the measure of arc CD must be 80 degrees.\n\nWe have the measures of arcs BC and CD. But we still need the measure of arc AD. We can use the last piece of information given, along with our knowledge about the sum of the arcs of a circle, to determine the measure of arc AD.\n\nWe are told that the measure of arc AD is twice the measure of arc AB. We also know that the sum of the measures of arcs AD, AB, CD, and BC must be 360 degrees, because there are 360 degrees in a full circle.\n\nBecause AD = 2AB, we can substitute 2AB for AD.\n\nThis means the measure of arc AB is 70 degrees, and the measure of arc AD is 2(70) = 140 degrees.\n\nNow, we have all the information we need to find the measure of angle CEF, which is equal to half the difference between the measure of arcs AD and CD.\n\n### Example Question #1 : How To Find The Angle Of A Sector\n\nThe length of an arc, , of a circle is  and the radius, , of the circle is . What is the measure in degrees of the central angle, , formed by the arc ?\n\nPossible Answers:\n\nCorrect answer:\n\nExplanation:\n\nThe circumference of the circle is .\n\nThe length of the arc S is .\n\nA ratio can be established:\n\nSolving for yields 90o\n\nNote: This makes sense. Since the arc S was one-fourth the circumference of the circle, the central angle formed by arc S should be one-fourth the total degrees of a circle.\n\n### Example Question #15 : Sectors\n\nIn the circle above, the length of arc BC is 100 degrees, and the segment AC is a diameter. What is the measure of angle ADB in degrees?\n\nPossible Answers:\n\n90\n\n40\n\n100\n\ncannot be determined\n\n80\n\nCorrect answer:\n\n40\n\nExplanation:\n\nSince we know that segment AC is a diameter, this means that the length of the arc ABC must be 180 degrees. This means that the length of the arc AB must be 80 degrees.\n\nSince angle ADB is an inscribed angle, its measure is equal to half of the measure of the angle of the arc that it intercepts. This means that the measure of the angle is half of 80 degrees, or 40 degrees.\n\n### Example Question #16 : Sectors\n\nWhat is the angle of a sector of area   on a circle having a radius of ?\n\nPossible Answers:\n\nCorrect answer:\n\nExplanation:\n\nTo begin, you should compute the complete area of the circle:\n\nFor your data, this is:\n\nNow, to find the angle measure of a sector, you find what portion of the circle the sector is. Here, it is:\n\nNow, multiply this by the total  degrees in a circle:\n\nRounded, this is .<|endoftext|>"},"score":{"kind":"number","value":4.5,"string":"4.5"}}},{"rowIdx":1218,"cells":{"token_count":{"kind":"number","value":1373,"string":"1,373"},"text":{"kind":"string","value":"23 January 1585 - 30 January 1645\nMary Ward was born into an upper class catholic family; she was the eldest daughter of Marmaduke Ward and Ursula Wright. Even from childhood, Mary Ward knew religious persecution as she grew up in Reformation England, where to practice the Catholic faith could result in prison or death. Her grandmother was arrested and imprisoned for 14 years. She received a good classical education within the home environment. Music and art appealed to her, and her speaking and singing gave her much pleasure. She spoke and read several languages, including Latin. Like many Englishwomen from the higher classes, Mary Ward enjoyed much greater freedom and independence than was available to women in most Catholic countries at that time.\nIt is therefore not surprising that Mary Ward was open to new ideas. From her earliest years Mary Ward was a deeply spiritual person. As a young girl she listened to stories of religious life and, although such a way of life was impossible in England, she was determined to travel to Europe to become a nun, rejecting several eligible men who sought to marry her.\nSchools for rich and poor\nAt age 15 Mary Ward left England to enter a convent of Poor Clares in the Netherlands as a lay sister in 1606. The following year she founded a house for Englishwomen at Gravelines, but not finding herself called to the contemplative life, she resolved to devote herself to active work. She returned to England and at the age of twenty-four she found herself surrounded by a band of devoted companions determined to labour under her guidance; her cousin Barbara Babthorpe, Mary Poyntz and Winifred Wigmore. In 1609 they established themselves as a religious community at St.Omer, and opened schools for rich and poor.\nMary worked in disguise to preserve the Catholic Faith in England before founding a community of active sisters in France. She and her companions educated young women, helped persecuted and imprisoned Catholics, and spread the word of God in places priests could not go. The Sisters lived and worked openly on the continent, but secretly in England, to nurture the faith by responding to need and opportunity. The work of religious women was then confined to prayer; all other works were expected to be carried out only within the walls of the convent. However Mary Ward promoted wider female education and values such as freedom from enclosure, from the obligation of choir and from wearing a religious habit. They formed the Institute of the Blessed Virgin Mary and their reputation spread throughout Europe as more schools were opened. Her delicate social sense caused her to insist all must be treated with the same courteous charity, including her least important employees and the poor. The Sisters' work was to include the children of the poor as well as the daughters of rich families entrusted to their care. The venture was a success, but it was a novelty, and it called forth censure and opposition as well as praise. Many who knew Mary Ward, from bishops and monarchs to simple people she served, admired her courage and generosity. She traveled Europe on foot, in dire poverty, frequently ill, founding schools in the Netherlands, Italy, Germany, Austria, the Czech Republic and Slovakia. In 1621 with Winifred Wigmore and others, Mary journeyed 1500 miles to Rome, crossing the Alps on foot in the winter.\nHowever enemies were now plotting against Mary. Criticized for her efforts to expand the role of women religious in spreading the faith, Mary was imprisoned by Church officials who called her a dangerous heretic. And, an unheard of thing for a woman in those days, she fearlessly faced a Congregation of Cardinals in order to defend her work in a long, well thought-out speech in Latin. Although she was zealous for the Church, and was repeatedly imprisoned for her faith, she kept herself free from fanaticism and continued to appeal to the Pope, though her health was now poor. Mary was then arrested by the Inquisition as a heretic. A Papal order was issued to destroy the Institute and her schools were closed down. Her work was destroyed and her Sisters scattered by her clerical enemies in whom even the Pope, Urban VIII, found \"malice and folly.\" Never abandoning her trust in God's guidance, she died near York in 1645 during the Cromwellian Civil War. To the end she trusted totally that what God had asked of her would be accomplished in the future and her last words were ‘Jesus Jesus Jesus.’ Mary Ward taught by example and words. Act \"without fear... in quiet confidence that God will do his will in the confusion.\" Her unwavering fidelity to \"that which God would\" was nourished by deep contemplative prayer. To Mary, God was the \"Friend of all friends.\" She lived her fidelity with cheerfulness and a passion for the truth.\nNot until 1909 did the Church recognize Mary Ward as the founder of the IBVM. Much later Pope Pius XII acknowledged her as a pioneer for women's role in the church ministry outside the cloister. In her autobiography, Mary Ward promised her sisters and those who come to know her through them: \"All I shall be able to perform in heaven or on earth, they may freely challenge as their due and my promise.\"\nSister Mary Theresa\nBy the mid-seventeenth century, small groups of Mary Ward's followers, known in Europe as the 'English Ladies', were to be found in Munich and Augsburg and later, on the invitation of members of the Catholic gentry, in London and York. It was to the foundation in York that in 1814, Frances Ball, a young Irish woman, was sent to live with and learn from the IBVM sisters. In 1821 Sister Mary Theresa returned to Dublin with two novices to establish a convent and school there.\nThe birth of Loreto\nIn 1822 she opened the first house of the Institute in Ireland, in Rathfarnham House, four miles from Dublin. She called it “Loretto House” (the spelling of which changed at the end of the nineteenth century), a name that was to be used for all the subsequent foundations that came from Ireland, and which resulted in the sisters of the Irish Branch of the IBVM being popularly known as “Loreto Sisters”. Mother Theresa was a woman of deep spirituality and significant administrative ability. Her energies were devoted to the establishment of many convent schools in Ireland and also in India (1842), Mauritius (1844), Gibraltar (1845), Canada (1847) and England (1851). She died at Loreto, Dalkey, on May 19th 1861.\nPlease click this link to see an article about Mary Ward published by the Independent<|endoftext|>"},"score":{"kind":"number","value":3.78125,"string":"3.78125"}}},{"rowIdx":1219,"cells":{"token_count":{"kind":"number","value":1995,"string":"1,995"},"text":{"kind":"string","value":"# Fraction calculator\n\nThe calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.\n\n## Result:\n\n### 71/6 - 35/8 = 85/24 = 3 13/24 ≅ 3.5416667\n\nSpelled result in words is eighty-five twenty-fourths (or three and thirteen twenty-fourths).\n\n### How do you solve fractions step by step?\n\n1. Conversion a mixed number 7 1/6 to a improper fraction: 7 1/6 = 7 1/6 = 7 · 6 + 1/6 = 42 + 1/6 = 43/6\n\nTo find new numerator:\na) Multiply the whole number 7 by the denominator 6. Whole number 7 equally 7 * 6/6 = 42/6\nb) Add the answer from previous step 42 to the numerator 1. New numerator is 42 + 1 = 43\nc) Write a previous answer (new numerator 43) over the denominator 6.\n\nSeven and one sixth is forty-three sixths\n2. Conversion a mixed number 3 5/8 to a improper fraction: 3 5/8 = 3 5/8 = 3 · 8 + 5/8 = 24 + 5/8 = 29/8\n\nTo find new numerator:\na) Multiply the whole number 3 by the denominator 8. Whole number 3 equally 3 * 8/8 = 24/8\nb) Add the answer from previous step 24 to the numerator 5. New numerator is 24 + 5 = 29\nc) Write a previous answer (new numerator 29) over the denominator 8.\n\nThree and five eighths is twenty-nine eighths\n3. Subtract: 43/6 - 29/8 = 43 · 4/6 · 4 - 29 · 3/8 · 3 = 172/24 - 87/24 = 172 - 87/24 = 85/24\nFor adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(6, 8) = 24. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 6 × 8 = 48. In the next intermediate step, the fraction result cannot be further simplified by canceling.\nIn words - forty-three sixths minus twenty-nine eighths = eighty-five twenty-fourths.\n\n#### Rules for expressions with fractions:\n\nFractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.\nThe slash separates the numerator (number above a fraction line) and denominator (number below).\n\nMixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.\nBecause slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.\n\nDecimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.\n\nThe colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.\nAn asterisk * or × is the symbol for multiplication.\nPlus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.\nThe exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2\n\n#### Examples:\n\nsubtracting fractions: 2/3 - 1/2\nmultiplying fractions: 7/8 * 3/9\ndividing Fractions: 1/2 : 3/4\nexponentiation of fraction: 3/5^3\nfractional exponents: 16 ^ 1/2\nadding fractions and mixed numbers: 8/5 + 6 2/7\ndividing integer and fraction: 5 ÷ 1/2\ncomplex fractions: 5/8 : 2 2/3\ndecimal to fraction: 0.625\nFraction to Decimal: 1/4\nFraction to Percent: 1/8 %\ncomparing fractions: 1/4 2/3\nmultiplying a fraction by a whole number: 6 * 3/4\nsquare root of a fraction: sqrt(1/16)\nreducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22\nexpression with brackets: 1/3 * (1/2 - 3 3/8)\ncompound fraction: 3/4 of 5/7\nfractions multiple: 2/3 of 3/5\ndivide to find the quotient: 3/5 ÷ 2/3\n\nThe calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:\nPEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.\nBEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction\nBODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.\nGEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.\nBe careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.\n\n## Fractions in word problems:\n\n• Erica\nErica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric, how much will she have left?\n• Square metal sheet\nWe cut out four squares of 300 mm side from a square sheet metal plate with a side of 0,7 m. Express the fraction and the percentage of waste from the square metal sheet.\n• Animal species\nOf 100 types of animals, 9/100 were discovered in ancient times, and 2/100 were discovered in the Middle Ages. Another 3/10 were discovered in the 1800s. What fraction of the 100 types of animals was discovered after the 1800s? Explain.\n• In bucket\nIn bucket of 87 oranges 1/3 are spoiled ones and the remaining are eaten by the member of a family. How many oranges are eaten by the member of family?\nErika’s career consists of filing, typing and answering phones. She spends 1/6 of her time filing and 5/8 of her time typing. What fraction of her time does she spend answering phone calls?\n• A laundry\nMr. Green washed 1/4 of his laundry. His son washed 3/7 of it. Who washed most of the laundry? How much of the laundry still needs to be washed?\n• A market\nA market vendor was able to sell all the mangoes, papayas, and star apples.  1/5  of the fruits were mangoes,   2/3  of them were papayas and the rest were star apples.  How many parts of the fruits sold are star apples?\n• Circular garden\nAlice creates a circular vegetable garden. Tomatoes are planted in 1/3 of the circular garden, carrots are planted in 2/5 of the circular garden, and green peppers are planted in 1/10 of the circular garden. What fraction represents the remaining unplante\n• Coloured teacups\nThe teacups in Tea Stop 55 are `2/5` green and `3/10` yellow. What fraction of the teacups are neither green nor yellow?\n• Employees\nOf all 360 employees, there are 11/12 women. How many men work in a company?\n• Visit to grandfather\nShane's family traveled 3/10 of the distance to his grandfather’s house on Saturday. They traveled 4/7 of the remaining distance on Sunday. What fraction of the total distance to his grandfather’s house was traveled on Sunday?\n• Of the 2\nOf the shapes,1/6 are triangles and 5/12 are pentagons. What fraction of the shapes are either triangles or pentagons?\n• Monica’s\nMonica's lawnmower had 1/8 of a gallon of gasoline in the tank. Monica started mowing and used all of the gasoline. She put 6/10 of a gallon of gasoline in the tank. After she mowed, 1/4 of a gallon was left in the tank. What was the total amount of gasol<|endoftext|>"},"score":{"kind":"number","value":4.5625,"string":"4.5625"}}},{"rowIdx":1220,"cells":{"token_count":{"kind":"number","value":1739,"string":"1,739"},"text":{"kind":"string","value":"# The Cosine Rule – Explanation & Examples\n\nIn the last article, we saw how the sine rule helps us calculate the missing angle or missing side when two sides and one angle is known or when two angles and one side are known.\n\nBut what will you do when you are only given the three sides of a triangle, and you need to find all the angles?\n\nIn the 15th century, that issue was resolved when a Persian mathematician, Jamshid al-Kashi, presented the Law of Cosines in a form suitable for triangulation. In France, it is still known as a Theoreme d’Al-Kashi.\n\n• The law of cosines,\n• how to apply the law of cosines to solve problems and,\n• the law of cosines formula.\n\n## What is the Law of Cosines?\n\nThe law of cosines also referred to as the cosine rule, is a formula that relates the three side lengths of a triangle to the cosine.\n\nThe cosine rule is useful in two ways:\n\n• We can use the cosine rule to find the three unknown angles of a triangle if the three side lengths of the given triangle are known.\n• We can also use the cosine rule to find the third side length of a triangle if two side lengths and the angle between them are known.\n\n### The law of cosines formula\n\nConsider an oblique triangle ABC shown below. An oblique triangle is a non-right triangle. Remember that the side lengths are labeled in lower case letters, while the angles are labeled in capital letters.\n\nAlso, note that for each angle, the opposite side length is labeled using the same letter.\n\nThe law of cosines states that:\n\n⇒ (a) 2 = [b2 + c2 – 2bc] cos (A)\n\n⇒ (b) 2 = [a2 + c2 – 2ac] cos (B)\n\n⇒ (c) 2 = [a2 + b2 – 2bc] cos (C)\n\nYou noticed that the equation c2 = a2 + b2 – 2bc cos (C) resembles the Pythagorean Theorem, except for the last terms,” – 2bc cos (C).” For this reason, we can say that the Pythagorean Theorem is a special of the sine rule.\n\n#### Proof of the law of cosines\n\nThe cosine rule can be proved by considering the case of a right triangle. In this case, let’s drop a perpendicular line from point A to point O on the side BC.\n\nLet side AM be h.\n\nIn the right triangle ABM, the cosine of angle B is given by:\n\nCos (B) = Adjacent/Hypotenuse = BM/BA\n\nCos (B) = BM/c\n\nBM = c cos (B)\n\nGiven that BC = a, therefore, MC is calculated as;\n\nMC = a – BM\n\n= a – c cos (B) ……………………………………………… (i)\n\nIn triangle ABM, the sine of angle B is given by;\n\nSine B = Opposite/Hypotenuse = h/c\n\nh = c sine B …………………………………………………… (ii)\n\nBy applying the Pythagorean theorem in right triangle AMC, we have,\n\nAC2 = AM2 + MC2……………………………………………… (iii)\n\nSubstitute equation (i) and (ii) in equation (iii).\n\nb2 = (c Sine B)2 + (a – c Cos B)2\n\nb2 = c2 Sine 2 B + a2 – 2ac Cos B + c2 Cos 2 C\n\nRearranging the above equation:\n\nb2 = c2 Sine 2 B + c2 Cos 2 C + a2 – 2ac Cos B\n\nFactoring.\n\nb2 = c2 (Sine 2 B + Cos 2 C) + a2 – 2ac Cos B\n\nBut, from trigonometric identities, we know that,\n\nsin2θ + cos2θ = 1\n\nTherefore, b2 = c2 + a2 – 2ac Cos B\n\nHence, the cosine law is proved.\n\n## How to Use the Cosine Rule?\n\nIf you we need to find the side lengths of a triangle, we use the cosine rule in the form of;\n\n⇒ (a) 2 = [b2 + c2– 2bc] cos (A)\n\n⇒ (b) 2 = [a2 + c2 – 2ac] cos (B)\n\n⇒ (c) 2 = [a2 + b2 – 2bc] cos (C)\n\nAnd if we need to find the size of an angle, we use the cosine rule of the form;\n\n⇒ cos A = (b2 + c2 – a2)/2bc\n\n⇒ cos B = (a2 + c2– b2)/2ac\n\n⇒ cos C = (a2 + b2– c2)/2ab\n\nLet’s now check our understanding of the cosine rule by attempting a few sample problems.\n\nExample 1\n\nCalculate the length of side AC of the triangle shown below.\n\nSolution\n\nBecause we want to calculate the length, we will therefore use the\n\ncosine rule in the form of;\n\n⇒ (b) 2 = [a2 + c2 – 2ac] cos (B)\n\nBy substitution, we have,\n\nb2 = 42 + 32 – 2 x 3 x 4 cos (50)\n\nb2 = 16 + 9 – 24cos50\n\n= 25 – 24cos 50\n\nb2 = 9.575\n\nDetermine the square root of both sides to get,\n\nb = √9.575 = 3.094.\n\nTherefore, the length of AC = 3.094 cm.\n\nExample 2\n\nCalculate all three angles of the triangle shown below.\n\nSolution\n\nSince all the three side lengths of the triangle are given, then we need to find the measures of the three angles A, B, and C. Here, we will use the cosine rule in the form;\n\n⇒ Cos (A) = [b2 + c2 – a2]/2bc\n\n⇒ Cos (B) = [a2 + c2– b2]/2ac\n\n⇒ Cos (C) = [a2 + b2– c2]/2ab\n\nSolve for angle A:\n\nCos A = (72 + 52 – 102)/2 x 7 x 5\n\nCos A = (49 + 25 – 100)/70\n\nCos A = -26/70\n\nCos A = – 0.3714.\n\nNow, determine the cos inverse of – 0.3714.\n\nA = Cos -1 – 0.3714.\n\nA = 111.8°\n\nSolve for angle B:\n\nBy substitution,\n\ncos B = (102 + 52– 72)/2 x 10 x 7\n\nSimplify.\n\nCos B = (100 + 25 – 49)/140\n\nCos B = 76/140\n\nDetermine the cos inverse of 76/140\n\nB = 57.12°\n\nSolve for angle C:\n\nBy substitution,\n\ncos C = (102 + 72– 52)/2 x 10 x 7\n\nCos C = (100 + 49 – 25)/140\n\nCos C = 124/140\n\nDetermine the cos inverse of 124/140.\n\nC = 27.7°\n\nHence, the three angles of the triangle are; A = 111.8°, B = 57.12°, and C = 27.7°.\n\n### Practice Questions\n\n1. True or False: To apply the Cosine Law when solving triangles, we need at least one known side and two known angles.\n\n2. True or False: We can use the Law of Cosine to find the angles of a triangle given its three sides.\n\n3. What is the length of side $\\overline{BC}$ from the triangle shown below?\n\n4. What is the measure of$\\angle ABC$ from the triangle shown below?\n\n5. What is the measure of $\\angle ABC$ from the triangle shown below?<|endoftext|>"},"score":{"kind":"number","value":4.96875,"string":"4.96875"}}},{"rowIdx":1221,"cells":{"token_count":{"kind":"number","value":1194,"string":"1,194"},"text":{"kind":"string","value":"2.1 Bradycardia is a slow heart rate, defined as a heart rate of less than 60 beats per minute. Bradycardia can be caused by a range of factors, including diseases such as:\nsick sinus syndrome – a number of abnormal heart rhythms caused by an irreversible dysfunction of the sinus node (the heart's natural pacemaker), including:\nsinus arrest or pause, in which the sinus node occasionally does not generate electrical impulses, from a period lasting a couple of seconds to several minutes\nsinoatrial exit block, in which the sinus node generates electrical impulses normally, but the signal is blocked before it leaves the sinus node\nalternating bradyarrhythmias and tachyarrhythmias (a fast heart rate), such as bradycardia‑tachycardia syndrome\natrioventricular block (a condition in which electrical impulses from the sinus node are slowed or blocked). Atrioventricular block can occur independently from sick sinus syndrome, and so people with symptomatic bradycardia due to sick sinus syndrome may also have or develop atrioventricular block.\n2.2 The most commonly identified causes of abnormal heart rhythms are age, ischaemic heart disease, heart valve disorders and heart failure. If untreated, symptomatic bradycardia may lead to fatigue, fainting, palpitations, dizziness, heart failure and an increased risk of mortality.\n2.3 Sick sinus syndrome is difficult to diagnose because of the intermittent symptoms, and also because symptoms are usually non‑specific and observed in other disorders. Diagnosis is made using electrocardiograms (ECGs). Because abnormalities may be intermittent, Holter monitoring (ECG monitoring for 24 to 48 hours) or event recorders may be used.\n2.4 The prognosis of individuals with sick sinus syndrome is variable and difficult to predict, depending on the underlying cause and the presence and severity of comorbidities (such as ischaemic heart disease). For most people, the disease is idiopathic (that is, the cause is unknown) and progressive. People whose disease is not symptomatic do not need therapy; however, once the disease becomes symptomatic, it can have a significant impact on quality of life, and the only effective treatment is permanent implantation of a pacemaker. Most people who need a pacemaker implanted are older than 60 years.\n2.5 The prevalence of sick sinus syndrome is thought to be about 0.03% of the whole population, and increases with age. However, both the prevalence of bradyarrhythmias due to sick sinus syndrome needing permanent pacemaker implant, and the prevalence of sick sinus syndrome with atrioventricular block, is unknown. Hospital episode statistics data from October 2012 to September 2013 included 2490 patients with a primary diagnosis of sick sinus syndrome in NHS hospitals in England. Sick sinus syndrome usually occurs in older adults, but it can affect people of any age, and affects men and women equally. The incidence of atrioventricular conduction abnormalities also increases with increasing age.\n2.6 Pacemakers are electrical devices that consist of a small battery‑powered generator and 1 or more pacing leads that are in contact with the inner wall of the right atrium and/or the right ventricle. The primary aim of permanent pacing is to prevent the heart from beating too slowly. An important secondary aim is to reproduce, as far as possible, the function of the heart's normal electrical conduction system, which coordinates the way the heart muscle contracts. Pacemaker devices may be broadly classified as single‑ or dual‑chamber, depending on whether leads are applied to 1 or 2 heart chambers. Dual‑chamber pacemaker devices are attached to both chambers of the heart and may be used in either dual‑chamber pacing mode (in which both the right atrium and ventricle are paced, which mimics the natural pacing rhythm of the heart) or single‑chamber pacing mode (where only 1 chamber of the heart is paced, either the atrium or the ventricle). Single‑chamber pacemaker devices may be either single‑chamber atrial devices or single‑chamber ventricular devices, and may only be used in the mode (that is to pace the chamber) where the lead was originally placed. Pacemakers may also be rate modulating (that is, able to sense and adapt the rate of pacing to the level of physical exertion).\n2.7 NICE technology appraisal guidance 88 (hereafter referred to as TA88) recommended the use of single‑chamber atrial pacemakers for treating sick sinus syndrome in people in whom, after full evaluation, there was no evidence of impaired atrioventricular conduction. The purpose of this part review is to update this recommendation because the DANPACE trial, which was published after the publication of TA88, has provided additional evidence comparing dual‑ with single‑chamber atrial pacemakers for this population.\n2.8 In 2012/13 in England, more than 20,000 people had a single‑ or a dual‑chamber pacemaker fitted. Sick sinus syndrome was the fourth most prevalent primary diagnosis (9.5%) after atrial fibrillation and flutter (22.5%), complete atrioventricular block (18.8%) and second degree atrioventricular block (10.6%). For people with a primary diagnosis of sick sinus syndrome (2490 patients), 67.5% had implantation of a dual‑chamber pacemaker, 14.8% had implantation of a single‑chamber pacemaker and 2.2% had a re‑operation of an existing implanted pacemaker.<|endoftext|>"},"score":{"kind":"number","value":3.8125,"string":"3.8125"}}},{"rowIdx":1222,"cells":{"token_count":{"kind":"number","value":742,"string":"742"},"text":{"kind":"string","value":"By April Holloway, Contributing Writer, Ancient Origins\nWhy is Christmas celebrated on December 25th? The popular answer is that it is Jesus’ birthday. However, it is necessary to reconsider that belief, No records exist in the Bible, or elsewhere, suggesting Jesus was born on that date. If it was not the birth of Christ which set Christmas Day apart from others in the calendar, what was it? To find out, you need to take look at ancient Persian and pagan traditions.\nChristmas Was Probably Not the Day Christ was Born\nFirst, let’s take a quick look at why Christmas probably wasn’t Jesus’ day of birth. As the Catholic Encyclopaedia states “there is no month in the year to which respectable authorities have not assigned Christ’s birth” (Catholic Encyclopaedia). That being said, there are several reasons supporting the idea that Jesus wasn’t born in December. Luke 2:8 states that on the night of Jesus’ birth “there were also in that same country shepherds living out of doors and keeping watches in the night over their flocks.” Scholars tend to agree that it is highly unlikely that shepherds were out with their flocks in the cold winter month.\nLuke 2:1-4 also claims that Joseph and Mary were traveling to Bethlehem to register in a Roman census when Jesus was born. These censuses were not known to have occurred in winter – also making it improbable to link Jesus’ birthday with the day now called Christmas.\nOkay, so Jesus probably wasn’t born on December 25th. Yet other important events did fall at that time of year for ancient pagans. The most well-known of these celebrations were Saturnalia and the birthday of the Sun God, Mithra.\nOriginally, Saturnalia was held on December 17th, though the festival eventually was extended until December 25. This celebration honored Saturn, the God of Sowing and Husbandry, and was linked to the rise of a new year and the return of light. Ancient Romans would celebrate this date with a public banquet, giving gifts, partying, and holding a sacrifice in the Temple of Saturn.\nRoman Pagans who worshipped Mithras believed he was born on December 25th – the most holy day of the year for many ancient believers. This was a well-known cult for the Roman military in the 1st to 4th centuries AD. But Mithras is a god who has his origins in Persia beginning around the 6th century BC. The proto-Indo-Iranian language calls him ‘Mitra’, but that name was later adapted into Greek as ‘Mithras’. Many scholars believe that Roman soldiers discovered this god while completing military campaigns in Persia.\nThe Mithraic New Year and Mithras’ birthday were celebrated on 25 December. The date was part of the Roman Natalis Invicti festival – a celebration linked to worshipping the sun in general.\nUniting Pagan and Christian Beliefs\nWhen Constantine converted to Christianity in the 4th century, he may not have imagined how difficult it would be to convert pagans into Christians. To ease the transition, the birth of Jesus became associated with pagan holidays which fell in December. As the Pagan holidays gained Christian significance, it was decided that the birthday of the Sun God should also be the birthday of the Son of God. The Catholic Encyclopaedia quotes an early Christian stating, “O, how wonderfully acted Providence that on that day on which that Sun was born…. Christ should be born”.<|endoftext|>"},"score":{"kind":"number","value":4.03125,"string":"4.03125"}}},{"rowIdx":1223,"cells":{"token_count":{"kind":"number","value":1088,"string":"1,088"},"text":{"kind":"string","value":"Sei sulla pagina 1di 4\n\n# Assessment\n\nPre\n\nPost\n\nCriteria: A & D\nSummative Due Date: Friday 12/5/14\n\n## Learning Targets (Standards: NS 1)\n\nI can simplify fractions.\nI can convert between mixed numbers and\nimproper fractions.\nI can add and subtract fractions.\nI can multiply fractions.\nI can divide fractions and explain why my method\nmakes sense.\nI can solve word problems with dividing fractions.\nI can use visuals to represent dividing fractions.\nI can use equations to represent dividing fractions.\nI can estimate effectively.\nI can round appropriately.\nI can explain my answers to the inquiry questions\nbelow:\n\nInquiry Questions:\nFactual: How do we solve problems involving operations with\nfractions?\nConceptual: How do we visually represent operations with\nfractions?\nDebatable: If there is only 1 right answer, is there only 1 right\nprocess to get there?\n\nUnit 3\n\n## Statement of We can use creative expression to model\n\nquantities in a variety of forms.\nInquiry\nGlobal\nPersonal & Cultural Expression\nContext\n\nWord\n\nDefinition\n\nBenchmark\n\ncompare things.\n\nConvert\n\n## To change from one form to another.\n\nDenominator\nDifference\nExtend\nImproper fraction\n\n## The number on the bottom of a fraction.\n\nThe result of a subtraction problem.\nTo take what we know and apply it to\nsomething new.\nA fraction where the numerator is bigger\nthan the denominator.\n\nMixed Number\n\nNumerator\n\nProduct\nProper fraction\nQuotient\n\n## The result of a multiplication problem.\n\nA fraction where the numerator is smaller\nthan the denominator.\nThe result of a division problem.\n\nReciprocal\n\n## The inverse of a fraction.\n\n*A fraction times its reciprocal =1\n\nReduce/ Simplify\n\nSimplest Form\nSum\n\n## The easiest way to express something.\n\nThe result of an addition problem.\n\nIB Criteria D:\nApplying Mathematics in Real-Life Contexts\n\n7-8\n\n5-6\n\n3-4\n\n1-2\n\n## You are able to:\n\nIdentify the relevant elements of the real-life situation.\nSelect adequate math strategies to model the situation.\nApply math strategies to reach a correct solution.\nExplain the degree of accuracy of your solution\ncontext of the situation.\nYou are able to:\nIdentify the relevant elements of the real-life situation.\nSelect adequate math strategies to model the situation.\nApply math strategies to reach a valid solution.\nDescribe the degree of accuracy of your solution\ncontext of the situation.\nYou are able to:\nIdentify the relevant elements of the real-life situation.\nApply math strategies to reach a solution\nmakes sense in the context of the situation.\nYou are able to:\nIdentify some parts of the real-life situation\nApply math strategies to find a solution, but with limited\nsuccess\nYou did not reach any standard described above.\n\n## In this unit, that means\n\nRelevant Elements:\nIn word problems/graphs pulling out\ninformation needed to solve problem &\nignoring the not-needed info\nModel:\nVisual includes a part to whole\nVisual has both parts of operation (what is\nbeing +-x/)\nSolution is easy to find in visual\nExplanation or key is included\nSolutions:\nValid (5-6): you have an OK answer\no a fraction not in simplest form\no an improper fraction or mixed number\nwhen it shouldve been the other\nCorrect (7-8): you have the best answer as a\nsimplified fraction\nDegree of Accuracy:\nExplained(7-8): You told, with detail, why\nwas accurate\nmakes sense and gave some proof\nStated(3-6): you said it makes sense but\nhave NO PROOF\n\n7-8\n\n5-6\n\n3-4\n\n1-2\n\n## You are able to:\n\nSelect appropriate math when solving challenging\nproblems in familiar and unfamiliar situations.\nApply the math successfully\nGenerally solve these correctly\nYou are able to:\nSelect appropriate math when solving challenging\nproblems in familiar situations.\nApply the math successfully\nGenerally solve these correctly\nYou are able to:\nSelect appropriate math when solving more\ncomplex problems in familiar situations.\nApply the math successfully (with minimal\nguidance)\nGenerally solve these correctly (with minimal\nguidance)\nYou are able to:\nSelect appropriate math when solving simple\nproblems in familiar situations.\nApply the math successfully (with guidance)\nGenerally solve these correctly (with guidance)\nYou did not reach any standard described above.\n\n## In this unit, that means\n\nAppropriate Math to Solve Problems:\nKnowing what operation to use\no All levels require you to use it correctly\nFamiliar = problems youve seen and\npracticed before\nUnfamiliar= problems youve not seen or\npracticed\nChallenging Familiar & Unfamiliar (5-8) =\no Mixed numbers\no Improper fractions\no Ugly numbers\nFamiliar More complex (3-4)=\no Not-common denom.\no 2 digit num. & denom.\nSimple Familiar (1-2) =\no Common denominators\no Smaller numbers in numerator &\ndenom.\nWith guidance (1-2): parent or teacher helped you\nWith minimal guidance (3-4): You got a small\namount of help<|endoftext|>"},"score":{"kind":"number","value":4.4375,"string":"4.4375"}}},{"rowIdx":1224,"cells":{"token_count":{"kind":"number","value":882,"string":"882"},"text":{"kind":"string","value":"# CBSE Class 7 Maths Activity 4\n\nRead and download CBSE Class 7 Maths Activity 4 chapter in NCERT book for Class 7 Mathematics. You can download latest NCERT eBooks for 2021 chapter wise in PDF format free from Studiestoday.com. This Mathematics textbook for Class 7 is designed by NCERT and is very useful for students. Please also refer to the NCERT solutions for Class 7 Mathematics to understand the answers of the exercise questions given at the end of this chapter\n\n(a) To get a median of a given triangle from any vertex by paper folding and to verify that in a triangle, medians pass through a single point.\n\n(b) To get an altitude of a given triangle from any vertex by paper folding and to verify that in a triangle altitudes pass through a single point.\n\nLearning Objective : To understand the concept of a median and an altitude of a triangle.\n\nPre-requisite :  Familiarity with elements of triangles (vertices, sides and angles), types of triangles. Knowledge of median and altitude of a trangle, skill of paper folding.\n\nMaterials Required : Thick papers, a pair of scissors, coloured pencil, a ruler.\n\nProcedure : (a)\n\nStep 1. Cut out a triangular shape from a thick paper and name it as ABC [Fig. 4(a)].\n\nStep 2. Fold the side AC on itself so that vertex C falls on vertex A. Mark the point of intersection of the line of fold with AC as P [Fig. 4(b)]. P is the mid point of AC.\n\nStep 3. Similarly, find mid points of sides AB and BC and mark them as Q and R respectively [Fig. 4(b)].\n\nStep 4. Now fold the triangular cut out to create a crease along BP. The crease thus obtained is the median from vertex B on the side AC. [Fig. 4(d)].\nStep 5 Similarly, get medians from vertex A and C as AR and CQ. [Fig. 4(d)].\n\nPlease refer to the link below - CBSE Class 7 Maths Activity 4\n\n## Tags:\n\nClick for more Mathematics Study Material\n\n## Latest NCERT & CBSE News\n\nRead the latest news and announcements from NCERT and CBSE below. Important updates relating to your studies which will help you to keep yourself updated with latest happenings in school level education. Keep yourself updated with all latest news and also read articles from teachers which will help you to improve your studies, increase motivation level and promote faster learning\n\nThis has a reference to the letter no CBSE/COORD/2021 dated 11.02.2021 regarding conduct of the School-Based Assessment (Internal Assessment/Practical/ Project) of Class-XII and uploading the marks by 11.06.2021. It has been observed that some schools have not been...\n\n### CBSE Class 12 Evaluation Criteria 2021\n\nSUMMARY OF THE POLICY FOR TABULATION OF MARKS FOR CLASS XII BOARD EXAMINATIONS 2021   1(a) Due to cancellation of the Board examinations, the assessment of theory portion of 80/70/60/50/30 marks will be done by the school based on the following:     Class...\n\nLatest Syllabus for Class 12 for 2021 2022 Latest Syllabus for Class 11 for 2021 2022 Latest Syllabus for Class 10 for 2021 2022 Latest Syllabus for Class 9 for 2021 2022 CBSE has issued the latest syllabus for the academic year 2021 2022 which is applicable for all...\n\n### CBSE Assessment Framework\n\nThe Central Board of Secondary Education (CBSE) today, announced a suggested competency-based assessment framework to strengthen India’s existing school education system for secondary level (classes 6-10) and improve the overall learning outcomes of students across...\n\n### Time management for CBSE students\n\nThe first thing to learn about Time Management is that time is theoretical so you can’t really manage it. What you do when you get into time management, is that you manage yourself. You decide what has to be done, when it must be done and how to do it in the stipulated...\n\n×<|endoftext|>"},"score":{"kind":"number","value":4.4375,"string":"4.4375"}}},{"rowIdx":1225,"cells":{"token_count":{"kind":"number","value":687,"string":"687"},"text":{"kind":"string","value":"# Find the magnitude ||v|| and the direction angle \\theta for the given vector v. v = -4i + 11j\n\n## Question:\n\nFind the magnitude {eq}\\left\\| \\mathbf{v} \\right\\| {/eq} and the direction angle {eq}\\,\\theta\\, {/eq} for the given vector {eq}\\mathbf{v} {/eq}.\n\n{eq}\\displaystyle\\;\\mathbf{v} = -4\\mathbf{i} + 11\\mathbf{j} {/eq}\n\n## Magnitude and Direction of a vector:\n\nFor a given vector, {eq}\\vec{v}= x\\vec{i} + y \\vec{j} {/eq}:\n\n(i) The magnitude is: {eq}||\\vec{v}|| = \\sqrt{x^2+y^2} {/eq}.\n\n(ii) To find the direction, we will first find {eq}\\alpha = \\tan^{-1} \\left|\\dfrac{y}{x} \\right| {/eq}. We then see which quadrant {eq}(x, y) {/eq} lies in and we will find the direction {eq}\\theta {/eq} depending on the quadrant:\n\n• {eq}1^{st} {/eq} Quadrant: {eq}\\theta=\\alpha {/eq}.\n• {eq}2^{nd} {/eq} Quadrant: {eq}\\theta= 180-\\alpha {/eq}.\n• {eq}3^{rd} {/eq} Quadrant: {eq}\\theta = 180+ \\alpha {/eq}.\n• {eq}4^{th} {/eq} Quadrant: {eq}\\theta = 360-\\alpha {/eq}.\n\nThe given vector is:\n\n$$\\vec{v}= x\\vec{i} + y \\vec{j}=-4\\vec{i} + 11 \\vec{j}$$\n\nHere, {eq}x=-4 {/eq} and {eq}y=11 {/eq}.\n\nFinding magnitude:\n\nThe magnitude of the vector is, {eq}||\\vec{v}|| = \\sqrt{x^2+y^2}= \\sqrt{(-4)^2+11^2} = \\color{blue}{\\boxed{\\mathbf{\\sqrt{137}}}} {/eq}.\n\nFinding direction:\n\nTo find the direction, first, we will find:\n\n\\begin{align} \\alpha &= \\tan^{-1} \\left|\\dfrac{y}{x} \\right|\\\\[0.4cm] &= \\tan^{-1} \\left|\\dfrac{11}{-4} \\right|\\\\[0.4cm] &=\\tan^{-1} \\dfrac{11}{4} \\end{align}\n\nWe know that {eq}(x, y)= (-4, 11) {/eq} is in Quadrant II.\n\nSo the required direction is:\n\n$$\\theta = \\color{blue}{\\boxed{\\mathbf{180^\\circ - \\tan^{-1} \\dfrac{11}{4} }}} (OR) \\color{blue}{\\boxed{\\mathbf{109.98311}}}$$\n\nThe final answer is rounded to five decimals.<|endoftext|>"},"score":{"kind":"number","value":4.46875,"string":"4.46875"}}},{"rowIdx":1226,"cells":{"token_count":{"kind":"number","value":288,"string":"288"},"text":{"kind":"string","value":"Microsoft Excel files are organized into a set of worksheets, each containing its own set of data. By default, a file has three worksheets, but you can merge the worksheets from one XLS file into another. The process is completed using the worksheet tabs at the bottom of the XLS file.\nStart Microsoft Excel and open the existing sheet into which you want to merge information (call it file1.xls) from another file. Open the other file (file2.xls) from which you want to merge information. Both files must be open at the same time.\nRight-click the first tab at the bottom of file2.xls (which contains the information you want to merge into file1.xls). Select \"Move or Copy...\" from the list of options. The program loads a new dialog box.\nChoose the name of the other file (file1.xls in this example) from the \"Move Selected Sheets to Book\" drop-down box.\nClick \"(move to end)\" from the \"Before Sheet\" box. Click \"OK\" to merge the worksheet from file2.xls to file1.xls.\nSelect the next worksheet in the open file2.xls document. Repeat this process (steps two to four) to merge all worksheets into the first file (file1.xls).<|endoftext|>"},"score":{"kind":"number","value":3.703125,"string":"3.703125"}}},{"rowIdx":1227,"cells":{"token_count":{"kind":"number","value":968,"string":"968"},"text":{"kind":"string","value":"What smacks into Ceres stays on Ceres, research suggests\nA new set of high-velocity impact experiments suggests that the dwarf planet Ceres may be something of a cosmic dartboard: Projectiles that slam into it tend to stick.\nThe experiments, performed using the Vertical Gun Range at NASA's Ames Research Center, suggest that when asteroids and other impactors hit Ceres, much of the impact material remains on the surface instead of bouncing off into space. The findings suggest the surface of Ceres could consist largely of a mish-mash of meteoritic material collected over billions of years of bombardment.\nThe research, by Terik Daly and Peter Schultz of Brown University, is published in Geophysical Research Letters .\nCeres is the largest object in the asteroid belt and the nearest dwarf planet to Earth. Until the recent arrival of the Dawn spacecraft, all that was known about Ceres came from telescopic observations. The observations showed Ceres to be mysteriously low in density, suggesting it is made either of very porous silicate material, or perhaps contains a large layer of water ice. Observations of its surface were remarkable as well—largely for being unremarkable.\n\"It's really bland in the telescopic observations,\" said Daly, a Ph.D. student at Brown and the study's lead author. \"It's like someone took a single color of spray paint and sprayed the whole thing. When we think about what might have caused this homogeneous surface, our thoughts turn to impact processes.\"\nAnd to understand impact processes, the researchers turned to NASA's Vertical Gun Range, a cannon with a 14-foot barrel that can launch projectiles at up to 16,000 miles per hour. For this work, Daly and Schultz wanted to simulate impacts into low-density surfaces that mimic the two broad possibilities for the composition of Ceres's surface: porous silicate or icy.\n\"The idea was to look at those two end-member cases, because we really don't know yet exactly what Ceres is like,\" Daly said.\nFor the porous silicate case, the researchers launched impactors into a powdered pumice. For the icy case, they used two targets: snow, and snow covered by a thin veneer of fluffy silicate material, simulating the possibility the Ceres's ice sits below a silicate layer. They then blasted these targets with pebble-sized bits of basalt and aluminum, simulating both stony and metallic meteorites.\nThe study showed that in all cases, large proportions of the impact material remained in and around the impact crater. This was especially true in the icy case, Daly said.\n\"We show that when you have a vertical impact into snow—an analog for the porous ice we think might be just beneath the surface of Ceres—you can have about 77 percent of the impactor's mass stay in or near the crater.\"\nThe results were a bit of a surprise, said Schultz, who has studied impact processes for many years as professor of earth, environmental, and planetary sciences at Brown.\n\"This is really contrary to previous estimates for small bodies,\" Schultz said. \"The thought was that you'd eject more material that you'd collect, but we show you can really deliver a ton of material.\"\nThe impact speeds used in the experiments were similar to speeds thought to be common in asteroid belt collisions. The findings suggest that a majority of impacts on porous bodies like Ceres cause an accumulation of impact material on the surface.\n\"People have thought that perhaps if an impact was unusually slow, then you could deliver this much material,\" Schultz said. \"But what we're saying is that for a typical, average-speed impact in the asteroid belt, you're delivering a ton of material.\"\nOver billions of years of such impacts, Ceres may have accumulated quite a bit of non-native material, Daly and Schultz said, much of it mixing together to create the relatively nondesdescript surface seen from telescopes. The researchers are hopeful that as the Dawn spacecraft scans the surface at much higher resolution, it might be able to pick out individual patches of this delivered material. That would help confirm the relevance of these experiments to celestial bodies, the researchers say.\nThe results have implications for missions that aim to return asteroid samples to Earth. Unless the landing sites are carefully chosen, the researchers say, those missions could end up with samples that aren't representative of the object's original material. To get that, it might be necessary to find an area where there has been a relatively recent impact.\n\"You can't do this like the old claw crane from the arcade,\" Schultz said. \"You can't just reach down and grab whatever's there. You may need to find a fresh impact where perhaps the native stuff has been churned up.\"<|endoftext|>"},"score":{"kind":"number","value":3.6875,"string":"3.6875"}}},{"rowIdx":1228,"cells":{"token_count":{"kind":"number","value":621,"string":"621"},"text":{"kind":"string","value":"In our school the children eat twice a day, before going out to play. To ensure good concentration it is important for children to drink lots of water, so please encourage your child to bring a bottle of water.\nWhat is a healthy lunch?\n• It is full of goodness like protein, vitamins and calcium.\n• Contains food with no sugar, because too much sugar is bad for your teeth.\n• Contains no colouring or additives.\n• Gives us strength and energy\n• Makes us fit and healthy\n• Helps our brain power\nWhat could be in a healthy lunch?\n• Sandwiches or rolls with cheese, meat or other fillings.\n• Pitta bread, crackers\n• Fruit (peeled and chopped for small children)\n• Vegetables (washed and chopped))\n• Yoghurt (easy to open)\nWhat drinks could we include in a healthy lunch?\n• Fruit juices\n• Diluted drinks\nPlease click on the links below for more information and some ideas as to what we can include in our children’s lunch boxes.\nThe children of Rath N.S. are very keen to have healthy lunches and show a good understanding of what that means. We hope these guidelines will assist everybody in making healthy choices.\nRegular physical activity is an important part of getting healthy and staying healthy. Encouraging children and teens to be active from a young age sets good habits early on and helps them develop the skills they need to stay active throughout their lives.\nRegular physical activity helps children and teens:\n- with healthy growth and development\n- build strong bones and muscles\n- improve balance and develop skills\n- maintain and develop flexibility\n- achieve and maintain a healthy weight\n- improve cardiovascular fitness\n- reduce stress and feel more relaxed\n- improve posture\n- boost confidence and self-esteem\n- have fun with their friends and make new ones\nChildren who don’t get enough physical activity are at a greater risk of becoming overweight or obese. This makes it harder for them to be active and keep up in sport or play.\nBeing overweight can also make children more prone to conditions such as asthma, flat feet and joint sprains. In the long term, it can contribute to conditions such as high blood pressure and cholesterol, heart disease, Type 2 Diabetes and liver disease.\nHow much activity?\nChildren and teens should get at least 60 minutes of physical activity every day, including vigorous activities that make them ‘huff and puff’. They can get even more health benefits from doing a couple of hours of physical activity every day.\nTo help children and teens be active every day, they need opportunities for sport, play and exercise at school, after school and on weekends.\nPhysical activity doesn’t all have to happen at the one time. It can be accumulated throughout the day; by walking or cycling to and from school, being involved in activities at school, active play at home or taking part in organised sport after school and on weekends.<|endoftext|>"},"score":{"kind":"number","value":3.890625,"string":"3.890625"}}},{"rowIdx":1229,"cells":{"token_count":{"kind":"number","value":831,"string":"831"},"text":{"kind":"string","value":"# Introductory Lesson – Area Model with Multiplication of\n\nIntroductory Lesson – Area Model with Multiplication of\nDecimals\nStandard: MCC5.NBT.7 Add, subtract, multiply, and divide decimals to\nhundredths, using concrete models or drawings and strategies based on place\nvalue, properties of operations, and/or the relationship between addition and\nsubtraction; relate the strategy to a written method and explain the reasoning\nused.\nEQ: What strategies can I use to multiply and divide decimals efficiently?\nHow do I use an area model to help me multiply decimals?\nMaterials: Base –Ten Grid Paper\nActivator: Review with students how to use an area model when multiplying with\nwhole numbers. See example below:\nThis model was intended to help link the calculations of the standard algorithm\nfor multi-digit multiplication to the distributive property and reveal the partial\nproducts as areas of smaller rectangles. The students produced area models for\nmulti-digit whole-number problems such as 57 x 14, shading and labeling four\npartial products in the model (500 + 200 + 70 + 28). You also want students to\nrecognize that the distributive property was at work. Ex. (50 + 7) x (10 + 4).\nLesson Part 1: Using the grid paper, model this problem: 5.7 x 1.4. Explain to\nstudents that indeed, the decimal multiplication problem 5.7 x 1.4 could be\nshown with the same base-10 diagram as the whole number multiplication 57 x\n14, but with dimensions that are labeled as one-tenth the size of those in the\nwhole number problem.\nTell students that now that a square of 100 little boxes is 1 whole, a row of 10\nsquares is .1, and one tiny square is .01. You can do this by showing a key.\nStudents need to note that the multipliers are the dimensions of the rectangle\nand the area shows the product. Further, in order to find the total product, one\nsums the areas of the smaller rectangles. In the diagram below, the smaller\nrectangles have the following dimensions: (1 x 5), (1 x .7), (.4 x 5), (.4 x .7) and\nrespective areas of 5, 0.7, 2.0, and 0.28 square units, for a total of 7.98 square\nunits. Finally, we want students to make the connection that the smaller regions\nhave the same dimensions and the same areas as the partial products\nproduced using the distributive property as this calculation shows:\n5.7 x 1.4 = (5 + .7) x (1 + .4) = (5 x 1) + (5 x .4) + (.7 x 1) + (.7 x .4)\n= 5 + 2.0 + 0.7 + 0.28 = 7.98.\nLesson Part 2: Pass out grid paper and have students set up the new problem:\n2.6 x 4.3. Allow students to use two different color crayons; one color for the\nwidth (4.3) and one for the length (2.6). Allow them to work in groups. Give\nplenty of grid paper for mistakes.\nOnce everyone has their grid labeled, then see if they can find the partial\nproducts.\nFind and color the product of 4 x 2\nFind and color the product of .3 x 2\nFind and color the product of .6 x 4\nFind and color the product of .6 x .3\nThis also can look like this:\n(4 + .3) x (2 + .6) – Use distributive property to get answer….\nSummarizer: Students are given a problem (2.3 x 4.1) and they will need to do it\nindependently using grid paper.<|endoftext|>"},"score":{"kind":"number","value":4.59375,"string":"4.59375"}}},{"rowIdx":1230,"cells":{"token_count":{"kind":"number","value":1079,"string":"1,079"},"text":{"kind":"string","value":"Lead Hazard Awareness Project: Fighting Lead-contaminated Soil and Dust\nPhiladelphia’s smelters are shut down, and cars no longer run on leaded gasoline. But the lead they released still clings to the soil surface, along with flakes of exterior lead paint. The result: lead is in the dirt that sticks to shoes and hands after work or play in bare soil. Lead is in the dust that the wind picks up and blows through open windows.(Lead paint is also a source of toxic dust.)\nChildren are most at risk of lead poisoning during early childhood, when lead dust is accidentally swallowed after normal hand-to-mouth behavior. Adults can inhale fine lead dust created by vigorous gardening, yard work, or home renovation without dust control.\nTesting for lead\n- Because both lead and its health effects are invisible, all children should be tested for lead before age one and again at age two. Ask your doctor.\n- Resources for testing yard soil and house dust are listed at the end of this brochure.\nStrategies for lead safety\n- In the yard and neighborhood, cover bare soil and wet down dust on walks and porch.\n- Keep lead-contaminated soil and dust out of your home. Less getting in means less to clean up. A bonus: Blocking lead dust also blocks pollen and other air pollutants that contribute to asthma.\nWhere to start: Powerfully protective habits\n- Make a rule: Remove all shoes and boots at the door. This is the single most powerful way to reduce lead in house dust. Keep indoor shoes or slippers handy.\n- Ban smoking in the house. Tobacco is a source of lead in both dust and children.\n- Do not allow children or pets to play or dig in bare soil anywhere in the city, including parks. Walk dogs on leash or on grass, and consider keeping cats indoors. (Pets can be lead-poisoned too.)\n- Before working or playing outdoors, wash your hands and your children’s hands with regular soap and water. Fine dust sticks tightly to the natural oils in skin, and washing can cut hand dust in half.\n- After playing or working outdoors or on floors, wash hands (everyone!) before meals. Hand sanitizers do not remove lead.\n- If your job or hobby exposes you to lead dust, change your clothes at work and bring them home in a plastic bag. Wash separately.\n- Feed small children frequent, small, nutritious meals. A full stomach absorbs up to 90 percent less lead.\n- Read, talk, sing to your child--engaging in irresistibly interesting activities with a loved adult actually helps the developing brain fight off lead.\nLead in soil and dust: cover it up, wet it down, keep it out, clean up what’s left\n1. Cover it up:\n- Cover bare soil with a lead-binding layer of compost followed by heavy, water-permeable landscape cloth. Cover the cloth with patio stones, six inches of mulch, or raised garden beds (see Resources). Pay special attention to the soil within 10 feet of painted house walls.\n- Cover smaller bare spots—for example, under swings—with rubber mats.\n- Engage your neighbors in a cover-the-bare-soil project. Neighborhood soil and dust control is known to increase lead safety for everyone.\n2. Wet it down:\n- In warm, dry weather, hose down your porch, sidewalk, and the adjacent street as well as any bare soil patches you see. Lead binds preferentially to the small soil particles that are picked up by a strong breeze.\n- If you see nearby excavation or construction without dust control, close your windows and storm windows and notify your neighbors and the city’s Air Management Services (215-685-7580).\n3. Keep it out:\n- If you cool with window fans, place a 2-inch pleated electrostatic air filter behind each fan to block both outdoor dust and paint dust in the window trough.\n- For child safety, be sure to extend the sides of the fan firmly within the window track and lower the sash on the fan’s top.\n- Change the filter when it gets dirty or if the fan motor slows.\n- Place an exhaust fan in the north or northeast window of a separate room.\n- If you have a central air (HVAC) system, ask your service technician about installing the thickest pleated electrostatic filter that will not interfere with air flow.\n- If you have window air conditioners, try to place them in north or northeast windows, out of the prevailing wind. Because AC filters cannot capture the tiny dust particles that carry the most lead, consider setting a portable air cleaner nearby.\n4. Clean up what’s left:\n- Lead-safe cleaning has two special requirements: to avoid stirring up loose dust and to remove the super-fine dust that sticks tightly to surfaces.\n- Vacuum carpets with either a sealed HEPA vacuum cleaner or a good household vacuum equipped with a HEPA filter. If possible, old threadbare carpets should be discarded or replaced. Worn spots can be covered with an area rug.<|endoftext|>"},"score":{"kind":"number","value":3.671875,"string":"3.671875"}}},{"rowIdx":1231,"cells":{"token_count":{"kind":"number","value":1960,"string":"1,960"},"text":{"kind":"string","value":"## Tuesday, August 25, 2020\n\n### Lesson 6, Rule of Thirds, Part 2 of 2, Serlio\n\nThe posts in this series  Lessons 1-7  are :\n\nhttps://www.jgrarchitect.com/2020/04/lessons.html\n\nhttps://www.jgrarchitect.com/2020/04/practical-geometry-lessons-2.html\n\nhttps://www.jgrarchitect.com/2020/04/practical-geometry-lesson-3.html\n\nhttps://www.jgrarchitect.com/2020/04/practical-geometry-lesson-4.html\n\nhttps://www.jgrarchitect.com/2020/04/practical-geometry-lesson-4b-old-first.html\n\nhttps://www.jgrarchitect.com/2020/06/practical-geometry-lessons-lesson-5.html\n\nhttps://www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-1_21.html\n\nhttps://www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-2-serlio.html\n\nhttps://www.jgrarchitect.com/2020/09/lesson-7-how-to-layout-frame-with-lines.html\n\n## Friday, August 21, 2020\n\n### Lesson 6: The Rule of Thirds, Part 1of 2\n\nThe Rule of Thirds is what artists call the grid that appears on your cell phone. It helps you compose and edit.\n\nA variation of this is used in Practical Geometry.\n\nSebastiano Serlio used this diagram in his book, On Architecture, published  in 1545.  He writes simple instructions for the reader; he says to construct the 'lines'.\n\nNote that the triangle (with its base at the bottom of the drawing) intersects the diagonals at the the upper corners of the door.  The width of the square is divided into thirds.\n\nCheck how the division into thirds in the square above this drawing  lines up with those intersections.  Serlio is using a a variation of the Rule of Thirds.\n\nLike Owen Biddle (see Lesson 5) Serlio sets out basic Geometry as used in construction in Book 1.\n\nThen he explains how to solve problems.  He does not show how he knows where to draw the lines shown above. He assumes the reader knows.\n\nHere are the instructions:\n\nDraw a square;\n\nAdd the diagonals to your square. Where they cross in the center. You have point 1.\n\nDivide one side of your square in half. Now you have  points 1 and 2.\nWith 2 points you can draw a line.\n\nAdd diagonals in each new rectangle.\n\nAdd the diagonals from the square.\n\nIf you were drawing this for a construction project on wood, on masonry, or on paper, you would not have separate squares.  All lines would be on your first square.  I have drawn each step without the extra lines for clarity.\n\nDo you see that the center line does not pass through the intersection of the diagonals? If you were the builder you would know that your diagonals will match when the line in centered. In this diagram they don't. So you would move  your center line.\n\nThis is the diagram for Serlio's drawing for the door.\n\nFor the Rule of Thirds (as we know it today) add the diagonals for the rectangles on both sides of the square.\n\nNote that you have intersections (4 points) not just where the lines  divide the square into smaller squares, but where the diagonals cross those lines.  2 points above the horizontal center line and 2 points below. Or: 2 on the right side of the vertical center line and 2 on the left.\n\nI have deliberately not added black points where the lines cross. You who are reading this will see it more clearly if you find those points yourself.\n\nConnect those new points and extend the lines across the square.\nYou have drawn the Rule of Thirds.\n\nSimilar diagonals could be drawn from the left to the right side and vice versa.\n\nI drew all the diagonals on graph paper to make it easier to follow.  The next lines to add would be the diagonals of the small squares.\nThe line does not come back to its beginning until it has continued through the complete pattern\n\nThe posts in this series  Lessons 1-7  are :\n\nhttps://www.jgrarchitect.com/2020/04/lessons.html\n\nhttps://www.jgrarchitect.com/2020/04/practical-geometry-lessons-2.html\n\nhttps://www.jgrarchitect.com/2020/04/practical-geometry-lesson-3.html\n\nhttps://www.jgrarchitect.com/2020/04/practical-geometry-lesson-4.html\n\nhttps://www.jgrarchitect.com/2020/04/practical-geometry-lesson-4b-old-first.html\n\nhttps://www.jgrarchitect.com/2020/06/practical-geometry-lessons-lesson-5.html\n\nhttps://www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-1_21.html\n\nhttps://www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-2-serlio.html\n\nhttps://www.jgrarchitect.com/2020/09/lesson-7-how-to-layout-frame-with-lines.html\n\n## Thursday, August 6, 2020\n\n### Owen Biddle's 'Young Carpenter's Assistant' , Plate I, G\n\nA note on Owen Biddle's Plate I, Diagram G. in his pattern book for beginning carpenters. *\n\nI wrote about Diagram G on this post: https://www.jgrarchitect.com/2020/06/practical-geometry-lessons-lesson-5.html\n\nI said that Biddle was not just introducing his 'carpenter assistant' to geometry; in Diagram G Biddle was explaining how to layout a square corner to work out a structural detail, cut a board, or set a frame on site.\n\nSince then I have explored the theoretical geometry of that diagram.\n\nThe number of right angles which can be drawn in a circle is infinite. The rule always works. That understanding is part of why geometry is seen as mystical or sacred.\n\nThis 'squaring the circle' diagram is from\nRobert Lawlor's Sacred Geometry*. (page 77, diagram 7.5)\nIt uses a geometry similar geometry to Biddle's diagram G: a diameter and an angle. Here the diameters are evenly spaced and the same angle  is used at every point on the circumference. But the angle is not 90*. It is not a 'square angle'.\nThis is decorative, not structural.\nThe shapes do not close. The line continues for 5 rotations. It does not create a square, but seeks to define the perimeter of a circle with straight lines.\n,\nI am often told that I work with Sacred Geometry, that the geometric patterns I recover are theoretical, mystical, and sacred. I agree they are geometry. No, they are not sacred. They are practical. They are geometry used in construction.\n\nHere is how Biddle's diagram comes about:\n\nBegin with  a point  - A\n\nChoose a radius - A-B,  and draw a circle. Using the daisy wheel find the diameter - B- A- C, dotted and dashed line.\n\nPick a point on the circumference of the circle - D.\n\nHere I have chosen 3 different D's  at random.\n\nConnect B-D and D-C.\n\nEach diagram will have a 90* (right) angle at the intersection of  B-D-C.\n\nWherever the D is placed. the angle will be 90*.\n\nBiddle's Diagram G begins with my line B-D.\nIt describes how to find my 90* angle of B-D-C. (his a-b-c) The answer is to find the diameter of a circle (a-d-c) that intersects a. That will give c. That will give the 90* the carpenter needs.\n\nBy Hound and Eye* has a very similar diagram for drawing a right angle .\nThe book is a  guide to furniture design, full of practical geometry. Each geometric problem is described step by step; practice work sheets are included.\nThis pattern is the beginning of a handmade try square.\n\n*Owen Biddle's The Young Carpenter's Assistant, 1805, Philadelphia. Dover Publishing  reprint,  See my Bibliography for more information.\n\n*Robert Lawlor, Sacred Geometry, Philosophy and Practice, 1982, Thames and Hudson, London.\n\n*Geo.R. Walker & Jim Tolpin, By Hound and Eye, A Plain & Easy Guide to Designing Furniture with No Further Trouble, 2013,  Lost Art Press, Kentucky The diagram shown above is from page 57.\n\nThis pattern is 4 overlapping hexagons.\nMy granddaughter, who is 7, watched me add the images to this post.\nShe wanted us to 'square the circle'. I did, using right angles where the diameters met the circumference. That produced these overlapping 6 hexagons, not squares.\n\nShe watched closely and observed that accurate work was not easy: my lines did not always cross exactly in the center of the circle. When we finished she asked me to erase all the diameters. This is the result. Maybe she will show me later what she added to the copy I printed for her.<|endoftext|>"},"score":{"kind":"number","value":4.40625,"string":"4.40625"}}},{"rowIdx":1232,"cells":{"token_count":{"kind":"number","value":3472,"string":"3,472"},"text":{"kind":"string","value":"What was “new” about the New South?\nThe following will discuss what, if anything was new about the New South that emerged in the United States after 1877. Prior to the American Civil War the old South had predominantly been an agrarian economy in which blacks were slaves who had worked on the cotton plantations, factories, or had been domestic servants. Cotton had been the major commodity of the economy, which had mainly been exported to Britain. The American Civil War had been fought over the issue of slavery and whether the Southern States had the right to cede from the United States to preserve the institution of slavery (Hobsbawm, 1975 p.184). The Civil War brought social and economic changes to the South. Its cotton exports had been drastically reduced, its agricultural and industrial output declined sharply, whilst much of its infrastructure was destroyed. During the civil war President Lincoln had proclaimed the emancipation of all slaves, whilst blacks had fought with distinction on the Union side. The devastation brought to the South by the civil war meant that a period of reconstruction was needed afterwards. Leading white Southerners such as Henry Grady called for a New South. The blacks that were freed, as a result of the Confederate States losing the civil war, also anticipated a New South. The blacks in the Southern States expected their lives to be better following the Union’s victory and the era of reconstruction. In many respects strong arguments can be made that their lives got worse rather than better. Du Bois for one contended that blacks “had fought slavery to save democracy and then lost democracy in a new and vaster slavery” (Du Bois, 1935 Chapter 1).\nThe result of the American Civil War in theory was that the four and a half million blacks in the United States were all free and equal with the white population. However, the end of the Reconstruction made those equal rights a mockery in the New South (Brogan, 1999, p.348).\nThat the New South was not a new place for the better for its black population was due to the way in which the American Civil War ended. Lincoln’s assassination was the South’s revenge for losing the War. Lincoln’s successor, Andrew Johnson was less capable of ensuring that the South changed in ways that benefited its black population. From his presidency onwards, the North did very little to ensure Southern blacks had any meaningful rights (Brogan, 1999, p.348). Southern blacks were only able to exercise their political rights whilst the Union forces remained in the South, those rights ceased to exist in reality once the South was left to run itself. The suppression of Southern blacks was arguably worse once they had been formally freed than when they had been slaves. Racial discrimination, the fear of violence and poverty meant that the New South was no better than the Old South had been (Hobsbawm, 1975, p.143). Neither the South in general or its repressed black population in particular, gained as much from the United States rapid industrialisation from the 1870s onwards as the North did (Hobsbawm, 1987, p.35).\nIn the New South there was a strong desire amongst the defeated Confederate States to make its black population subject to its strict political and economic controls for as long as possible. The abolition of slavery had not seen the end of the cotton plantations. However, jobs and better pay were given to the whites rather than blacks. Blacks were given the lowest paid jobs and they could be punished for not taking them. For many blacks the newness of the New South was the increased harshness of the discrimination they were subjected to. Whilst the whites in the New South had been unable to defeat the Union during the American Civil War, they were in a position to make life very unpleasant for the black population of the New South. Much discrimination was given legality through the ‘Black Codes’ of the Southern legislatures that severely restricted the rights of former slaves. Slavery had, in many respects, been restored in a less obvious form (Brogan, 1999, p.352). Those blacks that tried to exercise their legal rights found legal and political obstacles placed in front of them, which effectively deprived them of all those rights. They also faced violence and intimidation on a regular basis (Bradbury & Temperley, 1998, p.153). The Southern states were able to prevent the Constitutional Amendments that abolished slavery and gave freed slaves their rights having a positive impact as they were responsible for their enforcement, rather than the national government (Murphy et al, 2001, p.315). States such as Louisiana had no intention of giving blacks any rights on the grounds it was unconstitutional to do so (Du Bois, 1935, p.454). A series of measures which were known as Jim Crow laws were used by the Southern States to segregate and repress their black populations. Although they claimed the segregated services were of equal quality, this was a sham to excuse neglecting their black communities (Cobb, 1992). Overall Jim Crow Laws delayed the economic development of the New South, whilst they institutionalised racial discrimination and segregation. The cost of providing segregated services lowered the quality of education, housing, and transport in the New South. Segregation had even been endorsed by the Supreme Court as long as services were of equal quality, which few bothered to check. Such discrimination was contrary to the way Henry Grady believed the New South should have developed. Grady argued that the best way to industrialise the New South was to treat blacks as equal partners rather than inferiors. Therefore social justice and equality were just as important as capital and machinery in building the New South (Mauk & Oakland, 1995 p. 108). Grady believed that the New South would be the perfect democracy as long blacks were treated equally. The civil war had been an opportunity for the South to stop its outdated reliance on slavery and cotton (Harris, 1890 p. 15). Segregation, as well as being morally questionable, kept the South relatively poor and backward in relation to the rest of the country (Hobsbawm, 1975 p.184).\nPoverty was a new feature of the New South. Poverty paradoxically enough had not been an issue for blacks in the South when they had been slaves. Although, they had no freedom, slaves were provided with basic levels of accommodation and food, on the logical basis that unhealthy slaves did not work as well as healthy ones. Southern slave owners had generally treated their slaves well enough for their numbers to increase at the same rate as the white population (Bradbury & Temperley, 1998 p. 153). Defenders of slavery had maintained that it kept the Southern states economically competitive, kept the black population at subsistence, whilst ensuring that all white men could find paid work (Brogan, 1999, p.371). Poverty, as freed slaves found to their cost, was as restrictive of their freedom as actual shackles had been. Freed slaves had to compete with whites to gain jobs. Poverty was closely linked with racial discrimination, in that whites were given better jobs and better working conditions, even when there were better-qualified blacks to do the jobs. Discrimination in the provision of education, housing and medical care also contributed to keep the blacks repressed and in poverty (Cobb, 1992). Blacks were disenfranchised by their poverty, whereas loopholes were used to ensure that poor whites kept the vote (Hobsbawm, 1987, p.24).\nAnother new feature of the New South was the increased levels of urbanisation. Cities such as New Orleans and Birmingham increased in size during the reconstruction era. The urbanisation of the New South was result of the industrial expansion encouraged by the Southern states and the migration of people trying to escape rural poverty. Migrating to the cities did not reduce racial discrimination and it barely increased opportunities for black people. Birmingham was the only city to achieve industrialisation on a major scale in the New South. The South was economically held back by its deliberately uneducated blacks and its under educated whites (Brogan, 1999, p.372). Southern blacks had also migrated to northern cities such as New York to increase their opportunities and to escape racial discrimination. The North was still prone to such discrimination even if it did give greater opportunity and blacks faced lower threats of violence. The Southern states had been motivated to enact the ‘Black Codes’ to restrict migration to both Southern and Northern cities (Brogan, 1999, p.363).\nUnemployment was a more obvious problem in the New South than it had been in the old South. Unemployment and low paid employment in a country with no public welfare provision was a serious problem, especially for blacks that were discriminated against and could not afford the basic necessities of life (Hobsbawm, 1987, p.103). Employers and plantation owners in the New South as a whole tended to keep the relationship between poor blacks and poor whites as unfriendly as possible. Factory and plantation owners feared that that if black and white workers had a good relationship they would form effective trade union movements and threaten the profits of the owners (Lewis, 1994). Discrimination in favour of white workers alienated blacks from them, whilst owners and employers kept control of their workers by threatening to use black workers as strike breakers. Such tactics were effective at preventing the emergence of trade unions but did nothing to improve race relations in the New South (Brogan, 1999, p. 371).\nThe creation of Birmingham, Alabama was a symbol of all that was new in the New South. The place had not existed before 1871, and calling it Birmingham after one of the most industrialised cities in Britain was a statement of intent. Birmingham, Alabama was to be the industrial heart of the New South (Vann Woodward, 1951). Henry Grady himself cited Birmingham as the best example of his plans for a New South, yet historians have argued as to whether the development of Birmingham was similar to the industrial development envisaged by the plantation owners prior to the civil war (Lewis, 1994). Post civil war reconstruction gave the Southern States the opportunity as well as the need to reconstruct their economy. Falling prices for raw cotton meant that plantation owners switched their attention to manufacturing finished cotton products in new cotton mills. Attempts were also made to diversify the Southern economy away from cotton by developing coal, steal, and iron production. During the reconstruction period the Federal government had tried to enhance the economic prospects of the South by having the railroads rebuilt and extended to improve the transport links with the rest of the United States. Southern plantation owners, investors from the rest of the United States, as well as foreign investors funded industrial development. One feature of the New South did not change from the old South; it was still economically weaker than the North (Spiller et al, 2005 p. 80). The economic changes of the New South only benefited a few plantation and factory owners, some of who became much wealthier than they had been before the civil war (Hobsbawm, 1987 p. 24).\nA new feature of the New South was the high level of violence directed against the black population by white racists. In the immediate post-civil war period the formation of the Ku Klux Klan demonstrated the popularity for white supremacist ideas in the Southern states. The Ku Klux Klan added murderous intentions to their racist outlooks. The emergence of the Ku Klux Klan led to many thousands of lynchings and murders throughout the New South. Blacks found it very difficult to protect them-selves from racially motivated violence on such a large scale. They received no meaningful levels of protection from the police, the courts or the state authorities, which often sympathised with white supremacist views and were therefore unwilling to take action against the Ku Klux Klan or individual racists. Racism and prejudices were built into the ‘Black Codes’ that made a mockery of the post civil war Constitutional Amendments. The Federal courts and governments were unwilling to intervene in the affairs of the New South, as far as the Federal governments was concerned the Constitutional Amendments were fully operative in the South. Nobody in Washington DC seemed to be bothered to act upon the plentiful evidence of racial murders and discrimination in the New South. Between 1887 and 1917 official United States government figures showed that 2,734 blacks were murdered in racially motivated crimes, the vast majority in the New South. Before that period the death toll had been even higher, and only the presence of the Union army before 1877 had prevented further bloodshed (Murphy et al, 2001 p. 320).\nIn some respects there were few new aspects in the New South. The combined effects of the Black Codes and Jim Crow laws meant that the New South restricted the freedoms of freed slaves to such an extent that slavery might as well have been retained. Economic, social, and political restrictions meant that insignificant numbers of blacks could vote in elections, own their land or gain education in the Southern States (Cobb, 1992). Low wages, unemployment, high rents, and direct discrimination were as effective as the Black Codes at keeping black people poor and powerless (Du Bois, 1935 p. 454). It is no wonder that many blacks believed that after reconstruction the New South made their lives worse than before. For them the only difference the old and new South was that they were underpaid for working on other peoples’ land and in other peoples’ factories rather than being paid at all. Only a small number of freed blacks had been able to make successes of their lives before the Jim Crow laws began to restrict opportunities. Only 4,000 freed slaves managed to purchase land in the New South, and most of them could not buy enough land to run successful farms (Murphy et al, 2001 p. 316). The New South was not a content place; the whites still fumed at their defeat in the civil war and re-imposed a quasi slavery upon the nominally free blacks (Hobsbawm, 1975 p. 143).\nTherefore, there were new aspects to the New South, although those aspects were not all positive or progressive in their nature. The Southern States were changed socially and economically as a result of the American Civil War. The economic consequences of the civil war were apparently severe. Agricultural and industrial outputs had been reduced, whilst the infrastructure of the Southern States had been badly damaged in the war. The war had disrupted the export of raw cotton which, had been the basis of the old South’s economy. Plantation owners had claimed that their plantations would be unprofitable with the abolition of slavery, a claim that proved unfounded due to the low wages they paid to white and black workers alike. The freed slaves found that life in the New South was in fact harsher in some respects than slavery. This was due to the increased racism and discrimination that was a new feature or perhaps at least a more obvious feature of the New South. The idea of the New South was promoted by the likes of Grady, as well as the new industrial centres such as Birmingham, Alabama and Atlanta. Overall in the period after 1877 industrial output in the Southern States did increase with the development of cotton mills, coal, steel, and iron production, although it still lagged behind the rest of the United States. Industrial development did not improve the lives of most people in the New South, just factory and plantation owners and the profits of outside investors. The legacy of the civil war was a long and bitter one, with the Southern whites repressing the blacks to compensate for defeat and demonstrate their alleged supremacy.\nBradbury M & Temperley H, (1998) Introduction to American Studies 3rd edition, Longman, London\nBrogan H, (1999) The Penguin History of the USA, Penguin, London\nCobb J C, (1992) The Most Southern Place on Earth: The Mississippi Delta & The Origins of Regional Identity, Oxford University Press, Oxford, and New York\nDu Bois W E B, (1935) Black Reconstruction in America, London\nHarris J C, (1890) Henry W. Grady: His Life, Writings, and Speeches, Cassell Publishing Co, New York\nHobsbawm E, (1975) The Age of Capital 1848-1875, Weidenfeld & Nicholson, London\nHobsbawm E, (1987) The Age of Empire 1875-1914, Weidenfeld & Nicholson, London\nLewis (1994) the emergence of Birmingham as a case study of continuity between the antebellum planter class and industrialization in the new south” agricultural history (spring 1994) p. 62-79\nMauk, D & Oakland, J (1995) – American Civilization Routledge, London\nMurphy D, Cooper K & Waldron M, (2001). United States1776-1992 Collins, London\nNew south plantation kingdom -‘the new south’ writings and speeches of Henry Grady, (1971)\nThe beehive press savannah, Georgia\nSpiller J, Clancy T, Young S, and Mosley S (2005) The United States 1763 – 2001, Routledge, London<|endoftext|>"},"score":{"kind":"number","value":4.375,"string":"4.375"}}},{"rowIdx":1233,"cells":{"token_count":{"kind":"number","value":421,"string":"421"},"text":{"kind":"string","value":"WHAT IS TMJ?\nTMJ stands for temporomandibular joint. It is the joint that connects your jaw (mandible) to your skull (temporal bones), and it is very distinctive compared to the other major joints in the human body. The temporomandibular joint is complex and unique in its structure. First, you have to move both joints at the same time and cannot choose to only move one joint. The temporomandibular joint has two distinct movements unlike any other joint in the body. The initial motion is a hinge movement that is followed by a translational (sliding) movement where the jaw comes out of the area where it usually is so it is like an intentional dislocation.\nWe use the TMJ more frequently than any other joint in our body. Every time we chew, talk, bite down or swallow we put the TMJ to work. The exact cause of a person’s TMJ disorder (TMD) is often difficult to determine, which is why it’s important to receive the correct treatment plan.\nWHAT ARE TEMPOROMANDIBULAR DISORDERS?\nTemporomandibular Disorders (TMD) are a complex and poorly understood set of conditions. TMD refers to problems related to pain, inflammation and improper functioning of:\n- Jaw joints (Temporomandibular joints)\nA condition in either the hard or soft tissues or both, including disc dislocation and arthritis/degeneration. This joint-related condition usually results from inflammation, disease or degeneration of the hard or soft tissues within the TMJ. Inflammation, disc dislocation (also called internal derangement) and degenerative arthritis are the most common arthrogenous disorders of the TMJ.\n- Muscles surrounding your jaw joints\nA condition affecting the muscles, usually brought on by overwork or tension in the jaw and supporting muscles.\n- Associated structures in your head, neck and face (including but not limited to your teeth, nerves and blood vessels)<|endoftext|>"},"score":{"kind":"number","value":3.96875,"string":"3.96875"}}},{"rowIdx":1234,"cells":{"token_count":{"kind":"number","value":373,"string":"373"},"text":{"kind":"string","value":"Normalization (image processing)\nIn image processing, normalization is a process that changes the range of pixel intensity values. Applications include photographs with poor contrast due to glare, for example. Normalization is sometimes called contrast stretching or histogram stretching. In more general fields of data processing, such as digital signal processing, it is referred to as dynamic range expansion.\nThe purpose of dynamic range expansion in the various applications is usually to bring the image, or other type of signal, into a range that is more familiar or normal to the senses, hence the term normalization. Often, the motivation is to achieve consistency in dynamic range for a set of data, signals, or images to avoid mental distraction or fatigue. For example, a newspaper will strive to make all of the images in an issue share a similar range of grayscale.\nNormalization transforms an n-dimensional grayscale image with intensity values in the range (Min,Max), into a new image with intensity values in the range (newMin,newMax).\nFor example, if the intensity range of the image is 50 to 180 and the desired range is 0 to 255 the process entails subtracting 50 from each of pixel intensity, making the range 0 to 130. Then each pixel intensity is multiplied by 255/130, making the range 0 to 255.\nNormalization might also be non linear, this happens when there isn't a linear relationship between and . An example of non-linear normalization is when the normalization follows a sigmoid function, in that case, the normalized image is computed according to the formula\nWhere defines the width of the input intensity range, and defines the intensity around which the range is centered.\nAuto-normalization in image processing software typically normalizes to the full dynamic range of the number system specified in the image file format.<|endoftext|>"},"score":{"kind":"number","value":3.703125,"string":"3.703125"}}},{"rowIdx":1235,"cells":{"token_count":{"kind":"number","value":1398,"string":"1,398"},"text":{"kind":"string","value":"In 2008, scientists working in Denisova Cave—a cold site in Siberia’s Altai Mountains—uncovered a strange pinky bone, broader than a typical human’s. The DNA within that bone revealed that its owner belonged to an entirely new group of ancient hominins, distinct from Homo sapiens or Neanderthals. That group became known as the Denisovans.\nResearchers have since decoded the Denisovan genome. But still, no one can say what they looked like. Every known Denisovan fossil would fit in your palm—that pinky, three teeth, and a remarkable bone sliver from a Denisovan-Neanderthal hybrid. And all of these remains came from the same cave.\nBut now, an international team of scientists has announced the identification of another Denisovan fossil, from a site 1,500 miles away. It’s the right half of a jawbone, found some 10,700 feet above sea level in a cave in China’s Xiahe County, on the eastern edge of the Tibetan plateau. The Xiahe mandible, as it is now known, is not only the first Denisovan fossil to be found outside Denisova Cave, but also the very first Denisovan fossil to be found at all. It just took four decades for anyone to realize that.\nThe mandible was discovered by a local monk in 1980 and donated to Lanzhou University. There, it lay unstudied until 2010, when a team led by Fahu Chen and Dongju Zhang—a climatologist and an archaeologist, respectively—began examining it in earnest. The world learned about the existence of the Denisovans at around that time, and though fossils had only been recovered from Siberia, it was clear that these hominins likely existed throughout much of East Asia. Smatterings of Denisovan DNA still persist in the genes of living people in this region and beyond, and how else could it have made it into the genomes of modern Tibetans or Melanesians? Still, “I never imagined that [the Xiahe mandible] could be a Denisovan,” Zhang says.\n“If it was one, we’d be so lucky,” she adds.\nThe mandible itself is very thick and sturdy. It has no chin, which rules out modern humans. The teeth within it are exceptionally large, and different in shape and size from those of Neanderthals, Homo erectus, and other known hominins.\nThe molecules in the specimen were especially telling. The team couldn’t detect any traces of ancient DNA, but it did find the next best thing—fragments of ancient collagen proteins, still lurking in one of the teeth. These fragments closely resemble the proteins of Denisovans, more so than those of Neanderthals, modern humans, or other great apes.\nBut Katerina Douka of the Max Planck Institute for the Science of Human History notes that methods for analyzing ancient proteins are relatively new, and less well tested than those for studying ancient DNA. Researchers should use both techniques on other specimens to check that they give the same results, Douka says. But for now, based on the data that exist, she agrees that the Xiahe mandible most likely belonged to a Denisovan.\n“It confirms that the Denisovans were perhaps widely distributed through East Asia,” Zhang says. For years, scientists had suspected as much. After all, people across East Asia and Melanesia (the region that includes New Guinea and its neighboring islands) have Denisovan DNA in their genes. This pattern—the product of ancient sexual encounters between Denisovans and humans—shouldn’t be possible if the Denisovans were just confined to a small Siberian cave. Instead, it seemed that they were already living in much of East Asia by the time ancient humans also spread through the region.\nIndeed, the Xiahe mandible, which is 160,000 years old, is by far the earliest hominin fossil from the Tibetan plateau. Researchers used to think that Homo sapiens was unique in adapting to the Himalayas, but the Denisovans were successfully living on the roof of the world at least 120,000 years earlier. They must also have adapted to extremely thin air—after all, the mandible was found in a cave that’s some 8,000 feet higher above sea level than Denisova itself. “Their presence that high up is truly astonishing,” Douka says.\nThis helps to explain a remarkable finding from 2014. Back then, Emilia Huerta-Sanchez and her colleagues showed that most Tibetan people carry a mutated version of the EPAS1 gene, which helps them cope with high-altitude air that has 40 percent less oxygen than what most people breathe. And that mutation, the team showed, came from Denisovans. By having sex with these hominins, ancient Tibetans picked up a useful genetic trait that their descendants still benefit from.\nThat result was surprising, because Denisova Cave is so far from Tibet, and so much lower in altitude. The new mandible resolves that discrepancy. Although it’s unclear whether its owner had the same EPAS1 variant that the other Denisovans did, it at least shows that Denisovans were in the right part of the world. “I was thrilled that they found a Denisovan-like jawbone at high altitude,” Huerta-Sanchez says.\n“The new discovery is an important step in understanding the Denisovans, but the big question still remains to be solved,” says Yousuke Kaifu of the National Museum of Nature and Science, Tokyo. And that is: If Denisovans were spread throughout Asia, why do Melanesians have so much more of their DNA than anyone else—5 percent, compared with just 0.2 percent in East Asians, and nothing in other groups?\nTo answer that question, scientists will need to find more Denisovan bones. Douka and her colleagues have started a project called Finder to do exactly that, by rapidly analyzing small, unidentifiable slivers from various sites in Asia. More intact specimens might also be lying around in museum collections. For example, the Xiahe team notes that its mandible has many similarities to the Penghu 1 mandible, which was fished out of the ocean near Taiwan in 2008. (“I agree that there are some similarities,” says Kaifu, who led the team that analyzed Penghu 1 in 2015.)\nChina has a long list of similar hominin fossils that have been hard to assign to other species. “Some of those may already be Denisovans,” Zhang says.<|endoftext|>"},"score":{"kind":"number","value":3.671875,"string":"3.671875"}}},{"rowIdx":1236,"cells":{"token_count":{"kind":"number","value":1274,"string":"1,274"},"text":{"kind":"string","value":"# Determinants of Matrix 4×4 with Examples\n\nHer you will learn how to find determinants of matrix 4×4 with example.\n\nLet’s begin –\n\n## Determinants of Matrix 4×4\n\nTo evaluate the determinant of a square matrix of order 4 we follow the same procedure as discussed in previous post in evaluating the determinant of a square matrix of order 3.\n\nIf A = $$\\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\\\ a_{21} & a_{22} & a_{23} & a_{24} \\\\ a_{31} & a_{32} & a_{33} & a_{34} \\\\ a_{41} & a_{42} & a_{43} & a_{44} \\end{bmatrix}$$ is a square matrix of order 4,\n\nthen | A | = $$a_{11}\\begin{vmatrix} a_{22} & a_{23} & a_{24} \\\\ a_{32} & a_{33} & a_{34} \\\\ a_{42} & a_{43} & a_{44} \\end{vmatrix}$$ – $$a_{12}\\begin{vmatrix} a_{21} & a_{23} & a_{24} \\\\ a_{31} & a_{33} & a_{34} \\\\ a_{41} & a_{43} & a_{44} \\end{vmatrix}$$ + $$a_{13}\\begin{vmatrix} a_{21} & a_{22} & a_{24} \\\\ a_{31} & a_{32} & a_{34} \\\\ a_{41} & a_{42} & a_{44} \\end{vmatrix}$$ – $$a_{14}\\begin{vmatrix} a_{21} & a_{22} & a_{23} \\\\ a_{31} & a_{32} & a_{33} \\\\ a_{41} & a_{42} & a_{43} \\end{vmatrix}$$\n\n## Example :\n\nFind the determinant of A = $$\\begin{bmatrix} 1 & 2 & -1 & 3 \\\\ 2 & 1 & -2 & 3\\\\ 3 & 1 & 2 & 1 \\\\ 1 & -1 & 0 & 2 \\end{bmatrix}$$.\n\nSolution : | A | = $$\\begin{vmatrix} 1 & 2 & -1 & 3 \\\\ 2 & 1 & -2 & 3\\\\ 3 & 1 & 2 & 1 \\\\ 1 & -1 & 0 & 2 \\end{vmatrix}$$\n\n$$\\implies$$ | A | = $$1\\begin{vmatrix} 1 & -2 & 3 \\\\ 1 & 2 & 1 \\\\ -1 & 0 & 2 \\end{vmatrix}$$ – $$2\\begin{vmatrix} 2 & -2 & 3 \\\\ 3 & 2 & 1 \\\\ 1 & 0 & 2 \\end{vmatrix}$$ + $$(-1)\\begin{vmatrix} 2 & 1 & 3 \\\\ 3 & 1 & 1 \\\\ 1 & -1 & 2 \\end{vmatrix}$$ – $$3\\begin{vmatrix} 2 & 1 & -2 \\\\ 3 & 1 & 2 \\\\ 1 & -1 & 0 \\end{vmatrix}$$\n\n| A | =  (1){$$(1)\\begin{vmatrix} 2 & 1 \\\\ 0 & 2 \\end{vmatrix}$$ – $$(-2)\\begin{vmatrix} 1 & 1 \\\\ -1 & 2 \\end{vmatrix}$$ + $$(3)\\begin{vmatrix} 1 & 2 \\\\ -1 & 0 \\end{vmatrix}$$}\n\n– (2){$$(2)\\begin{vmatrix} 2 & 1 \\\\ 0 & 2 \\end{vmatrix}$$ – $$(-2)\\begin{vmatrix} 3 & 1 \\\\ 1 & 2 \\end{vmatrix}$$ + $$(3)\\begin{vmatrix} 3 & 2 \\\\ 1 & 0 \\end{vmatrix}$$}\n\n+ (-1){$$(2)\\begin{vmatrix} 1 & 1 \\\\ -1 & 2 \\end{vmatrix}$$ – $$(1)\\begin{vmatrix} 3 & 1 \\\\ 1 & 2 \\end{vmatrix}$$ + $$(3)\\begin{vmatrix} 3 & 1 \\\\ 1 & -1 \\end{vmatrix}$$}\n\n– (3){$$(2)\\begin{vmatrix} 1 & 2 \\\\ -1 & 0 \\end{vmatrix}$$ – $$(1)\\begin{vmatrix} 3 & 2 \\\\ 1 & 0 \\end{vmatrix}$$ + $$(-2)\\begin{vmatrix} 3 & 1 \\\\ 1 & -1 \\end{vmatrix}$$}\n\n$$\\implies$$ | A | = 1{(1)(4 – 0) – (-2)(2 + 1) + (3)(0 + 2)} – 2{(2)(4 – 0) – (-2)(6 – 1) + (3)(0 – 2)} – (-1){(2)(2 + 1) – (1)(6 – 1) + (3)(-3 – 1)} – 3{(2)(0 + 2) – (1)(0 – 2) + (-2)(-3 – 1)}\n\n$$\\implies$$ | A | = 1(16) – 2(12) + (-1)(-11) – 3(14) = -39<|endoftext|>"},"score":{"kind":"number","value":4.46875,"string":"4.46875"}}},{"rowIdx":1237,"cells":{"token_count":{"kind":"number","value":691,"string":"691"},"text":{"kind":"string","value":"Causes of Vertigo\nVertigo, or dizziness, usually results from a disorder in the peripheral vestibular system (i.e., structures of the inner ear). Dizziness also may occur as a result of a disorder in the central vestibular system (i.e., vestibular nerve, brainstem, and cerebellum). In some cases, the cause of vertigo is unknown.\nPeripheral vestibular disorders include the following:\n- Benign paroxysmal positional vertigo (BPPV; most common peripheral disorder; may be accompanied by hearing loss, reduced cognitive function, and facial muscle weakness)\n- Cogan's syndrome (inflammation of connective tissue in the cornea; results in vertigo, ringing in the ears [tinnitus], and loss of hearing)\n- Ménière disease (fluctuating pressure of inner ear fluid [endolymph]; results in severe vertigo, ringing in the ears [tinnitus], and progressive hearing loss)\n- Ototoxicity (i.e., ear poisoning)\n- Vestibular neuritis (inflammation of vestibular nerve cells; may be caused by viral infection)\nBenign paroxysmal positional vertigo (BPPV) usually affects one of the sensing tubes in the inner ear called the posterior semicircular canal. BPPV occurs when debris made up of calcium carbonate and protein (called otoliths or ear crystals) builds up in and moves around in the posterior semicircular canal. BPPV also can affect the anterior canal or the horizontal canal.\nWhen the head is moved in certain ways (e.g., turning in bed, looking up, bending over), the calcium crystals move around and trigger inner ear sensors, causing a brief sensation of spinning. Inner ear degeneration (usually occurs in elderly patients), head trauma, and inner ear infection (e.g., otitis media, labyrinthitis) can cause BPPV.\nSome medications and environmental chemicals (e.g., lead, mercury, tin) can cause ototoxicity (i.e., ear poisoning), which may result in damage to the inner ear or the 8th cranial nerve (acoustic nerve) and cause vertigo. The damage can be permanent or temporary.Long-term use or high doses of certain antibiotics (e.g., aminoglycosides [streptomycin, gentamicin]) and antineoplastics (e.g., cisplatin, carboplatin) can cause permanent ototoxicity.\nMedications that may cause temporary ototoxicity include the following:\n- Anticonvulsants (e.g., phenytoin, carbamazepine)\n- Antidepressants (e.g., clomipramine, amoxapine)\n- Antihypertensives (e.g., labetalol, enalapril)\n- Loop diuretics (e.g., bumetanide, furosemide)\n- Pain relievers (e.g., aspirin)\n- Prescription and over-the-counter cold medicines\n- Quinine (e.g., chloroquine, quinidine)\nAlcohol, even in small amounts, can cause temporary vertigo in some people.<|endoftext|>"},"score":{"kind":"number","value":3.84375,"string":"3.84375"}}},{"rowIdx":1238,"cells":{"token_count":{"kind":"number","value":462,"string":"462"},"text":{"kind":"string","value":"      Â\n          Â\n\n# How to Factor an Equation with 2 Terms?\n\nAn equation in mathematics is a relation that shows the equality between the variables and constants. An equation can be identity equation or inconsistent equation or conditional equation. An identity equation is satisfied by all possible values of the variables. A conditional statement is the one which has a restricted Set of values of the variables satisfying the given equation. For example, 4x + 8 = 12, Here only x = 1 can be the only possible value satisfying the equation. Last is the inconsistent equation that has no solutions. The number of terms in the equation determines the type of the equation like if we have 2 terms in the equation, then it is called as a binomial equation and so on. Let us learn how to factor an equation with 2 terms considering some examples of it:\n\nExample 1: Suppose we have a binomial equation as: 4x2 – 8 x = 0, then what is the value of 'x'?\n\nSolution: In the given equation we have to factorize the left hand side of the equation to get the value of 'x' as follows:\n\n4x2 – 8x = 0,\n\nOr 4x (x – 2) = 0,\n\nOr x (x – 2) = 0,\n\nOr x = 0 and x – 2 = 0,\n\nOr x = 1 and x = 2,\n\nHere we notice that for an equation of degrees 2 of the variable, we get two possible answers or values of the variable. Likewise we get more number of solutions for higher degrees of variables.\n\nExample 2: Suppose we have an equation as: 5x3 + 125x2 = 0. Then what is the value of 'x'?\n\nSolution: The given equation contains two terms on the left side and has to be factored:\n\n5x2 (x + 25) = 0,\n\nOr x2 = 0 and x = -25,\n\nOr x = 0, x =0 and x = -25.\n\n• ## Common Factors Calculator\n\nMath Topics\nTop Scorers in Worksheets<|endoftext|>"},"score":{"kind":"number","value":4.78125,"string":"4.78125"}}},{"rowIdx":1239,"cells":{"token_count":{"kind":"number","value":2398,"string":"2,398"},"text":{"kind":"string","value":"When linux is running, many activities are happening at the same time. These activities are called processes . Usually, the kernel has many processes running, and each user logged in will also have several processes running. For example, you might wish to run a C program that takes a long time to finish. It's possible in linux to run your program in the background, while freeing up the prompt in your shell to get on with some other work (like analysing your data). So, in linux, several programs can run at once, and each process can run at a different priority level . In this example, you might wish to give the C program a lower priority than your data analysis program.\n1. Jobs: background and foreground\nAll shells under linux have this capability, it is called job control , and it lets the user switch back and forth between processes, putting processes into the background, suspending processes, killing processes and changing the priority at which any of these processes are handled.\nWhen ever you launch a new process by typing a command in your shell, you can automatically put this process into the background by appending your command with a & .\nOf all jobs running at one time, at most one job is the foreground job. All other jobs will be suspended jobs or background jobs. The foreground job ties up the shell prompt - so if you want access to your shell in this instance, this foreground job must be put into the background. To do this you must first suspend the forground job, tying CtrlZ in your shell (you don't need to hit return).\nYou can always find out what jobs you have launched and are running in the current shell by typing the command jobs . Again, an example may help.\n|Let's say the user phillipa types jobs at the prompt. She is running 3 processes in this directory. One is called emacs (an editor), one is called xmakemol, and another is called xmgrace (she has suspended this job). The output of jobs is as follows;|\n - 21682 Running emacs carts2hexfrac.f\n 24410 Running xmakemol -f bulk-opt.xyz\n + 25174 Suspended xmgrace\n|The first entry in each line gives the job number .The second entry gives an indication of the most recently stopped (or started) job - in other words, the current job . The current job is signified by a + . Also indicated in this column is the previous job (the one that was current before the current job was started/stopped. The previous job is signified by a - .|\n|The third column gives the status of each job. In this case you can see that two are running and one is suspended. A suspended job will stay in this state until it is restarted or killed .|\n|The final entry on each line gives the actual process.|\nThere are several linux commands regarding jobs, so you need a way to refer to each job. This can be done by referring to the job number , symbollically given as %n for job number n. You can also refer to the current job with just % .\nWe've already seen an example of moving the foreground job to the background. You can also move a background job to the foreground by typing fg %n (where n is a number). You can also cause any suspended job (of which there might be several) to execute in the background by typing bg %n (where n is a number).\nYou can kill any job (so it will cease to exist as a process) by typing kill %n (where n is a number). The kill command comes with plenty of options - you'll find out more later.\nPutting it into Practice:\n|Part I Suppose you'd like to start a program called emacs from your shell. You'd also like to put this job into the background. You can then run this job from the prompt by typing|\n|After you hit return, a window should pop up, and you should have control of your shell prompt as normal.|\n|Now, check what processes you have active at the moment by typing jobs . Note the job number of your emacs process.|\n|You should now kill your emacs process by entering kill %n (where n is your job number). The emacs window should vanish.|\n|Part II Try launching emacs from your shell again, this time without the ampersand ( no & at the end ).|\n|The emacs window will again pop up, but this time you will not have control of the shell prompt, since emacs is now running as a foreground job.|\n|You can always kill a foreground process by typing Ctrl C. Or, you can background this process by the usual route of suspending ( Ctrl Z followed by bg ).|\nBack to top\n2. Redirection of input/output\nMost jobs need input and output options. By convention in linux, the default input is the standard input and the default output is the standard output . The standard input is defined as the input from the prompt at your terminal. Similarly, the standard output is defined as the output to your prompt at your terminal. It would be a huge pain if you could only use standard input and standard output - imagine a program that needs 10,000 numbers as input - you wouldn't want to feed them in by hand! Instead, under linux, there is the oportunity to redirect input and output. Input redirection is indicated by a < in front of the input file name, and output redirection is indicated by a > in front of a file name.\nAnother useful redirection is >> . This will append output to a file, rather than write over that file with the output.\n|Suppose you have a program called averages , and the data you wish to average is collected in a file called ifile . Let's say you also would like to put your output into a file called ofile for safekeeping. You'd also like to put this job into the background. You can then run this job from the prompt by typing|\naverages < ifile > ofile &\nBack to top\n3. Finding the process id\nLinux assigns each process an identification number called a process id (PID) when the process is created. There are several ways for a user to determine a PID for a given process.\nFirst, if you launch a background process by issuing a command to your shell, after you hit return, you should see two items displayed in the standard output of your xterm. The first one will be in square brackets (this is the job number), and the second is a number (this is the PID).\n|The user harry launches an emacs process by typing|\n|Immediately after hitting return, the following output is sent to his display|\n|This tells harry that this job number is 5, and the PID for this emacs process is 18408|\nSometimes a process is not explcitly launched by the user (e.g. a process can be launched by a shell script executed by the user), and therefore the above method can not be used to determine the PID. Another way to find the PID is to use the command ps . This command reports the process status for ALL processes running on your machine. This may be a long list, since even if you are not running any jobs yourself, the kernel runs many background processes (known as daemons). ps (stands for Process Status) comes with many options, the most useful being:-e produce information about every process\n-f produce a full listing\n-l produce a long listing\na helpful option is -ef which gives two of these options.\n|The user harry wants a status report of all processes on his machine. To do this he types|\n|Immediately after hitting return, output is sent to his display, a sample of this is shown below:|\nUID PID PPID C STIME TTY TIME CMD\nroot 0 0 0 Sep 18 ? 0:17 sched\nroot 1 0 0 Sep 18 ? 0:54 /etc/init -\ndaemon 156 1 0 Sep 18 ? 0:00 /usr/lib/nfs/statd\nThe headings on this table of output have the following meanings\nUID the user who owns the process\nPID the process id, a unique identifier assigned to each process\nPPID the parent process id, the process that spawned the current process\nC this field is obsolete\nSTIME the start time for the current process\nTTY the controlling terminal for the current process\nTIME the amount of CPU time accumulated by the current process\nCMD the command used to invoke the process\nAnother way to find out PIDs is to use the command top . As output this command gives a continually updated listing of active processes in real time. It is run ONLY as a foreground process. The output can be sorted in terms of CPU usage (this is the default), memory usage and runtime. top has an interactive element - while it is running, you can type many options, a sample of these are given below:space key immediately updates the display\nn or # change the number of processes to show\nM sort display by memory usage\nk kill a job\nThe last option is important. To kill a job in top, after typing k, you will be prompted for the PID of the process you wish to kill. Type in the PID of your desired process and hit return. You will then be prompted for the signal with which to send this kill command (a signal notifies a process of any abnormal external event). To make sure the process really dies, use the signal \"9\" (type 9 when prompted). This is known as a \"hard kill\".\nPutting it into Practice:\n|Start by typing top at your prompt and hit return. As output, you should see a list of processes under various headings (these heading should be self-explanatory). One of those headings, PID, gives the process id for each process.|\n|While top is running, type k in the window.|\n|You will be prompted for a process id. Find the PID of the top process itself (it should be up the top of the list), type in this PID and hit return.|\n|You will now be prompted for the signal. Type 9 and hit return.|\n|You top session should terminate immediately!|\nFrom running your top process in the previous example, most of the headings in the output display should make sense. One heading, PRI deserves a little more explanation. The number under PRI gives the priority of the job. For regular users, you can set this number between 1 (highest priority) and 19 (lowest priority) - more about this later. The default priority is 10. System administrators can set the priority to be a negative number (even higher priority), all the way to -20.\nBack to top\n4. Killing jobs by PIDNow that you know how to get a PID by using the ps command, you can kill jobs on the command line, even if these processes don't show up when you type jobs. You can simply kill a process by using the kill command in the following way:\nkill -signal PID\nusually, the signal number will be 9 (as explained above)\nFor example, if user harry wants to kill an emacs process, knowing that the PID of this process is 6743, at his shell prompt, he can type\nkill -9 6743Back to top<|endoftext|>"},"score":{"kind":"number","value":3.703125,"string":"3.703125"}}},{"rowIdx":1240,"cells":{"token_count":{"kind":"number","value":738,"string":"738"},"text":{"kind":"string","value":"This is interesting!\nEXTRACT (12 Literacies)\n01 Play: the capacity to experiment with one’s surroundings as a form of problem-solving. Having a strong sense of play can be helpful when you pick up a new piece of technology that you’ve never used before, when you’re trying to write an essay and your outline isn’t functioning as you’d hoped, and when you’re designing anything at all, from a dress to a web page to a concert’s program.\n02 Performance: the ability to adopt alternative identities for the purpose of improvisation and discovery. Being able to move fluidly and effectively between roles can help you when you’re exploring online communities, when you’re trying to decide what actions are ethical, and when you’re shuffling between home, work and school.\n03 Simulation: the ability to interpret and construct dynamic models of real-world processes. Being able to interpret, manipulate and create simulations can help you understand innumerable complex systems, like ecologies and computer networks – and make you better at playing video games!\n04 Appropriation: the ability to meaningfully sample and remix media content. Being able to remix media content (and knowing when doing so is appropriate) can help you understand literary works, music, and art; it can also help lead you to a deeper understanding of copyright and cultural clashes.\n05 Multitasking: the ability to scan one’s environment and shift focus as needed to salient details. Being a good multitasker is required in our new media landscape – and that includes learning when it isn’t good to multitask.\n06 Distributed Cognition: the ability to interact meaningfully with tools that expand mental capacities. That can mean something as simple as using a ruler or calculator, or something as complex as efficiently using Wikipedia on your iPhone to access information on the fly.\n07 Collective Intelligence: the ability to pool knowledge and compare notes with others toward a common goal. This ability is key to open source projects. Being able to pool knowledge with others can allow us to solve challenges far more complex than the individual mind can process.\n08 Judgment: the ability to evaluate the reliability and credibility of different information sources. If you’re worried about your students using Wikipedia at inappropriate times and taking everything they read on the internet as gospel truth, you’re worried that they aren’t exercising good judgment. But judgment also includes knowing when sources are appropriate for your use: for instance, sometimes Wikipedia might be the appropriate resource to use.\n09 Transmedia Navigation: the ability to follow the flow of stories and information across multiple media. Anyone who needs to do research needs a good understanding of transmedia navigation – how to follow threads through video, still photography, written work, music, online sources etc.\n10 Networking — the ability to search for, synthesize, and disseminate information. Writing something isn’t enough without the ability to circulate it to the communities where it will matter.\n11 Negotiation — the ability to travel across diverse communities, discerning and respecting multiple perspectives, and grasping and following alternative norms. We now need to know how to live in multiple communities – from the hyperlocal to the global and from those composed of people like us to those consisting of people very different from us.\n12 Visualization – the ability to translate information into visual models and understand the information visual models are communicating. VIsualization has become a key way we cope with large data sets and make sense of the complexity of our environment.<|endoftext|>"},"score":{"kind":"number","value":3.75,"string":"3.75"}}},{"rowIdx":1241,"cells":{"token_count":{"kind":"number","value":2541,"string":"2,541"},"text":{"kind":"string","value":"# Normal Distributions\n\n## Fitting a bell curve to a histogram\n\nEstimated13 minsto complete\n%\nProgress\nPractice Normal Distributions\n\nMEMORY METER\nThis indicates how strong in your memory this concept is\nProgress\nEstimated13 minsto complete\n%\nNormal Distributions\n\n### Normal Distributions\n\nPreviously you've spent some time learning about probability distributions. A distribution, itself, is simply a description of the possible values of a random variable and the possible occurrences of these values. Remember that probability distributions show you all the possible values of your variable (\\begin{align*}X\\end{align*}), and the probability associated with each of these values \\begin{align*}(P(X))\\end{align*}. You were also introduced to the concept of binomial distributions, or distributions of experiments where there are a fixed number of successes in \\begin{align*}X\\end{align*} (random variable) trials, and each trial is independent of the other. In addition, you were introduced to binomial distributions in order to compare them with multinomial distributions. Remember that multinomial distributions involve experiments where the number of possible outcomes is greater than 2, and the probability is calculated for each outcome for each trial.\n\nIn this first concept on probability distributions, you are going to begin by learning about normal distributions. A normal distribution curve can be easily recognized by its shape. The first 2 diagrams above show examples of normal distributions. What shape do they look like? Do they look like a bell to you? Compare the first 2 diagrams above to the third diagram. A normal distribution is called a bell curve because its shape is comparable to a bell. It has this shape because the majority of the data is concentrated at the middle and slowly decreases symmetrically on either side. This gives it a shape similar to a bell.\n\nActually, the normal distribution curve was first called a Gaussian curve after a very famous mathematician, Carl Friedrich Gauss. He lived between 1777 and 1855 in Germany. Gauss studied many aspects of mathematics. One of these was probability distributions, and in particular, the bell curve. It is interesting to note that Gauss also spoke about global warming and postulated the eventual finding of Ceres, the planet residing between Mars and Jupiter. A neat fact about Gauss is that he was also known to have beautiful handwriting.\n\nYou previously learned about discrete random variables. Remember that discrete random variables are ones that have a finite number of values within a certain range. In other words, a discrete random variable cannot take on all values within an interval. For example, say you were counting from 1 to 10. You would count 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. These are discrete values. 3.5 would not count as a discrete value within the limits of 1 to 10. For a normal distribution, however, you are working with continuous variables. Continuous variables, unlike discrete variables, can take on any value within the limits of the variable. So, for example, if you were counting from 1 to 10, 3.5 would count as a value for the continuous variable. Lengths, temperatures, ages, and heights are all examples of continuous variables. Compare these to discrete variables, such as the number of people attending your class, the number of correct answers on a test, or the number of tails on a coin flip. You can see how a continuous variable would take on an infinite number of values, whereas a discrete variable would take on a finite number of values. As you may know, you can actually see this when you graph discrete and continuous data.\n\n#### Interpreting Graphs\n\nLook at the 2 graphs below. The first graph is a graph of the height of a child as he or she ages. The second graph is the cost of a gallon of gasoline as the years progress. Which graph represents discrete data? Which graph represents continuous data?\n\nIf you look at the first graph, the data points are joined, because as the child ages from birth to age 1, for example, his height also increases. As he continues to age, he continues to grow. The data is said to be continuous and, therefore, you can connect the points on the graph. For the second graph, the price of a gallon of gas at the end of each year is recorded. In 1930, a gallon of gas cost \\begin{align*}10 \\cancel{\\text{c}}\\end{align*}. You would not have gone in and paid \\begin{align*}10.2 \\cancel{\\text{c}}\\end{align*} or \\begin{align*}9.75 \\cancel{\\text{c}}\\end{align*}. The data is, therefore, discrete, and the data points cannot be connected.\n\nLet’s look at a few problems to show how histograms approximate normal distribution curves.\n\n#### Understanding Data Distributions\n\nJillian takes a survey of the heights of all of the students in her high school. There are 50 students in her school. She prepares a histogram of her results. Is the data normally distributed?\n\nIf you take a normal distribution curve and place it over Jillian’s histogram, you can see that her data does not represent a normal distribution.\n\nIf the histogram were actually shaped like a normal distribution, it would have a shape like the curve below:\n\n#### Interpreting Distributions\n\nThomas did a survey similar to Jillian’s in his school. His high school had 100 students. Is his data normally distributed?\n\nIf you take a normal distribution curve and place it over Thomas’s histogram, you can see that his data also does not represent a normal distribution.\n\n### Example\n\n#### Example 1\n\nJoanne posted a problem to her friends on FaceBook. She told her friends that her grade 12 math project was to measure the lifetimes of the batteries used in different toys. She surveyed people in her neighborhood and asked them, on average, how many hours their typical battery lasts. Her results are shown below:\n\n\\begin{align*}& 98 && 108 && 107 && 79 && 100 && 112 && 97 && 79 && 41 && 127\\\\ &135 && 100 && 92 && 80 && 66 && 62 && 119 && 118 && 56 && 112\\\\ &99 && 83 && 86 && 62 && 127 && 155 && 107 && 140 && 144 && 122\\\\ &110 && 116 && 134 && 102 && 133 && 157 && 100 && 96 && 55 && 132\\\\ &126 && 171 && 169 && 146 && 122 && 74 && 70 && 82 && 84 && 93\\end{align*}\n\nIs her data normally distributed? Where is the center of the distribution?\n\nIf you take a normal distribution curve and place it over Joanne’s histogram, you can see that her data appears to come from a normal distribution.\n\nThis means that the data fits a normal distribution with a mean around 105. Using the TI-84 calculator, you can actually find the mean of this data to be 105.7.\n\nWhat Joanne’s data does tell us is that the mean score (105.7) is at the center of the distribution, and the data from all of the other scores (times) are spread from that mean. You will be learning much more about standard normal distributions in a later Concept. But for now, remember the 2 key points about a standard normal distribution. The first key point is that the data represented is continuous. The second key point is that the data is centered at the mean and is symmetrically distributed on either side of that mean.\n\n### Review\n\n1. The following data was collected on a recent 25-point math quiz. Does the data represent a normal distribution? Can you determine anything from the data? \\begin{align*}& 20 && 17 && 22 && 23 && 25\\\\ & 14 && 15 && 14 && 17 && 9\\\\ & 18 && 2 && 11 && 18 && 19\\\\ & 14 && 21 && 19 && 20 && 18\\\\ & 16 && 13 && 14 && 10 && 12\\end{align*}\n2. A recent blockbuster movie was rated PG, with an additional violence warning. The manager of a movie theater did a survey of moviegoers to see what ages were attending the movie in an attempt to see if people were adhering to the warnings. Is his data normally distributed? Do moviegoers at the theater regularly adhere to warnings? \\begin{align*}& 17 && 9 && 20 && 27 && 16\\\\ & 15 && 14 && 24 && 19 && 14\\\\ & 19 && 7 && 21 && 18 && 12\\\\ & 5 && 10 && 15 && 23 && 14\\\\ & 17 && 13 && 13 && 12 && 14\\end{align*}\n3. The heights of coniferous trees were measured in a local park in a regular inspection. Is the data normally distributed? Are there areas of the park that seem to be in danger? The measurements are all in feet. \\begin{align*}& 22.8 && 9.7 && 23.2 && 21.2 && 23.5\\\\ & 18.2 && 7.0 && 8.8 && 25.7 && 19.4\\\\ & 25.0 && 8.8 && 23.0 && 23.2 && 20.1\\\\ & 23.1 && 18.5 && 21.7 && 21.7 && 9.1\\\\ & 4.3 && 7.8 && 3.4 && 20.0 && 8.5\\end{align*}\n\nDetermine if the points representing each of the following data sets can be connected when graphed.\n\n1. The number of students enrolled in a college each semester\n2. The weight of a baby seal each day as it grows\n3. The number of coins a coin collector owns each week\n4. The speed of a rocket each second as it accelerates\n5. The amount of water in a swimming pool each minute as it is drained\n6. The number of employees a company has each month as it expands\n7. The thickness of a glacier each year as it melts\n\n### Notes/Highlights Having trouble? Report an issue.\n\nColor Highlighted Text Notes\n\n### Vocabulary Language: English Spanish\n\nTermDefinition\ncontinuous variable Variable that takes on any value within the limits of the variable\ndiscrete values discrete values are data where a finite number of values exist between any two values.\ndistribution A distribution is a description of the possible values of a random variable and the possible occurrences of these values.\nnormal distribution curve A normal distribution curve is a symmetrical curve that shows the highest frequency in the center with an identical curve on either side of the center.\ncontinuous variables A continuous variable is a variable that takes on any value within the limits of the variable.\nempirical rule The empirical rule states that for data that is normally distributed, approximately 68% of the data will fall within one standard deviation of the mean, approximately 95% of the data will fall within two standard deviations of the mean, and approximately 99.7% of the data will fall within three standard deviations of the mean.\nnormal probability plot A normal probability plot is a graph is a plot of the z -scores of the data as quantiles against the actual data values. If a distribution is normal, this plot will be linear.\nnormal quartile plot normal quartile plot is another name for a normal probability plot.\nstandard normal distribution The standard normal distribution, $\\phi(x)=\\frac{1}{\\sqrt{2 \\pi}}e^{- \\frac{1}{2}x^2}$ is a normal distribution with mean of 0 and a standard deviation of 1.\nz-score The z -score of a value is the number of standard deviations between the value and the mean of the set.<|endoftext|>"},"score":{"kind":"number","value":4.625,"string":"4.625"}}},{"rowIdx":1242,"cells":{"token_count":{"kind":"number","value":2473,"string":"2,473"},"text":{"kind":"string","value":"William Bradford (Mayflower passenger) facts for kids\nWilliam Bradford (1590 – 1657) was a passenger on the Mayflower in 1620. He travelled to the New World to live in religious freedom. He became the second Governor of Plymouth Colony and served for over 30 years. Bradford kept a journal of the history of the early life in Plymouth Colony. It is called Of Plymouth Plantation.\nBradford's father died was he was one year old. When he was four years old he was sent to live with his grandfather. Two years later, his grandfather died and he returned to live with his mother and stepfather. A year later, in 1597, his mother died. Bradford became an orphan at age 7. He was sent to live with two uncles.\nHis uncles wanted Bradford to help on the farm but Bradford was ill and was not able to work. Bradford started to read many books. One of the books was the Bible. This may be where he started learning of a religion that was different from the church of England.\nWhen Bradford was 12 years old he went with a friend to hear the pastor Richard Clyfton. Clyfton was a minister who did not believe in the Church of England. Bradford believed in Clyfton’s words. His uncles told him not to go to that church but he went anyway.\nKing James I of England started to punish the people who did not follow the Church of England. They were called Separatists. By 1607 many members of the Separatists were arrested. Brewster was made to pay a fine for going to the church. Some members were put into prison and others were watched night and day. The people also learned that other Separatists in London had been put into prison and left to starve to death.\nThe Separatists decided to leave England for the Dutch Republic (where religious freedom was permitted). William Bradford went with them. It was illegal to leave England. The group was betrayed by an English sea captain who brought them to the English authorities. Most of the congregation, and Bradford, were put into prison. By the summer of 1608, the Separatists, and William Bradford, escaped to Leiden, Netherlands. Bradford was 18.\nIn the Dutch Republic\nWilliam Bradford moved to Amsterdam in August 1608. He lived with the Brewsters in Stink Alley. The Separatists had spent most of their money leaving England and were poor. After nine months, the congregation moved to the smaller city of Leiden.\nIn 1613, Bradford married Dorothy May, the daughter of a rich English couple living in Amsterdam. In 1617, the Bradfords had their first child, John Bradford.\nBy 1617, the Separatists decided to travel to Virginia Colony in North America. The Separatists could practice their religion in the Dutch Republic, they wanted their children to keep their English customs and language. The colonists made a business arrangement with the Merchant Adventurers. These were a group of men who were willing to pay for the Pilgrims to go to the New World with the promise that the pilgrims would pay them back with furs and other goods they got from the colony. By July 1620, Robert Cushman and John Carver had made the plans and fifty Separatists left Delftshaven on the ship Speedwell.\nMany families were split as some Separatists stayed behind in the Netherlands, planning to make the voyage to the New World after the colony had been established. William and Dorothy Bradford left their three-year-old son John with Dorothy's parents in Amsterdam, possibly because he was too frail to make the voyage.\nThe Speedwell planned to meet with the Mayflower and both ships would travel to Virginia Colony. The Speedwell proved unsafe and the passengers were put on the Mayflower. This made the ship very crowded. Not all the passengers were Separatists. There were about 50 colonists who went because they had certain skills. These skills would prove useful on the Mayflower and at the colony. Bradford said they hugged each other and cried and said goodbye to family and friends for the last time. They knew they were pilgrims and lifted their eyes to heaven, their dearest country and this comforted them.\nThe Mayflower Voyage\nThe Mayflower left Plymouth, England on 16 September 1620. There were 102 passengers and 30–40 crew. On 19 November 1620, the Mayflower spotted land. The Mayflower was supposed to land in Virginia Colony, but the ship was too damaged and they were forced to land at Cape Cod now called Provincetown Harbor.\nThey landed on November 21. They wrote the Mayflower Compact, which made rules on how they would live and treat each other. Bradford signed the Mayflower Compact.\nAnchored and first explorations at Plymouth Colony\nUp to this time, Bradford, aged 30, had never been a leader in the colony. When the Mayflower anchored in the Harbor, the men went out to explore the land for a place where they could live. Bradford volunteered to be a member of the exploration parties. In November and December, these parties went out three times on foot and by boat.\nOne of the earliest exploration missions included William Bradford, John Carver, Myles Standish, Edward Winslow, John Howland, Richard Warren, Stephen Hopkins and Edward Doty. Of this group Hopkins was the most experienced, having seen Indians during his time at Jamestown. According to Bradford, they set off in below freezing weather, many of the men already being ill, \"and when the set sail the salt spray froze on their coats, as if they had been glazed\". During their trip they saw Indians who looked like they were cutting up a large whale. When the Indians saw the men they ran away. The men camped and protected themselves from the cold and the Indians.\nThey located a place that would be Plymouth Colony in December. During the first expedition on foot, Bradford was caught up in a deer trap made by Native Americans and brought upside down. During the third exploration, on 6 December 1620, Bradford and the other men found Plymouth Bay. A winter storm nearly sank their boat as they came near the bay. The explorers were very cold and started to get sick. The waves were high but they landed on Clark's Island.\nDuring this time, they searched areas all around the bay and found a place to live and build their homes. The location had a large hill. This hill is now called Burial Hill. It was a good place to build a fort for safety against any possible enemies. There were many brooks providing water. They would later find out that the site had been the location of a Native American village known as Patuxet; and that is why a large area had already been cleared for planting of food. The Patuxet tribe of Indians, between 1616 and 1619, had all died from bad sickness. Bradford later wrote that bones of the dead were easy to find in many places.\nThe Mayflower arrived in Plymouth Bay on 20 December 1620. The settlers began building the colony's first house on 25 December.\nOn January 11, 1621, as Bradford was helping to build houses when he was got a great pain in his hipbone and fell. Bradford was taken to the \"common house\" (the only house that was finished) and they thought he would die that night, but he did not.\nBradford got well but many of the pilgrims were not so fortunate. During the months of February and March 1621 sometimes two or three people died every day. By the end of the winter, half of the 100 settlers had died. They buried their dead in secret so the Native Americans would not see how many were dying. They buried them in Cole's Hill and tried to also hide the graves from the animals.\nThere were only a small number of men who were healthy. Everyone, including the children had to work and care for the sick. One of these was Captain Myles Standish. He was a soldier who had been hired by the settlers to protect the people of the colony. Standish cared for Bradford during his illness. They became good friends. Bradford had no military experience and when he became governor he would come to trust Captain Standish's advice on military matters.\nWilliam was chosen by the people to be Governor of Plymouth Colony. He kept journals (a written record) which are very important even today because they are the only history of how the Mayflower passengers and their families lived in Plymouth Colony. He died before he could finish the second book.\nEarly service as governor\nOn March 16, the settlers had their first meeting with the Native Americans who lived in the region. Samoset, an Indian walked into the village of Plymouth and greeted them in English. Samoset had learned some English from the traders who had visited the area he was from. This soon led to a visit by Massasoit, the leader of the Pokanoket tribe. At that meeting, Massassoit made a treaty with Governor John Carver. The treaty said that they would be friends and protect each other against unfriendly Indians.\nBradford wrote down what was said. He would soon become governor and the treaty that was very important was the promise to help each other. This agreement was not liked by Massasoit's enemies in other tribes.\nIn April 1621, Governor Carver became sick while working in the fields on a hot day. He died a few days later. The settlers of Plymouth then chose Bradford as the new governor. Bradford would remain governor most of his life.\nThe elected leadership of Plymouth Colony was at first a governor and an assistant governor. The assistant governor for the first three years of the colony's history was Isaac Allerton. In 1624, it was changed to five assistants. This was called the \"governor's council\". These men gave advice to the governor and had a vote on important matters. They helped Bradford in managing the Colony. Assistants during the early years of the colony included Thomas Prence, Stephen Hopkins, John Alden, and John Howland.\nWilliam Bradford married Dorothy in Amsterdam, Holland on December 10, 1613. They had one son. Dorothy fell from the deck of the Mayflower into the water and drowned a short time after the ship landed. Her husband was on an exploring mission. John was born in Leiden, Holland and died in 1620.\nHis second wife was Alice Southwarth in Plymouth. They had three children. She died in Plymouth in March 1670 and was buried on Burial Hill in Plymouth near her husband.\nBradford's wrote Of Plymouth Plantation. It was a history about Plymouth colony. It is often called a journal. It was a collection of what he saw and what he thought was important. This was written in two books. The first book was written in 1630 but the second was never finished. During the years 1646 and 1650, he wrote about the colony's history up to 1646.\nIn Of Plymouth Plantation, Bradford saw what was happening to be like what happened in the Bible. As Philip Gould writes, \"Bradford hoped to show the workings of God to future generations.\" The book was not published until 1656, the year before his death, and it was well liked. Bradford has been called the father of American history. Many American authors have put what Bradford said in their books. Thomas Prence did so in his book named A Chronological History of New-England in the Form of Annals. Today it is considered very important and studied in American schools and colleges. Bradford's journal was included in another work entitled Mourt's Relation. This was written in part by Edward Winslow, and made into a book in England in 1622.\nImages for kids\nProvincetown, Massachusetts memorial to Pilgrims who died at sea or on board the Mayflower in Cape Cod Harbor in November/December 1620\nWilliam Bradford (Mayflower passenger) Facts for Kids. Kiddle Encyclopedia.<|endoftext|>"},"score":{"kind":"number","value":3.8125,"string":"3.8125"}}},{"rowIdx":1243,"cells":{"token_count":{"kind":"number","value":1949,"string":"1,949"},"text":{"kind":"string","value":"+0\n\n# Probability puzzles.\n\n+1\n141\n4\n+90639\n\nHere are some more probability puzzles.\n\n1) If you break a stick in two places at random, forming three pieces, what is the probability of being able to form a triangle with the pieces?\n\n2) If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form a triangle with them?\n\n3) If you break a stick in two places at random, what is the probability of being able to form an acute triangle — where each angle is less than 90 degrees — with the pieces?\n\n4) If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form an acute triangle with the sticks?\n\nThe probabilities are, respectively, 1/4, 1/2, ln(8)−2ln⁡(8)−2 and 1−π/41−π/4.\n\nThe full answers are here but I doubt they will be available for very long.\n\nhttps://fivethirtyeight.com/features/how-do-you-like-these-rectangles/\n\nMelody  Oct 4, 2017\nSort:\n\n#1\n+90639\n0\n\nQuestion 1.\n\n1) If you break a stick in two places at random, forming three pieces, what is the probability of being able to form a triangle with the pieces?\n\nI will use a technique known as probablity contour mapping.\n\nLet the stick be one unit long.\n\nBreak it so the the three peices are  x, 1-y and y-x  units long where y>x\n\nNote that when you add these up you get 1.      x+(1-y)+(y-x)=1\n\nObviously both x and y must be less than 1 unit long.\n\nNow to form a triangle the sum of any two sides must be greater than the third side.\n\nso we have\n\n1)      $$x+(1-y)>y-x\\\\ x+1-y>y-x\\\\ -2y>-2x-1\\\\ y 2) \\(x+(y-x)>1-y\\\\ y>1-y\\\\ y>0.5$$\n\n3)    $$(1-y)+(y-x)>x\\\\ 1-x>x\\\\ x<0.5$$\n\nYou plot this on a number plane and get the following region:\n\nBut so far I have only considered where y>x,  maybe x>y\n\nthen I get the inverse region. That is the given region reflected about the line y=x.\n\nSo the total region is\n\n1/4 of the entire region is shaded so the probablility that the three peices can form a triangle is 1/4\n\nMelody  Oct 6, 2017\n#2\n+90639\n0\n\nNumber 2\n\nthis answer is copied, I have not done it myself yet.\n\nFor No. 2: What if, instead of pieces from one stick, we pick up three sticks of random length somewhere between zero and one? This is a little trickier. When we plotted the first problem, it could be collapsed to two dimensions, because we were only really worried about two sticks — the length of our third piece was automatically determined by the length of our first two pieces. But this problem is in three dimensions, so the solution needs to be plotted not in a one-by-one square but rather in a one-by-one-by-one cube.\n\nTo help solve the problem, consider what wouldn’t solve it: a violation of our triangle inequalities. Suppose, for example, that x>y+zx>y+z, which makes it impossible to build a triangle. In that formulation, those points are contained in a pyramid bounded by the planes y=0y=0, z=0z=0, x=1x=1 and x=y+zx=y+z. There are three such pyramids in this cube, one for each of the ways the triangle inequality can be violated.\n\nEach of those pyramids has a volume of ⅙. (A pyramid has a volume equal to the area of its base times its height, all divided by three. Our pyramid in question has height 1 and a triangular base with area ½.). Therefore, there is a ½ chance we can’t make a triangle, and a ½ chance we can. And so we have the answer to the second problem.\n\nMelody  Oct 6, 2017\n#3\n+90639\n0\n\nNumber 3\n\nNo. 3: Getting tougher still (as though that were possible)! Now it’s time for calculus. Guy D. Moore explains this one for us:\n\nThe problem asks us to ensure that three pieces form an acute triangle. Consider three pieces with lengths  $$x>y>z$$\n\nFirst, think about a right triangle. The formula for that, as our middle school teachers drilled into our heads, is $$x^2=y^2+z^2$$. (Otherwise known as the Pythagorean Theorem.) To have an acute triangle, all angles must be less than 90 degrees, so we tweak that formula: $$x^2 From this we get that \\(y and \\((1-x)^2 , which is the same as \\(\\begin{equation*}x<\\frac{1-2y^2}{2(1-y)}\\end{equation*}$$\n\nSince we’re dealing with pieces of the same stick and not three separate sticks, we can return to plotting in two dimensions, not three. And our mirrored-triangle plot is useful again since our answer lies within those two original triangles. This time, though, we need to draw two new three-pointed shapes within those two triangles. The area of those shapes will be our answer — the probability of an acute triangle.\n\nSo to calculate our new shapes, we need to cut pieces out of our original triangles. The area of one of those pieces is expressed in an integral (which is the calculus part of the solution). That integral is:\n\n$$\\begin{equation*}\\int_0^{1/2}\\frac{1-2y}{2-2y}dy=\\frac{1-\\ln(2)}{2}\\end{equation*}$$\n\nThere are six shapes, each with the same area, cut out of our one-by-one square, leaving:\n\n$$\\begin{equation*}1-3(1-\\ln(2))=\\ln(8)-2\\end{equation*}$$\n\nIn that equation, “ln” is the natural log, which equals an implied probability of acute triangle-formation of about 7.9 percent. (Who knew that natural logs are a great way to solve stick problems?) Guy also provided this illustration of the curvy areas we calculated:\n\nMelody  Oct 6, 2017\n#4\n+90639\n0\n\nNo. 4:\n\nIf you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form an acute triangle with the sticks?\n\nWe’re back to three dimensions again for the final question. This solution furthers the solution from problem No. 2, the way that solution No. 3 furthers solution No. 1. Laurent explained his solution this way:\n\nWe’ll solve this problem the same way we solved No. 2, but we’ll replace the triangle inequalities with the acute triangle inequalities. As in No. 2, we end up with a three-dimensional volume rather than a two-dimensional area. For simplicity again, we’ll assume that cc is the largest length, which accounts for one-third of all possibilities.\n\nLaurent provided a lovely illustration of this volume:\n\nOur answer will ultimately be three times the area of this shape (this shape only accounts for stick c being longest, and two identical shapes will be generated for stick b being longest and stick a being longest).\n\nOur solution lies in the filled-in parts of that shape. While this lookscomplicated, the curved surface inside that area has the equation $$c^2 = a^2 + b^2$$, which is, conveniently, the equation of a right circular cone!\n\nSo we can calculate the volume of the region of interest by subtraction. It’s ⅓ of the volume of the cube minus ¼ of the volume of the cone. (One-third because we’re considering only one out of three scenarios, the one where c is longest. And ¼ because the cone’s base is ¼ of a circle.)\n\nThe total probability is three times this volume, because we must account for the remaining identical pieces.\n\nThe final answer is $$3(1/3 − 1/4 ( \\pi/3 ) ) = 1 − \\pi/4$$or about 0.2146.\n\nSo the probability of forming an acute triangle with three randomly chosen lengths is about 21.5 percent.\n\nMelody  Oct 6, 2017\n\n### 27 Online Users\n\nWe use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details<|endoftext|>"},"score":{"kind":"number","value":4.4375,"string":"4.4375"}}},{"rowIdx":1244,"cells":{"token_count":{"kind":"number","value":1670,"string":"1,670"},"text":{"kind":"string","value":"Despite decades of taking measurements, scientists cannot agree on how long neutrons live. Neutrons are stable inside atoms, but on their own they decay in about 15 minutes, more or less, into a few other particles. Exactly how much more or less is the sticking point. Each experiment seems to yield a different answer.\nThe lack of resolution is frustrating. Understanding the lifetime of the neutron is important not only for knowledge’s sake but also to answer other more fundamental questions about new physics beyond the known particles and processes in the universe, says Jeffrey Nico, who leads a neutron lifetime experiment at the National Institute of Standards and Technology (NIST) in Gaithersburg, Md. “We can’t leave this disagreement just hanging out there.”\nTo measure how long neutrons last before decaying, scientists look for either the disappearance of neutrons or the appearance of their decay products. Neutrons disappear through a process called neutron beta decay, transforming themselves into protons by emitting an electron (shedding a negative charge to become positively charged) and an antineutrino. One method of measuring how long neutrons last before decaying entails trapping a group of neutrons inside a physical bottle and counting how many remain after various lengths of time. Another, called the beam method, is to produce a tightly confined spray, or beam, of neutrons and surround it with a “proton trap” to count the protons created when neutrons decay.\nBeam experiments have been going on for more than 30 years, with the leaders in the field located at NIST. Their best, latest beam results, published in 2013, measured the neutron lifetime at 887.7 seconds, plus or minus 3.1 seconds. Bottle experiments, in contrast, go back only about 15 years but have already posted results with higher precision than beam findings. The best result to date from a bottle experiment came in 2008 from a collaboration between the Petersburg Nuclear Physics Institute and the Joint Institute for Nuclear Research in Russia, along with the Institut Max von Laue Paul Langevin (ILL) in France. That team measured the neutron lifetime at 878.5 seconds, plus or minus one second.\nA difference of about nine seconds may not sound like much, but it is larger than the estimated error bars on the experiments, meaning those error bars are wrong on some—maybe all—of the results. “The discrepancy is embarrassing,” says University of Tennessee physicist Geoffrey Greene, who worked on the NIST project. “One or more of us screwed up, and we just have to find out which ones.”\nAnd although nine seconds is not long, it is enough to make a difference in some calculations that rely on the neutron lifetime, such as predictions of how the first atomic nuclei formed. Protons and neutrons started out as free particles in the hot, dense baby universe. Only after the cosmos had cooled sufficiently—within in the first 20 minutes after the big bang—did they combine to form nuclei in a process called nucleosynthesis. “Essentially all of the helium in the universe was made at that epoch,” Nico says. To know how many neutrons were available to form nuclei, scientists must know how long neutrons last before decaying. “For big bang nucleosynthesis, right now the neutron lifetime is the input with the largest uncertainty on it, so by improving the neutron lifetime, you can improve those predictions,” Nico adds.\nIf nucleosynthesis predictions turn out not to match evidence from astrophysical observations of the abundance of helium, for example, exotic physics just might be at work. One possibility is dark matter—the invisible, but plentiful, substance in the universe thought to be made of some types of undiscovered particles. “All sorts of dark matter candidates might actually play a role in big bang nucleosynthesis,” says theoretical physicist Susan Gardner of the University of Kentucky. These particles might have interacted with protons and neutrons or participated in reactions somehow to alter the number of nuclei that formed.\nUnderstanding neutron beta decay is also important for understanding one of the four fundamental forces of nature—the weak force. This force is responsible for nuclear fusion as well as radioactive decay—such as neutron beta decay. “Neutron decay is one of the simplest examples of the weak interaction between light particles like electrons and heavy particles like quarks [the ingredients of neutrons],” Greene says. “That’s what we’re trying to get at by studying neutron decay.” The standard model of particle physics describes neutron decay very well, but scientists wonder if it is the full description. If measurements of neutron decay ever deviated from standard model predictions, they might point us toward a new, deeper level of physics.\nAnother possibility is that new physics is behind the trouble scientists are having with neutron lifetimes. The discrepancy between bottle and beam measurements is curious, scientists point out. Bottle experiments tend to find a shorter neutron lifetime than the beam tests do. Perhaps neutrons occasionally decay by a process other than beta decay, which means they might turn into something other than protons. That would create a number of “missing” protons in the beam experiments, which would lead them to measure a longer lifetime than bottle experiments. (It would seem as if fewer neutrons are decaying in the allotted time, when really they are simply decaying into additional, unseen products.) “Is there some kind of new physics that could explain getting a different value for these two methods?” Greene asks. “The fact that we’re seeing a longer lifetime is consistent with the notion, and that would be tremendously exciting, but I think it’s more likely somebody’s screwed up their experiment.”\nNeutron decay experiments are painstakingly complex. For beam experiments, the main challenge is to make sure to correctly tally both the neutrons in the beam, and the protons that result from neutron decays. “Conceptually it’s fairly straightforward, but the trick is counting those particles absolutely,” Nico says. Additionally, the researchers must measure the length of the proton trap precisely to get accurate measurements.\nThe difficulty in bottle experiments comes from the bottle—particularly the possibility that neutrons interact somehow with the bottle walls , which can be made of various materials, such as oil-coated copper. “When the neutron touches the wall, in the ideal case they should be totally reflected with no losses, but there can be contaminations on the wall,” says Peter Geltenbort of ILL, a member of the leading bottle experiment team. That group is working on building a larger bottle, and comparing the results with experiments using smaller bottles. “The idea is if you compare different bottles, you can then do an extrapolation to an infinitely large bottle and bring the precision down to an order of 0.3 or 0.4 seconds.”\nAnother tack is to eliminate the walls of the bottle altogether. A few teams are working on “bottles” made out of magnetic and gravitational fields, as opposed to tangible materials, to trap neutrons. Although neutrons don’t have an electrical charge, their spin gives them what’s called a magnetic moment, making them act like tiny magnets inside a magnetic field. “We put more than 5,000 individual magnets together to form a trap, and the neutrons are effectively levitated by the presence of this strong magnetic field,” says Chen-Yu Liu, a physicist at Indiana University Bloomington who is working on a magnetic-gravitational field bottle experiment at the Los Alamos National Laboratory.\nUltimately, both the beam and bottle experiment teams hope they can reconcile their results. “I think it’s universally acknowledged that both methods are fundamentally sound,” Greene says. “It’s a matter of the devil is in the details.” And if neutrons persist in their befuddling behavior, it just might mean the universe is a bit more complicated than we thought.<|endoftext|>"},"score":{"kind":"number","value":4.09375,"string":"4.09375"}}},{"rowIdx":1245,"cells":{"token_count":{"kind":"number","value":410,"string":"410"},"text":{"kind":"string","value":"History of the Bude Canal\nThe idea of the Bude Canal was conceived in 1774 by Cornishman, John Edyvean. Its main purpose would be to transport lime-rich sea-sand inland, to use as a manure on farms where soils were acidic and unproductive. Other cargoes, such as coal, culm, slate, timber, iron and bricks could also be carried and farm produce exported.\nHowever, his scheme proved too ambitious and it was abandoned until 1817, when renewed local interest produced a canal survey and report. This proposed a 19ft wide canal, consisting of a two mile stretch navigable by barge from Bude to Helebridge; plus another 33 miles of narrow tub-boat canal, where wheeled ‘youb-boats’ would be pulled by horses, then winched up a system of ‘inclined planes’ to get over the hills.\nAlso proposed was the building of a sea lock, a breakwater and moving the mouth of the River need, whilst water would be supplied from the specially formed Lower Tamar Lake. Costs were estimated at £90,000.\nThe project was eventually agreed and work began on the 23rd July, 1819 under the Bude Harbour & Canal Company. Four years later, much of the line was opened and in 1825 all that was ever to be built was completed at the final cost of £120,000.\nAlthough never a great success for the shareholders, the canal was indispensable to inland communities and enabled their farms to grow and prosper. However, trade declined drastically when the railway reached Bude in 1898 and the Canal closed in 2901. Most sections were sold back to the landowners.\nIn 2006 the Bude Canal Regeneration Project began work on a £5 million facelift, aimed to restore the first two miles of the canal from the sea lock to Helebridge. This was completed in 2009.\nRead more about the history of the Bude Canal.<|endoftext|>"},"score":{"kind":"number","value":3.96875,"string":"3.96875"}}},{"rowIdx":1246,"cells":{"token_count":{"kind":"number","value":734,"string":"734"},"text":{"kind":"string","value":"Every NGO and every project needs to have a budget. A budget describes the money that an organisation plans to raise and spend for a set purpose over a given period of time.\n\"Failing to plan is planning to fail.\"\nThe budget plays an important role in every stage of NGO work: planning new work, raising funds, implementing activities, monitoring work and evaluation.\nThe programme cycle and the financial cycle go hand in hand.\nHow to prepare a budget (PLAN)\nWriting a budget involves answering a number of questions, that can only be answered by programme staff and finance staff working together: They start with plans and goals, not numbers:\n- What objectives are we trying to achieve?\n- What activities will be involved in achieving these objectives?\n- What resources will we need to carry out these activities?\n- What will these resources cost?\n- Where will the funds come from? What will the source be?\n- Is the result realistic?\nThis might look like hard work, but we actually work through these questions every day, when we handle our own personal money. You can see a short illustrated example here: Rudi goes to the cinema.\nA good format to start with is a detailed budget worksheet that can later be summarised or re-coded to match donor formats as needed. After the budget has been prepared it needs to be approved by the Board and / or donors.\nSee an example budget worksheet.\nImplementing the budget (DO)\n- The person with the responsibility for implementing a budget is called a ‘budget holder’, and is usually a programme or project manager, not a finance person.\n- They use the budget to guide the implementation, checking BEFORE spending that items are budgeted for, and allocating expenditure to appropriate budget lines.\n- Budget holders should focus on delivering the project objectives within the total budget rather than on spending all the cash outlined in the budget.\n- Over or under spending within reasonable limits is not a problem as long as there are good explanations (but check your grant conditions, as some can be more restrictive than others).\nMonitoring the budget (REVIEW)\nOnce the budget has been agreed and the activity implemented, the process is completed by comparing the plan (budget) with the eventual outcome (‘actual’), to see if there is anything we have learnt or could do differently next time. This is covered another section of the Guide.\nSee the section of the Guide on financial reporting.\nWhat makes a good budget?\nBudgets should be clear, so that other people can pick them up and understand them easily. Always add notes to explain any estimates or assumptions you make. You should use the same list of accounts codes in your budget that you use to keep your accounts.\nSee Top tips 3 for more advice on making effective budgets.\nAs well as a budget, NGOs normally need a cash flow forecast. This predicts when money will arrive in your bank account and when it will leave. It is very important to make sure that you always have enough cash available. For instance, even if a donor has agreed to make you a grant, you cannot pay salaries until the cash actually arrives.\nSee Top tips 6 for more on how to create a cash flow forecast.\nSee here for an example cash flow forecast\nMango's Financial Managment Essentials Handbook includes a whole chapter on budgeting and some sample formats.\nWant to learn more?\nMango’s Budgeting Essentials training course is available as a 5-week eWorkshop.<|endoftext|>"},"score":{"kind":"number","value":3.703125,"string":"3.703125"}}},{"rowIdx":1247,"cells":{"token_count":{"kind":"number","value":1332,"string":"1,332"},"text":{"kind":"string","value":"Courses\nCourses for Kids\nFree study material\nOffline Centres\nMore\nStore\n\n# Graphs and Graphical Representation\n\nReviewed by:\nLast updated date: 09th Aug 2024\nTotal views: 389.7k\nViews today: 10.89k\n\n## What are Graphs and Graphical Representation?\n\nGraphical representation refers to the use of charts and graphs to visually analyze and display, interpret numerical value, clarify the qualitative structures. The data is represented by a variety of symbols such as line charts, bars, circles, ratios. Through this, greater insight is stuck in the mind while analyzing the information.\n\nGraphs can easily illustrate the behavior, highlight changes, and can study data points that may sometimes be overlooked. The type of data presentation depends upon the type of data being used.\n\n## Graphical Representation of Data\n\nThe graphical representation is simply a way of analyzing numerical data. It comprises a relation between data, information, and ideas in a diagram. Anything portrayed in a graphical manner is easy to understand and is also termed as the most important learning technique. The graphical presentation is always dependent on the type of information conveyed. There are different types of graphical representation. These are as follows:\n\n1. ### Line Graphs:\n\nAlso denoted as linear graphs are used to examine continuous data and are also useful in predicting future events in time.\n\n1. ### Histograms:\n\nThis graph uses bars to represent the information. The bars represent the frequency of numerical data. All intervals are equal and hence, the width of each bar is also equal.\n\n1. ### Bar Graphs:\n\nThese are used to display the categories and compare the data using solid bars. These bars represent the quantities.\n\n1. ### Frequency Table:\n\nThis table shows the frequency of data that falls within that given time interval.\n\n1. ### Line Plot:\n\nIt shows the frequency of data on a given line number.\n\n1. ### Circle Graph:\n\nIt is also known as a pie chart and shows the relationship between the parts of the whole. The circle consists of 100% and other parts shown are in different proportions.\n\n1. ### Scatter Plot:\n\nThe diagram shows the relationship between two sets of data. Each dot represents individual information of the data.\n\n1. ### Venn Diagram:\n\nIt consists of overlapping circles, each depicting a set. The inner-circle made is a graphical representation.\n\n1. ### Stem and Leaf Plot:\n\nThe data is organized from the least value to the highest value. The digits of the least place value form the leaf and that of the highest place value form the stem.\n\n1. ### Box and Whisker Plot:\n\nThe data is summarised by dividing it into four parts. Box and whisker show the spread and median of the data.\n\n### Graphical Presentation of Data - Definition\n\nIt is a way of analyzing numerical data. It is a sort of chart which shows statistical data in the form of lines or curves which are plotted on the surface. It enables studying the cause and effect relationships between two variables. It helps to measure the extent of change in one variable when another variable changes.\n\n### Principles of Graphical Representation\n\nThe variables in the graph are represented using two lines called coordinate axes. The horizontal and vertical axes are denoted by x and y respectively. Their point of intersection is called an origin ‘O’. Considering x-axes, the distance from the origin to the right will take a positive value, and the distance from the origin to the left will take a negative value. Taking the same procedure on y-axes. The points above origin will take the positive values and the points below origin will take negative values. As discussed in the earlier section about the types of graphical representation. There are four most widely used graphs namely histogram, pie diagram, frequency polygon, and ogive frequency graph.\n\n### Rules for Graphical Representation of Data\n\nThere are certain rules to effectively represent the information in graphical form. Certain rules are discussed below:\n\n1. Title: One has to make sure that a suitable title is given to the graph which indicates the presentation subject.\n\n2. Scale: It should be used efficiently to represent data in an accurate manner.\n\n3. Measurement unit: It is used to calculate the distance between the box\n\n4. Index: Differentiate appropriate colors, shades, and design I graph for a better understanding of the information conveyed.\n\n5. Data sources: Include the source of information at the bottom graph wherever necessary. It adds to the authenticity of the information.\n\n6. Keep it simple: Construct the graph in an easy to understand manner and keep it simple for the reader to understand. Looking at the graph the information portrayed is easily understandable.\n\n### Importance of Graphical Representation of Data\n\nSome of the importance and advantages of using graphs to interpret data are listed below:\n\n1. The graph is easiest to understand as the information portrayed is in facts and figures. Any information depicted in facts, figures, comparison grabs our attention, due to which they are memorizable for the long term.\n\n2. It allows us to relate and compare data for different time periods.\n\n3. It is used in statistics to determine the mean, mode, and median of different data.\n\n4. It saves a lot of time as it covers most of the information in facts and figures. This in turn compacts the information.\n\n## FAQs on Graphs and Graphical Representation\n\nAns: These graphical presentations of data are vital components in analyzing the information. Data visualization is one of the most fundamental approaches to data representation. Its advantages include the following points:\n\n1. Facilitates and improves learning\n\n2. Flexibility of use\n\n3. Understands content\n\n4. Increase structure thinking\n\n5. Supports creative thinking\n\n6. Portrays the whole picture\n\n7. Improves communication\n\nWith advantages, certain disadvantages are also linked to the graphical representation. The disadvantages concern the high cost of human effort, the process of selecting the most appropriate graphical and tabular presentation, creative thinking, greater design to interpret information, visualizing data, and as human resource is used. The potential for human bias plays a huge role.\n\nQ2. What is the Graphical Representation of Data in Statistics?\n\nGraphs are powerful data evaluation tools. They provide a quick visual summary of the information. In statistics information depicted is of mean, mode, and median. Box plots, histograms are used to depict the information. These graphs provide information about ranges, shapes, concentration, extreme values, etc. It studies information between different sets and trends whether increasing or decreasing. Since graphical methods are qualitative, they are not only the basis of comparison and information.<|endoftext|>"},"score":{"kind":"number","value":4.5625,"string":"4.5625"}}},{"rowIdx":1248,"cells":{"token_count":{"kind":"number","value":1473,"string":"1,473"},"text":{"kind":"string","value":"# Division law with integer bases\n\nLesson\n\nThe method to divide power terms is similar to the multiplication law, however in this case we subtract the powers from one another, rather than add them. Let's look at an expanded example to see why this is the case.\n\nIf we wanted to simplify the expression $a^6\\div a^2$a6÷​a2, we could write it as:\n\nWe can see that there are six $a$as being divided by two $a$as to give a result of four $a$as, and notice that $4$4 is the difference of the powers in the original expression.\n\nSo, in our example above,\n\n $a^6\\div a^2$a6÷​a2 $=$= $a^{6-2}$a6−2 $=$= $a^4$a4\n\nLet's look at another specific example. Say we wanted to find the value of $2^7\\div2^3$27÷​23. By evaluating each term in the quotient separately we would have\n\n $2^7\\div2^3$27÷​23 $=$= $128\\div8$128÷​8 $=$= $16$16\n\nAlternatively, by first expanding the terms in the original expression we can arrive at a simplified version of the expression on our way to the final value.\n\n $2^7\\div2^3$27÷​23 $=$= $\\left(2\\times2\\times2\\times2\\times2\\times2\\times2\\right)\\div\\left(2\\times2\\times2\\right)$(2×2×2×2×2×2×2)÷​(2×2×2) $=$= $2^4$24 $=$= $16$16\n\nNotice in the second line we have identified that $2^7\\div2^3=2^4$27÷​23=24.\n\n### The exponent law of division\n\nWe can avoid having to write each expression in expanded form by using the division law (which is also known as the quotient law).\n\nThe division law\n\n$\\frac{a^m}{a^n}=a^{m-n}$aman=amn, where $a$a is any number,\n\nThat is, when dividing terms with a common base:\n\n• Keep the same base\n• Find the difference in the power.\n\nOf course, we can also write the division law in the form $a^m\\div a^n=a^{m-n}$am÷​an=amn.\n\n### Dividing terms with common bases\n\nAs with using the multiplication (or product) law, we can only apply the division (or quotient) law to terms with the same bases (just like we can only add and subtract like terms in algebra). We can simplify $\\frac{9^8}{9^3}$9893 because the numerator and denominator have the same base: $9$9.\n\nCareful!\n\nWe cannot simplify $\\frac{8^5}{7^3}$8573 because the two terms do not have the same base (one has a base of $8$8 and the other has a base of $7$7).\n\n#### Worked example\n\nSimplify the following by first writing it in expanded form: $\\frac{3^7}{3^2}$3732.\n\nThink: We want to first write the expression in expanded form, so that we can then cancel out common factors from the numerator and denominator.\n\nDo:\n\n $\\frac{3^7}{3^2}$3732​ $=$= $\\frac{3\\times3\\times3\\times3\\times3\\times3\\times3}{3\\times3}$3×3×3×3×3×3×33×3​ $=$= $\\frac{3\\times3\\times3\\times3\\times3}{1}$3×3×3×3×31​ $=$= $3^5$35\n\nReflect: We can see that the answer matches what we would expect if we used the division law. That is, $\\frac{3^7}{3^2}=3^{7-2}$3732=372 giving us the final answer of $3^5$35.\n\n### Dividing other terms\n\nWe can also simplify expressions involving multiplication as well, such as $\\frac{4\\times3^6}{2\\times3^4}$4×362×34. We follow the same steps as when we are multiplying expressions. That is, we can treat the problem in two parts. Let's take a look at an example.\n\n#### Worked example\n\nSimplify $\\left(8\\times5^9\\right)\\div\\left(2\\times5^4\\right)$(8×59)÷​(2×54) using exponent laws.\n\nThink: First, let's write the expression as a fraction. To make the simplifications easier, we can then split the fraction up into two fractions using the fact $\\frac{a\\times b}{c\\times d}=\\frac{a}{c}\\times\\frac{b}{d}$a×bc×d=ac×bd.\n\nDo:\n\n $\\left(8\\times5^9\\right)\\div\\left(2\\times5^4\\right)$(8×59)÷​(2×54) $=$= $\\frac{8\\times5^9}{2\\times5^4}$8×592×54​ $=$= $\\frac{8}{2}\\times\\frac{5^9}{5^4}$82​×5954​ $=$= $4\\times\\frac{5^9}{5^4}$4×5954​ $=$= $4\\times5^5$4×55 $=$= $4\\times5^5$4×55\n\nReflect: Combining the steps, we get $\\left(8\\times5^9\\right)\\div\\left(2\\times5^4\\right)=4\\times5^5$(8×59)÷​(2×54)=4×55 and as this process becomes more familiar we can reduce the amount of intermediate steps we take to arrive at the solution.\n\n#### Practice questions\n\n##### Question 1\n\nConvert the following to fraction form and evaluate using exponent laws: $4^8\\div4^5$48÷​45.\n\n##### Question 2\n\nSimplify the following, giving your answer in exponential form: $\\frac{\\left(-7\\right)^{11}}{\\left(-7\\right)^5}$(7)11(7)5.\n\n##### Question 3\n\nEvaluate the following using exponent laws: $5\\times\\frac{4^5}{4^3}$5×4543.\n\n### Outcomes\n\n#### 9D.NA1.03\n\nDerive, through the investigation and examination of patterns, the exponent rules for multiplying and dividing monomials, and apply these rules in expressions involving one and two variables with positive exponents<|endoftext|>"},"score":{"kind":"number","value":5,"string":"5"}}},{"rowIdx":1249,"cells":{"token_count":{"kind":"number","value":443,"string":"443"},"text":{"kind":"string","value":"Вы находитесь на странице: 1из 1\n\n# Percentage\n\nExample 1. When the cost of an article is decreased by 75%, it cost becomes Rs. 1050. Find the\ncost of the article, before the discount was given.\nSolution. The discount of the article is 75 %. After discount is given, let the new cost be x% of the\noriginal cost (before the discount was given).\nx% = (100 - 75)% = 25%.\n100× 1050\n∴ Original price = = Rs. 4200.\n25\nExample 2. The ratio of cost of two articles is 2 : 5. Then, what percent is the cost of 1 st article is\nthe 2nd article?\nSolution. Let the cost of first and second article be Rs. 2x and Rs. 5x.\n2 x ×100\n∴ Required per cent = = 40%.\n5 x ×1\nExample 3. What principle is lent, if the payable interest after 5 years is Rs. 365, at the rate of 10\n% per annum?\nSI ×100\nSolution. The principal is calculated by the formula .\nR×T\nWe have,\nSI = Rs. 365\nR = 10 % pa\nT = 5 years\n365× 100\nP= = Rs. 730\n5× 10\n∴ Rs. 730 is lent.\nExample 4. Rakesh, at simple interest, borrows a sum at 5 % pa and for 10 years. If the interest\nafter 10 years is 5 times the principal, then find the interest paid.\nSolution. Let principal is Rs. x. Let the interest be Rs. y. Then we have,\nRs. (x + y) = 5x\n Rs. y = 5x - x (transposing +x to right side, changing to -x)\n Rs. y = Rs. 4x (Express y as x terms).\n Interest = 4x.<|endoftext|>"},"score":{"kind":"number","value":4.5625,"string":"4.5625"}}},{"rowIdx":1250,"cells":{"token_count":{"kind":"number","value":382,"string":"382"},"text":{"kind":"string","value":"# 180 Days of Math for Third Grade Day 92 Answers Key\n\nBy accessing our 180 Days of Math for Third Grade Answers Key Day 92 regularly, students can get better problem-solving skills.\n\n## 180 Days of Math for Third Grade Answers Key Day 92\n\nDirections: Solve each problem.\n\nQuestion 1.\n\nBy subtracting 23 from 45 we get 22.\n\nQuestion 2.\n4 × 6 =\nExplanation:\nBy performing the multiplication operation we can find the product of 4 and 6.\n4 × 6 = 24\n\nQuestion 3.\n5 × 6 =\nExplanation:\nBy performing the multiplication operation we can find the product of 5 and 6.\n5 × 6 = 30\n\nQuestion 4.\nWhat is the numeral for one hundred ninety-seven?\nAnswer: The numeral for one hundred ninety-seven is 197.\n\nQuestion 5.\nCircle the smallest number.\n451, 145, 514\nAnswer: The smallest number among the three numbers is 145.\n\nQuestion 6.\nFill in the missing number.\n48, ________, 32, 24, 16\n32 – 24 = 8\nSo, the pattern is 8.\n48 – 8 = 40\nSo, the missing number is 40.\n\nQuestion 7.\n__________ cm = 1 m\nConvert from meters to centimeters.\n1 meter = 100 cm\n100 cm = 1 m\n\nQuestion 8.\nWhat time is shown on the clock?\n\nQuestion 9.\nWhat is the range of the number of cousins?\n\n Name Number of Cousins Anne 15 Marq 1 Jesus 3 Kelly 4 Tami 7\n\nAnswer: The range of the number of cousins is 1 to 15.\n\nQuestion 10.\nWhat is 10¢ less than 3 quarters?<|endoftext|>"},"score":{"kind":"number","value":4.4375,"string":"4.4375"}}},{"rowIdx":1251,"cells":{"token_count":{"kind":"number","value":618,"string":"618"},"text":{"kind":"string","value":"Influenza, commonly called “the flu,” is an illness caused by RNA viruses that infect the respiratory tract of many animals, birds, and humans. Influenza, or the “flu,” is an illness of the breathing system (respiratory system) and muscles caused by a virus.\nCauses of Flu\nInfluenza A usually occurs in early winter and early spring. Influenza B can appear at any time of the year. H1N1, or Swine Flu, is another type of Influenza. Most people catch the flu when they breathe in tiny droplets from coughs or sneezes of someone who has the flu. It is also spread when you touch a something with the virus on it and then touch your mouth, nose, or eyes. Symptoms appear 1 – 7 days later (usually within 2 – 3 days). The flu spreads easily. It often strikes a community all at once. Students or workers become sick within 2 or 3 weeks of the flu’s arrival in a school or workplace. Sometimes people confuse colds and flu. They do share some of the same symptoms. Most people get a cold several times each year. But they usually get the flu once every few years.\nSymptoms of Flu\nThe first symptoms are a fever between 102 and 106 °F. The infection results in the person getting fever, cough, headache, and malaise (tired, no energy); some people also may develop a sore throat, nausea, vomiting, and diarrhea. Then breathing symptoms begin to increase. The symptom is usually not a dry cough. Most people also develop a sore throat and headache. Runny nose and sneezing are common. It is a clear, watery nasal discharge.\nTreatment of Flu\nAcetaminophen (Tylenol) and ibuprofen (Advil, Motrin) help lower fever. Sometimes doctors suggest you use both types of medicine. Do NOT Use Aspirin. A fever does not need to come all the way down to normal. Most people will feel better when their temperature drops by even 1 degree. Over-the-counter cold medicines may make some of your symptoms better. Cough drops will help with your sore throat.\nA vaccine is available to prevent the flu, its effectiveness varies according to the degree of match between the viral strains used to prepare the vaccine and those strains actually in circulation in a given year.\nShould see a doctor if you have flu-like symptoms and you:-\n- are 65 or over\n- are pregnant\n- have a long-term medical condition such as diabetes, heart disease, lung disease, kidney or neurological disease\n- have a weakened immune system\nHow Long does Flu lasts?\nThe typical course of the flu is seven to ten days.” While you’re waiting for the virus to pass, prevent it from getting worse by drinking plenty of fluids, washing hands frequently, and eating vitamin D-fortified foods like orange juice and yogurt.<|endoftext|>"},"score":{"kind":"number","value":4.125,"string":"4.125"}}},{"rowIdx":1252,"cells":{"token_count":{"kind":"number","value":398,"string":"398"},"text":{"kind":"string","value":"We spent a bit of time talking about how great 0 is. We looked at a product like a * b = 12 and determined, after a long list of possibilities, that there was an infinite number of combinations of values for a and b that would make this true. However, if a * b = 0, we know that a or b must be 0. Many students have been struggling with finding the zeros of questions like these:\nI think they now understand why you can just take the opposite of the constant term in every case.\nWe consolidated the process of solving a quadratic equation that can be factored and then followed up with a lot of practice questions.\nThen we hit one with a common factor and I, of course, did what my students suggested and factored it without taking the common factor out first.\nWe talked a lot about this one. If you are factoring, but not solving, you must take the common factor out either at the beginning or the end. They saw pretty quickly that taking it out at the beginning made their solution much easier. We talked about why you could divide by 3 when you are solving, but not when you are just factoring. After class a student asked me about using the box method for example 2a. I had not tried one that contained a common factor and it is not as obvious as you might think (well, it was not obvious to me, anyway). I will go over it with him tomorrow, stressing that taking the common factor out first will make it all work much more nicely.\nPart (c) gave us the opportunity to look at a difference of squares. I drew the corresponding tiles on the whiteboard and they all remembered factoring these types of quadratics. Part (d) was gave them the tools to deal with a -1 coefficient of the variable squared. We finished with this one:\nToday's homework was the second box of the handout from yesterday.<|endoftext|>"},"score":{"kind":"number","value":4.03125,"string":"4.03125"}}},{"rowIdx":1253,"cells":{"token_count":{"kind":"number","value":650,"string":"650"},"text":{"kind":"string","value":"Factoring and Multiples\nPrime Numbers\nNumbers\n\n# Is it that 3 consecutive whole number can be prime factors Explain?\n\n###### Wiki User\n\n3 consecutive numbers cannot be prime factors. Any three consecutive numbers would include at least one even number. The only even Prime number is 2, and (2,3,4) doesn't qualify.\n\n๐Ÿ™\n0\n๐Ÿคจ\n0\n๐Ÿ˜ฎ\n0\n๐Ÿ˜‚\n0\n\n## Related Questions\n\nA prime number has no factors except itself and 1.\n\nPrime number next to any prime number is called consecutive prime number. Eg:- 2,3,5,7 are prime numbers\n\nIt is a prime number because its only factors are 1 and itself.\n\n3 consecutive numbers cannot be prime factors. Any three consecutive numbers would include at least one even number. The only even prime number is 2, and (2,3,4) doesn't qualify.\n\nAll numbers have factors. Some factors are prime numbers. We call these prime factors. The factors of 6 are 1, 2, 3 and 6. The prime factors of 6 are 2 and 3.\n\nNo. It doesn't matter what factors you choose at the beginning, but all the factors should be prime at the end.\n\nThe only even prime number is 2 because it has only two factors which are itself and one.\n\nA prime number is a positive integer that has exactly two different factors: 1 and itself. This is why 1 is not a prime number: it has the factors 1 and itself, but they are the same - it does not have exactly two different factors.\n\nA factor tree shows the prime factors of the working number and can also be used as a visual demonstration to show the prime factors of the working number.\n\n75 is a composite number because it has more than two factors whereas prime numbers have only two factors\n\n47 and 53 are prime numbers because they only have two factors.\n\nPrime numbers have only two factors: one and themselves. The only factors of 17 are 1 and 17. 17 is a prime number.\n\nA prime number has only two factors: 1 and the number itself. A composite number has more than two factors.\n\nYes, because a composite number is any number that has more prime factors other than one and itself . The prime factors of 33 are 3 and 11.\n\n37 is a prime number because it has only 2 factors which are itself and one\n\nA pair of prime numbers are always relatively prime, whether they are consecutive or not. This is so because \"relatively prime\" means they have no common factors.\n\nYes it can be prime. A prime number is any number that only has one and itself as factors. By definition even numbers have a factor of two. Therefore the only even number that is prime is the number 2.\n\n###### Math and ArithmeticPrime NumbersFactoring and MultiplesComposite Numbers\n\nCopyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.<|endoftext|>"},"score":{"kind":"number","value":4.46875,"string":"4.46875"}}},{"rowIdx":1254,"cells":{"token_count":{"kind":"number","value":525,"string":"525"},"text":{"kind":"string","value":"# How To Solve Quadratic Equation Graphical Method Ideas\n\nHow To Solve Quadratic Equation Graphical Method Ideas. We can solve a quadratic equation by factoring, completing the square, using the quadratic formula or using the graphical method. A(8)(a) solve quadratic equations having real solutions by factoring, taking square roots, completing the square, and applying the quadratic formula.\n\nTo solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. %straight line will give the roots of the given quadratic equation. Resource objective(s) given a quadratic equation, the student will use graphical methods to solve the equation.\n\n### You Can Solve Quadratic Equations By Graphing, Factoring, Completing The Square, & The Quadratic Formula.\n\nThen, factor the expression, and set each set of parentheses equal to 0 as separate equations. Transform the equation so that the quadratic term and the linear term equal a constant. We can solve the quadratic equation ax 2 + bx + c = d through graphing using the following steps:\n\n### %Parabola And A Straight Line.\n\nWe can solve a quadratic equation by factoring, completing the square, using the quadratic formula or using the graphical method. Another way of solving a quadratic equation is to solve it graphically. A x 2 + b x + c = 0, w h e r e a ≠ 0.\n\n### By Using The Quadratic Formula 4.\n\nSimplest form of solving quadratic equations using formula method These are the solutions of the equation. One of the ways we can solve a quadratic equation is by factoring.\n\n### There Are Three Basic Methods To Solve A Quadratic Equation In Mathematics:\n\nTo solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. By using the trial and error method. ( ) steps to solve an equation by completing the square:\n\n### The Values Of The Variable Satisfying The Given Quadratic Equation Are Called Its Roots.\n\nYou know by now how to solve a quadratic equation using factoring. In this explainer, we will see how we can also use a graphical method to solve a quadratic equation. Quadratic functions :are written as 𝑓 ;= 2− −6 or = 2− −6.f(x) is zf of x and means that the y value is dependent upon the value of x.once you have an x value and you substitute it into the equation, the value of f(x) will result.f(x) is really just a another way of\n\nTags :<|endoftext|>"},"score":{"kind":"number","value":4.625,"string":"4.625"}}},{"rowIdx":1255,"cells":{"token_count":{"kind":"number","value":576,"string":"576"},"text":{"kind":"string","value":"Tracking the destruction of the world's forests is to become much easier for scientists and forest managers, thanks to a software tool unveiled by search-engine giant Google's philanthropic arm today.\nThe software, which uses Google's computing resources to extract scientific information from decades of satellite images of forests, was demonstrated at the UN climate conference in Copenhagen. \"We hope this technology will help stop the destruction of the world's rapidly disappearing forests,\" said a statement on the Google.org blog.\n\"Emissions from tropical deforestation are comparable to the emissions of all of the EU, and are greater than those of all cars, trucks, planes, ships and trains worldwide. According to the Stern Review – the report prepared for the British government in 2006 on the economics of climate change by Lord Nicholas Stern – protecting the world's standing forests is a highly cost-effective way to cut carbon emissions and mitigate climate change.\"\nThe UN mechanism to reduce deforestation is called Reducing Emissions from Deforestation and Forest Degradation in Developing Countries (Redd), a system whereby richer countries would provide financial incentives to protect forests in poorer nations. For Redd to be successful, however, countries need ways to accurately monitor and report on the state of their forests.\nIn Google.org's prototype software, environmental authorities or NGOs interested in monitoring forests start with satellite images of their area and track how the size and shape of the tree cover has changed over time. The software can processes the images to extract useful scientific and tracking information about how much the forests have changed.\nFor the analysis, the Google.org team worked with Greg Asner of Carnegie Institution for Science and Carlos Souza of Imazon. Technology developed by Asner and Souza is used in Latin America to track changes in forest cover – but mainstream use of the models has been slow due to lack of access to high-quality satellite images and the computer power needed to carry out the analysis.\nGoogle.org's solution is to enhance the Asner and Souza models using its own computing power. \"What if we could gather together all of the earth's raw satellite imagery data – petabytes of historical, present and future data – and make it easily available on this platform? We decided to find out, by working with Greg and Carlos to re-implement their software online, on top of a prototype platform we've built that gives them easy access to terabytes of satellite imagery and thousands of computers in our data centres,\" it wrote.\nColby Loucks, deputy director of the conservation science program at WWF-US said: \"A cost-effective and transparent approach for monitoring deforestation is needed to help pave the way for a global Redd program. If Google's system can be expanded to cover forests globally and access near real-time imagery, it can potentially be a powerful tool that helps tropical countries monitor forest loss.\"<|endoftext|>"},"score":{"kind":"number","value":3.703125,"string":"3.703125"}}},{"rowIdx":1256,"cells":{"token_count":{"kind":"number","value":1349,"string":"1,349"},"text":{"kind":"string","value":"# College Algebra Tutorial 25\n\nCollege Algebra\nTutorial 25: Slope of a Line\n\nLearning Objectives\n\nAfter completing this tutorial, you should be able to:\n1. Know when a line rises, falls, is horizontal or is vertical.\n2. Find the slope given two points on the line.\n\nIntroduction\n\nIn this tutorial we will explore the slope of the line.  Basically, the slope of the line measures the steepness of the line.  We will start with some visuals to help you understand what the slope of a straight line tells us about the line.  Next, we will practice finding the slope given two points on the line.  Let's see what you can do with slopes.\n\nTutorial\n\nSlope\n\nThe slope of a line measures the steepness of the line.\n\nMost of you are probably familiar with associating slope with \"rise over run\".\n\nRise means how many units you move up or down from point to point.  On the graph that would be a change in the y values.\n\nRun means how far left or right you move from point to point.  On the graph, that would mean a change of x values.\n\nPositive Slope:\n\nNote that when a line has a positive slope it rises up left to right.\n\nNegative Slope:\n\nNote that when a line has a negative slope it falls left to right.\n\nZero Slope:\n\nslope = 0\n\nNote that when a line is horizontal the slope is 0.\n\nUndefined Slope:\n\nslope = undefined\n\nNote that when the line is vertical the slope is undefined.\n\nSlope Formula Given Two Points\n\nGiven two points  and\n\nThe subscripts just indicate that these are two different points.  It doesn't matter which one you call point 1 and which one you call point 2 as long as you are consistent throughout that problem.\n\nNote that we use the letter m to represent slope.\n\nExample 1: Find the slope of the straight line that passes through (7, 5) and (5, 1) or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical.\n\n*Plug in x and y values into slope formula\n\n*Simplify\n\nThe slope of the line is 2.\n\nSince the slope is positive, the line would rise (left to right).\n\nExample 2: Find the slope of the straight line that passes through (1, -4) and (-1, 3) or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical.\n\n*Plug in x and y values into slope formula\n\n*Simplify\n\nMake sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2.  1 - (-1) is not the same as 1 - 1.\n\nThe slope of the line is -7/2.\n\nSince the slope is negative, the line would fall (left to right).\n\nExample 3: Find the slope of the straight line that passes through (4, 1) and (-2, 1) or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical.\n\n*Plug in x and y values into slope formula\n\n*Simplify\n\nMake sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2.  4 - (-2) is not the same as 4 - 2.\n\nThe slope of the line is 0.\n\nSince the slope is zero, the line would be horizontal.\n\nExample 4: Find the slope of the straight line that passes through (-2, 3) and (-2, 5) or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical.\n\n*Plug in x and y values into slope formula\n\n*Simplify\n\nMake sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2.  -2 - (-2) is not the same as -2 - 2.\n\nThe slope of the line is undefined.\n\nSince the slope is undefined, the line would be vertical.\n\nPractice Problems\n\nThese are practice problems to help bring you to the next level.  It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it.  Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument.  In fact there is no such thing as too much practice.\n\nTo get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that  problem.  At the link you will find the answer as well as any steps that went into finding that answer.\n\nPractice Problems 1a - 1d: Find the slope of the straight line that passes through the given points or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical.\n\n1a. (2, -5) and (-1, -1)\n1b. (7, 1) and (7, 2)\n\n1c. (1, 7) and (-2, -2)\n1d.  (-1, -5) and (-4, -5)\n\nNeed Extra Help on these Topics?\n\nThe following are webpages that can assist you in the topics that were covered on this page:\n\nThis website helps you with the slope of a line.\n\nhttp://www.purplemath.com/modules/slope.htm\nThis webpage helps you with slope.\n\nhttp://www.math.com/school/subject2/lessons/S2U4L2DP.html\nThis website covers slopes and y-intercept.\n\nGo to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.\n\nVideos at this site were created and produced by Kim Seward and Virginia Williams Trice.\nLast revised on Jan. 16, 2010 by Kim Seward.<|endoftext|>"},"score":{"kind":"number","value":4.90625,"string":"4.90625"}}},{"rowIdx":1257,"cells":{"token_count":{"kind":"number","value":1056,"string":"1,056"},"text":{"kind":"string","value":"Class 12 Maths\n\n# Inverse Trignometric Functions\n\n## NCERT Solution\n\n### Exercise 2.1 Part 2\n\n##### Find the principal values of the following:\n\nQuestion 9: text(cos)^(-1)((-1)/(sqrt2))\n\nSolution: Let text(cos)^(-1)((-1)/(sqrt2))=θ\n\nSo, text(cos)θ=(-1)/(sqrt2)\n\nOr, text(cos)θ=text(cos)(π-(π)/(4))\n\nOr, text(cos)θ=text(cos)(3π)/(4)\n\nSince value of text(cos)^(-1)x lies between 0 and π. Hence, principal value of text(cos)^(-1)((-1)/(sqrt2)) is (3π)/(4)\n\nQuestion 10: text(cosec)^(-1)(-sqrt2)\n\nSolution: Let text(cosec)^(-1)(-sqrt2)=θ\n\nSo, text(cosec)^(-1)θ=-sqrt2\n\nOr, text(cosec)^(-1)θ=cosec(-(π)/(4))\n\nSince value of text(cosec)^(-1)x is [(-π)/(2), (π)/(2)]-(0). Hence, principal value of text(cosec)^(-1)(-sqrt2) is (-π)/(4)\n\nQuestion 11: text(tan)^(-1)+text(cos)^(-1)(-1/2)+text(sin)^(-1)(-1/2)\n\nSolution: Let text(tan)^(-1)(1)=θ_1\n\nOr, text(tan)θ_1=1=text(tan)(π)/(4)\n\nOr, θ_1=(π)/4\n\nLet text(cos)^(-1)(-1/2)=θ_2\n\nOr, text(cos)θ_2=-1/2=-text(cos)((π)/(3))\n\nOr, text(cos)θ_2=text(cos)(π-(π)/(3))\n\nOr, text(cos)θ_2=text(cos)((2π)/(3))\n\nOr, θ_2=(2π)/3\n\nLet text(sin)^(-1)(-1/2)=θ_3\n\nOr, text(sin)θ_3=-1/2\n\nOr, text(sin)θ_3=-text(sin)((π)/(6))=text(sin)(-(π)/(6))\n\nOr, θ_3=-(π)/(6)\n\nNow, text(tan)^(-1)+text(cos)^(-1)(-1/2)+text(sin)^(-1)(-1/2)\n\n=θ_1+θ_2+θ_3\n\n=(π)/(4)+(2π)/(3)-(π)/(6)\n\n=(9π)/(12)=(3π)/4\n\nQuestion 12: text(cos)^(-1)(1/2)+2text(sin)^(-1)(1/2)\n\nSolution: Let text(cos)^(-1)(1/2)=θ\n\nOr, text(cos)θ=1/2=text(cos)((π)/(3))\n\nOr, θ=(π)/(3)\n\nLet text(sin)^(-1)(1/2)=θ_1\n\nOr, text(sin)θ_1=1/2=text(sin)(π)/(6)\n\nOr, θ_1=(π)/(6)\n\nSo, text(cos)^(-1)(1/2)+2text(sin)^(-1)(1/2)\n\nθ+2θ_1\n\n=(π)/(3)+(2π)/(6)\n\n=(2π+2π)/(6)=(2π)/(3)\n\nQuestion 13: If text(sin)^(-1)x=y then\n\n1. 0≤y≤π\n2. -(π)/(2)≤y≤(π)/(2)\n3. 0"},"score":{"kind":"number","value":4.59375,"string":"4.59375"}}},{"rowIdx":1258,"cells":{"token_count":{"kind":"number","value":1163,"string":"1,163"},"text":{"kind":"string","value":"What does the recently photographed black hole have to do with climate change?\nThe image is a VERY cool representation of an amazing scientific achievement. But more importantly, it reminds us of the value of science and scientific process.\nOn April 10, the first mesmerizing image of a black hole was released to the public. The black hole, named M87-star, is located in the center of Messier 87 galaxy some 318 quintillion miles (53 million light-years) away from Earth.\nThat’s 318,000,000,000,000,000,000 miles!\nHow’d did scientists do that?\nThey focused mega-telescopes at 8 different sites around the world at the black hole at the same time for 10 days last year and recorded the images. Then, hundreds of researchers worked tirelessly to develop computer algorithms (equations) to stitch those images together and produce the first ever picture of a black hole (read more here).\nThe image is a VERY cool representation of an amazing scientific achievement. More importantly, it reminds us of the value of science and scientific process. This achievement will help us better understand physics and allow researchers to better test long-standing theories, including Einstein’s theory of relativity.\nThe problem with most scientific writing, though, is that often it isn’t easily digestible to the everyday reader. The terminology, details, and backstory take years to learn, and scientists write articles and papers for other scientists to advance their field. This is an important exercise because it pushes scientific achievement, but it comes at a cost to the public’s ability to understand the work.\nMore than ever, we rely on news outlets to report on scientific findings in an easy to follow format. Despite reporters best efforts, sometimes headlines can be misleading or content can be misconstrued. This happens because scientists often hedge their bets when reporting findings, using words like “may be”, “appears to be”, or “correlated with.” The scientific process teaches us to do this because science is an iterative process that doesn’t produce a single “correct answer.” Instead, an experiment yields an observation that either fits or doesn’t fit with our expectations. We then use what we learn from that observation to better refine our expectations, and run more experiments. Repeating this process several times gives us much more confidence in predictions, and it helps us better apply our science.\nThis is why every few years you hear conflicting reports about coffee, red wine, or chocolate. One study may find a positive health benefit of those items, while another may find a negative one. These studies aren’t inherently wrong because they disagree. Instead, they show us that food and health have a complicated relationship, and underlying factors—like a family history of heart disease—are probably more important predictors of health than the food itself.\nThe iterative nature of science is what gives it its credibility, and when scientists reach a consensus, we should pay attention. That’s why climate change is such a pressing issue. 98% of climate experts have concluded that climate change is human-caused, and without drastic measures to reduce our carbon emissions, we are in for warming on a global scale that hasn’t been seen on earth in millions of years. It’s an uncomfortable and scary situation, but scientists have also given us powerful recommendations to slow this rate of change and better plan for future climate events.\nSo, how can we all better support and understand science, especially when it contributes to the public good?\nWe all need to communicate more effectively, and believe scientists when they reach a consensus. Scientists are already working to better share their science to the public, especially through news sources, blogs, and social media. The good news is that Americans are already great at scientific literacy, meaning they can work through scientific problems using the scientific process. By bridging this communication divide on both sides, we can all make better decisions to protect our future.\nAbout the Author\nDr. Danielle Begley-Millier\nDanielle is an ecologist with an extensive background in forest research, specifically focusing on ecological interactions. She comes to Teatown from Penn State University where she received her doctoral degree in wildlife in 2018. During her time at Penn State, Danielle managed a large research project aimed at understanding the effects of both white-tailed deer and soil nutrition on forest plant communities. She is responsible for the implementation of Teatown’s land stewardship and environmental research programs, as well as the supervision of select volunteers and interns as they participate in these programs. In addition to her Ph.D., Danielle earned her master’s degree in biology from Wright State University in 2013.\nInterested in getting involved? Check out these upcoming citizen science and stewardship opportunities:\nUpcoming Events › Citizen Science\nUpcoming Events › Citizen Science View as calendar\nAre you interested in learning about the pests that threaten New York’s forests, and would you like to lend a helping hand in monitoring their spread? Teatown is looking for dedicated volunteers and community scientists to participate in a forest pest monitoring program. For adults.Find out more »\nGardening with native plants in our suburban landscape has many environmental benefits. Join landscape designer Amanda Bayley, for a presentation on how to beautify your own yard with native plants. She will present on ecoregions, review the native plant communities that are part of Westchester’s ecoregion, and go over local plants to garden with that are best suited for wet, dry, sunny, and shady areas.Find out more »<|endoftext|>"},"score":{"kind":"number","value":4.09375,"string":"4.09375"}}},{"rowIdx":1259,"cells":{"token_count":{"kind":"number","value":315,"string":"315"},"text":{"kind":"string","value":"# How do you find the intercepts for y=-3x+3?\n\nJun 14, 2015\n\nThe intercept with the $y$ axis is where $x = 0$, hence $y = - 3 \\cdot 0 + 3 = 3$. That is $\\left(0 , 3\\right)$\n\nThe intercept with the $x$ axis is where $y = 0$, hence $0 = - 3 x + 3$, hence $x = 1$. That is $\\left(1 , 0\\right)$\n\n#### Explanation:\n\nTo find where $y = - 3 x + 3$ intercepts the $y$ axis, substitute $x = 0$ into the equation to find:\n\n$y = \\left(- 3 \\cdot 0\\right) + 3 = 0 + 3 = 3$.\n\nThat is $\\left(0 , 3\\right)$\n\nTo find where $y = - 3 x + 3$ intercepts the $x$ axis, substitute $y = 0$ into the equation to find:\n\n$0 = - 3 x + 3$\n\nAdd #3x to both sides to get:\n\n$3 x = 3$\n\nDivide both sides by $3$ to get:\n\n$x = 1$.\n\nThat is $\\left(1 , 0\\right)$\n\ngraph{-3x+3 [-9.5, 10.5, -3, 7]}<|endoftext|>"},"score":{"kind":"number","value":4.6875,"string":"4.6875"}}},{"rowIdx":1260,"cells":{"token_count":{"kind":"number","value":213,"string":"213"},"text":{"kind":"string","value":"Our aim of teaching mathematics at our school are to ensure that all pupils:\nbecome fluent in the fundamentals of mathematics, including through varied and frequent practice with increasingly complex problems over time, so that pupils develop conceptual understanding and the ability to recall and apply knowledge rapidly and accurately.\nChildren are encouraged to reason mathematically by following a line of enquiry, conjecturing relationships and generalisations, and developing an argument, justification or proof using mathematical language\nThey are taught to solve problems by applying their mathematics to a variety of routine and non-routine problems with increasing sophistication, including breaking down problems into a series of simpler steps and persevering in seeking solutions\nWe stress the importance of knowing multiplication tables. By the end of year 4, pupils should have memorised their multiplication tables up to and including the 12 multiplication table.\nAll pupils are taught to develop efficient strategies for mental and written calculations which is clearly outlined within our school policy.\nPupils should read and spell mathematical vocabulary, at a level consistent with their increasing word reading and spelling knowledge.<|endoftext|>"},"score":{"kind":"number","value":3.796875,"string":"3.796875"}}},{"rowIdx":1261,"cells":{"token_count":{"kind":"number","value":420,"string":"420"},"text":{"kind":"string","value":"# CBSE Class 10-Mathematics: Chapter –15 Probability Part 3\n\nGlide to success with Doorsteptutor material for CBSE : fully solved questions with step-by-step explanation- practice your way to success.\n\nQuestion 12:\n\nA number is chosen at random from the numbers What is the probability that\n\ncan take values\n\nTo set\n\nProbability\n\nQuestion 12:\n\nA number is selected from the number and then a second number is randomly selected from the numbers . What is the probability that the product of the two numbers will be less than ?\n\nNumber can be selected in three wavs and corresponding to each such wav there are three wavs of selecting number y. Therefore. Two numbers can be selected in wavs as listed below:\n\nFavourable number of elementary events\n\nHence, required probability\n\nQuestion 14:\n\nIn the adjoining figure a dart is thrown at the dart board and lands in the interior of the circle. What is the probability that the dart will land in the shaded region.\n\nWe have\n\nUsing Pythagoras Theorem is A ABC. We have\n\nArea of the circle\n\nArea of shaded region circle – Area of rectangle ABCD\n\nHence\n\nQuestion 15:\n\nIn the fig points A. B.C and D are the centres of four circles.each having a radius of 1 unit. If a point is chosen at random from the interior of a square ABCD. What is the probability that the point will be chosen from the shaded region.\n\nRadius of the circle is 1 unit\n\nArea of the circle =Area of a sector\n\nSide of the square ABCD\n\nArea of square\n\nArea of square Area of sectors\n\nProbability\n\nQuestion 16:\n\nIn the adjoining figure ABCD is a square with sides of length units points are the mid points of the sides BC & CD respectively. If a point is selected at random from the interior of the square what is the probability that the point will be chosen from the interior of the square what is the probability that the point will be chosen from the interior of the triangles APO.<|endoftext|>"},"score":{"kind":"number","value":4.53125,"string":"4.53125"}}},{"rowIdx":1262,"cells":{"token_count":{"kind":"number","value":969,"string":"969"},"text":{"kind":"string","value":"# Multiples of 6\n\nThe Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, … i.e. M6 = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, …}. Multiples of 6 are the series of numbers obtained by multiplying 6 with the natural numbers.\n\n6 × 1 = 6\n\n6 × 2 = 12\n\n6 × 3 = 18\n\n6 × 4 = 24\n\n6 × 5 = 30\n\n6 × 6 = 36\n\n6 × 7 = 42\n\n6 × 8 = 48\n\n6 × 9 = 54\n\n6 × 10 = 60\n\nAnd so on.\n\nFrom this series, we observe that the multiples of 6 can be derived by multiplying 6 by natural numbers in ascending order. These multiples form an arithmetic sequence with a common difference of 6.\n\n## Properties and Patterns of Multiples of 6\n\nMultiples of 6 possess several interesting properties that make them unique. Here are a few notable ones:\n\n1.    Divisibility by 6: Every multiple of 6 is divisible by 6 without leaving a remainder. This property stems from the fact that when we divide any multiple of 6 by 6, the quotient is always an integer.\n\n2.    Divisibility by 2 and 3: Since 6 is divisible by both 2 and 3, all multiples of 6 are also divisible by both 2 and 3. Divisibility by 2 implies that all multiples of 6 are even numbers, while divisibility by 3 suggests that the sum of the digits of any multiple of 6 is also divisible by 3.\n\n3.    Repeating Pattern of Digits: When we examine the units digit of multiples of 6, a fascinating pattern emerges. The units digit repeats in a cycle of 6, 2, 8, 4, 0, and again 6, as we proceed to higher multiples. For instance, the units digit of 6 is 6, the units digit of 12 is 2, and so on.\n\n********************\n\n10 Math Problems officially announces the release of Quick Math Solver and 10 Math ProblemsApps on Google Play Store for students around the world.\n\n********************\n\n********************\n\n## Applications of Multiples of 6\n\nThe study of multiples of 6 finds applications in various fields, including mathematics, science, and everyday life:\n\n1.    LCM: One of the applications of multiples of 6 is in finding the lowest common multiple (LCM) of two or more numbers. The LCM is the lowest multiple that two or more numbers have in common. For example, to find the LCM of 6 and 4, we need to find the multiples of both numbers and identify the lowest multiple they have in common. The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, … etc. The multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 … etc. The lowest multiple that they have in common is 12. Therefore, the LCM of 6 and 4 is 12.\n\n2.    Timekeeping: The division of an hour into 60 minutes and each minute into 60 seconds is based on the concept of multiples. By dividing 60 by 6, we obtain 10, which corresponds to the number of minutes in one-tenth of an hour.\n\n3.    Music: In music theory, multiples of 6 play a crucial role in understanding rhythm. The time signatures and beats per measure are based on multiples of 6, allowing musicians to create harmonious compositions.\n\n## Conclusion:\n\nThe Multiples of 6 are the numbers obtained by multiplying 6 with natural numbers. The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, … etc. The multiples of 6 have various properties, such as divisibility by 6, divisibility by 2 and 3, repeating pattern of digits, etc. The multiples of 6 have several applications in mathematics, such as finding the LCM, timekeeping, and music.<|endoftext|>"},"score":{"kind":"number","value":4.5625,"string":"4.5625"}}},{"rowIdx":1263,"cells":{"token_count":{"kind":"number","value":726,"string":"726"},"text":{"kind":"string","value":"# Multiplying Fractions And Mixed Numbers Worksheet 6th Grade\n\nThis multiplication worksheet is focused on training students the best way to psychologically increase complete figures. Students are able to use personalized grids to suit specifically a single question. The worksheets also includedecimals and fractions, and exponents. There are also multiplication worksheets with a distributed home. These worksheets can be a need to-have for the mathematics school. They could be utilized in class to discover ways to mentally grow total line and numbers them up. Multiplying Fractions And Mixed Numbers Worksheet 6th Grade.\n\n## Multiplication of entire numbers\n\nIf you want to improve your child’s math skills, you should consider purchasing a multiplication of whole numbers worksheet. These worksheets can assist you master this fundamental principle. It is possible to go for one digit multipliers or two-digit and about three-digit multipliers. Capabilities of 10 may also be an excellent solution. These worksheets will enable you to training extended practice and multiplication reading through the figures. Also, they are the best way to aid your youngster comprehend the necessity of learning the different kinds of entire figures.\n\n## Multiplication of fractions\n\nExperiencing multiplication of fractions with a worksheet may help professors strategy and get ready classes successfully. Utilizing fractions worksheets will allow instructors to swiftly evaluate students’ idea of fractions. Pupils can be pushed to complete the worksheet in a a number of time as well as then tag their techniques to see where by they need more training. Individuals can be helped by phrase conditions that associate maths to true-daily life circumstances. Some fractions worksheets incorporate types of comparing and contrasting numbers.\n\n## Multiplication of decimals\n\nOnce you increase two decimal numbers, ensure that you group of people them vertically. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. For instance, 01 x (11.2) by 2 can be equal to 01 by 2.33 by 11.2 except if the item has decimal areas of lower than two. Then, this product is round towards the nearby total variety.\n\n## Multiplication of exponents\n\nA math concepts worksheet for Multiplication of exponents will allow you to training dividing and multiplying phone numbers with exponents. This worksheet may also supply things that requires individuals to flourish two various exponents. You will be able to view other versions of the worksheet, by selecting the “All Positive” version. Apart from, also you can get into special recommendations on the worksheet by itself. When you’re done, you are able to click on “Produce” and the worksheet will probably be acquired.\n\n## Department of exponents\n\nThe essential tip for section of exponents when multiplying phone numbers is usually to subtract the exponent inside the denominator through the exponent in the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. By way of example, \\$23 separated by 4 will the same 27. This method is not always accurate, however. This method can cause uncertainty when multiplying numbers that happen to be too large or too small.\n\n## Linear capabilities\n\nYou’ve probably noticed that the cost was \\$320 x 10 days if you’ve ever rented a car. So the total rent would be \\$470. A linear purpose of this kind offers the develop f(by), in which ‘x’ is the number of times the vehicle was rented. Furthermore, they have the shape f(x) = ax b, where ‘b’ and ‘a’ are genuine figures.<|endoftext|>"},"score":{"kind":"number","value":4.5,"string":"4.5"}}},{"rowIdx":1264,"cells":{"token_count":{"kind":"number","value":416,"string":"416"},"text":{"kind":"string","value":"When the Spanish conquerors arrived first to these lands; they could not explain themselves how Peruvian \"Indians\" (ignorant, wild, without any ability of logical reasoning, one more animal species according to conquerors) could have built such a greatness. Their religious fanaticism led them to believe that all that was simply work of demons or malign spirits. Still today, many people believe in the inability of ancient Quechuas to create such a wonder, so they suggest that they were made by beings of some other worlds, extraterrestrial beings with superior technology that made all that possible. However, our history and archaeology demonstrate that those objects of admiration are an undeniable work of the Incas, Quechuas, Andean people or however pre-Hispanic inhabitants of this corner of the world would be named.\nThe imperial city Cusco, meaning ‘navel of the earth,’ was laid out in the form of a puma, the animal that symbolized the Inca dynasty. The belly of the puma was the main plaza, the river Tullumayo formed its spine, and the hill of Sacsayhuaman its head.\nOne of the most imposing architectonic complexes inherited from the Incan Society is Sacsayhuaman, which because of several of its qualities is considered as one of the best monuments that mankind built on the earth's surface.\nThe wall or rampart is the most impressive section, built with enormous carved limestone boulders, this construction has a broken line that faces to the main plaza called Chuquipampa which is a slope with 25 angles and 60 walls.The biggest carved boulder of the first wall weighs about 70 tons and like all of the other rocks was brought from a quarry called Sisicancha, three kilometers away and where there are still rocks that were transported part of the way. Each wall is made up of 10 fronts with the most important ones known as Rumipunco, tiupunku, Achuanpunku and Viracocha punku.<|endoftext|>"},"score":{"kind":"number","value":3.859375,"string":"3.859375"}}},{"rowIdx":1265,"cells":{"token_count":{"kind":"number","value":631,"string":"631"},"text":{"kind":"string","value":"Sign in or start a free trial to avail of this feature.\n3. Combine Strings of Text\nWe often need to combine multiple cells of text into one cell and to do this we use '&'. Find out how to use this very useful command correctly in this lesson\n&: Chain text strings together\nLOWER: Change cell text to lowercase\nALT + I , C: Insert column\nF2: Jump back into cell formula\nCombining text strings together in Excel is very easy and one of the most common commands you'll perform when manipulating data.\nIn this lesson, we're going to take the first and last name for each user and create a username, which will be the lowercase first name, followed by an underscore, followed by a lowercase last name.\nWe’ll start by creating a new column with Alt + I C, and we'll widen this using the mouse.\nI'll entitle this column Username and I'll start by writing equals and including the first name.\nTo chain the first name to the last name, we use the ampersand sign, which on most keyboards is Shift + 7.\nAnd then to include the last name in this string, I'll simply select it, which is H2, and press Enter.\nAnd as you can see, the ampersand sign chains the first name and the last name entries together in one cell.\nHowever, we still have a lot of work to do.\nOur first task will be to change the last name to lowercase.\nSo let's re-enter the formula with F2, and we replace H2 with the function “lower” and then the cell H2 passed into this function.\nIf I now press Enter, you can see that we now have the first name and the last name both in lowercase.\nAll we have to do now is include the underscore.\nSo let's return to our formula again and press F2 to jump back inside.\nWhen you're chaining strings and you'd like to include a character such as an underscore, you simply wrap that character n double quotes and then connect it to the other elements using the ampersand sign.\nSo I'll move in position to the first ampersand sign, open double quotes, write an underscore, close double quotes, and then write another ampersand sign to connect the underscore to the last name.\nIf I now press enter, you can see we now have the user name perfectly created for Kaci Rawlins.\nUsing the ampersand sign, we can combine strings of text in almost any way we want.\nTo fill out the username for the remaining entries, we can just select the cell and double click the bottom right hand corner with our mouse.\nTo get some more practice at chaining strings together, try to create a new column called Full Name that separates the first name and the last name with a space and has both values in the capitalized format.\nI'll leave my answer in the after file which can be downloaded underneath this video.<|endoftext|>"},"score":{"kind":"number","value":3.6875,"string":"3.6875"}}},{"rowIdx":1266,"cells":{"token_count":{"kind":"number","value":1441,"string":"1,441"},"text":{"kind":"string","value":"In the human digestive system, large organic masses are broken down into smaller particles that the body can use as fuel. This is a complex process. The breakdown of the nutrients requires the coordination of several enzymes secreted from specialized cells within the mouth, stomach, intestines, and liver. The major organs or structures that coordinate digestion within the human body include the mouth, esophagus, stomach, small and large intestines, and liver.\nIn the human body, the mouth (oral cavity) is a specialized organ for receiving food and breaking up large organic masses. In the mouth, food is changed mechanically by biting and chewing. Humans have four kinds of teeth: incisors are chisel-shaped teeth in the front of the mouth for biting; canines are pointed teeth for tearing; and premolars and molars are flattened, ridged teeth for grinding, pounding, and crushing food.\nIn the mouth, food is moistened by saliva, a sticky fluid that binds food particles together into a soft mass. Three pairs of salivary glands—parotid, submaxillary, and sublingual—secrete saliva into the mouth. The saliva contains an enzyme called amylase, which digests starch molecules into smaller molecules of the disaccharide maltose.\nDuring chewing, the tongue moves food about and manipulates it into a mass called a bolus. The bolus is pushed back into the pharynx (throat) and is forced through the opening to the esophagus.\nThe esophagus is a thick-walled muscular tube located behind the windpipe that extends through the neck and chest to the stomach. The bolus of food moves through the esophagus by peristalsis: a rhythmic series of muscular contractions that propels the bolus along. The contractions are assisted by the pull of gravity.\nThe esophagus joins the stomach at a point just below the diaphragm. A valvelike ring of muscle called the cardiac sphincter surrounds the opening to the stomach. The sphincter relaxes as the bolus passes through and then quickly closes.\nThe stomach is an expandable pouch located high in the abdominal cavity. Layers of stomach muscle contract and churn the bolus of food with gastric juices to form a soupy liquid called chyme.\nThe stomach stores food and prepares it for further digestion. In addition, the stomach plays a role in protein digestion. Gastric glands called chief cells secrete pepsinogen. Pepsinogen is converted to the enzyme pepsin in the presence of hydrochloric acid. Hydrochloric acid is secreted by parietal cells in the stomach lining. The pepsin then digests large proteins into smaller proteins called peptides. To protect the stomach lining from the acid, a third type of cell secretes mucus that lines the stomach cavity. An overabundance of acid due to mucus failure may lead to an ulcer.\nThe soupy mixture called chyme spurts from the stomach through a sphincter into the small intestine. An adult’s small intestine is about 23 feet long and is divided into three sections: the first 10 to 12 inches form the duodenum; the next 10 feet form the jejunum; and the final 12 feet form the ileum. The inner surface of the small intestine contains numerous fingerlike projections called villi (the singular is villus). Each villus has projections of cells called microvilli to increase the surface area.\nMost chemical digestion takes place in the duodenum. In this region, enzymes digest nutrients into simpler forms that can be absorbed. Intestinal enzymes are supplemented by enzymes from the pancreas, a large, glandular organ near the stomach. In addition, bile enters the small intestine from the gallbladder to assist in fat digestion.\nThe enzymes functioning in carbohydrate digestion include amylase (for starch), maltase (for maltose), sucrase (for sucrose), and lactase (for lactose). For fats, the principal enzyme is lipase. Before lipase can act, the large globules of fat must be broken into smaller droplets by bile. Bile is a mixture of salts, pigments, and cholesterol that is produced by the liver and stored in the gallbladder, a saclike structure underneath the liver.\nProtein digestion is accomplished by several enzymes, including two pancreatic enzymes: trypsin and chymotrypsin. Peptides are broken into smaller peptides, and peptidases reduce the enzymes to amino acids. Nucleases digest nucleic acids into nucleotides in the small intestine also.\nMost absorption in the small intestine occurs in the jejunum. The products of digestion enter cells of the villi, move across the cells, and enter blood vessels called capillaries. Diffusion accounts for the movement of many nutrients, but facilitated diffusion is responsible for the movement of glucose and amino acids. The products of fat digestion pass as small droplets of fat into lacteals, which are branches of the lymphatic system.\nAbsorption is completed in the final part of the small intestine, the ileum. Substances that have not been digested or absorbed then pass into the large intestine.\nThe small intestine joins the large intestine in the lower-right abdomen of the body. The two organs meet at a blind sac called the cecum and a small fingerlike organ called the appendix. Evolutionary biologists believe the cecum and appendix are vestiges of larger organs that may have been functional in human ancestors.\nThe large intestine is also known as the colon. It is divided into ascending, transverse, and descending portions, each about one foot in length. The colon’s chief functions are to absorb water and to store, process, and eliminate the residue following digestion and absorption. The intestinal matter remaining after water has been reclaimed is known as feces. Feces consist of nondigested food (such as cellulose), billions of mostly harmless bacteria, bile pigments, and other materials. The feces are stored in the rectum and passed out through the anus to complete the digestion process.\nThe liver has an important function in processing the products of human digestion. For example, cells of the liver remove excess glucose from the bloodstream and convert the glucose to a polymer called glycogen for storage.\nThe liver also functions in amino acid metabolism. In a process called deamination, it converts some amino acids to compounds that can be used in energy metabolism. In doing so, the liver removes the amino groups from amino acids and uses the amino groups to produce urea. Urea is removed from the body in the urine (see Chapter 26). Fats are processed into two-carbon units that can enter the Krebs cycle for energy metabolism. The liver also stores vitamins and minerals, forms many blood proteins, synthesizes cholesterol, and produces bile for fat digestion.<|endoftext|>"},"score":{"kind":"number","value":4.21875,"string":"4.21875"}}},{"rowIdx":1267,"cells":{"token_count":{"kind":"number","value":671,"string":"671"},"text":{"kind":"string","value":"When art enthusiasts first come across Teyjah McAren’s modern cave paintings, they often have questions about the meaning of the symbols they see. Visitors also want to know more about the origins of rock art and petroglyphs and the ancient peoples who painted them in the distant past. Learn more with our helpful guide to the language, symbols and heritage of cave art.\nAncient peoples around the world made art on natural geological surfaces such as cave walls, cliff faces, bedrock and boulders. Rock art includes pictographs, petroglyphs, engravings, petroforms, and geoglyphs.\nA hieroglyph is a form of picture writing which was the form of writing used in ancient Egypt. Each cluster of symbols represented a word or idea rather than just a sound like the alphabet.\nA petroglyph is a rock art design that has been created by carving or engraving. Other terms for petroglyphs are “rock art” and “rock writing”.\nA pictograph is a rock art design that was drawn or painted on a rock surface. Art materials could include powdered minerals, charcoal, plant substances, or blood.\nA geoglyph is rock art made on the ground. Geoglyphs include intaglios, which are made by removing stones or portions of the ground; petroforms (also called rock alignments) made by rearranging stones to form sculptures like Inukshuks); and gravel geoglyphs, which are shaped from gravel-covered areas.\nRock art was created by innovative creators using a wide range of methods and materials. One way was through pecking the stone with tools to give it a rough, pitted surface. Rubbing or wearing down the stone was used to create lines.\nSymbols and images could be incised, gouged out in intaglio technique, chipped out with a stone, sculpted, overpainted or crosshatched.\nSome rock art was painted with charcoal and earth pigments mixed with grease, gum, or water. Often, engraving and painting techniques were sometimes combined.\nSome believe that petroglyphs were created to mark tribal territory, indicate water, or show activities conducted in those areas. Other petroglyphs seem to record important events, such as a successful hunt or the migration of a tribe. Some symbols recorded good hunting sites. Others tracked the movement of the sun, the moon, planets, and stars. Some petroglyphs seem to relate to a spiritual life, religious ceremonies, sacred beings and the activities of shamans. Some rock art hints at epic legends and mythical protagonists.\nPetroglyphs can be associated with particular tribes or styles, but exact interpretation is exceedingly challenging. As these early cultures died out, their messages may have been lost, or the meaning altered with each successive generation. Even present-day members from different tribes may read the symbols differently. By learning more about these ancient cultures, we may gain insight into the mysterious signs they have left to us through the epochs.\nRock Art is found on almost every continent, in caves on boulders or rocky outcrops, on ancient monuments and other structures such as Stonehenge.<|endoftext|>"},"score":{"kind":"number","value":4,"string":"4"}}},{"rowIdx":1268,"cells":{"token_count":{"kind":"number","value":132,"string":"132"},"text":{"kind":"string","value":"# How do you solve \\sqrt { 9v - 15} = \\sqrt { 5v - 3}?\n\nDec 13, 2016\n\n$v = 3$\n\n#### Explanation:\n\n$\\sqrt{9 v - 15} = \\sqrt{5 v - 3}$\n\nSquare both sides.\n\n$9 v - 15 = 5 v - 3$\n\nSubtract $5 v$ from both sides.\n\n$4 v - 15 = - 3$\n\nAdd $15$ to both sides.\n\n$4 v = 12$\n\nDivide both sides by $4$.\n\n$v = 3$<|endoftext|>"},"score":{"kind":"number","value":4.4375,"string":"4.4375"}}},{"rowIdx":1269,"cells":{"token_count":{"kind":"number","value":344,"string":"344"},"text":{"kind":"string","value":"The Arctic experienced its second-warmest year on record between October 2017 and September 2018, according to the annual Arctic Report Card. And average air temperatures in this frigid region have hit record or near-record levels every year since 2014.\nTemperatures in the Arctic are now increasing at roughly twice the rate of the global average, researchers say in the report released by the US National Oceanic and Atmospheric Administration (NOAA).\nWarming is causing changes across the entire Arctic, and those changes are building, said Emily Osborne, the programme manager at the NOAA Arctic Research Program in Silver Spring, Maryland. She helped to present results from the report on 11 December during a meeting of the American Geophysical Union in Washington DC.\nWild reindeer and caribou populations, for example, have declined by more than 50% since the 1990s. And more than 99% of sea ice is now considered relatively new, meaning that it hasn’t lasted for more than four summers without melting. It’s now the thinnest and most susceptible to warming temperatures that it's been in 30 years.\nHarmful algal blooms are seven times more prevalent off the Alaskan coast than they were 40 years ago. And the Arctic Ocean now contains the highest concentration of microplastics of any ocean basin in the world, according to the report.\nThe microplastics and algal bloom findings really surprised Donald Perovich, a geophysicist at Dartmouth College in Hanover, New Hampshire, who worked on the report card. “Even here — the farthest point on Earth— we’re still seeing impacts from lower latitudes,” he says.<|endoftext|>"},"score":{"kind":"number","value":3.796875,"string":"3.796875"}}},{"rowIdx":1270,"cells":{"token_count":{"kind":"number","value":717,"string":"717"},"text":{"kind":"string","value":"# High School Math : How to find the length of the diagonal of a square\n\n## Example Questions\n\n### Example Question #1 : How To Find The Length Of The Diagonal Of A Square\n\nWhat is the length of a diagonal of a square with a side length ? Round to the nearest tenth.\n\nExplanation:\n\nA square is comprised of two 45-45-90 right triangles. The hypotenuse of a 45-45-90 right triangle follows the rule below, where  is the length of the sides.\n\nIn this instance, is equal to 6.\n\n### Example Question #1 : How To Find The Length Of The Diagonal Of A Square\n\nA square has sides of . What is the length of the diagonal of this square?\n\nExplanation:\n\nTo find the diagonal of the square, we effectively cut the square into two triangles.\n\nThe pattern for the sides of a is .\n\nSince two sides are equal to , this triangle will have sides of .\n\nTherefore, the diagonal (the hypotenuse) will have a length of .\n\n### Example Question #3 : How To Find The Length Of The Diagonal Of A Square\n\nA square has sides of . What is the length of the diagonal of this square?\n\nExplanation:\n\nTo find the diagonal of the square, we effectively cut the square into two triangles.\n\nThe pattern for the sides of a is .\n\nSince two sides are equal to , this triangle will have sides of .\n\nTherefore, the diagonal (the hypotenuse) will have a length of .\n\n### Example Question #233 : Geometry\n\nWhat is the length of the diagonal of a square with a side length of ?\n\nExplanation:\n\nTo find the diagonal of a square, we must use the side length to create a 90 degree triangle with side lengths of , and a hypotenuse which is equal to the diagonal.\n\nPythagorean’s Theorem states , where a and b are the legs and c is the hypotenuse.\n\nTake  and  and plug them into the equation for  and :\n\nAfter squaring the numbers, add them together:\n\nOnce you have the sum, take the square root of both sides:\n\nSimplify to find the answer: , or .\n\n### Example Question #5 : How To Find The Length Of The Diagonal Of A Square\n\nWhat is the length of the diagonal of a 7-by-7 square? (Round to the nearest tenth.)\n\nExplanation:\n\nTo find the diagonal of a square we must use the side lengths to create a 90 degree triangle with side lengths of 7 and a hypotenuse which is equal to the diagonal.\n\nWe can use the Pythagorean Theorem here to solve for the hypotenuse of a right triangle.\n\nThe Pythagorean Theorem states , where a and b are the sidelengths and c is the hypotenuse.\n\nPlug the side lengths into the equation as  and :\n\nSquare the numbers:\n\nAdd the terms on the left side of the equation together:\n\nTake the square root of both sides:\n\nTherefore the length of the diagonal is 9.9.\n\n### Example Question #1 : Squares\n\nThe perimeter of a square is 48. What is the length of its diagonal?\n\nExplanation:\n\nPerimeter = side * 4\n\n48 = side * 4\n\nSide = 12\n\nWe can break up the square into two equal right triangles. The diagonal of the sqaure is then the hypotenuse of these two triangles.\n\nTherefore, we can use the Pythagorean Theorem to solve for the diagonal:<|endoftext|>"},"score":{"kind":"number","value":4.59375,"string":"4.59375"}}},{"rowIdx":1271,"cells":{"token_count":{"kind":"number","value":776,"string":"776"},"text":{"kind":"string","value":"A paper published this week in Nature Geoscience has released the first measurements of greenhouse gases from permafrost under Arctic lakes. But while the study confirms those gases locked away in ice for thousands of years are seeping free, it concludes the amounts are not yet large.\n\"It's a lit fuse, but the length of that fuse is very long,\" said lead author Katey Walter Anthony of the University of Alaska.\n\"According to the model projections, we're getting ready for the part where it starts to explode. But it hasn't happened yet.\"\nScientists have long known that permafrost contains vast quantities of carbon in dead plants and other organic material, about twice as much as the entire atmosphere. Now, that permafrost is melting more quickly as the Arctic warms up faster than anywhere else on Earth.\nThe melt often takes place in Arctic lakes where liquid water thaws long-frozen soil. The material released is digested by tiny bugs and turned into carbon dioxide and methane.\n\"I've been walking on these lakes when they were frozen for a very long time,\" said Walter Anthony.\n\"I would go out after the ice formed, look at the lake ice surface and see bubbles. I observed that bubbles are most dense and largest along the edge, where the margins were expanding, where the permafrost was thawing.\"\nHer observation led to three questions: Were the bubbles generated by melting permafrost? If so, was the permafrost releasing ancient, long-dormant carbon? And, if so, how much?\nResearchers looked at lakes in Alaska and Siberia, as well as data from Canada. They used aerial photographs and other information to measure how the area had changed over the last 60 years.\nThey found that, across the Arctic, the amount of gas being released from a lake was directly related to its expansion. The more permafrost was melted around the water's edge, the bigger the lake became, and the more greenhouse gases were released.\nThe team captured some of those gases and subjected them to radiocarbon dating. They found the gases had been generated from carbon stored for anywhere between 10,000 and 30,000 years.\nIt was the answer to the final question that most surprised Walter Anthony.\nClimate scientists have long predicted a spike in the release of such long-dormant carbon. But if that's true, it isn't happening yet.\n\"Today, the permafrost-carbon feedback is pretty small,\" she said.\nModels suggest that over the next 90 years, greenhouse gas releases from permafrost will be 100 times higher than the levels Walter Anthony measured.\n\"I believe it will (happen), but when are we going to start seeing that?\" she asked.\nThe increasingly warmer Arctic may eventually reach a permafrost-carbon tipping point, Walter Anthony suggested.\nThe stakes are high. Scientists estimate there are more than 1,400 petagrams of old carbon stored in permafrost. Each petagram is a billion tonnes.\n\"There's a lot of interest in what the fate of that permafrost carbon is.\"\nThursday, Aug. 25, 2016 at 7:46 AM\n— Follow Bob Weber on Twitter at @row1960\noriginal story HERE\nSign up for the Global Warming Blog for free by clicking here. In your email you will receive critical news, research and the warning signs for the next global warming disaster.\nClick here to learn how global warming has become irreversible and what you can do to protect your family and assets.\nTo share this blog post: Go to the original shorter version of this post. Look to lower right for the large green Share button.\nTo view our current agreement or disagreement with this blog article, click here.<|endoftext|>"},"score":{"kind":"number","value":3.953125,"string":"3.953125"}}},{"rowIdx":1272,"cells":{"token_count":{"kind":"number","value":1876,"string":"1,876"},"text":{"kind":"string","value":"# 2.10 Extension: Fitting functions to quadratic data\n\nLesson\n\nIn our lesson from Algebra 1 on lines of best fit we saw how to fit data to a linear model to make predictions and interpret patterns in the data.  Now, we'll look at how we might do the same for data that more closely resembles a quadratic curve.\n\n## Using technology to fit quadratic regression models\n\nLet's consider the following data set:\n\n$x$x $4$4 $4.8$4.8 $5.1$5.1 $6$6 $7.1$7.1 $8.2$8.2 $9.4$9.4\n$y$y $19.4$19.4 $20.4$20.4 $20.2$20.2 $19.1$19.1 $18$18 $14.9$14.9 $10$10\n\nAs we can see from the scatter plot, the data appears to be following a quadratic pattern better than a linear one.\n\nInstead of trying to fit the data to a linear function, let's have our calculator fit a quadratic regression model to the data.\n\nAs you can see, when we go to choose a model for regression, there are many to choose from. Your knowledge of functions will help you make the best choice, but this lesson will focus on quadratics.\n\nAnd here we have the equation of the quadratic function fitted to the data.\n\nWe can see the value of the coefficient of determination, $r^2$r2, is very strong (it's very close to $1$1).\n\nGiven the coefficients $a$a, $b$b, and $c$c we have the quadratic function $y=-0.52x^2+5.27x+6.86$y=0.52x2+5.27x+6.86\n\n#### Practice questions\n\n##### Question 1\n\nThe scatter diagram shows the number of fish in pond as a function of time.\n\n1. Which type of model would be appropriate for this data?\n\nLinear\n\nA\n\nB\n\nLinear\n\nA\n\nB\n2. Will the coefficient of $x^2$x2 in the quadratic model be positive or negative?\n\nPositive\n\nA\n\nNegative\n\nB\n\nPositive\n\nA\n\nNegative\n\nB\n\n##### Question 2\n\nA social researcher claims that the longer people stay in their job, the less satisfaction they gain from their work. She asked a sample of people how many years they had been employed in their current job and to rate their level of satisfaction out of $10$10. The results are presented in the table.\n\nNumber of years employed Satisfaction rating\n$2$2 $9$9\n$3$3 $5$5\n$5$5 $4$4\n$6$6 $2$2\n$8$8 $5$5\n$10$10 $7$7\n$12$12 $8$8\n$13$13 $7$7\n$15$15 $9$9\n1. Create a scatter plot for the data collected.\n\n2. Does the scatter plot support the social worker’s claims?\n\nYes\n\nA\n\nNo\n\nB\n\nYes\n\nA\n\nNo\n\nB\n3. Which form of equation would be best suited to model the relationship between the number of years employed and satisfaction with the job?\n\n$y=mx+b$y=mx+b\n\nA\n\n$y=a^x+b$y=ax+b\n\nB\n\n$y=ax^2+bx+c$y=ax2+bx+c\n\nC\n\n$y=mx+b$y=mx+b\n\nA\n\n$y=a^x+b$y=ax+b\n\nB\n\n$y=ax^2+bx+c$y=ax2+bx+c\n\nC\n4. The social researcher herself has been employed in her current job for $4$4 years and rates her satisfaction with her work a $10$10 out of $10$10. What is the difference between the satisfaction rating approximated by the model $y=0.5x^2-6x+20$y=0.5x26x+20 and her actual rating?\n\n##### Question 3\n\nNine data points have been plotted below with a quadratic curve of best fit.\n\n1. Predict the $y$y-value of a point with an $x$x-value of $13$13.\n\n2. Which of the following points would be predicted by the quadratic curve of best fit?\n\n$\\left(3,4\\right)$(3,4)\n\nA\n\n$\\left(14,10\\right)$(14,10)\n\nB\n\n$\\left(2,9\\right)$(2,9)\n\nC\n\n$\\left(15,15\\right)$(15,15)\n\nD\n\n$\\left(3,4\\right)$(3,4)\n\nA\n\n$\\left(14,10\\right)$(14,10)\n\nB\n\n$\\left(2,9\\right)$(2,9)\n\nC\n\n$\\left(15,15\\right)$(15,15)\n\nD\n\n## Interpolation and extrapolation from a model\n\nGiven a set of data relating two variables $x$x and $y$y, we can use a model to best estimate how the dependent variable changes in response to the independent variable $x$x. A model allows us to go one step further and make predictions about other possible ordered pairs that fit this relationship.\n\n#### Exploration\n\nSay we gathered several measurements on the population $P$P of a small town $t$t years after an earthquake. We can then plot the data on the $xy$xy-plane as shown below.\n\n Population of a town measured at several instances.\n\nWe can fit a model through the observed data to make predictions about the population at certain times after the earthquake. One plausible model might look like this function:\n\n A curve modeling population of a town over time.\n\nTo make a prediction on the population, say two years after the earthquake, we first identify the point on the curve when $t=2$t=2. Then we find the corresponding value of $P$P. As you can see below, the model predicts that two years after the earthquake, the population of the town was $250$250.\n\n A predicted population of $250$250 when $t=2$t=2.\n\nA prediction which is made within the observed data set is called an interpolation. Roughly speaking, we've gathered data between $t=0.3$t=0.3 and $t=2.3$t=2.3 so a prediction at $t=2$t=2 would be classified as an interpolation.\n\nIf we predict the population six years after the earthquake, we find that the population is roughly $16$16. A prediction outside the observed data set such as this one is called an extrapolation.\n\n A predicted population of $16$16 when $t=6$t=6.\n\nHow reliable are these predictions? Well, any model that fits the observed data will make reliable predictions from interpolations since the model roughly passes through the center of the data points. We can say that the model follows the trend of the observed data.\n\nHowever, extrapolations are generally unreliable since we make assumptions about how the relationship continues outside of collected data. Sometimes extrapolation can be made more reliable if we have additional information about the relationship.\n\nConsider if we were to use the following quadratic model to fit the data. We can see that interpolating doesn't change much from the previous model, but the predicted values from extrapolation are very different.\n\n Polynomial curve modeling population of a town over time.\n\nWith further information, say like government funding and support aid, the population of the town might increase after a certain point and so the polynomial curve may be an appropriate model to use.\n\nRemember!\n\nA prediction made within the observed data is called interpolation.\n\nA prediction made outside the observed data is called extrapolation.\n\nGenerally, extrapolation is less reliable than interpolation since the model makes assumptions about the relationship outside the observed data set.\n\n#### Practice question\n\n##### Question 4\n\nThe height of a particular projectile $y$y in meters was measured at different times $t$t in seconds and the following curve of best fit was drawn.\n\n1. Using the curve of best fit, what is the predicted height of the projectile after $1$1 second?\n\n2. Using the curve of best fit, what is the predicted height of the projectile after $4$4 seconds?\n\n3. The height at a given time has the following relationship $y=-5t^2+100$y=5t2+100. Which of the following statements about interpolation and extrapolation is true?\n\nExtrapolation is less reliable since we make assumptions about the relationship outside of the observed data.\n\nA\n\nExtrapolation is less reliable since we should use a line of best fit instead.\n\nB\n\nExtrapolation is less reliable since we make assumptions about the relationship outside of the observed data.\n\nA\n\nExtrapolation is less reliable since we should use a line of best fit instead.\n\nB<|endoftext|>"},"score":{"kind":"number","value":4.625,"string":"4.625"}}},{"rowIdx":1273,"cells":{"token_count":{"kind":"number","value":728,"string":"728"},"text":{"kind":"string","value":"Developmental Delay and Disabilities\nDevelopmental delay refers to the children’s physical and mental functioning level which is not on par with children of their age group. In other words, these children do not reach their developmental milestone at the expected age. Developmental delay is not only delay in physical development and hence, may not be easily detected by parents or caregivers. Developmental delay and disabilities in children include: delay in speech and language development, delay in cognition development, delay in motor development, delay in emotional and social skills, Global Developmental Delay (delay in most of the areas), and Pervasive Developmental Disorder\n|Delay in speech and language development||Children’s ability in speech and language is lower than children in their age group as they take a longer time to reach the expected speech and language level. For instance, they have no meaningful speech at all at the age of 2, have difficulty at speaking in sentence level at the age of 3, and have “baby talk” or still mumbling after the age of 3.|\n|Delay in cognition development||Cognition delay refers to children’s intelligence or cognition that has not reached the expected developmental milestone. These children receive equal education with peers but parents and teachers may find their learning is much slower. In addition, they have difficulty in understanding abstract concepts, in memorizing and recalling information as well as poor organizational skills.|\n|Delay in motor development||Motor development delay includes delay in gross and fine motor skills as well as body coordination. Gross motor skills include walking, running, crawling and any movement that involves large group of muscles. Fine motor skills include skills of manipulating hands and wrist such as eye-hand coordination, finger movement and manipulation.|\n|Delay in emotional and social skills||Children with such delay would encounter difficulty with emotion regulation and socializing. They often fail to establish and maintain friendship. For instance, they have difficulty in maintaining conversations; they are passive in play, and they prefer to be alone. On top of that, they might have inappropriate emotional responses and behaviors, i.e. laugh or burst out in tears in inappropriate context. In terms of emotion, they tend to get anxious and panic easily.|\n|Global Developmental Delay||Children have delay in most of their skill areas. It is common to see children having delay in more than a specific area. For instance, a child could have motor development and speech development at the same time.|\n|Pervasive Developmental Disorder||This is a specific developmental disability as it affects children’s ability in social interaction, communication and their interest. According to the DSM-IV-TR, Pervasive Developmental Disorders include Autistic Disorder, Asperger’s Disorder, Rett’s Disorder, and Childhood Disintegrative Disorder.|\nDevelopmental delay, especially speech and cognition delay, is closely related with Learning Disabilities. In many cases, parents seek for professional consultation and help only when they found out that their children are slow in academic learning (often reported by school teachers). Unfortunately, by then these children have already reached school age and most of them have missed the critical time for early intervention (0-6 years old). The developmental delay, especially speech and cognition delay, has already affected their school performance.\nIf parents or main caregivers had detected signs of developmental delay earlier, these children’s functional skills could be improved with early intervention. Hence, parents or caregivers are encouraged to be observant and caring in understanding children’s development milestone.<|endoftext|>"},"score":{"kind":"number","value":3.90625,"string":"3.90625"}}},{"rowIdx":1274,"cells":{"token_count":{"kind":"number","value":1558,"string":"1,558"},"text":{"kind":"string","value":"# Parabolas Set 4\n\nGo back to  'SOLVED EXAMPLES'\n\nExample – 7\n\nThe normals at three points A, B and C on a parabola intersect at O. F is the focus of the parabola. Prove that\n\n$FA\\cdot FB \\cdot FC = \\frac{l}{4} \\cdot F{O^2}$\n\nwhere l is the length of the latus-rectum.\n\nSolution: Assume the parabola to be $${y^2} = 4ax,$$ and the point O to be (h, k). F is the point (a, 0).\n\nAny normal to is $${y^2}= 4ax$$ is\n\n$y = mx - 2am -a{m^3}$\n\nand since this passes through O(h, k), we have\n\n$k = mh - 2am -a{m^3}...\\left( 1 \\right)$\n\nThe feet of the three normal correspond to A, B and C. Thus, if $${m_1},\\,{m_2}{\\;\\rm{and }}\\;{\\rm{ }}{m_3}$$ and are the roots of (1), the co-ordinates of A, B and C are\n\n$(am_1^2,\\, -2a{m_1}),\\,(am_2^2,\\, - 2a{m_2}){\\;\\rm{and}}\\;{\\rm{ }}(am_3^2,\\, - 2a{m_3})$\n\nWe have now,\n\n$FA = \\sqrt {{{(a -am_1^2)}^2} + {{(2a{m_1})}^2}}$\n\n$= a(1 + m_1^2)$\n\nSimilarly, $$FB = a(1 + m_2^2){\\;\\rm{ and}}\\;{\\rm{ }}FC = a(1 + m_3^2)$$.\n\nThus,\n\n\\begin{align}&FA \\cdot FB \\cdot FC= {a^3}(1 + m_1^2)(1 + m_2^2)(1 + m_3^2)\\\\\\\\&= {a^3}\\left( {1 +m_1^2 + m_2^2 + m_3^2 + m_1^2m_2^2 + m_2^2m_3^2 + m_3^2m_1^2 +{{({m_1}{m_2}{m_3})}^2}} \\right) \\qquad \\quad \\dots \\left( 2 \\right)\\end{align}\n\nFrom(1), we know the values of\n\n\\left.\\begin{align}{m_1} + {m_2} +{m_3} = 0\\\\{m_1}{m_2} +{m_2}{m_3} + {m_3}{m_1} = \\left( {\\frac{{2a - h}}{a}} \\right)\\\\{m_1}{m_2}{m_3} = -\\frac{k}{a}\\end{align}\\right\\} \\qquad \\quad \\dots\\left( 3 \\right)\n\nOur task is to express the relation (2) in terms of the known quantities given by(3). This can be done as follows :\n\n\\begin{align}&\\qquad \\qquad \\qquad \\qquad m_1^2+ m_2^2 + m_3^3 = {({m_1} + {m_2} + {m_3})^2} - 2({m_1}{m_2} + {m_2}{m_3} +{m_3}{m_1})\\\\\\\\&m_1^2m_2^2 + m_2^2m_3^2 + m_3^2m_1^2 = {({m_1}{m_2} + {m_2}{m_3} +{m_3}{m_1})^2} - 2{m_1}{m_2}{m_3}({m_1} + {m_2} + {m_3})\\end{align}\n\nSubstituting the appropriate values gives\n\n\\begin{align}FA\\cdot FB \\cdot FC &= {a^3}\\left\\{ {1 + {{\\left( {\\frac{{2a - h}}{a}} \\right)}^2}- 2\\left( {\\frac{{2a - h}}{a}} \\right) + \\frac{{{k^2}}}{{{a^2}}}} \\right\\}\\\\\\\\ &=a\\{ {(h - a)^2} + {k^2}\\} \\end{align}\n\nwhich evidently equals $$a \\cdot F{O^2}.$$ Since $$l = 4a,$$ we get the desired result:\n\n$FA \\cdot FB \\cdot FC = \\frac{l}{4} \\cdot F{O^2}$\n\nExample - 8\n\nNormals are drawn form the point P with slopes $${m_1},{\\rm{ }}{m_2},{m_3}$$ to the parabola $${y^2} = 4x.$$ If the locus of P with $${m_1}{m_2} = \\alpha$$ is a part of the parabola itself,then find $$\\alpha$$ .\n\nSolution: If we let P be the point (h, k), we have\n\n${m^3} + (2 - h)m + k = 0 \\qquad \\quad \\dots \\left( 1 \\right)$\n\nThus,\n\n\\begin{align}&\\qquad \\;{m_1}{m_2}{m_3}= - k\\\\\\\\&\\Rightarrow \\quad {m_3} = - \\frac{k}{\\alpha }\\left( {{\\rm{since}}\\;{m_1}{m_2} = \\alpha } \\right)\\end{align}\n\nSubstituting this back into (1), we obtain\n\n\\begin{align}& \\qquad \\qquad - \\frac{{{k^3}}}{{{\\alpha ^3}}} - \\frac{k}{\\alpha }(2 - h) + k = 0\\\\\\\\&\\Rightarrow \\qquad \\;\\;{k^2} = {\\alpha ^2}h - 2{\\alpha ^2} + {\\alpha ^3} \\qquad \\qquad \\dots \\left( 2\\right)\\end{align}\n\nAlso, since P lies on the parabola itself, we have\n\n${k^2} =4h...\\left( 3 \\right)$\n\nFrom(2) and (3), we have\n\n$\\begin{array}{l}{\\alpha^2} = 4{\\rm{ }}\\;and\\;{\\rm{ }}{\\alpha ^3} - 2{\\alpha ^2} = 0\\\\\\\\\\Rightarrow \\alpha = 2\\end{array}$\n\nLearn from the best math teachers and top your exams\n\n• Live one on one classroom and doubt clearing\n• Practice worksheets in and after class for conceptual clarity\n• Personalized curriculum to keep up with school<|endoftext|>"},"score":{"kind":"number","value":4.46875,"string":"4.46875"}}},{"rowIdx":1275,"cells":{"token_count":{"kind":"number","value":5548,"string":"5,548"},"text":{"kind":"string","value":"NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics\n\n# NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics\n\nEdited By Sumit Saini | Updated on Jul 18, 2022 12:03 PM IST\n\nNCERT solutions for Class 11 Maths chapter 15 exercise 15.1 mainly deals with the mean deviation topics and other exercises include important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range, etc. In Class 9 and 10 Maths NCERT syllabus, students have already read the basic Statistics like mean, mode, and median. In this NCERT book Class 11 chapter applications of Statistics with advanced concepts are discussed.\n\nJEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy\n\nExercise 15.1 Class 11 Maths have easy questions. Solving NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 is a must to solve upcoming exercises. Direct questions are asked from this exercise as well in Boards. Also students can refer to the following exercise of NCERT for further information about upcoming exercises.\n\n## Statistics Class 11 Chapter 15-Exercise: 15.1\n\nMean ( $\\overline{x}$ ) of the given data:\n\n$\\overline{x} = \\frac{1}{8}\\sum_{i=1}^{8}x_i = \\frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10$\n\nThe respective absolute values of the deviations from mean, $|x_i - \\overline{x}|$ are\n\n6, 3, 2, 1, 0, 2, 3, 7\n\n$\\therefore$ $\\sum_{i=1}^{8}|x_i - 10| = 24$\n\n$\\therefore$ $M.D.(\\overline{x}) = \\frac{1}{n}\\sum_{i=1}^{n}|x_i - \\overline{x}|$\n\n$= \\frac{24}{8} = 3$\n\nHence, the mean deviation about the mean is 3.\n\nMean ( $\\overline{x}$ ) of the given data:\n\n$\\\\ \\overline{x} = \\frac{1}{8}\\sum_{i=1}^{8}x_i = \\frac{38+70+48+40+42+55+63+46+54+44}{10} \\\\ = \\frac{500}{10} = 50$\n\nThe respective absolute values of the deviations from mean, $|x_i - \\overline{x}|$ are\n\n12, 20, 2, 10, 8, 5, 13, 4, 4, 6\n\n$\\therefore$ $\\sum_{i=1}^{8}|x_i - 50| = 84$\n\n$\\therefore$ $M.D.(\\overline{x}) = \\frac{1}{n}\\sum_{i=1}^{n}|x_i - \\overline{x}|$\n\n$= \\frac{84}{10} = 8.4$\n\nHence, the mean deviation about the mean is 8.4.\n\nQuestion:3. Find the mean deviation about the median.\n\nNumber of observations, n = 12, which is even.\n\nArranging the values in ascending order:\n\n10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.\n\nNow, Median (M)\n\n$\\\\ = \\frac{(\\frac{12}{2})^{th} observation + (\\frac{12}{2}+ 1)^{th} observation}{2} \\\\ = \\frac{13 + 14}{2} = \\frac{27}{2}= 13.5$\n\nThe respective absolute values of the deviations from median, $|x_i - M|$ are\n\n3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5\n\n$\\therefore$ $\\sum_{i=1}^{8}|x_i - 13.5| = 28$\n\n$\\therefore$ $M.D.(M) = \\frac{1}{12}\\sum_{i=1}^{n}|x_i - M|$\n\n$= \\frac{28}{12} = 2.33$\n\nHence, the mean deviation about the median is 2.33.\n\nQuestion:4. Find the mean deviation about the median.\n\nNumber of observations, n = 10, which is even.\n\nArranging the values in ascending order:\n\n36, 42, 45, 46, 46, 49, 51, 53, 60, 72\n\nNow, Median (M)\n\n$\\\\ = \\frac{(\\frac{10}{2})^{th} observation + (\\frac{10}{2}+ 1)^{th} observation}{2} \\\\ = \\frac{46 + 49}{2} = \\frac{95}{2}= 47.5$\n\nThe respective absolute values of the deviations from median, $|x_i - M|$ are\n\n11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5\n\n$\\therefore$ $\\sum_{i=1}^{8}|x_i - 47.5| = 70$\n\n$\\therefore$ $M.D.(M) = \\frac{1}{10}\\sum_{i=1}^{n}|x_i - M|$\n\n$= \\frac{70}{10} = 7$\n\nHence, the mean deviation about the median is 7.\n\nQuestion:5 Find the mean deviation about the mean.\n\n $x_i$ $f_i$ $f_ix_i$ $|x_i - \\overline{x}|$ $f_i|x_i - \\overline{x}|$ 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 $\\sum{f_i}$ = 25 $\\sum f_ix_i$ = 350 $\\sum f_i|x_i - \\overline{x}|$ =158\n\n$N = \\sum_{i=1}^{5}{f_i} = 25 ; \\sum_{i=1}^{5}{f_ix_i} = 350$\n\n$\\overline{x} = \\frac{1}{N}\\sum_{i=1}^{n}f_ix_i = \\frac{350}{12} = 14$\n\nNow, we calculate the absolute values of the deviations from mean, $|x_i - \\overline{x}|$ and\n\n$\\sum f_i|x_i - \\overline{x}|$ = 158\n\n$\\therefore$ $M.D.(\\overline{x}) = \\frac{1}{25}\\sum_{i=1}^{n}f_i|x_i - \\overline{x}|$\n\n$= \\frac{158}{25} = 6.32$\n\nHence, the mean deviation about the mean is 6.32\n\nQuestion:6. Find the mean deviation about the mean.\n\n $x_i$ $f_i$ $f_ix_i$ $|x_i - \\overline{x}|$ $f_i|x_i - \\overline{x}|$ 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 $\\sum{f_i}$ = 80 $\\sum f_ix_i$ = 4000 $\\sum f_i|x_i - \\overline{x}|$ =1280\n\n$N = \\sum_{i=1}^{5}{f_i} = 80 ; \\sum_{i=1}^{5}{f_ix_i} = 4000$\n\n$\\overline{x} = \\frac{1}{N}\\sum_{i=1}^{n}f_ix_i = \\frac{4000}{80} = 50$\n\nNow, we calculate the absolute values of the deviations from mean, $|x_i - \\overline{x}|$ and\n\n$\\sum f_i|x_i - \\overline{x}|$ = 1280\n\n$\\therefore$ $M.D.(\\overline{x}) = \\frac{1}{80}\\sum_{i=1}^{5}f_i|x_i - \\overline{x}|$\n\n$= \\frac{1280}{80} = 16$\n\nHence, the mean deviation about the mean is 16\n\nQuestion:7. Find the mean deviation about the median.\n\n $x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 5 8 8 2 16 7 6 14 0 0 9 2 16 2 4 10 2 18 3 6 12 2 20 5 10 15 6 26 8 48\n\nNow, N = 26 which is even.\n\nMedian is the mean of $\\dpi{100} 13^{th}$ and $\\dpi{100} 14^{th}$ observations.\n\nBoth these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.\n\nTherefore, Median, M $\\dpi{100} = \\frac{13^{th} observation + 14^{th} observation}{2} = \\frac{7 + 7}{2} = \\frac{14 }{2} = 7$\n\nNow, we calculate the absolute values of the deviations from median, $|x_i - M|$ and\n\n$\\sum f_i|x_i - M|$ = 84\n\n$\\therefore$ $M.D.(M) = \\frac{1}{26}\\sum_{i=1}^{6}|x_i - M|$\n\n$= \\frac{84}{26} = 3.23$\n\nHence, the mean deviation about the median is 3.23\n\n $x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 15 3 3 13.5 40.5 21 5 8 7.5 37.5 27 6 14 1.5 9 30 7 21 1.5 10.5 35 8 29 6.5 52\n\nNow, N = 30, which is even.\n\nMedian is the mean of $15^{th}$ and $\\dpi{100} 16^{th}$ observations.\n\nBoth these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.\n\nTherefore, Median, M $\\dpi{100} = \\frac{15^{th} observation + 16^{th} observation}{2} = \\frac{30 + 30}{2} = 30$\n\nNow, we calculate the absolute values of the deviations from median, $|x_i - M|$ and\n\n$\\sum f_i|x_i - M|$ = 149.5\n\n$\\therefore$ $M.D.(M) = \\frac{1}{29}\\sum_{i=1}^{5}|x_i - M|$\n\n$= \\frac{149.5}{29} = 5.1$\n\nHence, the mean deviation about the median is 5.1\n\nQuestion:9. Find the mean deviation about the mean.\n\n Income per day in Rs Number of persons\n\n Income per day Number of Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \\overline{x}|$ $f_i|x_i - \\overline{x}|$ 0 -100 4 50 200 308 1232 100 -200 8 150 1200 208 1664 200-300 9 250 2250 108 972 300-400 10 350 3500 8 80 400-500 7 450 3150 92 644 500-600 5 550 2750 192 960 600-700 4 650 2600 292 1168 700-800 3 750 2250 392 1176 $\\sum{f_i}$ =50 $\\sum f_ix_i$ =17900 $\\sum f_i|x_i - \\overline{x}|$ =7896\n\n$N = \\sum_{i=1}^{8}{f_i} = 50 ; \\sum_{i=1}^{8}{f_ix_i} = 17900$\n\n$\\overline{x} = \\frac{1}{N}\\sum_{i=1}^{8}f_ix_i = \\frac{17900}{50} = 358$\n\nNow, we calculate the absolute values of the deviations from mean, $|x_i - \\overline{x}|$ and\n\n$\\sum f_i|x_i - \\overline{x}|$ = 7896\n\n$\\therefore$ $M.D.(\\overline{x}) = \\frac{1}{50}\\sum_{i=1}^{8}f_i|x_i - \\overline{x}|$\n\n$= \\frac{7896}{50} = 157.92$\n\nHence, the mean deviation about the mean is 157.92\n\nQuestion:10. Find the mean deviation about the mean.\n\n Height in cms Number of person\n\n Height in cms Number of Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \\overline{x}|$ $f_i|x_i - \\overline{x}|$ 95 -105 9 100 900 25.3 227.7 105 -115 13 110 1430 15.3 198.9 115-125 26 120 3120 5.3 137.8 125-135 30 130 3900 4.7 141 135-145 12 140 1680 14.7 176.4 145-155 10 150 1500 24.7 247 $\\sum{f_i}$ =100 $\\sum f_ix_i$ =12530 $\\sum f_i|x_i - \\overline{x}|$ =1128.8\n\n$N = \\sum_{i=1}^{6}{f_i} = 100 ; \\sum_{i=1}^{6}{f_ix_i} = 12530$\n\n$\\overline{x} = \\frac{1}{N}\\sum_{i=1}^{6}f_ix_i = \\frac{12530}{100} = 125.3$\n\nNow, we calculate the absolute values of the deviations from mean, $|x_i - \\overline{x}|$ and\n\n$\\sum f_i|x_i - \\overline{x}|$ = 1128.8\n\n$\\therefore$ $M.D.(\\overline{x}) = \\frac{1}{100}\\sum_{i=1}^{6}f_i|x_i - \\overline{x}|$\n\n$= \\frac{1128.8}{100} = 11.29$\n\nHence, the mean deviation about the mean is 11.29\n\n Marks Number of girls\n\n Marks Number of Girls $f_i$ Cumulative Frequency c.f. Mid Points $x_i$ $|x_i - M|$ $f_i|x_i - M|$ 0-10 6 6 5 22.85 137.1 10-20 8 14 15 12.85 102.8 20-30 14 28 25 2.85 39.9 30-40 16 44 35 7.15 114.4 40-50 4 48 45 17.15 68.6 50-60 2 50 55 27.15 54.3 $\\sum f_i|x_i - M|$ =517.1\n\nNow, N = 50, which is even.\n\nThe class interval containing $\\dpi{80} \\left (\\frac{N}{2} \\right)^{th}$ or $\\dpi{100} 25^{th}$ item is 20-30. Therefore, 20-30 is the median class.\n\nWe know,\n\nMedian $\\dpi{100} = l + \\frac{\\frac{N}{2}- C}{f}\\times h$\n\nHere, l = 20, C = 14, f = 14, h = 10 and N = 50\n\nTherefore, Median $\\dpi{100} = 20 + \\frac{25 - 14}{14}\\times 10 = 20 + 7.85 = 27.85$\n\nNow, we calculate the absolute values of the deviations from median, $|x_i - M|$ and\n\n$\\sum f_i|x_i - M|$ = 517.1\n\n$\\therefore$ $M.D.(M) = \\frac{1}{50}\\sum_{i=1}^{6}f_i|x_i - M|$\n\n$= \\frac{517.1}{50} = 10.34$\n\nHence, the mean deviation about the median is 10.34\n\n Age (in years) Number\n\n[ Hint Convert the given data into continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval]\n\n Age (in years) Number $f_i$ Cumulative Frequency c.f. Mid Points $x_i$ $|x_i - M|$ $f_i|x_i - M|$ 15.5-20.5 5 5 18 20 100 20.5-25.5 6 11 23 15 90 25.5-30.5 12 23 28 10 120 30.5-35.5 14 37 33 5 70 35.5-40.5 26 63 38 0 0 40.5-45.5 12 75 43 5 60 45.5-50.5 16 91 48 10 160 50.5-55.5 9 100 53 15 135 $\\sum f_i|x_i - M|$ =735\n\nNow, N = 100, which is even.\n\nThe class interval containing $\\dpi{80} \\left (\\frac{N}{2} \\right)^{th}$ or $50^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.\n\nWe know,\n\nMedian $\\dpi{100} = l + \\frac{\\frac{N}{2}- C}{f}\\times h$\n\nHere, l = 35.5, C = 37, f = 26, h = 5 and N = 100\n\nTherefore, Median $\\dpi{100} = 35.5 + \\frac{50 - 37}{26}\\times 5 = 35.5 + 2.5 = 38$\n\nNow, we calculate the absolute values of the deviations from median, $|x_i - M|$ and\n\n$\\sum f_i|x_i - M|$ = 735\n\n$\\therefore$ $M.D.(M) = \\frac{1}{100}\\sum_{i=1}^{8}f_i|x_i - M|$\n\n$= \\frac{735}{100} = 7.35$\n\nHence, the mean deviation about the median is 7.35\n\n## More About NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1\n\nThe NCERT Class 11 Maths chapter Statistics is very scoring chapters and questions are asked on the expected concepts only. It does not have much application in other chapters hence if one is not good at other chapters can strengthen this chapter to score well in the exam. Exercise 15.1 Class 11 Maths Mainly discusses the mean deviation about mean and median. Hence NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 can be seen as a scoring exercise from the exam perspective.\n\n## Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1\n\n• The Class 11 Maths chapter 15 exercise is prepared by expert faculties with rich experience.\n\n• Exercise 15.1 Class 11 Maths can help one enhance their marks in the exam as one question of 5 marks is mostly asked from this.\n\n• Class 11 Maths chapter 15 exercise 15.1 solutions provided here are comprehensive in manner.\n\nAlso see-\n\nJEE Main Highest Scoring Chapters & Topics\nJust Study 40% Syllabus and Score upto 100%\n\n## Subject Wise NCERT Exemplar solutions\n\nHappy learning!!!\n\n1. Which topics are covered in Exercise 15.1 Class 11 Maths?\n\nExercise 15.1 Class 11 Maths includes mean deviation about mean and median.\n\n2. Is this chapter 15 related to other chapters ?\n\nNo, mostly concepts are new and can be understood if other chapters are not read.\n\n3. What is the difficulty level of the questions asked in this chapter ?\n\nQuestions are easier to moderate level of difficulty if concepts are memorized.\n\n4. Is it necessary to remember the formulas ?\n\nYes, some basic formulas must be remembered to solve the questions.\n\n5. How many questions are there in the Exercise 15.1 Class 11 Maths ?\n\nTotal 12 questions are discussed in this exercise.\n\n## Upcoming School Exams\n\n#### National Means Cum-Merit Scholarship\n\nApplication Date:05 August,2024 - 06 September,2024\n\nExam Date:19 September,2024 - 19 September,2024\n\nExam Date:20 September,2024 - 20 September,2024\n\nExam Date:26 September,2024 - 26 September,2024\n\nApplication Date:30 September,2024 - 30 September,2024\n\nGet answers from students and experts\n\nA block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is\n\n Option 1) Option 2) Option 3) Option 4)\n\nA person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :\n\n Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg\n\nAn athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range\n\n Option 1) Option 2) Option 3) Option 4)\n\nA particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point\n\n Option 1) Option 2) Option 3) Option 4)\n\nIn the reaction,\n\n Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .\n\nHow many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?\n\n Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2\n\nIf we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will\n\n Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.\n\nWith increase of temperature, which of these changes?\n\n Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.\n\nNumber of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is\n\n Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023\n\nA pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is\n\n Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9<|endoftext|>"},"score":{"kind":"number","value":4.59375,"string":"4.59375"}}},{"rowIdx":1276,"cells":{"token_count":{"kind":"number","value":302,"string":"302"},"text":{"kind":"string","value":"The amount of data in the world is vast and is increasing exponentially. It is easy to become overwhelmed and lose sight of the goal of data: to answer questions we have about the world in a specific, concise manner. The goal of this course is to help craft answerable questions—and then answer them. In order to do this, we will be using a programming language (“R”) to help us organize data, make clean, clear graphs, and help with appropriate analysis of the data. This course will serve two main goals. The first is an introductory statistics course: gain knowledge of the basic statistical tests, how to interpret their results in a reasonable manner, and understand what those tests are doing at a conceptual level. The second is to learn the computational language of R: how to sort, shape, and handle data, create simulations and interpret the results, and build clean, clear graphical representations of the data presented. This course is taught at the introductory level and has no prerequisites, but does require a significant amount of time and energy outside of the classroom as we are working towards the two aforementioned goals at once. This course is appropriate for students who plan to seriously create and analyze their own statistics for their work. It may be taken alone, or as a sequel to Presentation of Statistics. There is some overlap between the two courses, but their focus and goals are different. Students who take Presentation of Statistics first will get a broader skill set and a more gentle introduction.<|endoftext|>"},"score":{"kind":"number","value":3.8125,"string":"3.8125"}}},{"rowIdx":1277,"cells":{"token_count":{"kind":"number","value":1595,"string":"1,595"},"text":{"kind":"string","value":"# Finding the inverse of a function\n\n• Jun 23rd 2011, 05:42 AM\nIBstudent\nFinding the inverse of a function\nWhat is the inverse of f(x) = x/x-1 ?\nthis is my attempt to solve it:\ny = x / x-1\ny(x-1) = x\nhence x(x-1) is the inverse.....\nhowever, substituting with x=5 gives......\n5/4=1.25\nso pluging this is the inverse function should give 5, right?\nlets see:\n1.25(1.25-1) = 1.25*.25= 0.3125. WRONG!!\n\ncan anyone tell me where I went wrong?\nthanks\n• Jun 23rd 2011, 05:57 AM\ne^(i*pi)\nRe: Finding the inverse of a function\nExpand $y(x-1) = x$\n\nThis gives $yx - y = x$. Now you can get all your x terms on the LHS and factor\n• Jun 23rd 2011, 06:12 AM\nProve It\nRe: Finding the inverse of a function\nAs an alternative, write $\\displaystyle \\frac{x}{x - 1} = \\frac{x - 1 + 1}{x - 1} = 1 + \\frac{1}{x - 1}$ so that your original function is\n\n$\\displaystyle y = 1 + \\frac{1}{x - 1}$.\n\nNow evaluating the inverse function should be much easier.\n• Jun 23rd 2011, 06:53 AM\nIBstudent\nRe: Finding the inverse of a function\nQuote:\n\nOriginally Posted by Prove It\nAs an alternative, write $\\displaystyle \\frac{x}{x - 1} = \\frac{x - 1 + 1}{x - 1} = 1 + \\frac{1}{x - 1}$ so that your original function is\n\n$\\displaystyle y = 1 + \\frac{1}{x - 1}$.\n\nNow evaluating the inverse function should be much easier.\n\nhow? why is (x+1-1)/x-1 = 1+ 1/x-1?\n• Jun 23rd 2011, 07:19 AM\nProve It\nRe: Finding the inverse of a function\nQuote:\n\nOriginally Posted by IBstudent\nhow? why is (x+1-1)/x-1 = 1+ 1/x-1?\n\nMy method is perfectly clear if you understand that $\\displaystyle \\frac{a + b}{c} = \\frac{a}{c} + \\frac{b}{c}$, but anyway, $\\displaystyle \\frac{x - 1 + 1}{x - 1} = \\frac{x - 1}{x - 1} + \\frac{1}{x - 1} = 1 + \\frac{1}{x - 1}$\n• Jun 23rd 2011, 07:51 AM\nIBstudent\nRe: Finding the inverse of a function\nahh I see\nThanks :-)\n• Jun 23rd 2011, 08:38 AM\ntopsquark\nRe: Finding the inverse of a function\nQuote:\n\nOriginally Posted by IBstudent\nhow? why is (x+1-1)/x-1 = 1+ 1/x-1?\n\n1. Another way to find the inverse function is to switch the roles of x and y. For example:\n$y = \\frac{x}{x - 1} \\to x = \\frac{y}{y - 1}$\nNow solve for y and you get your inverse function.\n\n2. Please be more careful in how you write you problem. For example:\n$y = x/x - 1$\nis really\n$y = (x/x) - 1 = 1 - 1 = 0$\n\nWhat you meant to write is y = x/(x - 1). Please use parentheses in the future.\n\n-Dan\n• Jun 23rd 2011, 09:09 PM\nIBstudent\nRe: Finding the inverse of a function\nokay so 1/(x-1) = y-1\n(y-1)(x-1) = 1\nyx-y-x+1 = 1\ny(x-1)=x\nahhhh I'm confused...:S\nis it x/(x-1) or x^2-x that is the inverse?\n• Jun 23rd 2011, 09:55 PM\nmr fantastic\nRe: Finding the inverse of a function\nQuote:\n\nOriginally Posted by IBstudent\nokay so 1/(x-1) = y-1\n(y-1)(x-1) = 1\nyx-y-x+1 = 1\ny(x-1)=x\nahhhh I'm confused...:S\nis it x/(x-1) or x^2-x that is the inverse?\n\nIt is basic algebra.\n\n1/(x-1) = y-1\n\n=> x - 1 = 1/(y - 1) => x = .....\n• Jun 23rd 2011, 10:53 PM\nIBstudent\nRe: Finding the inverse of a function\nI know that, so, the inverse of x/(x-1) is x/(x-1) ?!\nthats weird\n• Jun 24th 2011, 12:43 AM\nAckbeet\nRe: Finding the inverse of a function\nQuote:\n\nOriginally Posted by IBstudent\nI know that, so, the inverse of x/(x-1) is x/(x-1) ?!\nthats weird\n\nAny function that is symmetric with respect to a reflection about the line y = x will behave that way. Other examples: y = x and y = 1/x. I'm sure there are others, but they're not coming to mind.\n• Jun 24th 2011, 03:36 AM\nTKHunny\nRe: Finding the inverse of a function\nQuote:\n\nOriginally Posted by IBstudent\nI know that, so, the inverse of x/(x-1) is x/(x-1) ?!\nthats weird\n\nNot quite and not really.\n\nIf $f(x) = \\frac{x}{x-1}$ Note, first that $x \\ne 1$\n\nThen, $f^{-1}(x) = \\frac{x}{x-1}$, where again, $x \\ne 1$\n\nNotation is important. Don't just skip it.\n• Jun 24th 2011, 04:46 AM\nmr fantastic\nRe: Finding the inverse of a function\nQuote:\n\nOriginally Posted by Ackbeet\nAny function that is symmetric with respect to a reflection about the line y = x will behave that way. Other examples: y = x and y = 1/x. I'm sure there are others, but they're not coming to mind.\n\nTo expand on this, functions that are their own inverses are called involutes. It is simple to show that the following functions are involutes:\n\nf(x) = x.\nf(x) = -x + b, where b is a real number.\n\n$f(x) = \\frac{ax + b}{cx - a}$ where a, b and c are real numbers. f(x) = 1/x is included as a special case.\n\nThe question I'll pose for the interested reader is: Are the above functions the only involutes?<|endoftext|>"},"score":{"kind":"number","value":4.375,"string":"4.375"}}},{"rowIdx":1278,"cells":{"token_count":{"kind":"number","value":646,"string":"646"},"text":{"kind":"string","value":"Whiptail lizards live throughout the western United States, with numerous species concentrated in the Sonoran Desert and some, such as the western whiptail, ranging as far north as Idaho and Oregon. These lizards prefer dry climates and seek out habitats with sparse vegetation, such as desert grass, pine, sagebrush, scrub and oak. Depending on the species, they reproduce asexually, sexually or even both.\nMost whiptail lizards' bodies measure from 3 to 5 inches, not counting the tail. Depending on the species, they can be brown, tan, red, black, olive or gray, and their backs usually feature lighter spots or stripes. Their diet includes insects such as beetles, crickets, grasshoppers and termites, although they also sometimes eat spiders and scorpions. Whiptail lizards are most active on summer mornings and late afternoons, and their reproductive cycle usually takes place between April and August.\nSeveral species, such as the Sonoran spotted and desert grassland whiptails, consist of females only. They reproduce through an asexual process called parthenogenesis, which begins with a simulated mating ritual in which the females bite and mount each other. Scientists believe this behavior causes them to produce and lay unfertilized eggs. The young that hatch are genetic clones of their mothers. Females of a few parthenogenetic species, such as checkered and New Mexico whiptails, occasionally breed with males of other whiptail species.\nMost whiptails are not parthenogenetic; instead, reproduction occurs after males and females mate. As in the female-only courting ritual, the males mount and bite their partners. They use their hemipenes—a pair of reproductive organs at the tail’s base—to fertilize the eggs. After mating, females can hold on to the sperm for future use.\nWhether they reproduce sexually or asexually, female whiptail lizards bury their egg clutches. In the warm desert regions of southern California, Arizona and New Mexico, they can lay two or three clutches during a summer. In the northern parts of their range, such as Colorado and Idaho, they lay only one. Clutches average about three eggs, although depending on the species, the female could lay four, five, six or more at a time. Juveniles usually hatch during July or August, after two to two and a half months of incubation. The lizards become sexually mature when they’re 1 or 2 years old.\n- Arizona-Sonora Desert Museum: Whiptails (Cnemidophorus spp.)\n- University of Michigan Museum of Zoology: Animal Diversity Web: Cnemidophorus Sonorae\n- U.S. Fish and Wildlife Service: Bosque del Apache NWR: Lizards of Bosque del Apache\n- Idaho State University: Cnemidophorus tigris\n- U.S. Department of the Interior: Bureau of Land Management: California: Western Whiptail\n- Exotic Pet Vet: Reptiles: Reproduction “From Egg to Adult”<|endoftext|>"},"score":{"kind":"number","value":4.03125,"string":"4.03125"}}},{"rowIdx":1279,"cells":{"token_count":{"kind":"number","value":140,"string":"140"},"text":{"kind":"string","value":"Each of the questions or incomplete statements below is followed by five suggested answers or completions. Select the one that is best in each case.\nAll of the following statements about carbon dioxide are true EXCEPT:\nSelect an Answer\nIt can be prepared by the action of acid on limestone.\nIt is used to extinguish fires.\nIt dissolves in water at room temperature.\nIt sublimes rather than melts at and atmosphere pressure.\nIt is less dense than air at a given temperature and pressure.\nAir contains mostly and . The molar mass of carbon dioxide is , so carbon dioxide is more dense than air. The other statements are true and are not correct choices.<|endoftext|>"},"score":{"kind":"number","value":3.71875,"string":"3.71875"}}},{"rowIdx":1280,"cells":{"token_count":{"kind":"number","value":1167,"string":"1,167"},"text":{"kind":"string","value":"HOME Organize UNIT #1 UNIT #2 UNIT #3 UNIT #4 UNIT #5 UNIT #6 Resources Interaction\n\nTeacher Notes\n\nCONCEPT 1Prove the Laws of Sines.\n\nAs mentioned in the previous objective up until this time we have been working with right triangles but now we are starting to expand the use of trigonometry to all triangles, even oblique triangles.  In this objective we will show that if any three of the six measures of triangle are given (provided at least one measure is a side), then the other three measures can be found.\n\nOne of the considerations when doing this, is if the three pieces of information force congruence or not.  So for example if we are given SAS, ASA, AAS/SSA, SSS and some cases of AS1S2 we know that the information is enough to guarantee congruence in the triangle.\n\nThe first new relationship in this objective that we are going to derive is called the Law of Sines.  We will follow similar logic to how we handled our new area formula.\n\n The Law of Sines is written in more than one form... sometimes the side lengths are in the numerator position and the sine vallues is in the denominator. It seems to be simply a preference because obviously it makes no difference.\n\nThe Law of Sines is a proportion that compares the sine of the angle to the length of its opposite side.  Solving this type of problem can be quite easy but it does only work in certain situations because an angle and its corresponding opposite side need to be known to form the ratio thus we need ∠A & a or ∠B & b or ∠C or c.\n\nThe Law of Sines will work for the following situations:  ASA, AAS and AS1S2.  The first two cases are congruence cases so when we solve for the missing information we are guaranteed to all get the same answers.\n\nLet’s try a couple examples of the congruence cases of ASA and AAS.  Solve for all angles & sides in the Δ.\n\nNotice that because the Law of Sines is a proportional relationship that:\n\n Largest Angle is always opposite the Longest Side (& vice versa) Smallest Angle is always opposite the Shortest Side (& vice versa)\n\n Good Luck... Here comes some tricky stuff... Back in GCO.8 I spent time discussing the AS1S2 so that when I get here it isn't totally a new concept.\n\nThe other case that works for the Law of Sines is AS1S2 but if you remember in G.CO.8 we discussed some of the issues that are present in that case – not all cases of AS1S2 are congruence relationships.  Because of this we have to again discuss the case by case situations of AS1S2 and relate them to the Law of Sines.\n\nCASE #1 – AS1S2, when S2 is greater than S1.\n\nThis forms a triangle congruence relationship.  There is only one way for S2 to be placed to complete the triangle.  Because this is a congruence relationship a single answer will be present when solving – let us do an example.\n\nSolve for all angles & sides in the Δ, given that m∠A = 47°, a = 20 cm and c = 12 cm.\n\nCASE #2 – AS1S2, when S2 is less than S1. (Too Short)\n\nThis is a weird case but when establishing all things that could happen we need to include this.  It is possible that S2 is not long enough to close the triangle.  This of course is not a congruence relationship, it doesn’t even form a triangle.  So how will you know if it is this situation… the numbers will tell you .\n\nSolve for all angles & sides in the Δ, given that m∠A = 73°, a = 8 cm and c = 14 cm.\n\nCASE #3 – AS1S2, when S2 is less than S1. (1 Intersection)\n\nThis case is typically known as HL, which stands for Hypotenuse – Leg.  It gets this special name because it is the right triangle that locks this shape.  A way to understand why this forms a congruence relationship is because in a right triangle if you know two sides, you can use the Pythagorean Theorem to calculate the third side.  HL forms a triangle congruence relationship.  No example given here because when you solve for the angle it will reveal itself to be 90° and you will then be in a right triangle.\n\nNow the final case – Known as the AMBIGUOUS CASE.  Ambiguous basically means…to be uncertain or more than one correct value – it gets this name because there are two possible triangles and thus there are two sets of answers possible.  If you encounter this special case, you are to completely solve for BOTH triangles.\n\nCASE #4 – AS1S2, when S2 is less than S1. (2 Intersections)\n\n Well here it is.... the Ambiguous case... the art of success here is helping students to recognize when it exists. Most students become very skilled with the Law of Sines but recognizing this case is the difficult part. If they know the other cases of AS1S2, when S2 is less than S1 definately help to narrow down the possible things that can happen.\n\nThis is known as the AMBIGUOUS CASE because two different triangles can be formed by this information.  Because S2 is shorter it can swing to form two possible locations.  This does NOT form a triangle congruence relationship.<|endoftext|>"},"score":{"kind":"number","value":4.71875,"string":"4.71875"}}},{"rowIdx":1281,"cells":{"token_count":{"kind":"number","value":6918,"string":"6,918"},"text":{"kind":"string","value":"Skip to main content\n\n# Modeling, Functions, and Graphs\n\n## Section3.6Chapter Summary and Review\n\n### SubsectionKey Concepts\n\n1. #### Direct and Inverse Variation.\n\n• $$y$$ varies directly with $$x$$ if the ratio $$\\dfrac{y}{x}$$ is constant, that is, if $$y = kx.$$\n• $$y$$ varies directly with a power of $$x$$ if the ratio $$\\dfrac{y}{x^n}$$ is constant, that is, if $$y = kx^n$$ .\n• $$y$$ varies inversely with $$x$$ if the product $$xy$$ is constant, that is, if $$y =\\dfrac{k}{x}$$ .\n• $$y$$ varies inversely with a power of $$x$$ if the product $$x^ny$$ is constant, that is, if $$y =\\dfrac{k}{x^n}$$ .\n2. The graph of a direct variation passes through the origin. The graph of an inverse variation has a vertical asymptote at the origin.\n3. If $$y = kx^n\\text{,}$$ we say that $$y$$ scales as $$x^n\\text{.}$$\n4. $$n$$th roots: $$s$$ is called an $$n$$th root of $$b$$ if $$s^n = b\\text{.}$$\n5. #### Exponential Notation.\n\nThe absolute value has the following properties:\n\\begin{alignat*}{2} \\amp a^{-n}=\\frac{1}{x^n} \\amp\\hphantom{00000} \\amp a\\ne 0\\\\ \\amp a^{0}=1 \\amp \\amp a\\ne 0\\\\ \\amp a^{1/n}=\\sqrt[n]{a} \\amp \\amp n \\text{ an integer, } n\\gt 2\\\\ \\amp a^{m/n}=(a^{1/n})^m=(a^m)^{1/n}, \\amp \\amp a\\gt 0, ~n\\ne 0 \\end{alignat*}\n6. In particular, a negative exponent denotes a reciprocal, and a fractional exponent denotes a root.\n7. $$\\displaystyle a^{m/n} = \\sqrt[n]{a^m} =\\left(\\sqrt[n]{a}\\right)^m$$\n8. To compute $$a^{m/n}\\text{,}$$ we can compute the $$n$$th root first, or the $$m$$th power, whichever is easier.\n9. We cannot write down an exact decimal equivalent for an irrational number, but we can approximate an irrational number to as many decimal places as we like.\n10. The laws of exponents are valid for all exponents $$m$$ and $$n\\text{,}$$ and for $$b\\ne 0\\text{.}$$\n\n#### Laws of Exponents.\n\n1. $$\\displaystyle a^m\\cdot a^n=a^{m+n}$$\n2. $$\\displaystyle \\dfrac{a^m}{a^n}=a^{m-n}$$\n3. $$\\displaystyle \\left(a^m\\right)^n=a^{mn}$$\n4. $$\\displaystyle \\left(ab\\right)^n=a^n b^n$$\n5. $$\\displaystyle \\left(\\dfrac{a}{b}\\right)^n=\\dfrac{a^n}{b^n}$$\n11. A function of the form $$f (x) = kx^p\\text{,}$$ where $$k$$ and $$p$$ are constants, is called a power function.\n12. An allometric equation is a power function of the form $$\\text{variable} = k(\\text{mass})^p\\text{.}$$\n13. We can solve the equation $$x^n = b$$ by raising both sides to the $$\\dfrac{1}{n}$$ power\n14. We can solve the equation $$x^{1/n} = b$$ by raising both sides to the $$n$$th power.\n15. To solve the equation $$x^{m/n} = k\\text{,}$$ we raise both sides to the power $$n/m\\text{.}$$\n16. The graphs of power functions $$y = x^{m/n}\\text{,}$$ where $$m/n$$ is positive are all increasing for $$x\\ge 0\\text{.}$$ If $$m/n\\gt 1\\text{,}$$ the graph is concave up. If $$0\\lt m/n \\lt 1\\text{,}$$ the graph is concave down.\n17. The notation $$z = f (x, y)$$ indicates that $$z$$ is a function of two variables, $$x$$ and $$y\\text{.}$$\n18. We can use a table with rows and columns to display the output values for a function of two variables.\n19. #### Joint Variation.\n\n• We say that $$z$$ varies jointly with $$x$$ and $$y$$ if\n\\begin{equation*} z = kxy,~~~ k\\ne 0 \\end{equation*}\n• We say that $$z$$ varies directly with $$x$$ and inversely with $$y$$ if\n\\begin{equation*} z = k\\dfrac{x}{y},~~~ k\\ne 0, ~~ y\\ne 0 \\end{equation*}\n20. We can represent a function of two variables graphically by showing a set of graphs for several fixed values of one of the variables.\n21. #### Roots of Real Numbers.\n\n• Every positive number has two real-valued roots, one positive and one negative, if the index is even.\n• A negative number has no real-valued root if the index is even.\n• Every real number, positive, negative, or zero, has exactly one real-valued root if the index is odd.\n.\n\n### ExercisesChapter 3 Review Problems\n\n#### 1.\n\nThe distance s a pebble falls through a thick liquid varies directly with the square of the length of time $$t$$ it falls.\n1. If the pebble falls $$28$$ centimeters in $$4$$ seconds, express the distance it will fall as a function of time.\n2. Find the distance the pebble will fall in $$6$$ seconds.\n\n#### 2.\n\nThe volume, $$V\\text{,}$$ of a gas varies directly with the temperature, $$T\\text{,}$$ and inversely with the pressure, $$P\\text{,}$$ of the gas.\n1. If $$V = 40$$ when $$T = 300$$ and $$P = 30\\text{,}$$ express the volume of the gas as a function of the temperature and pressure of the gas.\n2. Find the volume when $$T = 320$$ and $$P = 40\\text{.}$$\n\n#### 3.\n\nThe demand for bottled water is inversely proportional to the price per bottle. If Droplets can sell $$600$$ bottles at $$$8$$ each, how many bottles can the company sell at$$$10$$ each?\n\n#### 4.\n\nThe intensity of illumination from a light source varies inversely with the square of the distance from the source. If a reading lamp has an intensity of $$100$$ lumens at a distance of $$3$$ feet, what is its intensity $$8$$ feet away?\n\n#### 5.\n\nA person’s weight, $$w\\text{,}$$ varies inversely with the square of his or her distance, $$r\\text{,}$$ from the center of the Earth.\n1. Express $$w$$ as a function of $$r\\text{.}$$ Let $$k$$ stand for the constant of variation.\n2. Make a rough graph of your function.\n3. How far from the center of the Earth must Neil be in order to weigh one-third of his weight on the surface? The radius of the Earth is about $$3960$$ miles.\n\n#### 6.\n\nThe period, $$T\\text{,}$$ of a pendulum varies directly with the square root of its length, $$L\\text{.}$$\n1. Express $$T$$ as a function of $$L\\text{.}$$ Let $$k$$ stand for the constant of variation.\n2. Make a rough graph of your function.\n3. If a certain pendulum is replaced by a new one four-fifths as long as the old one, what happens to the period?\n\n#### Exercise Group.\n\nIn Problems 7–10, $$y$$ varies directly or inversely with a power of $$x\\text{.}$$ Find the power of $$x$$ and the constant of variation, $$k\\text{.}$$ Write a formula for each function of the form $$y = kx^n$$ or $$y = \\dfrac{k}{x^n}\\text{.}$$\n##### 7.\n $$x$$ $$y$$ $$2$$ $$4.8$$ $$5$$ $$30.0$$ $$8$$ $$76.8$$ $$11$$ $$145.2$$\n##### 8.\n $$x$$ $$y$$ $$1.4$$ $$75.6$$ $$2.3$$ $$124.2$$ $$5.9$$ $$318.6$$ $$8.3$$ $$448.2$$\n##### 9.\n $$x$$ $$y$$ $$0.5$$ $$40.0$$ $$2.0$$ $$10.0$$ $$4.0$$ $$5.0$$ $$8.0$$ $$2.5$$\n##### 10.\n $$x$$ $$y$$ $$1.5$$ $$320.0$$ $$2.5$$ $$115.2$$ $$4.0$$ $$45.0$$ $$6.0$$ $$20.0$$\n\n#### Exercise Group.\n\nFor Problems 11–16, write without negative exponents and simplify.\n##### 11.\n1. $$\\displaystyle (-3)^{-4}$$\n2. $$\\displaystyle 4^{-3}$$\n##### 12.\n1. $$\\displaystyle \\left(\\dfrac{1}{3}\\right)^{-2}$$\n2. $$\\displaystyle \\dfrac{3}{5^{-2}}$$\n##### 13.\n1. $$\\displaystyle (3m)^{-5}$$\n2. $$\\displaystyle -7y^{-8}$$\n##### 14.\n1. $$\\displaystyle a^{-1}+ a^{-2}$$\n2. $$\\displaystyle \\dfrac{3q^{-9}}{r^{-2}}$$\n##### 15.\n1. $$\\displaystyle 6c^{-7}\\cdot (3)^{-1} c^4$$\n2. $$\\displaystyle \\dfrac{11z^{-7}}{3^{-2} z^{-5}}$$\n##### 16.\n1. $$\\displaystyle \\left(2d^{-2}k^3 \\right)^{-4}$$\n2. $$\\displaystyle \\dfrac{2w^3(w^{-2})^{-3}}{5w^{-5}}$$\n\n#### Exercise Group.\n\nFor Problems 17–20, write each power in radical form.\n##### 17.\n1. $$\\displaystyle 25m^{1/2}$$\n2. $$\\displaystyle 8n^{-1/3}$$\n##### 18.\n1. $$\\displaystyle (13d)^{2/3}$$\n2. $$\\displaystyle 6x^{2/5}y^{3/5}$$\n##### 19.\n1. $$\\displaystyle (3q)^{-3/4}$$\n2. $$\\displaystyle 7(uv)^{3/2}$$\n##### 20.\n1. $$\\displaystyle (a^2+b^2)^{0.5}$$\n2. $$\\displaystyle (16-x^2)^{0.25}$$\n\n#### Exercise Group.\n\nFor Problems 21–24, write each radical as a power with a fractional exponent.\n##### 21.\n1. $$\\displaystyle 2\\sqrt[3]{x^2}$$\n2. $$\\displaystyle \\dfrac{1}{4}\\sqrt[4]{x}$$\n##### 22.\n1. $$\\displaystyle z^2\\sqrt{z}$$\n2. $$\\displaystyle z\\sqrt[3]{z}$$\n##### 23.\n1. $$\\displaystyle \\dfrac{6}{\\sqrt[4]{b^3}}$$\n2. $$\\displaystyle \\dfrac{-1}{3\\sqrt[3]{b}}$$\n##### 24.\n1. $$\\displaystyle \\dfrac{-4}{(\\sqrt[4]{a})^2}$$\n2. $$\\displaystyle \\dfrac{2}{(\\sqrt{a})^3}$$\n\n#### Exercise Group.\n\nFor Problems 25–28, sketch graphs by hand for each function on the domain $$(0,\\infty)\\text{.}$$\n##### 25.\n$$y$$ varies directly with $$x^2\\text{.}$$ The constant of variation is $$k = 0.25\\text{.}$$\n##### 26.\n$$y$$ varies directly with $$x\\text{.}$$ The constant of variation is $$k = 1.5\\text{.}$$\n##### 27.\n$$y$$ varies inversely with $$x\\text{.}$$ The constant of variation is $$k = 2\\text{.}$$\n##### 28.\n$$y$$ varies inversely with $$x^2\\text{.}$$ The constant of variation is $$k = 4\\text{.}$$\n\n#### Exercise Group.\n\nFor Problems 29–30, write each function in the form $$y = kx^p\\text{.}$$\n##### 29.\n$$f(x)=\\dfrac{2}{3x^4}$$\n##### 30.\n$$g(x)=\\dfrac{8x^7}{29}$$\n\n#### Exercise Group.\n\nFor Problems 31–34,\n1. Evaluate each function for the given values.\n2. Graph the function.\n##### 31.\n$$Q(x)=4x^{5/2}$$\n $$x$$ $$16$$ $$\\dfrac{1}{4}$$ $$3$$ $$100$$ $$Q(x)$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$\n##### 32.\n$$T(w)=-3w^{2/3}$$\n $$w$$ $$27$$ $$\\dfrac{1}{8}$$ $$20$$ $$1000$$ $$T(w)$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$\n##### 33.\n$$f(x)=x^{0.3}$$\n $$x$$ $$0$$ $$1$$ $$5$$ $$10$$ $$20$$ $$50$$ $$70$$ $$100$$ $$f(x)$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$\n##### 34.\n$$g(x)=-x^{-0.7}$$\n $$x$$ $$0.1$$ $$0.2$$ $$0.5$$ $$1$$ $$2$$ $$5$$ $$8$$ $$10$$ $$g(x)$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$ $$\\hphantom{000}$$\n\n#### 35.\n\nAccording to the theory of relativity, the mass of an object traveling at velocity $$v$$ is given by the function\n\\begin{equation*} m=\\dfrac{M}{\\sqrt{1-\\dfrac{v^2}{c^2}}} \\end{equation*}\nwhere $$M$$ is the mass of the object at rest and $$c$$ is the speed of light. Find the mass of a man traveling at a velocity of $$0.7c$$ if his rest mass is $$80$$ kilograms.\n\n#### 36.\n\nThe cylinder of smallest surface area for a given volume has a radius and height both equal to $$\\sqrt[3]{\\dfrac{V}{\\pi}} \\text{.}$$ Find the dimensions of the tin can of smallest surface area with volume $$60$$ cubic inches.\n\n#### 37.\n\nMembership in the Wildlife Society has grown according to the function\n\\begin{equation*} M(t) = 30t^{3/4} \\end{equation*}\nwhere $$t$$ is the number of years since its founding in $$1970\\text{.}$$\n1. Sketch a graph of the function $$M(t)\\text{.}$$\n2. What was the society’s membership in $$1990\\text{?}$$\n3. In what year will the membership be $$810$$ people?\n\n#### 38.\n\nThe heron population in Saltmarsh Refuge is estimated by conservationists at\n\\begin{equation*} P(t) = 360t^{-2/3} \\end{equation*}\nwhere $$t$$ is the number of years since the refuge was established in $$1990\\text{.}$$\n1. Sketch a graph of the function $$P(t)\\text{.}$$\n2. How many heron were there in $$1995\\text{?}$$\n3. In what year will there be only $$40$$ heron left?\n\n#### 39.\n\nManufacturers of ships (and other complex products) find that the average cost of producing a ship decreases as more of those ships are produced. This relationship is called the experience curve, given by the equation\n\\begin{equation*} C = ax^{-b} \\end{equation*}\nwhere $$C$$ is the average cost per ship in millions of dollars and $$x$$ is the number of ships produced. The value of the constant $$b$$ depends on the complexity of the ship. (Source: Storch, Hammon, and Bunch, 1988)\n1. What is the significance of the constant of proportionality $$a\\text{?}$$\n2. For one kind of ship, $$b = \\dfrac{1}{8}\\text{,}$$ and the cost of producing the first ship is $$$12$$ million. Write the equation for $$C$$ as a function of $$x$$ using radical notation. 3. Compute the cost per ship when $$2$$ ships have been built. By what percent does the cost per ship decrease? By what percent does the cost per ship decrease from building $$2$$ ships to building $$4$$ ships? 4. By what percent does the average cost decrease from building $$n$$ ships to building $$2n$$ ships? (In the shipbuilding industry, the average cost per ship usually decreases by $$5$$ to $$10\\%$$ each time the number of ships doubles.) Hint. What is the value of $$C$$ if only one ship is built? #### 40. A population is in a period of supergrowth if its rate of growth, $$R\\text{,}$$ at any time is proportional to $$P^k\\text{,}$$ where $$P$$ is the population at that time and $$k$$ is a constant greater than $$1\\text{.}$$ Suppose $$R$$ is given by \\begin{equation*} R = 0.015 P^{1.2} \\end{equation*} where $$P$$ is measured in thousands and $$R$$ is measured in thousands per year. 1. Find $$R$$ when $$P = 20\\text{,}$$ when $$P = 40\\text{,}$$ and when $$P = 60\\text{.}$$ 2. What will the population be when its rate of growth is $$5000$$ per year? 3. Graph $$R$$ and use your graph to verify your answers to parts (a) and (b). #### Exercise Group. For Problems 41–50, solve ##### 41. $$6t^{-3} = \\dfrac{3}{500}$$ ##### 42. $$3.5 - 2.4p^{-2} = -6.1$$ ##### 43. $$\\sqrt[3]{x+1} = 2$$ ##### 44. $$x^{2/3}+2 = 6$$ ##### 45. $$(x-1)^{-3/2} = \\dfrac{1}{8}$$ ##### 46. $$(2x+1)^{-1/2} =\\dfrac{1}{3}$$ ##### 47. $$8\\sqrt[4]{x+6} =24$$ ##### 48. $$9.8 =7\\sqrt[3]{z-4}$$ ##### 49. $$\\dfrac{2}{3} (2y+1)^{0.2} = 6$$ ##### 50. $$1.3w^{0.3}+4.7 =5.2$$ #### Exercise Group. For Problems 51–54, solve each formula for the indicated variable. ##### 51. $$t=\\sqrt{\\dfrac{2v}{g}}~~~ \\text{,}$$ for $$g$$ ##### 52. $$q-1=2\\sqrt{\\dfrac{r^2-1}{3}}~~~ \\text{,}$$ for $$r$$ ##### 53. $$R=\\dfrac{1+\\sqrt{p^2+1}}{2}~~~ \\text{,}$$ for $$p$$ ##### 54. $$q=\\sqrt[3]{\\dfrac{1+r^2}{2}}~~~ \\text{,}$$ for $$r$$ #### Exercise Group. For Problems 55–60, simplify by applying the laws of exponents. ##### 55. $$(7t)^3 (7t)^{-1}$$ ##### 56. $$\\dfrac{36r^{-2}s}{9r^{-3}s^4}$$ ##### 57. $$\\dfrac{(2k^{-1})^{-4}}{4k^{-3}}$$ ##### 58. $$(2w^{-3})(2w^{-3})^{5}(-5w^2)$$ ##### 59. $$\\dfrac{8a^{-3/4}}{a^{-11/4}}$$ ##### 60. $$b^{2/3}(4b^{-2/3}-b^{1/3})$$ #### 61. When the Concorde landed at Heathrow Airport in London, the width, $$w\\text{,}$$ of the sonic boom felt on the ground is given in kilometers by the following formula: \\begin{equation*} w=4\\left(\\frac{Th}{m} \\right)^{1/2} \\end{equation*} where $$T$$ stands for the temperature on the ground in kelvins, $$h$$ is the altitude of the Concorde when it breaks the sound barrier, and $$m$$ is the drop in temperature for each gain in altitude of one kilometer. 1. Find the width of the sonic boom if the ground temperature was $$293$$ K, the altitude of the Concorde was $$15$$ kilometers, and the temperature drop was $$4$$ K per kilometer of altitude. 2. Graph $$w$$ as a function of $$h$$ if $$T = 293$$ and $$m = 4\\text{.}$$ #### 62. The manager of an office supply store must decide how many of each item in stock she should order. The Wilson lot size formula gives the most cost-efficient quantity, $$Q\\text{,}$$ as a function of the cost, $$C\\text{,}$$ of placing an order, the number of items, $$N\\text{,}$$ sold per week, and the weekly inventory cost, $$I\\text{,}$$ per item (cost of storage, maintenance, and so on). \\begin{equation*} Q=\\left(\\frac{2CN}{I} \\right)^{1/2} \\end{equation*} 1. How many reams of computer paper should she order if she sells on average $$80$$ reams per week, the weekly inventory cost for a ream is$$$0.20\\text{,}$$ and the cost of ordering, including delivery charges, is $$$25\\text{?}$$ 2. Graph $$Q$$ as a function of $$N$$ if $$C = 25$$ and $$I = 0.2\\text{.}$$ #### 63. Two businesswomen start a small company to produce saddle bags for bicycles. The number of saddle bags, $$q\\text{,}$$ they can produce depends on the amount of money, $$m\\text{,}$$ they invest and the number of hours of labor, $$w\\text{,}$$ they employ, according to the Cobb-Douglas formula \\begin{equation*} q= 0.6m^{1/4}w^{3/4} \\end{equation*} where $$m$$ is measured in thousands of dollars. 1. If the businesswomen invest$$$100,000$$ and employ $$1600$$ hours of labor in their first month of production, how many saddle bags can they expect to produce?\n2. With the same initial investment, how many hours of labor would they need in order to produce $$200$$ saddle bags?\n\n#### 64.\n\nA child who weighs $$w$$ pounds and is $$h$$ inches tall has a surface area (in square inches) given approximately by\n\\begin{equation*} S= 8.5 h^{0.35}w^{0.55} \\end{equation*}\n1. What is the surface area of a child who weighs $$60$$ pounds and is $$40$$ inches tall?\n2. What is the weight of a child who is $$50$$ inches tall and whose surface area is $$397$$ square inches?\n\n#### 65.\n\nThe cost, $$C\\text{,}$$ of insulating the ceiling in a building depends on the thickness of the insulation and the area of the ceiling. The table shows values of $$C = f (t, A)\\text{,}$$ where $$t$$ is the thickness of the insulation and $$A$$ is the area of the ceiling.\n Cost of Insulation (dollars) Area (sq m) Thickness(cm) $$100$$ $$200$$ $$300$$ $$400$$ $$500$$ $$600$$ $$4$$ $$72$$ $$144$$ $$216$$ $$288$$ $$300$$ $$432$$ $$5$$ $$90$$ $$180$$ $$270$$ $$360$$ $$450$$ $$540$$ $$6$$ $$108$$ $$216$$ $$324$$ $$432$$ $$540$$ $$648$$ $$7$$ $$126$$ $$252$$ $$378$$ $$504$$ $$630$$ $$756$$ $$8$$ $$144$$ $$288$$ $$432$$ $$576$$ $$720$$ $$864$$ $$9$$ $$162$$ $$324$$ $$486$$ $$648$$ $$810$$ $$972$$\n1. What does it cost to insulate a ceiling with an area of $$500$$ square meters with $$5$$ cm of insulation? Write your answer in function notation.\n2. Solve the equation $$864 = f (t, 600)$$ and interpret your answer.\n3. Consider the row corresponding to a thickness of $$4$$ cm. How does the cost of insulating the ceiling depend on the area of the ceiling?\n4. Consider the column corresponding to an area of $$100$$ square meters. How does the cost depend on the thickness of the insulation?\n5. Given that the cost varies jointly with the thickness of the insulation and the area of the ceiling, write an equation for cost as a function of area and thickness of insulation.\n6. Use your formula from part (e) to determine the cost of insulating a building with $$10$$ centimeters of insulation if the area of the ceiling is $$800$$ square meters.\n\n#### 66.\n\nThe volume, $$V\\text{,}$$ of a quantity of helium depends on both the temperature and the pressure of the gas. The table shows values of $$V = f (P, T )$$ for temperature in kelvins and pressure in atmospheres.\n Volume (cubic meters) Temperature (K) Pressure(atmospheres) $$100$$ $$150$$ $$200$$ $$250$$ $$300$$ $$350$$ $$1$$ $$18$$ $$27$$ $$36$$ $$45$$ $$54$$ $$63$$ $$2$$ $$9$$ $$13.5$$ $$18$$ $$22.5$$ $$27$$ $$31.5$$ $$3$$ $$6$$ $$9$$ $$12$$ $$15$$ $$18$$ $$21$$ $$4$$ $$4.5$$ $$6.75$$ $$9$$ $$11.25$$ $$13.5$$ $$15.75$$\n1. What is the volume of helium when the pressure is $$4$$ atmospheres and the temperature is $$350$$ K? Write your answer in function notation.\n2. Solve the equation $$15 = f (3, T )$$ and interpret your answer.\n3. Consider the row corresponding to $$2$$ atmospheres. How is the volume related to the absolute temperature?\n4. Consider the column corresponding to $$300$$ K. How is the volume related to the pressure?\n5. Given that the volume of the gas varies directly with temperature and inversely with pressure, write an equation for volume as a function of temperature and pressure.\n6. Use your formula from part (e) to determine the volume of the helium at $$50$$ K and pressure of $$0.4$$ atmospheres.\n\n#### 67.\n\nIn his hiking guidebook, Afoot and Afield in Los Angeles County, Jerry Schad notes that the number of people on a wilderness trail is inversely proportional to \"the square of the distance and the cube of the elevation gain from the nearest road.\"\n1. Choose variables and write a formula for this relationship.\n2. On a sunny Saturday afternoon, you count $$42$$ people enjoying the Rock Pool at Malibu Creek State Park. The Rock Pool is $$1.5$$ miles from the main parking lot, and the trail includes an elevation gain of $$250$$ feet. Calculate the constant of variation in your formula from part (a).\nHint: Convert the elevation gain to miles.\n3. Lookout Trail leads $$1.9$$ miles from the parking lot and involves an elevation gain of $$500$$ feet. How many people would you expect to encounter at the end of the trail?\n\n#### 68.\n\nA company’s monthly production, $$P\\text{,}$$ depends on the capital, $$C\\text{,}$$ the company has invested and the amount of labor, $$L\\text{,}$$ available each month. The Cobb-Douglas model for production assumes that $$P$$ varies jointly with $$C^a$$ and $$L^b\\text{,}$$ where $$a$$ and $$b$$ are positive constants less than $$1\\text{.}$$ The Aztech Chip Company invested $$625$$ units of capital and hired $$256$$ workers and produces $$8000$$ computer chips each month.\n1. Suppose that $$a = 0.25\\text{,}$$ $$b = 0.75\\text{.}$$ Find the constant of variation and a formula giving $$P$$ in terms of $$C$$ and $$L\\text{.}$$\n2. If Aztech increases its labor force to $$300$$ workers, what production level can they expect?\n3. If Aztech maintains its labor force at $$256$$ workers, what amount of capital outlay would be required for monthly production to reach $$16,000$$ computer chips?<|endoftext|>"},"score":{"kind":"number","value":4.6875,"string":"4.6875"}}},{"rowIdx":1282,"cells":{"token_count":{"kind":"number","value":703,"string":"703"},"text":{"kind":"string","value":"## Revision Notes of Ch 1 Real Numbers Class 10th Math\n\nTopics in the Chapter\n\n• Euclid’s Division Lemma\n• Euclid’s Division Algorithm\n• The Fundamental Theorem of Arithmetic\n• Revisiting Irrational Numbers\n• Revisiting Rational Numbers and Their Decimal Expansions\n\nEuclid’s Division Lemma\n\n→ For any two positive integers a and b, there exist unique integers q and r such that\na = bq + r, 0  ≤ r < b\nHere,  a, b, q and r are called dividend, divisor, quotient and remainder respectively.\n\n→ Example: Dividing 34 by 6, we get quotient as 6 and remainder as 4.\nSo we can write it as, 34 = 6×5 + 4\nHere, a = 34, b = 6, q = 5 and r = 4\na = bq + r, 0  ≤ r < b\n\nEuclid’s Division Algorithm\n\n→ Euclid’s Division Algorithm is application of Euclid’s Division Lemma means it is technique to compute HCF of two natural numbers.\n\n→ To obtain the HCF of two positive integers, say c and d, with c > d, we follow these steps:\nStep 1: Apply Euclid’s division lemma to c and d to obtain two whole numbers q and r.\nThus, c = dq + r, 0  ≤ r < d\n\nStep 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, again apply the division lemma to d and r.\n\nStep 3 : Continue the process till the remainder is zero. The last divisor when remainder becomes zero is the HCF of c and d\n\n→ Example: Use Euclid’s division algorithm to find the HCF of 420 and 130.\nHere, 420 > 130 so we can write,\nStep 1: 420 = 130 × 3 + 30 (remainder is 30 so we will apply division lemma again on 130 and 30)\nStep 2: 130 = 30 × 4 + 10 (remainder is 10 so we will apply division lemma again on 30 and 10)\nStep 3: 30 = 10 × 3 + 0\nNow, the remainder is 0 and the last divisor is 10. So, 10 is the HCF of 420 and 130.\n\nThe Fundamental Theorem of Arithmetic\n\nEvery composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.\n\n→ Let x be composite number which can be factorised as,\nx =  p1p2 ..... pn , where  p1p2 ..... pn are primes and written in ascending order, i.e., p1 ≤ p2 ≤ ..... ≤ pn. If we combine the same primes, we will get powers of primes. This is called prime factorization.\nFor example: 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13\n\nBy expressing any two numbers as their prime factors, their highest common factor (HCF) and lowest common multiple (LCM) can be calculated.<|endoftext|>"},"score":{"kind":"number","value":5,"string":"5"}}},{"rowIdx":1283,"cells":{"token_count":{"kind":"number","value":163,"string":"163"},"text":{"kind":"string","value":"Beth is beginning to teach her class the headings on the CAFE Menu: Comprehension, Accuracy, Fluency, and Expand Vocabulary. This lesson introducing the goals is used throughout the year to review when needed.\nWatch and Listen:\n- Beth Lawson teaches her students kinesthetic motions for the meaning of each heading. Is this something you will teach your students?\n- What value do you see in teaching students the goals and the meaning of each goal?\n- What adaptations, if any, would you make to teach this lesson to your students?\n- When might you review this with your class?\nCAFE Essential Elements:\n- Teach—6.1 Introduce classroom CAFE Menu to students\n- Teach—6.2 Classroom CAFE Menu as an instructional tool<|endoftext|>"},"score":{"kind":"number","value":3.78125,"string":"3.78125"}}},{"rowIdx":1284,"cells":{"token_count":{"kind":"number","value":675,"string":"675"},"text":{"kind":"string","value":"In this quick tutorial you'll learn how to draw a Stingray in 5 easy steps - great for kids and novice artists.\nThe images above represents how your finished drawing is going to look and the steps involved.\nBelow are the individual steps - you can click on each one for a High Resolution printable PDF version.\nAt the bottom you can read some interesting facts about the Stingray.\nMake sure you also check out any of the hundreds of drawing tutorials grouped by category.\nHow to Draw a Stingray - Step-by-Step Tutorial\nStep 1: Start by drawing the stingray's body. It looks like a pentagon with the bottom lines curled up. If you're having trouble, think of baseball's home plate upside down. Leave an opening between the curled lines at the bottom for the fins.\nStep 2: Now draw the fins at the bottom to connect the two curled lines. The fins look like a lowercase \"w\" that has been stretched apart.\nStep 3: Next, draw the tail beneath the fins. Stingrays have a very long and thin tail with a sharp point at the end. They are called stingrays because some species have venom in their tails.\nStep 4: Next, draw the eyes near the top of the body. You can also draw the spiracles above the eyes, which are shaped like crescent moons. Spiracles are openings, like gills, that help stingrays breathe.\nStep 5: Last, draw the spots and color your stingray. You can draw as many spots as you want. There are over 60 species of stingrays and some are covered in spots, while some have none. There is even a species of stingray with spots that look just like a leopard's! Activity: Have students color and draw spots on their stingrays. Let them share why they chose certain colors or spots (e.g. bright spots for a tropical environment or dark plain colors to blend in).\nInteresting Facts about Stingrays\nStingrays are fish with wide flat bodies made of cartilage, which is the same material that make up your ears. Stingrays have no bones. There are more than 60 species of stingrays. They are related to sharks. Stingrays got their names because of their tail. Their tail has a spine with a very sharp point. Some species have venom in their tails. The spine and the venom are dangerous to humans, although stingrays rarely attack humans.\nDid you know?\n- Stingrays live at the bottom of the ocean, in warm climates.\n- Stingrays have two fins that run alongside their bodies. To swim, some species flap their fins the same way birds flap their wings to fly. Other species use their whole bodies to move in a wavy motion.\n- Stingrays usually eat oysters, shrimp, crabs, clams, and mussels.\n- Stingrays can grow up to 6 ? feet.\n- In the wild, stingrays live to be about 15 to 25 years old.\n- Stingrays bury themselves in the sand when they want to hide from predators.\nStingrays use their tails as a form of defense. They whip their tails around when they attack. Some people use the tails of stingrays for the tips of weapons such as daggers.<|endoftext|>"},"score":{"kind":"number","value":3.921875,"string":"3.921875"}}},{"rowIdx":1285,"cells":{"token_count":{"kind":"number","value":1030,"string":"1,030"},"text":{"kind":"string","value":"This column, which first ran in our August 21 issue, has been revised and updated by Dr. Crutchfield to run again due to the ongoing Ebola crisis.\nEbola Fever is a viral infection. The virus causes the bleeding (hemorrhage) of internal organs and of the skin. It also causes the infected person to have a fever. It is also known as Ebola Hemorrhagic Fever.\nThere is no vaccination against it. Ebola Hemorrhagic Fever can kill up to 90 percent of those infected. It is named after the Ebola River, an area where one of the first cases were reported in 1976.\nThere are several sub-types of the virus. The Zaire sub-type/strain is considered the most deadly. As of this report, since 1976, there have been approximately 6,500 cases of documented Ebola with over 4,000 fatalities.\nCurrently, countries in West Africa (including Guinea, Sierra Leone and Liberia) are experiencing one of the worst outbreaks of the Ebola virus ever documented with approximately 4,000 deaths at the time this article was written. One projection has an estimate of 50-100,000 fatalities before the epidemic is under control. The World Health Organization calls the Ebola virus “one of the world’s most virulent diseases.”\nFruit bats are probably the natural host for the virus, and they can pass it on to other animals. The human infections are thought to have originated from contact with infected monkeys.\nEarly symptoms include a sudden fever, general fatigue and weakness, muscle pains, headaches, red eyes, a skin rash with bruise-like blisters, chest pain, difficulty breathing, difficulty swallowing, vomiting, diarrhea and internal bleeding. It can affect internal organs like the kidney and liver and eventually cause death.\nOne of the common and curious early symptoms is severe hiccups. The symptoms occur anywhere from two to 21 days after infection but most commonly appear eight to 12 days after exposure.\nBecause the early symptoms are commonly seen in many illnesses, diagnosis can be challenging, and it can often be misdiagnosed as another disease such as malaria, typhoid fever, meningitis, the plague, or even cholera. Ebola can be definitively confirmed by specific laboratory tests.\nAlthough there are experimental medicines to treat Ebola (such as ZMapp, Brincidofovir and Tekmira), none are FDA-approved yet. The World Health Organization has announced that in severe outbreaks it is not unethical to consider using experimental medications.\nCurrently, the treatment for patients is purely supportive: Maintain good hydration, support normal oxygen and blood pressure levels, treat symptoms, and prevent secondary infection. With early detection and supportive measures such as these, the mortality rate may be decreased to 40 percent or lower.\nEbola is contracted by coming directly into contact with blood, body secretions, or other body fluids from an infected person, or from exposure to a contaminated needle of an infected person. The vast majority of patients with Ebola know exactly where they caught it, from taking care of another person with the disease or who had symptoms such as fever, diarrhea, vomiting, bleeding and hiccups.\nThe good news is that the virus is sensitive to sanitizing measures such as chlorine and other medical disinfectants, heat treatment and detergents. Currently, there are some promising vaccines under research and development.\nThe chances of catching it on an airplane are thought to be very low. It is not spread by sneezing or coughing. There needs to be prolonged, frequent, and direct contact with body fluids or blood infected with the virus in order to catch it.\nIt is important to note: The infected person must be symptomatic in order to transmit the virus. Also, the corpses of persons who have died from Ebola infection can transmit the infection, so those caring for the body must exercise extreme caution and protective measures.\nOutbreaks in the past have been controlled, and once an area has gone “infection free” for 42 days (twice the incubation period), the infection is considered to be under control. With the ease of worldwide travel, the challenge for governments, international healthcare organizations, airlines (and other passenger transportation companies), industrial/commercial sanitation and cleaning organizations and medical professionals will be to keep the infection contained and from spreading globally.\nIf you think you have been exposed to the virus, or to a person with the virus, please contact your physician, local hospital or 911 at once.\nCharles E. Crutchfield III, MD is a board certified dermatologist and Clinical Professor of Dermatology at the University of Minnesota Medical School. He also has a private practice in Eagan, MN. He has been selected as one of the top 10 dermatologists in the U.S. by Black Enterprise magazine and one of the top 21 African American physicians in the U.S. by the Atlanta Post. Dr. Crutchfield is an active member of the Minnesota Association of Black Physicians, MABP.org.<|endoftext|>"},"score":{"kind":"number","value":4.0625,"string":"4.0625"}}},{"rowIdx":1286,"cells":{"token_count":{"kind":"number","value":487,"string":"487"},"text":{"kind":"string","value":"# How do you solve x^2 + 32x + 15 = -x^2 + 16x + 11 by completing the square?\n\nJun 22, 2015\n\nSimplify the given equation then apply the process of \"completing the squares\" to obtain\n$\\textcolor{w h i t e}{\\text{XXXX}}$$x = - 4 \\pm \\sqrt{14}$\n\n#### Explanation:\n\nGiven ${x}^{2} + 32 x + 15 = - {x}^{2} + 16 x + 11$\n\nFirst simplify to get all terms involving $x$ on the left side and a simple constant on the right:\n$\\textcolor{w h i t e}{\\text{XXXX}}$$2 {x}^{2} + 16 x = - 4$\nFurther simplify by dividing by 2\n$\\textcolor{w h i t e}{\\text{XXXX}}$${x}^{2} + 8 x = - 2$\n\nNow we are ready to begin completing the square.\n#\n\nNoting that the squared binomial ${\\left(x + a\\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$\n\nIf ${x}^{2} + 8 x$ are the first 2 terms of an expanded squared binomial,\nthen $2 a x = 8 x \\rightarrow a = 4 \\mathmr{and} {a}^{2} = 16$\nSo the completed square must be (after remembering that anything added to one side of an equation must also be added to the other)\n$\\textcolor{w h i t e}{\\text{XXXX}}$${x}^{2} + 8 x + 16 = - 2 + 16$\nor\n$\\textcolor{w h i t e}{\\text{XXXX}}$${\\left(x + 4\\right)}^{2} = 14$\n\nTaking the square root of both sides:\n$\\textcolor{w h i t e}{\\text{XXXX}}$$x + 4 = \\pm \\sqrt{14}$\nand\n$\\textcolor{w h i t e}{\\text{XXXX}}$$x = - 4 \\pm \\sqrt{14}$<|endoftext|>"},"score":{"kind":"number","value":4.84375,"string":"4.84375"}}},{"rowIdx":1287,"cells":{"token_count":{"kind":"number","value":3015,"string":"3,015"},"text":{"kind":"string","value":"Pairwise comparisons of proportions\n\nPairwise comparison means comparing all pairs of something. If I have three items A, B and C, that means comparing A to B, A to C, and B to C. Given n items, I can determine the number of possible pairs using the binomial coefficient:\n\n$$\\frac{n!}{2!(n – 2)!} = \\binom {n}{2}$$\n\nUsing the R statistical computing environment, we can use the choose function to quickly calculate this. For example, how many possible 2-item combinations can I “choose” from 10 items:\n\nchoose(10,2)\n[1] 45\n\n\n\nWe sometimes want to make pairwise comparisons to see where differences occur. Let’s say we go to 8 high schools in an area, survey 30 students at each school, and ask them whether or not they floss their teeth at least once a day. When finished we’ll have 8 proportions of students who answered “Yes”. An obvious first step would be to conduct a hypothesis test for any differences between these proportions. The null would be no difference between the proportions versus some difference. If we reject the null, we have evidence of differences. But where are the differences? This leads us to pairwise comparisons of proportions, where we make multiple comparisons. The outcome of these pairwise comparisons will hopefully tell us which schools have significantly different proportions of students flossing.\n\nMaking multiple comparisons leads to an increased chance of making a false discovery, i.e. rejecting a null hypothesis that should not have been rejected. When we run a hypothesis test, we always run a risk of finding something that isn’t there. Think of flipping a fair coin 10 times and getting 9 or 10 heads (or 0 or 1 heads). That’s improbable but not impossible. If it happened to us we may conclude the coin is unfair, but that would be the wrong conclusion if the coin truly was fair. It just so happened we were very unlucky to witness such an unusual event. As we said, the chance of this happening is low in a single trial, but we increase our chances of it happening by conducting multiple trials.\n\nThe probability of observing 0, 1, 9 or 10 heads when flipping a fair coin 10 times is about 2% which can be calculated in R as follows:\n\npbinom(q = 1, size = 10, prob = 0.5) * 2\n[1] 0.02148438\n\n\n\nTherefore the the probability of getting 2 – 8 heads is about 98%:\n\n1 - pbinom(q = 1, size = 10, prob = 0.5) * 2\n[1] 0.9785156\n\n\n\nThe probability of getting 2 – 8 heads in 10 trials is 98% multiplied by itself 10 times:\n\n(1 - pbinom(q = 1, size = 10, prob = 0.5) * 2)^10\n[1] 0.8047809\n\n\n\nTherefore the probability of getting 0, 1, 9, or 10 heads in 10 trials is now about 20%:\n\n1 - (1 - pbinom(q = 1, size = 10, prob = 0.5) * 2)^10\n[1] 0.1952191\n\n\n\nWe can think of this as doing multiple hypothesis tests. Flip 10 coins 10 times each, get the proportion of heads for each coin, and use 10 one-sample proportion tests to statistically determine if the results we got are consistent with a fair coin. In other words, do we get any p-values less than, say, 0.05?\n\nWe can simulate this in R. First we replicate 1,000 times the act of flipping 10 fair coins 10 times each and counting the number of heads using the rbinom function. This produces a 10 x 1000 matrix of results that we save as “coin.flips”. We then apply a function to each column of the matrix that runs 10 one-sample proportion tests using the prop.test function and saves a TRUE/FALSE value if any of the p-values are less than 0.05 (we talk more about the prop.test function below). This returns a vector we save as “results” that contains TRUE or FALSE for each replicate. R treats TRUE and FALSE as 0 or 1, so calling mean on results returns the proportion of TRUEs in the vector. We get about 20%, confirming our calculations. (If you run the code below you’ll probably get a slightly different but similar answer.)\n\ntrials <- 10\ncoin.flips <- replicate(1000, rbinom(n = 10, size = trials, prob = 0.5))\n\nmultHypTest <- function(x){\npvs <- sapply(x, function(x)prop.test(x = x, n = trials, p = 0.5)\\$p.value)\nany(pvs < 0.05)\n}\n\nresults <- apply(coin.flips,2,multHypTest)\nmean(results)\n[1] 0.206\n\n\n\nThat’s just for 10 trials. What about 15 or 20 or more? You can re-run the code above with trials set to a different value. We can also visualize it by plotting the probability of an unusual result (0, 1, 9, or 10 heads) versus the number trials. Notice how rapidly the probability of a false discovery increases with the number of trials.\n\ncurve(expr = 1 - (1 - pbinom(q = 1, size = 10, prob = 0.5) * 2)^x,\nxlim = c(1,50),\nxlab = \"Number of tests\",\nylab = \"Probability of 0, 1, 9, or 10 heads\")\n\n\n\nSo what does all of this tell us? It reveals that traditional significance levels such as 0.05 are too high when conducting multiple hypothesis tests. We need to either adjust our significance level or adjust our p-values. As we’ll see, the usual approach is to adjust the p-values using one of several methods for p-value adjustment.\n\nLet’s return to our example of examining the proportion of high school students (sample size 30 at each school) who floss at 8 different high schools. We’ll simulate this data as if the true proportion is 30% at each school (i.e., no difference). We use set.seed to make the data reproducible.\n\nset.seed(15)\nn <- 30\nk <- 8\nschool <- rep(1:k, each = n)\nfloss <- replicate(k, sample(x = c(\"Y\",\"N\"),\nsize = n,\nprob = c(0.3, 0.7),\nreplace = TRUE))\ndat <- data.frame(school, floss = as.vector(floss))\n\n\n\nWith our data generated, we can tabulate the number of Yes and No responses at each school:\n\nflossTab <- with(dat, table(school, floss))\nflossTab\nfloss\nschool N Y\n1 18 12\n2 19 11\n3 14 16\n4 19 11\n5 26 4\n6 15 15\n7 20 10\n8 21 9\n\n\n\nUsing prop.table we can determine the proportions. Specifying margin = 1 means proportions are calculated across the rows for each school. (We also round to two decimal places for presentation purposes.) The second column contains the proportion of students who answered Yes at each school.\n\nround(prop.table(flossTab, margin = 1),2)\nfloss\nschool N Y\n1 0.60 0.40\n2 0.63 0.37\n3 0.47 0.53\n4 0.63 0.37\n5 0.87 0.13\n6 0.50 0.50\n7 0.67 0.33\n8 0.70 0.30\n\n\n\nFirst we might want to run a test to see if we can statistically conclude that not all proportions are equal. We can do this with the prop.test function. The prop.test function requires that Yes (or “success”) counts be in the first column of a table and No (or “failure”) counts in the second column. Thus we switch the columns using subsetting brackets with a vector indicating column order.\n\nprop.test(flossTab[,c(\"Y\",\"N\")])\n\n8-sample test for equality of proportions without continuity correction\n\ndata: flossTab[, c(\"Y\", \"N\")]\nX-squared = 13.78, df = 7, p-value = 0.05524\nalternative hypothesis: two.sided\nsample estimates:\nprop 1 prop 2 prop 3 prop 4 prop 5 prop 6 prop 7 prop 8\n0.4000000 0.3666667 0.5333333 0.3666667 0.1333333 0.5000000 0.3333333 0.3000000\n\n\n\nThe p-value of 0.055 is borderline significant and indicates some evidence of differences among proportions. We generated the data so we know there actually is no difference! But if this were real data that we had spent considerable resources collecting, we might be led to believe (perhaps even want to believe) some differences indeed exist. That p-value is so close to significance! School #5, in particular, with a proportion of 13% looks far lower than school #3 with 53%. We could conclude this hypothesis test is significant at 0.10 level and proceed to pairwise comparisons.\n\nTo do that in R we use the pairwise.prop.test function which requires a table in the same format as prop.test, Yes counts in the first column and No counts in the second column:\n\npairwise.prop.test(x = flossTab[,c(\"Y\",\"N\")])\n\nPairwise comparisons using Pairwise comparison of proportions\n\ndata: flossTab[, c(\"Y\", \"N\")]\n\n1 2 3 4 5 6 7\n2 1.000 - - - - - -\n3 1.000 1.000 - - - - -\n4 1.000 1.000 1.000 - - - -\n5 1.000 1.000 0.073 1.000 - - -\n6 1.000 1.000 1.000 1.000 0.149 - -\n7 1.000 1.000 1.000 1.000 1.000 1.000 -\n8 1.000 1.000 1.000 1.000 1.000 1.000 1.000\n\n\n\nThis produces a table of 28 p-values since there are 28 possible pairs between 8 items. We interpret the table by using row and column numbers to find the p-value for a particular pair. For example the p-value of 0.073 at the intersection of row 5 and column 3 is the p-value for the two-sample proportion test between school #5 and school #3. It appears to be insignificant at the traditional 5% level. All other p-values are clearly insignificant. In fact, most are 1. This is due to the p-value adjustment that was made. The output tells us the “holm” method was used. We won’t get into the details of how this method works, but suffice to say it increases the p-values in an effort to adjust for the many comparisons being made. In this case, it does what it’s supposed to: it adjusts the p-values and allows us to make a good case there is no differences between schools, at least not at the 5% level, which would be the correct decision.\n\nWe can do pairwise comparisons without adjusted p-values by setting p.adjust.method = \"none\". Let’s do that and see what happens:\n\n# NOTE: This analysis is wrong!\npairwise.prop.test(x = flossTab[,c(\"Y\",\"N\")], p.adjust.method = \"none\")\n\nPairwise comparisons using Pairwise comparison of proportions\n\ndata: flossTab[, c(\"Y\", \"N\")]\n\n1 2 3 4 5 6 7\n2 1.0000 - - - - - -\n3 0.4376 0.2993 - - - - -\n4 1.0000 1.0000 0.2993 - - - -\n5 0.0410 0.0736 0.0026 0.0736 - - -\n6 0.6038 0.4345 1.0000 0.4345 0.0055 - -\n7 0.7888 1.0000 0.1927 1.0000 0.1270 0.2949 -\n8 0.5883 0.7842 0.1161 0.7842 0.2100 0.1876 1.0000\n\n\n\nNotice now we have significant differences for 3 pairs: (5,1), (5,3), and (6,5). Again we know this is wrong because we simulated the data. The truth is all schools have a floss rate of 30%. But we see that through random chance and not adjusting our p-values for multiple testing we got what look to be significant results. This illustrates the importance of using adjusted p-values when making multiple comparisons.\n\nThere are other p-value adjustment methods available. A common and conservative choice is the bonferroni method. It simply multiplies all p-values by the number of pairs. In our example that is 28. To see all p-value adjustment methods available in R enter ?p.adjust at the console.\n\nFor questions or clarifications regarding this article, contact the UVA Library StatLab: statlab@virginia.edu\n\nView the entire collection of UVA Library StatLab articles.\n\nClay Ford\nStatistical Research Consultant\nUniversity of Virginia Library\nOctober 20, 2016<|endoftext|>"},"score":{"kind":"number","value":4.40625,"string":"4.40625"}}},{"rowIdx":1288,"cells":{"token_count":{"kind":"number","value":912,"string":"912"},"text":{"kind":"string","value":"Typography is the art or process of printing with type. However, in modern usage typography includes all manner of non-printed letter forms such as websites, eBooks, electronic billboards, and even textiles. Through the use of type, a person can visually tell a story using little to no imagery. This post is intended to teach you about the anatomy of type and to help you better understand what to look for when choosing your next font.\nFirst, the basics …\nTypeface vs Font, What’s the difference?\nA typeface, also called a font family, is a set of fonts designed with a stylistic unity, each comprising a coordinate set of glyphs. A font is a complete character set of a typeface at a particular size, weight, and style.\nSerif vs Sans Serif, Which one to use?\nSerif letters are drawn with features at the ends of their strokes. The serifs are the little feet we see in fonts like Times. These are some of the oldest type designs. The feet along the baseline help guide the eye from left to right, making them very ‘readable’ fonts.\nSans Serif (french for “without serifs”) are letters drawn with straighter lines and no feet. Their larger letterforms make them very legible, but can cause greater eye strain when used in long runs of text. Helvetica is considered the quintessential sans serif font.\nThe Anatomy of a Letter\nThere is a standard set of terms to describe the parts of a character. These terms, and the parts of the letter they represent, are often referred to as “letter anatomy” or “typeface anatomy.” By breaking down letters into parts, a designer can better understand how type is created and altered and how to use it effectively.\n- Baseline – The baseline is the invisible line on which characters sit. While the baseline may differ from typeface to typeface, it is consistent within a typeface. Rounded letters such as “e” may extend slightly below the baseline.\n- Meanline – The meanline falls at the top of many lowercase letters such as “e,” “g” and “y.” It is also at the curve of letters like “h.”\n- X-Height – The x-height is the distance between the meanline and the baseline. It is referred to as the x-height because it is the height of a lowercase “x.” This height can vary greatly between typefaces.\n- Cap Height – The cap height is the distance from the baseline to the top of uppercase letters like “H” and “J.”\n- Ascender – The part of a character that extends above the meanline is known as an ascender. Note that this is the same as extending above the x-height.\n- Descender – The part of a character that extends below the baseline is known as a descender, such as the bottom stroke of a “y.”\n- Serif – Fonts are often divided into serif and sans serif. Serif fonts are distinguishable by the extra stroke at the ends of the character, known as a serif.\n- Stem – The vertical line of a “B” and the primary diagonal line of a “V” are known as the stem. The stem is often the main “body” of a letter.\n- Bar – The horizontal lines of an “E” are known as bars. Bars are horizontal or diagonal lines of a letter, also known as arms, and are open on at least one side.\n- Bowl – An open or closed circular line that creates an interior space, such as in “e” and “b.”\n- Counter – The inside of a bowl.\n- Leg – The bottom stroke of a letter, such as the base of an “L” or diagonal stroke of a “K.”\n- Shoulder – The curve at the beginning of a leg of a character, such as in an “m.”\nExamples of Great Type-Only Design\nWant to learn more? Visit Canva.com for an illustrated glossary of typographic terms.<|endoftext|>"},"score":{"kind":"number","value":4.0625,"string":"4.0625"}}},{"rowIdx":1289,"cells":{"token_count":{"kind":"number","value":474,"string":"474"},"text":{"kind":"string","value":"Development of the Heart\nThe heart is the first functional organism in a vertebrate embryo and there are 5 different stages to heart development.\nThe first stage of heart development involves the forming of the primitive heart tube. The primitive heart tube consisits of a single tube which is formed when cardiac precurser cells fuse together. Even at this early stage of development the heart consists of different regions and layers, these regions are:\n- bulbs ordis (which develops to form the aorta and ventricles)\nThe developing layers of the heart are:\n- cardiac jelly\n- cardiac mantle layers (which will form the myometrium and epicardium of the heart)\nThe heart first begins to beat during this stage of heart development, about 22-23 days after conception. This can be detected via vaginal ultra sound.\nThe second stage of heart development then occurs within 24 hours of the end of the first stage, this stage is known as heart looping. The tube shaped heart concorts into an \"S\" shape and bends to the right, this is known as d-looping. This process creates a new shape, which provides a primitive region for the ventricles to begin forming. This stage of development is initiated by activated heart-specific proteins.\nThe third stage of heart development is known as the two-chambered stage, consisting of one atrium and one ventricle. During this stage the cardiac jelly serves to act as a valve inbetween the atrium and the ventricle. The developing heart at this two-chambered stage is said to resemble the heart of a frog.\nThe fourth stage of heart development begins by the atria dividing. The heart at this point of development is therefore known as the three-chambered heart, consisting of two newly divided atria which sit ontop of the remaining ventricle, the third chamber. This three-chambered heart is said to resemble the heart of a snake or turtle.\nThe fifth and final stage of heart development occurs 10 weeks after conception. In this stage of development the single ventricle divides to form 2 new ventricles. This leaves the heart complete with all 4 chambers; 2 atria and 2 ventricles, and two large blood vessels to carry blood to and from the heart.<|endoftext|>"},"score":{"kind":"number","value":3.8125,"string":"3.8125"}}},{"rowIdx":1290,"cells":{"token_count":{"kind":"number","value":480,"string":"480"},"text":{"kind":"string","value":"Focus: In this unit (Unit 2 for schools using the CKHG series in Sequence grade-level order), students explore how the cultural and political traditions of ancient Greece and Rome have influenced Western society more profoundly than perhaps any other civilizations in world history. The political institutions of these two great civilizations—including the early forms of democracy established in Athens and several other city-states of ancient Greece, and the judicious power sharing articulated in the Roman Republic—have been incorporated into many subsequent societies.\nFor ancient Greece, topics of study include the beginnings of democratic government; limitations of Greek democracy; the “classical” ideal in art and life; Pericles and the Golden Age; Greek myths; the Persian Wars and the Peloponnesian War; the philosophers Socrates, Plato, and Aristotle; and, the spread of Greek culture. For ancient Rome, topics of study include government and society in the Roman Republic; the Punic Wars; Julius Caesar; Caesar Augustus, the Pax Romana, and law and administration in the Roman Empire; the significance of Virgil’s Aeneid; Christianity under Roman rule; Constantine; and the causes of the “decline and fall” of the Roman Empire.\nNumber of Lessons: 17\nLesson Time: 45 minutes each daily. Each lesson may be divided into shorter segments.\nAdditional Search Terms: social studies • geography • map skills • nonfiction • informational text • primary source documents • myths • Mediterranean region • city-state • Athens • Parthenon • Olympics • Apollo and Daphne • Orpheus and Eurydice • Narcissus and Echo • Thermopylae • Sparta • Alexandria • the Pantheon • Ovid • Pygmalion and Galatea myths • patricians and plebeians • consuls, tribunes, senators • Carthage • Hannibal • Shakespeare\nCKHG Grade Levels: CKHG units are correlated to topics at the grade levels specified in the Core Knowledge Sequence, which allows students in schools following the Sequence to build knowledge grade by grade. This particular unit falls in Grade 6 in the Core Knowledge Sequence. In other settings, individual CKHG units may be used as supplemental resources. In general, the content and presentation in the CKHG units for Grade 6 are appropriate for students in Grade 6 and up.<|endoftext|>"},"score":{"kind":"number","value":4.1875,"string":"4.1875"}}},{"rowIdx":1291,"cells":{"token_count":{"kind":"number","value":1187,"string":"1,187"},"text":{"kind":"string","value":"# Ray-Triangle Intersection¶\n\nImplementation of ray-triangle intersection algorithm.\n\n## Primitives¶\n\nThe first thing is to define the ray origin $O$ and direction $\\large ê$, where the unit vector $\\large ê = \\frac{\\vec{e}}{\\parallel \\vec{e} \\parallel}$.\n\nNext step is to define the 3D triangle in counter-clockwise order in relation to the direction of the face.\n\n## Intersection¶\n\nTo find the ray intersection, the next step is to define the triangle normal $\\hat{n}$, where:\n\n$$\\large \\hat{n} = \\frac{\\vec{AB} \\times \\vec{AC}}{\\parallel \\vec{AB} \\times \\vec{AC} \\parallel}$$\n\np.s.: to calculate ray-triangle intersection it is not necessary to normalize the normal vector.\n\nThe supporting plane is what the triangle lies on, sharing the same normal vector. Given the plane equation:\n\n$$\\large ax + by + cz + d = 0$$\n\nHaving the vector normal as $\\hat{n} = [a b c]^T$ and $P = [x y z]^T$ as any point on the plane, we can define $d$ as follows:\n\n$$\\large \\hat{n} \\cdot P + d = 0 \\quad \\therefore \\quad d = - \\hat{n} \\cdot P$$\n\np.s.: in this case any known point can be used. Lets use the point $A$ so $P = A$\n\nBefore finding the intersection point $P$ on the plane, we must calculate the parameter $t$. We start by looking at the parametric equation of a line segment, which has the same direction of $ê$ and origin from $O$:\n\n$$\\large P(P_x, P_y, P_z) = O + ê t$$\n\nwhere:\n\n$$\\large P_x = O_x + ê_x t \\\\ \\large P_y = O_y + ê_y t \\\\ \\large P_z = O_z + ê_z t$$\n\nUsing this concept on the plane equation, we have:\n\n$$\\large ax + by + cz + d = 0 \\\\ \\large aP_x + bP_y + cP_z + d = 0 \\\\ \\large a(O_x + ê_x t) + b(O_y + ê_y t) + c(O_z + ê_z t) + d = 0 \\\\ \\large aO_x + aê_x t + bO_y + bê_y t + cO_z + cê_z t + d = 0 \\\\ \\large (aê_x + bê_y + cê_z)t + (aO_x + bO_y + cO_z) + d = 0 \\\\ \\large (\\hat{n} \\cdot \\hat{e})t + \\hat{n} \\cdot O + d = 0 \\\\ \\large t = - \\frac{\\hat{n} \\cdot O + d}{\\hat{n} \\cdot \\hat{e}}$$\n\nTo figure out if the plane intersection point is inside or outside the triangle, we basically have to define the vector from each vertices to $P$ and cross it with its oriented edge segment (for each vertex). If the intersection point is outside the triangle, the resulting vector will be in the opposite direction from the normal one.\n\n$$\\large [(B - A) \\times (P - A)] \\cdot \\hat{n} \\geq 0 \\\\ \\large [(B - B) \\times (P - B)] \\cdot \\hat{n} \\geq 0 \\\\ \\large [(B - C) \\times (P - C)] \\cdot \\hat{n} \\geq 0$$\n\nIf all these conditionals are obeyed, we can conclude that the point $P$ is inside the triangle. Otherwise, the point is going to be outiside toward to the edges of the negative values.\n\n## 3D Intersection¶\n\nApply same model for each pixel of an image plane as the origin and the ray direction is based on the perspective camera model:\n\n## Barycentric coordinates¶\n\nThe barycentric coordinates will help us to interpolate in-between vertex values. To do that we have to calculate the area of all resulting triangles. Any triangle area can be calculated as follows:\n\n$$\\large \\text{Area}_{ABC} = \\frac{\\parallel (B - A) \\times (C - A) \\parallel}{2}$$\n\nNext step is to find the weight of each point so that we can use it to interpolate any desired values.\n\n$$\\large \\alpha = \\frac{\\text{Area}_{BCP}}{\\text{Area}_{ABC}} = \\frac{\\parallel (C - B) \\times (P - B) \\parallel}{\\parallel (B - A) \\times (C - A) \\parallel} \\\\\\\\ \\large \\beta = \\frac{\\text{Area}_{CAP}}{\\text{Area}_{ABC}} = \\frac{\\parallel (A - C) \\times (P - C) \\parallel}{\\parallel (B - A) \\times (C - A) \\parallel} \\\\\\\\ \\large \\gamma = \\frac{\\text{Area}_{ABP}}{\\text{Area}_{ABC}} = \\frac{\\parallel (B - A) \\times (P - A) \\parallel}{\\parallel (B - A) \\times (C - A) \\parallel}$$\n\nHave the weights we can interpolate any kind of value (color, for example) by using:\n\n$$\\large V = \\frac{\\alpha V_A + \\beta V_B + \\gamma V_C}{\\alpha + \\beta + \\gamma}$$<|endoftext|>"},"score":{"kind":"number","value":4.46875,"string":"4.46875"}}},{"rowIdx":1292,"cells":{"token_count":{"kind":"number","value":266,"string":"266"},"text":{"kind":"string","value":"Ecological and Evolutionary Effects of Crop Domestication\nHow does rapid evolution that occurs during plant cultivation influence ecological and evolution interactions between plants and their enemies?\nThe domestication of plants and animals is one of the most important activities in human history. It is also the world’s largest, longest lasting, and best replicated selection experiment. By comparing domesticated organisms to their wild relatives we can address numerous basic and applied questions in evolutionary-ecology. One hypothesis we are address is that domestication, and its associated changes in yield, should cause plants to evolve reduced resistance to enemies. We are testing this hypothesis, and other extensions of it, using a combination of large scale comparative experiments and meta-analyses. Future projects will explore the community level impacts of domestication using field experiments as well as continue to explore the impacts on the evolution of plant enemies.\nImpact of domestication on resistance to 2 generalist herbivores across 29 domestication events --> New Phytologist 2012\nDomestication slowing the rate of aphid rapid evolution --> Ecology Letters 2015\nMeta-Analysis on the impact of domestication on resistance and defense traits --> PRSLB 2016\nGeneral Synthesis of the Eco-Evolutionary Impacts of Domestication and Agricultural on wild species -- > PRSLB 2016<|endoftext|>"},"score":{"kind":"number","value":3.6875,"string":"3.6875"}}},{"rowIdx":1293,"cells":{"token_count":{"kind":"number","value":822,"string":"822"},"text":{"kind":"string","value":"The fact that we’re able to hear sounds in our environment and process them almost instantly is nothing short of miraculous. Without even having to think about it, we can hear noises both loud and soft, near and far away. The pathway that these sounds travel in our ears is intricate, and if any one of its parts isn’t working correctly, we can experience hearing loss.\nSound waves begin by entering your outer ear, called the pinna. It funnels these waves down your ear canal and into your middle ear, which are separated by the eardrum. The eardrum is a flexible membrane which begins to move when sound vibrations hit it, and that in turn starts to move three small bones in the middle ear: the hammer, anvil, and stirrup. These bones work together to amplify the sound waves and move them to your inner ear. The middle ear is where hearing loss begins to occur, so it’s important that all of these moving parts are free of damage.\nThe inner ear is full of fine hair-like cells, replete with nerve endings, within a spiral-shaped organ called the cochlea. These tiny hair cells collect information from sound vibrations coming in from the middle ear and transmit those vibrations into nerve impulses (via the auditory nerve) to your brain. The brain processes and interprets these signals as sounds, allowing us to hear the noises around us.\nIf you suffer from hearing loss, it means that one of the above sections isn’t working quite right. Our hearing evaluation is designed to diagnose which type of hearing loss is present, and enables us to identify the most effective solution for you.\nConductive Hearing Loss\nAlthough rather uncommon and typically temporary, a conductive hearing loss can occur in some patients. It’s caused when an issue in the outer or middle ear blocks sound from the inner ear. Treatment involves the use of medication or surgery while other individuals opt to use hearing aids to improve their hearing ability.\nConductive hearing loss can be caused by:\n- Ear infections\n- Benign tumors\n- Swimmer’s Ear\n- Foreign object in the ear\n- Fluid in the middle ear from colds or allergies\n- Absence or malformation of the outer ear, ear canal, or middle ear\n- Perforated eardrum\n- Impacted cerumen (earwax)\nSensorineural Hearing Loss\nThe most common type of hearing loss, sensorineural hearing loss occurs due to a problem with the inner ear or auditory nerve. It presents itself when either the auditory nerve or the hair-like cells in the cochlea have sustained damage or are missing. This results in the inability to send complete nerve signals to the brain.\nThis kind of hearing loss can be caused by the following:\n- Head trauma\n- Drugs that are toxic to hearing (ototoxicity)\n- Genetics or aging\n- Malformation of the inner ear\n- Exposure to loud noise\nMixed Hearing Loss\nWhen multiple parts of the ear’s anatomy are damaged, a mixed hearing loss can occur. In most cases, both the middle or outer ear along with the auditory nerve or inner ear have sustained an injury of some type or have encountered one of the conditions listed above. The conductive hearing loss present may be reversible while the sensorineural hearing loss is often permanent.\nAuditory Processing Disorders\nRather than a hearing impairment which affects the ability to detect sounds, Auditory Processing Disorder (APD) causes individuals to struggle with their ability to organize, analyze, and interpret noises around them. While all parts of the ear are functioning properly, those with APD find that the hurdle they encounter is in their brain. Often caused by a tumor, disease, injury, heredity, or an unknown cause, the auditory processing centers in the brain do not function normally. APD does not always include hearing loss and many times the treatments for this disorder versus a hearing impairment are dramatically different.<|endoftext|>"},"score":{"kind":"number","value":4.125,"string":"4.125"}}},{"rowIdx":1294,"cells":{"token_count":{"kind":"number","value":219,"string":"219"},"text":{"kind":"string","value":"Desalination is a process that takes away mineral components from saline water. More generally, desalination refers to the removal of salts and minerals from a target substance, as in soil desalination, which is an issue for agriculture.\nSaltwater is desalinated to produce water suitable for human consumption or irrigation. One by-product of desalination is salt. Desalination is used on many seagoing ships and submarines. Most of the modern interest in desalination is focused on cost-effective provision of fresh water for human use. Along with recycled wastewater, it is one of the few rainfall-independent water sources.\nDue to its energy consumption, desalinating sea water is generally more costly than fresh water from rivers or groundwater, water recycling and water conservation. However, these alternatives are not always available and depletion of reserves is a critical problem worldwide. Currently, approximately 1% of the world's population is dependent on desalinated water to meet daily needs, but the UN expects that 14% of the world's population will encounter water scarcity by 2025.<|endoftext|>"},"score":{"kind":"number","value":3.8125,"string":"3.8125"}}},{"rowIdx":1295,"cells":{"token_count":{"kind":"number","value":761,"string":"761"},"text":{"kind":"string","value":"# What is the percent of increase from $147 annually to$414?\n\n##### 1 Answer\nMay 23, 2016\n\n$\\textcolor{red}{\\text{The time span not defined as 1 year}}$\nFor 1 year only color(blue)(-> \"percentage is \" 187.5%)\nOr simple interest over whole period\n'.....................................................\nFor more than 1 year where $n$ is number of years\nand annually compounded interest.\n\n\" \"color(blue)(x%=log^(-1)((log(414)-log(144))/n + 2) - 100)\n\n#### Explanation:\n\n$\\textcolor{red}{\\text{You have not stated the number of years.}}$\n$\\textcolor{red}{\\text{The use of the word 'annually' implies more than 1 year}}$\n\nLet the number of years be $n$\n\n\" \"$144(1+x/100)^n=$414\n\n'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n$\\textcolor{b l u e}{\\text{Suppose the number of years is 1 then}}$\n\n\" \"144+(144xx ?/100)=414\n\n$\\textcolor{b r o w n}{\\text{Subtract 144 from both sides}}$\n\n\" \"144xx ?/100=270\n\n$\\textcolor{b r o w n}{\\text{Multiply both sides by } \\frac{100}{144}}$\n\ncolor(blue)(\" \"?%=270xx100/144 = 187.5%)\n\n'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\n$\\textcolor{b l u e}{\\text{Suppose \"n>1)color(magenta)(\" \"->\" Needs algebra}}$\n\n$\\textcolor{b r o w n}{\\text{Notice that } \\left(1 + \\frac{x}{100}\\right) \\to \\left(\\frac{100 + x}{100}\\right)}$\n\n$\\textcolor{b r o w n}{\\text{Taking logs of both sides}}$\n\n\" \"log($144)+nlog((100+x)/100)=log($414)\n\n$\\text{ } n \\log \\left(\\frac{100 + x}{100}\\right) = \\log \\left(414\\right) - \\log \\left(144\\right)$\n\n$\\text{ } \\log \\left(\\frac{100 + x}{100}\\right) = \\frac{\\log \\left(414\\right) - \\log \\left(144\\right)}{n}$\n\n$\\text{ } \\log \\left(100 + x\\right) - \\log \\left(100\\right) = \\frac{\\log \\left(414\\right) - \\log \\left(144\\right)}{n}$\n\n$\\text{ } \\log \\left(100 + x\\right) = \\frac{\\log \\left(414\\right) - \\log \\left(144\\right)}{n} + \\log \\left(100\\right)$\n\n'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\ncolor(brown)(\"But \"log(100)=2\n\n$\\text{ } \\log \\left(100 + x\\right) = \\frac{\\log \\left(414\\right) - \\log \\left(144\\right)}{n} + 2$\n\n$\\text{ } 100 + x = {\\log}^{- 1} \\left(\\frac{\\log \\left(414\\right) - \\log \\left(144\\right)}{n} + 2\\right)$\n\n\" \"color(blue)(x%=log^(-1)((log(414)-log(144))/n + 2) - 100)<|endoftext|>"},"score":{"kind":"number","value":4.71875,"string":"4.71875"}}},{"rowIdx":1296,"cells":{"token_count":{"kind":"number","value":1291,"string":"1,291"},"text":{"kind":"string","value":"## How to Solve\n\nThe puzzle can be played with any number of disks, although many toy versions have around seven to nine of them. The game seems impossible to many novices, yet is solvable with a simple algorithm. The number of moves required to solve a Tower of Hanoi puzzle is 2n -1, where n is the number of disks.\n\nI have discussed some common methods to solve 'towers of hanoi' puzzle game. Though complex methods are available, for the sake of simply solving them, these methods are enough.\n\nIterative solution\n\nThe following solution is a simple solution for the toy puzzle. Alternate moves between the smallest piece and a non-smallest piece. When moving the smallest piece, always move it to the next position in the same direction (to the right if the starting number of pieces is even, to the left if the starting number of pieces is odd). If there is no tower position in the chosen direction, move the piece to the opposite end, but then continue to move in the correct direction. For example, if you started with three pieces, you would move the smallest piece to the opposite end, then continue in the left direction after that. When the turn is to move the non-smallest piece, there is only one legal move. Doing this will complete the puzzle using the fewest number of moves to do so.\n\nIt should perhaps be noted that this can be rewritten as a strikingly elegant set of rules:\n\nSimpler statement of iterative solution\n\nAlternating between the smallest and the next-smallest disks, follow the steps for the appropriate case:\nFor an even number of disks:\n\n• make the legal move between pegs A and B\n• make the legal move between pegs A and C\n• make the legal move between pegs B and C\n• repeat until complete\nFor an odd number of disks:\n\n• make the legal move between pegs A and C\n• make the legal move between pegs A and B\n• make the legal move between pegs B and C\n• repeat until complete\nIn each case, a total of 2n-1 moves are made.\n\nEquivalent Iterative Solution\n\nAnother way to generate the unique optimal iterative solution, is to number the disks 1 through n, largest to smallest. If n is odd, the first move is from the Start to the Finish peg, and if n is even the first move is from the Start to the Using peg.\n\nNow, add the constraint that no odd disk may be placed directly on an odd disk, and no even disk may be placed directly on an even disk. With this extra constraint, and the obvious rule of never undoing your last move, there is only one move at every turn. The sequence of these unique moves is an optimal solution to the problem equivalent to the iterative solution described above.\n\nRecursive solution\n\nA key to solving this puzzle is to recognize that it can be solved by breaking the problem down into a collection of smaller problems and further breaking those problems down into even smaller problems until a solution is reached. The following procedure demonstrates this approach.\n\n• label the pegs A, B, C - these labels may move at different steps\n• let n be the total number of discs\n• number the discs from 1 (smallest, topmost) to n (largest, bottommost)\n\nTo move n discs from peg A to peg C:\n\n1. move n-1 discs from A to B. This leaves disc n alone on peg A\n2. move disc n from A to C\n3. move n-1 discs from B to C so they sit on disc n\n\nThe above is a recursive algorithm: to carry out steps 1 and 3, apply the same algorithm again for n-1. The entire procedure is a finite number of steps, since at some point the algorithm will be required for n = 1. This step, moving a single disc from peg A to peg B, is trivial. This approach can be given a rigorous mathematical formalism with the theory of dynamic programming, and is often used as an example of recursion when teaching programming.\n\nNon-recursive solution\n\nThe list of moves for a tower being carried from one peg onto another one, as produced by the recursive algorithm has many regularities. When counting the moves starting from 1, the ordinal of the disk to be moved during move m is the number of times m can be divided by 2. Hence every odd move involves the smallest disk. It can also be observed that the smallest disk traverses the pegs f, t, r, f, t, r, etc. for odd height of the tower and traverses the pegs f, r, t, f, r, t, etc. for even height of the tower. This provides the following algorithm, which is easier, carried out by hand, than the recursive algorithm.\nIn alternate moves:\n\n• move the smallest disk to the peg it has not recently come from.\n• move another disk legally (there will be one possibility only)\n\nFor the very first move, the smallest disk goes to peg t if h is odd and to peg r if h is even.\nAlso observe that:\n\n• Disks whose ordinals have even parity move in the same sense as the smallest disk.\n• Disks whose ordinals have odd parity move in opposite sense.\n• If h is even, the remaining third peg during successive moves is t, r, f, t, r, f, etc.\n• If h is odd, the remaining third peg during successive moves is r, t, f, r, t, f, etc.\n\nWith this knowledge, a set of disks in the middle of an optimal solution can be recovered with no more state information than the positions of each disk:\n\n• Call the moves detailed above a disk's 'natural' move.\n• Examine the smallest top disk that is not disk 0, and note what its only (legal) move would be: (if there is no such disc, then we are either at the first or last move).\n• If that move is the disk's 'natural' move, then the disc has not been moved since the last disc 0 move, and that move should be taken.\n• If that move is not the disk's 'natural' move, then move disk 0.\n\n•<|endoftext|>"},"score":{"kind":"number","value":4.40625,"string":"4.40625"}}},{"rowIdx":1297,"cells":{"token_count":{"kind":"number","value":611,"string":"611"},"text":{"kind":"string","value":"# Calculations topic – Dosage and unit conversions\n\n#### Dosage conversions\n\nDosages are sometimes written on prescriptions in a way that is unhelpful to the patient and as a pharmacist, we have to convert them into a more patient friendly version.\n\nAs an example:\nFolic acid solution, 3 mg each day\nSince the medicine is a liquid, it would be inappropriate for us to label the medicine as 3 mg each day.\nWe need to know the strength of the liquid, in this case 2.5 mg/5mL, and then calculate the volume to give.\n\nYou can use a stepwise process:\n\n2.5 mg in 5 mL Divide each side by 5\n0.5 mg in 1 mL Multiply each side by 6\n3 mg in 6 mL\n\nOr by a simple spotting technique where you can see that 3 mg is 20% (or 1/5) bigger than 2.5 mg so you will need to increase the volume in the same way\n\n5 mL + 20% = 6 ml\n\nYou would follow a similar process in other situations, for example\n\nWhen the strength of a product is in units/ mL, the dose is prescribed in units but the patient will be measuring out a volume\n\n#### Unit conversions\n\nWe tend in this country to use metric units in medicine.\n\nWhere this is the case, it is rare to need to convert except between different scales of units:\n\n##### Weight – units getting smaller\n x 1000 x1000 x1000 x1000 Kg —–> g —–> mg —–> microgram —–> nanogram\n##### Weight – units getting bigger\n ÷1000 ÷1000 ÷1000 ÷1000 nanogran —–> microgram —–> mg —–> g —–> kg\n\nSo for the same weight of product, the larger the unit the smaller the number\n\n#### Volumes\n\nThis works just the same way, as the weights but we only tend to change from:\n\n• L to mL, multiply the number by 1000\n• mL to L, divide the number by 1000\n\nA person’s height and weight – you may need to do this to calculate BMI\n\nMany people still measure their height and weight in imperial measurements, so\n\nThere are:\n\n• 12 inches in one foot\n• 3 feet in one yard\n• 1 inch is equivalent to 2.54 cm (to two decimal places)\n\nSo take an imperial height measurement, convert it to inches and then multiply the number of inches to get the measurement in cm\n\nThere are also:\n\n• 16 ounces in one pound\n• 14 pounds in one stone\n• 1 pound is equivalent to 0.453592 kg (to six decimal places)\n\nSo take an imperial weight measurement, convert it to pounds and then multiply the number of pounds by 0.453592 to get the measurement in kg<|endoftext|>"},"score":{"kind":"number","value":4.40625,"string":"4.40625"}}},{"rowIdx":1298,"cells":{"token_count":{"kind":"number","value":357,"string":"357"},"text":{"kind":"string","value":"In analyzing the data from physics experiments, we need to deal with measurement that are discrete or continuous in nature. Our aim is to make inferences about the models that we believe appropriately describe the physical situation, and/or, within a given model, to determine the values of relevant physics quantities. Thus, we need the probability rules that apply to uncertain variables, whether they are discrete or continuous. The rules for complete classes described in the preceding section clearly apply directly to discrete variables. With only slight changes, the same rules also apply to continuous variables because they may be thought of as a limit of discrete variables, as interval between possible discrete values goes to zero.\nFor a discrete variable , the expression , which is called a probability function, has the interpretation in terms of the probability of the proposition , where is true when the value of the variable is equal to . In the case of continuous variables, we use the same notation, but with the meaning of a probability density function (pdf). So , in terms of a proposition, is the probability , where is true when the value of the variable lies in the range of to . In general, the meaning is clear from the context; otherwise it should be stated. Probabilities involving more than one variable, like , have the meaning of the probability of a logical product; they are usually called joint probabilities.\nTable 1 summarizes useful formulae for discrete and continuous variables. The interpretation and use of these relations in Bayesian inference will be illustrated in the following sections.\n|discrete variables||continuous variables|\n|moment of order|\n|A function such that , or , is called improper. Improper|\n|functions are often used to describe relative beliefs about the possible values of a variable.|<|endoftext|>"},"score":{"kind":"number","value":3.984375,"string":"3.984375"}}},{"rowIdx":1299,"cells":{"token_count":{"kind":"number","value":973,"string":"973"},"text":{"kind":"string","value":"# Video: GCSE Mathematics Foundation Tier Pack 1 • Paper 1 • Question 1\n\nGCSE Mathematics Foundation Tier Pack 1 • Paper 1 • Question 1\n\n04:03\n\n### Video Transcript\n\nShown are four numbers. Three, negative three, eight, negative eight. Use two of these numbers to finish the calculation. Something plus something equals negative five.\n\nSo looking at the numbers, straightaway, we can actually rule out two pairs of numbers. And we can do that if we actually just consider the five in the answer. So we disregard the negative and just look at the five.\n\nWe want the two numbers when actually added together to give us an answer that has a five in it. So therefore, we cannot have the two pairs of numbers that are three and negative three or eight and negative eight. And that’s because neither of these will actually give us a result that involves a five. Okay, so that’s great. We’ve ruled out two pairs.\n\nNow, let’s see what other pairs are available. Well, the other pairs that are actually available to us are these four here. Remember that actually when it’s addition, we can actually have it either way around; it won’t affect it. We have three add eight, three add negative eight, negative three add eight, or negative three add negative eight.\n\nWell, straightaway, we can actually rule out three add eight. Well, it’s gonna be 11. So we know that that can’t be right because it won’t give us negative five. But we could also rule it out straightaway because if we think of it that if we’ve got a positive add a positive, then therefore our answer is gonna be positive. But we need our answer to be negative because it’s negative five.\n\nOkay, so let’s take a look at the other three pairs that we’ve got. So next, we have three add negative eight. Well, the first thing to note about three add negative eight is that when we actually have a plus and then a minus, then therefore what it actually become is a negative. So it actually become three minus eight.\n\nSo now that we know that the sum is gonna be three minus eight, let’s think about what this will look like on our number line. So we know we’re going left on our number line. And that’s because if you’re subtracting one positive number from another, then we know that we’re gonna go left on our number line. So we can see if we actually subtract eight from three, then we go down eight spaces.\n\nSo when we do that, what we’re gonna do is get to negative five, which is what we’re looking forward. So this is looking promising. So we can say that three add negative eight is equal to negative five and just to remind you it’s because three add negative eight turns into three subtract eight.\n\nOkay, great, let’s move on to the next couple just to double check. So now, we’re actually gonna stop on negative three. And if we’re gonna add on eight, we’re gonna move eight spaces to the right. And when we do that, we’re gonna get to positive five. So therefore, we can say that negative three add eight is gonna be equal to five. So this is incorrect cause this is not what we’re looking for.\n\nOkay, finally, we move on to our last pair. So now, for the last pair, we have negative three add negative eight. And as the same as the one previously, we know that if we’re gonna add a negative, it’s the same as subtracting the value. So it’s gonna be negative three minus eight. And therefore, as it’s negative three and minus eight, we’re gonna move eight spaces to the left. So we go from negative three, count eight spaces, and we get to negative 11, which again isn’t the value we’re looking for.\n\nOkay, great, so we now know that the pair of our numbers is definitely gonna be three and negative eight. And we can just double check that cause we’ve already shown that three add negative eight is equal to negative five. Well, as we’re dealing with addition, it shouldn’t matter which way around it is. But we just double check. We’ve got negative eight add three.\n\nSo therefore, as we’re adding on three, we’re gonna be moving to the right. We can see if you move three places to the right, we get to negative five. So therefore, we can definitely say that if there are four numbers, three, negative three, eight, and negative eight, and we need to use two of them to finish the calculation something plus something is equal to negative five, then the two numbers are negative eight and three.<|endoftext|>"},"score":{"kind":"number","value":4.6875,"string":"4.6875"}}}],"truncated":false,"partial":false},"paginationData":{"pageIndex":12,"numItemsPerPage":100,"numTotalItems":2000000,"offset":1200,"length":100}},"jwt":"eyJhbGciOiJFZERTQSJ9.eyJyZWFkIjp0cnVlLCJwZXJtaXNzaW9ucyI6eyJyZXBvLmNvbnRlbnQucmVhZCI6dHJ1ZX0sImlhdCI6MTc1NjQ4NTgzNSwic3ViIjoiL2RhdGFzZXRzL3Fpbmd5MjAyNC9EcmFmdFdlYi12MSIsImV4cCI6MTc1NjQ4OTQzNSwiaXNzIjoiaHR0cHM6Ly9odWdnaW5nZmFjZS5jbyJ9.CyLUmQI0Iluu9hqnTlrMbgsTTv8YNNunw_1Ekr15_QzwpF_Th1zhtFmsEepJSqemN2RtqPbW2vkfJKOLaoLcDA","displayUrls":true},"discussionsStats":{"closed":1,"open":0,"total":1},"fullWidth":true,"hasGatedAccess":true,"hasFullAccess":true,"isEmbedded":false,"savedQueries":{"community":[],"user":[]}}">
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The main purpose of the sculpture of the French Romanesque was to serve as a decorative element. There are very few direct references to sculptors of the time, like Gislebertus or Gillebert who probably trained in Cluny and who around 1135 carved most of the sculptures of the cathedral of St. Lazarus in Autun including its marvelous tympanum*. The history of the origins of medieval French sculpture is as important as that of the medieval French architecture: one reason for this is that, in the peak of the Romanesque, French sculpture reached a perfection similar to that of the archaic Greek sculpture. Later, with the passing of time, French Romanesque sculpture evolved into the lavish facades of the Gothic cathedrals covered with beautiful images. Little is known about the origins of French sculpture, perhaps even less than that of ancient Greek sculpture. The first known group of French Romanesque sculpture dates back to the early twelfth century, though its origins are not clearly known. There was probably a traditional school of French sculpture during the ninth and tenth centuries, but there is no trace of this. In the Carolingian period, especially during the reigns of Charlemagne and Louis the Pious, there were artists able to sculpt wonders in ivory and to carve metals with as much or more artistry than that of the Roman or Byzantine sculptors. From these wonderful works of the Carolingian sculpture, some of its techniques must have survived until the dark and confusing times immediately preceding the year 1000. The truth is that all known buildings built before the millennium have little or no sculptured decoration. Their facades were only decorated with blind arches and pilasters. These elements divided walls in sections resulting in surfaces with purely geometric value and a rigorous but very simple aesthetic. Figurative forms did not decorate Romanesque monuments until well into the eleventh century. The first dated artifact usually mentioned belonging to the French Romanesque sculpture is a lintel decorated with rough figures from the Cathedral of Saint Genis les Fonts in Roussillon, which bears an inscription from the year 1020. But this wasn’t a French site and this artifact appears to be a mere local art product. Instead, in the great royal monastery of Fleury in Saint-Benoît-sur-Loire, there is a tower built by Gauzelin in the year 1030 which bears ambitious historiated capitals* decorated with apocalyptic figures inspired by the Mozarabic Beatus. From this time on, the French sculpture progressed very fast until the end of XI century when great decorative works of sculpture appeared. These works included facade’s reliefs in an archaic style, but full of strength and dramatic intention, as well as capitals decorated with beautiful foliage, friezes located on the churches’ naves, and even some free standing statues. The French Romanesque schools of sculpture developed around three centers of great intellectual activity. One was the Royal Domain which began with the works of Gauzelin in Fleury; another school was formed at the court of Toulouse; and a third originated as a result of the great artistic works produced by the Order of Cluny. These schools mixed between each other too: there are important sculptures that cannot be assigned to any particular school, there are also monumental decorations so independent in style that make them look as if they were the work of foreign artists. Some sculptors signed their work on the abacus of a capital or in a separate plate that was later applied to the monument: we have said before how Gislebertus signed his facade at Autun, but all that’s known about these artists are their names. From the study of these commemorative inscriptions it appears that these artists were laymen, contrary to the Gothic painters and sculptors who were mostly theologians. The most widely used theme to decorate a church’s facade was the view of the Almighty as the Pantocrator, the Son of God. He appears sitting on a throne inside the almond-shaped halo of the rainbow (or mandorla*) accompanied by the four symbolic animals or Tetramorphs as they were seen by the ancient prophets and described in the Book of Revelation. It is a scene that, when there was enough space available, included the 24 elder kings, each carrying the offertory cup and a violin or zither to sing praises to the Almighty. This scene was sculpted in the tympanum of St-Pierre de Moissac from around 1130. Sometimes sculptors added two archangels raising their prayers plus the Apostles and the Virgin, because they also saw the Lord glorified in the act of ascending to Heaven. This complicated vision, appearance or Theophany* was often reduced to the halo with the Lord on his throne and the four symbolic animals: in this way He loses his prophetic and apocalyptic character to became a synthesis of the history of the King of Heaven, just simply as He manifested to the patriarchs and prophets and as He will appear to judge the living and the dead. Besides representing Jesus as a deified Pantocrator, French Romanesque artists also portrayed him during episodes of his human earthly life. During the Romanesque, the almost exclusive image of the Son of Man paralleled the importance or near monopoly that images of Mary will later have for Gothic sculpture and painting. In the Gothic period, episodes of the Virgin’s life based on the apocryphal Gospels replaced in large part the representations of her Son Jesus. In the Romanesque period, on the contrary, the dominant topic was the Christology: the assumptions of the Old Testament were understood to illustrate parallel episodes of the life of Jesus. The sacrifice of Abel alluded to the Golgotha, Abraham sacrificing Isaac, Samson dying while Dagon’s temple was crumbling, David as the pillar of the race from which the Messiah had to be born, all these episodes alluded to Jesus the King Redeemer. From the Gospels, artists were only interested in episodes that directly included Jesus. Mary was His Mother, the Theotokos*, and so was always represented carrying the Child in Her arms. Though Romanesque themes were identical in content, they were repeated with great stylistic variation. The French Romanesque sculpture reached a superb formal and technical perfection, as well as a great ability to compose complicated scenes within the peculiar spaces dictated by the design and disposition of the facades’ surfaces. French Romanesque sculpture was also peculiar for further advancing in the representation of the unreal. The great works of French Romanesque sculpture of the eleventh and the first half of the twelfth centuries (like the marvelous facades of Moissac, Vezelay and Autun), are full of expressionism, even more that’s shown in the works of many modern artists. The fact of overstepping the figures’ proportions by dislocating their members to prove with impossible gestures their heavenly and semi-divine character doesn’t imply that the figures of these Romanesque facades are a morbid or a degenerate form of art. At best they can be described as “baroque” Romanesque because they pushed the boundaries of moderation in their decorative style, a situation that leads to “baroquism” in all artistic styles. Vezelay is considered today one of the most sublime artistic Western monuments because of the extraordinary tympanum of its facade (from ca. 1120) in which a tall Pantocrator ascends to heaven between the Apostles. In Saint Lazarus of Autun, the tympanum describes an impressive Final Judgment with horrible demons and angels with endless bodies. It is believed that some Romanesque sculptors from Central France could have been involved in the decoration of the oldest parts of the first Gothic cathedrals. It is surprising to find in the Cathedral of Chartres rigid caryatids, incredibly tall, with their mantles’ folds falling vertical like column’s grooves while their faces show huge vitality. Other similar statues from the porch of the royal abbey of Saint-Denis have even more exaggerated proportions. Between them, there are figures of tall young queens with braids reaching their feet parallel to their dresses’ folds, they look so artificial that seem more columns than statues. In the region of Reims prevailed an aristocratic and refined style, which seems derived from the Carolingian art that evolved earlier in that collegiate cathedral. Other French Romanesque sculptors were characterized by a great naturalism, especially those of the Toulouse School. The initial literary humanism manifested in Toulouse by the troubadour poetry produced similar effects in the visual arts. Proof of this are the beautiful capitals of the cloister of San Pedro de Moissac, from late eleventh century, and the gentle relieves from the Gate Miégeville at the Basilica of Saint Sernin, also in Toulouse. In many of these relief decorations, the purely decorative ornamentation representing grape vines, acanthus leaves, and fretwork was sometimes accentuated by the use of trepan over intense black backgrounds. The Oriental influence was continuous during the Romanesque period, not only coming from Syria and Byzantium, but also from the Far East. There are some reliefs that decorate the nave of the cathedral of Bayeux that seem to be inspired by the art of India. *Historiated Capital: A capital which is decorated with figures of animals, birds, or humans, used either alone or combined with foliage. Sometimes these figures don’t have any meaning, although they may be symbolic or part of a narrative sequence. *Mandorla: Also an aureole, the radiance of luminous cloud which, in portrayals of sacred personages, surrounds the whole figure. *Theophany: From Ancient Greek meaning “appearance of a god”. Refers to the appearance of a deity to a human or other being. The term theophany also has acquired a specific usage for Christians and Jews with respect to the Bible: It refers to the manifestation of God to man; the sensible sign by which the presence of God is revealed. *Theotokos: Refers to the Greek title of Mary, the mother of Jesus used especially in the Eastern Orthodox, Oriental Orthodox, and Eastern Catholic Churches. Its literal English translations include “God-bearer”, “Birth-Giver of God” and “the one who gives birth to God.” *Tympanum: A semi-circular or triangular decorative wall surface over an entrance, bounded by a lintel and arch. It often contains sculpture or other imagery or ornaments. In Christian architecture a tympanum usually contains religious imagery, and thus are usually the most important location for monumental sculpture on the outside of the temple. In classical architecture, and in classicising styles from the Renaissance onwards, tympana were usually triangular; in Romanesque architecture they were in a semi-circular shape, and in Gothic architecture they had a more vertical shape coming to a point at the top.<|endoftext|>
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History of India conveys that the countries which came to India for TRADE, were such as Portuguese, the Danes, the Dutch, the French but only the British succeeded. Owing to luxuries and weak military position of the Mughal Raj, the British won in the wars with fulcrum of the local rulers. Delhi was under Mughul rule between 1605 -1858. Jahangir 1605-1627, Shah Jahan up to 1658, Aurangzeb up to 1707, Bahadur Shah II 1837- 1857. The last Mughal emperor Bahadur Shah II was deposed in 1858 by the British East India Company and exiled to Burma following the War of 1857 after the fall of Delhi to the “company troops”. A perusal of world history, it evinces that if any nation conquered the other nation, they enforced their own system of administration. Similarly, the British rulers changed the whole administration of India…except PUNJAB. A new “law & justice” enforced on BENGAL-MADRAS -U.P. & other territories which came under their controls. Hardly after 20 years of Maharaja Ranjit Singh’s death, had the whole continent of India passed into the hands of East India Company. India historically witnessed development of judicial system from the Mughul period to commercial British & than to amalgamate Politics of India. The British consolidated India and after taking direct charge, governed for a period of almost about two centuries. They promoted a different model of the British judicial system as Indian judicial system…. The OBJECTIVITY was completely different for subjects of Colonial countries than of… direct U.K. citizens. They facilitated different judicial model in post 1947 India. It is strongly believed that new Indian legal compositions could not achieve objectivity, neither identified them being a “COURT OF JUSTICE” nor as a “COURT OF LAW”. The judges, in pre-independence India … were the symbol of imperial power, addressed by colonial subjects a … His LORDSHIP or His MAJESTY. ADVOCATES & ETHICAL VALUES; Some of the ETHICAL standards desired to be maintained by Advocates in India included; being a member of the legal profession has a social duty to show the people a beacon of light by his conduct and actions, to play a vital role in the preservation of society and justice system. They are under an obligation to uphold the rule of law and ensure that the PUBLIC JUSTICE SYSTEM is enabled to function at its full potential. But new Indian culture has inculcated serious virus in this profession in competition to outwit each other by commercially higher & higher earning without segregation for genuine client & criminality of a client. The “Code of Criminal Procedure, 1973”, containing XXXVI chapters, 484 sections, was positioned to play a Principal role in … The criminal justice dispensation. The Cr. PC being a principal instrument highly relies upon system of EVIDENCE. … This is to be collected -synchronized -constructed and brought before Judicial Magistracy by the executive Police system, who in turn function under the… Political Governance. The British Era of 200 hundreds had provision for participation of Jury team in the superior courts. Jury team used to comprise 12 to 18-23 members for contribution/ constructing judicial verdict by using larger number of minds achieving ultimate objectivity of Justice dispensations. It is an established natural fact .. that an individual will have its own limitations of thought process in a given situation and a team of professionally competent citizens, enjoying impeccable social record, could play significant role in justice dispensation… The Executive & Judicial system in Punjab under the rule of Maharaja Ranjit Singh between 1799 – 1839 … proved to be one of the best models in the world. The Law & Order was so affective, that not a single individual qualified for an award of capital (death) punishment in his regime. JURY:… The Indian General public is not much aware of the values under this concept. .. The Jurors are selected from a jury pool formed for a specified period of time from list of citizens living in the jurisdiction of the court… A jury is a sworn body of people convened to render an impartial verdict (a finding of fact on a question) officially submitted to them by a court, or to set a penalty or judgment. In each court district where a grand jury is required, a group of citizens hold an inquiry on criminal complaints brought by the prosecutor deciding, whether a trial is warranted, in which case, an indictment is issued. .. Juries were formerly used in India up … until KM Nanavati v State of Maharashtra (1959), which was used for the abolition of jury trials. The famed case, had Mr K.M. Nanavati who was tried for the murder of his wife Sylvia’s paramour, Prem Ahuja. The Government encountered unprecedented media coverage and single case was referred for closing chapter of jury trial systems in India! The advantages of a jury… system were never highlighted judicially by “new young India” for their own personalized-individualistic wisdom. The Juries otherwise represented the common public and therefore were judging in line with generally accepted values of the society. Discussions among juries could lead to more thorough consideration of all aspects of the case. It is stated that it may have been difficult to corrupt.. 12 Jurors than to one/two judges. A few disadvantages of a jury system brought forward were technically not that significant for abolishing the JURY system, in reality there may be many other political reasons. The fault line should have been considered to be corrected in Jury system. … Juries allegedly could be swayed by the current prejudices in the society, which are not supported by law but this can happen more dangerously with single judge. … British had reasons for Jury system involving local team of Jurist to guide “English born” in better decision making but New India wanted to bring in new Governance chemistry. The mind of new Indian leadership may have found better scope and convenience to influence single/double judge than poaching for more than a dozen Jury members. Hence the case 1959 appears to be just merely an excuse for own agenda to misdirect public debate. The wish of Judicial working on the lines of British Lordships is good dream but .. English borne are heard to have displayed better – superior justice orientations. The mother of criminal Justice … based on Cr.PC has encountered numerous fault lines in practical ground operations in India. The statement u/s 161 is recorded by Police but is not admissible in the court. The district Police head is delegated Powers of District Magistrate in discharge of public duties while presenting Challan in the court. .. Police function under Political Governance system is another fact The u/s 164 provide for recording of statement before Magistrate but .. again Police continue to have major role to be played in facilitation of statement recordings. The role of FIR, event site map, Inquest report, arrest of accused, recovery of weapon, medical examination, case summary.. are all within domain of POLICE and… will always be found playing suspicious role for award of punishment to the culprit. The Punjab Police system initially had … Prosecuting Police Inspector different than Investigating Police Inspector. Now both jobs are designed to be carried out by one Police -Station House Officer. How an SHO is compelled to work in a particular political environment… is well known to everyone. The legal personalities known as advocate are sworn to facilitate JUSTICE assisting COURT for fulfillment of the holistic duty. … Society knows the actual levels reached by advocates defending worst to worst for the .. sake of MONEY though the concept of justice is lost at that very stage. There are hardly …. rarest of rare when an advocate is heard to have declined case on the basis of known criminality of a person. The techniques found out by them kills reformist actions of well meaning judicial authorities. India introduced summary trial u/s 138 under negotiable instrument act to be settled in less than a year’s time. This too has been abused to be reduced to other cases consuming time. One of sample case for this act took 11 years with delaying tactic invented by ….. The .. opposite counsel The abolishing of Jury system prima facie appears to have been bad action in 1961, the introduction of Cr.PC /1973 thought to be delivering JUSTICE also failed …allegedly by Police or by Political Governance but ….who are they in all institutions, our own known …friends…..relatives… put on vicious cycle of money…power race at the behest of ever increasing consumer demands of families and partially satisfying ego to down another family or family member of the society. .. Sir to lords to Majesties got their … cause list justifying trendzs of distortions in the ethic of society. PUNJAB was not known what all is being … talked-heard-seen moving fast to compromising so many for small life without enough resistance. …Worthy court will take notice for judgment on the material presented before them by the Police. The POLICE can help in minimizing the crime graph, but cannot stop them unless collective consciousness do not wake up to the HERITAGE VALUES.. discouraging younger generations not to follow PATH OF DESTRUCTION despite having numerous life challenges. The new generations are in much better conditions once faced by elders during JOURNEY OF LIFE; living -hiding in forest for the sake of certain principals taught by Guru Nanak Sahib to Guru Gobind Singh ji..to Baba Banda Singh Bahadur. Let this simple thought process be shared with co citizens for WAKE UP CALL, suggesting the worthy COURTS to be dispensing………………….better JUSTICE and assuming the work being COURT OF LAW & JUSTICE … for the sake of saving humanity. Article Posted on babaushahi.com : 2015-04-22 By Nirmal Singh Keerka, Punjab Heritage & Education Foundation Email : [email protected]<|endoftext|>
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Unitary Method: A method, in which the value of a unit quantity is first obtained to find the value of any other required quantity, is called unitary method. The basic rules applied in this method are: 1. To get more, we multiply 2. To get less, we divide In solving problems based on unitary method, we come across two types of variations: 1. Direct Variation 2. Indirect Variation Direct Variation: If two variables are so related to each other that an increase (or decrease) in the first causes an increase (or decrease) in the second, then the two variables are said to vary directly. They are directly proportion to each other. Examples: 1. More articles, more cost 2. More men, more will be the work done 3. More working hours, more will be the work done Rule for Solving Problems Based on Direct Variation: When, the value of a certain quantity of a variable is given then first ·we divide the given value by the given quantity to find the value of one unit. Now to find the value of a certain quantity, we multiply this quantity with the value of one unit. Example: 10 pens cost 150 Rupees. Therefore one pen would cost (150/10 = 15 Rupees). Therefore if you have to find the cost of 2 pens, it would be 2 multiplied by 15 which would be 30 Rupees. Indirect Variation: If two variables are so related to each other that an increase (or decrease) in the first causes a decrease (or increase) in the other, then the two variables are said to vary indirectly or are inversely proportional. Examples: 1. More men would take less time to complete the job. 2. High speed would reduce the time taken to reach the destination. Rule for Solving Problems Based on Direct Variation: When the value of a certain quantity of a variable is given, then first we multiply the given value by the given quantity to find the value of one unit. Now, to find the value of certain quantity we divide the value of one unity by this quantity. Example: Let us say that 10 men finish the work in 10 days. Therefore, one man can finish the work in 10 x 10 = 100 days. i.e. total work to be completed is 100 man days. Hence, if we were to put 50 men, then the work would get done in (100/50 = 2 days).<|endoftext|>
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Sie sind auf Seite 1von 3 # Chapter 27 Examples Gauss Law Example 1: Calculate the electric eld due to an innite line of charge. Use Gauss Law. Solution: Draw a gure. ~ dA ~ electric ux through the caps = 0, since E electric ux through the side surface is E = E From Gauss Law: And since qenclosed = h then E = E 2 r h = Thus: E 2 r h = E= qenclosed h = 0 0 h 0 Z ds = E S = E 2 r h E = qenclosed 0 2 0 r 2 Example 2: Calculate the electric eld due to an innite nonconducting sheet of charge. Solution: Draw a gure. We will use Gauss Law. 0 so E= Z ~ dA ~ = qenclosed E 20 0 (EA + EA) = A Example 3: Calculate the electric eld due to an innite conducting charged plate. Solution: Draw a gure. We will use Gauss Law. 0 so E= Z ~ dA ~ = qenclosed E 0 0 (EA + 0) = A Example 4: Calculate the electric eld due to two innite nonconducting oppositely charged sheets brought together to a distance a. Q1 : if the magnitude of the positive surface charge density + is twice larger than the negative charge density (i.e., + = 2 ). Q2 : if the magnitudes of charge densities are the same (i.e., + = ). Solution: Draw a gure. We will use Gauss Law. Using the principle of superposition: Q1 : Elef t Eright of + plane = = + + + /2 + + = = 20 20 20 2 0 40 + + /2 + + + = + = 20 20 20 20 40 + + + /2 3+ + = + = 20 20 20 20 40 of plane Ebetween Q2 : planes Ebetween planes ## + + = =0 20 20 20 20 + Eright of plane = + = =0 20 20 20 20 + = + = + = it is pointing 20 20 20 20 0 Elef t of + plane Example 5: Calculate the electric eld due to two innite conducting oppositely charged plates brought together to a distance a. Solution: Draw a gure. We will use Gauss Law. Using the principle of superposition: Eleft of + plane = 0 = Eright = of plane Ebetween planes + = 2 0 2 0 0 it is pointing 2 Example 6 Two coaxial cylinders of innite length are charged with +2 C/m on the inner and 1 C/m on the outer. The inner cylinder has radius 1 m and the outer has radius 2 m. What is the electric eld everywhere? Solution: Draw a gure. We will use Gauss Law. We will form coaxial cylinders for our Gaussian surfaces. We will make them 1 m in length, but by the symmetry resulting from having innite cylinders, the ux through the ends is zero. First, form a Gaussian surface that is a cylinder coaxial with the other two, and inside the inner cylinder. It encloses no charge, so E = 0. I.e., E = 0 for r < 1 m. Next, form a Gaussian surface that is a cylinder coaxial between the others with radius r. It encloses charge of +2 C, since it is 1 m in length. By Gauss Law, (2 C) (2 C) 3.6 1010 N m q 1 = = = E= 0 A 0 (2 r)(1 m) (8.85 1012 C2 /N m2 )2(3.14)(1 m) r r C I.e., for 1 m < r < 2 m, E = (3.6 1010 N m/C)/r. Last, form a Gaussian surface that is a cylinder coaxial and outside the outer cylinder, and again 1 m in length. It encloses net charge of (2 C + (1 C) = 1 C), so: E= q 1 (1 C) (1 C) 1.8 1010 N m = = = 0 A 0 (2 r)(1 m) (8.85 1012 C2 /N m2 )2(3.14)(1 m) r r C 2<|endoftext|>
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Desert biomes receive very little rain and cover about one-fifth of the planet's surface. They are divided into four sub-habitats based on their location, aridity, climate and temperature: arid deserts, semi-arid deserts, coastal deserts and cold deserts. ● Arid deserts are hot and dry and are located at low latitudes throughout the world. Temperatures are warm all year and hottest during the summer. Arid deserts receive little rainfall, and most rain that does fall usually evaporates. Arid deserts are located in North America, South America, Central America, Africa, Australia and Southern Asia. ● Semi-arid deserts are usually not as hot and dry as arid deserts. They have long, dry summers and cool winters with some rain. Semi arid deserts are found in North America, Europe, Asia, Newfoundland and Greenland. ● Coastal deserts are usually located on the western edges of continents at approximately 23°N and 23°S latitude, the Tropic of Cancer and the Tropic of Capricorn. Cold ocean currents run parallel to the coast, producing heavy fogs. Despite high humidity in coastal deserts, it rarely rains. ● Cold deserts have low temperatures and long winters and are found above the treelines of mountain ranges and in the Arctic and Antarctic. They experience more rain than other deserts. Many locations of the tundra are cold deserts. Desert animals include coyotes, kangaroo rats, spiders, meerkats, roadrunners, reptiles, toads, snakes, pronghorn, birds and bats. Dry and baron landscapes, deserts receive intense sunshine and little rain. They are places of extremes, with a greater range of temperatures throughout the day than any other habitats. Temperatures range from boiling in the middle of the day, to freezing at night. The two main types of deserts are true deserts (hot deserts) and semi-deserts. True deserts are located on either side of the tropics. Semi-deserts occur on every continent, usually far from the tropics. Semi-deserts receive at least twice as much rain each year than true deserts. Deserts are formed from large fluctuations in temperature between day and night which puts strain on rocks. The stress causes the rocks to break into pieces. Occasional downpours of rain cause flash floods. The rain falling on hot rocks can cause them to shatter. The rubble is strewn over the ground and further eroded by the wind. Wind-blown sand grains further break down stones, causing more sand. Rocks are smoothed down, and the wind sorts sand into deposits. The grains end up as sheets of sand. Other deserts are flat, stony plains where all the fine material has been blown away leaving an area of smooth stones. These deserts are called desert pavements and little further erosion takes place. Some deserts include rock outcrops, exposed bedrock and clays once deposited by flowing water. Oases occur where there are underground sources of water in the form of springs and seepages from aquifers. A unique desert is the Gobi desert in Asia which is located across China and stretches up to the Siberian Mountains where winters are very cold. Despite the cold winters, the mountains block rain-clouds from reaching the desert area. A variety of plants and animals live in desert habitats. Plants tend to be tough and wiry with small or no leaves. Some plants germinate, bloom and die in the course of a few weeks after rainfall. Some long-lived plants survive for years with deep roots that tap into underground moisture. Most animals are nocturnal, coming above ground or out of the shade at night when temperatures are cooler. Reptiles, insects and small birds are the most common animals in true deserts. Mammals are more common in semi-deserts, where plant life is more plentiful. Animals of the deserts are adapted to dry and arid conditions. They are efficient at conserving water, extracting most of their needs from their food and concentrating their urine. The addax antelope, dik-dik, Grant's gazelle and oryx never need to drink. The thorny devil in Australia sucks water through channels in its body located from its feet to its mouth. The camel minimizes its water loss by producing concentrated urine and dry dung, and is able to lose 40% of its body weight through water loss without dying of dehydration. Camels have humps of fatty tissue that concentrate body fat in one area, minimizing the insulating effect fat would have if distributed over their whole bodies. Birds are able to fly to areas of greater food availability as the desert blooms after local rainfall, and can fly to faraway waterholes. Carnivores obtain much of their water needs from the body fluids of their prey. Flies, beetles, ants, termites, locusts, millipedes, scorpions and spiders have hard cuticles which are impervious to water and many lay their eggs underground where their young develop away from the surface temperature extremes. Some arthropods make use of the ephemeral pools that form after rain and complete their life cycle in a matter of days. Reptiles do not sweat, so they shelter during the heat of the day. In the first part of the night, as the ground radiates the heat absorbed during the day, they emerge and search for prey. Some snakes move sidewards to navigate high sand-dunes. Even amphibians have adapted to desert habitats, spending the hot dry months in deep burrows where they shed their skins numerous times to create cocoons around them to retain moisture. Some animals remain in a state of dormancy for long periods, becoming active again when the rare rains fall. They then reproduce rapidly while conditions are favorable before returning to dormancy. Deserts habitats have been the least affected by human activities, remaining relatively untouched. Threats do include extraction of oil from the sand and grazing farm animals that deplete desert plants, threatening wildlife that rely on those plants. Desertification can be caused by tilling for agriculture, overgrazing and deforestation. "Dead zone" is a more common term for hypoxia, which refers to a reduced level of oxygen in the water. Hypoxic zones are areas in the ocean of such low oxygen concentration that animal life suffocates and dies, and as a result are sometimes called "dead zones." One of the largest dead zones forms in the Gulf of Mexico every spring. Each spring as farmers fertilize their lands preparing for crop season, rain washes fertilizer off the land and into streams and rivers. Less oxygen dissolved in the water is often referred to as a “dead zone” because most marine life either dies, or, if they are mobile such as fish, leave the area. Habitats that would normally be teeming with life become, essentially, biological deserts. Hypoxic zones can occur naturally, but scientists are concerned about the areas created or enhanced by human activity. There are many physical, chemical, and biological factors that combine to create dead zones, but nutrient pollution is the primary cause of those zones created by humans. Excess nutrients that run off land or are piped as wastewater into rivers and coasts can stimulate an overgrowth of algae, which then sinks and decomposes in the water. The decomposition process consumes oxygen and depletes the supply available to healthy marine life. Dead zones occur in many areas, particularly along the East Coast, the Gulf of Mexico, and the Great Lakes, but there is no part of the world that is immune. The second largest dead zone in the world is located in the U.S., in the northern Gulf of Mexico. There is strong evidence that global sea level is now rising at an increased rate and will continue to rise during this century. A warming climate can cause seawater to expand and ice over land to melt, both of which can cause a rise in sea level. While studies show that sea levels changed little from AD 0 until 1900, sea levels began to climb in the 20th century. The two major causes of global sea-level rise are thermal expansion caused by the warming of the oceans (since water expands as it warms) and the loss of land-based ice (such as glaciers) due to increased melting. First, as the oceans warm due to an increasing global temperature, seawater expands—taking up more space in the ocean basin and causing a rise in water level. The second mechanism is the melting of ice over land, which then adds water to the ocean. Records and research show that sea level has been steadily rising at a rate of 0.04 to 0.1 inches per year since 1900. Since 1992, new methods of satellite altimetry (the measurement of elevation or altitude) indicate a rate of rise of 0.12 inches per year. This is a significantly larger rate than the sea-level rise averaged over the last several thousand years. Even though American consumers throw away about 80 billion pounds of food a year, only about half are aware that food waste is a problem. Even more, researchers have identified that most people perceive benefits to throwing food away, some of which have limited basis in fact. A recent study found that 53 percent of respondents said they were aware that food waste is a problem. This is about 10 percent higher than a previous study, which indicates awareness of the problem could be growing. But it is still amazingly low. If we can increase awareness of the problem, consumers are more likely to increase purposeful action to reduce food waste. You don’t change your behavior if you don’t realize there’s a problem in the first place. Generally, people consider three things regarding food waste. They perceive there are practical benefits, such as a reduced risk of foodborne illness, but at the same time they feel guilty about wasting food. They also know that their behaviors and how they manage their household influences how much food they waste. How Americans Think About Food Waste: Perceived benefits: 68 percent believe that throwing away food after the package date has passed reduces the chance of foodborne illness, and 59 percent believe some food waste is necessary to be sure meals are fresh and flavorful. Feelings of guilt: 77 percent feel a general sense of guilt when throwing away food. At the same time, only 58 percent understand that throwing away food is bad for the environment, and only 42 percent believe wasted food is a major source of wasted money. Control: 51 percent believe it would be difficult to reduce household food waste and 42 percent say they don’t have enough time to worry about it. Still, 53 percent admit they waste more food when they buy in bulk or purchase large quantities during sales. At the same time, 87 percent think they waste less food than similar households. Many people feel they derive some type of benefit by throwing food away, but many of those benefits are not real. For example, they misunderstand “Sell by” and “Use by” dates on food packages. Only in rare circumstances is that date about food safety. Food waste is the largest source of municipal solid waste in the U.S. and the most destructive type of household waste in terms of greenhouse gas emissions. Consumers can help by reducing food waste. Wildlife preservation is informed management of the natural environment to protect and benefit plants and animals. Extinction may occur due to natural causes. However, the actions of people and the growth of human population have all too quickly created a threat to the well being of wildlife. There have been declines in the numbers of some species and extinction of others. The need for conservation was created by human beings. About 2 million years ago, when Homo sapiens first appeared on the earth, their world was biologically rich. Millions of species of plants and animals flourished...from the single celled to the complex. The first humans enjoyed a lush and beautiful environment filled with brilliant color and variety. Every ecosystem harbored life in many forms...from forest to meadow, wetland to desert. These early people chose to decorate their dwellings with paintings of the wildlife that made up their environment. As they evolved and developed belief systems, they used the plants and animals that surrounded them in their rituals. Nature was integrated into their culture. It has played an important part in the way modern man thinks and behaves today. We bring nature into our daily lives. If you have a companion animal, or even a house plant, if you enjoy a landscape painting or a piece of nature photography, or if you visit a park or a nature preserve, you are recognizing the importance of natural elements in your life. The difference we perceive in the range of natural settings, from the beauty of a garden to the desolation of a vacant lot, is determined by the kinds of organisms that each contains and the communities they form. ALL THINGS CONTRIBUTE Few of us would prefer an environment of concrete buildings and asphalt paving to gorgeous coastlines, majestic mountains or peaceful forests. Our pleasure in life would be diminished if only one bird sang, or merely a handful of fish lived in the sea. But our aesthetic appreciation of the wildlife that fills our earth is only one reason to preserve the variety and abundance of species. All living things contribute to the ecology and are vital to its health and continuation. Despite our advances in technology, we as human beings still rely on our environment to provide many of the things necessary to our survival. The earth's biodiversity supports all life, including that of humans. Our food, medicines, energy sources, textiles and building materials are all derived directly or indirectly from living organisms. Our way of life is inextricably linked to the natural world. Plants convert the energy of the sun through photosynthesis into the energy that sustains all life on this planet. Everything we eat can be traced to either a plant or to an animal that lived by eating plants. For this reason, the vegetation on this planet is necessary to our survival. Maintaining a variety of plant forms is crucial. Although the food we consume represents only about 100 kinds of plants, there are countless others we might utilize. As our population increases and land for agricultural use dwindles, we will have to look for other food crops and new ways to grow them. It is important to preserve a variety of plant species with their future use in mind. Almost all of our medicines come from living organisms: some directly as from bacteria or fungi or plants, others are now synthetically made but were originally discovered in their natural form. In China and other parts of the world, medicinal plants in their original form are used as treatment for all kinds of illness. Many of our manufactured pharmaceuticals offer a more controlled use of these plants, but are none the less dependent upon them. Science hopes to identify even more organisms beneficial to the treatment of disease. We have only scratched the surface of the vast number of plant species to be studied. A great discovery could still be found that might change the lives of millions. The study of living things advances our knowledge in all areas. By observing the behavior of the great apes anthropologists learn about prehistoric man. By studying the movements of the creatures and plants of the earth engineers can learn about mechanics. Yet there are organisms that have yet to be scientifically studied. For example, fungi exist in countless numbers and forms. They can be used to preserve food, to produce medicine such as antibiotics without which many lives would be lost and much of the food we eat depends on them. We would have no bread if not for yeast to make it rise, no wine without fermentation. The importance of the organisms around us gains some perspective when we see the practical and economic applications of those organisms. Yet we have explored only a fraction of the species of existing fungi. There are secrets yet to be learned and benefits yet to be gained. If even one species is lost we may have missed a vital opportunity to improve our lives. The one species that perishes might have had the potential to feed entire populations, to cure disease or to provide invaluable knowledge. We must also see beyond our own needs. There is a much larger picture and many ecological reasons to preserve species. Scientists refer to the role played by living things as "ecosystem services." Communities of microbes, plants and animals, along with nonliving environmental features such as soil and water, constitute an ecosystem. Ecosystem services are provided by many species including those that prevent soil erosion or affect the quality of the air, or convert the energy from the sun into food, or influence the climate, and other functions vital to the ecosystem as a whole. Optimally, the earth is self-perpetuating, but its continued ability to be a healthy environment for humans is dependent upon the species that sustain its ecosystems. The forests, wetlands, prairies and deserts are all necessary to its well being. If we continue to allow species to die out, it will become increasingly difficult for these ecosystems to operate successfully and it may become difficult for all living things to survive. The very climate of the earth is dependent on the vital ecosystems that comprise it. The earth's forests perform the vital task of photosynthesis, which removes carbon dioxide from the atmosphere as plants make food. If the forests are cleared and not replaced, our atmosphere will change. TAKING IT FOR GRANTED There is dramatic evidence that the earth's ecology is badly stressed. We have taken the importance of the ecosystem for granted and we are blind and deaf to the signs of the strain. Because plants that hold soil in their roots have been eliminated, about one-fifth of all the topsoil in the world has eroded and is lost. The consequences of this loss are fewer plants, fewer productive farms and therefore less food for animals and humans alike. Understanding and maintaining natural communities is the key to sustaining life on earth. No species is unimportant. They are all part of the system. DOING THE RIGHT THING Beyond the questions of ecology and economics is the ethical issue. What right do we have as one single species to destroy other living things. Human beings began to destroy the other organisms in their environment when they began to practice agriculture more than 10,000 years ago. There were no more than several million people then. With our exploding population the rate of consumption has proportionately increased...about 40 percent of the net biological productivity (what is produced by all living organisms) on the land. We are already taking a disproportionate share of the bounty of the earth. Ecologists believe that we need to respect the value of other organisms and preserve them before we increase that share. These organisms deserve our respect. They support our very lives on the planet. With the development of ever more efficient weapons, humans have been able to kill wildlife with growing efficiency. Hunters have caused several species of animals to perish. For agriculture, industry and for living space we have cleared the forests, drained the wetlands, and dammed the rivers. This encroachment on the environment has negatively impacted vast amounts of plant and animal habitat. What hasn't been destroyed has been disrupted, and the natural processes altered. This affects the diversity and size of wildlife populations in these habitats. Some are no longer connected to their ecosystems. Various species became extinct before there were humans on the earth, but new species developed to replace them. The variety of life continued. Now, however, when people kill off a species there is little hope that it will be replaced. The variety of life is decreasing. Many species of wildlife are gone forever. In North America alone such extinction includes the Carolina parakeet, the passenger pigeon, the California grizzly bear and a birch tree that once flourished in Virginia. An increased interest in conservation began in the late nineteenth century. Many governments passed laws to protect and set aside national parks and reserves for wildlife. It was these efforts that saved the American bison, the pronghorn and many rare plants found in Hawaii and in the Galapagos. Yet several hundred species of animals and thousands of species of plants are still at risk. These include well-loved animals like the Giant Pandas, the Asiatic lion, the Bengal tiger, the blue whale, the mountain gorilla, the whooping crane, the California condor, the Florida panther and all the Asian rhinoceroses. The St. Helena redwood, the black cabbage tree, the Ozark chestnut and several kinds of California manzanitas face extinction as well. Wetland conservation is aimed at protecting and preserving areas where water exists at or near the earth's surface, such as swamps, marshes and bogs. Wetlands cover at least 6% of the earth and have become a focal issue for conservation due to the 'ecosystem services' they provide. More than three billion people, around half the world’s population, obtain their basic water needs from inland freshwater wetlands. The same number of people rely on rice as their staple food, a crop grown largely in natural and artificial wetlands. In some parts of the world, such as the Kilombero wetland in Tanzania, almost the entire local population relies on wetland cultivation for their livelihoods. In addition to food, wetlands supply fiber, fuel and medicinal plants. They also provide valuable ecosystems for birds and other aquatic creatures, help reduce the damaging impact of floods, control pollution and regulate the climate. From economic importance, to esthetics, the reasons for conserving wetlands have become numerous over the past few decades. The main functions performed by wetlands are water filtration, water storage, biological productivity, and habitat for wildlife. Wetlands aid in water filtration by removing excess nutrients, slowing the water allowing particulates to settle out of the water which can then be absorbed into plant roots. Studies have shown that up to 92% of phosphorus and 95% of nitrogen can be removed from passing water through a wetland. Wetlands also let pollutants settle and stick to soil particles, up to 70% of sediments in runoff. Some wetland plants have even been found with accumulations of heavy metals more than 100,000 times that of the surrounding waters' concentration. Without these functions, the waterways would continually increase their nutrient and pollutant load, leading to an isolated deposit of high concentrations further down the line. An example of such a situation is the Mississippi River’s dead zone, an area where nutrient excess has led to large amounts of surface algae which use up the oxygen and create hypoxic conditions (very low levels of oxygen). Wetlands can even filter out and absorb harmful bacteria from the water. Their complex food chain hosts various microbes and bacteria, which invertebrates feed on. These invertebrates can filter up to 90% of bacteria out of the water this way. Wetlands can store approximately 1-1.5 million gallons of floodwater per acre. When you combine that with the approximate total acres of wetlands in the United States (107.7 million acres), you get an approximate total of 107.7 - 161.6 million million gallons of floodwater US wetlands can store. By storing and slowing water, wetlands allow groundwater to be recharged. And combining the ability of wetlands to store and slow down water with their ability to filter out sediments, wetlands serve as strong erosion buffers. Through wetlands ability to absorb nutrients, they are able to be highly biologically productive (able to produce biomass quickly). Freshwater wetlands are even comparable to tropical rainforests in plant productivity. Their ability to efficiently create biomass may become important to the development of alternative energy sources. While wetlands only cover around 5% of the Conterminous United States’s land surface, they support 31% of the plant species. They also support, through feeding and nesting, up to ½ of the native North American bird species. Nearly all wetland conservation work is done through one of 4 channels. They consist of easements, land purchase, revolving land and monetary funding. In locations where wildlife habitat has been degraded and the land is for sale, wetland conservation organizations will seek to acquire it. Once purchased, the habitat will be restored and easements will be placed on land to perpetually protect resource values. People commonly perceive mountains as pyramid-shaped masses that steadily narrow as they slope upward. But researchers have found they actually have four principal shapes. Not only are pyramid-shaped mountains in the minority, but most ranges increase in area at higher elevations. Besides reshaping the mountains in our mind's eye, these findings could lead scientists to reconsider conservation strategies for mountain species. The four principal shapes of mountain ranges include: diamond, pyramid, inverted pyramid and hourglass. For all the range shapes except pyramid, land availability can be greater at higher elevations than it is farther down the mountainside. Yet, people's idea that land area steadily shrinks as a mountain rises is so entrenched that it has come to guide conservation plans and research. This needs to change. A majority of mountain ranges studied (39 percent), such as the Rocky Mountains, are diamond-shaped. This means that land-area increases from the bottom until the mid-elevation range before contracting quickly. Hourglass-shaped mountain ranges such as the Himalayas make up 23 percent of ranges. Land area in these types rises slightly then decreases at mid-elevations before increasing sharply at higher elevations. The nearby Kunlun Mountains of China are representative of the 6 percent of ranges worldwide that take the form of inverse pyramids which gradually expand in area as elevation increases before, like the hourglass ranges, suddenly widening toward their peaks. As mountain species move to higher elevations to escape rising global temperatures, they are expected to face a consistent loss of territory – as well as an increase in resource competition. That all but ensures their eventual extinction. But while this risk exists in pyramid-shaped ranges, many species in other range types might in fact benefit from seeking higher altitudes if they move to an elevation with more land area than the one they left. Research is needed to more precisely identify those elevation zones where species will encounter territory losses and potentially become more threatened as they move upward. The limited resources that exist for conservation could then be targeted to those species. Animals that could benefit from an increase in elevation may still face other threats – habitat loss, food availability and exposure to existing animals and diseases, for instance. Even the range shapes themselves provide unique areas of concern. Hourglass-shaped ranges such as the Himalayas present a "bottleneck" at mid-elevation that could become overwhelmed with species moving upslope from more expansive lower elevations. Not every elevation holds equal value for limited conservation resources. Some gradients, and some portions of gradients, will be more important than others. Protecting land within an elevational bottleneck, for example, is critical. That is where species will be greatly pressured, and often long before they reach the mountaintop. Our oceans are filled with items that do not belong there. Huge amounts of consumer plastics, metals, rubber, paper, textiles, derelict fishing gear, vessels, and other lost or discarded items enter the marine environment every day, making marine debris one of the most widespread pollution problems facing the world's oceans and waterways. Marine debris is defined as any persistent solid material that is manufactured or processed and directly or indirectly, intentionally or unintentionally, disposed of or abandoned into the marine environment or the Great Lakes. It is a global problem, and it is an everyday problem. There is no part of the world left untouched by debris and its impacts. Marine debris is a threat to our environment, navigation safety, the economy, and human health. Most of all, marine debris is preventable. Anything man-made, including litter and fishing gear, can become marine debris once lost or thrown into the marine environment. The most common materials that make up marine debris are plastics, glass, metal, paper, cloth, rubber, and wood. Glass, metal, and rubber are similar to plastic in that they are used for a wide range of products. While they can be worn away - broken down into smaller and smaller fragments - they generally do not biodegrade entirely. As these materials are used commonly in our society, their occurrence as marine debris is overwhelming. Debris typically comes from both land-based and ocean-based sources. Plastics are used in many aspects of daily life and are a big part of our waste stream. Derelict fishing gear refers to nets, lines, crab/shrimp pots, and other recreational or commercial fishing equipment that has been lost, abandoned, or discarded in the marine environment. Thousands of abandoned and derelict vessels litter ports, waterways and estuaries, creating a threat to navigation, recreation, and the environment. How does marine debris move and where does it go? Wind, gyres, and ocean currents all impact how marine debris gets around. Floatable marine debris items, once they enter the ocean, are carried via oceanic currents and atmospheric winds. Factors that impact currents and winds, such as El Niño and seasons, also affect the movement of marine debris in the ocean. Debris items can be carried far from their origin, which makes it difficult to determine exactly where an item came from. Oceanic features can also help trap items in debris accumulation zones, often referred to in the media and marine debris community as “garbage patches.” Wildlife entanglement and ingestion, economic costs, and habitat damage are some impacts of marine debris. Marine debris is an eyesore along shorelines around the world. It degrades the beauty of the coastal environment and, in many cases, may cause economic loss if an area is a popular tourist destination. Would you want to swim at a beach littered in trash? Coastal communities may not have the resources to continually clean up debris. Marine debris can scour, break, smother, and otherwise damage important marine habitat, such as coral reefs. Many of these habitats serve as the basis of marine ecosystems and are critical to the survival of many other species. Wildlife Entanglement and Ghostfishing One of the most notable types of impacts from marine debris is wildlife entanglement. Derelict nets, ropes, line, or other fishing gear, packing bands, rubber bands, balloon string, six-pack rings, and a variety of marine debris can wrap around marine life. Entanglement can lead to injury, illness, suffocation, starvation, and even death. Many animals, such as sea turtles, seabirds, and marine mammals, have been known to ingest marine debris. The debris item may be mistaken for food and ingested, an animal's natural food (e.g. fish eggs) may be attached to the debris, or the debris item may have been ingested accidentally with other food. Debris ingestion may lead to loss of nutrition, internal injury, intestinal blockage, starvation, and even death. Vessel Damage and Navigation Hazards Marine debris can be quite large and difficult to see in the ocean, if it's floating below the water surface. Encounters with marine debris at sea can result in costly vessel damage, either to its structure or through a tangled propeller or clogged intake. Alien Species Transport If a marine organism attaches to debris, it can travel hundreds of miles and land on a shoreline where it is non-native. Invasive species can have a devastating impact on local ecosystems and can be costly to eradicate. Marine debris, especially large and heavy debris, can crush and damage coral. Coral reef ecosystems are complex, dynamic, and sensitive systems. Although they are geologically robust and have persisted through major climactic shifts, they are however, sensitive to small environmental perturbations over the short-term. Natural And Human Influences Slight changes in one component of the ecosystem affect the health of other components. Changes may be attributed to a number of causes but generally fall into two categories, natural disturbances and anthropogenic disturbances. Distinguishing between natural and anthropogenic disturbance is not always simple because the impacts of human actions may not be seen until well after the action has occurred, or may not be seen until it is coupled with a natural disturbance. Also, some events that appear to be natural may have been influenced by human actions. Impacts may be direct or indirect and may be compounded where several occur. For these reasons, it is often difficult to make cause-and-effect linkages when reef degradation is observed. Coral reef ecosystems are naturally variable and experience natural disturbances that vary on both temporal and spatial scales. Natural disturbance events that affect coral reefs include tropical storms, outbreaks of a coral predators, disease, extended periods of elevated or low water temperatures, and extremely low tides. Although these events disturb the reefs and may kill a significant amount of coral, they are part of a natural cycle that reefs experience and the reef ecosystem may benefit in other ways. The destruction caused by a hurricane, for example, opens space for reef organisms that had been excluded by larger and longer lived corals. Hurricanes also flush out accumulated sediment within the reef and create more substrate for organisms to settle and grow on. A healthy reef ecosystem will eventually recover from natural disturbance events. However, when these natural disturbances occur to a reef system that has been impacted by human activities, the reef system may have a reduced or even no capacity to rebound. A natural disturbance acting synergistically with accumulated human impacts may result in destruction that is not reversed in the same time frame it naturally would occur. Coral reefs around the world have experienced major recent natural disturbances. These natural events may have been influenced by human activities. A recent World Resources Institute report estimates that nearly 60 percent of the world's reefs are threatened by increasing human activity. The expanding human population and its activities may impact coral reef health in a number of ways. Development, urbanization, and agriculture lead to increases in freshwater runoff, polluted runoff, sedimentation, and nutrient inputs. Growing industry and automobile usage cause an increase in emissions contributing to the green house effect and chemical deposition from air to water. Commercial and private vessel traffic mean the possibility of fuel leaks or spills, vessel groundings, and anchor damage. Harvest of reef resources is also taking a toll on the health of coral reef ecosystems. Overfishing on reefs leads to an unbalanced ecosystem, allowing more competitive or less desirable organisms to become dominant. Fishing methods such as the use of explosives and poisons severely harm reefs and reef organisms. Harvest of coral skeleton for souvenirs depletes healthy corals or substrate where coral larvae might have settled. Increased tourism in areas of coral reef habitat contributes to increased pressure from scuba diving, recreational fishing, and vessel traffic. In general, oil spills can affect animals and plants in two ways: from the oil itself and from the response or cleanup operations. Understanding both types of impacts can help spill responders minimize overall impacts to ecological communities and help them to recover much more quickly. Spilled oil can harm living things because its chemical constituents are poisonous. This can affect organisms both from internal exposure to oil through ingestion or inhalation and from external exposure through skin and eye irritation. Oil can also smother some small species of fish or invertebrates and coat feathers and fur, reducing birds' and mammals' ability to maintain their body temperatures. What Creatures Are Most Affected by Oil Spills? Since most oils float, the creatures most affected by oil are animals like sea otters and seabirds that are found on the sea surface or on shorelines if the oil comes ashore. During most oil spills, seabirds are harmed and killed in greater numbers than other kinds of creatures. Sea otters can easily be harmed by oil, since their ability to stay warm depends on their fur remaining clean. If oil remains on a beach for a while, other creatures, such as snails, clams, and terrestrial animals may suffer. What Measures Are Taken When an Animal Comes in Contact with Oil? Most states have regulations about the specific procedures to follow. Untrained people should not try to capture any oiled bird or animal. At most U.S. spills, a bird and/or mammal rehabilitation center is set up to care for oiled animals. What Type of Spilled Oil Causes the Most Harm? The type of oil spilled matters because different types of oil behave differently in the environment, and animals and birds are affected differently by different types of oil. However, it's not so easy to say which kind is worst. First, we should distinguish between "light" and "heavy" oils. Fuel oils, such as gasoline and diesel fuel, are very "light" oils. Light oils are very volatile (they evaporate relatively quickly), so they usually don't remain for long in the aquatic or marine environment (typically no longer than a few days). If they spread out on the water, as they do when they are accidentally spilled, they will evaporate relatively quickly. However, while they are present, light oils present two significant hazards. First, some can ignite or explode. Second, many light oils, such as gasoline and diesel, are also considered to be toxic. They can kill animals or plants that they touch, and they also are dangerous to humans who breathe their fumes or get them on their skin. In contrast, very "heavy" oils (like bunker oils, which are used to fuel ships) look black and may be sticky for a time until they weather sufficiently, but even then they can persist in the environment for months or even years if not removed. While these oils can be very persistent, they are generally significantly less acutely toxic than lighter oils. Instead, the short-term threat from heavy oils comes from their ability to smother organisms. Over the long-term, some chronic health effects like tumors may result in some organisms. Also, if heavy oils get onto the feathers of birds, the birds may die of hypothermia (they lose the ability to keep themselves warm). We observe this same effect if sea otters become oiled. After days or weeks, some heavy oils will harden, becoming very similar to an asphalt road surface. In this hardened state, heavy oils will probably not harm animals or plants that come in contact with them. In between light and heavy oils are many different kinds of medium oils, which will last for some amount of time in the environment and will have different degrees of toxicity. Ultimately, the effects of any oil depend on where it is spilled, where it goes, and what animals and plants, or people, it affects. Being at the top of the food chain is no guarantee of a species survival. Not only are many of these so-called apex predators susceptible to human impacts, they also are slow to recover from them, which makes these animals vulnerable despite their high-ranking ecosystem status. Ecologists and conservation biologists have repeatedly sounded the alarm about the global decline of apex predators — a group that includes gray wolves, spotted owls, bald eagles, cheetahs, killer whales and sea otters. However, restoration practitioners have met with limited success despite major efforts to recover some of the world’s most charismatic megafauna. Recovery of apex predators is key because they often provide fundamental services such as disease regulation, the maintenance of biodiversity, and carbon sequestration. To recover apex predators, we must first appreciate that the pathway to predator recovery may differ markedly from the pathway predators initially followed to decline. New research, conducted by Adrian Stier at UC Santa Barbara’s National Center for Ecological Analysis and Synthesis, examines the big picture with regard to predator and ecosystem recovery. Stier worked on the study with colleagues at the National Oceanic and Atmospheric Administration, Oregon State University and University of Florida. The scientists’ comprehensive literature review revealed that full recovery of apex predator populations is currently the exception rather than the rule. In addition to well-known considerations, such as continued exploitation and slow life histories of these species, several under appreciated factors complicate predator recoveries. Not all predator species are equivalent, so we need to tailor successful recovery strategies based on how these animals are connected to the surrounding ecosystem. The ‘when’ is just as important as ‘what’ with respect to timing predator recoveries. This means designing adaptive sequences of management strategies that embrace key environmental and species interactions as they emerge. A good example of a successful restoration project is the reintroduction of wolves to the ecosystem in and around Yellowstone National Park. However, reintroducing wolves has not recreated an ecosystem that looks the same as it did pre-1920 when wolves were abundant. While wolves have contributed to a reduced elk population in recent years, lower elk numbers have not been sufficient to restore willows, the region’s dominant woody vegetation on which elk and other animals feed. This in turn has likely limited the recovery of the beaver population, which uses willow as building material for dams in small streams. Sometimes just reintroducing a species isn’t enough. An ecosystem can morph into a different-looking system that can be relatively stable, and adding in these top predators doesn’t necessarily cause that system to recover back into its original state. Then again, that may not always be the ultimate goal. Researchers point out that the recovery of apex predators isn’t always well-received, and reintroducing them in an artificial way can be controversial. Conservation needs can be balanced. We have the opportunity to identify efficient win-win solutions that offer dual prosperity to these majestic carnivores and the human systems within which they are embedded. Kelp forests grow predominantly on the Pacific Coast, from Alaska and Canada to the waters of Baja California. Tiered like a terrestrial rainforest with a canopy and several layers below, the kelp forests of the eastern Pacific coast are dominated by two canopy-forming, brown macroalgae species, giant kelp (Macrocystis pyrifera) and bull kelp (Nereocystis leutkeana). Conditions Required for Growth Kelp forests grow along rocky coastlines in depths of about 2 m to more than 30 m (6 to 90+ ft). Kelp favors nutrient-rich, cool waters that range in temperature from 5o to 20o C (42o to 72o F). These brown algae communities live in clear water conditions through which light penetrates easily. Kelp recruits most successfully in regions of upwelling (regions where the ocean layers overturn, bringing cool, nutrient-rich bottom waters to the surface) and regions with continuously cold, high-nutrient waters. Because the amount of dissolved inorganic nitrogen decreases significantly in marine waters warmer than 20oC, kelp experiences reduced or negative growth rates in warm water. Kelp survival is positively correlated with the strength of the substrate. The larger and stronger the rock on which it is anchored, the greater the chance of kelp survival. Winter storms and high-energy environments easily uproot the kelp and can wash entire plants ashore. Unique Characteristics of Kelp Plants Instead of tree-like roots that extend into the substrate, kelp has "anchors" called holdfasts that grip onto rocky substrates. From the holdfasts, kelp plants grow toward the water's surface. Gas bladders called pneumatocysts, another unique feature of kelp, keep the upper portions of the algae afloat. A giant kelp plant has a pneumatocyst at the base of each blade. In contrast, a bull kelp plant has only one pneumatocyst that supports several blades near the water's surface. Giant kelp is a perennial (it lives for several years) while bull kelp is an annual (it completes its life cycle in one year). Both types of kelp have a two-stage life cycle. They exist in their earliest life stages as spores, released with millions of others from the parent kelp, the sporophyte. The spores grow into a tiny male or female plant called a gametophyte, which produces either sperm or eggs. After fertilization occurs, the embryos may grow into mature plants (sporophytes), completing the life cycle. Giant kelp can live up to seven years. Factors such as the severity of winter storms may affect its life span. Its average growth (in spring) is 27 cm/day (~10 inches/day), yet it may grow up to 61 cm/day (2 ft/day). The average growth of bull kelp is 10 cm/day (~4 inches/day). The Kelp Forest Ecosystem A host of invertebrates, fish, marine mammals, and birds exist in kelp forest environs. From the holdfasts to the surface mats of kelp fronds, the array of habitats on the kelp itself may support thousands of invertebrate individuals, including polychaetes, amphipods, decapods, and ophiuroids. California sea lions, harbor seals, sea otters, and whales may feed in the kelp or escape storms or predators in the shelter of kelp. On rare occasions gray whales have been spotted seeking refuge in kelp forests from predatory killer whales. All larger marine life, including birds and mammals, may retreat to kelp during storms or high-energy regimes because the kelp helps to weaken currents and waves. Perhaps the most familiar image of kelp forests is a picture of a sea otter draped in strands of kelp, gripping a sea urchin on its belly. Both sea otters (Enhydra lutris) and sea urchins (Strongylocentrotus spp.) play critical roles in the stable equilibrium ecosystem. Sea urchins graze kelp and may reach population densities large enough to destroy kelp forests at the rate of 30 feet per month. Urchins move in "herds," and enough urchins may remain in the "barrens" of a former kelp forest to negate any attempt at regrowth. Sea otters, playing a critical role in containing the urchin populations, prey on urchins and thus control the numbers of kelp grazers. The Bureau of Land Management’s current approach to managing our public lands is allowing companies to lease most of America’s public lands for oil and gas development – with over 90 percent of public lands open to leasing – undermining conservation efforts and cheating taxpayers, according to data detailed in The Wilderness Society’s report, No Exit: Fixing the BLM’s Indiscriminate Oil & Gas Leasing. The Wilderness Society found the Bureau of Land Management’s current policies for oil and gas leasing are outdated and out of step with the agency’s guiding principles. The BLM rarely closes lands to oil and gas leasing in its resource management plans, despite the risks to wildlife, recreation, cultural and wilderness resources, while ignoring important opportunities to protect other values. There is almost no effort to protect some public lands from oil and gas leasing. 90 percent of U.S. public lands and mineral resources are available for leasing, even if BLM has found they have no actual potential for oil and gas development. The agency’s Handbook on Planning for Fluid Mineral Resources has not been overhauled in more than twenty-five years. The current approach to leasing is in conflict with the agency’s guiding management principle, the multiple use mandate. The BLM is required to manage public lands for a range of uses such as conservation, wildlife management and recreation, but the agency routinely defaults to keeping lands open to leasing, which precludes all other uses. When public lands with low energy development potential are leased to oil and gas companies, taxpayers lose out on revenue. These lands are routinely purchased for well below-market value, and can be held for a nominal annual fee for the duration of the 10-year lease term without yielding a meaningful return from development. Oil and gas companies often extend the terms of the leases they hold indefinitely through “suspensions,” which can last decades, with no annual fees. Furthermore, pervasive leasing creates roadblocks for supporting other resources, such as recreation, wilderness values, and fish and wildlife habitat. Conservation efforts are thwarted by BLM’s current policies, as speculative leases prevent the proactive management of environmentally valuable areas. Protective designations for these other values are difficult to obtain – creating a double standard which improperly favors oil and gas over other multiple uses. In the Bighorn Basin Resource Management Plan in Wyoming, the BLM considered whether to manage 43 inventoried units, totaling over 476,000 acres, to protect their wilderness characteristics. But ultimately, none of these lands are being managed to protect wilderness characteristics, primarily because they contain speculative oil and gas leases. In the White River Resource Management Plan Amendment in Colorado, the BLM expressly acknowledged that undeveloped leases on low-potential lands effectively prevented management to protect wilderness characteristics. Greater sage-grouse habitat in Idaho is open to oil and gas leasing under the federal management plan for sage-grouse in Idaho, even though no productive oil and gas wells have ever been drilled in Idaho and 100% of the most important habitat does not have high or even moderate oil and gas potential. No other ecosystem in America removes as much carbon from the atmosphere as prairie grasslands. Some carbon that is produced by our giant industrial complex is recycled into the fertile soils that have become a breadbasket for the entire world. The rolling acres of grassland stretching across the center of the United States are a classic American image. Early European settlers of this eco-region were so impressed by these endless grasslands that they compared them to the ocean, and named their wagons "prairie schooners" after large ships of the time. Less than 4 percent of this once vast prairie grassland survives today. It is fascinating to note that 80% of prairie plant life is underground. Long tentacled root systems survive grazing, fire and flood to sprout each spring and renew an amazing cycle of life that, due to its low lying subtlety, is often over looked. The prairie grasslands begin with the Great Plains at the eastern edge of the Rocky Mountains and extend all the way to the Appalachian Mountains in the eastern part of the country. The Rocky Mountains prevent moist air from moving over the Great Plains, and this "rain shadow" helps to keep the prairie grasslands extremely dry. However, it is not just the lack of rain that makes the prairie a harsh place to live. Twelve thousand years ago, retreating glaciers left behind a flat landscape open to extreme heat in the summer and extreme cold in the winter. The lack of geographic barriers or cover means that the wind runs rampant across the plains, leading to the "black blizzards" of the 1930s Dust Bowl and continuously endangering agriculture. Despite these extremes, many plants and animals such as wildflowers, pronghorn antelope, mule deer, prairie dogs, and coyotes make their homes in the prairie grasslands. In addition, small, isolated wetlands dot the dry prairies, providing much-needed water and aquatic habitat for birds. In the Northern Great Plains, these wetlands formed as the glaciers receded and left round, sunken areas behind them. Rain and groundwater fill these depressions during certain times of year, creating scattered wetland habitat known as "prairie potholes." The Prairie Pothole Region in the Northern Great Plains contains 5-8 million small wetlands and some of the most important freshwater resources in North America. Bullrushes, sedges, and cattails grow on the edges of these potholes because they prefer standing water, and these plants in turn provide food and shelter for other species, such as birds. More than half of the migratory waterfowl in North America depend on prairie potholes for their survival. THREATS TO PRAIRIE GRASSLANDS Human activity has damaged many Great Plains habitats, primarily through agricultural and livestock activity in the region. For example, only 40-50% of the original prairie pothole wetlands remain intact and undrained today. Climate change will affect the prairie grasslands ecoregion by pushing temperatures higher and decreasing rainfall in certain areas. Climate records reveal that while the average annual temperatures in the United States have increased about 1°F (0.6°C) over the past hundred years, average temperatures on the central and northern Great Plains have risen by at least 2°F (1.1°C). In some areas, such as North and South Dakota and portions of Montana, average temperatures have increased as much as 5.5°F (3.1 °C). In addition to rapidly rising temperatures, patterns of rainfall have changed over the same time period so that the eastern areas of Montana, Wyoming, and Colorado have suffered a decrease in precipitation of 10%. Climate models predict that this increased drought in some areas will cause wetlands to relocate or disappear. Climate change will challenge wetlands in particular, because most wetlands in the plains occur where the effects of climate change are predicted to be most severe. These findings imply that climatically drier portions of the Prairie Pothole Region, including areas that migratory birds rely on, are especially vulnerable. However, higher temperatures and decreased precipitation will make life harder for the entire region. For over one hundred years, gas and oil production on public lands has caused harm to species and ecosystems and contaminated air, soil, and water. The manufacturing and drilling of oil results in public lands becoming fragmented, driving wildlife away and harming habitats. At the same time, fires, oil disasters and other pollutants result in the contamination of water reserves, both on the surface and underground. By building roads to connect to drilling sites, human activity in previously unharmed areas skyrockets, leading to littering, increased poaching, roadkill, and fires. What’s more, it becomes easier for foreign species to take over and overwhelm the native fauna and flora. Perhaps most importantly, by allowing the gas and oil industry to develop further our reliance on fossil fuels is strengthened, producing greenhouse gasses and facilitating global warming. Massive environmental value is hidden in our oceans and public lands, ranging from clean water to clean air, and natural ecosystems providing essential habitat for some of our most endangered species. But fossil fuel is valuable monetarily, which is why the government is selling public lands to anyone willing to pay the highest bid. Corruption and greed plagues the decision making process of how to best manage our public lands and waters. Our climate is at a crucial point. Unless we overcome our dependence on fossil fuels by 2050, we will be facing extreme phenomena such as flooded coasts, human health disasters and massive extinctions of wild species. Climate change is happening now, not tomorrow. With global warming set to boost the rate at which wildlife is pushed to extinction, there is no better use for our oceans and public lands than providing safe haven to species and protection of their ecosystems. Despite the alarming messages, the government keeps sacrificing these habitats to the oil and gas industry, to which they have leased over 67 million acres – 55 times more land than the Grand Canyon National Park. More than 25 percent of all greenhouse emissions in the country comes from these leases, while some of our most valued lands are being destroyed. Both our national and natural heritage pay a high toll. The nation’s public lands are industrialized, coastlines and pristine rivers are contaminated, underprivileged communities are undermined, and wildlife is pushed closer towards extinction. For every new fossil fuel lease, the world is burdened with additional climate disruption. There is nothing rational in a policy that allows for the destruction of natural heritage so that more climate pollution can be produced. The federal fossil fuel leases in our oceans and on our public lands are unacceptable and need to stop. If we did this, we would spare the atmosphere of 450 billion tons of pollution. Vast areas of public lands, wildlife habitat and oceans would be saved in the process. The Potential Greenhouse Gas Emissions of U.S. Federal Fossil Fuels report showed that by putting an end to the federal fossil fuel leases, we would prevent 450 billion tons of carbon pollutants from becoming potential greenhouse emissions. This comprises over 25 percent of the total emissions that are permitted, should the world adopt a target to prevent global warming from surpassing 2 degrees Celsius which would cause catastrophic consequences to humans and natural ecosystems worldwide. It is impossible to stop climate change with the regulation of tailpipes and smokestacks alone; extracted fossil fuels are intended to be burned, so any policy aspiring to counter climate change should limit the fossil fuel supply. There is no better place to start this process than our oceans and public lands, which harbor significant biological and ecological value. The aquatic biome includes habitats around the world dominated by water. Aquatic ecosystems are divided into two main groups based on their salinity—freshwater habitats and marine habitats. ● Freshwater habitats are aquatic habitats with low levels of salt, less than one percent. They include rivers, lakes, streams, ponds, swamps, wetlands, bogs and lagoons. ● Marine habitats are aquatic habitats with salt concentrations of more than one percent. They include oceans, seas and coral reefs. Some habitats exist where saltwater and freshwater mix together. These include mud flats, mangroves and salt marshes. Aquatic ecosystems support a diverse assortment of animals including fishes, amphibians, reptiles, mammals, birds and invertebrates. When evaporated sea water falls as rain, it flows down mountain streams creating rivers and lakes. Rain water feeds freshwater rivers, which then flows back into the sea. Streams, rivers and lakes are home to countless animal species. The two main types of freshwater habitat are rivers and lakes. Lakes are often fed by streams or rivers. They can also be enclosed areas where species live that are found nowhere else on the planet. Rivers usually contain large animals able to cope with strong currents, as well as animals such as crabs and birds that feed on the fish within the water. Freshwater rivers provide habitat to a wide variety of species including fish, amphibians, reptiles, insects, birds and mammals. An extraordinary number of fish species inhabit streams and rivers. Freshwater lakes are also home to a vast variety of wildlife. Some species spend their entire lives in one area. Others visit momentarily to eat and drink. Many species are specially adapted to life in particular lakes. Large mammals, including zebras, primates, giraffes and deer, visit lakes to drink. Many freshwater habitats have been drastically affected by human activities. Chemicals and pesticides contaminate the water, as well as waste water. Animals and plants that inhabit the water can be affected, as are the animals that eat them. Oceans create the largest habitat in the world. Countless animal species inhabit the planet's oceans which cover over 75% of the earth. The two main types of ocean habitat are coastal, inshore habitats found around land, and open ocean habitats that stretch around the planet. More animal species live in the rich, shallower waters than the deep sea, though animals live throughout the oceans. The ocean landscape is as vast and varied as on land, featuring underwater continental shelves, mountains, valleys, volcanoes, trenches and plains. Warmer, coastal waters around the globe are home to the majority of species. These areas feature more food sources than the deep ocean. Smaller aquatic animals often inhabit the shallower regions. Coastal waters provide them with a variety of places to hide, with fewer large predators. Larger animals tend to prefer deeper regions beneath the waves along the continental shelves. Plankton -- microscopic plants and animals, fish eggs and animals in their larvae form -- provide a plentiful food source for many marine animals. Tiny fishes and crustaceans, to the largest animal on the planet, the blue whale, feed on this vital food source. The two largest threats to ocean habitats are over-fishing and pollution. Pollution from the land and air accumulate in the sea with devastating effects to many plant and animal species. Over-fishing threatens many species with extinction. Coral reefs are the richest habitats on the earth. Found along the coastlines, they provide habitat to countless plant and animal species including fish, reptiles, invertebrates, echinoderms and crustaceans. Coral reefs are located in the tropical and sub-tropical coastal regions where it is always warm, day and night, year-round. The two main types of coral reef habitats are soft coral reefs and hard coral reefs. Soft corals are animals that move through the water, eventually settling. Hard corals are the reef-building corals that are hard coral shells left behind when corals die. The largest coral reefs are located along the south-west coast of Africa, in the Caribbean and all around Australia, south-east Asia and the coastal regions of the South Pacific Ocean. So rich in life and biodiversity, coral reefs are home to an incredible variety animal species able to survive together with little competition for food. Animal species that inhabit coral reefs vary tremendously in shape, size and color. Sea urchins, starfish and crustaceans are invertebrates that call coral reefs home. Sea snakes hunt small fish and eels in the coral reefs. Eels and seahorses are among the many fish species. Sharks do not live permanently in coral reefs, but visit often in search of prey. Sea turtles also make frequent trips to coral reefs in search of food. The threats to coral reefs and coastline wildlife include commercial fishing, pollution and storms. Dredging involves dragging fishing nets across the sea bed, destroying coral reefs in the process. Many animal species that inhabit coral reefs are on the brink of extinction. Sea storms, such as tsunamis, can also reek havoc on coral reef environments. Wetlands are found throughout the world, often in more temperate regions where vegetation grows quickly. These large areas of water contain a wealth of plants and are broken up by small islands of land. Wetlands include swamps, marshes, fens and bogs. Many wildlife species are specifically adapted to wetland environments, including fish, amphibians, birds, mammals, reptiles and insects. The two main types of shallow watery areas are swamps and wetlands. Swamps are usually located in forested areas. Trees, such as mangrove trees, survive in salt-water conditions and require ample space for their roots. Wetlands are usually near large rivers or estuaries that flood when river banks burst from a lot of rain. Mangrove swamps are one of the richest habitats on the planet. Numerous animals species live above and below the water's surface. Many animal species that live in mangrove forests are found nowhere else on earth. The mangrove tree's enormous roots provide shelter to small fishes, amphibians and reptiles and provide a way for the animals to get in and out of the water. Larger animals have ample fish to feed on. Large aquatic birds such as heron spear fish with long beaks in wetland habitats. Salt-water swamps contain snapping turtles, crabs, crocodiles and alligators. Amphibians and reptiles inhabit the water's edge. Many insects live, and lay their eggs, in wetland habitats...providing food for frogs and lizards. The main threats to wetlands are deforestation and pollution. The animals in wetland habitats are specifically adapted to their environment and are vulnerable to toxins in the water and air. Islands form when land breaks away from large land masses or volcanoes erupt on the sea floor. They are found throughout the world. Their isolated nature results in unique wildlife species, often different from their counterparts living in mainland habitats. Some island animal species have developed completely separately from mainland species. Numerous habitats including forests, wetlands, deserts and tundra can be found on different islands. Limited in size and resources, ecosystems on islands are fragile and easily disturbed. Human activity and the introduction of new species on islands has caused much harm, making many species endangered or extinct. With nowhere else for them to go, the loss of habitat or food sources is particularly damaging to island animals. Lemurs live only on the island of Madagascar, the tree kangaroo only in Papua New Guinea, the kiwi only in New Zealand and the orangutan only on the Indonesian islands of Borneo and Sumatra. Separated from the mainland, these species have adapted to their isolated environments. The kiwi and the kakapo birds have adapted to a flightless lifestyle since there were no large predators on the islands to flee from. The introduction of predators by humans threatens their survival. Orangutans suffer from mass deforestation in south-east Asia and the exotic pet trade. A breaking point has been reached in conserving the fragile habitats of islands. Without immediate action to save these precious ecosystems, many species will be lost forever. Marine conservation, also known as marine resources conservation, is the protection and preservation of ecosystems in oceans and seas. Marine conservation focuses on limiting human-caused damage to marine ecosystems, and on restoring damaged marine ecosystems. Marine conservation also focuses on preserving vulnerable marine species. Marine conservation is the study of conserving physical and biological marine resources and ecosystem functions. This is a relatively new discipline. Marine conservationists rely on a combination of scientific principles derived from marine biology, oceanography and fisheries science, as well as on human factors such as demand for marine resources and marine law, economics and policy in order to determine how to best protect and conserve marine species and ecosystems. Coral reefs are the epicenter for immense amounts of biodiversity, and are a key player in the survival of an entire ecosystem. They provide various marine animals with food, protection and shelter which keep generations of species alive. Unfortunately, because of human impact of coral reefs, these ecosystems are becoming increasingly degraded and in need of conservation. The biggest threats include overfishing, destructive fishing practices and sedimentation and pollution from land-based sources. This in conjunction with increased carbon in oceans, coral bleaching, and diseases, results in no pristine reefs left anywhere in the world. In fact, up to 88% of coral reefs in Southeast Asia are now threatened, with 50% of those reefs at either "high" or "very high" risk of disappearing which directly effects biodiversity and survival of species dependent on coral. In island nations such as Samoa, Indonesia and the Philippines, many fisherman are unable to catch as many fish as they used to, so they are increasingly using cyanide and dynamite in fishing, which further degrades the coral reef ecosystem. This perpetuation of bad habits simply leads to the further decline of coral reefs and therefore perpetuates the problem. One solution to stopping this cycle is to educate the local community about why conservation of marine spaces that include coral reefs is important. Once the local communities understand the personal stakes at risk then they will actually fight to preserve the reefs. Conserving coral reefs has many economic, social, and ecological benefits, not only for the people who live on these islands, but for people throughout the world as well. Although humans cause the greatest threat to our marine environment, humans also have the ability to create effective management plans that will be the key to successful marine conservation. One of the best marine conservation tools simply stems from smarter individualist choices we make. Strategies and techniques for marine conservation tend to combine theoretical disciplines, such as population biology, with practical conservation strategies, such as setting up protected areas, as with marine protected areas (MPAs) or Voluntary Marine Conservation Areas. Other techniques include restoring the populations of endangered species through artificial means. International laws and treaties related to marine conservation include the 1966 Convention on Fishing and Conservation of Living Resources of the High Seas. United States laws related to marine conservation include the 1972 Marine Mammal Protection Act, as well as the 1972 Marine Protection, Research and Sanctuaries Act which established the National Marine Sanctuaries program. In 2010, the Scottish Parliament enacted new legislation for the protection of marine life with the Marine (Scotland) Act 2010. The provisions in the Act include: marine planning, marine licensing, marine conservation, seal conservation and enforcement. The world’s oceans are on the verge of collapse. The overexploitation of fish has tripled since the 1970s, rapidly depleting the seas of fish. About 90 percent of the world’s fish have now been fully or overfished, and a 17 percent increase in production is expected by 2025, according to the UN Food and Agriculture Organization (FAO). The UN's The State of World Fisheries and Aquaculture (SOFIA) says that the state of the world's marine “resources” is not improving. Almost a third of commercial fish stocks are now fished at biologically unsustainable levels, triple the level of 1974. Some 31.4 percent of the commercial wild fish stocks regularly monitored by FAO have been overfished. The situation in the Mediterranean and Black Sea - where 59% of assessed stocks are fished at biologically unsustainable levels - is alarming. This is especially true for larger fish such as hake, mullet, sole and sea breams. In the Eastern Mediterranean, the possible expansion of invasive fish species associated to climate change is a concern. Globally, fish provide 6.7 percent of all protein consumed by humans. Some 57 million people are engaged in the primary fish production sectors, a third of them in aquaculture. Fishery products account for one percent of all global merchandise trade in value terms, representing more than nine percent of total agricultural exports. The depletion of the oceans' fish starts with consumer demand. You can make a difference by eliminating your consumption of seafood. The average person can save 225 fish and 151 shellfish a year by cutting seafood from their diet. Well over 900 plants and animals are endangered, and hundreds more are threatened. Many of the reasons certain animals are disappearing forever are because of human activities. FIVE MAJOR CAUSES The mnemonic HIPPO represents the five major causes of declining wildlife biodiversity: H - Habitat Loss I - Invasive Species P - Pollution and Pesticides P - Population Growth (human) and the Pet Trade O - Over-hunting and Over-collecting Habitat Loss results from human activities and land development. Many animal species are in decline because their environment is no longer able to fulfill their basic requirements. All species require food, water, shelter, space and the ability to find a mate and have children. Some species require small habitats, while others need large areas to successfully survive. Animal agriculture is the leading cause of habitat loss and deforestation. Invasive Species are plants and animals transported from one country or region to another and introduced into the wild. While most do not survive in a foreign world, some assimilate into their new world and thrive. Often they out-compete native plants and animals for their niche in the ecosystem, upsetting the balance of nature. Pollution and Pesticides, in forms of garbage and trash, air and water pollution, soil contamination and noise and light pollution, harm ecosystems and wildlife. Pesticides are toxic and harm more than their target. Pollution harms the environment and animals. Population Growth and the Pet Trade threaten countless animal species. As humans take more and more wilderness areas for agriculture, housing and industry, less land is available for wildlife. Native animals are often forced into less suitable habitats and can decline or disappear forever. Many “pets”, including fish, reptiles, spiders, birds, rodents and exotic mammals, are harvested from the wild. Over-hunting and Over-collecting has impacted many endangered species, reeking havoc on ecosystems and eliminating entire species forever. HOTSPOTS & COLDSPOTS ● Biodiversity Hotspots are regions with large numbers of species that do not live anywhere else in the world, where habitat destruction has occurred at alarming rates. Many organizations and agencies focus on saving these hotspots in an effort to do the greatest good and save the most species. Hotspots make up less than 2% of the planet. ● Coldspots, over 98% of the earth, are areas that have less species diversity but they need just as much help as areas with lots of biodiversity. In fact, some biodiversity coldspots are home to very rare plants and animals. Protecting these areas before too much destruction occurs prevents us from having to work backwards. THE DOMINO EFFECT All plants and animals have many complex intertwining links with other living things around them. Hippopotamus have birds that feed off the insects that grow on them. If the hippo were to become extinct, so would the birds…leading to further destruction as other species depend on the birds. This is referred to as Chains of Extinction, or the Domino Effect. A keystone species is a plant or animal that plays a crucial role in how an ecosystem functions. Without the keystone species, the ecosystem would be dramatically different or would not be able to survive. While all species in a habitat rely on each other, keystone species have a huge impact on their environment. Their disappearance would start a domino effect, leading to other species in the ecosystem also disappearing. An indicator species is a plant or animal species humans focus on to gather information about an ecosystem. Their presence or absence in an environment can be a signal that all is well, or something is not right. Certain types of plants or animals may exist in a very specific area. If the species begins to disappear, this ecoregion may be shrinking and action may need to be taken to save the environment. Indicator species can tell humans about the health of the environment. Many are extremely sensitive to pollution or human interference and serve as a “miner's canary”. UMBRELLA SPECIES An umbrella species is a plant or animal species that has a wide range and requirements for living as high or higher than other animals in the habitat. If the umbrella species' requirements are met, then so are the needs of many other species in its ecosystem. The Monarch butterfly is an example of an umbrella species because of its lengthy migrations across North America, covering lots of ecosystems. Any protections given to the Monarch will also “umbrella” many other species and habitats. Often times umbrella species are used by organizations and agencies to capture the public's attention for support for conservation efforts. These flagship species - such as pandas, whales, tigers, gorillas and butterflies - are species that the public finds captivating and are interested in helping. When the flagship species is helped, so are species in their ecosystems that the general public may find less appealing. Surviving in an environment of continuous threat and stress is a serious challenge for most living species. Living organisms, in whatever form, need to adapt to changes in the weather, climate and all sorts of changes in the environment. Add to this the natural calamities in the form of floods, storms, fires and volcanic bursts and their aftermath. When new lifeforms enter their ecosystems, pressure on existing species mount. Dangers can be parasitic or predatory in nature. Challenges to adaptation can be in the form of diseases or the very complexity of biological changes themselves. After millions of years of adapting to their environments, animals faced a new kind of threat - the advent of human beings. The effect of humans on the planet has been profound and has threatened the existence of all kinds of organisms to a degree that has caused scientists to believe that the Earth is beginning to witness extinction on a mass scale. Being a a part of nature, the dangers arising from human actions are extensions of natural threats. But the threats from mankind are within human control and can be curbed by changes in behavior. Humans are in a position to realize the consequences of their actions on the environment and can easily make changes in behavior that would affect the future health of the planet in a positive way. Human Threats To Animals Destruction of habitat and fragmentation: Destruction and fragmentation of animal habitats for the purpose of agriculture, urban development, building of hydro-electric projects and other self-serving uses are major threats to the Earth’s wildlife. Effects on global climate: Large scale emissions from fossil fuel burning and excessive flaring of gas have wrought serious damage to the Earth's atmosphere, especially the ozone layer, causing climate changes in many parts of the globe. Introduction of new and invasive species: Loss of endemic species have been caused by introduction of new species into their ancient habitats. Hunting and poaching: Hunting animals for sport and poaching them for profit as part of the illicit wildlife trade are among the major threats to wildlife. Effects of pollutants: Industrial wastes, fertilizers, and pesticides have infiltrated and consumed entire habitats of all forms of living creatures and organisms. Accidents: Loss of habitat causing animals to venture onto freeways and other manmade obstacles is also a common occurrence. More birds are being killed by a growing traffic in aviation and from colliding into windows.<|endoftext|>
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  TOPICS Exercise - 3.1 Question-1 :-  How will you describe the position of a table lamp on your study table to another person? Solution :- ``` To describe the position of a table lamp on the study table, we have two take two lines, a perpendicular and horizontal. Considering the table as a plane and taking perpendicular line as Y axis and horizontal as X axis. Take one corner of table as origin where both X and Y axes intersect each other. Now, the length of table is Y axis and breadth is X axis. From The origin, join the line to the lamp and mark a point. Calculate the distance of this point from both X and Y axes and then write it in terms of coordinates. Let the distance of point from X axis is x and from Y axis is y then the the position of the table lamp in terms of coordinates is (x,y). ``` Question-2 :-  (Street Plan) : A city has two main roads which cross each other at the centre of the city. These two roads are along the North-South direction and East-West direction. All the other streets of the city run parallel to these roads and are 200 m apart. There are 5 streets in each direction. Using 1cm = 200 m, draw a model of the city on your notebook. Represent the roads/streets by single lines. There are many cross- streets in your model. A particular cross-street is made by two streets, one running in the North - South direction and another in the East - West direction. Each cross street is referred to in the following manner : If the 2nd street running in the North - South direction and 5th in the East - West direction meet at some crossing, then we will call this cross-street (2, 5). Using this convention, find: (i) how many cross - streets can be referred to as (4, 3). (ii) how many cross - streets can be referred to as (3, 4). Solution :- ``` (i) Only one street can be referred to as (4, 3) as we see from the figure. (ii) Only one street can be referred to as (3, 4) as we see from the figure. ``` CLASSES<|endoftext|>
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Assessment for learning: a concrete example The transcript below was taken from a science lesson with 13/14 year olds on Particles. You can also download it [PDF]. What do you notice about how learning is influenced by the way the teacher asks and responds to questions? What does the teacher do to help identify students’ understanding and move their learning forward? Add your thoughts to the comments You’ll need to take account of the fact that this transcript is from a class of students who will be at a different stage in their learning and in their approach to learning than would be the case with younger or older students. The underlying principles, however, are the same. The transcript begins some 40 minutes into the lesson, just as the students have completed a practical where they mixed salt with water and then evaporated the water to retrieve the salt. Teacher: So I want you to think about what we just did. Let’s just recap. So we started with…..? Jake: Salt solution. Teacher: Salt solution? Jake: Well salt and then salt solution ‘cos we put in water. Teacher: Okay. And what happened when the salt went into the water? Hayley? Hayley: It sort of disappeared. Teacher points to another student. Bijan: Went invisible in the water but it was still there. Teacher: How do we know that Bijan? Bijan: The water is salty. So it’s there. You just can’t see it. Teacher: Any other evidence? Go on Jake. Jake: Well we got it back when we heated it. Teacher: Who else found that? That the salt was in the evaporating dish at the end. Most of you. Good. So can you just explain what you think happened in your books. Just a few words to help you get the explanation clear. This one … dissolve. Dissolving. Can you use that please? And this one…evaporate or evaporating. Where did the evaporating come in Bianca? Bianca: With the water. Teacher: With the water? Which bit as we had water all the way through? Bianca: Last bit. We heated it and that made the salt… made the water evaporate. Teacher: Okay so you heated the salt solution with your Bunsen and that evaporated the water, leaving the salt from the solution behind. Everyone okay with that? Okay 5 minutes to make notes. Talk to one another first if you aren’t sure and talk to me if you are both unsure. Okay. Go. Students start task. Teacher circulates reading what the students are writing in their books After 5 minutes, the teacher demonstrates mixing a small beaker of peas into a large beaker of rice. Teacher: Let’s imagine that this is the water (holds up beaker of rice) and this is the salt (holds up beaker of peas) and we mix them (pours peas into rice and stirs it with a spoon). What do we have now? Sian? Sian: A mixture. Teacher: of… ? Sian: Rice and peas. Teacher: Yep but if the rice is the water and the peas the salt? Sian: Don’t get it sir. Teacher: Who can help Sian? Bianca? Bianca: It’s the salt solution. Teacher: Exactly. In our model, the rice and peas together are the same as the salt solution. So what I want you to discuss in your groups is how can we get the salt – the peas – back out? And also why we can’t see the salt – the peas – when they’re like this (holds up the beaker). Teacher circulates listening to what the students are saying. Samira: Is it that the peas hide cos they’re in the middle and the rice is on the outside? Lucy: Yeah maybe. But water is see-through so it wouldn’t work like that with the salt and water. Samira: Maybe it’s because rice is a solid. Would it work if we made the rice liquid? Jo: Don’t think it goes liquid just soft when you cook it. Teacher: Think of the rice and peas as particles. So the rice are the water particles and the peas the salt particles. So you mix the particles (points to the beaker of pea-rice mixture on the front bench). Can we get the peas out? Maybe not in the same way we get the salt out but can we separate the peas? Samira: You could tip it all out and pick out the peas. Jo: Or use a sieve thing. Teacher: Okay. So let’s go back to the question. Why can’t we see the salt when its in solution? Jo: Because it’s mixed in with the water. It’s in the solution. Samira: Is it also because there’s only a little bit of the peas and so they get spread through. Lucy: Or maybe it’s the size. Oh no cos’ the peas are bigger. Is the salt bigger sir? Teacher: It is Lucy but listen again to what Samira just said. Samira: Oh it’s spread out. The peas there and the salt here get spread out so you don’t notice them. You only see the liquid. Jo: So if there was more salt would you see it then cos there wouldn’t be enough water particles to hide it? Teacher nods. And moves away to another group. The lesson ends with a group discussion based on the 2 questions: How do you separate the peas from the rice/pea mixture? Why can’t we see the salt when its in solution? © National STEM Learning Centre<|endoftext|>
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs! #### Online Quiz (WorksheetABCD) Questions Per Quiz = 2 4 6 8 10 ### Middle/High School Algebra, Geometry, and Statistics (AGS)2.12 Solving Inequations - II Example: Solve 2(x+3) < 5 (x-3). Solution: 2x+6 < 5x-15. Add -6 to both sides, we get 2x + 6 - 6 < 5x - 15 -6 2x < 5x - 21 Add -5x to both sides, we get 2x - 5x < 5x -21 -5x -3x < -21 Divide both sides by -3. -3x/-3 > -21/-3 x > 7 \ Any number > 7 is a solution. Verification: Substituting x = 8 > 7 L.H.S. = 2(x+3) = 2(8+3) = 2*11 = 22 R.H.S. = 5(x-3) = 5(8-3) = 5 * 5 = 25 \ 22 < 25 \ 8 > 7 is a root. Similarly we can show that any number greater than 7 is a root. Directions: Solve the following inequality. Also write at least ten examples of your own. Q 1: Solve 2(x + 7) < 9(x - 7).x < 11x < 5x > 11x > 4 Q 2: Solve 2(x + 4) < 6(x - 4).x > 0x < 0x > 8x < 8 Q 3: Solve 5(x + 2) < 7(x - 2).x > 20x < 15x > 12x < 12 Q 4: Solve 4(x + 3) < 7(x - 3).x > 1x < 6x < 11x > 11 Q 5: Solve 5(x + 4) < 9(x - 4).x > 14x < 13x > 15x < 14 Q 6: Solve 3(x + 3) < 6(x - 3).x > 9x < 9x > 12x < 6 Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!<|endoftext|>
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While Richard ‘Strongbow’ de Clare was the first Norman to gain a foothold in Ireland, by 1171, King Henry of England had gotten involved and he and his knights carved out huge sections of eastern and southern Ireland for themselves. Other knights marched north and established a northern bastion at Carrickfergus, which became the seat of the Earl of Ulster. Over time, however, Anglo-Norman rule ebbed and flowed. In fact, as the centuries progressed, it ebbed more than flowed, such that by 1500, the descendants of the original conquerors were almost completely assimilated into the native Irish clans. It reached a point such that Henry VIII offered amnesty to all lords in Ireland regardless of ethnicity, provided they surrendered their lands to him (to receive them back immediately by royal charter). Unfortunately for Ireland, after two hundred years of being mostly ignored by the English crown, the Tudors decided that the time had come to ‘pacify’ and ‘Anglicize’ the island to bring it under more direct English control. The country’s offenses were remaining Catholic while England had gone Protestant, and the continued existence of clans and kingdoms outside of the standardized English system. The “Old English” families, as the former Anglo-Norman families were called, were viewed as no better than the native Irish. All were stripped of power and forced off their lands by new rulers and imported settlers from England, Scotland, and Wales, who were, of course, Protestant as well. “The first and most important result of the conquest was the disarmament of the native Irish lordships and the establishment of central government control for the first time over the whole island; Irish culture, law and language were replaced; and many Irish lords lost their lands and hereditary authority. Thousands of English, Scottish and Welsh settlers were introduced into the country and the administration of justice was enforced according to English common law and statutes of the Parliament of Ireland. As the 16th century progressed, the religious question grew in significance. Rebels such as James Fitzmaurice Fitzgerald and Hugh O’Neill sought and received help from Catholic powers in Europe, justifying their actions on religious grounds . . . Under James I, Catholics were barred from all public office … the Gaelic Irish and Old English increasingly defined themselves as Catholic in opposition to the Protestant New English … By the end of the resulting Cromwellian conquest of Ireland in the 1650s, the “New English” Protestants dominated the country, and after the Glorious Revolution of 1688 their descendants went on to form the Protestant Ascendancy.” https://en.wikipedia.org/wiki/Tudor_conquest_of_Ireland Far more than the initial Norman conquest of Ireland, it is in the Cromwellian conquest where the roots of the deep resentment of the Irish people towards the English lie, as well as the source of the campaign for independence that marked the eighteenth through twentieth centuries.<|endoftext|>
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# How do you find the interior angle of a triangle with exterior angles? Because the interior angles of a triangle add to 180°, and angles c+d also add to 180°: ## How do you find the interior angle of a triangle with exterior angles? Because the interior angles of a triangle add to 180°, and angles c+d also add to 180°: 1. The interior angles of a triangle add to 180°:a + b + c = 180° 2. Angles c and d make a straight angle, which is 180°:d + c = 180° 3. So d + c equals a + b + c:d + c = a + b + c. 4. Subtract c from both sides:d = a + b. ## Which angle is complementary to? Two angles are called complementary when their measures add to 90 degrees. Two angles are called supplementary when their measures add up to 180 degrees. ## Why do angles in a triangle add to 180? A triangle’s angles add up to 180 degrees because one exterior angle is equal to the sum of the other two angles in the triangle. In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the third angle to make a 180 angle. ## How are interior and exterior angles related? In summary, we learned that an interior angle is an angle inside a shape, while an exterior angle is an angle made by the side of a shape and a line drawn out from an adjacent side. The sum of the interior angles of a triangle is always 180. An exterior angle is equal to the sum of the non-adjacent interior angles. ## What is the angle in a triangle? The angles inside a triangle are called interior angles. The three interior angles in a triangle will always add up to 180°. At each corner the exterior and interior angles are on a straight line, so at each corner these two angles add up to 180°. ## Which set of angles measures would determine a triangle? Note that the right angle is shown with a corner mark and does not need to be labeled 90°. The sum of the measures of the three interior angles of a triangle is always 180°. This fact can be applied to find the measure of the third angle of a triangle, if you are given the other two. ## Why is the sum of the exterior angles of a triangle 360? Properties. An exterior angle of a triangle is equal to the sum of the opposite interior angles. If the equivalent angle is taken at each vertex, the exterior angles always add to 360° In fact, this is true for any convex polygon, not just triangles. ## Can exterior angles of a triangle be a straight angle? AN EXTERIOR ANGLE OF A TRIANGLE CANNOT BE A STRAIGHT LINE BECAUSE A TRIANGLE HAS 180 DEGREES IN ADDING ALL THE THREE ANGLES OF A TRIANGLE. ## How do you prove exterior angles of a triangle equal 360? Let the exterior angles of A, B and C be ∠a, ∠b and ∠c respectively. Thus sum of exterior angles of a triangle is 360°. Let the exterior angles of A, B and C be ∠a, ∠b and ∠c respectively. Thus sum of exterior angles of a triangle is 360°. ## What set of angles could be the interior angles of a triangle? The interior angles of a triangle are the three angles on the inside of a triangle. These three angles always sum to 180 degrees. ## Which set of three angles could represent the interior angles of a triangle 49 59 72? Answer: 49 degrees, 59 degrees, 72 degrees is a set of interior angles of a triangle. ## Which set of three angles could represent the interior angles of a triangle? Answer: 19° ,70° and 91° only one represent the interior angle of triangle. ## Which set of three angles could represent the interior angles of a triangle 26 51 103? Answer. up to 180, and 26+51+103=180. Hope I was able to help! ## What are interior and exterior angles of a triangle? The angles on the inside are called Interior angles. The sum of the interior angles of a triangle is always 180 degrees. The exterior angle is the angle between any side of a shape, and a line extended from the next side. The sum of an exterior angle and its adjacent interior angle is also 180 degrees. ## What 3 angles make a triangle? In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). A triangle has three angles, one at each vertex, bounded by a pair of adjacent sides. ## What is the measure of AngleSRT? What is the measure of AngleSRT? Triangle S R T. Angle S is 53 degrees and angle T is 48 degrees. exterior angle ## Which angles form a linear pair quizlet? Angle 1 and angle 2 are a linear pair if and only if they are adjacent angles with common sides that form apposite rays. ## How many exterior angles are there in a triangle? 6 exterior angles ## What is the measure of BAC? A BAC describes the amount of alcohol in a person’s blood, expressed as weight of alcohol per unit of volume of blood. For example, 0.08 percent BAC indicates 80 mg of alcohol per 100 ml of blood. For most legal purposes, however, a blood sample is not necessary to determine a person’s BAC. ## What is the measure of angle mop? Angle MOP is an exterior angle of ∆ MNO, measure of angle M = 70° and measure of angle O = 65° . ## Which of the following can be angles of a triangle? Sum of the three angles = 180 degrees. Simplify. So, the three angles of a triangle are 60°, 48°, and 72°. ## How do you prove the exterior angles of a triangle? Angle Sum Property Of A Triangle & Exterior Angle Theorem 1. Theorem 1: Angle sum property of triangle states that the sum of interior angles of a triangle is 180°. 2. Proof: 3. Theorem 2: If any side of a triangle is extended, then the exterior angle so formed is the sum of the two opposite interior angles of the triangle. ## Which set of three angles could represent the interior angles of a triangle 26? -Only the 26 °, 51 °, 103 ° set add exactly up to 180° and therefore satisfies the triangle’s interior angles condition. ## What is the measure of SRT Msrt degrees? Answer Expert Verified If this is the correct attachment of the problem, the measure of SRT in degrees is 90 degrees. Lines TW and SV are perpendicular lines that meet on point R. These lines are called perpendicular lines because they intersect at a right angle. hexagon 180°<|endoftext|>
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In this article, the inhabitants of the thirteen colonies that supported the American Revolution are primarily referred to as "Americans," with occasional references to "Patriots," "Whigs," "Rebels" or "Revolutionaries." Colonists who supported the British in opposing the Revolution are usually referred to as "Loyalists" or "Tories." The geographical area of the thirteen colonies that both groups shared is often referred to simply as "America." The American Revolutionary War (1775–1783), also known as the American War of Independence, began as a war between the Kingdom of Great Britain and thirteen united former British colonies on the North American continent, and ended in a global war between several European great powers. The war was the culmination of the political American Revolution, whereby the colonists rejected the right of the Parliament of Great Britain to govern them without representation. In 1775, revolutionaries gained control of each of the thirteen colonial governments, set up the Second Continental Congress, and formed a Continental Army. Petitions to the king to intervene with the parliament on their behalf resulted in Congress being declared traitors and the states in rebellion the following year. The Americans responded by formally declaring their independence as a new nation, the United States of America, claiming sovereignty and rejecting any allegiance to the British monarchy. In 1777 the Continentals captured a British army, leading to France entering the war on the side of the Americans in early 1778, and evening the military strength with Britain. Spain and the Dutch Republic – French allies – also went to war with Britain over the next two years. Throughout the war, the British were able to use their naval superiority to capture and occupy coastal cities, but control of the countryside (where 90% of the population lived) largely eluded them due to their relatively small land army. French involvement proved decisive, with a French naval victory in the Chesapeake leading to the surrender of a second British army at Yorktown in 1781. In 1783, the Treaty of Paris ended the war and recognized the sovereignty of the United States over the territory bounded by what is now Canada to the north, Florida to the south, and the Mississippi River to the west. Combatants before 1778 Edit American armies and militiasEdit At the outset of the war, the Thirteen Colonies lacked a professional army and navy. Each colony provided for its own defenses through the use of local militia. Militiamen were lightly armed, slightly trained, and usually did not have uniforms. Their units served for only a few weeks or months at a time, were reluctant to go very far from home, and were thus generally unavailable for extended operations. Militia lacked the training and discipline of regular soldiers but were more numerous and could overwhelm regular troops as at the battles of Concord, Bennington and Saratoga, and the siege of Boston. Both sides used partisan warfare but the Americans were particularly effective at suppressing Loyalist activity when British regulars were not in the area. Seeking to coordinate military efforts, the Continental Congress established (on paper) a regular army in June 1775, and appointed George Washington as commander-in-chief. The development of the Continental Army was always a work in progress, and Washington used both his regulars and state militia throughout the war. The United States Marine Corps traces its institutional roots to the Continental Marines of the war, formed at Tun Tavern in Philadelphia, by a resolution of the Continental Congress on November 10, 1775, a date regarded and celebrated as the birthday of the Marine Corps. At the beginning of 1776, Washington's army had 20,000 men, with two-thirds enlisted in the Continental Army and the other third in the various state militias. At the end of the American Revolution in 1783, both the Continental Navy and Continental Marines were disbanded. About 250,000 men served as regulars or as militiamen for the Revolutionary cause in the eight years of the war, but there were never more than 90,000 total men under arms at one time. Armies were small by European standards of the era; the greatest number of men that Washington personally commanded in the field at any one time was fewer than 17,000. This could be attributed to tactical preferences, but it also could be because of lack of powder on the American side. By comparison, Duffy notes that Frederick the Great usually commanded from 23,000 to 50,000 in battle. - Main article: Loyalist (American Revolution) Historians have estimated that approximately 40-45% of the colonists actively supported the rebellion while 15-20% of the population of the thirteen colonies remained loyal to the British Crown. The remaining 35-45% attempted to remain neutral. At least 25,000 Loyalists fought on the side of the British. Thousands served in the Royal Navy. On land, Loyalist forces fought alongside the British in most battles in North America. Many Loyalists fought in partisan units, especially in the Southern theater. The British military encountered many difficulties in maximizing the use of Loyalist factions. British historian Jeremy Black wrote, “In the American war it was clear to both royal generals and revolutionaries that organized and significant Loyalist activity would require the presence of British forces.” In the South, the use of Loyalists presented the British with “major problems of strategic choice” since while it was necessary to widely disperse troops in order to defend Loyalist areas, it was also recognized that there was a need for “the maintenance of large concentrated forces able” to counter major attacks from the American forces. In addition, the British were forced to ensure that their military actions would not “offend Loyalist opinion”, eliminating such options as attempting to “live off the country’, destroying property for intimidation purposes, or coercing payments from colonists (“laying them under contribution”). British armies and auxiliariesEdit Early in 1775, the British Army consisted of about 36,000 men worldwide, but wartime recruitment steadily increased this number. Great Britain had a difficult time appointing general officers, however. General Thomas Gage, in command of British forces in North America when the rebellion started, was criticized for being too lenient (perhaps influenced by his American wife). General Jeffrey Amherst, 1st Baron Amherst turned down an appointment as commander in chief due to an unwillingness to take sides in the conflict. Similarly, Admiral Augustus Keppel turned down a command, saying "I cannot draw the sword in such a cause." William Howe and John Burgoyne were both members of parliament who opposed military solutions to the American rebellion. Howe and Henry Clinton both made statements that they were not willing participants in the war, but were following orders. Over the course of the war, Great Britain signed treaties with various German states, which supplied about 30,000 soldiers. Germans made up about one-third of the British troop strength in North America. Hesse-Kassel contributed more soldiers than any other state, and German soldiers came to be known as "Hessians" to the Americans. Rebel propagandists called German soldiers "foreign mercenaries," and they are scorned as such in the Declaration of Independence. By 1779, the number of British and German troops stationed in North America was over 60,000, although these were spread from Canada to Florida. About 10,000 Loyalist Americans under arms for the British are included in these figures. African American — slave and free — served on both sides during the war. The British actively recruited slaves belonging to Patriot masters. Because of manpower shortages, George Washington lifted the ban on black enlistment in the Continental Army in January 1776. Small all-black units were formed in Rhode Island and Massachusetts; many were slaves promised freedom for serving. Another all-black unit came from Haiti with French forces. At least 5,000 black soldiers fought for the Revolutionary cause and more than 20,000 black soldiers fought on the British side. Most Native Americans east of the Mississippi River were affected by the war, and many communities were divided over the question of how to respond to the conflict. Though a few tribes were on friendly terms with the Americans, most Native Americans opposed the United States, since native lands were threatened by expanding American settlement. An estimated 13,000 warriors fought on the British side; the largest group, the Iroquois Confederacy, fielded about 1,500 men. War in the north, 1775–1780 Edit - Main article: Boston campaign Before the war, Boston had been the scene of much revolutionary activity, leading to the Massachusetts Government Act that ended home rule as a punishment in 1774. Popular resistance to these measures, however, compelled the newly appointed royal officials in Massachusetts to resign or to seek refuge in Boston. Lieutenant General Thomas Gage, the British North American commander-in chief, commanded four regiments of British regulars (about 4,000 men) from his headquarters in Boston, but the countryside was in the hands of the Revolutionaries. On the night of April 18, 1775, General Gage sent 700 men to seize munitions stored by the colonial militia at Concord, Massachusetts. Riders including Paul Revere alerted the countryside, and when British troops entered Lexington on the morning of April 19, they found 77 minutemen formed up on the village green. Shots were exchanged, killing several minutemen. The British moved on to Concord, where a detachment of three companies was engaged and routed at the North Bridge by a force of 500 minutemen. As the British retreated back to Boston, thousands of militiamen attacked them along the roads, inflicting great damage before timely British reinforcements prevented a total disaster. With the Battles of Lexington and Concord, the war had begun. The militia converged on Boston, bottling up the British in the city. About 4,500 more British soldiers arrived by sea, and on June 17, 1775, British forces under General William Howe seized the Charlestown peninsula at the Battle of Bunker Hill. The Americans fell back, but British losses were so heavy that the attack was not followed up. The siege was not broken, and Gage was soon replaced by Howe as the British commander-in-chief. In July 1775, newly appointed General Washington arrived outside Boston to take charge of the colonial forces and to organize the Continental Army. Realizing his army's desperate shortage of gunpowder, Washington asked for new sources. Arsenals were raided and some manufacturing was attempted; 90% of the supply (2 million pounds) was imported by the end of 1776, mostly from France. The standoff continued throughout the fall and winter. In early March 1776, heavy cannons that the patriots had captured at Fort Ticonderoga were brought to Boston by Colonel Henry Knox, and placed on Dorchester Heights. Since the artillery now overlooked the British positions, Howe's situation was untenable, and the British fled on March 17, 1776, sailing to their naval base at Halifax, Nova Scotia. Washington then moved most of the Continental Army to fortify New York City. - Main article: Invasion of Canada (1775) Three weeks after the siege of Boston began, a troop of militia volunteers led by Ethan Allen and Benedict Arnold captured Fort Ticonderoga, a strategically important point on Lake Champlain between New York and the Province of Quebec. After that action they also raided Fort St. John's, not far from Montreal, which alarmed the population and the authorities there. In response, Quebec's governor Guy Carleton began fortifying St. John's, and opened negotiations with the Iroquois and other Indian tribes for their support. These actions, combined with lobbying by both Allen and Arnold and the fear of a British attack from the north, eventually convinced the Congress to authorize an invasion of Quebec, with the goal of driving the British military from that province. (Quebec was at that time frequently referred to as Canada, as most of its territory included the former French Province of Canada.) Two Quebec-bound expeditions were undertaken. On September 28, 1775, Brigadier General Richard Montgomery marched north from Fort Ticonderoga with about 1,700 militiamen, besieging and capturing Fort St. Jean on November 2 and then Montreal on November 13. General Carleton escaped to Quebec City and began preparing that city for an attack. The second expedition, led by Colonel Arnold, went through the wilderness of what is now northern Maine. It was a logistical nightmare, with 300 men turning back, and another 200 perishing due to the difficult conditions. By the time Arnold reached Quebec City in early November, he had but 600 of his original 1,100 men. Montgomery's force joined Arnold's, and they attacked Quebec City on December 31, but were defeated by Carleton in a battle that ended with Montgomery dead, Arnold wounded, and over 400 Americans taken prisoner. The remaining Americans held on outside Quebec City until the spring of 1776, suffering from poor camp conditions and smallpox, and then withdrew when a squadron of British ships under Captain Charles Douglas arrived to relieve the siege. Another attempt was made by the Americans to push back towards Quebec, but they failed at Trois-Rivières on June 8, 1776. Carleton then launched his own invasion and defeated Arnold at the Battle of Valcour Island in October. Arnold fell back to Fort Ticonderoga, where the invasion had begun. While the invasion ended as a disaster for the Americans, Arnold's efforts in 1776 delayed a full-scale British counteroffensive until the Saratoga campaign of 1777. The invasion cost the Americans their base of support in British public opinion, "So that the violent measures towards America are freely adopted and countenanced by a majority of individuals of all ranks, professions, or occupations, in this country." It gained them at best limited support in the population of Quebec, which, while somewhat supportive early in the invasion, became less so later during the occupation, when American policies against suspected Loyalists became harsher, and the army's hard currency ran out. Two small regiments of Canadiens were recruited during the operation, and they were with the army on its retreat back to Ticonderoga. New York and New JerseyEdit - Main article: New York and New Jersey campaign Having withdrawn his army from Boston, General Howe now focused on capturing New York City. To defend the city, General Washington divided his 20,000 soldiers between Long Island and Manhattan. While British troops were assembling on Staten Island for the campaign, Washington had the newly issued Declaration of American Independence read to his men. No longer was there any possibility of compromise. On August 27, 1776, after landing about 22,000 men on Long Island, the British drove the Americans back to Brooklyn Heights in the largest battle of the entire Revolution. Howe then laid siege to fortifications there. In a feat considered by many historians to be one of his most impressive actions as Commander in Chief, Washington personally directed the withdrawal of his entire remaining army and all their supplies across the East River in one night without discovery by the British or losing a single man. On September 15, Howe landed about 12,000 men on lower Manhattan, quickly taking control of New York City. The Americans withdrew to Harlem Heights, where they skirmished the next day but held their ground. When Howe moved to encircle Washington's army in October, the Americans again fell back, and a battle at White Plains was fought on October 28. Once more Washington retreated, and Howe returned to Manhattan and captured Fort Washington in mid November, taking about 2,000 prisoners (with an additional 1,000 having been captured during the battle for Long Island). Thus began the infamous "prison ships" system the British maintained in New York for the remainder of the war, in which more American soldiers and sailors died of neglect than died in every battle of the entire war, combined. General Lord Cornwallis continued to chase Washington's army through New Jersey, until the Americans withdrew across the Delaware River into Pennsylvania in early December. With the campaign at an apparent conclusion for the season, the British entered winter quarters. Although Howe had missed several opportunities to crush the diminishing American army, he had killed or captured over 5,000 Americans. The outlook of the Continental Army was bleak. "These are the times that try men's souls," wrote Thomas Paine, who was with the army on the retreat. The army had dwindled to fewer than 5,000 men fit for duty, and would be reduced to 1,400 after enlistments expired at the end of the year. Congress had abandoned Philadelphia in despair, although popular resistance to British occupation was growing in the countryside. Washington decided to take the offensive, stealthily crossing the Delaware on Christmas night and capturing nearly 1,000 Hessians at the Battle of Trenton on December 26, 1776. Cornwallis marched to retake Trenton but was outmaneuvered by Washington, who successfully attacked the British rearguard at Princeton on January 3, 1777. Washington then entered winter quarters at Morristown, New Jersey, having given a morale boost to the American cause. New Jersey militia continued to harass British and Hessian forces throughout the winter, forcing the British to retreat to their base in and around New York City. At every stage the British strategy assumed a large base of Loyalist supporters would rally to the King given some military support. In February 1776 Clinton took 2,000 men and a naval squadron to invade North Carolina, which he called off when he learned the Loyalists had been crushed at the Battle of Moore's Creek Bridge. In June he tried to seize Charleston, South Carolina, the leading port in the South, hoping for a simultaneous rising in South Carolina. It seemed a cheap way of waging the war but it failed as the naval force was defeated by the forts and because no local Loyalists attacked the town from behind. The loyalists were too poorly organized to be effective, but as late as 1781 senior officials in London, misled by Loyalist exiles, placed their confidence in their rising. Saratoga and PhiladelphiaEdit When the British began to plan operations for 1777, they had two main armies in North America: Carleton's army in Quebec, and Howe's army in New York. In London, Lord George Germain approved campaigns for these armies which, because of miscommunication, poor planning, and rivalries between commanders, did not work in conjunction. Although Howe successfully captured Philadelphia, the northern army was lost in a disastrous surrender at Saratoga. Both Carleton and Howe resigned after the 1777 campaign. - Main article: Saratoga campaign The first of the 1777 campaigns was an expedition from Quebec led by General John Burgoyne. The goal was to seize the Lake Champlain and Hudson River corridor, effectively isolating New England from the rest of the American colonies. Burgoyne's invasion had two components: he would lead about 10,000 men along Lake Champlain towards Albany, New York, while a second column of about 2,000 men, led by Barry St. Leger, would move down the Mohawk River valley and link up with Burgoyne in Albany, New York. Burgoyne set off in June, and recaptured Fort Ticonderoga in early July. Thereafter, his march was slowed by Americans who literally knocked down trees in his path. A detachment was sent out to seize supplies but was decisively defeated in the Battle of Bennington by American militia in August, depriving Burgoyne of nearly 1,000 men. Meanwhile, St. Leger — half of his force Native Americans led by Sayenqueraghta — had laid siege to Fort Stanwix. American militiamen and their Native American allies marched to relieve the siege but were ambushed and scattered at the Battle of Oriskany. When a second relief expedition approached, this time led by Benedict Arnold, St. Leger's Indian support abandoned him, forcing him to break off the siege and return to Quebec. Burgoyne's army had been reduced to about 6,000 men by the loss at Bennington and the need to garrison Ticonderoga, and he was running short on supplies. Despite these setbacks, he determined to push on towards Albany. An American army of 8,000 men, commanded by the General Horatio Gates, had entrenched about 10 miles (16 km) south of Saratoga, New York. Burgoyne tried to outflank the Americans but was checked at the first battle of Saratoga in September. Burgoyne's situation was desperate, but he now hoped that help from Howe's army in New York City might be on the way. It was not: Howe had instead sailed away on his expedition to capture Philadelphia. American militiamen flocked to Gates' army, swelling his force to 11,000 by the beginning of October. After being badly beaten at the second battle of Saratoga, Burgoyne surrendered on October 17. Saratoga was the turning point of the war. Revolutionary confidence and determination, suffering from Howe's successful occupation of Philadelphia, was renewed. More importantly, the victory encouraged France to make an open alliance with the Americans, after two years of semi-secret support. For the British, the war had now become much more complicated. - Main article: Philadelphia campaign Having secured New York City in 1776, General Howe concentrated on capturing Philadelphia, the seat of the Revolutionary government, in 1777. He moved slowly, landing 15,000 troops in late August at the northern end of Chesapeake Bay. Washington positioned his 11,000 men between Howe and Philadelphia but was driven back at the Battle of Brandywine on September 11, 1777. The Continental Congress once again abandoned Philadelphia, and on September 26, Howe finally outmaneuvered Washington and marched into the city unopposed. Washington unsuccessfully attacked the British encampment in nearby Germantown in early October and then retreated to watch and wait. After repelling a British attack at White Marsh, Washington and his army encamped at Valley Forge in December 1777, about 20 miles (32 km) from Philadelphia, where they stayed for the next six months. Over the winter, 2,500 men (out of 10,000) died from disease and exposure. The next spring, however, the army emerged from Valley Forge in good order, thanks in part to a training program supervised by Baron von Steuben, who introduced the most modern Prussian methods of organization and tactics. General Clinton replaced Howe as British commander-in-chief. French entry into the war had changed British strategy, and Clinton abandoned Philadelphia in order to reinforce New York City, now vulnerable to French naval power. Washington shadowed Clinton on his withdrawal and forced a strategic victory at the battle at Monmouth on June 28, 1778, the last major battle in the north. Clinton's army went to New York City in July, arriving just before a French fleet under Admiral d'Estaing arrived off the American coast. Washington's army returned to White Plains, New York, north of the city. Although both armies were back where they had been two years earlier, the nature of the war had now changed. An international war, 1778–1783 Edit Template:Refimprove In 1778, the war over the rebellion in North America became international; spreading not only to Europe, but to the European colonies, chiefly in India. After learning of the American victory in Saratoga, France signed the Treaty of Alliance with the United States on February 6, 1778. Spain entered the war as an ally of France in June 1779, a renewal of the Bourbon Family Compact. Unlike France, however, Spain initially refused to recognize the independence of the United States — Spain was not keen on encouraging similar anti-colonial rebellions in the Spanish Empire. Both countries had quietly provided assistance to the Americans since the beginning of the war, hoping to dilute British power. So too had the Netherlands, eventually brought into open war at the end of 1780. In London King George III gave up hope of subduing America by more armies while Britain had a European war to fight. "It was a joke," he said, "to think of keeping Pennsylvania." There was no hope of recovering New England. But the King was determined "never to acknowledge the independence of the Americans, and to punish their contumacy by the indefinite prolongation of a war which promised to be eternal." His plan was to keep the 30,000 men garrisoned in New York, Rhode Island, Quebec, and Florida; other forces would attack the French and Spanish in the West Indies. To punish the Americans the King planned to destroy their coasting-trade, bombard their ports; sack and burn towns along the coast (like New London, Connecticut), and turn loose the Native Americans to attack civilians in frontier settlements. These operations, the King felt, would inspire the Loyalists; would splinter the Congress; and "would keep the rebels harassed, anxious, and poor, until the day when, by a natural and inevitable process, discontent and disappointment were converted into penitence and remorse" and they would beg to return to his authority. The plan meant destruction for the Loyalists and loyal Native Americans, and indefinite prolongation of a costly war, as well as the risk of disaster as the French and Spanish were assembling an armada to invade the British isles and seize London. The British planned to re-subjugate the rebellious colonies after dealing with their European allies. When the war began, the British had overwhelming naval superiority over the American colonists. The Royal Navy had over 100 ships of the line and many frigates and smaller craft, although this fleet was old and in poor condition, a situation which would be blamed on Lord Sandwich, the First Lord of the Admiralty. During the first three years of the war, the Royal Navy was primarily used to transport troops for land operations and to protect commercial shipping. The American colonists had no ships of the line, and relied extensively on privateering to harass British shipping. The privateers caused worry disproportionate to their material success although those operating out of French channel ports before and after France joined the war caused significant embarrassment to the Royal Navy and inflamed Anglo-French relations. During the war, about 55,000 American seamen served aboard the privateers. The American privateers had almost 1,700 ships, and they captured 2,283 enemy ships. The Continental Congress authorized the creation of a small Continental Navy in October, 1775, which was primarily used for commerce raiding. John Paul Jones became the first great American naval hero, capturing HMS Drake on April 24, 1778, the first victory for any American military vessel in British waters. French entry into the war meant that British naval superiority was now contested. The Franco-American alliance began poorly, however, with failed operations at Rhode Island in 1778 and Savannah, Georgia, in 1779. Part of the problem was that France and the United States had different military priorities: France hoped to capture British possessions in the West Indies before helping to secure American independence. While French financial assistance to the American war effort was already of critical importance, French military aid to the Americans would not show positive results until the arrival in July 1780 of a large force of soldiers led by the Comte de Rochambeau. Spain entered the war on the side of the Americans with the goal of recapturing Gibraltar and Minorca, which had been lost to the British in 1704. Gibraltar was besieged for more than three years, but the British garrison stubbornly resisted for years and was finally resupplied after Admiral Rodney's victory in the "Moonlight Battle" (January, 1780). Further Franco-Spanish efforts to capture Gibraltar were unsuccessful. One notable success took place on February 5, 1782 when Spanish and French forces captured Minorca, which Spain retained after the war. Ambitious plans for an invasion of England had to be abandoned. West Indies and Gulf CoastEdit Template:Seealso There was much action in the West Indies, with several islands changing hands, especially in the Lesser Antilles. At the Battle of the Saintes in April 1782, a victory by Rodney's fleet over the French Admiral de Grasse frustrated the hopes of France and Spain to take Jamaica and other colonies from the British. On May 8, 1782, Count Bernardo de Gálvez, the Spanish governor of Louisiana, captured the British naval base at New Providence in the Bahamas. Nevertheless, except for the French retention of the small island of Tobago, sovereignty in the West Indies was returned to the status quo ante bellum in the 1783 peace treaty. On the Gulf Coast, Gálvez seized three British Mississippi River outposts in 1779: Manchac, Baton Rouge, and Natchez. Gálvez then captured Mobile in 1780 and forced the surrender of the British outpost at Pensacola in 1781. His actions led to Spain acquiring East and West Florida in the peace settlement. India and the NetherlandsEdit The military action in North America and the Caribbean helped spark a conflict between Britain and France over India, in the form of the Second Anglo-Mysore War (1780-1784). The two chief combatants were Tipu Sultan, ruler of the Kingdom of Mysore and a key French ally, and the British government of Madras. In 1780, the British struck against the United Provinces of the Netherlands in order to preempt Dutch involvement in the League of Armed Neutrality, a declaration of several European powers that they would conduct neutral trade during the war. Britain was not willing to allow the Netherlands to openly give aid to the American rebels. Agitation by Dutch radicals and a friendly attitude towards the United States by the Dutch government — both influenced by the American Revolution — also encouraged the British to attack. The Fourth Anglo-Dutch War lasted into 1784 and was disastrous to the Dutch mercantile economy. It effectively ended the last Dutch pretence to being a global power, and paved the way for the Batavian Republic. - Main article: Southern theater of the American Revolutionary War During the first three years of the American Revolutionary War, the primary military encounters were in the north. After French entry into the war, the British turned their attention to the southern colonies, where they hoped to regain control by recruiting Loyalists. This southern strategy also had the advantage of keeping the Royal Navy closer to the Caribbean, where the British needed to defend their possessions against the French and Spanish. On December 29, 1778, an expeditionary corps from Clinton's army in New York captured Savannah, Georgia. An attempt by French and American forces to retake Savannah failed on October 9, 1779. Clinton then besieged Charleston, capturing it on May 12, 1780. With relatively few casualties, Clinton had seized the South's biggest city and seaport, paving the way for what seemed like certain conquest of the South. The remnants of the southern Continental Army began to withdraw to North Carolina but were pursued by Lt. Colonel Banastre Tarleton, who defeated them at the Waxhaws on May 29, 1780. With these events, organized American military activity in the region collapsed, though the war was carried on by partisans such as Francis Marion. Cornwallis took over British operations, while Horatio Gates arrived to command the American effort. On August 16, 1780, Gates was defeated at the Battle of Camden, setting the stage for Cornwallis to invade North Carolina. Cornwallis' victories quickly turned, however. One wing of his army was utterly defeated at the Battle of Kings Mountain on October 7, 1780. Tarleton was decisively defeated at the Battle of Cowpens on January 17, 1781, by American General Daniel Morgan. General Nathanael Greene, Gates' replacement, proceeded to wear down the British in a series of battles, each of them tactically a victory for the British but giving no strategic advantage to the victors. Greene summed up his approach in a motto that would become famous: "We fight, get beat, rise, and fight again." Unable to capture or destroy Greene's army, Cornwallis moved north to Virginia. In March 1781, General Washington dispatched General Lafayette to defend Virginia. The young Frenchman skirmished with Cornwallis, avoiding a decisive battle while gathering reinforcements. Cornwallis was unable to trap Lafayette, and so he moved his forces to Yorktown, Virginia, in July so the Royal Navy could return his army to New York. Northern and western FrontierEdit West of the Appalachian Mountains and along the border with Quebec, the American Revolutionary War was an "Indian War". Most Native Americans supported the British. Like the Iroquois Confederacy, tribes such as the Cherokees and the Shawnees split into factions. The British supplied their native allies with muskets and gunpowder and advised raids against civilian settlements, especially in New York, Kentucky, and Pennsylvania. Joint Iroquois-Loyalist attacks in the Wyoming Valley and at Cherry Valley in 1778 provoked Washington to send the Sullivan Expedition into western New York during the summer of 1779. There was little fighting as Sullivan systematically destroyed the Native American winter food supplies, forcing them to flee permanently to British bases in Quebec and the Niagara Falls area. In the Ohio Country and the Illinois Country, the Virginia frontiersman George Rogers Clark attempted to neutralize British influence among the Ohio tribes by capturing the outposts of Kaskaskia and Vincennes in the summer of 1778. When General Henry Hamilton, the British commander at Detroit, retook Vincennes, Clark returned in a surprise march in February 1779 and captured Hamilton himself. In March 1782, Pennsylvania militiamen killed about a hundred neutral Native Americans in the Gnadenhütten massacre. In one of the last major encounters of the war, a force of 200 Kentucky militia was defeated at the Battle of Blue Licks in August 1782. Yorktown and the Surrender of Cornwallis Edit The northern, southern, and naval theaters of the war converged in 1781 at Yorktown, Virginia. In early September, French naval forces defeated a British fleet at the Battle of the Chesapeake, cutting off Cornwallis' escape. Washington hurriedly moved American and French troops from New York, and a combined Franco-American force of 17,000 men commenced the Siege of Yorktown in early October. For several days, the French and Americans bombarded the British defenses. Cornwallis' position quickly became untenable, and he surrendered his entire army of 7,000 men on October 19, 1781. With the surrender at Yorktown, King George lost control of Parliament to the peace party, and there were no further major military activities on land. The British had 30,000 garrison troops occupying New York City, Charleston, and Savannah. The war continued at sea between the British and the French fleets in the West Indies. Treaty of Paris Edit - Main article: Treaty of Paris (1783) In London as political support for the war plummeted after Yorktown, Prime Minister Lord North resigned in March 1782. In April 1782, the Commons voted to end the war in America. Preliminary peace articles were signed in Paris at the end of November, 1782; the formal end of the war did not occur until the Treaty of Paris was signed on September 3, 1783, and the United States Congress of the Confederation ratified the treaty on January 14, 1784. The last British troops left New York City on November 25, 1783. Britain negotiated the Paris peace treaty without consulting her Native American allies and ceded all Native American territory between the Appalachian Mountains and the Mississippi River to the United States. Full of resentment, Native Americans reluctantly confirmed these land cessions with the United States in a series of treaties, but the fighting would be renewed in conflicts along the frontier in the coming years, the largest being the Northwest Indian War. Costs of the war Edit The total loss of life resulting from the American Revolutionary War is unknown. As was typical in the wars of the era, disease claimed more lives than battle. Historian Joseph Ellis suggests that Washington's decision to have his troops inoculated against the smallpox epidemic was one of his most important decisions. An estimated 25,000 American Revolutionaries died during active military service. About 8,000 of these deaths were in battle; the other 17,000 deaths were from disease, including about 8,000 - 12,000 who died while prisoners of war, most in rotting prison ships in New York. The number of Revolutionaries seriously wounded or disabled by the war has been estimated from 8,500 to 25,000. The total American military casualty figure was therefore as high as 50,000. About 171,000 seamen served for the British during the war; about 25 to 50 percent of them had been pressed into service. About 1,240 were killed in battle, while 18,500 died from disease. The greatest killer was scurvy, a disease known at the time to be easily preventable by issuing lemon juice to sailors. About 42,000 British sailors deserted during the war. Approximately 1,200 Germans were killed in action and 6,354 died from illness or accident. About 16,000 of the remaining German troops returned home, but roughly 5,500 remained in the United States after the war for various reasons, many eventually becoming American citizens. No reliable statistics exist for the number of casualties among other groups, including Loyalists, British regulars, Native Americans, French and Spanish troops, and civilians. The British spent about £80 million and ended with a national debt of £250 million, which it easily financed at about £9.5 million a year in interest. The French spent 1.3 billion livres (about £56 million). Their total national debt was £187 million, which they could not easily finance; over half the French national revenue went to debt service in the 1780s. The debt crisis became a major enabling factor of the French Revolution as the government was unable to raise taxes without public approval. The United States spent $37 million at the national level plus $114 million by the states. This was mostly covered by loans from France and the Netherlands, loans from Americans, and issuance of more and more paper money (which became "not worth a continental.") The U.S. finally solved its debt problem in the 1790s with the arrival of Alexander Hamilton and his National Bank. Template:POV-section The war of American independence could be described as a civil war within the Thirteen Colonies that escalated to a major war between European powers. It has also been argued that after the Declaration of Independence on July 4, 1776, it was a war between two different nations, America and Great Britain, as the Revolutionaries legitimately controlled the governments of all thirteen colonies. Whether or not people have the right to self determine territorial independence by democratic means remains a contentious issue. During the war the Americans benefited greatly from international assistance. In addition, Britain had significant military disadvantages. Distance was a major problem: most troops and supplies had to be shipped across the Atlantic Ocean. The British usually had logistical problems whenever they operated away from port cities, while the Americans had local sources of manpower and food and were more familiar with (and acclimated to) the territory. Additionally, ocean travel meant that British communications were always about two months out of date: by the time British generals in America received their orders from London, the military situation had usually changed. Suppressing a rebellion in America also posed other problems. Since the colonies covered a large area and had not been united before the war, there was no central area of strategic importance. In Europe, the capture of a capital often meant the end of a war; in America, when the British seized cities such as New York and Philadelphia, the war continued unabated. Furthermore, the large size of the colonies meant that the British lacked the manpower to control them by force. Once any area had been occupied, troops had to be kept there or the Revolutionaries would regain control, and these troops were thus unavailable for further offensive operations. The British had sufficient troops to defeat the Americans on the battlefield but not enough to simultaneously occupy the colonies. This manpower shortage became critical after French and Spanish entry into the war, because British troops had to be dispersed in several theaters, where previously they had been concentrated in America. The British also had the difficult task of fighting the war while simultaneously retaining the allegiance of Loyalists. Loyalist support was important, since the goal of the war was to keep the colonies in the British Empire, but this imposed numerous military limitations. Early in the war, the Howe brothers served as peace commissioners while simultaneously conducting the war effort, a dual role which may have limited their effectiveness. Additionally, the British could have recruited more slaves and Native Americans to fight the war, but this would have alienated many Loyalists, even more so than the controversial hiring of German mercenaries. The need to retain Loyalist allegiance also meant that the British were unable to use the harsh methods of suppressing rebellion they employed in Ireland and Scotland. Even with these limitations, many potentially neutral colonists were nonetheless driven into the ranks of the Revolutionaries because of the war. This combination of factors led ultimately to the downfall of British rule in America and the rise of the revolutionaries' own independent nation, the United States of America. See also Edit - Battles of the American Revolutionary War - Diplomacy in the American Revolutionary War - History of the United States of America - Intelligence in the American Revolutionary War - List of British Forces in the American Revolutionary War - List of Continental Forces in the American Revolutionary War - List of revolutions and rebellions - War of 1812 To avoid duplication, notes for sections with a link to a "Main article" will be found in the linked article. - ↑ British writers generally favor "American War of Independence", "American Rebellion", or "War of American Independence". See Omohundro Institute of Early American History and Culture, Bibliography at http://revolution.h-net.msu.edu/bib.html for usage in titles. - ↑ Black, War for America: The Fight for Independence, 1775–1783, p. 59. On militia see Boatner, p. 707, and Russell F. Weigley, The American Way of War (1973), ch. 2. - ↑ Crocker III, H. W. (2006). Don't Tread on Me. New York: Crown Forum. p. 51. ISBN 9781400053636. - ↑ Boatner, p. 264 says the largest force Washington commanded was "under 17,000"; Christopher Duffy (The Military Experience in the Age of Reason, 1715–1789, estimates Washington's maximum was "only 13,000 troops". - ↑ Calhoon, "Loyalism and neutrality" in Greene and Pole, A Companion to the American Revolution (2000) p.235 - ↑ Savas and Dameron p. xli - ↑ Black p. 12 - ↑ Black pg. 13-14 - ↑ Black p. 14 - ↑ Ketchum, 76 - ↑ Ketchum, 77 - ↑ Black, pp. 27–29; Boatner, pp. 424–26. - ↑ Weintraub, p. 240; figure for 1780. - ↑ Revolutionary all-black units: Kaplan and Kaplan, pp. 64–69. - ↑ American Revolution — African Americans In The Revolutionary Period. - ↑ James H. Merrell, "Indians and the New Republic" in The Blackwell Encyclopedia of the American Revolution, p. 393; Boatner, p. 545. - ↑ Higginbotham, p. 75–77. - ↑ Orlando W. Stephenson, "The Supply of Gunpowder in 1776," American Historical Review, Vol. 30, No. 2 (Jan. 1925), pp. 271–281 in JSTOR. - ↑ Arthur S. Lefkowitz, "The Long Retreat: The Calamitous American Defense of New Jersey 1776, 1998. Retrieved September 10, 2007. - ↑ Rockingham to Burke September 1776, Watson The Reign of George III p. 203. - ↑ McCullough - ↑ Stiles, Henry Reed. "Letters from the prisons and prison-ships of the revolution." Thomson Gale, December 31, 1969. ISBN 978-1432812225 - ↑ Dring, Thomas and Greene, Albert. "Recollections of the Jersey Prison Ship" (American Experience Series, No 8). Applewood Books. November 1, 1986. ISBN 978-0918222923 - ↑ Taylor, George. "Martyrs To The Revolution In The British Prison-Ships In The Wallabout Bay." (originally printed 1855) Kessinger Publishing, LLC. October 2, 2007. ISBN 978-0548592175. - ↑ Banks, James Lenox. "Prison ships in the Revolution: New facts in regard to their management." 1903. - ↑ Hawkins, Christopher. "The life and adventures of Christopher Hawkins, a prisoner on board the 'Old Jersey' prison ship during the War of the Revolution." Holland Club. 1858. - ↑ Andros, Thomas. "The old Jersey captive: Or, A narrative of the captivity of Thomas Andros...on board the old Jersey prison ship at New York, 1781. In a series of letters to a friend." W. Peirce. 1833. - ↑ Lang, Patrick J.. "The horrors of the English prison ships, 1776 to 1783, and the barbarous treatment of the American patriots imprisoned on them." Society of the Friendly Sons of Saint Patrick, 1939. - ↑ Onderdonk. Henry. "Revolutionary Incidents of Suffolk and Kings Counties; With an Account of the Battle of Long Island and the British Prisons and Prison-Ships at New York." Associated Faculty Press, Inc. June, 1970. ISBN 978-0804680752. - ↑ West, Charles E.. "Horrors of the prison ships: Dr. West's description of the wallabout floating dungeons, how captive patriots fared." Eagle Book Printing Department, 1895. - ↑ Higginbotham, pp. 188–98. - ↑ George Athan Billias. George Washington's Generals and Opponents: Their Exploits and Leadership (1994); Higginbotham, pp. 175–188. - ↑ George Otto Trevelyan, George the Third and Charles Fox: The Concluding Part of the American Revolution. (1912), vol. 1, p. 4. - ↑ Trevelyan, George the Third and Charles Fox vol. 1, p. 5. - ↑ Privateers or Merchant Mariners help win the Revolutionary War - ↑ Privateers - ↑ Higginbotham, pp. 331–46. - ↑ Number of British troops still in America: Piers Mackesy, The War for America: 1775–1783, p. 435. - ↑ Smallpox epidemic: Elizabeth Anne Fenn, Pox Americana: The Great Smallpox Epidemic of 1775–82, p. 275. A great number of these smallpox deaths occurred outside the theater of war — in Mexico or among Native Americans west of the Mississippi River. Washington and inoculation: Ellis, His Excellency: George Washington, p. 87. - ↑ American dead and wounded: John Shy, A People Numerous and Armed, pp. 249–50. The lower figure for number of wounded comes from Chambers, p. 849. - ↑ British seamen: Mackesy, p. 6, 176. - ↑ Robert and Isabelle Tombs, That Sweet Enemy: The French and the British from the Sun King to the Present (2007), p. 179. - ↑ Merrill Jensen, The New Nation (1950), p. 379. - ↑ Black, p. 39; Don Higginbotham, "The War for Independence, to Saratoga", in The Blackwell Encyclopedia of the American Revolution, p. 298, 306. - ↑ Higginbotham, p. 298, 306; Black, p. 29, 42. - ↑ Harsh methods: Black, pp. 14–16; slaves and Indians: Black, p. 35, 38. Neutrals into Revolutionaries: Black, p. 16. - Black, Jeremy. War for America: The Fight for Independence, 1775–1783. (2001). Analysis from a noted British military historian. - Boatner, Mark Mayo, III. Encyclopedia of the American Revolution. 1966; revised 1974. ISBN 0-8117-0578-1. Military topics, references many secondary sources - Chambers, John Whiteclay II, ed. in chief. The Oxford Companion to American Military History. Oxford University Press, 1999. ISBN 0-19-507198-0. - Duffy, Christopher. The Military Experience in the Age of Reason, 1715–1789. (1987). ISBN 0-689-11993-3. - Ellis, Joseph J. His Excellency: George Washington. (2004). ISBN 1-4000-4031-0. - Fenn, Elizabeth Anne. Pox Americana: The Great Smallpox Epidemic of 1775–82. New York: Hill and Wang, 2001. ISBN 0-8090-7820-1. - Greene, Jack P. and J.R. Pole, eds. The Blackwell Encyclopedia of the American Revolution. Malden, Massachusetts: Blackwell, 1991; reprint 1999. ISBN 1-55786-547-7. Collection of essays focused on political and social history. - Higginbotham, Don. The War of American Independence: Military Attitudes, Policies, and Practice, 1763–1789. Northeastern University Press, 1983. ISBN 0-930350-44-8. Overview of military topics; online in ACLS History E-book Project. - Kaplan, Sidney and Emma Nogrady Kaplan. The Black Presence in the Era of the American Revolution. Amherst, Massachusetts: The University of Massachusetts Press, 1989. ISBN 0-87023-663-6. - Ketchum, Richard M (1997). Saratoga: Turning Point of America's Revolutionary War. Henry Holt and Company, Inc. ISBN 0-8050-4681-X. - Mackesy, Piers. The War for America: 1775–1783. London, 1964. Reprinted University of Nebraska Press, 1993, ISBN 0-8032-8192-7. Highly regarded examination of British strategy and leadership. online edition - McCullogh, David. 1776. New York: Simon & Schuster, 2005. - Savas, Theodore P. and Dameron, J. David. A Guide to the Battles of the American Revolution. New York: Savas Beatie LLC, 2006. ISBN 10: 1-932714-12-X. - Shy, John. A People Numerous and Armed: Reflections on the Military Struggle for American Independence. New York: Oxford University Press, 1976 (ISBN 0-19-502013-8); revised University of Michigan Press, 1990 (ISBN 0-472-06431-2). Collection of essays. - J. Steven Watson; The Reign of George III, 1760–1815. 1960. Standard history of British politics. online edition - Weintraub, Stanley: Iron Tears; America's Battle for Freedom, Britain's Quagmire: 1775-1783. New York: Free Press, 2005 (a division of Simon and Schuster). ISBN 0-7432-2687-9 An account of the British politics on the conduct of the war. - Template:Cite web Further reading Edit These are some of the standard works about the war in general which are not listed above; books about specific campaigns, battles, units, and individuals can be found in those articles. - Bancroft, George. History of the United States of America, from the discovery of the American continent. (1854–78), vol. 7–10. - Bobrick, Benson. Angel in the Whirlwind: The Triumph of the American Revolution. Penguin, 1998 (paperback reprint). - Fremont-Barnes, Gregory, and Richard A. Ryerson, eds. The Encyclopedia of the American Revolutionary War: A Political, Social, and Military History (ABC-CLIO, 2006) 5 volume paper and online editions; 1000 entries by 150 experts, covering all topics - George Athan Billias. George Washington's Generals and Opponents: Their Exploits and Leadership (1994) scholarly studies of key generals on each side - Hibbert, Christopher. Redcoats and Rebels: The American Revolution through British Eyes. New York: Norton, 1990. ISBN 0-393-02895-X. - Jensen, Merrill. The Founding of a Nation: A History of the American Revolution 1763–1776. (2004) - Kwasny, Mark V. Washington's Partisan War, 1775–1783. Kent, Ohio: 1996. ISBN 0-87338-546-2. Militia warfare. - Middlekauff, Robert. The Glorious Cause: The American Revolution, 1763–1789. Oxford University Press, 1984; revised 2005. ISBN 0-19-516247-1. online edition - Savas, Theodore P., and Dameron, J. David. A Guide to the Battles of the American Revolution. New York, 2006. - Symonds, Craig L. A Battlefield Atlas of the American Revolution (1989), newly drawn maps - Ward, Christopher. The War of the Revolution. 2 volumes. New York: Macmillan, 1952. History of land battles in North America. - Weintraub, Stanley. Iron Tears: America's Battle for Freedom, Britain's Quagmire: 1775–1783. Free Press, 2004. Examination of the British political viewpoint. - Wood, W. J. Battles of the Revolutionary War, 1775–1781. ISBN 0-306-81329-7 (2003 paperback reprint). Analysis of tactics of a dozen battles, with emphasis on American military leadership. - Men-at-Arms series: short (48pp), very well illustrated descriptions: - Marko Zlatich, Peter Copeland. General Washington's Army (1): 1775–78 (1994); Zlatich. General Washington's Army (2): 1779–83 (1994); Rene Chartrand. The French Army in the American War of Independence (1994); Robin May, The British Army in North America 1775–1783 (1993) - The Partisan in War, a treatise on light infantry tactics written by Colonel Andreas Emmerich in 1789. - Chronology of the American Revolutionary War - Library of Congress Guide to the American Revolution - Battlefield atlas of the American Revolution West Point Atlas - American Revolutionary War History Resources - Entry to US Army Center for Military History, a huge bibliography - Political bibliography from Omohundro Institute of Early American History and Culture - Spain's role in the American Revolution from the Atlantic to the Pacific Ocean - African-American soldiers in the Revolution - American Revolution & Independence - Liberty — The American Revolution from PBS - American Revolutionary War 1775–1783 in the News - Haldimand Collection Haldimand Collection, 232 series fully indexed; extensive military correspondence of British generals - Africans in America from PBS - Approval of the American victory in England Unique arch inscription commemorates "Liberty in N America Triumphant MDCCLXXXIII" - The Spanish and Latin American Contribution to the American Revolutionary War ar:حرب الاستقلال الأمريكية bn:আমেরিকান বিপ্লব be:Вайна за незалежнасць Злучаных Штатаў Амерыкі be-x-old:Вайна за незалежнасьць ЗША br:Brezel dieubidigezh ar Stadoù Unanet bg:Американска война за независимост ca:Guerra d'Independència dels Estats Units cs:Americká válka za nezávislost cy:Rhyfel Annibyniaeth America da:Amerikanske uafhængighedskrig de:Amerikanischer Unabhängigkeitskrieg es:Guerra de Independencia de los Estados Unidos eo:Usona Revolucio fa:انقلاب آمریکا fr:Guerre d'indépendance des États-Unis d'Amérique ga:Cogadh Réabhlóideach Mheiriceá ko:미국 독립 전쟁 hr:Američki rat za neovisnost id:Perang Revolusi Amerika ia:Guerra de Independentia del Statos Unite is:Bandaríska frelsisstríðið it:Guerra di indipendenza americana he:מלחמת העצמאות של ארצות הברית kn:ಅಮೇರಿಕದ ಕ್ರಾಂತಿಕಾರಿ ಯುದ್ಧ lb:Amerikaneschen Onofhängegkeetskrich hu:Amerikai függetlenségi háború nl:Amerikaanse Onafhankelijkheidsoorlog ja:アメリカ独立戦争 no:Den amerikanske uavhengighetskrigen pl:Wojna o niepodległość Stanów Zjednoczonych pt:Guerra da Independência dos Estados Unidos da América ro:Războiul de independenţă al Statelor Unite ale Americii ru:Война за независимость США simple:American Revolutionary War sk:Americká vojna za nezávislosť sl:Ameriška vojna za neodvisnost sr:Амерички рат за независност fi:Yhdysvaltain vapaussota sv:Amerikanska revolutionskriget tr:Amerikan Bağımsızlık Savaşı zh:美國獨立戰爭<|endoftext|>
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NGC 7027, as it’s called, is one of the youngest and brightest (yet relatively unknown) planetary nebula located in our galaxy. Amazingly, it’s also quite small (comparatively). The actual nebula stretches out more than 14,000 AU’s (1 AU is the distance between the Earth and the sun), an act that likely started only 600 years ago. Most nebulae of this kind generally measure in at about 1 light-year (or a bit larger) in diameter. It’s absolutely fascinating to me that nebulae, formed in the same fashion (from the same general size and type of star), can create structures that look absolutely nothing alike. This is a great example of the great diversity — a planetary nebula, located approximately 3,000 light-years from Earth (in the Cygnus constellation). These nebulae form after sun-like stars exhaust their supply of hydrogen found in their core, causing the stars to balloon in size whilst the process of helium consumption begins. Throughout this process, the outer-layers of material peel off (similar to an onion), forming a planetary nebula (a fate our sun will experience eventually.) Larger Image can be found at NASA<|endoftext|>
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On Monday, January 16, people gathered across the nation to honor the late Martin Luther King Jr. Columbus Alive featured research by Derek Alderman, head of the Department of Geography, in examining the road it took to make this day a federal holiday. With such longstanding traditions, one might easily forget the hard-won fight to cement Martin Luther King Jr. Day as a federal holiday. As outlined by the King Center, the first legislation for MLK Day was introduced in 1968, but not signed into law until 1983. And it wasn’t until 1999 that all 50 states agreed to enact the holiday. Similarly, with nearly 900 US streets named for King as of 2014—according to research by Alderman—one might not realize the complex process of establishing those geographical markers. In the case of Columbus, naming Martin Luther King Jr. Boulevard, which runs from East Spring Street and Hamilton Avenue to Mt. Vernon and St. Clair avenues, brought up issues of race, money and community preservation.<|endoftext|>
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# Degrees To Gons Conversion ° 1 ° = 1.11111111111111 gon Want to convert from gons to degrees instead? Disclaimer: We've spent hundreds of hours building and testing our calculators and conversion tools. However, we cannot be held liable for any damages or losses (monetary or otherwise) arising out of or in connection with their use. Full disclaimer. ## How to convert degrees to gons (° to gon) The formula for converting degrees to gons is: gon = ° × 1.11111111111111. To calculate the degree value in gons first substitute the degree value into the preceding formula, and then perform the calculation. If we wanted to calculate 1 degree in gons we follow these steps: gon = ° × 1.11111111111111 gon = 1 × 1.11111111111111 gon = 1.11111111111111 In other words, 1 degree is equal to 1.11111111111111 gons. ## Example Conversion Let's take a look at an example. The step-by-step process to convert 4 degrees to gons is: 1. Understand the conversion formula: gon = ° × 1.11111111111111 2. Substitute the required value. In this case we substitute 4 for ° so the formula becomes: gon = 4 × 1.11111111111111 3. Calculate the result using the provided values. In our example the result is: 4 × 1.11111111111111 = 4.44444444444444 gon In summary, 4 degrees is equal to 4.44444444444444 gons. ## Converting gons to degrees In order to convert the other way around i.e. gons to degrees, you would use the following formula: ° = gon × 0.9. To convert gons to degrees first substitute the gon value into the above formula, and then execute the calculation. If we wanted to calculate 1 gon in degrees we follow these steps: ° = gon × 0.9 ° = 1 × 0.9 ° = 0.9 Or in other words, 1 gon is equal to 0.9 degrees. ## Conversion Unit Definitions ### What is a Degree? A degree is a unit of measure for angles, which is used to quantify the size of an angle. One degree is defined as 1/360th of a full rotation, which is a complete revolution around a circle. To better understand this definition, it's important to note that a circle has 360 degrees. Therefore, an angle that spans one quarter of a circle (i.e., 90 degrees) is a right angle, while an angle that spans half a circle (i.e., 180 degrees) is a straight angle. When measuring angles, degrees are usually denoted using the degree symbol (°), which is placed after the numerical value of the angle. For example, an angle of 45 degrees would be written as 45°. ### What is a Gon? The gon is a unit of measure for angles, which is used to quantify the size of an angle. One gon is defined as 1/400th of a full rotation, which is a complete revolution around a circle. To better understand this definition, it's important to note that a circle has 400 gons. Therefore, an angle that spans one quarter of a circle (i.e., 100 gons) is a right angle, while an angle that spans half a circle (i.e., 200 gons) is a straight angle. While the gon is not as commonly used as the degree, it is still used in some fields, such as surveying and geodesy, particularly in Europe and Russia. The advantage of using the gon over the degree is that it allows for easier calculation and conversion of angles, particularly when dealing with right angles and other angles that are a multiple of 100 gons. ## Degrees To Gons Conversion Table Below is a lookup table showing common degrees to gons conversion values. Degree (°)Gon 1 °1.11111111111111 gon 2 °2.22222222222222 gon 3 °3.33333333333333 gon 4 °4.44444444444444 gon 5 °5.55555555555555 gon 6 °6.66666666666666 gon 7 °7.77777777777777 gon 8 °8.88888888888888 gon 9 °9.99999999999999 gon 10 °11.1111111111111 gon 11 °12.22222222222221 gon 12 °13.33333333333332 gon 13 °14.44444444444443 gon ## Other Common DegreeConversions Below is a table of common conversions from degrees to other angle units. ConversionResult 1 degree in circles0.00277777777777778 circle 1 degree in gradians1.11111111111111 grad 1 degree in arcminutes60 arcmin 1 degree in radians0.0174532925 rad 1 degree in revolutions0.0027777777777 rev 1 degree in arcseconds3600 arcsec 1 degree in turns0.0027777777777 tr<|endoftext|>
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Definition of Anchor Tag in Html5 In an HTML document <a></a> tags or anchor tags are used for creating links. These tags are called anchor tags because they can be used as anchors on the page. But this is not in practice in the modern web development any more. They are not anchors any more they are just links. To discuss more about the anchor tags or <a> tags see the code below. Anchor Tag in Html Example/Hyperlink Tag Html5 <!DOCTYPE html> <html lang='en'> <head> <meta charset="UTF-8" /> <title> HTML Hyperlinks </title> </head> <body> <p> To understand more look at this link of <a href="https://howtocode.pk/" target=”_blank”>my home page</a>. Just the text inside the anchor tags is the part of the link. </p> </body> </html> To understand more look at this link of How to code. Just the text inside the anchor tags is the part of the link. Here in the above code you can see that we are having a paragraph of text. In the text we are using the <a> anchor tag for a link to our home page. As you can see that the text between the anchor tags is displayed in the output. There are two attributes that we are using in this <a> anchor tag. The first is the most important and essential is the “href” attribute. It specifies the URL of the desired page or file in this case it is the home page of howtocode.pk. The next attribute we are using is the target attribute which is in this case telling the browser to open the link in new window. The anchor tag is an inline tag. It can contain other inline tags as well as the block level tags in it. It means that if you want to make any tag a link simply put it into the anchor or hyperlink tag. To elaborate more we have the following code for you. HTML Code Example for Anchor tags <!DOCTYPE html> <html lang='en'> <head> <meta charset="UTF-8" /> <title> HTML Hyperlinks </title> </head> <body> <h2><a href="https://howtocode.pk/" target="_blank"> Block level elements in the anchor tag</a> </h2> </a> </body> </html> In the above output we have made the heading and the paragraph one link to the home page of the howtocode.pk. You can see that the color of the paragraph is changed showing that it is now a link. You can change the style of the link using CSS. To understand more look at this link of my home page. Just the text inside the anchor tags is the part of the link.<|endoftext|>
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Use this fantastic collection of teaching, display and activity resources in your Early Years setting. This huge pack includes a wide range of English and Maths materials, along with printables to help you organise your classroom. Join Teaching Packs to download these resources today! - Six Traditional Stories – A collection of stories that you can share with your class using a large display… or print them for shared / independent reading activities. - Six Story Sequencing Activities – Can your children cut out the pictures and put them into the correct order to retell the stories? Could they use them to retell the story to an audience or write a word / sentence to accompany each image? - Ten Sets of Alphabet Lines – Sets of printable strips that show the letters of the alphabet. Available in precursive and cursive fonts. - Four Pages of Letter and Phoneme Cards – A handy collection of cards, showing different letters and phonemes, that can be used for word building activities. - Twenty-Six Handwriting Sheets – A selection of printable pages that you can use to develop your children’s handwriting skills. Available in precursive and cursive fonts. - One Alphabet Matching Activity – Can your children match the pairs of shoes to review their knowledge of upper and lower case letters? Available with precursive and cursive fonts. - One ‘Wallaby Words’ Activity – Cut out the wallaby pictures and challenge your children to read the words and sort them into real / imaginary words by placing them into the correct wallaby’s pouch. - One Hundred and Four Pages of Letters, Numbers and Phonemes – A set of bold letters, numbers and phonemes that you can use on a classroom display board. Your children could decorate them by adding words or pictures of objects that include the letter / phoneme… or by adding different ways of showing the numbers. - One ‘I Spy…’ Resource – Ask your children to draw (or write the names of) objects that they can see beginning with a particular letter / phoneme. The boxes have been left blank so that you can add your own choice of letters / phonemes. - One Making Patterns Resource – Use the included colour, shape and emoji cards to create a repeating pattern. Why not laminate these resources so they can be used again and again? - Five Patterns Challenges – A set of printable worksheets that challenge your children to continue the colour, shape and number patterns. - Six Picture Patterns Activities – A selection of animals and objects with spaces for your children to colour / decorate a repeating pattern. - Six ‘Stepping Stone Patterns’ Printables – Can your children decorate the ‘stepping stones’ using a repeating to help the characters reach their target? - Two ‘Follow the Lines’ Printables – Simple activities to help your children to improve their pencil control. - Twelve Amazing Mazes – A collection of printable mazes based on a variety of themes. Can your children make their way through from the start to the finish? - Seventy-Five Colouring Pages – A HUGE colouring booklet with pages for your children to decorate. Includes a range of animals, people, vehicles, places, events and more! - Ten Number Posters – A set of posters that show the numbers 1 to 20 with spots for children to count. You could also print these at a smaller scale for use as number cards. - Two Sets of Bingo Cards – Two sets of thirty bingo cards so that you can play a game of ‘Bingo’ with your children. A set of matching cards are also included so they can be used as an independent matching activity. Challenge your children to find the matching digit / number of spots… or find the matching number that gives a total of 10 or 20. - Ten Sets of Number Lines – Sets of number lines (with numbers from 0 to 10) that your children can refer to during Maths activities. Themes include superheroes, rainforests, football, robots and more. A blank version is also available. - Three Pages of Number Strips – A collection of printable ‘strips’ that include the numbers 1-10, 1-20 and 1-50. Use them for counting, recognising numbers, identifying numbers before / after and more! - Two Sets of Number Jigsaws – Cut out the number cards and then cut them again along the diagonal lines. Can your children match the pieces of each number together again? - Two ‘Missing Numbers’ Activity Sheets – These number strips have missing numbers. Can your children write in the missing numbers? Available in two versions (from 1-10 and 1-20). - Sixty-One ‘Star Number’ Resources – A collection of printable pages that help your children to understand different ways of writing numbers up to twenty (and the numbers that come before and after each one). - Ten ‘Addition to 10 / 20’ Printables – Review your children’s addition skills with these simple addition sheets - Ten ‘Ordering Size’ Activities – Ten pages of images that your children can cut out and place in order of size. - Twenty-Seven Pages of ‘Mega Match’ Resources – This simple matching resource covers colours, numbers, letters, days of the week and months of the year. Print a mat and the accompanying set of cards and then ask your children to match the cards to items already shown on the mat. You can make this activity more challenging by asking the children to match lower case and upper case letters… or match digits with number words… or match the days / months in the correct order! - Fifteen Days, Months and Seasons Posters – Colourful posters (available in a variety of fonts) that show the names of the days / months. Use them on a display board or cut them up and use them for ordering activities. - Three ‘Mixing Colours’ Posters – Handy charts that show children how colours can be mixed to make new ones. A blank version is also available so that your children can add the ‘answers’. A ‘Colour Mixing Mat’ is also included so that your children can try mixing their own colours. - One Hundred Home Learning Activities – A huge set of activities that families can explore together at home, based on twenty different themes… or use them as classroom activities with your children! - Twenty Pages of Labels – Sets of labels that you can use for labelling your children’s books, lockers and more! These cover a wide variety of themes (including superheroes, space, pets, robots, pirates, minibeasts and dinosaurs). Watch the video below to view samples of resources from the Early Years Essentials Pack. All payments are handled securely. Questions? Visit our Help Pages. Reviews from our members: Submit your review This has got a great variety of early years activities. It saved me so much time as they were all in one pack and my son loves the activities. This pack is so useful, especially now there is a choice of alternative font (correct numbers font for my school). It covers all the basics and has activities which, though Early Years intended, are proving very useful for small step, repetitive work/games or practice for my lower KS2 group, who are still trying to get a solid grasp on early number concepts. This has been such a useful resource and saved me a lot of time too! I especially loved the stories which has been shared with my class many times both for shared and personal reading time, thank you 🙂 Just loved how much was in this pack. How colourful and well set out it was. There are so many things I can use and place up in my classroom.<|endoftext|>
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Question Video: The Derivative of an Inverse Sine Function | Nagwa Question Video: The Derivative of an Inverse Sine Function | Nagwa # Question Video: The Derivative of an Inverse Sine Function Mathematics • Higher Education Find (d/dπ‘₯) sin⁻¹ (π‘₯/π‘Ž), where π‘Ž β‰  0. 02:52 ### Video Transcript Find d by dπ‘₯ of inverse sin of π‘₯ over π‘Ž, where π‘Ž is not equal to zero. In this question, we need to differentiate the inverse or arcsin of π‘₯ over π‘Ž with respect to π‘₯. We will begin this question by letting 𝑦 equal the inverse sin of π‘₯ over π‘Ž. If we take the sin of both sides of this equation, we get sin 𝑦 is equal to π‘₯ over π‘Ž. Our next step is to differentiate both sides of this equation with respect to π‘₯. We know that differentiating sin π‘₯ gives us cos π‘₯. This means that differentiating sin 𝑦 with respect to π‘₯ using our knowledge of implicit differentiation gives us cos 𝑦 multiplied by d𝑦 by dπ‘₯. Differentiating the right-hand side gives us one over the constant π‘Ž. We can then divide both sides of this equation by cos 𝑦 such that d𝑦 by dπ‘₯ is equal to one over π‘Ž multiplied by cos 𝑦. We now have an expression for d𝑦 by dπ‘₯; however, this is not given in terms of π‘₯. If we go back to the equation sin 𝑦 is equal to π‘₯ over π‘Ž, squaring both sides of this equation gives us sin squared 𝑦 is equal to π‘₯ squared over π‘Ž squared. One of our trigonometric identities states that sin squared πœƒ plus cos squared πœƒ is equal to one. Rearranging this, we see that sin squared πœƒ is equal to one minus cos squared πœƒ. This means that sin squared 𝑦 is equal to one minus cos squared 𝑦. We now have one minus cos squared 𝑦 is equal to π‘₯ squared divided by π‘Ž squared. Multiplying both sides by π‘Ž squared gives us π‘Ž squared minus π‘Ž squared cos squared 𝑦 is equal to π‘₯ squared. We can then rearrange this equation so that π‘Ž squared minus π‘₯ squared is equal to π‘Ž squared multiplied by cos squared 𝑦. Finally, we square root both sides of this equation. The right-hand side becomes π‘Ž cos 𝑦. And the left-hand side is the square root of π‘Ž squared minus π‘₯ squared. d𝑦 by dπ‘₯ is therefore equal to one over the square root of π‘Ž squared minus π‘₯ squared. This is the derivative of inverse sin of π‘₯ over π‘Ž. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions<|endoftext|>
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# High School Math : Sectors ## Example Questions ### Example Question #5 : How To Find The Length Of An Arc Find the circumference of the following sector: Possible Answers: Correct answer: Explanation: The formula for the circumference of a sector is , where  is the radius of the sector and  is the fraction of the sector. Plugging in our values, we get: ### Example Question #1 : How To Find The Length Of An Arc In the circle above, the angle A in radians is What is the length of arc A? Possible Answers: Correct answer: Explanation: Circumference of a Circle = Arc Length ### Example Question #41 : Geometry Solve for . (Figure not drawn to scale). Possible Answers: Correct answer: Explanation: We can solve for the angle, , by using the below relationship. In the figure, intercepted arc is given. ### Example Question #11 : Sectors A pie has a diameter of 12". A piece is cut out, having a surface area of 4.5π. What is the angle of the cut? Possible Answers: 12.5° 90° 45° 4.5° 25° Correct answer: 45° Explanation: This is simply a matter of percentages. We first have to figure out what percentage of the surface area is represented by 4.5π. To do that, we must calculate the total surface area. If the diameter is 12, the radius is 6. Don't be tricked by this! A = π * 6 * 6 = 36π Now, 4.5π is 4.5π/36π percentage or 0.125 (= 12.5%) To figure out the angle, we must take that percentage of 360°: 0.125 * 360 = 45° ### Example Question #12 : Sectors Eric is riding a Ferris wheel. The Ferris wheel has 18 compartments, numbered in order clockwise. If compartment 1 is at 0 degrees and Eric enters compartment 13, what angle is he at? Possible Answers: 260 300 180 280 240 Correct answer: 240 Explanation: 12 compartments further means 240 more degrees. 240 is the answer. 360/12 = 240 degrees ### Example Question #42 : Geometry Solve for . Possible Answers: Correct answer: Explanation: The measure of an arc of an inscribed angle is twice the size of the inscribed angle. ### Example Question #13 : Sectors In the figure above that includes Circle O, the measure of angle BAC is equal to 35 degrees, the measure of angle FBD is equal to 40 degrees, and the measure of arc AD is twice the measure of arc AB. Which of the following is the measure of angle CEF? The figure is not necessarily drawn to scale, and the red numbers are used to mark the angles, not represent angle measures. Possible Answers: Correct answer: Explanation: The measure of angle CEF is going to be equal to half of the difference between the measures two arcs that it intercepts, namely arcs AD and CD. Thus, we need to find the measure of arcs AD and CD. Let's look at the information given and determine how it can help us figure out the measures of arcs AD and CD. Angle BAC is an inscribed angle, which means that its meausre is one-half of the measure of the arc that it incercepts, which is arc BC. Thus, since angle BAC is 35 degrees, the measure of arc BC must be 70 degrees. We can use a similar strategy to find the measure of arc CD, which is the arc intercepted by the inscribed angle FBD. Because angle FBD has a measure of 40 degrees, the measure of arc CD must be 80 degrees. We have the measures of arcs BC and CD. But we still need the measure of arc AD. We can use the last piece of information given, along with our knowledge about the sum of the arcs of a circle, to determine the measure of arc AD. We are told that the measure of arc AD is twice the measure of arc AB. We also know that the sum of the measures of arcs AD, AB, CD, and BC must be 360 degrees, because there are 360 degrees in a full circle. Because AD = 2AB, we can substitute 2AB for AD. This means the measure of arc AB is 70 degrees, and the measure of arc AD is 2(70) = 140 degrees. Now, we have all the information we need to find the measure of angle CEF, which is equal to half the difference between the measure of arcs AD and CD. ### Example Question #1 : How To Find The Angle Of A Sector The length of an arc, , of a circle is  and the radius, , of the circle is . What is the measure in degrees of the central angle, , formed by the arc ? Possible Answers: Correct answer: Explanation: The circumference of the circle is . The length of the arc S is . A ratio can be established: Solving for yields 90o Note: This makes sense. Since the arc S was one-fourth the circumference of the circle, the central angle formed by arc S should be one-fourth the total degrees of a circle. ### Example Question #15 : Sectors In the circle above, the length of arc BC is 100 degrees, and the segment AC is a diameter. What is the measure of angle ADB in degrees? Possible Answers: 90 40 100 cannot be determined 80 Correct answer: 40 Explanation: Since we know that segment AC is a diameter, this means that the length of the arc ABC must be 180 degrees. This means that the length of the arc AB must be 80 degrees. Since angle ADB is an inscribed angle, its measure is equal to half of the measure of the angle of the arc that it intercepts. This means that the measure of the angle is half of 80 degrees, or 40 degrees. ### Example Question #16 : Sectors What is the angle of a sector of area   on a circle having a radius of ? Possible Answers: Correct answer: Explanation: To begin, you should compute the complete area of the circle: For your data, this is: Now, to find the angle measure of a sector, you find what portion of the circle the sector is. Here, it is: Now, multiply this by the total  degrees in a circle: Rounded, this is .<|endoftext|>
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23 January 1585 - 30 January 1645 Mary Ward was born into an upper class catholic family; she was the eldest daughter of Marmaduke Ward and Ursula Wright. Even from childhood, Mary Ward knew religious persecution as she grew up in Reformation England, where to practice the Catholic faith could result in prison or death. Her grandmother was arrested and imprisoned for 14 years. She received a good classical education within the home environment. Music and art appealed to her, and her speaking and singing gave her much pleasure. She spoke and read several languages, including Latin. Like many Englishwomen from the higher classes, Mary Ward enjoyed much greater freedom and independence than was available to women in most Catholic countries at that time. It is therefore not surprising that Mary Ward was open to new ideas. From her earliest years Mary Ward was a deeply spiritual person. As a young girl she listened to stories of religious life and, although such a way of life was impossible in England, she was determined to travel to Europe to become a nun, rejecting several eligible men who sought to marry her. Schools for rich and poor At age 15 Mary Ward left England to enter a convent of Poor Clares in the Netherlands as a lay sister in 1606. The following year she founded a house for Englishwomen at Gravelines, but not finding herself called to the contemplative life, she resolved to devote herself to active work. She returned to England and at the age of twenty-four she found herself surrounded by a band of devoted companions determined to labour under her guidance; her cousin Barbara Babthorpe, Mary Poyntz and Winifred Wigmore. In 1609 they established themselves as a religious community at St.Omer, and opened schools for rich and poor. Mary worked in disguise to preserve the Catholic Faith in England before founding a community of active sisters in France. She and her companions educated young women, helped persecuted and imprisoned Catholics, and spread the word of God in places priests could not go. The Sisters lived and worked openly on the continent, but secretly in England, to nurture the faith by responding to need and opportunity. The work of religious women was then confined to prayer; all other works were expected to be carried out only within the walls of the convent. However Mary Ward promoted wider female education and values such as freedom from enclosure, from the obligation of choir and from wearing a religious habit. They formed the Institute of the Blessed Virgin Mary and their reputation spread throughout Europe as more schools were opened. Her delicate social sense caused her to insist all must be treated with the same courteous charity, including her least important employees and the poor. The Sisters' work was to include the children of the poor as well as the daughters of rich families entrusted to their care. The venture was a success, but it was a novelty, and it called forth censure and opposition as well as praise. Many who knew Mary Ward, from bishops and monarchs to simple people she served, admired her courage and generosity. She traveled Europe on foot, in dire poverty, frequently ill, founding schools in the Netherlands, Italy, Germany, Austria, the Czech Republic and Slovakia. In 1621 with Winifred Wigmore and others, Mary journeyed 1500 miles to Rome, crossing the Alps on foot in the winter. However enemies were now plotting against Mary. Criticized for her efforts to expand the role of women religious in spreading the faith, Mary was imprisoned by Church officials who called her a dangerous heretic. And, an unheard of thing for a woman in those days, she fearlessly faced a Congregation of Cardinals in order to defend her work in a long, well thought-out speech in Latin. Although she was zealous for the Church, and was repeatedly imprisoned for her faith, she kept herself free from fanaticism and continued to appeal to the Pope, though her health was now poor. Mary was then arrested by the Inquisition as a heretic. A Papal order was issued to destroy the Institute and her schools were closed down. Her work was destroyed and her Sisters scattered by her clerical enemies in whom even the Pope, Urban VIII, found "malice and folly." Never abandoning her trust in God's guidance, she died near York in 1645 during the Cromwellian Civil War. To the end she trusted totally that what God had asked of her would be accomplished in the future and her last words were ‘Jesus Jesus Jesus.’ Mary Ward taught by example and words. Act "without fear... in quiet confidence that God will do his will in the confusion." Her unwavering fidelity to "that which God would" was nourished by deep contemplative prayer. To Mary, God was the "Friend of all friends." She lived her fidelity with cheerfulness and a passion for the truth. Not until 1909 did the Church recognize Mary Ward as the founder of the IBVM. Much later Pope Pius XII acknowledged her as a pioneer for women's role in the church ministry outside the cloister. In her autobiography, Mary Ward promised her sisters and those who come to know her through them: "All I shall be able to perform in heaven or on earth, they may freely challenge as their due and my promise." Sister Mary Theresa By the mid-seventeenth century, small groups of Mary Ward's followers, known in Europe as the 'English Ladies', were to be found in Munich and Augsburg and later, on the invitation of members of the Catholic gentry, in London and York. It was to the foundation in York that in 1814, Frances Ball, a young Irish woman, was sent to live with and learn from the IBVM sisters. In 1821 Sister Mary Theresa returned to Dublin with two novices to establish a convent and school there. The birth of Loreto In 1822 she opened the first house of the Institute in Ireland, in Rathfarnham House, four miles from Dublin. She called it “Loretto House” (the spelling of which changed at the end of the nineteenth century), a name that was to be used for all the subsequent foundations that came from Ireland, and which resulted in the sisters of the Irish Branch of the IBVM being popularly known as “Loreto Sisters”. Mother Theresa was a woman of deep spirituality and significant administrative ability. Her energies were devoted to the establishment of many convent schools in Ireland and also in India (1842), Mauritius (1844), Gibraltar (1845), Canada (1847) and England (1851). She died at Loreto, Dalkey, on May 19th 1861. Please click this link to see an article about Mary Ward published by the Independent<|endoftext|>
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# Fraction calculator The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression. ## Result: ### 71/6 - 35/8 = 85/24 = 3 13/24 ≅ 3.5416667 Spelled result in words is eighty-five twenty-fourths (or three and thirteen twenty-fourths). ### How do you solve fractions step by step? 1. Conversion a mixed number 7 1/6 to a improper fraction: 7 1/6 = 7 1/6 = 7 · 6 + 1/6 = 42 + 1/6 = 43/6 To find new numerator: a) Multiply the whole number 7 by the denominator 6. Whole number 7 equally 7 * 6/6 = 42/6 b) Add the answer from previous step 42 to the numerator 1. New numerator is 42 + 1 = 43 c) Write a previous answer (new numerator 43) over the denominator 6. Seven and one sixth is forty-three sixths 2. Conversion a mixed number 3 5/8 to a improper fraction: 3 5/8 = 3 5/8 = 3 · 8 + 5/8 = 24 + 5/8 = 29/8 To find new numerator: a) Multiply the whole number 3 by the denominator 8. Whole number 3 equally 3 * 8/8 = 24/8 b) Add the answer from previous step 24 to the numerator 5. New numerator is 24 + 5 = 29 c) Write a previous answer (new numerator 29) over the denominator 8. Three and five eighths is twenty-nine eighths 3. Subtract: 43/6 - 29/8 = 43 · 4/6 · 4 - 29 · 3/8 · 3 = 172/24 - 87/24 = 172 - 87/24 = 85/24 For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(6, 8) = 24. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 6 × 8 = 48. In the next intermediate step, the fraction result cannot be further simplified by canceling. In words - forty-three sixths minus twenty-nine eighths = eighty-five twenty-fourths. #### Rules for expressions with fractions: Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part. The slash separates the numerator (number above a fraction line) and denominator (number below). Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2. Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3. Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45. The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3. An asterisk * or × is the symbol for multiplication. Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses. The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2 #### Examples: subtracting fractions: 2/3 - 1/2 multiplying fractions: 7/8 * 3/9 dividing Fractions: 1/2 : 3/4 exponentiation of fraction: 3/5^3 fractional exponents: 16 ^ 1/2 adding fractions and mixed numbers: 8/5 + 6 2/7 dividing integer and fraction: 5 ÷ 1/2 complex fractions: 5/8 : 2 2/3 decimal to fraction: 0.625 Fraction to Decimal: 1/4 Fraction to Percent: 1/8 % comparing fractions: 1/4 2/3 multiplying a fraction by a whole number: 6 * 3/4 square root of a fraction: sqrt(1/16) reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22 expression with brackets: 1/3 * (1/2 - 3 3/8) compound fraction: 3/4 of 5/7 fractions multiple: 2/3 of 3/5 divide to find the quotient: 3/5 ÷ 2/3 The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are: PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction. GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction. Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right. ## Fractions in word problems: • Erica Erica bought 3 1/2 yards of fabric. If she uses 2/3 of the fabric, how much will she have left? • Square metal sheet We cut out four squares of 300 mm side from a square sheet metal plate with a side of 0,7 m. Express the fraction and the percentage of waste from the square metal sheet. • Animal species Of 100 types of animals, 9/100 were discovered in ancient times, and 2/100 were discovered in the Middle Ages. Another 3/10 were discovered in the 1800s. What fraction of the 100 types of animals was discovered after the 1800s? Explain. • In bucket In bucket of 87 oranges 1/3 are spoiled ones and the remaining are eaten by the member of a family. How many oranges are eaten by the member of family? Erika’s career consists of filing, typing and answering phones. She spends 1/6 of her time filing and 5/8 of her time typing. What fraction of her time does she spend answering phone calls? • A laundry Mr. Green washed 1/4 of his laundry. His son washed 3/7 of it. Who washed most of the laundry? How much of the laundry still needs to be washed? • A market A market vendor was able to sell all the mangoes, papayas, and star apples.  1/5  of the fruits were mangoes,   2/3  of them were papayas and the rest were star apples.  How many parts of the fruits sold are star apples? • Circular garden Alice creates a circular vegetable garden. Tomatoes are planted in 1/3 of the circular garden, carrots are planted in 2/5 of the circular garden, and green peppers are planted in 1/10 of the circular garden. What fraction represents the remaining unplante • Coloured teacups The teacups in Tea Stop 55 are `2/5` green and `3/10` yellow. What fraction of the teacups are neither green nor yellow? • Employees Of all 360 employees, there are 11/12 women. How many men work in a company? • Visit to grandfather Shane's family traveled 3/10 of the distance to his grandfather’s house on Saturday. They traveled 4/7 of the remaining distance on Sunday. What fraction of the total distance to his grandfather’s house was traveled on Sunday? • Of the 2 Of the shapes,1/6 are triangles and 5/12 are pentagons. What fraction of the shapes are either triangles or pentagons? • Monica’s Monica's lawnmower had 1/8 of a gallon of gasoline in the tank. Monica started mowing and used all of the gasoline. She put 6/10 of a gallon of gasoline in the tank. After she mowed, 1/4 of a gallon was left in the tank. What was the total amount of gasol<|endoftext|>
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# The Cosine Rule – Explanation & Examples In the last article, we saw how the sine rule helps us calculate the missing angle or missing side when two sides and one angle is known or when two angles and one side are known. But what will you do when you are only given the three sides of a triangle, and you need to find all the angles? In the 15th century, that issue was resolved when a Persian mathematician, Jamshid al-Kashi, presented the Law of Cosines in a form suitable for triangulation. In France, it is still known as a Theoreme d’Al-Kashi. • The law of cosines, • how to apply the law of cosines to solve problems and, • the law of cosines formula. ## What is the Law of Cosines? The law of cosines also referred to as the cosine rule, is a formula that relates the three side lengths of a triangle to the cosine. The cosine rule is useful in two ways: • We can use the cosine rule to find the three unknown angles of a triangle if the three side lengths of the given triangle are known. • We can also use the cosine rule to find the third side length of a triangle if two side lengths and the angle between them are known. ### The law of cosines formula Consider an oblique triangle ABC shown below. An oblique triangle is a non-right triangle. Remember that the side lengths are labeled in lower case letters, while the angles are labeled in capital letters. Also, note that for each angle, the opposite side length is labeled using the same letter. The law of cosines states that: ⇒ (a) 2 = [b2 + c2 – 2bc] cos (A) ⇒ (b) 2 = [a2 + c2 – 2ac] cos (B) ⇒ (c) 2 = [a2 + b2 – 2bc] cos (C) You noticed that the equation c2 = a2 + b2 – 2bc cos (C) resembles the Pythagorean Theorem, except for the last terms,” – 2bc cos (C).” For this reason, we can say that the Pythagorean Theorem is a special of the sine rule. #### Proof of the law of cosines The cosine rule can be proved by considering the case of a right triangle. In this case, let’s drop a perpendicular line from point A to point O on the side BC. Let side AM be h. In the right triangle ABM, the cosine of angle B is given by: Cos (B) = Adjacent/Hypotenuse = BM/BA Cos (B) = BM/c BM = c cos (B) Given that BC = a, therefore, MC is calculated as; MC = a – BM = a – c cos (B) ……………………………………………… (i) In triangle ABM, the sine of angle B is given by; Sine B = Opposite/Hypotenuse = h/c h = c sine B …………………………………………………… (ii) By applying the Pythagorean theorem in right triangle AMC, we have, AC2 = AM2 + MC2……………………………………………… (iii) Substitute equation (i) and (ii) in equation (iii). b2 = (c Sine B)2 + (a – c Cos B)2 b2 = c2 Sine 2 B + a2 – 2ac Cos B + c2 Cos 2 C Rearranging the above equation: b2 = c2 Sine 2 B + c2 Cos 2 C + a2 – 2ac Cos B Factoring. b2 = c2 (Sine 2 B + Cos 2 C) + a2 – 2ac Cos B But, from trigonometric identities, we know that, sin2θ + cos2θ = 1 Therefore, b2 = c2 + a2 – 2ac Cos B Hence, the cosine law is proved. ## How to Use the Cosine Rule? If you we need to find the side lengths of a triangle, we use the cosine rule in the form of; ⇒ (a) 2 = [b2 + c2– 2bc] cos (A) ⇒ (b) 2 = [a2 + c2 – 2ac] cos (B) ⇒ (c) 2 = [a2 + b2 – 2bc] cos (C) And if we need to find the size of an angle, we use the cosine rule of the form; ⇒ cos A = (b2 + c2 – a2)/2bc ⇒ cos B = (a2 + c2– b2)/2ac ⇒ cos C = (a2 + b2– c2)/2ab Let’s now check our understanding of the cosine rule by attempting a few sample problems. Example 1 Calculate the length of side AC of the triangle shown below. Solution Because we want to calculate the length, we will therefore use the cosine rule in the form of; ⇒ (b) 2 = [a2 + c2 – 2ac] cos (B) By substitution, we have, b2 = 42 + 32 – 2 x 3 x 4 cos (50) b2 = 16 + 9 – 24cos50 = 25 – 24cos 50 b2 = 9.575 Determine the square root of both sides to get, b = √9.575 = 3.094. Therefore, the length of AC = 3.094 cm. Example 2 Calculate all three angles of the triangle shown below. Solution Since all the three side lengths of the triangle are given, then we need to find the measures of the three angles A, B, and C. Here, we will use the cosine rule in the form; ⇒ Cos (A) = [b2 + c2 – a2]/2bc ⇒ Cos (B) = [a2 + c2– b2]/2ac ⇒ Cos (C) = [a2 + b2– c2]/2ab Solve for angle A: Cos A = (72 + 52 – 102)/2 x 7 x 5 Cos A = (49 + 25 – 100)/70 Cos A = -26/70 Cos A = – 0.3714. Now, determine the cos inverse of – 0.3714. A = Cos -1 – 0.3714. A = 111.8° Solve for angle B: By substitution, cos B = (102 + 52– 72)/2 x 10 x 7 Simplify. Cos B = (100 + 25 – 49)/140 Cos B = 76/140 Determine the cos inverse of 76/140 B = 57.12° Solve for angle C: By substitution, cos C = (102 + 72– 52)/2 x 10 x 7 Cos C = (100 + 49 – 25)/140 Cos C = 124/140 Determine the cos inverse of 124/140. C = 27.7° Hence, the three angles of the triangle are; A = 111.8°, B = 57.12°, and C = 27.7°. ### Practice Questions 1. True or False: To apply the Cosine Law when solving triangles, we need at least one known side and two known angles. 2. True or False: We can use the Law of Cosine to find the angles of a triangle given its three sides. 3. What is the length of side $\overline{BC}$ from the triangle shown below? 4. What is the measure of$\angle ABC$ from the triangle shown below? 5. What is the measure of $\angle ABC$ from the triangle shown below?<|endoftext|>
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2.1 Bradycardia is a slow heart rate, defined as a heart rate of less than 60 beats per minute. Bradycardia can be caused by a range of factors, including diseases such as: sick sinus syndrome – a number of abnormal heart rhythms caused by an irreversible dysfunction of the sinus node (the heart's natural pacemaker), including: sinus arrest or pause, in which the sinus node occasionally does not generate electrical impulses, from a period lasting a couple of seconds to several minutes sinoatrial exit block, in which the sinus node generates electrical impulses normally, but the signal is blocked before it leaves the sinus node alternating bradyarrhythmias and tachyarrhythmias (a fast heart rate), such as bradycardia‑tachycardia syndrome atrioventricular block (a condition in which electrical impulses from the sinus node are slowed or blocked). Atrioventricular block can occur independently from sick sinus syndrome, and so people with symptomatic bradycardia due to sick sinus syndrome may also have or develop atrioventricular block. 2.2 The most commonly identified causes of abnormal heart rhythms are age, ischaemic heart disease, heart valve disorders and heart failure. If untreated, symptomatic bradycardia may lead to fatigue, fainting, palpitations, dizziness, heart failure and an increased risk of mortality. 2.3 Sick sinus syndrome is difficult to diagnose because of the intermittent symptoms, and also because symptoms are usually non‑specific and observed in other disorders. Diagnosis is made using electrocardiograms (ECGs). Because abnormalities may be intermittent, Holter monitoring (ECG monitoring for 24 to 48 hours) or event recorders may be used. 2.4 The prognosis of individuals with sick sinus syndrome is variable and difficult to predict, depending on the underlying cause and the presence and severity of comorbidities (such as ischaemic heart disease). For most people, the disease is idiopathic (that is, the cause is unknown) and progressive. People whose disease is not symptomatic do not need therapy; however, once the disease becomes symptomatic, it can have a significant impact on quality of life, and the only effective treatment is permanent implantation of a pacemaker. Most people who need a pacemaker implanted are older than 60 years. 2.5 The prevalence of sick sinus syndrome is thought to be about 0.03% of the whole population, and increases with age. However, both the prevalence of bradyarrhythmias due to sick sinus syndrome needing permanent pacemaker implant, and the prevalence of sick sinus syndrome with atrioventricular block, is unknown. Hospital episode statistics data from October 2012 to September 2013 included 2490 patients with a primary diagnosis of sick sinus syndrome in NHS hospitals in England. Sick sinus syndrome usually occurs in older adults, but it can affect people of any age, and affects men and women equally. The incidence of atrioventricular conduction abnormalities also increases with increasing age. 2.6 Pacemakers are electrical devices that consist of a small battery‑powered generator and 1 or more pacing leads that are in contact with the inner wall of the right atrium and/or the right ventricle. The primary aim of permanent pacing is to prevent the heart from beating too slowly. An important secondary aim is to reproduce, as far as possible, the function of the heart's normal electrical conduction system, which coordinates the way the heart muscle contracts. Pacemaker devices may be broadly classified as single‑ or dual‑chamber, depending on whether leads are applied to 1 or 2 heart chambers. Dual‑chamber pacemaker devices are attached to both chambers of the heart and may be used in either dual‑chamber pacing mode (in which both the right atrium and ventricle are paced, which mimics the natural pacing rhythm of the heart) or single‑chamber pacing mode (where only 1 chamber of the heart is paced, either the atrium or the ventricle). Single‑chamber pacemaker devices may be either single‑chamber atrial devices or single‑chamber ventricular devices, and may only be used in the mode (that is to pace the chamber) where the lead was originally placed. Pacemakers may also be rate modulating (that is, able to sense and adapt the rate of pacing to the level of physical exertion). 2.7 NICE technology appraisal guidance 88 (hereafter referred to as TA88) recommended the use of single‑chamber atrial pacemakers for treating sick sinus syndrome in people in whom, after full evaluation, there was no evidence of impaired atrioventricular conduction. The purpose of this part review is to update this recommendation because the DANPACE trial, which was published after the publication of TA88, has provided additional evidence comparing dual‑ with single‑chamber atrial pacemakers for this population. 2.8 In 2012/13 in England, more than 20,000 people had a single‑ or a dual‑chamber pacemaker fitted. Sick sinus syndrome was the fourth most prevalent primary diagnosis (9.5%) after atrial fibrillation and flutter (22.5%), complete atrioventricular block (18.8%) and second degree atrioventricular block (10.6%). For people with a primary diagnosis of sick sinus syndrome (2490 patients), 67.5% had implantation of a dual‑chamber pacemaker, 14.8% had implantation of a single‑chamber pacemaker and 2.2% had a re‑operation of an existing implanted pacemaker.<|endoftext|>
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By April Holloway, Contributing Writer, Ancient Origins Why is Christmas celebrated on December 25th? The popular answer is that it is Jesus’ birthday. However, it is necessary to reconsider that belief, No records exist in the Bible, or elsewhere, suggesting Jesus was born on that date. If it was not the birth of Christ which set Christmas Day apart from others in the calendar, what was it? To find out, you need to take look at ancient Persian and pagan traditions. Christmas Was Probably Not the Day Christ was Born First, let’s take a quick look at why Christmas probably wasn’t Jesus’ day of birth. As the Catholic Encyclopaedia states “there is no month in the year to which respectable authorities have not assigned Christ’s birth” (Catholic Encyclopaedia). That being said, there are several reasons supporting the idea that Jesus wasn’t born in December. Luke 2:8 states that on the night of Jesus’ birth “there were also in that same country shepherds living out of doors and keeping watches in the night over their flocks.” Scholars tend to agree that it is highly unlikely that shepherds were out with their flocks in the cold winter month. Luke 2:1-4 also claims that Joseph and Mary were traveling to Bethlehem to register in a Roman census when Jesus was born. These censuses were not known to have occurred in winter – also making it improbable to link Jesus’ birthday with the day now called Christmas. Okay, so Jesus probably wasn’t born on December 25th. Yet other important events did fall at that time of year for ancient pagans. The most well-known of these celebrations were Saturnalia and the birthday of the Sun God, Mithra. Originally, Saturnalia was held on December 17th, though the festival eventually was extended until December 25. This celebration honored Saturn, the God of Sowing and Husbandry, and was linked to the rise of a new year and the return of light. Ancient Romans would celebrate this date with a public banquet, giving gifts, partying, and holding a sacrifice in the Temple of Saturn. Roman Pagans who worshipped Mithras believed he was born on December 25th – the most holy day of the year for many ancient believers. This was a well-known cult for the Roman military in the 1st to 4th centuries AD. But Mithras is a god who has his origins in Persia beginning around the 6th century BC. The proto-Indo-Iranian language calls him ‘Mitra’, but that name was later adapted into Greek as ‘Mithras’. Many scholars believe that Roman soldiers discovered this god while completing military campaigns in Persia. The Mithraic New Year and Mithras’ birthday were celebrated on 25 December. The date was part of the Roman Natalis Invicti festival – a celebration linked to worshipping the sun in general. Uniting Pagan and Christian Beliefs When Constantine converted to Christianity in the 4th century, he may not have imagined how difficult it would be to convert pagans into Christians. To ease the transition, the birth of Jesus became associated with pagan holidays which fell in December. As the Pagan holidays gained Christian significance, it was decided that the birthday of the Sun God should also be the birthday of the Son of God. The Catholic Encyclopaedia quotes an early Christian stating, “O, how wonderfully acted Providence that on that day on which that Sun was born…. Christ should be born”.<|endoftext|>
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Sei sulla pagina 1di 4 # Assessment Pre Post Criteria: A & D Summative Due Date: Friday 12/5/14 ## Learning Targets (Standards: NS 1) I can simplify fractions. I can convert between mixed numbers and improper fractions. I can add and subtract fractions. I can multiply fractions. I can divide fractions and explain why my method makes sense. I can solve word problems with dividing fractions. I can use visuals to represent dividing fractions. I can use equations to represent dividing fractions. I can estimate effectively. I can round appropriately. I can explain my answers to the inquiry questions below: Inquiry Questions: Factual: How do we solve problems involving operations with fractions? Conceptual: How do we visually represent operations with fractions? Debatable: If there is only 1 right answer, is there only 1 right process to get there? Unit 3 ## Statement of We can use creative expression to model quantities in a variety of forms. Inquiry Global Personal & Cultural Expression Context Word Definition Benchmark compare things. Convert ## To change from one form to another. Denominator Difference Extend Improper fraction ## The number on the bottom of a fraction. The result of a subtraction problem. To take what we know and apply it to something new. A fraction where the numerator is bigger than the denominator. Mixed Number Numerator Product Proper fraction Quotient ## The result of a multiplication problem. A fraction where the numerator is smaller than the denominator. The result of a division problem. Reciprocal ## The inverse of a fraction. *A fraction times its reciprocal =1 Reduce/ Simplify Simplest Form Sum ## The easiest way to express something. The result of an addition problem. IB Criteria D: Applying Mathematics in Real-Life Contexts 7-8 5-6 3-4 1-2 ## You are able to: Identify the relevant elements of the real-life situation. Select adequate math strategies to model the situation. Apply math strategies to reach a correct solution. Explain the degree of accuracy of your solution context of the situation. You are able to: Identify the relevant elements of the real-life situation. Select adequate math strategies to model the situation. Apply math strategies to reach a valid solution. Describe the degree of accuracy of your solution context of the situation. You are able to: Identify the relevant elements of the real-life situation. Apply math strategies to reach a solution makes sense in the context of the situation. You are able to: Identify some parts of the real-life situation Apply math strategies to find a solution, but with limited success You did not reach any standard described above. ## In this unit, that means Relevant Elements: In word problems/graphs pulling out information needed to solve problem & ignoring the not-needed info Model: Visual includes a part to whole Visual has both parts of operation (what is being +-x/) Solution is easy to find in visual Explanation or key is included Solutions: Valid (5-6): you have an OK answer o a fraction not in simplest form o an improper fraction or mixed number when it shouldve been the other Correct (7-8): you have the best answer as a simplified fraction Degree of Accuracy: Explained(7-8): You told, with detail, why was accurate makes sense and gave some proof Stated(3-6): you said it makes sense but have NO PROOF 7-8 5-6 3-4 1-2 ## You are able to: Select appropriate math when solving challenging problems in familiar and unfamiliar situations. Apply the math successfully Generally solve these correctly You are able to: Select appropriate math when solving challenging problems in familiar situations. Apply the math successfully Generally solve these correctly You are able to: Select appropriate math when solving more complex problems in familiar situations. Apply the math successfully (with minimal guidance) Generally solve these correctly (with minimal guidance) You are able to: Select appropriate math when solving simple problems in familiar situations. Apply the math successfully (with guidance) Generally solve these correctly (with guidance) You did not reach any standard described above. ## In this unit, that means Appropriate Math to Solve Problems: Knowing what operation to use o All levels require you to use it correctly Familiar = problems youve seen and practiced before Unfamiliar= problems youve not seen or practiced Challenging Familiar & Unfamiliar (5-8) = o Mixed numbers o Improper fractions o Ugly numbers Familiar More complex (3-4)= o Not-common denom. o 2 digit num. & denom. Simple Familiar (1-2) = o Common denominators o Smaller numbers in numerator & denom. With guidance (1-2): parent or teacher helped you With minimal guidance (3-4): You got a small amount of help<|endoftext|>
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# CBSE Class 7 Maths Activity 4 Read and download CBSE Class 7 Maths Activity 4 chapter in NCERT book for Class 7 Mathematics. You can download latest NCERT eBooks for 2021 chapter wise in PDF format free from Studiestoday.com. This Mathematics textbook for Class 7 is designed by NCERT and is very useful for students. Please also refer to the NCERT solutions for Class 7 Mathematics to understand the answers of the exercise questions given at the end of this chapter (a) To get a median of a given triangle from any vertex by paper folding and to verify that in a triangle, medians pass through a single point. (b) To get an altitude of a given triangle from any vertex by paper folding and to verify that in a triangle altitudes pass through a single point. Learning Objective : To understand the concept of a median and an altitude of a triangle. Pre-requisite :  Familiarity with elements of triangles (vertices, sides and angles), types of triangles. Knowledge of median and altitude of a trangle, skill of paper folding. Materials Required : Thick papers, a pair of scissors, coloured pencil, a ruler. Procedure : (a) Step 1. Cut out a triangular shape from a thick paper and name it as ABC [Fig. 4(a)]. Step 2. Fold the side AC on itself so that vertex C falls on vertex A. Mark the point of intersection of the line of fold with AC as P [Fig. 4(b)]. P is the mid point of AC. Step 3. Similarly, find mid points of sides AB and BC and mark them as Q and R respectively [Fig. 4(b)]. Step 4. Now fold the triangular cut out to create a crease along BP. The crease thus obtained is the median from vertex B on the side AC. [Fig. 4(d)]. Step 5 Similarly, get medians from vertex A and C as AR and CQ. [Fig. 4(d)]. Please refer to the link below - CBSE Class 7 Maths Activity 4 ## Tags: Click for more Mathematics Study Material ## Latest NCERT & CBSE News Read the latest news and announcements from NCERT and CBSE below. Important updates relating to your studies which will help you to keep yourself updated with latest happenings in school level education. Keep yourself updated with all latest news and also read articles from teachers which will help you to improve your studies, increase motivation level and promote faster learning This has a reference to the letter no CBSE/COORD/2021 dated 11.02.2021 regarding conduct of the School-Based Assessment (Internal Assessment/Practical/ Project) of Class-XII and uploading the marks by 11.06.2021. It has been observed that some schools have not been... ### CBSE Class 12 Evaluation Criteria 2021 SUMMARY OF THE POLICY FOR TABULATION OF MARKS FOR CLASS XII BOARD EXAMINATIONS 2021   1(a) Due to cancellation of the Board examinations, the assessment of theory portion of 80/70/60/50/30 marks will be done by the school based on the following:     Class... Latest Syllabus for Class 12 for 2021 2022 Latest Syllabus for Class 11 for 2021 2022 Latest Syllabus for Class 10 for 2021 2022 Latest Syllabus for Class 9 for 2021 2022 CBSE has issued the latest syllabus for the academic year 2021 2022 which is applicable for all... ### CBSE Assessment Framework The Central Board of Secondary Education (CBSE) today, announced a suggested competency-based assessment framework to strengthen India’s existing school education system for secondary level (classes 6-10) and improve the overall learning outcomes of students across... ### Time management for CBSE students The first thing to learn about Time Management is that time is theoretical so you can’t really manage it. What you do when you get into time management, is that you manage yourself. You decide what has to be done, when it must be done and how to do it in the stipulated... ×<|endoftext|>
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# Find the magnitude ||v|| and the direction angle \theta for the given vector v. v = -4i + 11j ## Question: Find the magnitude {eq}\left\| \mathbf{v} \right\| {/eq} and the direction angle {eq}\,\theta\, {/eq} for the given vector {eq}\mathbf{v} {/eq}. {eq}\displaystyle\;\mathbf{v} = -4\mathbf{i} + 11\mathbf{j} {/eq} ## Magnitude and Direction of a vector: For a given vector, {eq}\vec{v}= x\vec{i} + y \vec{j} {/eq}: (i) The magnitude is: {eq}||\vec{v}|| = \sqrt{x^2+y^2} {/eq}. (ii) To find the direction, we will first find {eq}\alpha = \tan^{-1} \left|\dfrac{y}{x} \right| {/eq}. We then see which quadrant {eq}(x, y) {/eq} lies in and we will find the direction {eq}\theta {/eq} depending on the quadrant: • {eq}1^{st} {/eq} Quadrant: {eq}\theta=\alpha {/eq}. • {eq}2^{nd} {/eq} Quadrant: {eq}\theta= 180-\alpha {/eq}. • {eq}3^{rd} {/eq} Quadrant: {eq}\theta = 180+ \alpha {/eq}. • {eq}4^{th} {/eq} Quadrant: {eq}\theta = 360-\alpha {/eq}. The given vector is: $$\vec{v}= x\vec{i} + y \vec{j}=-4\vec{i} + 11 \vec{j}$$ Here, {eq}x=-4 {/eq} and {eq}y=11 {/eq}. Finding magnitude: The magnitude of the vector is, {eq}||\vec{v}|| = \sqrt{x^2+y^2}= \sqrt{(-4)^2+11^2} = \color{blue}{\boxed{\mathbf{\sqrt{137}}}} {/eq}. Finding direction: To find the direction, first, we will find: \begin{align} \alpha &= \tan^{-1} \left|\dfrac{y}{x} \right|\\[0.4cm] &= \tan^{-1} \left|\dfrac{11}{-4} \right|\\[0.4cm] &=\tan^{-1} \dfrac{11}{4} \end{align} We know that {eq}(x, y)= (-4, 11) {/eq} is in Quadrant II. So the required direction is: $$\theta = \color{blue}{\boxed{\mathbf{180^\circ - \tan^{-1} \dfrac{11}{4} }}} (OR) \color{blue}{\boxed{\mathbf{109.98311}}}$$ The final answer is rounded to five decimals.<|endoftext|>
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Microsoft Excel files are organized into a set of worksheets, each containing its own set of data. By default, a file has three worksheets, but you can merge the worksheets from one XLS file into another. The process is completed using the worksheet tabs at the bottom of the XLS file. Start Microsoft Excel and open the existing sheet into which you want to merge information (call it file1.xls) from another file. Open the other file (file2.xls) from which you want to merge information. Both files must be open at the same time. Right-click the first tab at the bottom of file2.xls (which contains the information you want to merge into file1.xls). Select "Move or Copy..." from the list of options. The program loads a new dialog box. Choose the name of the other file (file1.xls in this example) from the "Move Selected Sheets to Book" drop-down box. Click "(move to end)" from the "Before Sheet" box. Click "OK" to merge the worksheet from file2.xls to file1.xls. Select the next worksheet in the open file2.xls document. Repeat this process (steps two to four) to merge all worksheets into the first file (file1.xls).<|endoftext|>
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What smacks into Ceres stays on Ceres, research suggests A new set of high-velocity impact experiments suggests that the dwarf planet Ceres may be something of a cosmic dartboard: Projectiles that slam into it tend to stick. The experiments, performed using the Vertical Gun Range at NASA's Ames Research Center, suggest that when asteroids and other impactors hit Ceres, much of the impact material remains on the surface instead of bouncing off into space. The findings suggest the surface of Ceres could consist largely of a mish-mash of meteoritic material collected over billions of years of bombardment. The research, by Terik Daly and Peter Schultz of Brown University, is published in Geophysical Research Letters . Ceres is the largest object in the asteroid belt and the nearest dwarf planet to Earth. Until the recent arrival of the Dawn spacecraft, all that was known about Ceres came from telescopic observations. The observations showed Ceres to be mysteriously low in density, suggesting it is made either of very porous silicate material, or perhaps contains a large layer of water ice. Observations of its surface were remarkable as well—largely for being unremarkable. "It's really bland in the telescopic observations," said Daly, a Ph.D. student at Brown and the study's lead author. "It's like someone took a single color of spray paint and sprayed the whole thing. When we think about what might have caused this homogeneous surface, our thoughts turn to impact processes." And to understand impact processes, the researchers turned to NASA's Vertical Gun Range, a cannon with a 14-foot barrel that can launch projectiles at up to 16,000 miles per hour. For this work, Daly and Schultz wanted to simulate impacts into low-density surfaces that mimic the two broad possibilities for the composition of Ceres's surface: porous silicate or icy. "The idea was to look at those two end-member cases, because we really don't know yet exactly what Ceres is like," Daly said. For the porous silicate case, the researchers launched impactors into a powdered pumice. For the icy case, they used two targets: snow, and snow covered by a thin veneer of fluffy silicate material, simulating the possibility the Ceres's ice sits below a silicate layer. They then blasted these targets with pebble-sized bits of basalt and aluminum, simulating both stony and metallic meteorites. The study showed that in all cases, large proportions of the impact material remained in and around the impact crater. This was especially true in the icy case, Daly said. "We show that when you have a vertical impact into snow—an analog for the porous ice we think might be just beneath the surface of Ceres—you can have about 77 percent of the impactor's mass stay in or near the crater." The results were a bit of a surprise, said Schultz, who has studied impact processes for many years as professor of earth, environmental, and planetary sciences at Brown. "This is really contrary to previous estimates for small bodies," Schultz said. "The thought was that you'd eject more material that you'd collect, but we show you can really deliver a ton of material." The impact speeds used in the experiments were similar to speeds thought to be common in asteroid belt collisions. The findings suggest that a majority of impacts on porous bodies like Ceres cause an accumulation of impact material on the surface. "People have thought that perhaps if an impact was unusually slow, then you could deliver this much material," Schultz said. "But what we're saying is that for a typical, average-speed impact in the asteroid belt, you're delivering a ton of material." Over billions of years of such impacts, Ceres may have accumulated quite a bit of non-native material, Daly and Schultz said, much of it mixing together to create the relatively nondesdescript surface seen from telescopes. The researchers are hopeful that as the Dawn spacecraft scans the surface at much higher resolution, it might be able to pick out individual patches of this delivered material. That would help confirm the relevance of these experiments to celestial bodies, the researchers say. The results have implications for missions that aim to return asteroid samples to Earth. Unless the landing sites are carefully chosen, the researchers say, those missions could end up with samples that aren't representative of the object's original material. To get that, it might be necessary to find an area where there has been a relatively recent impact. "You can't do this like the old claw crane from the arcade," Schultz said. "You can't just reach down and grab whatever's there. You may need to find a fresh impact where perhaps the native stuff has been churned up."<|endoftext|>
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In our school the children eat twice a day, before going out to play. To ensure good concentration it is important for children to drink lots of water, so please encourage your child to bring a bottle of water. What is a healthy lunch? • It is full of goodness like protein, vitamins and calcium. • Contains food with no sugar, because too much sugar is bad for your teeth. • Contains no colouring or additives. • Gives us strength and energy • Makes us fit and healthy • Helps our brain power What could be in a healthy lunch? • Sandwiches or rolls with cheese, meat or other fillings. • Pitta bread, crackers • Fruit (peeled and chopped for small children) • Vegetables (washed and chopped)) • Yoghurt (easy to open) What drinks could we include in a healthy lunch? • Fruit juices • Diluted drinks Please click on the links below for more information and some ideas as to what we can include in our children’s lunch boxes. The children of Rath N.S. are very keen to have healthy lunches and show a good understanding of what that means. We hope these guidelines will assist everybody in making healthy choices. Regular physical activity is an important part of getting healthy and staying healthy. Encouraging children and teens to be active from a young age sets good habits early on and helps them develop the skills they need to stay active throughout their lives. Regular physical activity helps children and teens: - with healthy growth and development - build strong bones and muscles - improve balance and develop skills - maintain and develop flexibility - achieve and maintain a healthy weight - improve cardiovascular fitness - reduce stress and feel more relaxed - improve posture - boost confidence and self-esteem - have fun with their friends and make new ones Children who don’t get enough physical activity are at a greater risk of becoming overweight or obese. This makes it harder for them to be active and keep up in sport or play. Being overweight can also make children more prone to conditions such as asthma, flat feet and joint sprains. In the long term, it can contribute to conditions such as high blood pressure and cholesterol, heart disease, Type 2 Diabetes and liver disease. How much activity? Children and teens should get at least 60 minutes of physical activity every day, including vigorous activities that make them ‘huff and puff’. They can get even more health benefits from doing a couple of hours of physical activity every day. To help children and teens be active every day, they need opportunities for sport, play and exercise at school, after school and on weekends. Physical activity doesn’t all have to happen at the one time. It can be accumulated throughout the day; by walking or cycling to and from school, being involved in activities at school, active play at home or taking part in organised sport after school and on weekends.<|endoftext|>
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# Introductory Lesson – Area Model with Multiplication of Introductory Lesson – Area Model with Multiplication of Decimals Standard: MCC5.NBT.7 Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. EQ: What strategies can I use to multiply and divide decimals efficiently? How do I use an area model to help me multiply decimals? Materials: Base –Ten Grid Paper Activator: Review with students how to use an area model when multiplying with whole numbers. See example below: This model was intended to help link the calculations of the standard algorithm for multi-digit multiplication to the distributive property and reveal the partial products as areas of smaller rectangles. The students produced area models for multi-digit whole-number problems such as 57 x 14, shading and labeling four partial products in the model (500 + 200 + 70 + 28). You also want students to recognize that the distributive property was at work. Ex. (50 + 7) x (10 + 4). Lesson Part 1: Using the grid paper, model this problem: 5.7 x 1.4. Explain to students that indeed, the decimal multiplication problem 5.7 x 1.4 could be shown with the same base-10 diagram as the whole number multiplication 57 x 14, but with dimensions that are labeled as one-tenth the size of those in the whole number problem. Tell students that now that a square of 100 little boxes is 1 whole, a row of 10 squares is .1, and one tiny square is .01. You can do this by showing a key. Students need to note that the multipliers are the dimensions of the rectangle and the area shows the product. Further, in order to find the total product, one sums the areas of the smaller rectangles. In the diagram below, the smaller rectangles have the following dimensions: (1 x 5), (1 x .7), (.4 x 5), (.4 x .7) and respective areas of 5, 0.7, 2.0, and 0.28 square units, for a total of 7.98 square units. Finally, we want students to make the connection that the smaller regions have the same dimensions and the same areas as the partial products produced using the distributive property as this calculation shows: 5.7 x 1.4 = (5 + .7) x (1 + .4) = (5 x 1) + (5 x .4) + (.7 x 1) + (.7 x .4) = 5 + 2.0 + 0.7 + 0.28 = 7.98. Lesson Part 2: Pass out grid paper and have students set up the new problem: 2.6 x 4.3. Allow students to use two different color crayons; one color for the width (4.3) and one for the length (2.6). Allow them to work in groups. Give plenty of grid paper for mistakes. Once everyone has their grid labeled, then see if they can find the partial products. Find and color the product of 4 x 2 Find and color the product of .3 x 2 Find and color the product of .6 x 4 Find and color the product of .6 x .3 This also can look like this: (4 + .3) x (2 + .6) – Use distributive property to get answer…. Summarizer: Students are given a problem (2.3 x 4.1) and they will need to do it independently using grid paper.<|endoftext|>
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Lead Hazard Awareness Project: Fighting Lead-contaminated Soil and Dust Philadelphia’s smelters are shut down, and cars no longer run on leaded gasoline. But the lead they released still clings to the soil surface, along with flakes of exterior lead paint. The result: lead is in the dirt that sticks to shoes and hands after work or play in bare soil. Lead is in the dust that the wind picks up and blows through open windows.(Lead paint is also a source of toxic dust.) Children are most at risk of lead poisoning during early childhood, when lead dust is accidentally swallowed after normal hand-to-mouth behavior. Adults can inhale fine lead dust created by vigorous gardening, yard work, or home renovation without dust control. Testing for lead - Because both lead and its health effects are invisible, all children should be tested for lead before age one and again at age two. Ask your doctor. - Resources for testing yard soil and house dust are listed at the end of this brochure. Strategies for lead safety - In the yard and neighborhood, cover bare soil and wet down dust on walks and porch. - Keep lead-contaminated soil and dust out of your home. Less getting in means less to clean up. A bonus: Blocking lead dust also blocks pollen and other air pollutants that contribute to asthma. Where to start: Powerfully protective habits - Make a rule: Remove all shoes and boots at the door. This is the single most powerful way to reduce lead in house dust. Keep indoor shoes or slippers handy. - Ban smoking in the house. Tobacco is a source of lead in both dust and children. - Do not allow children or pets to play or dig in bare soil anywhere in the city, including parks. Walk dogs on leash or on grass, and consider keeping cats indoors. (Pets can be lead-poisoned too.) - Before working or playing outdoors, wash your hands and your children’s hands with regular soap and water. Fine dust sticks tightly to the natural oils in skin, and washing can cut hand dust in half. - After playing or working outdoors or on floors, wash hands (everyone!) before meals. Hand sanitizers do not remove lead. - If your job or hobby exposes you to lead dust, change your clothes at work and bring them home in a plastic bag. Wash separately. - Feed small children frequent, small, nutritious meals. A full stomach absorbs up to 90 percent less lead. - Read, talk, sing to your child--engaging in irresistibly interesting activities with a loved adult actually helps the developing brain fight off lead. Lead in soil and dust: cover it up, wet it down, keep it out, clean up what’s left 1. Cover it up: - Cover bare soil with a lead-binding layer of compost followed by heavy, water-permeable landscape cloth. Cover the cloth with patio stones, six inches of mulch, or raised garden beds (see Resources). Pay special attention to the soil within 10 feet of painted house walls. - Cover smaller bare spots—for example, under swings—with rubber mats. - Engage your neighbors in a cover-the-bare-soil project. Neighborhood soil and dust control is known to increase lead safety for everyone. 2. Wet it down: - In warm, dry weather, hose down your porch, sidewalk, and the adjacent street as well as any bare soil patches you see. Lead binds preferentially to the small soil particles that are picked up by a strong breeze. - If you see nearby excavation or construction without dust control, close your windows and storm windows and notify your neighbors and the city’s Air Management Services (215-685-7580). 3. Keep it out: - If you cool with window fans, place a 2-inch pleated electrostatic air filter behind each fan to block both outdoor dust and paint dust in the window trough. - For child safety, be sure to extend the sides of the fan firmly within the window track and lower the sash on the fan’s top. - Change the filter when it gets dirty or if the fan motor slows. - Place an exhaust fan in the north or northeast window of a separate room. - If you have a central air (HVAC) system, ask your service technician about installing the thickest pleated electrostatic filter that will not interfere with air flow. - If you have window air conditioners, try to place them in north or northeast windows, out of the prevailing wind. Because AC filters cannot capture the tiny dust particles that carry the most lead, consider setting a portable air cleaner nearby. 4. Clean up what’s left: - Lead-safe cleaning has two special requirements: to avoid stirring up loose dust and to remove the super-fine dust that sticks tightly to surfaces. - Vacuum carpets with either a sealed HEPA vacuum cleaner or a good household vacuum equipped with a HEPA filter. If possible, old threadbare carpets should be discarded or replaced. Worn spots can be covered with an area rug.<|endoftext|>
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## Tuesday, August 25, 2020 ### Lesson 6, Rule of Thirds, Part 2 of 2, Serlio The posts in this series  Lessons 1-7  are : https://www.jgrarchitect.com/2020/04/lessons.html https://www.jgrarchitect.com/2020/04/practical-geometry-lessons-2.html https://www.jgrarchitect.com/2020/04/practical-geometry-lesson-3.html https://www.jgrarchitect.com/2020/04/practical-geometry-lesson-4.html https://www.jgrarchitect.com/2020/04/practical-geometry-lesson-4b-old-first.html https://www.jgrarchitect.com/2020/06/practical-geometry-lessons-lesson-5.html https://www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-1_21.html https://www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-2-serlio.html https://www.jgrarchitect.com/2020/09/lesson-7-how-to-layout-frame-with-lines.html ## Friday, August 21, 2020 ### Lesson 6: The Rule of Thirds, Part 1of 2 The Rule of Thirds is what artists call the grid that appears on your cell phone. It helps you compose and edit. A variation of this is used in Practical Geometry. Sebastiano Serlio used this diagram in his book, On Architecture, published  in 1545.  He writes simple instructions for the reader; he says to construct the 'lines'. Note that the triangle (with its base at the bottom of the drawing) intersects the diagonals at the the upper corners of the door.  The width of the square is divided into thirds. Check how the division into thirds in the square above this drawing  lines up with those intersections.  Serlio is using a a variation of the Rule of Thirds. Like Owen Biddle (see Lesson 5) Serlio sets out basic Geometry as used in construction in Book 1. Then he explains how to solve problems.  He does not show how he knows where to draw the lines shown above. He assumes the reader knows. Here are the instructions: Draw a square; Add the diagonals to your square. Where they cross in the center. You have point 1. Divide one side of your square in half. Now you have  points 1 and 2. With 2 points you can draw a line. Add diagonals in each new rectangle. Add the diagonals from the square. If you were drawing this for a construction project on wood, on masonry, or on paper, you would not have separate squares.  All lines would be on your first square.  I have drawn each step without the extra lines for clarity. Do you see that the center line does not pass through the intersection of the diagonals? If you were the builder you would know that your diagonals will match when the line in centered. In this diagram they don't. So you would move  your center line. This is the diagram for Serlio's drawing for the door. For the Rule of Thirds (as we know it today) add the diagonals for the rectangles on both sides of the square. Note that you have intersections (4 points) not just where the lines  divide the square into smaller squares, but where the diagonals cross those lines.  2 points above the horizontal center line and 2 points below. Or: 2 on the right side of the vertical center line and 2 on the left. I have deliberately not added black points where the lines cross. You who are reading this will see it more clearly if you find those points yourself. Connect those new points and extend the lines across the square. You have drawn the Rule of Thirds. Similar diagonals could be drawn from the left to the right side and vice versa. I drew all the diagonals on graph paper to make it easier to follow.  The next lines to add would be the diagonals of the small squares. The line does not come back to its beginning until it has continued through the complete pattern The posts in this series  Lessons 1-7  are : https://www.jgrarchitect.com/2020/04/lessons.html https://www.jgrarchitect.com/2020/04/practical-geometry-lessons-2.html https://www.jgrarchitect.com/2020/04/practical-geometry-lesson-3.html https://www.jgrarchitect.com/2020/04/practical-geometry-lesson-4.html https://www.jgrarchitect.com/2020/04/practical-geometry-lesson-4b-old-first.html https://www.jgrarchitect.com/2020/06/practical-geometry-lessons-lesson-5.html https://www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-1_21.html https://www.jgrarchitect.com/2020/08/lesson-6-rule-of-thirds-part-2-serlio.html https://www.jgrarchitect.com/2020/09/lesson-7-how-to-layout-frame-with-lines.html ## Thursday, August 6, 2020 ### Owen Biddle's 'Young Carpenter's Assistant' , Plate I, G A note on Owen Biddle's Plate I, Diagram G. in his pattern book for beginning carpenters. * I wrote about Diagram G on this post: https://www.jgrarchitect.com/2020/06/practical-geometry-lessons-lesson-5.html I said that Biddle was not just introducing his 'carpenter assistant' to geometry; in Diagram G Biddle was explaining how to layout a square corner to work out a structural detail, cut a board, or set a frame on site. Since then I have explored the theoretical geometry of that diagram. The number of right angles which can be drawn in a circle is infinite. The rule always works. That understanding is part of why geometry is seen as mystical or sacred. This 'squaring the circle' diagram is from Robert Lawlor's Sacred Geometry*. (page 77, diagram 7.5) It uses a geometry similar geometry to Biddle's diagram G: a diameter and an angle. Here the diameters are evenly spaced and the same angle  is used at every point on the circumference. But the angle is not 90*. It is not a 'square angle'. This is decorative, not structural. The shapes do not close. The line continues for 5 rotations. It does not create a square, but seeks to define the perimeter of a circle with straight lines. , I am often told that I work with Sacred Geometry, that the geometric patterns I recover are theoretical, mystical, and sacred. I agree they are geometry. No, they are not sacred. They are practical. They are geometry used in construction. Here is how Biddle's diagram comes about: Begin with  a point  - A Choose a radius - A-B,  and draw a circle. Using the daisy wheel find the diameter - B- A- C, dotted and dashed line. Pick a point on the circumference of the circle - D. Here I have chosen 3 different D's  at random. Connect B-D and D-C. Each diagram will have a 90* (right) angle at the intersection of  B-D-C. Wherever the D is placed. the angle will be 90*. Biddle's Diagram G begins with my line B-D. It describes how to find my 90* angle of B-D-C. (his a-b-c) The answer is to find the diameter of a circle (a-d-c) that intersects a. That will give c. That will give the 90* the carpenter needs. By Hound and Eye* has a very similar diagram for drawing a right angle . The book is a  guide to furniture design, full of practical geometry. Each geometric problem is described step by step; practice work sheets are included. This pattern is the beginning of a handmade try square. *Owen Biddle's The Young Carpenter's Assistant, 1805, Philadelphia. Dover Publishing  reprint,  See my Bibliography for more information. *Robert Lawlor, Sacred Geometry, Philosophy and Practice, 1982, Thames and Hudson, London. *Geo.R. Walker & Jim Tolpin, By Hound and Eye, A Plain & Easy Guide to Designing Furniture with No Further Trouble, 2013,  Lost Art Press, Kentucky The diagram shown above is from page 57. This pattern is 4 overlapping hexagons. My granddaughter, who is 7, watched me add the images to this post. She wanted us to 'square the circle'. I did, using right angles where the diameters met the circumference. That produced these overlapping 6 hexagons, not squares. She watched closely and observed that accurate work was not easy: my lines did not always cross exactly in the center of the circle. When we finished she asked me to erase all the diameters. This is the result. Maybe she will show me later what she added to the copy I printed for her.<|endoftext|>
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What was “new” about the New South? The following will discuss what, if anything was new about the New South that emerged in the United States after 1877. Prior to the American Civil War the old South had predominantly been an agrarian economy in which blacks were slaves who had worked on the cotton plantations, factories, or had been domestic servants. Cotton had been the major commodity of the economy, which had mainly been exported to Britain. The American Civil War had been fought over the issue of slavery and whether the Southern States had the right to cede from the United States to preserve the institution of slavery (Hobsbawm, 1975 p.184). The Civil War brought social and economic changes to the South. Its cotton exports had been drastically reduced, its agricultural and industrial output declined sharply, whilst much of its infrastructure was destroyed. During the civil war President Lincoln had proclaimed the emancipation of all slaves, whilst blacks had fought with distinction on the Union side. The devastation brought to the South by the civil war meant that a period of reconstruction was needed afterwards. Leading white Southerners such as Henry Grady called for a New South. The blacks that were freed, as a result of the Confederate States losing the civil war, also anticipated a New South. The blacks in the Southern States expected their lives to be better following the Union’s victory and the era of reconstruction. In many respects strong arguments can be made that their lives got worse rather than better. Du Bois for one contended that blacks “had fought slavery to save democracy and then lost democracy in a new and vaster slavery” (Du Bois, 1935 Chapter 1). The result of the American Civil War in theory was that the four and a half million blacks in the United States were all free and equal with the white population. However, the end of the Reconstruction made those equal rights a mockery in the New South (Brogan, 1999, p.348). That the New South was not a new place for the better for its black population was due to the way in which the American Civil War ended. Lincoln’s assassination was the South’s revenge for losing the War. Lincoln’s successor, Andrew Johnson was less capable of ensuring that the South changed in ways that benefited its black population. From his presidency onwards, the North did very little to ensure Southern blacks had any meaningful rights (Brogan, 1999, p.348). Southern blacks were only able to exercise their political rights whilst the Union forces remained in the South, those rights ceased to exist in reality once the South was left to run itself. The suppression of Southern blacks was arguably worse once they had been formally freed than when they had been slaves. Racial discrimination, the fear of violence and poverty meant that the New South was no better than the Old South had been (Hobsbawm, 1975, p.143). Neither the South in general or its repressed black population in particular, gained as much from the United States rapid industrialisation from the 1870s onwards as the North did (Hobsbawm, 1987, p.35). In the New South there was a strong desire amongst the defeated Confederate States to make its black population subject to its strict political and economic controls for as long as possible. The abolition of slavery had not seen the end of the cotton plantations. However, jobs and better pay were given to the whites rather than blacks. Blacks were given the lowest paid jobs and they could be punished for not taking them. For many blacks the newness of the New South was the increased harshness of the discrimination they were subjected to. Whilst the whites in the New South had been unable to defeat the Union during the American Civil War, they were in a position to make life very unpleasant for the black population of the New South. Much discrimination was given legality through the ‘Black Codes’ of the Southern legislatures that severely restricted the rights of former slaves. Slavery had, in many respects, been restored in a less obvious form (Brogan, 1999, p.352). Those blacks that tried to exercise their legal rights found legal and political obstacles placed in front of them, which effectively deprived them of all those rights. They also faced violence and intimidation on a regular basis (Bradbury & Temperley, 1998, p.153). The Southern states were able to prevent the Constitutional Amendments that abolished slavery and gave freed slaves their rights having a positive impact as they were responsible for their enforcement, rather than the national government (Murphy et al, 2001, p.315). States such as Louisiana had no intention of giving blacks any rights on the grounds it was unconstitutional to do so (Du Bois, 1935, p.454). A series of measures which were known as Jim Crow laws were used by the Southern States to segregate and repress their black populations. Although they claimed the segregated services were of equal quality, this was a sham to excuse neglecting their black communities (Cobb, 1992). Overall Jim Crow Laws delayed the economic development of the New South, whilst they institutionalised racial discrimination and segregation. The cost of providing segregated services lowered the quality of education, housing, and transport in the New South. Segregation had even been endorsed by the Supreme Court as long as services were of equal quality, which few bothered to check. Such discrimination was contrary to the way Henry Grady believed the New South should have developed. Grady argued that the best way to industrialise the New South was to treat blacks as equal partners rather than inferiors. Therefore social justice and equality were just as important as capital and machinery in building the New South (Mauk & Oakland, 1995 p. 108). Grady believed that the New South would be the perfect democracy as long blacks were treated equally. The civil war had been an opportunity for the South to stop its outdated reliance on slavery and cotton (Harris, 1890 p. 15). Segregation, as well as being morally questionable, kept the South relatively poor and backward in relation to the rest of the country (Hobsbawm, 1975 p.184). Poverty was a new feature of the New South. Poverty paradoxically enough had not been an issue for blacks in the South when they had been slaves. Although, they had no freedom, slaves were provided with basic levels of accommodation and food, on the logical basis that unhealthy slaves did not work as well as healthy ones. Southern slave owners had generally treated their slaves well enough for their numbers to increase at the same rate as the white population (Bradbury & Temperley, 1998 p. 153). Defenders of slavery had maintained that it kept the Southern states economically competitive, kept the black population at subsistence, whilst ensuring that all white men could find paid work (Brogan, 1999, p.371). Poverty, as freed slaves found to their cost, was as restrictive of their freedom as actual shackles had been. Freed slaves had to compete with whites to gain jobs. Poverty was closely linked with racial discrimination, in that whites were given better jobs and better working conditions, even when there were better-qualified blacks to do the jobs. Discrimination in the provision of education, housing and medical care also contributed to keep the blacks repressed and in poverty (Cobb, 1992). Blacks were disenfranchised by their poverty, whereas loopholes were used to ensure that poor whites kept the vote (Hobsbawm, 1987, p.24). Another new feature of the New South was the increased levels of urbanisation. Cities such as New Orleans and Birmingham increased in size during the reconstruction era. The urbanisation of the New South was result of the industrial expansion encouraged by the Southern states and the migration of people trying to escape rural poverty. Migrating to the cities did not reduce racial discrimination and it barely increased opportunities for black people. Birmingham was the only city to achieve industrialisation on a major scale in the New South. The South was economically held back by its deliberately uneducated blacks and its under educated whites (Brogan, 1999, p.372). Southern blacks had also migrated to northern cities such as New York to increase their opportunities and to escape racial discrimination. The North was still prone to such discrimination even if it did give greater opportunity and blacks faced lower threats of violence. The Southern states had been motivated to enact the ‘Black Codes’ to restrict migration to both Southern and Northern cities (Brogan, 1999, p.363). Unemployment was a more obvious problem in the New South than it had been in the old South. Unemployment and low paid employment in a country with no public welfare provision was a serious problem, especially for blacks that were discriminated against and could not afford the basic necessities of life (Hobsbawm, 1987, p.103). Employers and plantation owners in the New South as a whole tended to keep the relationship between poor blacks and poor whites as unfriendly as possible. Factory and plantation owners feared that that if black and white workers had a good relationship they would form effective trade union movements and threaten the profits of the owners (Lewis, 1994). Discrimination in favour of white workers alienated blacks from them, whilst owners and employers kept control of their workers by threatening to use black workers as strike breakers. Such tactics were effective at preventing the emergence of trade unions but did nothing to improve race relations in the New South (Brogan, 1999, p. 371). The creation of Birmingham, Alabama was a symbol of all that was new in the New South. The place had not existed before 1871, and calling it Birmingham after one of the most industrialised cities in Britain was a statement of intent. Birmingham, Alabama was to be the industrial heart of the New South (Vann Woodward, 1951). Henry Grady himself cited Birmingham as the best example of his plans for a New South, yet historians have argued as to whether the development of Birmingham was similar to the industrial development envisaged by the plantation owners prior to the civil war (Lewis, 1994). Post civil war reconstruction gave the Southern States the opportunity as well as the need to reconstruct their economy. Falling prices for raw cotton meant that plantation owners switched their attention to manufacturing finished cotton products in new cotton mills. Attempts were also made to diversify the Southern economy away from cotton by developing coal, steal, and iron production. During the reconstruction period the Federal government had tried to enhance the economic prospects of the South by having the railroads rebuilt and extended to improve the transport links with the rest of the United States. Southern plantation owners, investors from the rest of the United States, as well as foreign investors funded industrial development. One feature of the New South did not change from the old South; it was still economically weaker than the North (Spiller et al, 2005 p. 80). The economic changes of the New South only benefited a few plantation and factory owners, some of who became much wealthier than they had been before the civil war (Hobsbawm, 1987 p. 24). A new feature of the New South was the high level of violence directed against the black population by white racists. In the immediate post-civil war period the formation of the Ku Klux Klan demonstrated the popularity for white supremacist ideas in the Southern states. The Ku Klux Klan added murderous intentions to their racist outlooks. The emergence of the Ku Klux Klan led to many thousands of lynchings and murders throughout the New South. Blacks found it very difficult to protect them-selves from racially motivated violence on such a large scale. They received no meaningful levels of protection from the police, the courts or the state authorities, which often sympathised with white supremacist views and were therefore unwilling to take action against the Ku Klux Klan or individual racists. Racism and prejudices were built into the ‘Black Codes’ that made a mockery of the post civil war Constitutional Amendments. The Federal courts and governments were unwilling to intervene in the affairs of the New South, as far as the Federal governments was concerned the Constitutional Amendments were fully operative in the South. Nobody in Washington DC seemed to be bothered to act upon the plentiful evidence of racial murders and discrimination in the New South. Between 1887 and 1917 official United States government figures showed that 2,734 blacks were murdered in racially motivated crimes, the vast majority in the New South. Before that period the death toll had been even higher, and only the presence of the Union army before 1877 had prevented further bloodshed (Murphy et al, 2001 p. 320). In some respects there were few new aspects in the New South. The combined effects of the Black Codes and Jim Crow laws meant that the New South restricted the freedoms of freed slaves to such an extent that slavery might as well have been retained. Economic, social, and political restrictions meant that insignificant numbers of blacks could vote in elections, own their land or gain education in the Southern States (Cobb, 1992). Low wages, unemployment, high rents, and direct discrimination were as effective as the Black Codes at keeping black people poor and powerless (Du Bois, 1935 p. 454). It is no wonder that many blacks believed that after reconstruction the New South made their lives worse than before. For them the only difference the old and new South was that they were underpaid for working on other peoples’ land and in other peoples’ factories rather than being paid at all. Only a small number of freed blacks had been able to make successes of their lives before the Jim Crow laws began to restrict opportunities. Only 4,000 freed slaves managed to purchase land in the New South, and most of them could not buy enough land to run successful farms (Murphy et al, 2001 p. 316). The New South was not a content place; the whites still fumed at their defeat in the civil war and re-imposed a quasi slavery upon the nominally free blacks (Hobsbawm, 1975 p. 143). Therefore, there were new aspects to the New South, although those aspects were not all positive or progressive in their nature. The Southern States were changed socially and economically as a result of the American Civil War. The economic consequences of the civil war were apparently severe. Agricultural and industrial outputs had been reduced, whilst the infrastructure of the Southern States had been badly damaged in the war. The war had disrupted the export of raw cotton which, had been the basis of the old South’s economy. Plantation owners had claimed that their plantations would be unprofitable with the abolition of slavery, a claim that proved unfounded due to the low wages they paid to white and black workers alike. The freed slaves found that life in the New South was in fact harsher in some respects than slavery. This was due to the increased racism and discrimination that was a new feature or perhaps at least a more obvious feature of the New South. The idea of the New South was promoted by the likes of Grady, as well as the new industrial centres such as Birmingham, Alabama and Atlanta. Overall in the period after 1877 industrial output in the Southern States did increase with the development of cotton mills, coal, steel, and iron production, although it still lagged behind the rest of the United States. Industrial development did not improve the lives of most people in the New South, just factory and plantation owners and the profits of outside investors. The legacy of the civil war was a long and bitter one, with the Southern whites repressing the blacks to compensate for defeat and demonstrate their alleged supremacy. Bradbury M & Temperley H, (1998) Introduction to American Studies 3rd edition, Longman, London Brogan H, (1999) The Penguin History of the USA, Penguin, London Cobb J C, (1992) The Most Southern Place on Earth: The Mississippi Delta & The Origins of Regional Identity, Oxford University Press, Oxford, and New York Du Bois W E B, (1935) Black Reconstruction in America, London Harris J C, (1890) Henry W. Grady: His Life, Writings, and Speeches, Cassell Publishing Co, New York Hobsbawm E, (1975) The Age of Capital 1848-1875, Weidenfeld & Nicholson, London Hobsbawm E, (1987) The Age of Empire 1875-1914, Weidenfeld & Nicholson, London Lewis (1994) the emergence of Birmingham as a case study of continuity between the antebellum planter class and industrialization in the new south” agricultural history (spring 1994) p. 62-79 Mauk, D & Oakland, J (1995) – American Civilization Routledge, London Murphy D, Cooper K & Waldron M, (2001). United States1776-1992 Collins, London New south plantation kingdom -‘the new south’ writings and speeches of Henry Grady, (1971) The beehive press savannah, Georgia Spiller J, Clancy T, Young S, and Mosley S (2005) The United States 1763 – 2001, Routledge, London<|endoftext|>
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WHAT IS TMJ? TMJ stands for temporomandibular joint. It is the joint that connects your jaw (mandible) to your skull (temporal bones), and it is very distinctive compared to the other major joints in the human body. The temporomandibular joint is complex and unique in its structure. First, you have to move both joints at the same time and cannot choose to only move one joint. The temporomandibular joint has two distinct movements unlike any other joint in the body. The initial motion is a hinge movement that is followed by a translational (sliding) movement where the jaw comes out of the area where it usually is so it is like an intentional dislocation. We use the TMJ more frequently than any other joint in our body. Every time we chew, talk, bite down or swallow we put the TMJ to work. The exact cause of a person’s TMJ disorder (TMD) is often difficult to determine, which is why it’s important to receive the correct treatment plan. WHAT ARE TEMPOROMANDIBULAR DISORDERS? Temporomandibular Disorders (TMD) are a complex and poorly understood set of conditions. TMD refers to problems related to pain, inflammation and improper functioning of: - Jaw joints (Temporomandibular joints) A condition in either the hard or soft tissues or both, including disc dislocation and arthritis/degeneration. This joint-related condition usually results from inflammation, disease or degeneration of the hard or soft tissues within the TMJ. Inflammation, disc dislocation (also called internal derangement) and degenerative arthritis are the most common arthrogenous disorders of the TMJ. - Muscles surrounding your jaw joints A condition affecting the muscles, usually brought on by overwork or tension in the jaw and supporting muscles. - Associated structures in your head, neck and face (including but not limited to your teeth, nerves and blood vessels)<|endoftext|>
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Normalization (image processing) In image processing, normalization is a process that changes the range of pixel intensity values. Applications include photographs with poor contrast due to glare, for example. Normalization is sometimes called contrast stretching or histogram stretching. In more general fields of data processing, such as digital signal processing, it is referred to as dynamic range expansion. The purpose of dynamic range expansion in the various applications is usually to bring the image, or other type of signal, into a range that is more familiar or normal to the senses, hence the term normalization. Often, the motivation is to achieve consistency in dynamic range for a set of data, signals, or images to avoid mental distraction or fatigue. For example, a newspaper will strive to make all of the images in an issue share a similar range of grayscale. Normalization transforms an n-dimensional grayscale image with intensity values in the range (Min,Max), into a new image with intensity values in the range (newMin,newMax). For example, if the intensity range of the image is 50 to 180 and the desired range is 0 to 255 the process entails subtracting 50 from each of pixel intensity, making the range 0 to 130. Then each pixel intensity is multiplied by 255/130, making the range 0 to 255. Normalization might also be non linear, this happens when there isn't a linear relationship between and . An example of non-linear normalization is when the normalization follows a sigmoid function, in that case, the normalized image is computed according to the formula Where defines the width of the input intensity range, and defines the intensity around which the range is centered. Auto-normalization in image processing software typically normalizes to the full dynamic range of the number system specified in the image file format.<|endoftext|>
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In 2008, scientists working in Denisova Cave—a cold site in Siberia’s Altai Mountains—uncovered a strange pinky bone, broader than a typical human’s. The DNA within that bone revealed that its owner belonged to an entirely new group of ancient hominins, distinct from Homo sapiens or Neanderthals. That group became known as the Denisovans. Researchers have since decoded the Denisovan genome. But still, no one can say what they looked like. Every known Denisovan fossil would fit in your palm—that pinky, three teeth, and a remarkable bone sliver from a Denisovan-Neanderthal hybrid. And all of these remains came from the same cave. But now, an international team of scientists has announced the identification of another Denisovan fossil, from a site 1,500 miles away. It’s the right half of a jawbone, found some 10,700 feet above sea level in a cave in China’s Xiahe County, on the eastern edge of the Tibetan plateau. The Xiahe mandible, as it is now known, is not only the first Denisovan fossil to be found outside Denisova Cave, but also the very first Denisovan fossil to be found at all. It just took four decades for anyone to realize that. The mandible was discovered by a local monk in 1980 and donated to Lanzhou University. There, it lay unstudied until 2010, when a team led by Fahu Chen and Dongju Zhang—a climatologist and an archaeologist, respectively—began examining it in earnest. The world learned about the existence of the Denisovans at around that time, and though fossils had only been recovered from Siberia, it was clear that these hominins likely existed throughout much of East Asia. Smatterings of Denisovan DNA still persist in the genes of living people in this region and beyond, and how else could it have made it into the genomes of modern Tibetans or Melanesians? Still, “I never imagined that [the Xiahe mandible] could be a Denisovan,” Zhang says. “If it was one, we’d be so lucky,” she adds. The mandible itself is very thick and sturdy. It has no chin, which rules out modern humans. The teeth within it are exceptionally large, and different in shape and size from those of Neanderthals, Homo erectus, and other known hominins. The molecules in the specimen were especially telling. The team couldn’t detect any traces of ancient DNA, but it did find the next best thing—fragments of ancient collagen proteins, still lurking in one of the teeth. These fragments closely resemble the proteins of Denisovans, more so than those of Neanderthals, modern humans, or other great apes. But Katerina Douka of the Max Planck Institute for the Science of Human History notes that methods for analyzing ancient proteins are relatively new, and less well tested than those for studying ancient DNA. Researchers should use both techniques on other specimens to check that they give the same results, Douka says. But for now, based on the data that exist, she agrees that the Xiahe mandible most likely belonged to a Denisovan. “It confirms that the Denisovans were perhaps widely distributed through East Asia,” Zhang says. For years, scientists had suspected as much. After all, people across East Asia and Melanesia (the region that includes New Guinea and its neighboring islands) have Denisovan DNA in their genes. This pattern—the product of ancient sexual encounters between Denisovans and humans—shouldn’t be possible if the Denisovans were just confined to a small Siberian cave. Instead, it seemed that they were already living in much of East Asia by the time ancient humans also spread through the region. Indeed, the Xiahe mandible, which is 160,000 years old, is by far the earliest hominin fossil from the Tibetan plateau. Researchers used to think that Homo sapiens was unique in adapting to the Himalayas, but the Denisovans were successfully living on the roof of the world at least 120,000 years earlier. They must also have adapted to extremely thin air—after all, the mandible was found in a cave that’s some 8,000 feet higher above sea level than Denisova itself. “Their presence that high up is truly astonishing,” Douka says. This helps to explain a remarkable finding from 2014. Back then, Emilia Huerta-Sanchez and her colleagues showed that most Tibetan people carry a mutated version of the EPAS1 gene, which helps them cope with high-altitude air that has 40 percent less oxygen than what most people breathe. And that mutation, the team showed, came from Denisovans. By having sex with these hominins, ancient Tibetans picked up a useful genetic trait that their descendants still benefit from. That result was surprising, because Denisova Cave is so far from Tibet, and so much lower in altitude. The new mandible resolves that discrepancy. Although it’s unclear whether its owner had the same EPAS1 variant that the other Denisovans did, it at least shows that Denisovans were in the right part of the world. “I was thrilled that they found a Denisovan-like jawbone at high altitude,” Huerta-Sanchez says. “The new discovery is an important step in understanding the Denisovans, but the big question still remains to be solved,” says Yousuke Kaifu of the National Museum of Nature and Science, Tokyo. And that is: If Denisovans were spread throughout Asia, why do Melanesians have so much more of their DNA than anyone else—5 percent, compared with just 0.2 percent in East Asians, and nothing in other groups? To answer that question, scientists will need to find more Denisovan bones. Douka and her colleagues have started a project called Finder to do exactly that, by rapidly analyzing small, unidentifiable slivers from various sites in Asia. More intact specimens might also be lying around in museum collections. For example, the Xiahe team notes that its mandible has many similarities to the Penghu 1 mandible, which was fished out of the ocean near Taiwan in 2008. (“I agree that there are some similarities,” says Kaifu, who led the team that analyzed Penghu 1 in 2015.) China has a long list of similar hominin fossils that have been hard to assign to other species. “Some of those may already be Denisovans,” Zhang says.<|endoftext|>
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# Determinants of Matrix 4×4 with Examples Her you will learn how to find determinants of matrix 4×4 with example. Let’s begin – ## Determinants of Matrix 4×4 To evaluate the determinant of a square matrix of order 4 we follow the same procedure as discussed in previous post in evaluating the determinant of a square matrix of order 3. If A = $$\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}$$ is a square matrix of order 4, then | A | = $$a_{11}\begin{vmatrix} a_{22} & a_{23} & a_{24} \\ a_{32} & a_{33} & a_{34} \\ a_{42} & a_{43} & a_{44} \end{vmatrix}$$ – $$a_{12}\begin{vmatrix} a_{21} & a_{23} & a_{24} \\ a_{31} & a_{33} & a_{34} \\ a_{41} & a_{43} & a_{44} \end{vmatrix}$$ + $$a_{13}\begin{vmatrix} a_{21} & a_{22} & a_{24} \\ a_{31} & a_{32} & a_{34} \\ a_{41} & a_{42} & a_{44} \end{vmatrix}$$ – $$a_{14}\begin{vmatrix} a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ a_{41} & a_{42} & a_{43} \end{vmatrix}$$ ## Example : Find the determinant of A = $$\begin{bmatrix} 1 & 2 & -1 & 3 \\ 2 & 1 & -2 & 3\\ 3 & 1 & 2 & 1 \\ 1 & -1 & 0 & 2 \end{bmatrix}$$. Solution : | A | = $$\begin{vmatrix} 1 & 2 & -1 & 3 \\ 2 & 1 & -2 & 3\\ 3 & 1 & 2 & 1 \\ 1 & -1 & 0 & 2 \end{vmatrix}$$ $$\implies$$ | A | = $$1\begin{vmatrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\ -1 & 0 & 2 \end{vmatrix}$$ – $$2\begin{vmatrix} 2 & -2 & 3 \\ 3 & 2 & 1 \\ 1 & 0 & 2 \end{vmatrix}$$ + $$(-1)\begin{vmatrix} 2 & 1 & 3 \\ 3 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix}$$ – $$3\begin{vmatrix} 2 & 1 & -2 \\ 3 & 1 & 2 \\ 1 & -1 & 0 \end{vmatrix}$$ | A | =  (1){$$(1)\begin{vmatrix} 2 & 1 \\ 0 & 2 \end{vmatrix}$$ – $$(-2)\begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix}$$ + $$(3)\begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix}$$} – (2){$$(2)\begin{vmatrix} 2 & 1 \\ 0 & 2 \end{vmatrix}$$ – $$(-2)\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}$$ + $$(3)\begin{vmatrix} 3 & 2 \\ 1 & 0 \end{vmatrix}$$} + (-1){$$(2)\begin{vmatrix} 1 & 1 \\ -1 & 2 \end{vmatrix}$$ – $$(1)\begin{vmatrix} 3 & 1 \\ 1 & 2 \end{vmatrix}$$ + $$(3)\begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix}$$} – (3){$$(2)\begin{vmatrix} 1 & 2 \\ -1 & 0 \end{vmatrix}$$ – $$(1)\begin{vmatrix} 3 & 2 \\ 1 & 0 \end{vmatrix}$$ + $$(-2)\begin{vmatrix} 3 & 1 \\ 1 & -1 \end{vmatrix}$$} $$\implies$$ | A | = 1{(1)(4 – 0) – (-2)(2 + 1) + (3)(0 + 2)} – 2{(2)(4 – 0) – (-2)(6 – 1) + (3)(0 – 2)} – (-1){(2)(2 + 1) – (1)(6 – 1) + (3)(-3 – 1)} – 3{(2)(0 + 2) – (1)(0 – 2) + (-2)(-3 – 1)} $$\implies$$ | A | = 1(16) – 2(12) + (-1)(-11) – 3(14) = -39<|endoftext|>
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Causes of Vertigo Vertigo, or dizziness, usually results from a disorder in the peripheral vestibular system (i.e., structures of the inner ear). Dizziness also may occur as a result of a disorder in the central vestibular system (i.e., vestibular nerve, brainstem, and cerebellum). In some cases, the cause of vertigo is unknown. Peripheral vestibular disorders include the following: - Benign paroxysmal positional vertigo (BPPV; most common peripheral disorder; may be accompanied by hearing loss, reduced cognitive function, and facial muscle weakness) - Cogan's syndrome (inflammation of connective tissue in the cornea; results in vertigo, ringing in the ears [tinnitus], and loss of hearing) - Ménière disease (fluctuating pressure of inner ear fluid [endolymph]; results in severe vertigo, ringing in the ears [tinnitus], and progressive hearing loss) - Ototoxicity (i.e., ear poisoning) - Vestibular neuritis (inflammation of vestibular nerve cells; may be caused by viral infection) Benign paroxysmal positional vertigo (BPPV) usually affects one of the sensing tubes in the inner ear called the posterior semicircular canal. BPPV occurs when debris made up of calcium carbonate and protein (called otoliths or ear crystals) builds up in and moves around in the posterior semicircular canal. BPPV also can affect the anterior canal or the horizontal canal. When the head is moved in certain ways (e.g., turning in bed, looking up, bending over), the calcium crystals move around and trigger inner ear sensors, causing a brief sensation of spinning. Inner ear degeneration (usually occurs in elderly patients), head trauma, and inner ear infection (e.g., otitis media, labyrinthitis) can cause BPPV. Some medications and environmental chemicals (e.g., lead, mercury, tin) can cause ototoxicity (i.e., ear poisoning), which may result in damage to the inner ear or the 8th cranial nerve (acoustic nerve) and cause vertigo. The damage can be permanent or temporary.Long-term use or high doses of certain antibiotics (e.g., aminoglycosides [streptomycin, gentamicin]) and antineoplastics (e.g., cisplatin, carboplatin) can cause permanent ototoxicity. Medications that may cause temporary ototoxicity include the following: - Anticonvulsants (e.g., phenytoin, carbamazepine) - Antidepressants (e.g., clomipramine, amoxapine) - Antihypertensives (e.g., labetalol, enalapril) - Loop diuretics (e.g., bumetanide, furosemide) - Pain relievers (e.g., aspirin) - Prescription and over-the-counter cold medicines - Quinine (e.g., chloroquine, quinidine) Alcohol, even in small amounts, can cause temporary vertigo in some people.<|endoftext|>
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                  # How to Factor an Equation with 2 Terms? An equation in mathematics is a relation that shows the equality between the variables and constants. An equation can be identity equation or inconsistent equation or conditional equation. An identity equation is satisfied by all possible values of the variables. A conditional statement is the one which has a restricted Set of values of the variables satisfying the given equation. For example, 4x + 8 = 12, Here only x = 1 can be the only possible value satisfying the equation. Last is the inconsistent equation that has no solutions. The number of terms in the equation determines the type of the equation like if we have 2 terms in the equation, then it is called as a binomial equation and so on. Let us learn how to factor an equation with 2 terms considering some examples of it: Example 1: Suppose we have a binomial equation as: 4x2 – 8 x = 0, then what is the value of 'x'? Solution: In the given equation we have to factorize the left hand side of the equation to get the value of 'x' as follows: 4x2 – 8x = 0, Or 4x (x – 2) = 0, Or x (x – 2) = 0, Or x = 0 and x – 2 = 0, Or x = 1 and x = 2, Here we notice that for an equation of degrees 2 of the variable, we get two possible answers or values of the variable. Likewise we get more number of solutions for higher degrees of variables. Example 2: Suppose we have an equation as: 5x3 + 125x2 = 0. Then what is the value of 'x'? Solution: The given equation contains two terms on the left side and has to be factored: 5x2 (x + 25) = 0, Or x2 = 0 and x = -25, Or x = 0, x =0 and x = -25. • ## Common Factors Calculator Math Topics Top Scorers in Worksheets<|endoftext|>
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When linux is running, many activities are happening at the same time. These activities are called processes . Usually, the kernel has many processes running, and each user logged in will also have several processes running. For example, you might wish to run a C program that takes a long time to finish. It's possible in linux to run your program in the background, while freeing up the prompt in your shell to get on with some other work (like analysing your data). So, in linux, several programs can run at once, and each process can run at a different priority level . In this example, you might wish to give the C program a lower priority than your data analysis program. 1. Jobs: background and foreground All shells under linux have this capability, it is called job control , and it lets the user switch back and forth between processes, putting processes into the background, suspending processes, killing processes and changing the priority at which any of these processes are handled. When ever you launch a new process by typing a command in your shell, you can automatically put this process into the background by appending your command with a & . Of all jobs running at one time, at most one job is the foreground job. All other jobs will be suspended jobs or background jobs. The foreground job ties up the shell prompt - so if you want access to your shell in this instance, this foreground job must be put into the background. To do this you must first suspend the forground job, tying CtrlZ in your shell (you don't need to hit return). You can always find out what jobs you have launched and are running in the current shell by typing the command jobs . Again, an example may help. |Let's say the user phillipa types jobs at the prompt. She is running 3 processes in this directory. One is called emacs (an editor), one is called xmakemol, and another is called xmgrace (she has suspended this job). The output of jobs is as follows;| - 21682 Running emacs carts2hexfrac.f 24410 Running xmakemol -f bulk-opt.xyz + 25174 Suspended xmgrace |The first entry in each line gives the job number .The second entry gives an indication of the most recently stopped (or started) job - in other words, the current job . The current job is signified by a + . Also indicated in this column is the previous job (the one that was current before the current job was started/stopped. The previous job is signified by a - .| |The third column gives the status of each job. In this case you can see that two are running and one is suspended. A suspended job will stay in this state until it is restarted or killed .| |The final entry on each line gives the actual process.| There are several linux commands regarding jobs, so you need a way to refer to each job. This can be done by referring to the job number , symbollically given as %n for job number n. You can also refer to the current job with just % . We've already seen an example of moving the foreground job to the background. You can also move a background job to the foreground by typing fg %n (where n is a number). You can also cause any suspended job (of which there might be several) to execute in the background by typing bg %n (where n is a number). You can kill any job (so it will cease to exist as a process) by typing kill %n (where n is a number). The kill command comes with plenty of options - you'll find out more later. Putting it into Practice: |Part I Suppose you'd like to start a program called emacs from your shell. You'd also like to put this job into the background. You can then run this job from the prompt by typing| |After you hit return, a window should pop up, and you should have control of your shell prompt as normal.| |Now, check what processes you have active at the moment by typing jobs . Note the job number of your emacs process.| |You should now kill your emacs process by entering kill %n (where n is your job number). The emacs window should vanish.| |Part II Try launching emacs from your shell again, this time without the ampersand ( no & at the end ).| |The emacs window will again pop up, but this time you will not have control of the shell prompt, since emacs is now running as a foreground job.| |You can always kill a foreground process by typing Ctrl C. Or, you can background this process by the usual route of suspending ( Ctrl Z followed by bg ).| Back to top 2. Redirection of input/output Most jobs need input and output options. By convention in linux, the default input is the standard input and the default output is the standard output . The standard input is defined as the input from the prompt at your terminal. Similarly, the standard output is defined as the output to your prompt at your terminal. It would be a huge pain if you could only use standard input and standard output - imagine a program that needs 10,000 numbers as input - you wouldn't want to feed them in by hand! Instead, under linux, there is the oportunity to redirect input and output. Input redirection is indicated by a < in front of the input file name, and output redirection is indicated by a > in front of a file name. Another useful redirection is >> . This will append output to a file, rather than write over that file with the output. |Suppose you have a program called averages , and the data you wish to average is collected in a file called ifile . Let's say you also would like to put your output into a file called ofile for safekeeping. You'd also like to put this job into the background. You can then run this job from the prompt by typing| averages < ifile > ofile & Back to top 3. Finding the process id Linux assigns each process an identification number called a process id (PID) when the process is created. There are several ways for a user to determine a PID for a given process. First, if you launch a background process by issuing a command to your shell, after you hit return, you should see two items displayed in the standard output of your xterm. The first one will be in square brackets (this is the job number), and the second is a number (this is the PID). |The user harry launches an emacs process by typing| |Immediately after hitting return, the following output is sent to his display| |This tells harry that this job number is 5, and the PID for this emacs process is 18408| Sometimes a process is not explcitly launched by the user (e.g. a process can be launched by a shell script executed by the user), and therefore the above method can not be used to determine the PID. Another way to find the PID is to use the command ps . This command reports the process status for ALL processes running on your machine. This may be a long list, since even if you are not running any jobs yourself, the kernel runs many background processes (known as daemons). ps (stands for Process Status) comes with many options, the most useful being:-e produce information about every process -f produce a full listing -l produce a long listing a helpful option is -ef which gives two of these options. |The user harry wants a status report of all processes on his machine. To do this he types| |Immediately after hitting return, output is sent to his display, a sample of this is shown below:| UID PID PPID C STIME TTY TIME CMD root 0 0 0 Sep 18 ? 0:17 sched root 1 0 0 Sep 18 ? 0:54 /etc/init - daemon 156 1 0 Sep 18 ? 0:00 /usr/lib/nfs/statd The headings on this table of output have the following meanings UID the user who owns the process PID the process id, a unique identifier assigned to each process PPID the parent process id, the process that spawned the current process C this field is obsolete STIME the start time for the current process TTY the controlling terminal for the current process TIME the amount of CPU time accumulated by the current process CMD the command used to invoke the process Another way to find out PIDs is to use the command top . As output this command gives a continually updated listing of active processes in real time. It is run ONLY as a foreground process. The output can be sorted in terms of CPU usage (this is the default), memory usage and runtime. top has an interactive element - while it is running, you can type many options, a sample of these are given below:space key immediately updates the display n or # change the number of processes to show M sort display by memory usage k kill a job The last option is important. To kill a job in top, after typing k, you will be prompted for the PID of the process you wish to kill. Type in the PID of your desired process and hit return. You will then be prompted for the signal with which to send this kill command (a signal notifies a process of any abnormal external event). To make sure the process really dies, use the signal "9" (type 9 when prompted). This is known as a "hard kill". Putting it into Practice: |Start by typing top at your prompt and hit return. As output, you should see a list of processes under various headings (these heading should be self-explanatory). One of those headings, PID, gives the process id for each process.| |While top is running, type k in the window.| |You will be prompted for a process id. Find the PID of the top process itself (it should be up the top of the list), type in this PID and hit return.| |You will now be prompted for the signal. Type 9 and hit return.| |You top session should terminate immediately!| From running your top process in the previous example, most of the headings in the output display should make sense. One heading, PRI deserves a little more explanation. The number under PRI gives the priority of the job. For regular users, you can set this number between 1 (highest priority) and 19 (lowest priority) - more about this later. The default priority is 10. System administrators can set the priority to be a negative number (even higher priority), all the way to -20. Back to top 4. Killing jobs by PIDNow that you know how to get a PID by using the ps command, you can kill jobs on the command line, even if these processes don't show up when you type jobs. You can simply kill a process by using the kill command in the following way: kill -signal PID usually, the signal number will be 9 (as explained above) For example, if user harry wants to kill an emacs process, knowing that the PID of this process is 6743, at his shell prompt, he can type kill -9 6743Back to top<|endoftext|>
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This is interesting! EXTRACT (12 Literacies) 01 Play: the capacity to experiment with one’s surroundings as a form of problem-solving. Having a strong sense of play can be helpful when you pick up a new piece of technology that you’ve never used before, when you’re trying to write an essay and your outline isn’t functioning as you’d hoped, and when you’re designing anything at all, from a dress to a web page to a concert’s program. 02 Performance: the ability to adopt alternative identities for the purpose of improvisation and discovery. Being able to move fluidly and effectively between roles can help you when you’re exploring online communities, when you’re trying to decide what actions are ethical, and when you’re shuffling between home, work and school. 03 Simulation: the ability to interpret and construct dynamic models of real-world processes. Being able to interpret, manipulate and create simulations can help you understand innumerable complex systems, like ecologies and computer networks – and make you better at playing video games! 04 Appropriation: the ability to meaningfully sample and remix media content. Being able to remix media content (and knowing when doing so is appropriate) can help you understand literary works, music, and art; it can also help lead you to a deeper understanding of copyright and cultural clashes. 05 Multitasking: the ability to scan one’s environment and shift focus as needed to salient details. Being a good multitasker is required in our new media landscape – and that includes learning when it isn’t good to multitask. 06 Distributed Cognition: the ability to interact meaningfully with tools that expand mental capacities. That can mean something as simple as using a ruler or calculator, or something as complex as efficiently using Wikipedia on your iPhone to access information on the fly. 07 Collective Intelligence: the ability to pool knowledge and compare notes with others toward a common goal. This ability is key to open source projects. Being able to pool knowledge with others can allow us to solve challenges far more complex than the individual mind can process. 08 Judgment: the ability to evaluate the reliability and credibility of different information sources. If you’re worried about your students using Wikipedia at inappropriate times and taking everything they read on the internet as gospel truth, you’re worried that they aren’t exercising good judgment. But judgment also includes knowing when sources are appropriate for your use: for instance, sometimes Wikipedia might be the appropriate resource to use. 09 Transmedia Navigation: the ability to follow the flow of stories and information across multiple media. Anyone who needs to do research needs a good understanding of transmedia navigation – how to follow threads through video, still photography, written work, music, online sources etc. 10 Networking — the ability to search for, synthesize, and disseminate information. Writing something isn’t enough without the ability to circulate it to the communities where it will matter. 11 Negotiation — the ability to travel across diverse communities, discerning and respecting multiple perspectives, and grasping and following alternative norms. We now need to know how to live in multiple communities – from the hyperlocal to the global and from those composed of people like us to those consisting of people very different from us. 12 Visualization – the ability to translate information into visual models and understand the information visual models are communicating. VIsualization has become a key way we cope with large data sets and make sense of the complexity of our environment.<|endoftext|>
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# Normal Distributions ## Fitting a bell curve to a histogram Estimated13 minsto complete % Progress Practice Normal Distributions MEMORY METER This indicates how strong in your memory this concept is Progress Estimated13 minsto complete % Normal Distributions ### Normal Distributions Previously you've spent some time learning about probability distributions. A distribution, itself, is simply a description of the possible values of a random variable and the possible occurrences of these values. Remember that probability distributions show you all the possible values of your variable (\begin{align*}X\end{align*}), and the probability associated with each of these values \begin{align*}(P(X))\end{align*}. You were also introduced to the concept of binomial distributions, or distributions of experiments where there are a fixed number of successes in \begin{align*}X\end{align*} (random variable) trials, and each trial is independent of the other. In addition, you were introduced to binomial distributions in order to compare them with multinomial distributions. Remember that multinomial distributions involve experiments where the number of possible outcomes is greater than 2, and the probability is calculated for each outcome for each trial. In this first concept on probability distributions, you are going to begin by learning about normal distributions. A normal distribution curve can be easily recognized by its shape. The first 2 diagrams above show examples of normal distributions. What shape do they look like? Do they look like a bell to you? Compare the first 2 diagrams above to the third diagram. A normal distribution is called a bell curve because its shape is comparable to a bell. It has this shape because the majority of the data is concentrated at the middle and slowly decreases symmetrically on either side. This gives it a shape similar to a bell. Actually, the normal distribution curve was first called a Gaussian curve after a very famous mathematician, Carl Friedrich Gauss. He lived between 1777 and 1855 in Germany. Gauss studied many aspects of mathematics. One of these was probability distributions, and in particular, the bell curve. It is interesting to note that Gauss also spoke about global warming and postulated the eventual finding of Ceres, the planet residing between Mars and Jupiter. A neat fact about Gauss is that he was also known to have beautiful handwriting. You previously learned about discrete random variables. Remember that discrete random variables are ones that have a finite number of values within a certain range. In other words, a discrete random variable cannot take on all values within an interval. For example, say you were counting from 1 to 10. You would count 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. These are discrete values. 3.5 would not count as a discrete value within the limits of 1 to 10. For a normal distribution, however, you are working with continuous variables. Continuous variables, unlike discrete variables, can take on any value within the limits of the variable. So, for example, if you were counting from 1 to 10, 3.5 would count as a value for the continuous variable. Lengths, temperatures, ages, and heights are all examples of continuous variables. Compare these to discrete variables, such as the number of people attending your class, the number of correct answers on a test, or the number of tails on a coin flip. You can see how a continuous variable would take on an infinite number of values, whereas a discrete variable would take on a finite number of values. As you may know, you can actually see this when you graph discrete and continuous data. #### Interpreting Graphs Look at the 2 graphs below. The first graph is a graph of the height of a child as he or she ages. The second graph is the cost of a gallon of gasoline as the years progress. Which graph represents discrete data? Which graph represents continuous data? If you look at the first graph, the data points are joined, because as the child ages from birth to age 1, for example, his height also increases. As he continues to age, he continues to grow. The data is said to be continuous and, therefore, you can connect the points on the graph. For the second graph, the price of a gallon of gas at the end of each year is recorded. In 1930, a gallon of gas cost \begin{align*}10 \cancel{\text{c}}\end{align*}. You would not have gone in and paid \begin{align*}10.2 \cancel{\text{c}}\end{align*} or \begin{align*}9.75 \cancel{\text{c}}\end{align*}. The data is, therefore, discrete, and the data points cannot be connected. Let’s look at a few problems to show how histograms approximate normal distribution curves. #### Understanding Data Distributions Jillian takes a survey of the heights of all of the students in her high school. There are 50 students in her school. She prepares a histogram of her results. Is the data normally distributed? If you take a normal distribution curve and place it over Jillian’s histogram, you can see that her data does not represent a normal distribution. If the histogram were actually shaped like a normal distribution, it would have a shape like the curve below: #### Interpreting Distributions Thomas did a survey similar to Jillian’s in his school. His high school had 100 students. Is his data normally distributed? If you take a normal distribution curve and place it over Thomas’s histogram, you can see that his data also does not represent a normal distribution. ### Example #### Example 1 Joanne posted a problem to her friends on FaceBook. She told her friends that her grade 12 math project was to measure the lifetimes of the batteries used in different toys. She surveyed people in her neighborhood and asked them, on average, how many hours their typical battery lasts. Her results are shown below: \begin{align*}& 98 && 108 && 107 && 79 && 100 && 112 && 97 && 79 && 41 && 127\\ &135 && 100 && 92 && 80 && 66 && 62 && 119 && 118 && 56 && 112\\ &99 && 83 && 86 && 62 && 127 && 155 && 107 && 140 && 144 && 122\\ &110 && 116 && 134 && 102 && 133 && 157 && 100 && 96 && 55 && 132\\ &126 && 171 && 169 && 146 && 122 && 74 && 70 && 82 && 84 && 93\end{align*} Is her data normally distributed? Where is the center of the distribution? If you take a normal distribution curve and place it over Joanne’s histogram, you can see that her data appears to come from a normal distribution. This means that the data fits a normal distribution with a mean around 105. Using the TI-84 calculator, you can actually find the mean of this data to be 105.7. What Joanne’s data does tell us is that the mean score (105.7) is at the center of the distribution, and the data from all of the other scores (times) are spread from that mean. You will be learning much more about standard normal distributions in a later Concept. But for now, remember the 2 key points about a standard normal distribution. The first key point is that the data represented is continuous. The second key point is that the data is centered at the mean and is symmetrically distributed on either side of that mean. ### Review 1. The following data was collected on a recent 25-point math quiz. Does the data represent a normal distribution? Can you determine anything from the data? \begin{align*}& 20 && 17 && 22 && 23 && 25\\ & 14 && 15 && 14 && 17 && 9\\ & 18 && 2 && 11 && 18 && 19\\ & 14 && 21 && 19 && 20 && 18\\ & 16 && 13 && 14 && 10 && 12\end{align*} 2. A recent blockbuster movie was rated PG, with an additional violence warning. The manager of a movie theater did a survey of moviegoers to see what ages were attending the movie in an attempt to see if people were adhering to the warnings. Is his data normally distributed? Do moviegoers at the theater regularly adhere to warnings? \begin{align*}& 17 && 9 && 20 && 27 && 16\\ & 15 && 14 && 24 && 19 && 14\\ & 19 && 7 && 21 && 18 && 12\\ & 5 && 10 && 15 && 23 && 14\\ & 17 && 13 && 13 && 12 && 14\end{align*} 3. The heights of coniferous trees were measured in a local park in a regular inspection. Is the data normally distributed? Are there areas of the park that seem to be in danger? The measurements are all in feet. \begin{align*}& 22.8 && 9.7 && 23.2 && 21.2 && 23.5\\ & 18.2 && 7.0 && 8.8 && 25.7 && 19.4\\ & 25.0 && 8.8 && 23.0 && 23.2 && 20.1\\ & 23.1 && 18.5 && 21.7 && 21.7 && 9.1\\ & 4.3 && 7.8 && 3.4 && 20.0 && 8.5\end{align*} Determine if the points representing each of the following data sets can be connected when graphed. 1. The number of students enrolled in a college each semester 2. The weight of a baby seal each day as it grows 3. The number of coins a coin collector owns each week 4. The speed of a rocket each second as it accelerates 5. The amount of water in a swimming pool each minute as it is drained 6. The number of employees a company has each month as it expands 7. The thickness of a glacier each year as it melts ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish TermDefinition continuous variable Variable that takes on any value within the limits of the variable discrete values discrete values are data where a finite number of values exist between any two values. distribution A distribution is a description of the possible values of a random variable and the possible occurrences of these values. normal distribution curve A normal distribution curve is a symmetrical curve that shows the highest frequency in the center with an identical curve on either side of the center. continuous variables A continuous variable is a variable that takes on any value within the limits of the variable. empirical rule The empirical rule states that for data that is normally distributed, approximately 68% of the data will fall within one standard deviation of the mean, approximately 95% of the data will fall within two standard deviations of the mean, and approximately 99.7% of the data will fall within three standard deviations of the mean. normal probability plot A normal probability plot is a graph is a plot of the z -scores of the data as quantiles against the actual data values. If a distribution is normal, this plot will be linear. normal quartile plot normal quartile plot is another name for a normal probability plot. standard normal distribution The standard normal distribution, $\phi(x)=\frac{1}{\sqrt{2 \pi}}e^{- \frac{1}{2}x^2}$ is a normal distribution with mean of 0 and a standard deviation of 1. z-score The z -score of a value is the number of standard deviations between the value and the mean of the set.<|endoftext|>
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William Bradford (Mayflower passenger) facts for kids William Bradford (1590 – 1657) was a passenger on the Mayflower in 1620. He travelled to the New World to live in religious freedom. He became the second Governor of Plymouth Colony and served for over 30 years. Bradford kept a journal of the history of the early life in Plymouth Colony. It is called Of Plymouth Plantation. Bradford's father died was he was one year old. When he was four years old he was sent to live with his grandfather. Two years later, his grandfather died and he returned to live with his mother and stepfather. A year later, in 1597, his mother died. Bradford became an orphan at age 7. He was sent to live with two uncles. His uncles wanted Bradford to help on the farm but Bradford was ill and was not able to work. Bradford started to read many books. One of the books was the Bible. This may be where he started learning of a religion that was different from the church of England. When Bradford was 12 years old he went with a friend to hear the pastor Richard Clyfton. Clyfton was a minister who did not believe in the Church of England. Bradford believed in Clyfton’s words. His uncles told him not to go to that church but he went anyway. King James I of England started to punish the people who did not follow the Church of England. They were called Separatists. By 1607 many members of the Separatists were arrested. Brewster was made to pay a fine for going to the church. Some members were put into prison and others were watched night and day. The people also learned that other Separatists in London had been put into prison and left to starve to death. The Separatists decided to leave England for the Dutch Republic (where religious freedom was permitted). William Bradford went with them. It was illegal to leave England. The group was betrayed by an English sea captain who brought them to the English authorities. Most of the congregation, and Bradford, were put into prison. By the summer of 1608, the Separatists, and William Bradford, escaped to Leiden, Netherlands. Bradford was 18. In the Dutch Republic William Bradford moved to Amsterdam in August 1608. He lived with the Brewsters in Stink Alley. The Separatists had spent most of their money leaving England and were poor. After nine months, the congregation moved to the smaller city of Leiden. In 1613, Bradford married Dorothy May, the daughter of a rich English couple living in Amsterdam. In 1617, the Bradfords had their first child, John Bradford. By 1617, the Separatists decided to travel to Virginia Colony in North America. The Separatists could practice their religion in the Dutch Republic, they wanted their children to keep their English customs and language. The colonists made a business arrangement with the Merchant Adventurers. These were a group of men who were willing to pay for the Pilgrims to go to the New World with the promise that the pilgrims would pay them back with furs and other goods they got from the colony. By July 1620, Robert Cushman and John Carver had made the plans and fifty Separatists left Delftshaven on the ship Speedwell. Many families were split as some Separatists stayed behind in the Netherlands, planning to make the voyage to the New World after the colony had been established. William and Dorothy Bradford left their three-year-old son John with Dorothy's parents in Amsterdam, possibly because he was too frail to make the voyage. The Speedwell planned to meet with the Mayflower and both ships would travel to Virginia Colony. The Speedwell proved unsafe and the passengers were put on the Mayflower. This made the ship very crowded. Not all the passengers were Separatists. There were about 50 colonists who went because they had certain skills. These skills would prove useful on the Mayflower and at the colony. Bradford said they hugged each other and cried and said goodbye to family and friends for the last time. They knew they were pilgrims and lifted their eyes to heaven, their dearest country and this comforted them. The Mayflower Voyage The Mayflower left Plymouth, England on 16 September 1620. There were 102 passengers and 30–40 crew. On 19 November 1620, the Mayflower spotted land. The Mayflower was supposed to land in Virginia Colony, but the ship was too damaged and they were forced to land at Cape Cod now called Provincetown Harbor. They landed on November 21. They wrote the Mayflower Compact, which made rules on how they would live and treat each other. Bradford signed the Mayflower Compact. Anchored and first explorations at Plymouth Colony Up to this time, Bradford, aged 30, had never been a leader in the colony. When the Mayflower anchored in the Harbor, the men went out to explore the land for a place where they could live. Bradford volunteered to be a member of the exploration parties. In November and December, these parties went out three times on foot and by boat. One of the earliest exploration missions included William Bradford, John Carver, Myles Standish, Edward Winslow, John Howland, Richard Warren, Stephen Hopkins and Edward Doty. Of this group Hopkins was the most experienced, having seen Indians during his time at Jamestown. According to Bradford, they set off in below freezing weather, many of the men already being ill, "and when the set sail the salt spray froze on their coats, as if they had been glazed". During their trip they saw Indians who looked like they were cutting up a large whale. When the Indians saw the men they ran away. The men camped and protected themselves from the cold and the Indians. They located a place that would be Plymouth Colony in December. During the first expedition on foot, Bradford was caught up in a deer trap made by Native Americans and brought upside down. During the third exploration, on 6 December 1620, Bradford and the other men found Plymouth Bay. A winter storm nearly sank their boat as they came near the bay. The explorers were very cold and started to get sick. The waves were high but they landed on Clark's Island. During this time, they searched areas all around the bay and found a place to live and build their homes. The location had a large hill. This hill is now called Burial Hill. It was a good place to build a fort for safety against any possible enemies. There were many brooks providing water. They would later find out that the site had been the location of a Native American village known as Patuxet; and that is why a large area had already been cleared for planting of food. The Patuxet tribe of Indians, between 1616 and 1619, had all died from bad sickness. Bradford later wrote that bones of the dead were easy to find in many places. The Mayflower arrived in Plymouth Bay on 20 December 1620. The settlers began building the colony's first house on 25 December. On January 11, 1621, as Bradford was helping to build houses when he was got a great pain in his hipbone and fell. Bradford was taken to the "common house" (the only house that was finished) and they thought he would die that night, but he did not. Bradford got well but many of the pilgrims were not so fortunate. During the months of February and March 1621 sometimes two or three people died every day. By the end of the winter, half of the 100 settlers had died. They buried their dead in secret so the Native Americans would not see how many were dying. They buried them in Cole's Hill and tried to also hide the graves from the animals. There were only a small number of men who were healthy. Everyone, including the children had to work and care for the sick. One of these was Captain Myles Standish. He was a soldier who had been hired by the settlers to protect the people of the colony. Standish cared for Bradford during his illness. They became good friends. Bradford had no military experience and when he became governor he would come to trust Captain Standish's advice on military matters. William was chosen by the people to be Governor of Plymouth Colony. He kept journals (a written record) which are very important even today because they are the only history of how the Mayflower passengers and their families lived in Plymouth Colony. He died before he could finish the second book. Early service as governor On March 16, the settlers had their first meeting with the Native Americans who lived in the region. Samoset, an Indian walked into the village of Plymouth and greeted them in English. Samoset had learned some English from the traders who had visited the area he was from. This soon led to a visit by Massasoit, the leader of the Pokanoket tribe. At that meeting, Massassoit made a treaty with Governor John Carver. The treaty said that they would be friends and protect each other against unfriendly Indians. Bradford wrote down what was said. He would soon become governor and the treaty that was very important was the promise to help each other. This agreement was not liked by Massasoit's enemies in other tribes. In April 1621, Governor Carver became sick while working in the fields on a hot day. He died a few days later. The settlers of Plymouth then chose Bradford as the new governor. Bradford would remain governor most of his life. The elected leadership of Plymouth Colony was at first a governor and an assistant governor. The assistant governor for the first three years of the colony's history was Isaac Allerton. In 1624, it was changed to five assistants. This was called the "governor's council". These men gave advice to the governor and had a vote on important matters. They helped Bradford in managing the Colony. Assistants during the early years of the colony included Thomas Prence, Stephen Hopkins, John Alden, and John Howland. William Bradford married Dorothy in Amsterdam, Holland on December 10, 1613. They had one son. Dorothy fell from the deck of the Mayflower into the water and drowned a short time after the ship landed. Her husband was on an exploring mission. John was born in Leiden, Holland and died in 1620. His second wife was Alice Southwarth in Plymouth. They had three children. She died in Plymouth in March 1670 and was buried on Burial Hill in Plymouth near her husband. Bradford's wrote Of Plymouth Plantation. It was a history about Plymouth colony. It is often called a journal. It was a collection of what he saw and what he thought was important. This was written in two books. The first book was written in 1630 but the second was never finished. During the years 1646 and 1650, he wrote about the colony's history up to 1646. In Of Plymouth Plantation, Bradford saw what was happening to be like what happened in the Bible. As Philip Gould writes, "Bradford hoped to show the workings of God to future generations." The book was not published until 1656, the year before his death, and it was well liked. Bradford has been called the father of American history. Many American authors have put what Bradford said in their books. Thomas Prence did so in his book named A Chronological History of New-England in the Form of Annals. Today it is considered very important and studied in American schools and colleges. Bradford's journal was included in another work entitled Mourt's Relation. This was written in part by Edward Winslow, and made into a book in England in 1622. Images for kids Provincetown, Massachusetts memorial to Pilgrims who died at sea or on board the Mayflower in Cape Cod Harbor in November/December 1620 William Bradford (Mayflower passenger) Facts for Kids. Kiddle Encyclopedia.<|endoftext|>
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+0 # Probability puzzles. +1 141 4 +90639 Here are some more probability puzzles. 1) If you break a stick in two places at random, forming three pieces, what is the probability of being able to form a triangle with the pieces? 2) If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form a triangle with them? 3) If you break a stick in two places at random, what is the probability of being able to form an acute triangle — where each angle is less than 90 degrees — with the pieces? 4) If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form an acute triangle with the sticks? The probabilities are, respectively, 1/4, 1/2, ln(8)−2ln⁡(8)−2 and 1−π/41−π/4. The full answers are here but I doubt they will be available for very long. https://fivethirtyeight.com/features/how-do-you-like-these-rectangles/ Melody  Oct 4, 2017 Sort: #1 +90639 0 Question 1. 1) If you break a stick in two places at random, forming three pieces, what is the probability of being able to form a triangle with the pieces? I will use a technique known as probablity contour mapping. Let the stick be one unit long. Break it so the the three peices are  x, 1-y and y-x  units long where y>x Note that when you add these up you get 1.      x+(1-y)+(y-x)=1 Obviously both x and y must be less than 1 unit long. Now to form a triangle the sum of any two sides must be greater than the third side. so we have 1)      $$x+(1-y)>y-x\\ x+1-y>y-x\\ -2y>-2x-1\\ y 2) \(x+(y-x)>1-y\\ y>1-y\\ y>0.5$$ 3)    $$(1-y)+(y-x)>x\\ 1-x>x\\ x<0.5$$ You plot this on a number plane and get the following region: But so far I have only considered where y>x,  maybe x>y then I get the inverse region. That is the given region reflected about the line y=x. So the total region is 1/4 of the entire region is shaded so the probablility that the three peices can form a triangle is 1/4 Melody  Oct 6, 2017 #2 +90639 0 Number 2 this answer is copied, I have not done it myself yet. For No. 2: What if, instead of pieces from one stick, we pick up three sticks of random length somewhere between zero and one? This is a little trickier. When we plotted the first problem, it could be collapsed to two dimensions, because we were only really worried about two sticks — the length of our third piece was automatically determined by the length of our first two pieces. But this problem is in three dimensions, so the solution needs to be plotted not in a one-by-one square but rather in a one-by-one-by-one cube. To help solve the problem, consider what wouldn’t solve it: a violation of our triangle inequalities. Suppose, for example, that x>y+zx>y+z, which makes it impossible to build a triangle. In that formulation, those points are contained in a pyramid bounded by the planes y=0y=0, z=0z=0, x=1x=1 and x=y+zx=y+z. There are three such pyramids in this cube, one for each of the ways the triangle inequality can be violated. Each of those pyramids has a volume of ⅙. (A pyramid has a volume equal to the area of its base times its height, all divided by three. Our pyramid in question has height 1 and a triangular base with area ½.). Therefore, there is a ½ chance we can’t make a triangle, and a ½ chance we can. And so we have the answer to the second problem. Melody  Oct 6, 2017 #3 +90639 0 Number 3 No. 3: Getting tougher still (as though that were possible)! Now it’s time for calculus. Guy D. Moore explains this one for us: The problem asks us to ensure that three pieces form an acute triangle. Consider three pieces with lengths  $$x>y>z$$ First, think about a right triangle. The formula for that, as our middle school teachers drilled into our heads, is $$x^2=y^2+z^2$$. (Otherwise known as the Pythagorean Theorem.) To have an acute triangle, all angles must be less than 90 degrees, so we tweak that formula: $$x^2 From this we get that \(y and \((1-x)^2 , which is the same as \(\begin{equation*}x<\frac{1-2y^2}{2(1-y)}\end{equation*}$$ Since we’re dealing with pieces of the same stick and not three separate sticks, we can return to plotting in two dimensions, not three. And our mirrored-triangle plot is useful again since our answer lies within those two original triangles. This time, though, we need to draw two new three-pointed shapes within those two triangles. The area of those shapes will be our answer — the probability of an acute triangle. So to calculate our new shapes, we need to cut pieces out of our original triangles. The area of one of those pieces is expressed in an integral (which is the calculus part of the solution). That integral is: $$\begin{equation*}\int_0^{1/2}\frac{1-2y}{2-2y}dy=\frac{1-\ln(2)}{2}\end{equation*}$$ There are six shapes, each with the same area, cut out of our one-by-one square, leaving: $$\begin{equation*}1-3(1-\ln(2))=\ln(8)-2\end{equation*}$$ In that equation, “ln” is the natural log, which equals an implied probability of acute triangle-formation of about 7.9 percent. (Who knew that natural logs are a great way to solve stick problems?) Guy also provided this illustration of the curvy areas we calculated: Melody  Oct 6, 2017 #4 +90639 0 No. 4: If you select three sticks, each of random length (between 0 and 1), what is the probability of being able to form an acute triangle with the sticks? We’re back to three dimensions again for the final question. This solution furthers the solution from problem No. 2, the way that solution No. 3 furthers solution No. 1. Laurent explained his solution this way: We’ll solve this problem the same way we solved No. 2, but we’ll replace the triangle inequalities with the acute triangle inequalities. As in No. 2, we end up with a three-dimensional volume rather than a two-dimensional area. For simplicity again, we’ll assume that cc is the largest length, which accounts for one-third of all possibilities. Laurent provided a lovely illustration of this volume: Our answer will ultimately be three times the area of this shape (this shape only accounts for stick c being longest, and two identical shapes will be generated for stick b being longest and stick a being longest). Our solution lies in the filled-in parts of that shape. While this lookscomplicated, the curved surface inside that area has the equation $$c^2 = a^2 + b^2$$, which is, conveniently, the equation of a right circular cone! So we can calculate the volume of the region of interest by subtraction. It’s ⅓ of the volume of the cube minus ¼ of the volume of the cone. (One-third because we’re considering only one out of three scenarios, the one where c is longest. And ¼ because the cone’s base is ¼ of a circle.) The total probability is three times this volume, because we must account for the remaining identical pieces. The final answer is $$3(1/3 − 1/4 ( \pi/3 ) ) = 1 − \pi/4$$or about 0.2146. So the probability of forming an acute triangle with three randomly chosen lengths is about 21.5 percent. Melody  Oct 6, 2017 ### 27 Online Users We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details<|endoftext|>
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Despite decades of taking measurements, scientists cannot agree on how long neutrons live. Neutrons are stable inside atoms, but on their own they decay in about 15 minutes, more or less, into a few other particles. Exactly how much more or less is the sticking point. Each experiment seems to yield a different answer. The lack of resolution is frustrating. Understanding the lifetime of the neutron is important not only for knowledge’s sake but also to answer other more fundamental questions about new physics beyond the known particles and processes in the universe, says Jeffrey Nico, who leads a neutron lifetime experiment at the National Institute of Standards and Technology (NIST) in Gaithersburg, Md. “We can’t leave this disagreement just hanging out there.” To measure how long neutrons last before decaying, scientists look for either the disappearance of neutrons or the appearance of their decay products. Neutrons disappear through a process called neutron beta decay, transforming themselves into protons by emitting an electron (shedding a negative charge to become positively charged) and an antineutrino. One method of measuring how long neutrons last before decaying entails trapping a group of neutrons inside a physical bottle and counting how many remain after various lengths of time. Another, called the beam method, is to produce a tightly confined spray, or beam, of neutrons and surround it with a “proton trap” to count the protons created when neutrons decay. Beam experiments have been going on for more than 30 years, with the leaders in the field located at NIST. Their best, latest beam results, published in 2013, measured the neutron lifetime at 887.7 seconds, plus or minus 3.1 seconds. Bottle experiments, in contrast, go back only about 15 years but have already posted results with higher precision than beam findings. The best result to date from a bottle experiment came in 2008 from a collaboration between the Petersburg Nuclear Physics Institute and the Joint Institute for Nuclear Research in Russia, along with the Institut Max von Laue Paul Langevin (ILL) in France. That team measured the neutron lifetime at 878.5 seconds, plus or minus one second. A difference of about nine seconds may not sound like much, but it is larger than the estimated error bars on the experiments, meaning those error bars are wrong on some—maybe all—of the results. “The discrepancy is embarrassing,” says University of Tennessee physicist Geoffrey Greene, who worked on the NIST project. “One or more of us screwed up, and we just have to find out which ones.” And although nine seconds is not long, it is enough to make a difference in some calculations that rely on the neutron lifetime, such as predictions of how the first atomic nuclei formed. Protons and neutrons started out as free particles in the hot, dense baby universe. Only after the cosmos had cooled sufficiently—within in the first 20 minutes after the big bang—did they combine to form nuclei in a process called nucleosynthesis. “Essentially all of the helium in the universe was made at that epoch,” Nico says. To know how many neutrons were available to form nuclei, scientists must know how long neutrons last before decaying. “For big bang nucleosynthesis, right now the neutron lifetime is the input with the largest uncertainty on it, so by improving the neutron lifetime, you can improve those predictions,” Nico adds. If nucleosynthesis predictions turn out not to match evidence from astrophysical observations of the abundance of helium, for example, exotic physics just might be at work. One possibility is dark matter—the invisible, but plentiful, substance in the universe thought to be made of some types of undiscovered particles. “All sorts of dark matter candidates might actually play a role in big bang nucleosynthesis,” says theoretical physicist Susan Gardner of the University of Kentucky. These particles might have interacted with protons and neutrons or participated in reactions somehow to alter the number of nuclei that formed. Understanding neutron beta decay is also important for understanding one of the four fundamental forces of nature—the weak force. This force is responsible for nuclear fusion as well as radioactive decay—such as neutron beta decay. “Neutron decay is one of the simplest examples of the weak interaction between light particles like electrons and heavy particles like quarks [the ingredients of neutrons],” Greene says. “That’s what we’re trying to get at by studying neutron decay.” The standard model of particle physics describes neutron decay very well, but scientists wonder if it is the full description. If measurements of neutron decay ever deviated from standard model predictions, they might point us toward a new, deeper level of physics. Another possibility is that new physics is behind the trouble scientists are having with neutron lifetimes. The discrepancy between bottle and beam measurements is curious, scientists point out. Bottle experiments tend to find a shorter neutron lifetime than the beam tests do. Perhaps neutrons occasionally decay by a process other than beta decay, which means they might turn into something other than protons. That would create a number of “missing” protons in the beam experiments, which would lead them to measure a longer lifetime than bottle experiments. (It would seem as if fewer neutrons are decaying in the allotted time, when really they are simply decaying into additional, unseen products.) “Is there some kind of new physics that could explain getting a different value for these two methods?” Greene asks. “The fact that we’re seeing a longer lifetime is consistent with the notion, and that would be tremendously exciting, but I think it’s more likely somebody’s screwed up their experiment.” Neutron decay experiments are painstakingly complex. For beam experiments, the main challenge is to make sure to correctly tally both the neutrons in the beam, and the protons that result from neutron decays. “Conceptually it’s fairly straightforward, but the trick is counting those particles absolutely,” Nico says. Additionally, the researchers must measure the length of the proton trap precisely to get accurate measurements. The difficulty in bottle experiments comes from the bottle—particularly the possibility that neutrons interact somehow with the bottle walls , which can be made of various materials, such as oil-coated copper. “When the neutron touches the wall, in the ideal case they should be totally reflected with no losses, but there can be contaminations on the wall,” says Peter Geltenbort of ILL, a member of the leading bottle experiment team. That group is working on building a larger bottle, and comparing the results with experiments using smaller bottles. “The idea is if you compare different bottles, you can then do an extrapolation to an infinitely large bottle and bring the precision down to an order of 0.3 or 0.4 seconds.” Another tack is to eliminate the walls of the bottle altogether. A few teams are working on “bottles” made out of magnetic and gravitational fields, as opposed to tangible materials, to trap neutrons. Although neutrons don’t have an electrical charge, their spin gives them what’s called a magnetic moment, making them act like tiny magnets inside a magnetic field. “We put more than 5,000 individual magnets together to form a trap, and the neutrons are effectively levitated by the presence of this strong magnetic field,” says Chen-Yu Liu, a physicist at Indiana University Bloomington who is working on a magnetic-gravitational field bottle experiment at the Los Alamos National Laboratory. Ultimately, both the beam and bottle experiment teams hope they can reconcile their results. “I think it’s universally acknowledged that both methods are fundamentally sound,” Greene says. “It’s a matter of the devil is in the details.” And if neutrons persist in their befuddling behavior, it just might mean the universe is a bit more complicated than we thought.<|endoftext|>
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History of the Bude Canal The idea of the Bude Canal was conceived in 1774 by Cornishman, John Edyvean. Its main purpose would be to transport lime-rich sea-sand inland, to use as a manure on farms where soils were acidic and unproductive. Other cargoes, such as coal, culm, slate, timber, iron and bricks could also be carried and farm produce exported. However, his scheme proved too ambitious and it was abandoned until 1817, when renewed local interest produced a canal survey and report. This proposed a 19ft wide canal, consisting of a two mile stretch navigable by barge from Bude to Helebridge; plus another 33 miles of narrow tub-boat canal, where wheeled ‘youb-boats’ would be pulled by horses, then winched up a system of ‘inclined planes’ to get over the hills. Also proposed was the building of a sea lock, a breakwater and moving the mouth of the River need, whilst water would be supplied from the specially formed Lower Tamar Lake. Costs were estimated at £90,000. The project was eventually agreed and work began on the 23rd July, 1819 under the Bude Harbour & Canal Company. Four years later, much of the line was opened and in 1825 all that was ever to be built was completed at the final cost of £120,000. Although never a great success for the shareholders, the canal was indispensable to inland communities and enabled their farms to grow and prosper. However, trade declined drastically when the railway reached Bude in 1898 and the Canal closed in 2901. Most sections were sold back to the landowners. In 2006 the Bude Canal Regeneration Project began work on a £5 million facelift, aimed to restore the first two miles of the canal from the sea lock to Helebridge. This was completed in 2009. Read more about the history of the Bude Canal.<|endoftext|>
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Every NGO and every project needs to have a budget. A budget describes the money that an organisation plans to raise and spend for a set purpose over a given period of time. "Failing to plan is planning to fail." The budget plays an important role in every stage of NGO work: planning new work, raising funds, implementing activities, monitoring work and evaluation. The programme cycle and the financial cycle go hand in hand. How to prepare a budget (PLAN) Writing a budget involves answering a number of questions, that can only be answered by programme staff and finance staff working together: They start with plans and goals, not numbers: - What objectives are we trying to achieve? - What activities will be involved in achieving these objectives? - What resources will we need to carry out these activities? - What will these resources cost? - Where will the funds come from? What will the source be? - Is the result realistic? This might look like hard work, but we actually work through these questions every day, when we handle our own personal money. You can see a short illustrated example here: Rudi goes to the cinema. A good format to start with is a detailed budget worksheet that can later be summarised or re-coded to match donor formats as needed. After the budget has been prepared it needs to be approved by the Board and / or donors. See an example budget worksheet. Implementing the budget (DO) - The person with the responsibility for implementing a budget is called a ‘budget holder’, and is usually a programme or project manager, not a finance person. - They use the budget to guide the implementation, checking BEFORE spending that items are budgeted for, and allocating expenditure to appropriate budget lines. - Budget holders should focus on delivering the project objectives within the total budget rather than on spending all the cash outlined in the budget. - Over or under spending within reasonable limits is not a problem as long as there are good explanations (but check your grant conditions, as some can be more restrictive than others). Monitoring the budget (REVIEW) Once the budget has been agreed and the activity implemented, the process is completed by comparing the plan (budget) with the eventual outcome (‘actual’), to see if there is anything we have learnt or could do differently next time. This is covered another section of the Guide. See the section of the Guide on financial reporting. What makes a good budget? Budgets should be clear, so that other people can pick them up and understand them easily. Always add notes to explain any estimates or assumptions you make. You should use the same list of accounts codes in your budget that you use to keep your accounts. See Top tips 3 for more advice on making effective budgets. As well as a budget, NGOs normally need a cash flow forecast. This predicts when money will arrive in your bank account and when it will leave. It is very important to make sure that you always have enough cash available. For instance, even if a donor has agreed to make you a grant, you cannot pay salaries until the cash actually arrives. See Top tips 6 for more on how to create a cash flow forecast. See here for an example cash flow forecast Mango's Financial Managment Essentials Handbook includes a whole chapter on budgeting and some sample formats. Want to learn more? Mango’s Budgeting Essentials training course is available as a 5-week eWorkshop.<|endoftext|>
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Courses Courses for Kids Free study material Offline Centres More Store # Graphs and Graphical Representation Reviewed by: Last updated date: 09th Aug 2024 Total views: 389.7k Views today: 10.89k ## What are Graphs and Graphical Representation? Graphical representation refers to the use of charts and graphs to visually analyze and display, interpret numerical value, clarify the qualitative structures. The data is represented by a variety of symbols such as line charts, bars, circles, ratios. Through this, greater insight is stuck in the mind while analyzing the information. Graphs can easily illustrate the behavior, highlight changes, and can study data points that may sometimes be overlooked. The type of data presentation depends upon the type of data being used. ## Graphical Representation of Data The graphical representation is simply a way of analyzing numerical data. It comprises a relation between data, information, and ideas in a diagram. Anything portrayed in a graphical manner is easy to understand and is also termed as the most important learning technique. The graphical presentation is always dependent on the type of information conveyed. There are different types of graphical representation. These are as follows: 1. ### Line Graphs: Also denoted as linear graphs are used to examine continuous data and are also useful in predicting future events in time. 1. ### Histograms: This graph uses bars to represent the information. The bars represent the frequency of numerical data. All intervals are equal and hence, the width of each bar is also equal. 1. ### Bar Graphs: These are used to display the categories and compare the data using solid bars. These bars represent the quantities. 1. ### Frequency Table: This table shows the frequency of data that falls within that given time interval. 1. ### Line Plot: It shows the frequency of data on a given line number. 1. ### Circle Graph: It is also known as a pie chart and shows the relationship between the parts of the whole. The circle consists of 100% and other parts shown are in different proportions. 1. ### Scatter Plot: The diagram shows the relationship between two sets of data. Each dot represents individual information of the data. 1. ### Venn Diagram: It consists of overlapping circles, each depicting a set. The inner-circle made is a graphical representation. 1. ### Stem and Leaf Plot: The data is organized from the least value to the highest value. The digits of the least place value form the leaf and that of the highest place value form the stem. 1. ### Box and Whisker Plot: The data is summarised by dividing it into four parts. Box and whisker show the spread and median of the data. ### Graphical Presentation of Data - Definition It is a way of analyzing numerical data. It is a sort of chart which shows statistical data in the form of lines or curves which are plotted on the surface. It enables studying the cause and effect relationships between two variables. It helps to measure the extent of change in one variable when another variable changes. ### Principles of Graphical Representation The variables in the graph are represented using two lines called coordinate axes. The horizontal and vertical axes are denoted by x and y respectively. Their point of intersection is called an origin ‘O’. Considering x-axes, the distance from the origin to the right will take a positive value, and the distance from the origin to the left will take a negative value. Taking the same procedure on y-axes. The points above origin will take the positive values and the points below origin will take negative values. As discussed in the earlier section about the types of graphical representation. There are four most widely used graphs namely histogram, pie diagram, frequency polygon, and ogive frequency graph. ### Rules for Graphical Representation of Data There are certain rules to effectively represent the information in graphical form. Certain rules are discussed below: 1. Title: One has to make sure that a suitable title is given to the graph which indicates the presentation subject. 2. Scale: It should be used efficiently to represent data in an accurate manner. 3. Measurement unit: It is used to calculate the distance between the box 4. Index: Differentiate appropriate colors, shades, and design I graph for a better understanding of the information conveyed. 5. Data sources: Include the source of information at the bottom graph wherever necessary. It adds to the authenticity of the information. 6. Keep it simple: Construct the graph in an easy to understand manner and keep it simple for the reader to understand. Looking at the graph the information portrayed is easily understandable. ### Importance of Graphical Representation of Data Some of the importance and advantages of using graphs to interpret data are listed below: 1. The graph is easiest to understand as the information portrayed is in facts and figures. Any information depicted in facts, figures, comparison grabs our attention, due to which they are memorizable for the long term. 2. It allows us to relate and compare data for different time periods. 3. It is used in statistics to determine the mean, mode, and median of different data. 4. It saves a lot of time as it covers most of the information in facts and figures. This in turn compacts the information. ## FAQs on Graphs and Graphical Representation Ans: These graphical presentations of data are vital components in analyzing the information. Data visualization is one of the most fundamental approaches to data representation. Its advantages include the following points: 1. Facilitates and improves learning 2. Flexibility of use 3. Understands content 4. Increase structure thinking 5. Supports creative thinking 6. Portrays the whole picture 7. Improves communication With advantages, certain disadvantages are also linked to the graphical representation. The disadvantages concern the high cost of human effort, the process of selecting the most appropriate graphical and tabular presentation, creative thinking, greater design to interpret information, visualizing data, and as human resource is used. The potential for human bias plays a huge role. Q2. What is the Graphical Representation of Data in Statistics? Graphs are powerful data evaluation tools. They provide a quick visual summary of the information. In statistics information depicted is of mean, mode, and median. Box plots, histograms are used to depict the information. These graphs provide information about ranges, shapes, concentration, extreme values, etc. It studies information between different sets and trends whether increasing or decreasing. Since graphical methods are qualitative, they are not only the basis of comparison and information.<|endoftext|>
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# Division law with integer bases Lesson The method to divide power terms is similar to the multiplication law, however in this case we subtract the powers from one another, rather than add them. Let's look at an expanded example to see why this is the case. If we wanted to simplify the expression $a^6\div a^2$a6÷​a2, we could write it as: We can see that there are six $a$as being divided by two $a$as to give a result of four $a$as, and notice that $4$4 is the difference of the powers in the original expression. So, in our example above, $a^6\div a^2$a6÷​a2 $=$= $a^{6-2}$a6−2 $=$= $a^4$a4 Let's look at another specific example. Say we wanted to find the value of $2^7\div2^3$27÷​23. By evaluating each term in the quotient separately we would have $2^7\div2^3$27÷​23 $=$= $128\div8$128÷​8 $=$= $16$16 Alternatively, by first expanding the terms in the original expression we can arrive at a simplified version of the expression on our way to the final value. $2^7\div2^3$27÷​23 $=$= $\left(2\times2\times2\times2\times2\times2\times2\right)\div\left(2\times2\times2\right)$(2×2×2×2×2×2×2)÷​(2×2×2) $=$= $2^4$24 $=$= $16$16 Notice in the second line we have identified that $2^7\div2^3=2^4$27÷​23=24. ### The exponent law of division We can avoid having to write each expression in expanded form by using the division law (which is also known as the quotient law). The division law $\frac{a^m}{a^n}=a^{m-n}$aman=amn, where $a$a is any number, That is, when dividing terms with a common base: • Keep the same base • Find the difference in the power. Of course, we can also write the division law in the form $a^m\div a^n=a^{m-n}$am÷​an=amn. ### Dividing terms with common bases As with using the multiplication (or product) law, we can only apply the division (or quotient) law to terms with the same bases (just like we can only add and subtract like terms in algebra). We can simplify $\frac{9^8}{9^3}$9893 because the numerator and denominator have the same base: $9$9. Careful! We cannot simplify $\frac{8^5}{7^3}$8573 because the two terms do not have the same base (one has a base of $8$8 and the other has a base of $7$7). #### Worked example Simplify the following by first writing it in expanded form: $\frac{3^7}{3^2}$3732. Think: We want to first write the expression in expanded form, so that we can then cancel out common factors from the numerator and denominator. Do: $\frac{3^7}{3^2}$3732​ $=$= $\frac{3\times3\times3\times3\times3\times3\times3}{3\times3}$3×3×3×3×3×3×33×3​ $=$= $\frac{3\times3\times3\times3\times3}{1}$3×3×3×3×31​ $=$= $3^5$35 Reflect: We can see that the answer matches what we would expect if we used the division law. That is, $\frac{3^7}{3^2}=3^{7-2}$3732=372 giving us the final answer of $3^5$35. ### Dividing other terms We can also simplify expressions involving multiplication as well, such as $\frac{4\times3^6}{2\times3^4}$4×362×34. We follow the same steps as when we are multiplying expressions. That is, we can treat the problem in two parts. Let's take a look at an example. #### Worked example Simplify $\left(8\times5^9\right)\div\left(2\times5^4\right)$(8×59)÷​(2×54) using exponent laws. Think: First, let's write the expression as a fraction. To make the simplifications easier, we can then split the fraction up into two fractions using the fact $\frac{a\times b}{c\times d}=\frac{a}{c}\times\frac{b}{d}$a×bc×d=ac×bd. Do: $\left(8\times5^9\right)\div\left(2\times5^4\right)$(8×59)÷​(2×54) $=$= $\frac{8\times5^9}{2\times5^4}$8×592×54​ $=$= $\frac{8}{2}\times\frac{5^9}{5^4}$82​×5954​ $=$= $4\times\frac{5^9}{5^4}$4×5954​ $=$= $4\times5^5$4×55 $=$= $4\times5^5$4×55 Reflect: Combining the steps, we get $\left(8\times5^9\right)\div\left(2\times5^4\right)=4\times5^5$(8×59)÷​(2×54)=4×55 and as this process becomes more familiar we can reduce the amount of intermediate steps we take to arrive at the solution. #### Practice questions ##### Question 1 Convert the following to fraction form and evaluate using exponent laws: $4^8\div4^5$48÷​45. ##### Question 2 Simplify the following, giving your answer in exponential form: $\frac{\left(-7\right)^{11}}{\left(-7\right)^5}$(7)11(7)5. ##### Question 3 Evaluate the following using exponent laws: $5\times\frac{4^5}{4^3}$5×4543. ### Outcomes #### 9D.NA1.03 Derive, through the investigation and examination of patterns, the exponent rules for multiplying and dividing monomials, and apply these rules in expressions involving one and two variables with positive exponents<|endoftext|>
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Вы находитесь на странице: 1из 1 # Percentage Example 1. When the cost of an article is decreased by 75%, it cost becomes Rs. 1050. Find the cost of the article, before the discount was given. Solution. The discount of the article is 75 %. After discount is given, let the new cost be x% of the original cost (before the discount was given). x% = (100 - 75)% = 25%. 100× 1050 ∴ Original price = = Rs. 4200. 25 Example 2. The ratio of cost of two articles is 2 : 5. Then, what percent is the cost of 1 st article is the 2nd article? Solution. Let the cost of first and second article be Rs. 2x and Rs. 5x. 2 x ×100 ∴ Required per cent = = 40%. 5 x ×1 Example 3. What principle is lent, if the payable interest after 5 years is Rs. 365, at the rate of 10 % per annum? SI ×100 Solution. The principal is calculated by the formula . R×T We have, SI = Rs. 365 R = 10 % pa T = 5 years 365× 100 P= = Rs. 730 5× 10 ∴ Rs. 730 is lent. Example 4. Rakesh, at simple interest, borrows a sum at 5 % pa and for 10 years. If the interest after 10 years is 5 times the principal, then find the interest paid. Solution. Let principal is Rs. x. Let the interest be Rs. y. Then we have, Rs. (x + y) = 5x  Rs. y = 5x - x (transposing +x to right side, changing to -x)  Rs. y = Rs. 4x (Express y as x terms).  Interest = 4x.<|endoftext|>
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# 180 Days of Math for Third Grade Day 92 Answers Key By accessing our 180 Days of Math for Third Grade Answers Key Day 92 regularly, students can get better problem-solving skills. ## 180 Days of Math for Third Grade Answers Key Day 92 Directions: Solve each problem. Question 1. By subtracting 23 from 45 we get 22. Question 2. 4 × 6 = Explanation: By performing the multiplication operation we can find the product of 4 and 6. 4 × 6 = 24 Question 3. 5 × 6 = Explanation: By performing the multiplication operation we can find the product of 5 and 6. 5 × 6 = 30 Question 4. What is the numeral for one hundred ninety-seven? Answer: The numeral for one hundred ninety-seven is 197. Question 5. Circle the smallest number. 451, 145, 514 Answer: The smallest number among the three numbers is 145. Question 6. Fill in the missing number. 48, ________, 32, 24, 16 32 – 24 = 8 So, the pattern is 8. 48 – 8 = 40 So, the missing number is 40. Question 7. __________ cm = 1 m Convert from meters to centimeters. 1 meter = 100 cm 100 cm = 1 m Question 8. What time is shown on the clock? Question 9. What is the range of the number of cousins? Name Number of Cousins Anne 15 Marq 1 Jesus 3 Kelly 4 Tami 7 Answer: The range of the number of cousins is 1 to 15. Question 10. What is 10¢ less than 3 quarters?<|endoftext|>
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Influenza, commonly called “the flu,” is an illness caused by RNA viruses that infect the respiratory tract of many animals, birds, and humans. Influenza, or the “flu,” is an illness of the breathing system (respiratory system) and muscles caused by a virus. Causes of Flu Influenza A usually occurs in early winter and early spring. Influenza B can appear at any time of the year. H1N1, or Swine Flu, is another type of Influenza. Most people catch the flu when they breathe in tiny droplets from coughs or sneezes of someone who has the flu. It is also spread when you touch a something with the virus on it and then touch your mouth, nose, or eyes. Symptoms appear 1 – 7 days later (usually within 2 – 3 days). The flu spreads easily. It often strikes a community all at once. Students or workers become sick within 2 or 3 weeks of the flu’s arrival in a school or workplace. Sometimes people confuse colds and flu. They do share some of the same symptoms. Most people get a cold several times each year. But they usually get the flu once every few years. Symptoms of Flu The first symptoms are a fever between 102 and 106 °F. The infection results in the person getting fever, cough, headache, and malaise (tired, no energy); some people also may develop a sore throat, nausea, vomiting, and diarrhea. Then breathing symptoms begin to increase. The symptom is usually not a dry cough. Most people also develop a sore throat and headache. Runny nose and sneezing are common. It is a clear, watery nasal discharge. Treatment of Flu Acetaminophen (Tylenol) and ibuprofen (Advil, Motrin) help lower fever. Sometimes doctors suggest you use both types of medicine. Do NOT Use Aspirin. A fever does not need to come all the way down to normal. Most people will feel better when their temperature drops by even 1 degree. Over-the-counter cold medicines may make some of your symptoms better. Cough drops will help with your sore throat. A vaccine is available to prevent the flu, its effectiveness varies according to the degree of match between the viral strains used to prepare the vaccine and those strains actually in circulation in a given year. Should see a doctor if you have flu-like symptoms and you:- - are 65 or over - are pregnant - have a long-term medical condition such as diabetes, heart disease, lung disease, kidney or neurological disease - have a weakened immune system How Long does Flu lasts? The typical course of the flu is seven to ten days.” While you’re waiting for the virus to pass, prevent it from getting worse by drinking plenty of fluids, washing hands frequently, and eating vitamin D-fortified foods like orange juice and yogurt.<|endoftext|>
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We spent a bit of time talking about how great 0 is. We looked at a product like a * b = 12 and determined, after a long list of possibilities, that there was an infinite number of combinations of values for a and b that would make this true. However, if a * b = 0, we know that a or b must be 0. Many students have been struggling with finding the zeros of questions like these: I think they now understand why you can just take the opposite of the constant term in every case. We consolidated the process of solving a quadratic equation that can be factored and then followed up with a lot of practice questions. Then we hit one with a common factor and I, of course, did what my students suggested and factored it without taking the common factor out first. We talked a lot about this one. If you are factoring, but not solving, you must take the common factor out either at the beginning or the end. They saw pretty quickly that taking it out at the beginning made their solution much easier. We talked about why you could divide by 3 when you are solving, but not when you are just factoring. After class a student asked me about using the box method for example 2a. I had not tried one that contained a common factor and it is not as obvious as you might think (well, it was not obvious to me, anyway). I will go over it with him tomorrow, stressing that taking the common factor out first will make it all work much more nicely. Part (c) gave us the opportunity to look at a difference of squares. I drew the corresponding tiles on the whiteboard and they all remembered factoring these types of quadratics. Part (d) was gave them the tools to deal with a -1 coefficient of the variable squared. We finished with this one: Today's homework was the second box of the handout from yesterday.<|endoftext|>
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Factoring and Multiples Prime Numbers Numbers # Is it that 3 consecutive whole number can be prime factors Explain? ###### Wiki User 3 consecutive numbers cannot be prime factors. Any three consecutive numbers would include at least one even number. The only even Prime number is 2, and (2,3,4) doesn't qualify. ๐Ÿ™ 0 ๐Ÿคจ 0 ๐Ÿ˜ฎ 0 ๐Ÿ˜‚ 0 ## Related Questions A prime number has no factors except itself and 1. Prime number next to any prime number is called consecutive prime number. Eg:- 2,3,5,7 are prime numbers It is a prime number because its only factors are 1 and itself. 3 consecutive numbers cannot be prime factors. Any three consecutive numbers would include at least one even number. The only even prime number is 2, and (2,3,4) doesn't qualify. All numbers have factors. Some factors are prime numbers. We call these prime factors. The factors of 6 are 1, 2, 3 and 6. The prime factors of 6 are 2 and 3. No. It doesn't matter what factors you choose at the beginning, but all the factors should be prime at the end. The only even prime number is 2 because it has only two factors which are itself and one. A prime number is a positive integer that has exactly two different factors: 1 and itself. This is why 1 is not a prime number: it has the factors 1 and itself, but they are the same - it does not have exactly two different factors. A factor tree shows the prime factors of the working number and can also be used as a visual demonstration to show the prime factors of the working number. 75 is a composite number because it has more than two factors whereas prime numbers have only two factors 47 and 53 are prime numbers because they only have two factors. Prime numbers have only two factors: one and themselves. The only factors of 17 are 1 and 17. 17 is a prime number. A prime number has only two factors: 1 and the number itself. A composite number has more than two factors. Yes, because a composite number is any number that has more prime factors other than one and itself . The prime factors of 33 are 3 and 11. 37 is a prime number because it has only 2 factors which are itself and one A pair of prime numbers are always relatively prime, whether they are consecutive or not. This is so because "relatively prime" means they have no common factors. Yes it can be prime. A prime number is any number that only has one and itself as factors. By definition even numbers have a factor of two. Therefore the only even number that is prime is the number 2. ###### Math and ArithmeticPrime NumbersFactoring and MultiplesComposite Numbers Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.<|endoftext|>
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# How To Solve Quadratic Equation Graphical Method Ideas How To Solve Quadratic Equation Graphical Method Ideas. We can solve a quadratic equation by factoring, completing the square, using the quadratic formula or using the graphical method. A(8)(a) solve quadratic equations having real solutions by factoring, taking square roots, completing the square, and applying the quadratic formula. To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. %straight line will give the roots of the given quadratic equation. Resource objective(s) given a quadratic equation, the student will use graphical methods to solve the equation. ### You Can Solve Quadratic Equations By Graphing, Factoring, Completing The Square, & The Quadratic Formula. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Transform the equation so that the quadratic term and the linear term equal a constant. We can solve the quadratic equation ax 2 + bx + c = d through graphing using the following steps: ### %Parabola And A Straight Line. We can solve a quadratic equation by factoring, completing the square, using the quadratic formula or using the graphical method. Another way of solving a quadratic equation is to solve it graphically. A x 2 + b x + c = 0, w h e r e a ≠ 0. ### By Using The Quadratic Formula 4. Simplest form of solving quadratic equations using formula method These are the solutions of the equation. One of the ways we can solve a quadratic equation is by factoring. ### There Are Three Basic Methods To Solve A Quadratic Equation In Mathematics: To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. By using the trial and error method. ( ) steps to solve an equation by completing the square: ### The Values Of The Variable Satisfying The Given Quadratic Equation Are Called Its Roots. You know by now how to solve a quadratic equation using factoring. In this explainer, we will see how we can also use a graphical method to solve a quadratic equation. Quadratic functions :are written as 𝑓 ;= 2− −6 or = 2− −6.f(x) is zf of x and means that the y value is dependent upon the value of x.once you have an x value and you substitute it into the equation, the value of f(x) will result.f(x) is really just a another way of Tags :<|endoftext|>
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Tracking the destruction of the world's forests is to become much easier for scientists and forest managers, thanks to a software tool unveiled by search-engine giant Google's philanthropic arm today. The software, which uses Google's computing resources to extract scientific information from decades of satellite images of forests, was demonstrated at the UN climate conference in Copenhagen. "We hope this technology will help stop the destruction of the world's rapidly disappearing forests," said a statement on the Google.org blog. "Emissions from tropical deforestation are comparable to the emissions of all of the EU, and are greater than those of all cars, trucks, planes, ships and trains worldwide. According to the Stern Review – the report prepared for the British government in 2006 on the economics of climate change by Lord Nicholas Stern – protecting the world's standing forests is a highly cost-effective way to cut carbon emissions and mitigate climate change." The UN mechanism to reduce deforestation is called Reducing Emissions from Deforestation and Forest Degradation in Developing Countries (Redd), a system whereby richer countries would provide financial incentives to protect forests in poorer nations. For Redd to be successful, however, countries need ways to accurately monitor and report on the state of their forests. In Google.org's prototype software, environmental authorities or NGOs interested in monitoring forests start with satellite images of their area and track how the size and shape of the tree cover has changed over time. The software can processes the images to extract useful scientific and tracking information about how much the forests have changed. For the analysis, the Google.org team worked with Greg Asner of Carnegie Institution for Science and Carlos Souza of Imazon. Technology developed by Asner and Souza is used in Latin America to track changes in forest cover – but mainstream use of the models has been slow due to lack of access to high-quality satellite images and the computer power needed to carry out the analysis. Google.org's solution is to enhance the Asner and Souza models using its own computing power. "What if we could gather together all of the earth's raw satellite imagery data – petabytes of historical, present and future data – and make it easily available on this platform? We decided to find out, by working with Greg and Carlos to re-implement their software online, on top of a prototype platform we've built that gives them easy access to terabytes of satellite imagery and thousands of computers in our data centres," it wrote. Colby Loucks, deputy director of the conservation science program at WWF-US said: "A cost-effective and transparent approach for monitoring deforestation is needed to help pave the way for a global Redd program. If Google's system can be expanded to cover forests globally and access near real-time imagery, it can potentially be a powerful tool that helps tropical countries monitor forest loss."<|endoftext|>
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# College Algebra Tutorial 25 College Algebra Tutorial 25: Slope of a Line Learning Objectives After completing this tutorial, you should be able to: 1. Know when a line rises, falls, is horizontal or is vertical. 2. Find the slope given two points on the line. Introduction In this tutorial we will explore the slope of the line.  Basically, the slope of the line measures the steepness of the line.  We will start with some visuals to help you understand what the slope of a straight line tells us about the line.  Next, we will practice finding the slope given two points on the line.  Let's see what you can do with slopes. Tutorial Slope The slope of a line measures the steepness of the line. Most of you are probably familiar with associating slope with "rise over run". Rise means how many units you move up or down from point to point.  On the graph that would be a change in the y values. Run means how far left or right you move from point to point.  On the graph, that would mean a change of x values. Positive Slope: Note that when a line has a positive slope it rises up left to right. Negative Slope: Note that when a line has a negative slope it falls left to right. Zero Slope: slope = 0 Note that when a line is horizontal the slope is 0. Undefined Slope: slope = undefined Note that when the line is vertical the slope is undefined. Slope Formula Given Two Points Given two points  and The subscripts just indicate that these are two different points.  It doesn't matter which one you call point 1 and which one you call point 2 as long as you are consistent throughout that problem. Note that we use the letter m to represent slope. Example 1: Find the slope of the straight line that passes through (7, 5) and (5, 1) or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical. *Plug in x and y values into slope formula *Simplify The slope of the line is 2. Since the slope is positive, the line would rise (left to right). Example 2: Find the slope of the straight line that passes through (1, -4) and (-1, 3) or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical. *Plug in x and y values into slope formula *Simplify Make sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2.  1 - (-1) is not the same as 1 - 1. The slope of the line is -7/2. Since the slope is negative, the line would fall (left to right). Example 3: Find the slope of the straight line that passes through (4, 1) and (-2, 1) or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical. *Plug in x and y values into slope formula *Simplify Make sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2.  4 - (-2) is not the same as 4 - 2. The slope of the line is 0. Since the slope is zero, the line would be horizontal. Example 4: Find the slope of the straight line that passes through (-2, 3) and (-2, 5) or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical. *Plug in x and y values into slope formula *Simplify Make sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2.  -2 - (-2) is not the same as -2 - 2. The slope of the line is undefined. Since the slope is undefined, the line would be vertical. Practice Problems These are practice problems to help bring you to the next level.  It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it.  Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument.  In fact there is no such thing as too much practice. To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that  problem.  At the link you will find the answer as well as any steps that went into finding that answer. Practice Problems 1a - 1d: Find the slope of the straight line that passes through the given points or state that the slope is undefined.  Then indicate if the line through the points rises (left to right), falls (left to right), is horizontal, or is vertical. 1a. (2, -5) and (-1, -1) 1b. (7, 1) and (7, 2) 1c. (1, 7) and (-2, -2) 1d.  (-1, -5) and (-4, -5) Need Extra Help on these Topics? The following are webpages that can assist you in the topics that were covered on this page: This website helps you with the slope of a line. http://www.purplemath.com/modules/slope.htm This webpage helps you with slope. http://www.math.com/school/subject2/lessons/S2U4L2DP.html This website covers slopes and y-intercept. Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions. Videos at this site were created and produced by Kim Seward and Virginia Williams Trice. Last revised on Jan. 16, 2010 by Kim Seward.<|endoftext|>
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Class 12 Maths # Inverse Trignometric Functions ## NCERT Solution ### Exercise 2.1 Part 2 ##### Find the principal values of the following: Question 9: text(cos)^(-1)((-1)/(sqrt2)) Solution: Let text(cos)^(-1)((-1)/(sqrt2))=θ So, text(cos)θ=(-1)/(sqrt2) Or, text(cos)θ=text(cos)(π-(π)/(4)) Or, text(cos)θ=text(cos)(3π)/(4) Since value of text(cos)^(-1)x lies between 0 and π. Hence, principal value of text(cos)^(-1)((-1)/(sqrt2)) is (3π)/(4) Question 10: text(cosec)^(-1)(-sqrt2) Solution: Let text(cosec)^(-1)(-sqrt2)=θ So, text(cosec)^(-1)θ=-sqrt2 Or, text(cosec)^(-1)θ=cosec(-(π)/(4)) Since value of text(cosec)^(-1)x is [(-π)/(2), (π)/(2)]-(0). Hence, principal value of text(cosec)^(-1)(-sqrt2) is (-π)/(4) Question 11: text(tan)^(-1)+text(cos)^(-1)(-1/2)+text(sin)^(-1)(-1/2) Solution: Let text(tan)^(-1)(1)=θ_1 Or, text(tan)θ_1=1=text(tan)(π)/(4) Or, θ_1=(π)/4 Let text(cos)^(-1)(-1/2)=θ_2 Or, text(cos)θ_2=-1/2=-text(cos)((π)/(3)) Or, text(cos)θ_2=text(cos)(π-(π)/(3)) Or, text(cos)θ_2=text(cos)((2π)/(3)) Or, θ_2=(2π)/3 Let text(sin)^(-1)(-1/2)=θ_3 Or, text(sin)θ_3=-1/2 Or, text(sin)θ_3=-text(sin)((π)/(6))=text(sin)(-(π)/(6)) Or, θ_3=-(π)/(6) Now, text(tan)^(-1)+text(cos)^(-1)(-1/2)+text(sin)^(-1)(-1/2) =θ_1+θ_2+θ_3 =(π)/(4)+(2π)/(3)-(π)/(6) =(9π)/(12)=(3π)/4 Question 12: text(cos)^(-1)(1/2)+2text(sin)^(-1)(1/2) Solution: Let text(cos)^(-1)(1/2)=θ Or, text(cos)θ=1/2=text(cos)((π)/(3)) Or, θ=(π)/(3) Let text(sin)^(-1)(1/2)=θ_1 Or, text(sin)θ_1=1/2=text(sin)(π)/(6) Or, θ_1=(π)/(6) So, text(cos)^(-1)(1/2)+2text(sin)^(-1)(1/2) θ+2θ_1 =(π)/(3)+(2π)/(6) =(2π+2π)/(6)=(2π)/(3) Question 13: If text(sin)^(-1)x=y then 1. 0≤y≤π 2. -(π)/(2)≤y≤(π)/(2) 3. 0<y<π 4. -(π)/(2)<y<(π)/(2) Answer: (b) -(π)/(2)≤y≤(π)/(2) Explanation: Here, text(sin)^(-1)x=y Or, text(sin) y=x Since value of text(sin)^(-1)x lies between -(π)/(2) and (π)/(2) Hence, y=[-(π)/(2), (π)/(2)] Question 14: text(tan)^(-1)sqrt3-text(sec)^(-1)(-2) is equal to 1. π 2. -(π)/(3) 3. (π)/(3) 4. (2π)/(3) Answer: (b) -(π)/(3) Explanation: Let text(tan)^(-1)sqrt3=θ Or, text(tan)θ=sqrt3=text(tan)(π)/(3) Or, θ=(π)/(3) Now, let text(sec)^(-1)(-2)=θ_1 Or, text(sec)θ_1=text(sec)(π-(π)/(3))=text(sec)(2π)/(3) Or, θ_1=(2π)/(3) Now, text(tan)^(-1)sqrt3-text(sec)^(-1)(-2)=θ-θ_1 =(π)/(3)-(2π)/(3)=-(π)/(3)<|endoftext|>
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What does the recently photographed black hole have to do with climate change? The image is a VERY cool representation of an amazing scientific achievement. But more importantly, it reminds us of the value of science and scientific process. On April 10, the first mesmerizing image of a black hole was released to the public. The black hole, named M87-star, is located in the center of Messier 87 galaxy some 318 quintillion miles (53 million light-years) away from Earth. That’s 318,000,000,000,000,000,000 miles! How’d did scientists do that? They focused mega-telescopes at 8 different sites around the world at the black hole at the same time for 10 days last year and recorded the images. Then, hundreds of researchers worked tirelessly to develop computer algorithms (equations) to stitch those images together and produce the first ever picture of a black hole (read more here). The image is a VERY cool representation of an amazing scientific achievement. More importantly, it reminds us of the value of science and scientific process. This achievement will help us better understand physics and allow researchers to better test long-standing theories, including Einstein’s theory of relativity. The problem with most scientific writing, though, is that often it isn’t easily digestible to the everyday reader. The terminology, details, and backstory take years to learn, and scientists write articles and papers for other scientists to advance their field. This is an important exercise because it pushes scientific achievement, but it comes at a cost to the public’s ability to understand the work. More than ever, we rely on news outlets to report on scientific findings in an easy to follow format. Despite reporters best efforts, sometimes headlines can be misleading or content can be misconstrued. This happens because scientists often hedge their bets when reporting findings, using words like “may be”, “appears to be”, or “correlated with.” The scientific process teaches us to do this because science is an iterative process that doesn’t produce a single “correct answer.” Instead, an experiment yields an observation that either fits or doesn’t fit with our expectations. We then use what we learn from that observation to better refine our expectations, and run more experiments. Repeating this process several times gives us much more confidence in predictions, and it helps us better apply our science. This is why every few years you hear conflicting reports about coffee, red wine, or chocolate. One study may find a positive health benefit of those items, while another may find a negative one. These studies aren’t inherently wrong because they disagree. Instead, they show us that food and health have a complicated relationship, and underlying factors—like a family history of heart disease—are probably more important predictors of health than the food itself. The iterative nature of science is what gives it its credibility, and when scientists reach a consensus, we should pay attention. That’s why climate change is such a pressing issue. 98% of climate experts have concluded that climate change is human-caused, and without drastic measures to reduce our carbon emissions, we are in for warming on a global scale that hasn’t been seen on earth in millions of years. It’s an uncomfortable and scary situation, but scientists have also given us powerful recommendations to slow this rate of change and better plan for future climate events. So, how can we all better support and understand science, especially when it contributes to the public good? We all need to communicate more effectively, and believe scientists when they reach a consensus. Scientists are already working to better share their science to the public, especially through news sources, blogs, and social media. The good news is that Americans are already great at scientific literacy, meaning they can work through scientific problems using the scientific process. By bridging this communication divide on both sides, we can all make better decisions to protect our future. About the Author Dr. Danielle Begley-Millier Danielle is an ecologist with an extensive background in forest research, specifically focusing on ecological interactions. She comes to Teatown from Penn State University where she received her doctoral degree in wildlife in 2018. During her time at Penn State, Danielle managed a large research project aimed at understanding the effects of both white-tailed deer and soil nutrition on forest plant communities. She is responsible for the implementation of Teatown’s land stewardship and environmental research programs, as well as the supervision of select volunteers and interns as they participate in these programs. In addition to her Ph.D., Danielle earned her master’s degree in biology from Wright State University in 2013. Interested in getting involved? Check out these upcoming citizen science and stewardship opportunities: Upcoming Events › Citizen Science Upcoming Events › Citizen Science View as calendar Are you interested in learning about the pests that threaten New York’s forests, and would you like to lend a helping hand in monitoring their spread? Teatown is looking for dedicated volunteers and community scientists to participate in a forest pest monitoring program. For adults.Find out more » Gardening with native plants in our suburban landscape has many environmental benefits. Join landscape designer Amanda Bayley, for a presentation on how to beautify your own yard with native plants. She will present on ecoregions, review the native plant communities that are part of Westchester’s ecoregion, and go over local plants to garden with that are best suited for wet, dry, sunny, and shady areas.Find out more »<|endoftext|>
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# How do you find the intercepts for y=-3x+3? Jun 14, 2015 The intercept with the $y$ axis is where $x = 0$, hence $y = - 3 \cdot 0 + 3 = 3$. That is $\left(0 , 3\right)$ The intercept with the $x$ axis is where $y = 0$, hence $0 = - 3 x + 3$, hence $x = 1$. That is $\left(1 , 0\right)$ #### Explanation: To find where $y = - 3 x + 3$ intercepts the $y$ axis, substitute $x = 0$ into the equation to find: $y = \left(- 3 \cdot 0\right) + 3 = 0 + 3 = 3$. That is $\left(0 , 3\right)$ To find where $y = - 3 x + 3$ intercepts the $x$ axis, substitute $y = 0$ into the equation to find: $0 = - 3 x + 3$ Add #3x to both sides to get: $3 x = 3$ Divide both sides by $3$ to get: $x = 1$. That is $\left(1 , 0\right)$ graph{-3x+3 [-9.5, 10.5, -3, 7]}<|endoftext|>
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Our aim of teaching mathematics at our school are to ensure that all pupils: become fluent in the fundamentals of mathematics, including through varied and frequent practice with increasingly complex problems over time, so that pupils develop conceptual understanding and the ability to recall and apply knowledge rapidly and accurately. Children are encouraged to reason mathematically by following a line of enquiry, conjecturing relationships and generalisations, and developing an argument, justification or proof using mathematical language They are taught to solve problems by applying their mathematics to a variety of routine and non-routine problems with increasing sophistication, including breaking down problems into a series of simpler steps and persevering in seeking solutions We stress the importance of knowing multiplication tables. By the end of year 4, pupils should have memorised their multiplication tables up to and including the 12 multiplication table. All pupils are taught to develop efficient strategies for mental and written calculations which is clearly outlined within our school policy. Pupils should read and spell mathematical vocabulary, at a level consistent with their increasing word reading and spelling knowledge.<|endoftext|>
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# CBSE Class 10-Mathematics: Chapter –15 Probability Part 3 Glide to success with Doorsteptutor material for CBSE : fully solved questions with step-by-step explanation- practice your way to success. Question 12: A number is chosen at random from the numbers What is the probability that can take values To set Probability Question 12: A number is selected from the number and then a second number is randomly selected from the numbers . What is the probability that the product of the two numbers will be less than ? Number can be selected in three wavs and corresponding to each such wav there are three wavs of selecting number y. Therefore. Two numbers can be selected in wavs as listed below: Favourable number of elementary events Hence, required probability Question 14: In the adjoining figure a dart is thrown at the dart board and lands in the interior of the circle. What is the probability that the dart will land in the shaded region. We have Using Pythagoras Theorem is A ABC. We have Area of the circle Area of shaded region circle – Area of rectangle ABCD Hence Question 15: In the fig points A. B.C and D are the centres of four circles.each having a radius of 1 unit. If a point is chosen at random from the interior of a square ABCD. What is the probability that the point will be chosen from the shaded region. Radius of the circle is 1 unit Area of the circle =Area of a sector Side of the square ABCD Area of square Area of square Area of sectors Probability Question 16: In the adjoining figure ABCD is a square with sides of length units points are the mid points of the sides BC & CD respectively. If a point is selected at random from the interior of the square what is the probability that the point will be chosen from the interior of the square what is the probability that the point will be chosen from the interior of the triangles APO.<|endoftext|>
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# Multiples of 6 The Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, … i.e. M6 = {6, 12, 18, 24, 30, 36, 42, 48, 54, 60, …}. Multiples of 6 are the series of numbers obtained by multiplying 6 with the natural numbers. 6 × 1 = 6 6 × 2 = 12 6 × 3 = 18 6 × 4 = 24 6 × 5 = 30 6 × 6 = 36 6 × 7 = 42 6 × 8 = 48 6 × 9 = 54 6 × 10 = 60 And so on. From this series, we observe that the multiples of 6 can be derived by multiplying 6 by natural numbers in ascending order. These multiples form an arithmetic sequence with a common difference of 6. ## Properties and Patterns of Multiples of 6 Multiples of 6 possess several interesting properties that make them unique. Here are a few notable ones: 1.    Divisibility by 6: Every multiple of 6 is divisible by 6 without leaving a remainder. This property stems from the fact that when we divide any multiple of 6 by 6, the quotient is always an integer. 2.    Divisibility by 2 and 3: Since 6 is divisible by both 2 and 3, all multiples of 6 are also divisible by both 2 and 3. Divisibility by 2 implies that all multiples of 6 are even numbers, while divisibility by 3 suggests that the sum of the digits of any multiple of 6 is also divisible by 3. 3.    Repeating Pattern of Digits: When we examine the units digit of multiples of 6, a fascinating pattern emerges. The units digit repeats in a cycle of 6, 2, 8, 4, 0, and again 6, as we proceed to higher multiples. For instance, the units digit of 6 is 6, the units digit of 12 is 2, and so on. ******************** 10 Math Problems officially announces the release of Quick Math Solver and 10 Math ProblemsApps on Google Play Store for students around the world. ******************** ******************** ## Applications of Multiples of 6 The study of multiples of 6 finds applications in various fields, including mathematics, science, and everyday life: 1.    LCM: One of the applications of multiples of 6 is in finding the lowest common multiple (LCM) of two or more numbers. The LCM is the lowest multiple that two or more numbers have in common. For example, to find the LCM of 6 and 4, we need to find the multiples of both numbers and identify the lowest multiple they have in common. The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, … etc. The multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40 … etc. The lowest multiple that they have in common is 12. Therefore, the LCM of 6 and 4 is 12. 2.    Timekeeping: The division of an hour into 60 minutes and each minute into 60 seconds is based on the concept of multiples. By dividing 60 by 6, we obtain 10, which corresponds to the number of minutes in one-tenth of an hour. 3.    Music: In music theory, multiples of 6 play a crucial role in understanding rhythm. The time signatures and beats per measure are based on multiples of 6, allowing musicians to create harmonious compositions. ## Conclusion: The Multiples of 6 are the numbers obtained by multiplying 6 with natural numbers. The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, … etc. The multiples of 6 have various properties, such as divisibility by 6, divisibility by 2 and 3, repeating pattern of digits, etc. The multiples of 6 have several applications in mathematics, such as finding the LCM, timekeeping, and music.<|endoftext|>
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# Multiplying Fractions And Mixed Numbers Worksheet 6th Grade This multiplication worksheet is focused on training students the best way to psychologically increase complete figures. Students are able to use personalized grids to suit specifically a single question. The worksheets also includedecimals and fractions, and exponents. There are also multiplication worksheets with a distributed home. These worksheets can be a need to-have for the mathematics school. They could be utilized in class to discover ways to mentally grow total line and numbers them up. Multiplying Fractions And Mixed Numbers Worksheet 6th Grade. ## Multiplication of entire numbers If you want to improve your child’s math skills, you should consider purchasing a multiplication of whole numbers worksheet. These worksheets can assist you master this fundamental principle. It is possible to go for one digit multipliers or two-digit and about three-digit multipliers. Capabilities of 10 may also be an excellent solution. These worksheets will enable you to training extended practice and multiplication reading through the figures. Also, they are the best way to aid your youngster comprehend the necessity of learning the different kinds of entire figures. ## Multiplication of fractions Experiencing multiplication of fractions with a worksheet may help professors strategy and get ready classes successfully. Utilizing fractions worksheets will allow instructors to swiftly evaluate students’ idea of fractions. Pupils can be pushed to complete the worksheet in a a number of time as well as then tag their techniques to see where by they need more training. Individuals can be helped by phrase conditions that associate maths to true-daily life circumstances. Some fractions worksheets incorporate types of comparing and contrasting numbers. ## Multiplication of decimals Once you increase two decimal numbers, ensure that you group of people them vertically. The product must contain the same number of decimal places as the multiplicant if you want to multiply a decimal number with a whole number. For instance, 01 x (11.2) by 2 can be equal to 01 by 2.33 by 11.2 except if the item has decimal areas of lower than two. Then, this product is round towards the nearby total variety. ## Multiplication of exponents A math concepts worksheet for Multiplication of exponents will allow you to training dividing and multiplying phone numbers with exponents. This worksheet may also supply things that requires individuals to flourish two various exponents. You will be able to view other versions of the worksheet, by selecting the “All Positive” version. Apart from, also you can get into special recommendations on the worksheet by itself. When you’re done, you are able to click on “Produce” and the worksheet will probably be acquired. ## Department of exponents The essential tip for section of exponents when multiplying phone numbers is usually to subtract the exponent inside the denominator through the exponent in the numerator. However, if the bases of the two numbers are not the same, you can simply divide the numbers using the same rule. By way of example, \$23 separated by 4 will the same 27. This method is not always accurate, however. This method can cause uncertainty when multiplying numbers that happen to be too large or too small. ## Linear capabilities You’ve probably noticed that the cost was \$320 x 10 days if you’ve ever rented a car. So the total rent would be \$470. A linear purpose of this kind offers the develop f(by), in which ‘x’ is the number of times the vehicle was rented. Furthermore, they have the shape f(x) = ax b, where ‘b’ and ‘a’ are genuine figures.<|endoftext|>
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When the Spanish conquerors arrived first to these lands; they could not explain themselves how Peruvian "Indians" (ignorant, wild, without any ability of logical reasoning, one more animal species according to conquerors) could have built such a greatness. Their religious fanaticism led them to believe that all that was simply work of demons or malign spirits. Still today, many people believe in the inability of ancient Quechuas to create such a wonder, so they suggest that they were made by beings of some other worlds, extraterrestrial beings with superior technology that made all that possible. However, our history and archaeology demonstrate that those objects of admiration are an undeniable work of the Incas, Quechuas, Andean people or however pre-Hispanic inhabitants of this corner of the world would be named. The imperial city Cusco, meaning ‘navel of the earth,’ was laid out in the form of a puma, the animal that symbolized the Inca dynasty. The belly of the puma was the main plaza, the river Tullumayo formed its spine, and the hill of Sacsayhuaman its head. One of the most imposing architectonic complexes inherited from the Incan Society is Sacsayhuaman, which because of several of its qualities is considered as one of the best monuments that mankind built on the earth's surface. The wall or rampart is the most impressive section, built with enormous carved limestone boulders, this construction has a broken line that faces to the main plaza called Chuquipampa which is a slope with 25 angles and 60 walls.The biggest carved boulder of the first wall weighs about 70 tons and like all of the other rocks was brought from a quarry called Sisicancha, three kilometers away and where there are still rocks that were transported part of the way. Each wall is made up of 10 fronts with the most important ones known as Rumipunco, tiupunku, Achuanpunku and Viracocha punku.<|endoftext|>
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Sign in or start a free trial to avail of this feature. 3. Combine Strings of Text We often need to combine multiple cells of text into one cell and to do this we use '&'. Find out how to use this very useful command correctly in this lesson &: Chain text strings together LOWER: Change cell text to lowercase ALT + I , C: Insert column F2: Jump back into cell formula Combining text strings together in Excel is very easy and one of the most common commands you'll perform when manipulating data. In this lesson, we're going to take the first and last name for each user and create a username, which will be the lowercase first name, followed by an underscore, followed by a lowercase last name. We’ll start by creating a new column with Alt + I C, and we'll widen this using the mouse. I'll entitle this column Username and I'll start by writing equals and including the first name. To chain the first name to the last name, we use the ampersand sign, which on most keyboards is Shift + 7. And then to include the last name in this string, I'll simply select it, which is H2, and press Enter. And as you can see, the ampersand sign chains the first name and the last name entries together in one cell. However, we still have a lot of work to do. Our first task will be to change the last name to lowercase. So let's re-enter the formula with F2, and we replace H2 with the function “lower” and then the cell H2 passed into this function. If I now press Enter, you can see that we now have the first name and the last name both in lowercase. All we have to do now is include the underscore. So let's return to our formula again and press F2 to jump back inside. When you're chaining strings and you'd like to include a character such as an underscore, you simply wrap that character n double quotes and then connect it to the other elements using the ampersand sign. So I'll move in position to the first ampersand sign, open double quotes, write an underscore, close double quotes, and then write another ampersand sign to connect the underscore to the last name. If I now press enter, you can see we now have the user name perfectly created for Kaci Rawlins. Using the ampersand sign, we can combine strings of text in almost any way we want. To fill out the username for the remaining entries, we can just select the cell and double click the bottom right hand corner with our mouse. To get some more practice at chaining strings together, try to create a new column called Full Name that separates the first name and the last name with a space and has both values in the capitalized format. I'll leave my answer in the after file which can be downloaded underneath this video.<|endoftext|>
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In the human digestive system, large organic masses are broken down into smaller particles that the body can use as fuel. This is a complex process. The breakdown of the nutrients requires the coordination of several enzymes secreted from specialized cells within the mouth, stomach, intestines, and liver. The major organs or structures that coordinate digestion within the human body include the mouth, esophagus, stomach, small and large intestines, and liver. In the human body, the mouth (oral cavity) is a specialized organ for receiving food and breaking up large organic masses. In the mouth, food is changed mechanically by biting and chewing. Humans have four kinds of teeth: incisors are chisel-shaped teeth in the front of the mouth for biting; canines are pointed teeth for tearing; and premolars and molars are flattened, ridged teeth for grinding, pounding, and crushing food. In the mouth, food is moistened by saliva, a sticky fluid that binds food particles together into a soft mass. Three pairs of salivary glands—parotid, submaxillary, and sublingual—secrete saliva into the mouth. The saliva contains an enzyme called amylase, which digests starch molecules into smaller molecules of the disaccharide maltose. During chewing, the tongue moves food about and manipulates it into a mass called a bolus. The bolus is pushed back into the pharynx (throat) and is forced through the opening to the esophagus. The esophagus is a thick-walled muscular tube located behind the windpipe that extends through the neck and chest to the stomach. The bolus of food moves through the esophagus by peristalsis: a rhythmic series of muscular contractions that propels the bolus along. The contractions are assisted by the pull of gravity. The esophagus joins the stomach at a point just below the diaphragm. A valvelike ring of muscle called the cardiac sphincter surrounds the opening to the stomach. The sphincter relaxes as the bolus passes through and then quickly closes. The stomach is an expandable pouch located high in the abdominal cavity. Layers of stomach muscle contract and churn the bolus of food with gastric juices to form a soupy liquid called chyme. The stomach stores food and prepares it for further digestion. In addition, the stomach plays a role in protein digestion. Gastric glands called chief cells secrete pepsinogen. Pepsinogen is converted to the enzyme pepsin in the presence of hydrochloric acid. Hydrochloric acid is secreted by parietal cells in the stomach lining. The pepsin then digests large proteins into smaller proteins called peptides. To protect the stomach lining from the acid, a third type of cell secretes mucus that lines the stomach cavity. An overabundance of acid due to mucus failure may lead to an ulcer. The soupy mixture called chyme spurts from the stomach through a sphincter into the small intestine. An adult’s small intestine is about 23 feet long and is divided into three sections: the first 10 to 12 inches form the duodenum; the next 10 feet form the jejunum; and the final 12 feet form the ileum. The inner surface of the small intestine contains numerous fingerlike projections called villi (the singular is villus). Each villus has projections of cells called microvilli to increase the surface area. Most chemical digestion takes place in the duodenum. In this region, enzymes digest nutrients into simpler forms that can be absorbed. Intestinal enzymes are supplemented by enzymes from the pancreas, a large, glandular organ near the stomach. In addition, bile enters the small intestine from the gallbladder to assist in fat digestion. The enzymes functioning in carbohydrate digestion include amylase (for starch), maltase (for maltose), sucrase (for sucrose), and lactase (for lactose). For fats, the principal enzyme is lipase. Before lipase can act, the large globules of fat must be broken into smaller droplets by bile. Bile is a mixture of salts, pigments, and cholesterol that is produced by the liver and stored in the gallbladder, a saclike structure underneath the liver. Protein digestion is accomplished by several enzymes, including two pancreatic enzymes: trypsin and chymotrypsin. Peptides are broken into smaller peptides, and peptidases reduce the enzymes to amino acids. Nucleases digest nucleic acids into nucleotides in the small intestine also. Most absorption in the small intestine occurs in the jejunum. The products of digestion enter cells of the villi, move across the cells, and enter blood vessels called capillaries. Diffusion accounts for the movement of many nutrients, but facilitated diffusion is responsible for the movement of glucose and amino acids. The products of fat digestion pass as small droplets of fat into lacteals, which are branches of the lymphatic system. Absorption is completed in the final part of the small intestine, the ileum. Substances that have not been digested or absorbed then pass into the large intestine. The small intestine joins the large intestine in the lower-right abdomen of the body. The two organs meet at a blind sac called the cecum and a small fingerlike organ called the appendix. Evolutionary biologists believe the cecum and appendix are vestiges of larger organs that may have been functional in human ancestors. The large intestine is also known as the colon. It is divided into ascending, transverse, and descending portions, each about one foot in length. The colon’s chief functions are to absorb water and to store, process, and eliminate the residue following digestion and absorption. The intestinal matter remaining after water has been reclaimed is known as feces. Feces consist of nondigested food (such as cellulose), billions of mostly harmless bacteria, bile pigments, and other materials. The feces are stored in the rectum and passed out through the anus to complete the digestion process. The liver has an important function in processing the products of human digestion. For example, cells of the liver remove excess glucose from the bloodstream and convert the glucose to a polymer called glycogen for storage. The liver also functions in amino acid metabolism. In a process called deamination, it converts some amino acids to compounds that can be used in energy metabolism. In doing so, the liver removes the amino groups from amino acids and uses the amino groups to produce urea. Urea is removed from the body in the urine (see Chapter 26). Fats are processed into two-carbon units that can enter the Krebs cycle for energy metabolism. The liver also stores vitamins and minerals, forms many blood proteins, synthesizes cholesterol, and produces bile for fat digestion.<|endoftext|>
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When art enthusiasts first come across Teyjah McAren’s modern cave paintings, they often have questions about the meaning of the symbols they see. Visitors also want to know more about the origins of rock art and petroglyphs and the ancient peoples who painted them in the distant past. Learn more with our helpful guide to the language, symbols and heritage of cave art. Ancient peoples around the world made art on natural geological surfaces such as cave walls, cliff faces, bedrock and boulders. Rock art includes pictographs, petroglyphs, engravings, petroforms, and geoglyphs. A hieroglyph is a form of picture writing which was the form of writing used in ancient Egypt. Each cluster of symbols represented a word or idea rather than just a sound like the alphabet. A petroglyph is a rock art design that has been created by carving or engraving. Other terms for petroglyphs are “rock art” and “rock writing”. A pictograph is a rock art design that was drawn or painted on a rock surface. Art materials could include powdered minerals, charcoal, plant substances, or blood. A geoglyph is rock art made on the ground. Geoglyphs include intaglios, which are made by removing stones or portions of the ground; petroforms (also called rock alignments) made by rearranging stones to form sculptures like Inukshuks); and gravel geoglyphs, which are shaped from gravel-covered areas. Rock art was created by innovative creators using a wide range of methods and materials. One way was through pecking the stone with tools to give it a rough, pitted surface. Rubbing or wearing down the stone was used to create lines. Symbols and images could be incised, gouged out in intaglio technique, chipped out with a stone, sculpted, overpainted or crosshatched. Some rock art was painted with charcoal and earth pigments mixed with grease, gum, or water. Often, engraving and painting techniques were sometimes combined. Some believe that petroglyphs were created to mark tribal territory, indicate water, or show activities conducted in those areas. Other petroglyphs seem to record important events, such as a successful hunt or the migration of a tribe. Some symbols recorded good hunting sites. Others tracked the movement of the sun, the moon, planets, and stars. Some petroglyphs seem to relate to a spiritual life, religious ceremonies, sacred beings and the activities of shamans. Some rock art hints at epic legends and mythical protagonists. Petroglyphs can be associated with particular tribes or styles, but exact interpretation is exceedingly challenging. As these early cultures died out, their messages may have been lost, or the meaning altered with each successive generation. Even present-day members from different tribes may read the symbols differently. By learning more about these ancient cultures, we may gain insight into the mysterious signs they have left to us through the epochs. Rock Art is found on almost every continent, in caves on boulders or rocky outcrops, on ancient monuments and other structures such as Stonehenge.<|endoftext|>
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# How do you solve \sqrt { 9v - 15} = \sqrt { 5v - 3}? Dec 13, 2016 $v = 3$ #### Explanation: $\sqrt{9 v - 15} = \sqrt{5 v - 3}$ Square both sides. $9 v - 15 = 5 v - 3$ Subtract $5 v$ from both sides. $4 v - 15 = - 3$ Add $15$ to both sides. $4 v = 12$ Divide both sides by $4$. $v = 3$<|endoftext|>
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The Arctic experienced its second-warmest year on record between October 2017 and September 2018, according to the annual Arctic Report Card. And average air temperatures in this frigid region have hit record or near-record levels every year since 2014. Temperatures in the Arctic are now increasing at roughly twice the rate of the global average, researchers say in the report released by the US National Oceanic and Atmospheric Administration (NOAA). Warming is causing changes across the entire Arctic, and those changes are building, said Emily Osborne, the programme manager at the NOAA Arctic Research Program in Silver Spring, Maryland. She helped to present results from the report on 11 December during a meeting of the American Geophysical Union in Washington DC. Wild reindeer and caribou populations, for example, have declined by more than 50% since the 1990s. And more than 99% of sea ice is now considered relatively new, meaning that it hasn’t lasted for more than four summers without melting. It’s now the thinnest and most susceptible to warming temperatures that it's been in 30 years. Harmful algal blooms are seven times more prevalent off the Alaskan coast than they were 40 years ago. And the Arctic Ocean now contains the highest concentration of microplastics of any ocean basin in the world, according to the report. The microplastics and algal bloom findings really surprised Donald Perovich, a geophysicist at Dartmouth College in Hanover, New Hampshire, who worked on the report card. “Even here — the farthest point on Earth— we’re still seeing impacts from lower latitudes,” he says.<|endoftext|>
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# High School Math : How to find the length of the diagonal of a square ## Example Questions ### Example Question #1 : How To Find The Length Of The Diagonal Of A Square What is the length of a diagonal of a square with a side length ? Round to the nearest tenth. Explanation: A square is comprised of two 45-45-90 right triangles. The hypotenuse of a 45-45-90 right triangle follows the rule below, where  is the length of the sides. In this instance, is equal to 6. ### Example Question #1 : How To Find The Length Of The Diagonal Of A Square A square has sides of . What is the length of the diagonal of this square? Explanation: To find the diagonal of the square, we effectively cut the square into two triangles. The pattern for the sides of a is . Since two sides are equal to , this triangle will have sides of . Therefore, the diagonal (the hypotenuse) will have a length of . ### Example Question #3 : How To Find The Length Of The Diagonal Of A Square A square has sides of . What is the length of the diagonal of this square? Explanation: To find the diagonal of the square, we effectively cut the square into two triangles. The pattern for the sides of a is . Since two sides are equal to , this triangle will have sides of . Therefore, the diagonal (the hypotenuse) will have a length of . ### Example Question #233 : Geometry What is the length of the diagonal of a square with a side length of ? Explanation: To find the diagonal of a square, we must use the side length to create a 90 degree triangle with side lengths of , and a hypotenuse which is equal to the diagonal. Pythagorean’s Theorem states , where a and b are the legs and c is the hypotenuse. Take  and  and plug them into the equation for  and : After squaring the numbers, add them together: Once you have the sum, take the square root of both sides: Simplify to find the answer: , or . ### Example Question #5 : How To Find The Length Of The Diagonal Of A Square What is the length of the diagonal of a 7-by-7 square? (Round to the nearest tenth.) Explanation: To find the diagonal of a square we must use the side lengths to create a 90 degree triangle with side lengths of 7 and a hypotenuse which is equal to the diagonal. We can use the Pythagorean Theorem here to solve for the hypotenuse of a right triangle. The Pythagorean Theorem states , where a and b are the sidelengths and c is the hypotenuse. Plug the side lengths into the equation as  and : Square the numbers: Add the terms on the left side of the equation together: Take the square root of both sides: Therefore the length of the diagonal is 9.9. ### Example Question #1 : Squares The perimeter of a square is 48. What is the length of its diagonal? Explanation: Perimeter = side * 4 48 = side * 4 Side = 12 We can break up the square into two equal right triangles. The diagonal of the sqaure is then the hypotenuse of these two triangles. Therefore, we can use the Pythagorean Theorem to solve for the diagonal:<|endoftext|>
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A paper published this week in Nature Geoscience has released the first measurements of greenhouse gases from permafrost under Arctic lakes. But while the study confirms those gases locked away in ice for thousands of years are seeping free, it concludes the amounts are not yet large. "It's a lit fuse, but the length of that fuse is very long," said lead author Katey Walter Anthony of the University of Alaska. "According to the model projections, we're getting ready for the part where it starts to explode. But it hasn't happened yet." Scientists have long known that permafrost contains vast quantities of carbon in dead plants and other organic material, about twice as much as the entire atmosphere. Now, that permafrost is melting more quickly as the Arctic warms up faster than anywhere else on Earth. The melt often takes place in Arctic lakes where liquid water thaws long-frozen soil. The material released is digested by tiny bugs and turned into carbon dioxide and methane. "I've been walking on these lakes when they were frozen for a very long time," said Walter Anthony. "I would go out after the ice formed, look at the lake ice surface and see bubbles. I observed that bubbles are most dense and largest along the edge, where the margins were expanding, where the permafrost was thawing." Her observation led to three questions: Were the bubbles generated by melting permafrost? If so, was the permafrost releasing ancient, long-dormant carbon? And, if so, how much? Researchers looked at lakes in Alaska and Siberia, as well as data from Canada. They used aerial photographs and other information to measure how the area had changed over the last 60 years. They found that, across the Arctic, the amount of gas being released from a lake was directly related to its expansion. The more permafrost was melted around the water's edge, the bigger the lake became, and the more greenhouse gases were released. The team captured some of those gases and subjected them to radiocarbon dating. They found the gases had been generated from carbon stored for anywhere between 10,000 and 30,000 years. It was the answer to the final question that most surprised Walter Anthony. Climate scientists have long predicted a spike in the release of such long-dormant carbon. But if that's true, it isn't happening yet. "Today, the permafrost-carbon feedback is pretty small," she said. Models suggest that over the next 90 years, greenhouse gas releases from permafrost will be 100 times higher than the levels Walter Anthony measured. "I believe it will (happen), but when are we going to start seeing that?" she asked. The increasingly warmer Arctic may eventually reach a permafrost-carbon tipping point, Walter Anthony suggested. The stakes are high. Scientists estimate there are more than 1,400 petagrams of old carbon stored in permafrost. Each petagram is a billion tonnes. "There's a lot of interest in what the fate of that permafrost carbon is." Thursday, Aug. 25, 2016 at 7:46 AM — Follow Bob Weber on Twitter at @row1960 original story HERE Sign up for the Global Warming Blog for free by clicking here. In your email you will receive critical news, research and the warning signs for the next global warming disaster. Click here to learn how global warming has become irreversible and what you can do to protect your family and assets. To share this blog post: Go to the original shorter version of this post. Look to lower right for the large green Share button. To view our current agreement or disagreement with this blog article, click here.<|endoftext|>
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# 2.10 Extension: Fitting functions to quadratic data Lesson In our lesson from Algebra 1 on lines of best fit we saw how to fit data to a linear model to make predictions and interpret patterns in the data.  Now, we'll look at how we might do the same for data that more closely resembles a quadratic curve. ## Using technology to fit quadratic regression models Let's consider the following data set: $x$x $4$4 $4.8$4.8 $5.1$5.1 $6$6 $7.1$7.1 $8.2$8.2 $9.4$9.4 $y$y $19.4$19.4 $20.4$20.4 $20.2$20.2 $19.1$19.1 $18$18 $14.9$14.9 $10$10 As we can see from the scatter plot, the data appears to be following a quadratic pattern better than a linear one. Instead of trying to fit the data to a linear function, let's have our calculator fit a quadratic regression model to the data. As you can see, when we go to choose a model for regression, there are many to choose from. Your knowledge of functions will help you make the best choice, but this lesson will focus on quadratics. And here we have the equation of the quadratic function fitted to the data. We can see the value of the coefficient of determination, $r^2$r2, is very strong (it's very close to $1$1). Given the coefficients $a$a, $b$b, and $c$c we have the quadratic function $y=-0.52x^2+5.27x+6.86$y=0.52x2+5.27x+6.86 #### Practice questions ##### Question 1 The scatter diagram shows the number of fish in pond as a function of time. 1. Which type of model would be appropriate for this data? Linear A B Linear A B 2. Will the coefficient of $x^2$x2 in the quadratic model be positive or negative? Positive A Negative B Positive A Negative B ##### Question 2 A social researcher claims that the longer people stay in their job, the less satisfaction they gain from their work. She asked a sample of people how many years they had been employed in their current job and to rate their level of satisfaction out of $10$10. The results are presented in the table. Number of years employed Satisfaction rating $2$2 $9$9 $3$3 $5$5 $5$5 $4$4 $6$6 $2$2 $8$8 $5$5 $10$10 $7$7 $12$12 $8$8 $13$13 $7$7 $15$15 $9$9 1. Create a scatter plot for the data collected. 2. Does the scatter plot support the social worker’s claims? Yes A No B Yes A No B 3. Which form of equation would be best suited to model the relationship between the number of years employed and satisfaction with the job? $y=mx+b$y=mx+b A $y=a^x+b$y=ax+b B $y=ax^2+bx+c$y=ax2+bx+c C $y=mx+b$y=mx+b A $y=a^x+b$y=ax+b B $y=ax^2+bx+c$y=ax2+bx+c C 4. The social researcher herself has been employed in her current job for $4$4 years and rates her satisfaction with her work a $10$10 out of $10$10. What is the difference between the satisfaction rating approximated by the model $y=0.5x^2-6x+20$y=0.5x26x+20 and her actual rating? ##### Question 3 Nine data points have been plotted below with a quadratic curve of best fit. 1. Predict the $y$y-value of a point with an $x$x-value of $13$13. 2. Which of the following points would be predicted by the quadratic curve of best fit? $\left(3,4\right)$(3,4) A $\left(14,10\right)$(14,10) B $\left(2,9\right)$(2,9) C $\left(15,15\right)$(15,15) D $\left(3,4\right)$(3,4) A $\left(14,10\right)$(14,10) B $\left(2,9\right)$(2,9) C $\left(15,15\right)$(15,15) D ## Interpolation and extrapolation from a model Given a set of data relating two variables $x$x and $y$y, we can use a model to best estimate how the dependent variable changes in response to the independent variable $x$x. A model allows us to go one step further and make predictions about other possible ordered pairs that fit this relationship. #### Exploration Say we gathered several measurements on the population $P$P of a small town $t$t years after an earthquake. We can then plot the data on the $xy$xy-plane as shown below. Population of a town measured at several instances. We can fit a model through the observed data to make predictions about the population at certain times after the earthquake. One plausible model might look like this function: A curve modeling population of a town over time. To make a prediction on the population, say two years after the earthquake, we first identify the point on the curve when $t=2$t=2. Then we find the corresponding value of $P$P. As you can see below, the model predicts that two years after the earthquake, the population of the town was $250$250. A predicted population of $250$250 when $t=2$t=2. A prediction which is made within the observed data set is called an interpolation. Roughly speaking, we've gathered data between $t=0.3$t=0.3 and $t=2.3$t=2.3 so a prediction at $t=2$t=2 would be classified as an interpolation. If we predict the population six years after the earthquake, we find that the population is roughly $16$16. A prediction outside the observed data set such as this one is called an extrapolation. A predicted population of $16$16 when $t=6$t=6. How reliable are these predictions? Well, any model that fits the observed data will make reliable predictions from interpolations since the model roughly passes through the center of the data points. We can say that the model follows the trend of the observed data. However, extrapolations are generally unreliable since we make assumptions about how the relationship continues outside of collected data. Sometimes extrapolation can be made more reliable if we have additional information about the relationship. Consider if we were to use the following quadratic model to fit the data. We can see that interpolating doesn't change much from the previous model, but the predicted values from extrapolation are very different. Polynomial curve modeling population of a town over time. With further information, say like government funding and support aid, the population of the town might increase after a certain point and so the polynomial curve may be an appropriate model to use. Remember! A prediction made within the observed data is called interpolation. A prediction made outside the observed data is called extrapolation. Generally, extrapolation is less reliable than interpolation since the model makes assumptions about the relationship outside the observed data set. #### Practice question ##### Question 4 The height of a particular projectile $y$y in meters was measured at different times $t$t in seconds and the following curve of best fit was drawn. 1. Using the curve of best fit, what is the predicted height of the projectile after $1$1 second? 2. Using the curve of best fit, what is the predicted height of the projectile after $4$4 seconds? 3. The height at a given time has the following relationship $y=-5t^2+100$y=5t2+100. Which of the following statements about interpolation and extrapolation is true? Extrapolation is less reliable since we make assumptions about the relationship outside of the observed data. A Extrapolation is less reliable since we should use a line of best fit instead. B Extrapolation is less reliable since we make assumptions about the relationship outside of the observed data. A Extrapolation is less reliable since we should use a line of best fit instead. B<|endoftext|>
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Developmental Delay and Disabilities Developmental delay refers to the children’s physical and mental functioning level which is not on par with children of their age group. In other words, these children do not reach their developmental milestone at the expected age. Developmental delay is not only delay in physical development and hence, may not be easily detected by parents or caregivers. Developmental delay and disabilities in children include: delay in speech and language development, delay in cognition development, delay in motor development, delay in emotional and social skills, Global Developmental Delay (delay in most of the areas), and Pervasive Developmental Disorder |Delay in speech and language development||Children’s ability in speech and language is lower than children in their age group as they take a longer time to reach the expected speech and language level. For instance, they have no meaningful speech at all at the age of 2, have difficulty at speaking in sentence level at the age of 3, and have “baby talk” or still mumbling after the age of 3.| |Delay in cognition development||Cognition delay refers to children’s intelligence or cognition that has not reached the expected developmental milestone. These children receive equal education with peers but parents and teachers may find their learning is much slower. In addition, they have difficulty in understanding abstract concepts, in memorizing and recalling information as well as poor organizational skills.| |Delay in motor development||Motor development delay includes delay in gross and fine motor skills as well as body coordination. Gross motor skills include walking, running, crawling and any movement that involves large group of muscles. Fine motor skills include skills of manipulating hands and wrist such as eye-hand coordination, finger movement and manipulation.| |Delay in emotional and social skills||Children with such delay would encounter difficulty with emotion regulation and socializing. They often fail to establish and maintain friendship. For instance, they have difficulty in maintaining conversations; they are passive in play, and they prefer to be alone. On top of that, they might have inappropriate emotional responses and behaviors, i.e. laugh or burst out in tears in inappropriate context. In terms of emotion, they tend to get anxious and panic easily.| |Global Developmental Delay||Children have delay in most of their skill areas. It is common to see children having delay in more than a specific area. For instance, a child could have motor development and speech development at the same time.| |Pervasive Developmental Disorder||This is a specific developmental disability as it affects children’s ability in social interaction, communication and their interest. According to the DSM-IV-TR, Pervasive Developmental Disorders include Autistic Disorder, Asperger’s Disorder, Rett’s Disorder, and Childhood Disintegrative Disorder.| Developmental delay, especially speech and cognition delay, is closely related with Learning Disabilities. In many cases, parents seek for professional consultation and help only when they found out that their children are slow in academic learning (often reported by school teachers). Unfortunately, by then these children have already reached school age and most of them have missed the critical time for early intervention (0-6 years old). The developmental delay, especially speech and cognition delay, has already affected their school performance. If parents or main caregivers had detected signs of developmental delay earlier, these children’s functional skills could be improved with early intervention. Hence, parents or caregivers are encouraged to be observant and caring in understanding children’s development milestone.<|endoftext|>
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# Parabolas Set 4 Go back to  'SOLVED EXAMPLES' Example – 7 The normals at three points A, B and C on a parabola intersect at O. F is the focus of the parabola. Prove that $FA\cdot FB \cdot FC = \frac{l}{4} \cdot F{O^2}$ where l is the length of the latus-rectum. Solution: Assume the parabola to be $${y^2} = 4ax,$$ and the point O to be (h, k). F is the point (a, 0). Any normal to is $${y^2}= 4ax$$ is $y = mx - 2am -a{m^3}$ and since this passes through O(h, k), we have $k = mh - 2am -a{m^3}...\left( 1 \right)$ The feet of the three normal correspond to A, B and C. Thus, if $${m_1},\,{m_2}{\;\rm{and }}\;{\rm{ }}{m_3}$$ and are the roots of (1), the co-ordinates of A, B and C are $(am_1^2,\, -2a{m_1}),\,(am_2^2,\, - 2a{m_2}){\;\rm{and}}\;{\rm{ }}(am_3^2,\, - 2a{m_3})$ We have now, $FA = \sqrt {{{(a -am_1^2)}^2} + {{(2a{m_1})}^2}}$ $= a(1 + m_1^2)$ Similarly, $$FB = a(1 + m_2^2){\;\rm{ and}}\;{\rm{ }}FC = a(1 + m_3^2)$$. Thus, \begin{align}&FA \cdot FB \cdot FC= {a^3}(1 + m_1^2)(1 + m_2^2)(1 + m_3^2)\\\\&= {a^3}\left( {1 +m_1^2 + m_2^2 + m_3^2 + m_1^2m_2^2 + m_2^2m_3^2 + m_3^2m_1^2 +{{({m_1}{m_2}{m_3})}^2}} \right) \qquad \quad \dots \left( 2 \right)\end{align} From(1), we know the values of \left.\begin{align}{m_1} + {m_2} +{m_3} = 0\\{m_1}{m_2} +{m_2}{m_3} + {m_3}{m_1} = \left( {\frac{{2a - h}}{a}} \right)\\{m_1}{m_2}{m_3} = -\frac{k}{a}\end{align}\right\} \qquad \quad \dots\left( 3 \right) Our task is to express the relation (2) in terms of the known quantities given by(3). This can be done as follows : \begin{align}&\qquad \qquad \qquad \qquad m_1^2+ m_2^2 + m_3^3 = {({m_1} + {m_2} + {m_3})^2} - 2({m_1}{m_2} + {m_2}{m_3} +{m_3}{m_1})\\\\&m_1^2m_2^2 + m_2^2m_3^2 + m_3^2m_1^2 = {({m_1}{m_2} + {m_2}{m_3} +{m_3}{m_1})^2} - 2{m_1}{m_2}{m_3}({m_1} + {m_2} + {m_3})\end{align} Substituting the appropriate values gives \begin{align}FA\cdot FB \cdot FC &= {a^3}\left\{ {1 + {{\left( {\frac{{2a - h}}{a}} \right)}^2}- 2\left( {\frac{{2a - h}}{a}} \right) + \frac{{{k^2}}}{{{a^2}}}} \right\}\\\\ &=a\{ {(h - a)^2} + {k^2}\} \end{align} which evidently equals $$a \cdot F{O^2}.$$ Since $$l = 4a,$$ we get the desired result: $FA \cdot FB \cdot FC = \frac{l}{4} \cdot F{O^2}$ Example - 8 Normals are drawn form the point P with slopes $${m_1},{\rm{ }}{m_2},{m_3}$$ to the parabola $${y^2} = 4x.$$ If the locus of P with $${m_1}{m_2} = \alpha$$ is a part of the parabola itself,then find $$\alpha$$ . Solution: If we let P be the point (h, k), we have ${m^3} + (2 - h)m + k = 0 \qquad \quad \dots \left( 1 \right)$ Thus, \begin{align}&\qquad \;{m_1}{m_2}{m_3}= - k\\\\&\Rightarrow \quad {m_3} = - \frac{k}{\alpha }\left( {{\rm{since}}\;{m_1}{m_2} = \alpha } \right)\end{align} Substituting this back into (1), we obtain \begin{align}& \qquad \qquad - \frac{{{k^3}}}{{{\alpha ^3}}} - \frac{k}{\alpha }(2 - h) + k = 0\\\\&\Rightarrow \qquad \;\;{k^2} = {\alpha ^2}h - 2{\alpha ^2} + {\alpha ^3} \qquad \qquad \dots \left( 2\right)\end{align} Also, since P lies on the parabola itself, we have ${k^2} =4h...\left( 3 \right)$ From(2) and (3), we have $\begin{array}{l}{\alpha^2} = 4{\rm{ }}\;and\;{\rm{ }}{\alpha ^3} - 2{\alpha ^2} = 0\\\\\Rightarrow \alpha = 2\end{array}$ Learn from the best math teachers and top your exams • Live one on one classroom and doubt clearing • Practice worksheets in and after class for conceptual clarity • Personalized curriculum to keep up with school<|endoftext|>
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NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics # NCERT Solutions for Exercise 15.1 Class 11 Maths Chapter 15 - Statistics Edited By Sumit Saini | Updated on Jul 18, 2022 12:03 PM IST NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 mainly deals with the mean deviation topics and other exercises include important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range, etc. In Class 9 and 10 Maths NCERT syllabus, students have already read the basic Statistics like mean, mode, and median. In this NCERT book Class 11 chapter applications of Statistics with advanced concepts are discussed. JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy Exercise 15.1 Class 11 Maths have easy questions. Solving NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 is a must to solve upcoming exercises. Direct questions are asked from this exercise as well in Boards. Also students can refer to the following exercise of NCERT for further information about upcoming exercises. ## Statistics Class 11 Chapter 15-Exercise: 15.1 Mean ( $\overline{x}$ ) of the given data: $\overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10$ The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are 6, 3, 2, 1, 0, 2, 3, 7 $\therefore$ $\sum_{i=1}^{8}|x_i - 10| = 24$ $\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|$ $= \frac{24}{8} = 3$ Hence, the mean deviation about the mean is 3. Mean ( $\overline{x}$ ) of the given data: $\\ \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{38+70+48+40+42+55+63+46+54+44}{10} \\ = \frac{500}{10} = 50$ The respective absolute values of the deviations from mean, $|x_i - \overline{x}|$ are 12, 20, 2, 10, 8, 5, 13, 4, 4, 6 $\therefore$ $\sum_{i=1}^{8}|x_i - 50| = 84$ $\therefore$ $M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|$ $= \frac{84}{10} = 8.4$ Hence, the mean deviation about the mean is 8.4. Question:3. Find the mean deviation about the median. Number of observations, n = 12, which is even. Arranging the values in ascending order: 10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18. Now, Median (M) $\\ = \frac{(\frac{12}{2})^{th} observation + (\frac{12}{2}+ 1)^{th} observation}{2} \\ = \frac{13 + 14}{2} = \frac{27}{2}= 13.5$ The respective absolute values of the deviations from median, $|x_i - M|$ are 3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5 $\therefore$ $\sum_{i=1}^{8}|x_i - 13.5| = 28$ $\therefore$ $M.D.(M) = \frac{1}{12}\sum_{i=1}^{n}|x_i - M|$ $= \frac{28}{12} = 2.33$ Hence, the mean deviation about the median is 2.33. Question:4. Find the mean deviation about the median. Number of observations, n = 10, which is even. Arranging the values in ascending order: 36, 42, 45, 46, 46, 49, 51, 53, 60, 72 Now, Median (M) $\\ = \frac{(\frac{10}{2})^{th} observation + (\frac{10}{2}+ 1)^{th} observation}{2} \\ = \frac{46 + 49}{2} = \frac{95}{2}= 47.5$ The respective absolute values of the deviations from median, $|x_i - M|$ are 11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5 $\therefore$ $\sum_{i=1}^{8}|x_i - 47.5| = 70$ $\therefore$ $M.D.(M) = \frac{1}{10}\sum_{i=1}^{n}|x_i - M|$ $= \frac{70}{10} = 7$ Hence, the mean deviation about the median is 7. Question:5 Find the mean deviation about the mean. $x_i$ $f_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 5 7 35 9 63 10 4 40 4 16 15 6 90 1 6 20 3 60 6 18 25 5 125 11 55 $\sum{f_i}$ = 25 $\sum f_ix_i$ = 350 $\sum f_i|x_i - \overline{x}|$ =158 $N = \sum_{i=1}^{5}{f_i} = 25 ; \sum_{i=1}^{5}{f_ix_i} = 350$ $\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{350}{12} = 14$ Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and $\sum f_i|x_i - \overline{x}|$ = 158 $\therefore$ $M.D.(\overline{x}) = \frac{1}{25}\sum_{i=1}^{n}f_i|x_i - \overline{x}|$ $= \frac{158}{25} = 6.32$ Hence, the mean deviation about the mean is 6.32 Question:6. Find the mean deviation about the mean. $x_i$ $f_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 10 4 40 40 160 30 24 720 20 480 50 28 1400 0 0 70 16 1120 20 320 90 8 720 40 320 $\sum{f_i}$ = 80 $\sum f_ix_i$ = 4000 $\sum f_i|x_i - \overline{x}|$ =1280 $N = \sum_{i=1}^{5}{f_i} = 80 ; \sum_{i=1}^{5}{f_ix_i} = 4000$ $\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{4000}{80} = 50$ Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and $\sum f_i|x_i - \overline{x}|$ = 1280 $\therefore$ $M.D.(\overline{x}) = \frac{1}{80}\sum_{i=1}^{5}f_i|x_i - \overline{x}|$ $= \frac{1280}{80} = 16$ Hence, the mean deviation about the mean is 16 Question:7. Find the mean deviation about the median. $x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 5 8 8 2 16 7 6 14 0 0 9 2 16 2 4 10 2 18 3 6 12 2 20 5 10 15 6 26 8 48 Now, N = 26 which is even. Median is the mean of $\dpi{100} 13^{th}$ and $\dpi{100} 14^{th}$ observations. Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7. Therefore, Median, M $\dpi{100} = \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7$ Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and $\sum f_i|x_i - M|$ = 84 $\therefore$ $M.D.(M) = \frac{1}{26}\sum_{i=1}^{6}|x_i - M|$ $= \frac{84}{26} = 3.23$ Hence, the mean deviation about the median is 3.23 $x_i$ $f_i$ $c.f.$ $|x_i - M|$ $f_i|x_i - M|$ 15 3 3 13.5 40.5 21 5 8 7.5 37.5 27 6 14 1.5 9 30 7 21 1.5 10.5 35 8 29 6.5 52 Now, N = 30, which is even. Median is the mean of $15^{th}$ and $\dpi{100} 16^{th}$ observations. Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30. Therefore, Median, M $\dpi{100} = \frac{15^{th} observation + 16^{th} observation}{2} = \frac{30 + 30}{2} = 30$ Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and $\sum f_i|x_i - M|$ = 149.5 $\therefore$ $M.D.(M) = \frac{1}{29}\sum_{i=1}^{5}|x_i - M|$ $= \frac{149.5}{29} = 5.1$ Hence, the mean deviation about the median is 5.1 Question:9. Find the mean deviation about the mean. Income per day in Rs Number of persons Income per day Number of Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 0 -100 4 50 200 308 1232 100 -200 8 150 1200 208 1664 200-300 9 250 2250 108 972 300-400 10 350 3500 8 80 400-500 7 450 3150 92 644 500-600 5 550 2750 192 960 600-700 4 650 2600 292 1168 700-800 3 750 2250 392 1176 $\sum{f_i}$ =50 $\sum f_ix_i$ =17900 $\sum f_i|x_i - \overline{x}|$ =7896 $N = \sum_{i=1}^{8}{f_i} = 50 ; \sum_{i=1}^{8}{f_ix_i} = 17900$ $\overline{x} = \frac{1}{N}\sum_{i=1}^{8}f_ix_i = \frac{17900}{50} = 358$ Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and $\sum f_i|x_i - \overline{x}|$ = 7896 $\therefore$ $M.D.(\overline{x}) = \frac{1}{50}\sum_{i=1}^{8}f_i|x_i - \overline{x}|$ $= \frac{7896}{50} = 157.92$ Hence, the mean deviation about the mean is 157.92 Question:10. Find the mean deviation about the mean. Height in cms Number of person Height in cms Number of Persons $f_i$ Mid Points $x_i$ $f_ix_i$ $|x_i - \overline{x}|$ $f_i|x_i - \overline{x}|$ 95 -105 9 100 900 25.3 227.7 105 -115 13 110 1430 15.3 198.9 115-125 26 120 3120 5.3 137.8 125-135 30 130 3900 4.7 141 135-145 12 140 1680 14.7 176.4 145-155 10 150 1500 24.7 247 $\sum{f_i}$ =100 $\sum f_ix_i$ =12530 $\sum f_i|x_i - \overline{x}|$ =1128.8 $N = \sum_{i=1}^{6}{f_i} = 100 ; \sum_{i=1}^{6}{f_ix_i} = 12530$ $\overline{x} = \frac{1}{N}\sum_{i=1}^{6}f_ix_i = \frac{12530}{100} = 125.3$ Now, we calculate the absolute values of the deviations from mean, $|x_i - \overline{x}|$ and $\sum f_i|x_i - \overline{x}|$ = 1128.8 $\therefore$ $M.D.(\overline{x}) = \frac{1}{100}\sum_{i=1}^{6}f_i|x_i - \overline{x}|$ $= \frac{1128.8}{100} = 11.29$ Hence, the mean deviation about the mean is 11.29 Marks Number of girls Marks Number of Girls $f_i$ Cumulative Frequency c.f. Mid Points $x_i$ $|x_i - M|$ $f_i|x_i - M|$ 0-10 6 6 5 22.85 137.1 10-20 8 14 15 12.85 102.8 20-30 14 28 25 2.85 39.9 30-40 16 44 35 7.15 114.4 40-50 4 48 45 17.15 68.6 50-60 2 50 55 27.15 54.3 $\sum f_i|x_i - M|$ =517.1 Now, N = 50, which is even. The class interval containing $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $\dpi{100} 25^{th}$ item is 20-30. Therefore, 20-30 is the median class. We know, Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$ Here, l = 20, C = 14, f = 14, h = 10 and N = 50 Therefore, Median $\dpi{100} = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85$ Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and $\sum f_i|x_i - M|$ = 517.1 $\therefore$ $M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|$ $= \frac{517.1}{50} = 10.34$ Hence, the mean deviation about the median is 10.34 Age (in years) Number [ Hint Convert the given data into continuous frequency distribution by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval] Age (in years) Number $f_i$ Cumulative Frequency c.f. Mid Points $x_i$ $|x_i - M|$ $f_i|x_i - M|$ 15.5-20.5 5 5 18 20 100 20.5-25.5 6 11 23 15 90 25.5-30.5 12 23 28 10 120 30.5-35.5 14 37 33 5 70 35.5-40.5 26 63 38 0 0 40.5-45.5 12 75 43 5 60 45.5-50.5 16 91 48 10 160 50.5-55.5 9 100 53 15 135 $\sum f_i|x_i - M|$ =735 Now, N = 100, which is even. The class interval containing $\dpi{80} \left (\frac{N}{2} \right)^{th}$ or $50^{th}$ item is 35.5-40.5. Therefore, 35.5-40.5 is the median class. We know, Median $\dpi{100} = l + \frac{\frac{N}{2}- C}{f}\times h$ Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100 Therefore, Median $\dpi{100} = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38$ Now, we calculate the absolute values of the deviations from median, $|x_i - M|$ and $\sum f_i|x_i - M|$ = 735 $\therefore$ $M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|$ $= \frac{735}{100} = 7.35$ Hence, the mean deviation about the median is 7.35 ## More About NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1 The NCERT Class 11 Maths chapter Statistics is very scoring chapters and questions are asked on the expected concepts only. It does not have much application in other chapters hence if one is not good at other chapters can strengthen this chapter to score well in the exam. Exercise 15.1 Class 11 Maths Mainly discusses the mean deviation about mean and median. Hence NCERT solutions for Class 11 Maths chapter 15 exercise 15.1 can be seen as a scoring exercise from the exam perspective. ## Benefits of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.1 • The Class 11 Maths chapter 15 exercise is prepared by expert faculties with rich experience. • Exercise 15.1 Class 11 Maths can help one enhance their marks in the exam as one question of 5 marks is mostly asked from this. • Class 11 Maths chapter 15 exercise 15.1 solutions provided here are comprehensive in manner. Also see- JEE Main Highest Scoring Chapters & Topics Just Study 40% Syllabus and Score upto 100% ## Subject Wise NCERT Exemplar solutions Happy learning!!! 1. Which topics are covered in Exercise 15.1 Class 11 Maths? Exercise 15.1 Class 11 Maths includes mean deviation about mean and median. 2. Is this chapter 15 related to other chapters ? No, mostly concepts are new and can be understood if other chapters are not read. 3. What is the difficulty level of the questions asked in this chapter ? Questions are easier to moderate level of difficulty if concepts are memorized. 4. Is it necessary to remember the formulas ? Yes, some basic formulas must be remembered to solve the questions. 5. How many questions are there in the Exercise 15.1 Class 11 Maths ? Total 12 questions are discussed in this exercise. ## Upcoming School Exams #### National Means Cum-Merit Scholarship Application Date:05 August,2024 - 06 September,2024 Exam Date:19 September,2024 - 19 September,2024 Exam Date:20 September,2024 - 20 September,2024 Exam Date:26 September,2024 - 26 September,2024 Application Date:30 September,2024 - 30 September,2024 Get answers from students and experts A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is Option 1) Option 2) Option 3) Option 4) A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 : Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range Option 1) Option 2) Option 3) Option 4) A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point Option 1) Option 2) Option 3) Option 4) In the reaction, Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts . How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms? Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2 If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance. With increase of temperature, which of these changes? Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction. Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023 A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9<|endoftext|>
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The amount of data in the world is vast and is increasing exponentially. It is easy to become overwhelmed and lose sight of the goal of data: to answer questions we have about the world in a specific, concise manner. The goal of this course is to help craft answerable questions—and then answer them. In order to do this, we will be using a programming language (“R”) to help us organize data, make clean, clear graphs, and help with appropriate analysis of the data. This course will serve two main goals. The first is an introductory statistics course: gain knowledge of the basic statistical tests, how to interpret their results in a reasonable manner, and understand what those tests are doing at a conceptual level. The second is to learn the computational language of R: how to sort, shape, and handle data, create simulations and interpret the results, and build clean, clear graphical representations of the data presented. This course is taught at the introductory level and has no prerequisites, but does require a significant amount of time and energy outside of the classroom as we are working towards the two aforementioned goals at once. This course is appropriate for students who plan to seriously create and analyze their own statistics for their work. It may be taken alone, or as a sequel to Presentation of Statistics. There is some overlap between the two courses, but their focus and goals are different. Students who take Presentation of Statistics first will get a broader skill set and a more gentle introduction.<|endoftext|>
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# Finding the inverse of a function • Jun 23rd 2011, 05:42 AM IBstudent Finding the inverse of a function What is the inverse of f(x) = x/x-1 ? this is my attempt to solve it: y = x / x-1 y(x-1) = x hence x(x-1) is the inverse..... however, substituting with x=5 gives...... 5/4=1.25 so pluging this is the inverse function should give 5, right? lets see: 1.25(1.25-1) = 1.25*.25= 0.3125. WRONG!! can anyone tell me where I went wrong? thanks • Jun 23rd 2011, 05:57 AM e^(i*pi) Re: Finding the inverse of a function Expand $y(x-1) = x$ This gives $yx - y = x$. Now you can get all your x terms on the LHS and factor • Jun 23rd 2011, 06:12 AM Prove It Re: Finding the inverse of a function As an alternative, write $\displaystyle \frac{x}{x - 1} = \frac{x - 1 + 1}{x - 1} = 1 + \frac{1}{x - 1}$ so that your original function is $\displaystyle y = 1 + \frac{1}{x - 1}$. Now evaluating the inverse function should be much easier. • Jun 23rd 2011, 06:53 AM IBstudent Re: Finding the inverse of a function Quote: Originally Posted by Prove It As an alternative, write $\displaystyle \frac{x}{x - 1} = \frac{x - 1 + 1}{x - 1} = 1 + \frac{1}{x - 1}$ so that your original function is $\displaystyle y = 1 + \frac{1}{x - 1}$. Now evaluating the inverse function should be much easier. how? why is (x+1-1)/x-1 = 1+ 1/x-1? • Jun 23rd 2011, 07:19 AM Prove It Re: Finding the inverse of a function Quote: Originally Posted by IBstudent how? why is (x+1-1)/x-1 = 1+ 1/x-1? My method is perfectly clear if you understand that $\displaystyle \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$, but anyway, $\displaystyle \frac{x - 1 + 1}{x - 1} = \frac{x - 1}{x - 1} + \frac{1}{x - 1} = 1 + \frac{1}{x - 1}$ • Jun 23rd 2011, 07:51 AM IBstudent Re: Finding the inverse of a function ahh I see Thanks :-) • Jun 23rd 2011, 08:38 AM topsquark Re: Finding the inverse of a function Quote: Originally Posted by IBstudent how? why is (x+1-1)/x-1 = 1+ 1/x-1? 1. Another way to find the inverse function is to switch the roles of x and y. For example: $y = \frac{x}{x - 1} \to x = \frac{y}{y - 1}$ Now solve for y and you get your inverse function. 2. Please be more careful in how you write you problem. For example: $y = x/x - 1$ is really $y = (x/x) - 1 = 1 - 1 = 0$ What you meant to write is y = x/(x - 1). Please use parentheses in the future. -Dan • Jun 23rd 2011, 09:09 PM IBstudent Re: Finding the inverse of a function okay so 1/(x-1) = y-1 (y-1)(x-1) = 1 yx-y-x+1 = 1 y(x-1)=x ahhhh I'm confused...:S is it x/(x-1) or x^2-x that is the inverse? • Jun 23rd 2011, 09:55 PM mr fantastic Re: Finding the inverse of a function Quote: Originally Posted by IBstudent okay so 1/(x-1) = y-1 (y-1)(x-1) = 1 yx-y-x+1 = 1 y(x-1)=x ahhhh I'm confused...:S is it x/(x-1) or x^2-x that is the inverse? It is basic algebra. 1/(x-1) = y-1 => x - 1 = 1/(y - 1) => x = ..... • Jun 23rd 2011, 10:53 PM IBstudent Re: Finding the inverse of a function I know that, so, the inverse of x/(x-1) is x/(x-1) ?! thats weird • Jun 24th 2011, 12:43 AM Ackbeet Re: Finding the inverse of a function Quote: Originally Posted by IBstudent I know that, so, the inverse of x/(x-1) is x/(x-1) ?! thats weird Any function that is symmetric with respect to a reflection about the line y = x will behave that way. Other examples: y = x and y = 1/x. I'm sure there are others, but they're not coming to mind. • Jun 24th 2011, 03:36 AM TKHunny Re: Finding the inverse of a function Quote: Originally Posted by IBstudent I know that, so, the inverse of x/(x-1) is x/(x-1) ?! thats weird Not quite and not really. If $f(x) = \frac{x}{x-1}$ Note, first that $x \ne 1$ Then, $f^{-1}(x) = \frac{x}{x-1}$, where again, $x \ne 1$ Notation is important. Don't just skip it. • Jun 24th 2011, 04:46 AM mr fantastic Re: Finding the inverse of a function Quote: Originally Posted by Ackbeet Any function that is symmetric with respect to a reflection about the line y = x will behave that way. Other examples: y = x and y = 1/x. I'm sure there are others, but they're not coming to mind. To expand on this, functions that are their own inverses are called involutes. It is simple to show that the following functions are involutes: f(x) = x. f(x) = -x + b, where b is a real number. $f(x) = \frac{ax + b}{cx - a}$ where a, b and c are real numbers. f(x) = 1/x is included as a special case. The question I'll pose for the interested reader is: Are the above functions the only involutes?<|endoftext|>
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Whiptail lizards live throughout the western United States, with numerous species concentrated in the Sonoran Desert and some, such as the western whiptail, ranging as far north as Idaho and Oregon. These lizards prefer dry climates and seek out habitats with sparse vegetation, such as desert grass, pine, sagebrush, scrub and oak. Depending on the species, they reproduce asexually, sexually or even both. Most whiptail lizards' bodies measure from 3 to 5 inches, not counting the tail. Depending on the species, they can be brown, tan, red, black, olive or gray, and their backs usually feature lighter spots or stripes. Their diet includes insects such as beetles, crickets, grasshoppers and termites, although they also sometimes eat spiders and scorpions. Whiptail lizards are most active on summer mornings and late afternoons, and their reproductive cycle usually takes place between April and August. Several species, such as the Sonoran spotted and desert grassland whiptails, consist of females only. They reproduce through an asexual process called parthenogenesis, which begins with a simulated mating ritual in which the females bite and mount each other. Scientists believe this behavior causes them to produce and lay unfertilized eggs. The young that hatch are genetic clones of their mothers. Females of a few parthenogenetic species, such as checkered and New Mexico whiptails, occasionally breed with males of other whiptail species. Most whiptails are not parthenogenetic; instead, reproduction occurs after males and females mate. As in the female-only courting ritual, the males mount and bite their partners. They use their hemipenes—a pair of reproductive organs at the tail’s base—to fertilize the eggs. After mating, females can hold on to the sperm for future use. Whether they reproduce sexually or asexually, female whiptail lizards bury their egg clutches. In the warm desert regions of southern California, Arizona and New Mexico, they can lay two or three clutches during a summer. In the northern parts of their range, such as Colorado and Idaho, they lay only one. Clutches average about three eggs, although depending on the species, the female could lay four, five, six or more at a time. Juveniles usually hatch during July or August, after two to two and a half months of incubation. The lizards become sexually mature when they’re 1 or 2 years old. - Arizona-Sonora Desert Museum: Whiptails (Cnemidophorus spp.) - University of Michigan Museum of Zoology: Animal Diversity Web: Cnemidophorus Sonorae - U.S. Fish and Wildlife Service: Bosque del Apache NWR: Lizards of Bosque del Apache - Idaho State University: Cnemidophorus tigris - U.S. Department of the Interior: Bureau of Land Management: California: Western Whiptail - Exotic Pet Vet: Reptiles: Reproduction “From Egg to Adult”<|endoftext|>
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Each of the questions or incomplete statements below is followed by five suggested answers or completions. Select the one that is best in each case. All of the following statements about carbon dioxide are true EXCEPT: Select an Answer It can be prepared by the action of acid on limestone. It is used to extinguish fires. It dissolves in water at room temperature. It sublimes rather than melts at and atmosphere pressure. It is less dense than air at a given temperature and pressure. Air contains mostly and . The molar mass of carbon dioxide is , so carbon dioxide is more dense than air. The other statements are true and are not correct choices.<|endoftext|>
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HOME Organize UNIT #1 UNIT #2 UNIT #3 UNIT #4 UNIT #5 UNIT #6 Resources Interaction Teacher Notes CONCEPT 1Prove the Laws of Sines. As mentioned in the previous objective up until this time we have been working with right triangles but now we are starting to expand the use of trigonometry to all triangles, even oblique triangles.  In this objective we will show that if any three of the six measures of triangle are given (provided at least one measure is a side), then the other three measures can be found. One of the considerations when doing this, is if the three pieces of information force congruence or not.  So for example if we are given SAS, ASA, AAS/SSA, SSS and some cases of AS1S2 we know that the information is enough to guarantee congruence in the triangle. The first new relationship in this objective that we are going to derive is called the Law of Sines.  We will follow similar logic to how we handled our new area formula. The Law of Sines is written in more than one form... sometimes the side lengths are in the numerator position and the sine vallues is in the denominator. It seems to be simply a preference because obviously it makes no difference. The Law of Sines is a proportion that compares the sine of the angle to the length of its opposite side.  Solving this type of problem can be quite easy but it does only work in certain situations because an angle and its corresponding opposite side need to be known to form the ratio thus we need ∠A & a or ∠B & b or ∠C or c. The Law of Sines will work for the following situations:  ASA, AAS and AS1S2.  The first two cases are congruence cases so when we solve for the missing information we are guaranteed to all get the same answers. Let’s try a couple examples of the congruence cases of ASA and AAS.  Solve for all angles & sides in the Δ. Notice that because the Law of Sines is a proportional relationship that: Largest Angle is always opposite the Longest Side (& vice versa) Smallest Angle is always opposite the Shortest Side (& vice versa) Good Luck... Here comes some tricky stuff... Back in GCO.8 I spent time discussing the AS1S2 so that when I get here it isn't totally a new concept. The other case that works for the Law of Sines is AS1S2 but if you remember in G.CO.8 we discussed some of the issues that are present in that case – not all cases of AS1S2 are congruence relationships.  Because of this we have to again discuss the case by case situations of AS1S2 and relate them to the Law of Sines. CASE #1 – AS1S2, when S2 is greater than S1. This forms a triangle congruence relationship.  There is only one way for S2 to be placed to complete the triangle.  Because this is a congruence relationship a single answer will be present when solving – let us do an example. Solve for all angles & sides in the Δ, given that m∠A = 47°, a = 20 cm and c = 12 cm. CASE #2 – AS1S2, when S2 is less than S1. (Too Short) This is a weird case but when establishing all things that could happen we need to include this.  It is possible that S2 is not long enough to close the triangle.  This of course is not a congruence relationship, it doesn’t even form a triangle.  So how will you know if it is this situation… the numbers will tell you . Solve for all angles & sides in the Δ, given that m∠A = 73°, a = 8 cm and c = 14 cm. CASE #3 – AS1S2, when S2 is less than S1. (1 Intersection) This case is typically known as HL, which stands for Hypotenuse – Leg.  It gets this special name because it is the right triangle that locks this shape.  A way to understand why this forms a congruence relationship is because in a right triangle if you know two sides, you can use the Pythagorean Theorem to calculate the third side.  HL forms a triangle congruence relationship.  No example given here because when you solve for the angle it will reveal itself to be 90° and you will then be in a right triangle. Now the final case – Known as the AMBIGUOUS CASE.  Ambiguous basically means…to be uncertain or more than one correct value – it gets this name because there are two possible triangles and thus there are two sets of answers possible.  If you encounter this special case, you are to completely solve for BOTH triangles. CASE #4 – AS1S2, when S2 is less than S1. (2 Intersections) Well here it is.... the Ambiguous case... the art of success here is helping students to recognize when it exists. Most students become very skilled with the Law of Sines but recognizing this case is the difficult part. If they know the other cases of AS1S2, when S2 is less than S1 definately help to narrow down the possible things that can happen. This is known as the AMBIGUOUS CASE because two different triangles can be formed by this information.  Because S2 is shorter it can swing to form two possible locations.  This does NOT form a triangle congruence relationship.<|endoftext|>
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Skip to main content # Modeling, Functions, and Graphs ## Section3.6Chapter Summary and Review ### SubsectionKey Concepts 1. #### Direct and Inverse Variation. • $$y$$ varies directly with $$x$$ if the ratio $$\dfrac{y}{x}$$ is constant, that is, if $$y = kx.$$ • $$y$$ varies directly with a power of $$x$$ if the ratio $$\dfrac{y}{x^n}$$ is constant, that is, if $$y = kx^n$$ . • $$y$$ varies inversely with $$x$$ if the product $$xy$$ is constant, that is, if $$y =\dfrac{k}{x}$$ . • $$y$$ varies inversely with a power of $$x$$ if the product $$x^ny$$ is constant, that is, if $$y =\dfrac{k}{x^n}$$ . 2. The graph of a direct variation passes through the origin. The graph of an inverse variation has a vertical asymptote at the origin. 3. If $$y = kx^n\text{,}$$ we say that $$y$$ scales as $$x^n\text{.}$$ 4. $$n$$th roots: $$s$$ is called an $$n$$th root of $$b$$ if $$s^n = b\text{.}$$ 5. #### Exponential Notation. The absolute value has the following properties: \begin{alignat*}{2} \amp a^{-n}=\frac{1}{x^n} \amp\hphantom{00000} \amp a\ne 0\\ \amp a^{0}=1 \amp \amp a\ne 0\\ \amp a^{1/n}=\sqrt[n]{a} \amp \amp n \text{ an integer, } n\gt 2\\ \amp a^{m/n}=(a^{1/n})^m=(a^m)^{1/n}, \amp \amp a\gt 0, ~n\ne 0 \end{alignat*} 6. In particular, a negative exponent denotes a reciprocal, and a fractional exponent denotes a root. 7. $$\displaystyle a^{m/n} = \sqrt[n]{a^m} =\left(\sqrt[n]{a}\right)^m$$ 8. To compute $$a^{m/n}\text{,}$$ we can compute the $$n$$th root first, or the $$m$$th power, whichever is easier. 9. We cannot write down an exact decimal equivalent for an irrational number, but we can approximate an irrational number to as many decimal places as we like. 10. The laws of exponents are valid for all exponents $$m$$ and $$n\text{,}$$ and for $$b\ne 0\text{.}$$ #### Laws of Exponents. 1. $$\displaystyle a^m\cdot a^n=a^{m+n}$$ 2. $$\displaystyle \dfrac{a^m}{a^n}=a^{m-n}$$ 3. $$\displaystyle \left(a^m\right)^n=a^{mn}$$ 4. $$\displaystyle \left(ab\right)^n=a^n b^n$$ 5. $$\displaystyle \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$$ 11. A function of the form $$f (x) = kx^p\text{,}$$ where $$k$$ and $$p$$ are constants, is called a power function. 12. An allometric equation is a power function of the form $$\text{variable} = k(\text{mass})^p\text{.}$$ 13. We can solve the equation $$x^n = b$$ by raising both sides to the $$\dfrac{1}{n}$$ power 14. We can solve the equation $$x^{1/n} = b$$ by raising both sides to the $$n$$th power. 15. To solve the equation $$x^{m/n} = k\text{,}$$ we raise both sides to the power $$n/m\text{.}$$ 16. The graphs of power functions $$y = x^{m/n}\text{,}$$ where $$m/n$$ is positive are all increasing for $$x\ge 0\text{.}$$ If $$m/n\gt 1\text{,}$$ the graph is concave up. If $$0\lt m/n \lt 1\text{,}$$ the graph is concave down. 17. The notation $$z = f (x, y)$$ indicates that $$z$$ is a function of two variables, $$x$$ and $$y\text{.}$$ 18. We can use a table with rows and columns to display the output values for a function of two variables. 19. #### Joint Variation. • We say that $$z$$ varies jointly with $$x$$ and $$y$$ if \begin{equation*} z = kxy,~~~ k\ne 0 \end{equation*} • We say that $$z$$ varies directly with $$x$$ and inversely with $$y$$ if \begin{equation*} z = k\dfrac{x}{y},~~~ k\ne 0, ~~ y\ne 0 \end{equation*} 20. We can represent a function of two variables graphically by showing a set of graphs for several fixed values of one of the variables. 21. #### Roots of Real Numbers. • Every positive number has two real-valued roots, one positive and one negative, if the index is even. • A negative number has no real-valued root if the index is even. • Every real number, positive, negative, or zero, has exactly one real-valued root if the index is odd. . ### ExercisesChapter 3 Review Problems #### 1. The distance s a pebble falls through a thick liquid varies directly with the square of the length of time $$t$$ it falls. 1. If the pebble falls $$28$$ centimeters in $$4$$ seconds, express the distance it will fall as a function of time. 2. Find the distance the pebble will fall in $$6$$ seconds. #### 2. The volume, $$V\text{,}$$ of a gas varies directly with the temperature, $$T\text{,}$$ and inversely with the pressure, $$P\text{,}$$ of the gas. 1. If $$V = 40$$ when $$T = 300$$ and $$P = 30\text{,}$$ express the volume of the gas as a function of the temperature and pressure of the gas. 2. Find the volume when $$T = 320$$ and $$P = 40\text{.}$$ #### 3. The demand for bottled water is inversely proportional to the price per bottle. If Droplets can sell $$600$$ bottles at $$$8$$ each, how many bottles can the company sell at$$$10$$ each? #### 4. The intensity of illumination from a light source varies inversely with the square of the distance from the source. If a reading lamp has an intensity of $$100$$ lumens at a distance of $$3$$ feet, what is its intensity $$8$$ feet away? #### 5. A person’s weight, $$w\text{,}$$ varies inversely with the square of his or her distance, $$r\text{,}$$ from the center of the Earth. 1. Express $$w$$ as a function of $$r\text{.}$$ Let $$k$$ stand for the constant of variation. 2. Make a rough graph of your function. 3. How far from the center of the Earth must Neil be in order to weigh one-third of his weight on the surface? The radius of the Earth is about $$3960$$ miles. #### 6. The period, $$T\text{,}$$ of a pendulum varies directly with the square root of its length, $$L\text{.}$$ 1. Express $$T$$ as a function of $$L\text{.}$$ Let $$k$$ stand for the constant of variation. 2. Make a rough graph of your function. 3. If a certain pendulum is replaced by a new one four-fifths as long as the old one, what happens to the period? #### Exercise Group. In Problems 7–10, $$y$$ varies directly or inversely with a power of $$x\text{.}$$ Find the power of $$x$$ and the constant of variation, $$k\text{.}$$ Write a formula for each function of the form $$y = kx^n$$ or $$y = \dfrac{k}{x^n}\text{.}$$ ##### 7. $$x$$ $$y$$ $$2$$ $$4.8$$ $$5$$ $$30.0$$ $$8$$ $$76.8$$ $$11$$ $$145.2$$ ##### 8. $$x$$ $$y$$ $$1.4$$ $$75.6$$ $$2.3$$ $$124.2$$ $$5.9$$ $$318.6$$ $$8.3$$ $$448.2$$ ##### 9. $$x$$ $$y$$ $$0.5$$ $$40.0$$ $$2.0$$ $$10.0$$ $$4.0$$ $$5.0$$ $$8.0$$ $$2.5$$ ##### 10. $$x$$ $$y$$ $$1.5$$ $$320.0$$ $$2.5$$ $$115.2$$ $$4.0$$ $$45.0$$ $$6.0$$ $$20.0$$ #### Exercise Group. For Problems 11–16, write without negative exponents and simplify. ##### 11. 1. $$\displaystyle (-3)^{-4}$$ 2. $$\displaystyle 4^{-3}$$ ##### 12. 1. $$\displaystyle \left(\dfrac{1}{3}\right)^{-2}$$ 2. $$\displaystyle \dfrac{3}{5^{-2}}$$ ##### 13. 1. $$\displaystyle (3m)^{-5}$$ 2. $$\displaystyle -7y^{-8}$$ ##### 14. 1. $$\displaystyle a^{-1}+ a^{-2}$$ 2. $$\displaystyle \dfrac{3q^{-9}}{r^{-2}}$$ ##### 15. 1. $$\displaystyle 6c^{-7}\cdot (3)^{-1} c^4$$ 2. $$\displaystyle \dfrac{11z^{-7}}{3^{-2} z^{-5}}$$ ##### 16. 1. $$\displaystyle \left(2d^{-2}k^3 \right)^{-4}$$ 2. $$\displaystyle \dfrac{2w^3(w^{-2})^{-3}}{5w^{-5}}$$ #### Exercise Group. For Problems 17–20, write each power in radical form. ##### 17. 1. $$\displaystyle 25m^{1/2}$$ 2. $$\displaystyle 8n^{-1/3}$$ ##### 18. 1. $$\displaystyle (13d)^{2/3}$$ 2. $$\displaystyle 6x^{2/5}y^{3/5}$$ ##### 19. 1. $$\displaystyle (3q)^{-3/4}$$ 2. $$\displaystyle 7(uv)^{3/2}$$ ##### 20. 1. $$\displaystyle (a^2+b^2)^{0.5}$$ 2. $$\displaystyle (16-x^2)^{0.25}$$ #### Exercise Group. For Problems 21–24, write each radical as a power with a fractional exponent. ##### 21. 1. $$\displaystyle 2\sqrt[3]{x^2}$$ 2. $$\displaystyle \dfrac{1}{4}\sqrt[4]{x}$$ ##### 22. 1. $$\displaystyle z^2\sqrt{z}$$ 2. $$\displaystyle z\sqrt[3]{z}$$ ##### 23. 1. $$\displaystyle \dfrac{6}{\sqrt[4]{b^3}}$$ 2. $$\displaystyle \dfrac{-1}{3\sqrt[3]{b}}$$ ##### 24. 1. $$\displaystyle \dfrac{-4}{(\sqrt[4]{a})^2}$$ 2. $$\displaystyle \dfrac{2}{(\sqrt{a})^3}$$ #### Exercise Group. For Problems 25–28, sketch graphs by hand for each function on the domain $$(0,\infty)\text{.}$$ ##### 25. $$y$$ varies directly with $$x^2\text{.}$$ The constant of variation is $$k = 0.25\text{.}$$ ##### 26. $$y$$ varies directly with $$x\text{.}$$ The constant of variation is $$k = 1.5\text{.}$$ ##### 27. $$y$$ varies inversely with $$x\text{.}$$ The constant of variation is $$k = 2\text{.}$$ ##### 28. $$y$$ varies inversely with $$x^2\text{.}$$ The constant of variation is $$k = 4\text{.}$$ #### Exercise Group. For Problems 29–30, write each function in the form $$y = kx^p\text{.}$$ ##### 29. $$f(x)=\dfrac{2}{3x^4}$$ ##### 30. $$g(x)=\dfrac{8x^7}{29}$$ #### Exercise Group. For Problems 31–34, 1. Evaluate each function for the given values. 2. Graph the function. ##### 31. $$Q(x)=4x^{5/2}$$ $$x$$ $$16$$ $$\dfrac{1}{4}$$ $$3$$ $$100$$ $$Q(x)$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ ##### 32. $$T(w)=-3w^{2/3}$$ $$w$$ $$27$$ $$\dfrac{1}{8}$$ $$20$$ $$1000$$ $$T(w)$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ ##### 33. $$f(x)=x^{0.3}$$ $$x$$ $$0$$ $$1$$ $$5$$ $$10$$ $$20$$ $$50$$ $$70$$ $$100$$ $$f(x)$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ ##### 34. $$g(x)=-x^{-0.7}$$ $$x$$ $$0.1$$ $$0.2$$ $$0.5$$ $$1$$ $$2$$ $$5$$ $$8$$ $$10$$ $$g(x)$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ $$\hphantom{000}$$ #### 35. According to the theory of relativity, the mass of an object traveling at velocity $$v$$ is given by the function \begin{equation*} m=\dfrac{M}{\sqrt{1-\dfrac{v^2}{c^2}}} \end{equation*} where $$M$$ is the mass of the object at rest and $$c$$ is the speed of light. Find the mass of a man traveling at a velocity of $$0.7c$$ if his rest mass is $$80$$ kilograms. #### 36. The cylinder of smallest surface area for a given volume has a radius and height both equal to $$\sqrt[3]{\dfrac{V}{\pi}} \text{.}$$ Find the dimensions of the tin can of smallest surface area with volume $$60$$ cubic inches. #### 37. Membership in the Wildlife Society has grown according to the function \begin{equation*} M(t) = 30t^{3/4} \end{equation*} where $$t$$ is the number of years since its founding in $$1970\text{.}$$ 1. Sketch a graph of the function $$M(t)\text{.}$$ 2. What was the society’s membership in $$1990\text{?}$$ 3. In what year will the membership be $$810$$ people? #### 38. The heron population in Saltmarsh Refuge is estimated by conservationists at \begin{equation*} P(t) = 360t^{-2/3} \end{equation*} where $$t$$ is the number of years since the refuge was established in $$1990\text{.}$$ 1. Sketch a graph of the function $$P(t)\text{.}$$ 2. How many heron were there in $$1995\text{?}$$ 3. In what year will there be only $$40$$ heron left? #### 39. Manufacturers of ships (and other complex products) find that the average cost of producing a ship decreases as more of those ships are produced. This relationship is called the experience curve, given by the equation \begin{equation*} C = ax^{-b} \end{equation*} where $$C$$ is the average cost per ship in millions of dollars and $$x$$ is the number of ships produced. The value of the constant $$b$$ depends on the complexity of the ship. (Source: Storch, Hammon, and Bunch, 1988) 1. What is the significance of the constant of proportionality $$a\text{?}$$ 2. For one kind of ship, $$b = \dfrac{1}{8}\text{,}$$ and the cost of producing the first ship is $$$12$$ million. Write the equation for $$C$$ as a function of $$x$$ using radical notation. 3. Compute the cost per ship when $$2$$ ships have been built. By what percent does the cost per ship decrease? By what percent does the cost per ship decrease from building $$2$$ ships to building $$4$$ ships? 4. By what percent does the average cost decrease from building $$n$$ ships to building $$2n$$ ships? (In the shipbuilding industry, the average cost per ship usually decreases by $$5$$ to $$10\%$$ each time the number of ships doubles.) Hint. What is the value of $$C$$ if only one ship is built? #### 40. A population is in a period of supergrowth if its rate of growth, $$R\text{,}$$ at any time is proportional to $$P^k\text{,}$$ where $$P$$ is the population at that time and $$k$$ is a constant greater than $$1\text{.}$$ Suppose $$R$$ is given by \begin{equation*} R = 0.015 P^{1.2} \end{equation*} where $$P$$ is measured in thousands and $$R$$ is measured in thousands per year. 1. Find $$R$$ when $$P = 20\text{,}$$ when $$P = 40\text{,}$$ and when $$P = 60\text{.}$$ 2. What will the population be when its rate of growth is $$5000$$ per year? 3. Graph $$R$$ and use your graph to verify your answers to parts (a) and (b). #### Exercise Group. For Problems 41–50, solve ##### 41. $$6t^{-3} = \dfrac{3}{500}$$ ##### 42. $$3.5 - 2.4p^{-2} = -6.1$$ ##### 43. $$\sqrt[3]{x+1} = 2$$ ##### 44. $$x^{2/3}+2 = 6$$ ##### 45. $$(x-1)^{-3/2} = \dfrac{1}{8}$$ ##### 46. $$(2x+1)^{-1/2} =\dfrac{1}{3}$$ ##### 47. $$8\sqrt[4]{x+6} =24$$ ##### 48. $$9.8 =7\sqrt[3]{z-4}$$ ##### 49. $$\dfrac{2}{3} (2y+1)^{0.2} = 6$$ ##### 50. $$1.3w^{0.3}+4.7 =5.2$$ #### Exercise Group. For Problems 51–54, solve each formula for the indicated variable. ##### 51. $$t=\sqrt{\dfrac{2v}{g}}~~~ \text{,}$$ for $$g$$ ##### 52. $$q-1=2\sqrt{\dfrac{r^2-1}{3}}~~~ \text{,}$$ for $$r$$ ##### 53. $$R=\dfrac{1+\sqrt{p^2+1}}{2}~~~ \text{,}$$ for $$p$$ ##### 54. $$q=\sqrt[3]{\dfrac{1+r^2}{2}}~~~ \text{,}$$ for $$r$$ #### Exercise Group. For Problems 55–60, simplify by applying the laws of exponents. ##### 55. $$(7t)^3 (7t)^{-1}$$ ##### 56. $$\dfrac{36r^{-2}s}{9r^{-3}s^4}$$ ##### 57. $$\dfrac{(2k^{-1})^{-4}}{4k^{-3}}$$ ##### 58. $$(2w^{-3})(2w^{-3})^{5}(-5w^2)$$ ##### 59. $$\dfrac{8a^{-3/4}}{a^{-11/4}}$$ ##### 60. $$b^{2/3}(4b^{-2/3}-b^{1/3})$$ #### 61. When the Concorde landed at Heathrow Airport in London, the width, $$w\text{,}$$ of the sonic boom felt on the ground is given in kilometers by the following formula: \begin{equation*} w=4\left(\frac{Th}{m} \right)^{1/2} \end{equation*} where $$T$$ stands for the temperature on the ground in kelvins, $$h$$ is the altitude of the Concorde when it breaks the sound barrier, and $$m$$ is the drop in temperature for each gain in altitude of one kilometer. 1. Find the width of the sonic boom if the ground temperature was $$293$$ K, the altitude of the Concorde was $$15$$ kilometers, and the temperature drop was $$4$$ K per kilometer of altitude. 2. Graph $$w$$ as a function of $$h$$ if $$T = 293$$ and $$m = 4\text{.}$$ #### 62. The manager of an office supply store must decide how many of each item in stock she should order. The Wilson lot size formula gives the most cost-efficient quantity, $$Q\text{,}$$ as a function of the cost, $$C\text{,}$$ of placing an order, the number of items, $$N\text{,}$$ sold per week, and the weekly inventory cost, $$I\text{,}$$ per item (cost of storage, maintenance, and so on). \begin{equation*} Q=\left(\frac{2CN}{I} \right)^{1/2} \end{equation*} 1. How many reams of computer paper should she order if she sells on average $$80$$ reams per week, the weekly inventory cost for a ream is$$$0.20\text{,}$$ and the cost of ordering, including delivery charges, is $$$25\text{?}$$ 2. Graph $$Q$$ as a function of $$N$$ if $$C = 25$$ and $$I = 0.2\text{.}$$ #### 63. Two businesswomen start a small company to produce saddle bags for bicycles. The number of saddle bags, $$q\text{,}$$ they can produce depends on the amount of money, $$m\text{,}$$ they invest and the number of hours of labor, $$w\text{,}$$ they employ, according to the Cobb-Douglas formula \begin{equation*} q= 0.6m^{1/4}w^{3/4} \end{equation*} where $$m$$ is measured in thousands of dollars. 1. If the businesswomen invest$$$100,000$$ and employ $$1600$$ hours of labor in their first month of production, how many saddle bags can they expect to produce? 2. With the same initial investment, how many hours of labor would they need in order to produce $$200$$ saddle bags? #### 64. A child who weighs $$w$$ pounds and is $$h$$ inches tall has a surface area (in square inches) given approximately by \begin{equation*} S= 8.5 h^{0.35}w^{0.55} \end{equation*} 1. What is the surface area of a child who weighs $$60$$ pounds and is $$40$$ inches tall? 2. What is the weight of a child who is $$50$$ inches tall and whose surface area is $$397$$ square inches? #### 65. The cost, $$C\text{,}$$ of insulating the ceiling in a building depends on the thickness of the insulation and the area of the ceiling. The table shows values of $$C = f (t, A)\text{,}$$ where $$t$$ is the thickness of the insulation and $$A$$ is the area of the ceiling. Cost of Insulation (dollars) Area (sq m) Thickness(cm) $$100$$ $$200$$ $$300$$ $$400$$ $$500$$ $$600$$ $$4$$ $$72$$ $$144$$ $$216$$ $$288$$ $$300$$ $$432$$ $$5$$ $$90$$ $$180$$ $$270$$ $$360$$ $$450$$ $$540$$ $$6$$ $$108$$ $$216$$ $$324$$ $$432$$ $$540$$ $$648$$ $$7$$ $$126$$ $$252$$ $$378$$ $$504$$ $$630$$ $$756$$ $$8$$ $$144$$ $$288$$ $$432$$ $$576$$ $$720$$ $$864$$ $$9$$ $$162$$ $$324$$ $$486$$ $$648$$ $$810$$ $$972$$ 1. What does it cost to insulate a ceiling with an area of $$500$$ square meters with $$5$$ cm of insulation? Write your answer in function notation. 2. Solve the equation $$864 = f (t, 600)$$ and interpret your answer. 3. Consider the row corresponding to a thickness of $$4$$ cm. How does the cost of insulating the ceiling depend on the area of the ceiling? 4. Consider the column corresponding to an area of $$100$$ square meters. How does the cost depend on the thickness of the insulation? 5. Given that the cost varies jointly with the thickness of the insulation and the area of the ceiling, write an equation for cost as a function of area and thickness of insulation. 6. Use your formula from part (e) to determine the cost of insulating a building with $$10$$ centimeters of insulation if the area of the ceiling is $$800$$ square meters. #### 66. The volume, $$V\text{,}$$ of a quantity of helium depends on both the temperature and the pressure of the gas. The table shows values of $$V = f (P, T )$$ for temperature in kelvins and pressure in atmospheres. Volume (cubic meters) Temperature (K) Pressure(atmospheres) $$100$$ $$150$$ $$200$$ $$250$$ $$300$$ $$350$$ $$1$$ $$18$$ $$27$$ $$36$$ $$45$$ $$54$$ $$63$$ $$2$$ $$9$$ $$13.5$$ $$18$$ $$22.5$$ $$27$$ $$31.5$$ $$3$$ $$6$$ $$9$$ $$12$$ $$15$$ $$18$$ $$21$$ $$4$$ $$4.5$$ $$6.75$$ $$9$$ $$11.25$$ $$13.5$$ $$15.75$$ 1. What is the volume of helium when the pressure is $$4$$ atmospheres and the temperature is $$350$$ K? Write your answer in function notation. 2. Solve the equation $$15 = f (3, T )$$ and interpret your answer. 3. Consider the row corresponding to $$2$$ atmospheres. How is the volume related to the absolute temperature? 4. Consider the column corresponding to $$300$$ K. How is the volume related to the pressure? 5. Given that the volume of the gas varies directly with temperature and inversely with pressure, write an equation for volume as a function of temperature and pressure. 6. Use your formula from part (e) to determine the volume of the helium at $$50$$ K and pressure of $$0.4$$ atmospheres. #### 67. In his hiking guidebook, Afoot and Afield in Los Angeles County, Jerry Schad notes that the number of people on a wilderness trail is inversely proportional to "the square of the distance and the cube of the elevation gain from the nearest road." 1. Choose variables and write a formula for this relationship. 2. On a sunny Saturday afternoon, you count $$42$$ people enjoying the Rock Pool at Malibu Creek State Park. The Rock Pool is $$1.5$$ miles from the main parking lot, and the trail includes an elevation gain of $$250$$ feet. Calculate the constant of variation in your formula from part (a). Hint: Convert the elevation gain to miles. 3. Lookout Trail leads $$1.9$$ miles from the parking lot and involves an elevation gain of $$500$$ feet. How many people would you expect to encounter at the end of the trail? #### 68. A company’s monthly production, $$P\text{,}$$ depends on the capital, $$C\text{,}$$ the company has invested and the amount of labor, $$L\text{,}$$ available each month. The Cobb-Douglas model for production assumes that $$P$$ varies jointly with $$C^a$$ and $$L^b\text{,}$$ where $$a$$ and $$b$$ are positive constants less than $$1\text{.}$$ The Aztech Chip Company invested $$625$$ units of capital and hired $$256$$ workers and produces $$8000$$ computer chips each month. 1. Suppose that $$a = 0.25\text{,}$$ $$b = 0.75\text{.}$$ Find the constant of variation and a formula giving $$P$$ in terms of $$C$$ and $$L\text{.}$$ 2. If Aztech increases its labor force to $$300$$ workers, what production level can they expect? 3. If Aztech maintains its labor force at $$256$$ workers, what amount of capital outlay would be required for monthly production to reach $$16,000$$ computer chips?<|endoftext|>
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## Revision Notes of Ch 1 Real Numbers Class 10th Math Topics in the Chapter • Euclid’s Division Lemma • Euclid’s Division Algorithm • The Fundamental Theorem of Arithmetic • Revisiting Irrational Numbers • Revisiting Rational Numbers and Their Decimal Expansions Euclid’s Division Lemma → For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, 0  ≤ r < b Here,  a, b, q and r are called dividend, divisor, quotient and remainder respectively. → Example: Dividing 34 by 6, we get quotient as 6 and remainder as 4. So we can write it as, 34 = 6×5 + 4 Here, a = 34, b = 6, q = 5 and r = 4 a = bq + r, 0  ≤ r < b Euclid’s Division Algorithm → Euclid’s Division Algorithm is application of Euclid’s Division Lemma means it is technique to compute HCF of two natural numbers. → To obtain the HCF of two positive integers, say c and d, with c > d, we follow these steps: Step 1: Apply Euclid’s division lemma to c and d to obtain two whole numbers q and r. Thus, c = dq + r, 0  ≤ r < d Step 2 : If r = 0, d is the HCF of c and d. If r ≠ 0, again apply the division lemma to d and r. Step 3 : Continue the process till the remainder is zero. The last divisor when remainder becomes zero is the HCF of c and d → Example: Use Euclid’s division algorithm to find the HCF of 420 and 130. Here, 420 > 130 so we can write, Step 1: 420 = 130 × 3 + 30 (remainder is 30 so we will apply division lemma again on 130 and 30) Step 2: 130 = 30 × 4 + 10 (remainder is 10 so we will apply division lemma again on 30 and 10) Step 3: 30 = 10 × 3 + 0 Now, the remainder is 0 and the last divisor is 10. So, 10 is the HCF of 420 and 130. The Fundamental Theorem of Arithmetic Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. → Let x be composite number which can be factorised as, x =  p1p2 ..... pn , where  p1p2 ..... pn are primes and written in ascending order, i.e., p1 ≤ p2 ≤ ..... ≤ pn. If we combine the same primes, we will get powers of primes. This is called prime factorization. For example: 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13 By expressing any two numbers as their prime factors, their highest common factor (HCF) and lowest common multiple (LCM) can be calculated.<|endoftext|>
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Beth is beginning to teach her class the headings on the CAFE Menu: Comprehension, Accuracy, Fluency, and Expand Vocabulary. This lesson introducing the goals is used throughout the year to review when needed. Watch and Listen: - Beth Lawson teaches her students kinesthetic motions for the meaning of each heading. Is this something you will teach your students? - What value do you see in teaching students the goals and the meaning of each goal? - What adaptations, if any, would you make to teach this lesson to your students? - When might you review this with your class? CAFE Essential Elements: - Teach—6.1 Introduce classroom CAFE Menu to students - Teach—6.2 Classroom CAFE Menu as an instructional tool<|endoftext|>
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In this quick tutorial you'll learn how to draw a Stingray in 5 easy steps - great for kids and novice artists. The images above represents how your finished drawing is going to look and the steps involved. Below are the individual steps - you can click on each one for a High Resolution printable PDF version. At the bottom you can read some interesting facts about the Stingray. Make sure you also check out any of the hundreds of drawing tutorials grouped by category. How to Draw a Stingray - Step-by-Step Tutorial Step 1: Start by drawing the stingray's body. It looks like a pentagon with the bottom lines curled up. If you're having trouble, think of baseball's home plate upside down. Leave an opening between the curled lines at the bottom for the fins. Step 2: Now draw the fins at the bottom to connect the two curled lines. The fins look like a lowercase "w" that has been stretched apart. Step 3: Next, draw the tail beneath the fins. Stingrays have a very long and thin tail with a sharp point at the end. They are called stingrays because some species have venom in their tails. Step 4: Next, draw the eyes near the top of the body. You can also draw the spiracles above the eyes, which are shaped like crescent moons. Spiracles are openings, like gills, that help stingrays breathe. Step 5: Last, draw the spots and color your stingray. You can draw as many spots as you want. There are over 60 species of stingrays and some are covered in spots, while some have none. There is even a species of stingray with spots that look just like a leopard's! Activity: Have students color and draw spots on their stingrays. Let them share why they chose certain colors or spots (e.g. bright spots for a tropical environment or dark plain colors to blend in). Interesting Facts about Stingrays Stingrays are fish with wide flat bodies made of cartilage, which is the same material that make up your ears. Stingrays have no bones. There are more than 60 species of stingrays. They are related to sharks. Stingrays got their names because of their tail. Their tail has a spine with a very sharp point. Some species have venom in their tails. The spine and the venom are dangerous to humans, although stingrays rarely attack humans. Did you know? - Stingrays live at the bottom of the ocean, in warm climates. - Stingrays have two fins that run alongside their bodies. To swim, some species flap their fins the same way birds flap their wings to fly. Other species use their whole bodies to move in a wavy motion. - Stingrays usually eat oysters, shrimp, crabs, clams, and mussels. - Stingrays can grow up to 6 ? feet. - In the wild, stingrays live to be about 15 to 25 years old. - Stingrays bury themselves in the sand when they want to hide from predators. Stingrays use their tails as a form of defense. They whip their tails around when they attack. Some people use the tails of stingrays for the tips of weapons such as daggers.<|endoftext|>
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This column, which first ran in our August 21 issue, has been revised and updated by Dr. Crutchfield to run again due to the ongoing Ebola crisis. Ebola Fever is a viral infection. The virus causes the bleeding (hemorrhage) of internal organs and of the skin. It also causes the infected person to have a fever. It is also known as Ebola Hemorrhagic Fever. There is no vaccination against it. Ebola Hemorrhagic Fever can kill up to 90 percent of those infected. It is named after the Ebola River, an area where one of the first cases were reported in 1976. There are several sub-types of the virus. The Zaire sub-type/strain is considered the most deadly. As of this report, since 1976, there have been approximately 6,500 cases of documented Ebola with over 4,000 fatalities. Currently, countries in West Africa (including Guinea, Sierra Leone and Liberia) are experiencing one of the worst outbreaks of the Ebola virus ever documented with approximately 4,000 deaths at the time this article was written. One projection has an estimate of 50-100,000 fatalities before the epidemic is under control. The World Health Organization calls the Ebola virus “one of the world’s most virulent diseases.” Fruit bats are probably the natural host for the virus, and they can pass it on to other animals. The human infections are thought to have originated from contact with infected monkeys. Early symptoms include a sudden fever, general fatigue and weakness, muscle pains, headaches, red eyes, a skin rash with bruise-like blisters, chest pain, difficulty breathing, difficulty swallowing, vomiting, diarrhea and internal bleeding. It can affect internal organs like the kidney and liver and eventually cause death. One of the common and curious early symptoms is severe hiccups. The symptoms occur anywhere from two to 21 days after infection but most commonly appear eight to 12 days after exposure. Because the early symptoms are commonly seen in many illnesses, diagnosis can be challenging, and it can often be misdiagnosed as another disease such as malaria, typhoid fever, meningitis, the plague, or even cholera. Ebola can be definitively confirmed by specific laboratory tests. Although there are experimental medicines to treat Ebola (such as ZMapp, Brincidofovir and Tekmira), none are FDA-approved yet. The World Health Organization has announced that in severe outbreaks it is not unethical to consider using experimental medications. Currently, the treatment for patients is purely supportive: Maintain good hydration, support normal oxygen and blood pressure levels, treat symptoms, and prevent secondary infection. With early detection and supportive measures such as these, the mortality rate may be decreased to 40 percent or lower. Ebola is contracted by coming directly into contact with blood, body secretions, or other body fluids from an infected person, or from exposure to a contaminated needle of an infected person. The vast majority of patients with Ebola know exactly where they caught it, from taking care of another person with the disease or who had symptoms such as fever, diarrhea, vomiting, bleeding and hiccups. The good news is that the virus is sensitive to sanitizing measures such as chlorine and other medical disinfectants, heat treatment and detergents. Currently, there are some promising vaccines under research and development. The chances of catching it on an airplane are thought to be very low. It is not spread by sneezing or coughing. There needs to be prolonged, frequent, and direct contact with body fluids or blood infected with the virus in order to catch it. It is important to note: The infected person must be symptomatic in order to transmit the virus. Also, the corpses of persons who have died from Ebola infection can transmit the infection, so those caring for the body must exercise extreme caution and protective measures. Outbreaks in the past have been controlled, and once an area has gone “infection free” for 42 days (twice the incubation period), the infection is considered to be under control. With the ease of worldwide travel, the challenge for governments, international healthcare organizations, airlines (and other passenger transportation companies), industrial/commercial sanitation and cleaning organizations and medical professionals will be to keep the infection contained and from spreading globally. If you think you have been exposed to the virus, or to a person with the virus, please contact your physician, local hospital or 911 at once. Charles E. Crutchfield III, MD is a board certified dermatologist and Clinical Professor of Dermatology at the University of Minnesota Medical School. He also has a private practice in Eagan, MN. He has been selected as one of the top 10 dermatologists in the U.S. by Black Enterprise magazine and one of the top 21 African American physicians in the U.S. by the Atlanta Post. Dr. Crutchfield is an active member of the Minnesota Association of Black Physicians, MABP.org.<|endoftext|>
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# How do you solve x^2 + 32x + 15 = -x^2 + 16x + 11 by completing the square? Jun 22, 2015 Simplify the given equation then apply the process of "completing the squares" to obtain $\textcolor{w h i t e}{\text{XXXX}}$$x = - 4 \pm \sqrt{14}$ #### Explanation: Given ${x}^{2} + 32 x + 15 = - {x}^{2} + 16 x + 11$ First simplify to get all terms involving $x$ on the left side and a simple constant on the right: $\textcolor{w h i t e}{\text{XXXX}}$$2 {x}^{2} + 16 x = - 4$ Further simplify by dividing by 2 $\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + 8 x = - 2$ Now we are ready to begin completing the square. # Noting that the squared binomial ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$ If ${x}^{2} + 8 x$ are the first 2 terms of an expanded squared binomial, then $2 a x = 8 x \rightarrow a = 4 \mathmr{and} {a}^{2} = 16$ So the completed square must be (after remembering that anything added to one side of an equation must also be added to the other) $\textcolor{w h i t e}{\text{XXXX}}$${x}^{2} + 8 x + 16 = - 2 + 16$ or $\textcolor{w h i t e}{\text{XXXX}}$${\left(x + 4\right)}^{2} = 14$ Taking the square root of both sides: $\textcolor{w h i t e}{\text{XXXX}}$$x + 4 = \pm \sqrt{14}$ and $\textcolor{w h i t e}{\text{XXXX}}$$x = - 4 \pm \sqrt{14}$<|endoftext|>
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Pairwise comparisons of proportions Pairwise comparison means comparing all pairs of something. If I have three items A, B and C, that means comparing A to B, A to C, and B to C. Given n items, I can determine the number of possible pairs using the binomial coefficient: $$\frac{n!}{2!(n – 2)!} = \binom {n}{2}$$ Using the R statistical computing environment, we can use the choose function to quickly calculate this. For example, how many possible 2-item combinations can I “choose” from 10 items: choose(10,2) [1] 45 We sometimes want to make pairwise comparisons to see where differences occur. Let’s say we go to 8 high schools in an area, survey 30 students at each school, and ask them whether or not they floss their teeth at least once a day. When finished we’ll have 8 proportions of students who answered “Yes”. An obvious first step would be to conduct a hypothesis test for any differences between these proportions. The null would be no difference between the proportions versus some difference. If we reject the null, we have evidence of differences. But where are the differences? This leads us to pairwise comparisons of proportions, where we make multiple comparisons. The outcome of these pairwise comparisons will hopefully tell us which schools have significantly different proportions of students flossing. Making multiple comparisons leads to an increased chance of making a false discovery, i.e. rejecting a null hypothesis that should not have been rejected. When we run a hypothesis test, we always run a risk of finding something that isn’t there. Think of flipping a fair coin 10 times and getting 9 or 10 heads (or 0 or 1 heads). That’s improbable but not impossible. If it happened to us we may conclude the coin is unfair, but that would be the wrong conclusion if the coin truly was fair. It just so happened we were very unlucky to witness such an unusual event. As we said, the chance of this happening is low in a single trial, but we increase our chances of it happening by conducting multiple trials. The probability of observing 0, 1, 9 or 10 heads when flipping a fair coin 10 times is about 2% which can be calculated in R as follows: pbinom(q = 1, size = 10, prob = 0.5) * 2 [1] 0.02148438 Therefore the the probability of getting 2 – 8 heads is about 98%: 1 - pbinom(q = 1, size = 10, prob = 0.5) * 2 [1] 0.9785156 The probability of getting 2 – 8 heads in 10 trials is 98% multiplied by itself 10 times: (1 - pbinom(q = 1, size = 10, prob = 0.5) * 2)^10 [1] 0.8047809 Therefore the probability of getting 0, 1, 9, or 10 heads in 10 trials is now about 20%: 1 - (1 - pbinom(q = 1, size = 10, prob = 0.5) * 2)^10 [1] 0.1952191 We can think of this as doing multiple hypothesis tests. Flip 10 coins 10 times each, get the proportion of heads for each coin, and use 10 one-sample proportion tests to statistically determine if the results we got are consistent with a fair coin. In other words, do we get any p-values less than, say, 0.05? We can simulate this in R. First we replicate 1,000 times the act of flipping 10 fair coins 10 times each and counting the number of heads using the rbinom function. This produces a 10 x 1000 matrix of results that we save as “coin.flips”. We then apply a function to each column of the matrix that runs 10 one-sample proportion tests using the prop.test function and saves a TRUE/FALSE value if any of the p-values are less than 0.05 (we talk more about the prop.test function below). This returns a vector we save as “results” that contains TRUE or FALSE for each replicate. R treats TRUE and FALSE as 0 or 1, so calling mean on results returns the proportion of TRUEs in the vector. We get about 20%, confirming our calculations. (If you run the code below you’ll probably get a slightly different but similar answer.) trials <- 10 coin.flips <- replicate(1000, rbinom(n = 10, size = trials, prob = 0.5)) multHypTest <- function(x){ pvs <- sapply(x, function(x)prop.test(x = x, n = trials, p = 0.5)\$p.value) any(pvs < 0.05) } results <- apply(coin.flips,2,multHypTest) mean(results) [1] 0.206 That’s just for 10 trials. What about 15 or 20 or more? You can re-run the code above with trials set to a different value. We can also visualize it by plotting the probability of an unusual result (0, 1, 9, or 10 heads) versus the number trials. Notice how rapidly the probability of a false discovery increases with the number of trials. curve(expr = 1 - (1 - pbinom(q = 1, size = 10, prob = 0.5) * 2)^x, xlim = c(1,50), xlab = "Number of tests", ylab = "Probability of 0, 1, 9, or 10 heads") So what does all of this tell us? It reveals that traditional significance levels such as 0.05 are too high when conducting multiple hypothesis tests. We need to either adjust our significance level or adjust our p-values. As we’ll see, the usual approach is to adjust the p-values using one of several methods for p-value adjustment. Let’s return to our example of examining the proportion of high school students (sample size 30 at each school) who floss at 8 different high schools. We’ll simulate this data as if the true proportion is 30% at each school (i.e., no difference). We use set.seed to make the data reproducible. set.seed(15) n <- 30 k <- 8 school <- rep(1:k, each = n) floss <- replicate(k, sample(x = c("Y","N"), size = n, prob = c(0.3, 0.7), replace = TRUE)) dat <- data.frame(school, floss = as.vector(floss)) With our data generated, we can tabulate the number of Yes and No responses at each school: flossTab <- with(dat, table(school, floss)) flossTab floss school N Y 1 18 12 2 19 11 3 14 16 4 19 11 5 26 4 6 15 15 7 20 10 8 21 9 Using prop.table we can determine the proportions. Specifying margin = 1 means proportions are calculated across the rows for each school. (We also round to two decimal places for presentation purposes.) The second column contains the proportion of students who answered Yes at each school. round(prop.table(flossTab, margin = 1),2) floss school N Y 1 0.60 0.40 2 0.63 0.37 3 0.47 0.53 4 0.63 0.37 5 0.87 0.13 6 0.50 0.50 7 0.67 0.33 8 0.70 0.30 First we might want to run a test to see if we can statistically conclude that not all proportions are equal. We can do this with the prop.test function. The prop.test function requires that Yes (or “success”) counts be in the first column of a table and No (or “failure”) counts in the second column. Thus we switch the columns using subsetting brackets with a vector indicating column order. prop.test(flossTab[,c("Y","N")]) 8-sample test for equality of proportions without continuity correction data: flossTab[, c("Y", "N")] X-squared = 13.78, df = 7, p-value = 0.05524 alternative hypothesis: two.sided sample estimates: prop 1 prop 2 prop 3 prop 4 prop 5 prop 6 prop 7 prop 8 0.4000000 0.3666667 0.5333333 0.3666667 0.1333333 0.5000000 0.3333333 0.3000000 The p-value of 0.055 is borderline significant and indicates some evidence of differences among proportions. We generated the data so we know there actually is no difference! But if this were real data that we had spent considerable resources collecting, we might be led to believe (perhaps even want to believe) some differences indeed exist. That p-value is so close to significance! School #5, in particular, with a proportion of 13% looks far lower than school #3 with 53%. We could conclude this hypothesis test is significant at 0.10 level and proceed to pairwise comparisons. To do that in R we use the pairwise.prop.test function which requires a table in the same format as prop.test, Yes counts in the first column and No counts in the second column: pairwise.prop.test(x = flossTab[,c("Y","N")]) Pairwise comparisons using Pairwise comparison of proportions data: flossTab[, c("Y", "N")] 1 2 3 4 5 6 7 2 1.000 - - - - - - 3 1.000 1.000 - - - - - 4 1.000 1.000 1.000 - - - - 5 1.000 1.000 0.073 1.000 - - - 6 1.000 1.000 1.000 1.000 0.149 - - 7 1.000 1.000 1.000 1.000 1.000 1.000 - 8 1.000 1.000 1.000 1.000 1.000 1.000 1.000 This produces a table of 28 p-values since there are 28 possible pairs between 8 items. We interpret the table by using row and column numbers to find the p-value for a particular pair. For example the p-value of 0.073 at the intersection of row 5 and column 3 is the p-value for the two-sample proportion test between school #5 and school #3. It appears to be insignificant at the traditional 5% level. All other p-values are clearly insignificant. In fact, most are 1. This is due to the p-value adjustment that was made. The output tells us the “holm” method was used. We won’t get into the details of how this method works, but suffice to say it increases the p-values in an effort to adjust for the many comparisons being made. In this case, it does what it’s supposed to: it adjusts the p-values and allows us to make a good case there is no differences between schools, at least not at the 5% level, which would be the correct decision. We can do pairwise comparisons without adjusted p-values by setting p.adjust.method = "none". Let’s do that and see what happens: # NOTE: This analysis is wrong! pairwise.prop.test(x = flossTab[,c("Y","N")], p.adjust.method = "none") Pairwise comparisons using Pairwise comparison of proportions data: flossTab[, c("Y", "N")] 1 2 3 4 5 6 7 2 1.0000 - - - - - - 3 0.4376 0.2993 - - - - - 4 1.0000 1.0000 0.2993 - - - - 5 0.0410 0.0736 0.0026 0.0736 - - - 6 0.6038 0.4345 1.0000 0.4345 0.0055 - - 7 0.7888 1.0000 0.1927 1.0000 0.1270 0.2949 - 8 0.5883 0.7842 0.1161 0.7842 0.2100 0.1876 1.0000 Notice now we have significant differences for 3 pairs: (5,1), (5,3), and (6,5). Again we know this is wrong because we simulated the data. The truth is all schools have a floss rate of 30%. But we see that through random chance and not adjusting our p-values for multiple testing we got what look to be significant results. This illustrates the importance of using adjusted p-values when making multiple comparisons. There are other p-value adjustment methods available. A common and conservative choice is the bonferroni method. It simply multiplies all p-values by the number of pairs. In our example that is 28. To see all p-value adjustment methods available in R enter ?p.adjust at the console. For questions or clarifications regarding this article, contact the UVA Library StatLab: [email protected] View the entire collection of UVA Library StatLab articles. Clay Ford Statistical Research Consultant University of Virginia Library October 20, 2016<|endoftext|>
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Typography is the art or process of printing with type. However, in modern usage typography includes all manner of non-printed letter forms such as websites, eBooks, electronic billboards, and even textiles. Through the use of type, a person can visually tell a story using little to no imagery. This post is intended to teach you about the anatomy of type and to help you better understand what to look for when choosing your next font. First, the basics … Typeface vs Font, What’s the difference? A typeface, also called a font family, is a set of fonts designed with a stylistic unity, each comprising a coordinate set of glyphs. A font is a complete character set of a typeface at a particular size, weight, and style. Serif vs Sans Serif, Which one to use? Serif letters are drawn with features at the ends of their strokes. The serifs are the little feet we see in fonts like Times. These are some of the oldest type designs. The feet along the baseline help guide the eye from left to right, making them very ‘readable’ fonts. Sans Serif (french for “without serifs”) are letters drawn with straighter lines and no feet. Their larger letterforms make them very legible, but can cause greater eye strain when used in long runs of text. Helvetica is considered the quintessential sans serif font. The Anatomy of a Letter There is a standard set of terms to describe the parts of a character. These terms, and the parts of the letter they represent, are often referred to as “letter anatomy” or “typeface anatomy.” By breaking down letters into parts, a designer can better understand how type is created and altered and how to use it effectively. - Baseline – The baseline is the invisible line on which characters sit. While the baseline may differ from typeface to typeface, it is consistent within a typeface. Rounded letters such as “e” may extend slightly below the baseline. - Meanline – The meanline falls at the top of many lowercase letters such as “e,” “g” and “y.” It is also at the curve of letters like “h.” - X-Height – The x-height is the distance between the meanline and the baseline. It is referred to as the x-height because it is the height of a lowercase “x.” This height can vary greatly between typefaces. - Cap Height – The cap height is the distance from the baseline to the top of uppercase letters like “H” and “J.” - Ascender – The part of a character that extends above the meanline is known as an ascender. Note that this is the same as extending above the x-height. - Descender – The part of a character that extends below the baseline is known as a descender, such as the bottom stroke of a “y.” - Serif – Fonts are often divided into serif and sans serif. Serif fonts are distinguishable by the extra stroke at the ends of the character, known as a serif. - Stem – The vertical line of a “B” and the primary diagonal line of a “V” are known as the stem. The stem is often the main “body” of a letter. - Bar – The horizontal lines of an “E” are known as bars. Bars are horizontal or diagonal lines of a letter, also known as arms, and are open on at least one side. - Bowl – An open or closed circular line that creates an interior space, such as in “e” and “b.” - Counter – The inside of a bowl. - Leg – The bottom stroke of a letter, such as the base of an “L” or diagonal stroke of a “K.” - Shoulder – The curve at the beginning of a leg of a character, such as in an “m.” Examples of Great Type-Only Design Want to learn more? Visit Canva.com for an illustrated glossary of typographic terms.<|endoftext|>
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Development of the Heart The heart is the first functional organism in a vertebrate embryo and there are 5 different stages to heart development. The first stage of heart development involves the forming of the primitive heart tube. The primitive heart tube consisits of a single tube which is formed when cardiac precurser cells fuse together. Even at this early stage of development the heart consists of different regions and layers, these regions are: - bulbs ordis (which develops to form the aorta and ventricles) The developing layers of the heart are: - cardiac jelly - cardiac mantle layers (which will form the myometrium and epicardium of the heart) The heart first begins to beat during this stage of heart development, about 22-23 days after conception. This can be detected via vaginal ultra sound. The second stage of heart development then occurs within 24 hours of the end of the first stage, this stage is known as heart looping. The tube shaped heart concorts into an "S" shape and bends to the right, this is known as d-looping. This process creates a new shape, which provides a primitive region for the ventricles to begin forming. This stage of development is initiated by activated heart-specific proteins. The third stage of heart development is known as the two-chambered stage, consisting of one atrium and one ventricle. During this stage the cardiac jelly serves to act as a valve inbetween the atrium and the ventricle. The developing heart at this two-chambered stage is said to resemble the heart of a frog. The fourth stage of heart development begins by the atria dividing. The heart at this point of development is therefore known as the three-chambered heart, consisting of two newly divided atria which sit ontop of the remaining ventricle, the third chamber. This three-chambered heart is said to resemble the heart of a snake or turtle. The fifth and final stage of heart development occurs 10 weeks after conception. In this stage of development the single ventricle divides to form 2 new ventricles. This leaves the heart complete with all 4 chambers; 2 atria and 2 ventricles, and two large blood vessels to carry blood to and from the heart.<|endoftext|>
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Focus: In this unit (Unit 2 for schools using the CKHG series in Sequence grade-level order), students explore how the cultural and political traditions of ancient Greece and Rome have influenced Western society more profoundly than perhaps any other civilizations in world history. The political institutions of these two great civilizations—including the early forms of democracy established in Athens and several other city-states of ancient Greece, and the judicious power sharing articulated in the Roman Republic—have been incorporated into many subsequent societies. For ancient Greece, topics of study include the beginnings of democratic government; limitations of Greek democracy; the “classical” ideal in art and life; Pericles and the Golden Age; Greek myths; the Persian Wars and the Peloponnesian War; the philosophers Socrates, Plato, and Aristotle; and, the spread of Greek culture. For ancient Rome, topics of study include government and society in the Roman Republic; the Punic Wars; Julius Caesar; Caesar Augustus, the Pax Romana, and law and administration in the Roman Empire; the significance of Virgil’s Aeneid; Christianity under Roman rule; Constantine; and the causes of the “decline and fall” of the Roman Empire. Number of Lessons: 17 Lesson Time: 45 minutes each daily. Each lesson may be divided into shorter segments. Additional Search Terms: social studies • geography • map skills • nonfiction • informational text • primary source documents • myths • Mediterranean region • city-state • Athens • Parthenon • Olympics • Apollo and Daphne • Orpheus and Eurydice • Narcissus and Echo • Thermopylae • Sparta • Alexandria • the Pantheon • Ovid • Pygmalion and Galatea myths • patricians and plebeians • consuls, tribunes, senators • Carthage • Hannibal • Shakespeare CKHG Grade Levels: CKHG units are correlated to topics at the grade levels specified in the Core Knowledge Sequence, which allows students in schools following the Sequence to build knowledge grade by grade. This particular unit falls in Grade 6 in the Core Knowledge Sequence. In other settings, individual CKHG units may be used as supplemental resources. In general, the content and presentation in the CKHG units for Grade 6 are appropriate for students in Grade 6 and up.<|endoftext|>
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# Ray-Triangle Intersection¶ Implementation of ray-triangle intersection algorithm. ## Primitives¶ The first thing is to define the ray origin $O$ and direction $\large ê$, where the unit vector $\large ê = \frac{\vec{e}}{\parallel \vec{e} \parallel}$. Next step is to define the 3D triangle in counter-clockwise order in relation to the direction of the face. ## Intersection¶ To find the ray intersection, the next step is to define the triangle normal $\hat{n}$, where: $$\large \hat{n} = \frac{\vec{AB} \times \vec{AC}}{\parallel \vec{AB} \times \vec{AC} \parallel}$$ p.s.: to calculate ray-triangle intersection it is not necessary to normalize the normal vector. The supporting plane is what the triangle lies on, sharing the same normal vector. Given the plane equation: $$\large ax + by + cz + d = 0$$ Having the vector normal as $\hat{n} = [a b c]^T$ and $P = [x y z]^T$ as any point on the plane, we can define $d$ as follows: $$\large \hat{n} \cdot P + d = 0 \quad \therefore \quad d = - \hat{n} \cdot P$$ p.s.: in this case any known point can be used. Lets use the point $A$ so $P = A$ Before finding the intersection point $P$ on the plane, we must calculate the parameter $t$. We start by looking at the parametric equation of a line segment, which has the same direction of $ê$ and origin from $O$: $$\large P(P_x, P_y, P_z) = O + ê t$$ where: $$\large P_x = O_x + ê_x t \\ \large P_y = O_y + ê_y t \\ \large P_z = O_z + ê_z t$$ Using this concept on the plane equation, we have: $$\large ax + by + cz + d = 0 \\ \large aP_x + bP_y + cP_z + d = 0 \\ \large a(O_x + ê_x t) + b(O_y + ê_y t) + c(O_z + ê_z t) + d = 0 \\ \large aO_x + aê_x t + bO_y + bê_y t + cO_z + cê_z t + d = 0 \\ \large (aê_x + bê_y + cê_z)t + (aO_x + bO_y + cO_z) + d = 0 \\ \large (\hat{n} \cdot \hat{e})t + \hat{n} \cdot O + d = 0 \\ \large t = - \frac{\hat{n} \cdot O + d}{\hat{n} \cdot \hat{e}}$$ To figure out if the plane intersection point is inside or outside the triangle, we basically have to define the vector from each vertices to $P$ and cross it with its oriented edge segment (for each vertex). If the intersection point is outside the triangle, the resulting vector will be in the opposite direction from the normal one. $$\large [(B - A) \times (P - A)] \cdot \hat{n} \geq 0 \\ \large [(B - B) \times (P - B)] \cdot \hat{n} \geq 0 \\ \large [(B - C) \times (P - C)] \cdot \hat{n} \geq 0$$ If all these conditionals are obeyed, we can conclude that the point $P$ is inside the triangle. Otherwise, the point is going to be outiside toward to the edges of the negative values. ## 3D Intersection¶ Apply same model for each pixel of an image plane as the origin and the ray direction is based on the perspective camera model: ## Barycentric coordinates¶ The barycentric coordinates will help us to interpolate in-between vertex values. To do that we have to calculate the area of all resulting triangles. Any triangle area can be calculated as follows: $$\large \text{Area}_{ABC} = \frac{\parallel (B - A) \times (C - A) \parallel}{2}$$ Next step is to find the weight of each point so that we can use it to interpolate any desired values. $$\large \alpha = \frac{\text{Area}_{BCP}}{\text{Area}_{ABC}} = \frac{\parallel (C - B) \times (P - B) \parallel}{\parallel (B - A) \times (C - A) \parallel} \\\\ \large \beta = \frac{\text{Area}_{CAP}}{\text{Area}_{ABC}} = \frac{\parallel (A - C) \times (P - C) \parallel}{\parallel (B - A) \times (C - A) \parallel} \\\\ \large \gamma = \frac{\text{Area}_{ABP}}{\text{Area}_{ABC}} = \frac{\parallel (B - A) \times (P - A) \parallel}{\parallel (B - A) \times (C - A) \parallel}$$ Have the weights we can interpolate any kind of value (color, for example) by using: $$\large V = \frac{\alpha V_A + \beta V_B + \gamma V_C}{\alpha + \beta + \gamma}$$<|endoftext|>
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Ecological and Evolutionary Effects of Crop Domestication How does rapid evolution that occurs during plant cultivation influence ecological and evolution interactions between plants and their enemies? The domestication of plants and animals is one of the most important activities in human history. It is also the world’s largest, longest lasting, and best replicated selection experiment. By comparing domesticated organisms to their wild relatives we can address numerous basic and applied questions in evolutionary-ecology. One hypothesis we are address is that domestication, and its associated changes in yield, should cause plants to evolve reduced resistance to enemies. We are testing this hypothesis, and other extensions of it, using a combination of large scale comparative experiments and meta-analyses. Future projects will explore the community level impacts of domestication using field experiments as well as continue to explore the impacts on the evolution of plant enemies. Impact of domestication on resistance to 2 generalist herbivores across 29 domestication events --> New Phytologist 2012 Domestication slowing the rate of aphid rapid evolution --> Ecology Letters 2015 Meta-Analysis on the impact of domestication on resistance and defense traits --> PRSLB 2016 General Synthesis of the Eco-Evolutionary Impacts of Domestication and Agricultural on wild species -- > PRSLB 2016<|endoftext|>
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The fact that we’re able to hear sounds in our environment and process them almost instantly is nothing short of miraculous. Without even having to think about it, we can hear noises both loud and soft, near and far away. The pathway that these sounds travel in our ears is intricate, and if any one of its parts isn’t working correctly, we can experience hearing loss. Sound waves begin by entering your outer ear, called the pinna. It funnels these waves down your ear canal and into your middle ear, which are separated by the eardrum. The eardrum is a flexible membrane which begins to move when sound vibrations hit it, and that in turn starts to move three small bones in the middle ear: the hammer, anvil, and stirrup. These bones work together to amplify the sound waves and move them to your inner ear. The middle ear is where hearing loss begins to occur, so it’s important that all of these moving parts are free of damage. The inner ear is full of fine hair-like cells, replete with nerve endings, within a spiral-shaped organ called the cochlea. These tiny hair cells collect information from sound vibrations coming in from the middle ear and transmit those vibrations into nerve impulses (via the auditory nerve) to your brain. The brain processes and interprets these signals as sounds, allowing us to hear the noises around us. If you suffer from hearing loss, it means that one of the above sections isn’t working quite right. Our hearing evaluation is designed to diagnose which type of hearing loss is present, and enables us to identify the most effective solution for you. Conductive Hearing Loss Although rather uncommon and typically temporary, a conductive hearing loss can occur in some patients. It’s caused when an issue in the outer or middle ear blocks sound from the inner ear. Treatment involves the use of medication or surgery while other individuals opt to use hearing aids to improve their hearing ability. Conductive hearing loss can be caused by: - Ear infections - Benign tumors - Swimmer’s Ear - Foreign object in the ear - Fluid in the middle ear from colds or allergies - Absence or malformation of the outer ear, ear canal, or middle ear - Perforated eardrum - Impacted cerumen (earwax) Sensorineural Hearing Loss The most common type of hearing loss, sensorineural hearing loss occurs due to a problem with the inner ear or auditory nerve. It presents itself when either the auditory nerve or the hair-like cells in the cochlea have sustained damage or are missing. This results in the inability to send complete nerve signals to the brain. This kind of hearing loss can be caused by the following: - Head trauma - Drugs that are toxic to hearing (ototoxicity) - Genetics or aging - Malformation of the inner ear - Exposure to loud noise Mixed Hearing Loss When multiple parts of the ear’s anatomy are damaged, a mixed hearing loss can occur. In most cases, both the middle or outer ear along with the auditory nerve or inner ear have sustained an injury of some type or have encountered one of the conditions listed above. The conductive hearing loss present may be reversible while the sensorineural hearing loss is often permanent. Auditory Processing Disorders Rather than a hearing impairment which affects the ability to detect sounds, Auditory Processing Disorder (APD) causes individuals to struggle with their ability to organize, analyze, and interpret noises around them. While all parts of the ear are functioning properly, those with APD find that the hurdle they encounter is in their brain. Often caused by a tumor, disease, injury, heredity, or an unknown cause, the auditory processing centers in the brain do not function normally. APD does not always include hearing loss and many times the treatments for this disorder versus a hearing impairment are dramatically different.<|endoftext|>
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Desalination is a process that takes away mineral components from saline water. More generally, desalination refers to the removal of salts and minerals from a target substance, as in soil desalination, which is an issue for agriculture. Saltwater is desalinated to produce water suitable for human consumption or irrigation. One by-product of desalination is salt. Desalination is used on many seagoing ships and submarines. Most of the modern interest in desalination is focused on cost-effective provision of fresh water for human use. Along with recycled wastewater, it is one of the few rainfall-independent water sources. Due to its energy consumption, desalinating sea water is generally more costly than fresh water from rivers or groundwater, water recycling and water conservation. However, these alternatives are not always available and depletion of reserves is a critical problem worldwide. Currently, approximately 1% of the world's population is dependent on desalinated water to meet daily needs, but the UN expects that 14% of the world's population will encounter water scarcity by 2025.<|endoftext|>
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# What is the percent of increase from $147 annually to$414? ##### 1 Answer May 23, 2016 $\textcolor{red}{\text{The time span not defined as 1 year}}$ For 1 year only color(blue)(-> "percentage is " 187.5%) Or simple interest over whole period '..................................................... For more than 1 year where $n$ is number of years and annually compounded interest. " "color(blue)(x%=log^(-1)((log(414)-log(144))/n + 2) - 100) #### Explanation: $\textcolor{red}{\text{You have not stated the number of years.}}$ $\textcolor{red}{\text{The use of the word 'annually' implies more than 1 year}}$ Let the number of years be $n$ " "$144(1+x/100)^n=$414 '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Suppose the number of years is 1 then}}$ " "144+(144xx ?/100)=414 $\textcolor{b r o w n}{\text{Subtract 144 from both sides}}$ " "144xx ?/100=270 $\textcolor{b r o w n}{\text{Multiply both sides by } \frac{100}{144}}$ color(blue)(" "?%=270xx100/144 = 187.5%) '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Suppose "n>1)color(magenta)(" "->" Needs algebra}}$ $\textcolor{b r o w n}{\text{Notice that } \left(1 + \frac{x}{100}\right) \to \left(\frac{100 + x}{100}\right)}$ $\textcolor{b r o w n}{\text{Taking logs of both sides}}$ " "log($144)+nlog((100+x)/100)=log($414) $\text{ } n \log \left(\frac{100 + x}{100}\right) = \log \left(414\right) - \log \left(144\right)$ $\text{ } \log \left(\frac{100 + x}{100}\right) = \frac{\log \left(414\right) - \log \left(144\right)}{n}$ $\text{ } \log \left(100 + x\right) - \log \left(100\right) = \frac{\log \left(414\right) - \log \left(144\right)}{n}$ $\text{ } \log \left(100 + x\right) = \frac{\log \left(414\right) - \log \left(144\right)}{n} + \log \left(100\right)$ '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ color(brown)("But "log(100)=2 $\text{ } \log \left(100 + x\right) = \frac{\log \left(414\right) - \log \left(144\right)}{n} + 2$ $\text{ } 100 + x = {\log}^{- 1} \left(\frac{\log \left(414\right) - \log \left(144\right)}{n} + 2\right)$ " "color(blue)(x%=log^(-1)((log(414)-log(144))/n + 2) - 100)<|endoftext|>
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## How to Solve The puzzle can be played with any number of disks, although many toy versions have around seven to nine of them. The game seems impossible to many novices, yet is solvable with a simple algorithm. The number of moves required to solve a Tower of Hanoi puzzle is 2n -1, where n is the number of disks. I have discussed some common methods to solve 'towers of hanoi' puzzle game. Though complex methods are available, for the sake of simply solving them, these methods are enough. Iterative solution The following solution is a simple solution for the toy puzzle. Alternate moves between the smallest piece and a non-smallest piece. When moving the smallest piece, always move it to the next position in the same direction (to the right if the starting number of pieces is even, to the left if the starting number of pieces is odd). If there is no tower position in the chosen direction, move the piece to the opposite end, but then continue to move in the correct direction. For example, if you started with three pieces, you would move the smallest piece to the opposite end, then continue in the left direction after that. When the turn is to move the non-smallest piece, there is only one legal move. Doing this will complete the puzzle using the fewest number of moves to do so. It should perhaps be noted that this can be rewritten as a strikingly elegant set of rules: Simpler statement of iterative solution Alternating between the smallest and the next-smallest disks, follow the steps for the appropriate case: For an even number of disks: • make the legal move between pegs A and B • make the legal move between pegs A and C • make the legal move between pegs B and C • repeat until complete For an odd number of disks: • make the legal move between pegs A and C • make the legal move between pegs A and B • make the legal move between pegs B and C • repeat until complete In each case, a total of 2n-1 moves are made. Equivalent Iterative Solution Another way to generate the unique optimal iterative solution, is to number the disks 1 through n, largest to smallest. If n is odd, the first move is from the Start to the Finish peg, and if n is even the first move is from the Start to the Using peg. Now, add the constraint that no odd disk may be placed directly on an odd disk, and no even disk may be placed directly on an even disk. With this extra constraint, and the obvious rule of never undoing your last move, there is only one move at every turn. The sequence of these unique moves is an optimal solution to the problem equivalent to the iterative solution described above. Recursive solution A key to solving this puzzle is to recognize that it can be solved by breaking the problem down into a collection of smaller problems and further breaking those problems down into even smaller problems until a solution is reached. The following procedure demonstrates this approach. • label the pegs A, B, C - these labels may move at different steps • let n be the total number of discs • number the discs from 1 (smallest, topmost) to n (largest, bottommost) To move n discs from peg A to peg C: 1. move n-1 discs from A to B. This leaves disc n alone on peg A 2. move disc n from A to C 3. move n-1 discs from B to C so they sit on disc n The above is a recursive algorithm: to carry out steps 1 and 3, apply the same algorithm again for n-1. The entire procedure is a finite number of steps, since at some point the algorithm will be required for n = 1. This step, moving a single disc from peg A to peg B, is trivial. This approach can be given a rigorous mathematical formalism with the theory of dynamic programming, and is often used as an example of recursion when teaching programming. Non-recursive solution The list of moves for a tower being carried from one peg onto another one, as produced by the recursive algorithm has many regularities. When counting the moves starting from 1, the ordinal of the disk to be moved during move m is the number of times m can be divided by 2. Hence every odd move involves the smallest disk. It can also be observed that the smallest disk traverses the pegs f, t, r, f, t, r, etc. for odd height of the tower and traverses the pegs f, r, t, f, r, t, etc. for even height of the tower. This provides the following algorithm, which is easier, carried out by hand, than the recursive algorithm. In alternate moves: • move the smallest disk to the peg it has not recently come from. • move another disk legally (there will be one possibility only) For the very first move, the smallest disk goes to peg t if h is odd and to peg r if h is even. Also observe that: • Disks whose ordinals have even parity move in the same sense as the smallest disk. • Disks whose ordinals have odd parity move in opposite sense. • If h is even, the remaining third peg during successive moves is t, r, f, t, r, f, etc. • If h is odd, the remaining third peg during successive moves is r, t, f, r, t, f, etc. With this knowledge, a set of disks in the middle of an optimal solution can be recovered with no more state information than the positions of each disk: • Call the moves detailed above a disk's 'natural' move. • Examine the smallest top disk that is not disk 0, and note what its only (legal) move would be: (if there is no such disc, then we are either at the first or last move). • If that move is the disk's 'natural' move, then the disc has not been moved since the last disc 0 move, and that move should be taken. • If that move is not the disk's 'natural' move, then move disk 0. •<|endoftext|>
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# Calculations topic – Dosage and unit conversions #### Dosage conversions Dosages are sometimes written on prescriptions in a way that is unhelpful to the patient and as a pharmacist, we have to convert them into a more patient friendly version. As an example: Folic acid solution, 3 mg each day Since the medicine is a liquid, it would be inappropriate for us to label the medicine as 3 mg each day. We need to know the strength of the liquid, in this case 2.5 mg/5mL, and then calculate the volume to give. You can use a stepwise process: 2.5 mg in 5 mL Divide each side by 5 0.5 mg in 1 mL Multiply each side by 6 3 mg in 6 mL Or by a simple spotting technique where you can see that 3 mg is 20% (or 1/5) bigger than 2.5 mg so you will need to increase the volume in the same way 5 mL + 20% = 6 ml You would follow a similar process in other situations, for example When the strength of a product is in units/ mL, the dose is prescribed in units but the patient will be measuring out a volume #### Unit conversions We tend in this country to use metric units in medicine. Where this is the case, it is rare to need to convert except between different scales of units: ##### Weight – units getting smaller x 1000 x1000 x1000 x1000 Kg —–> g —–> mg —–> microgram —–> nanogram ##### Weight – units getting bigger ÷1000 ÷1000 ÷1000 ÷1000 nanogran —–> microgram —–> mg —–> g —–> kg So for the same weight of product, the larger the unit the smaller the number #### Volumes This works just the same way, as the weights but we only tend to change from: • L to mL, multiply the number by 1000 • mL to L, divide the number by 1000 A person’s height and weight – you may need to do this to calculate BMI Many people still measure their height and weight in imperial measurements, so There are: • 12 inches in one foot • 3 feet in one yard • 1 inch is equivalent to 2.54 cm (to two decimal places) So take an imperial height measurement, convert it to inches and then multiply the number of inches to get the measurement in cm There are also: • 16 ounces in one pound • 14 pounds in one stone • 1 pound is equivalent to 0.453592 kg (to six decimal places) So take an imperial weight measurement, convert it to pounds and then multiply the number of pounds by 0.453592 to get the measurement in kg<|endoftext|>
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In analyzing the data from physics experiments, we need to deal with measurement that are discrete or continuous in nature. Our aim is to make inferences about the models that we believe appropriately describe the physical situation, and/or, within a given model, to determine the values of relevant physics quantities. Thus, we need the probability rules that apply to uncertain variables, whether they are discrete or continuous. The rules for complete classes described in the preceding section clearly apply directly to discrete variables. With only slight changes, the same rules also apply to continuous variables because they may be thought of as a limit of discrete variables, as interval between possible discrete values goes to zero. For a discrete variable , the expression , which is called a probability function, has the interpretation in terms of the probability of the proposition , where is true when the value of the variable is equal to . In the case of continuous variables, we use the same notation, but with the meaning of a probability density function (pdf). So , in terms of a proposition, is the probability , where is true when the value of the variable lies in the range of to . In general, the meaning is clear from the context; otherwise it should be stated. Probabilities involving more than one variable, like , have the meaning of the probability of a logical product; they are usually called joint probabilities. Table 1 summarizes useful formulae for discrete and continuous variables. The interpretation and use of these relations in Bayesian inference will be illustrated in the following sections. |discrete variables||continuous variables| |moment of order| |A function such that , or , is called improper. Improper| |functions are often used to describe relative beliefs about the possible values of a variable.|<|endoftext|>
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# Video: GCSE Mathematics Foundation Tier Pack 1 • Paper 1 • Question 1 GCSE Mathematics Foundation Tier Pack 1 • Paper 1 • Question 1 04:03 ### Video Transcript Shown are four numbers. Three, negative three, eight, negative eight. Use two of these numbers to finish the calculation. Something plus something equals negative five. So looking at the numbers, straightaway, we can actually rule out two pairs of numbers. And we can do that if we actually just consider the five in the answer. So we disregard the negative and just look at the five. We want the two numbers when actually added together to give us an answer that has a five in it. So therefore, we cannot have the two pairs of numbers that are three and negative three or eight and negative eight. And that’s because neither of these will actually give us a result that involves a five. Okay, so that’s great. We’ve ruled out two pairs. Now, let’s see what other pairs are available. Well, the other pairs that are actually available to us are these four here. Remember that actually when it’s addition, we can actually have it either way around; it won’t affect it. We have three add eight, three add negative eight, negative three add eight, or negative three add negative eight. Well, straightaway, we can actually rule out three add eight. Well, it’s gonna be 11. So we know that that can’t be right because it won’t give us negative five. But we could also rule it out straightaway because if we think of it that if we’ve got a positive add a positive, then therefore our answer is gonna be positive. But we need our answer to be negative because it’s negative five. Okay, so let’s take a look at the other three pairs that we’ve got. So next, we have three add negative eight. Well, the first thing to note about three add negative eight is that when we actually have a plus and then a minus, then therefore what it actually become is a negative. So it actually become three minus eight. So now that we know that the sum is gonna be three minus eight, let’s think about what this will look like on our number line. So we know we’re going left on our number line. And that’s because if you’re subtracting one positive number from another, then we know that we’re gonna go left on our number line. So we can see if we actually subtract eight from three, then we go down eight spaces. So when we do that, what we’re gonna do is get to negative five, which is what we’re looking forward. So this is looking promising. So we can say that three add negative eight is equal to negative five and just to remind you it’s because three add negative eight turns into three subtract eight. Okay, great, let’s move on to the next couple just to double check. So now, we’re actually gonna stop on negative three. And if we’re gonna add on eight, we’re gonna move eight spaces to the right. And when we do that, we’re gonna get to positive five. So therefore, we can say that negative three add eight is gonna be equal to five. So this is incorrect cause this is not what we’re looking for. Okay, finally, we move on to our last pair. So now, for the last pair, we have negative three add negative eight. And as the same as the one previously, we know that if we’re gonna add a negative, it’s the same as subtracting the value. So it’s gonna be negative three minus eight. And therefore, as it’s negative three and minus eight, we’re gonna move eight spaces to the left. So we go from negative three, count eight spaces, and we get to negative 11, which again isn’t the value we’re looking for. Okay, great, so we now know that the pair of our numbers is definitely gonna be three and negative eight. And we can just double check that cause we’ve already shown that three add negative eight is equal to negative five. Well, as we’re dealing with addition, it shouldn’t matter which way around it is. But we just double check. We’ve got negative eight add three. So therefore, as we’re adding on three, we’re gonna be moving to the right. We can see if you move three places to the right, we get to negative five. So therefore, we can definitely say that if there are four numbers, three, negative three, eight, and negative eight, and we need to use two of them to finish the calculation something plus something is equal to negative five, then the two numbers are negative eight and three.<|endoftext|>
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