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520	# Prove by induction: $\sum_{i=0}^{n-1}\frac{i}{2^i}=2-\frac{n+1}{2^{n-1}}$ So far I have: Let $P(n)$: $\sum_{i=0}^{n-1}\frac{i}{2^i}$ for all n $\geq 1$. Basis: show $P(1)$: $$\sum_{i=0}^{n-1}\frac{i}{2^i}=0$$ $\therefore P(1)$ holds. Inductive step: Show $P(k) \to P(k+1)$. Assume $P(k)$: $\sum_{i=0}^{k-1}\frac{i}{2^i}=2-\frac{k+1}{2^{k-1}},$ for arbitrary $k\geq 1$ and show $P(k+1)$: $\sum_{i=0}^{k}\frac{i}{2^i}=2-\frac{k+2}{2^k}.$ So we show: \begin{align}\sum_{i=0}^{k}\frac{i}{2^i}&=\frac{k}{2^k}+\sum_{i=0}^{k-1}\frac{i}{2^i}=\frac{k}{2^k}+2-\frac{k+1}2^{k-1}\quad \text{by IH}\\[0.2cm]&=2+\frac{k}{2^k}-\frac{k+1}{2^{k-1}}\end{align} So I am confused now. I cannot solve for $2-\frac{n+1}{2^{n-1}}$ from this can I? Where did I go wrong? No, there is no mistake (apart from a typo, one line before the end), so you are almost there$$2+\frac{k}{2^k}-\frac{k+1}{2^{k-1}}=2+\frac{k-(2)(k+1)}{2^k}=2+\frac{-k-2}{2^k}=2-\frac{k+2}{2^k}$$ Also, in the last line you say that you need to solve for $2-\dfrac{n+1}{2^{n-1}}$, but as you said above, what you really need is to solve for $2-\dfrac{n+2}{2^{n}}$.<\|endoftext\|>	4.4375
490	Men and women adorned their heads in very different ways in Mesopotamia, situated in the region centered in modern-day Iraq near the Tigris and Euphrates Rivers between 3000 and 300 B.C.E. In the early years of civilization there, most men shaved their heads bald while women braided their long hair into elaborate styles pinned to the top of their heads. They also covered their hair with netting, scarves, or turbans. Elaborate hairstyles soon became important for both men and women in Mesopotamia. Men started to grow their hair longer and would wear it in waves. The king began to wear a full beard and long braided hair tied in a large bun at the nape of his neck. Women continued to wear their hair long, twisting it into large buns that covered the top of the head to the base of the neck and adorning it with ribbons and pins. The wealthiest people decorated their elaborate hairstyles with beautifully made jewelry of gold and silver. A royal tomb from Sumeria dating from 2500 B.C.E. included a golden helmet with a leather lining. The gold of the helmet was expertly formed to resemble the hairstyle popular for men of the time: waves around the face with a bun tied in the back. The same tomb contained jewels of the queen as well. One of the most impressive pieces is a headdress made of a wreath of golden leaves and blue lapis lazuli flowers with a golden fan topped with similar flowers in the back. In addition to these ornate headdresses, the king and queen also wore beautiful jewelry. Assyrian rule from 1380 to 612 B.C.E. altered hairstyles slightly. Men wore full beards and mustaches with longer curled hair. But some people with certain occupations, such as priests, doctors, and slaves, had specific hairstyles and headdresses, especially for special ceremonies. The king, for example, wore a tall hat made of alternating rows of patterned and plain bands topped with a pointed cone. Persians, who ruled Mesopotamia from 550 to 330 B.C.E. , continued to curl their hair but began to wear rounded and pointed hats, probably made of leather. Payne, Blanche. History of Costume: From the Ancient Egyptians to the Twentieth Century. New York: Harper and Row, 1965.<\|endoftext\|>	3.78125
1,082	Math Courses / Course / Chapter # Postulates & Theorems in Math: Definition & Applications Lesson Transcript Instructor: Laura Pennington Laura received her Master's degree in Pure Mathematics from Michigan State University, and her Bachelor's degree in Mathematics from Grand Valley State University. She has 20 years of experience teaching collegiate mathematics at various institutions. In mathematics, postulates and theorems are concepts that help guide how mathematical problems are handled. Understand more by reviewing the definitions and applications of postulates and theorems. Updated: 01/12/2022 ## What Is a Postulate? Suppose you decide to participate in a group research project for a new hair growth product that a company is testing. When you go to the group meeting, the first thing you have to do is fill out a questionnaire asking for your initial hair length before using the product. As you're all doing this, a woman in the group, Angie, states that she has the longest hair in the group. You take a quick glance around, and it looks like she does have the longest hair of all of the people in the group. Now, are you going to accept her statement as true, or are you going to whip out a tape measure and measure the length of everyone's hair to verify the truth of her statement? Most likely, you would accept her statement as true, because it's fairly obvious that she has the longest hair in the group. In mathematics, we call a statement like this a postulate. A postulate is a statement that is accepted as true without having to formally prove it. In the same way that it was fairly obvious that Angie's hair was the longest in the group, postulates in mathematics are usually easy to accept as true using simple mathematical reasoning. This is why they don't need to be proven formally. For example, a well-known postulate in mathematics is the segment addition postulate, which states the following: • Segment Addition Postulate: If a point, B, is drawn on a line segment AC, then AC is the sum of AB and BC. That is, AB + BC = AC. Looking at the image of this postulate, we see that it's obvious that if we split a line segment into two parts, then adding up those two parts will give us the original line segment. Therefore, we accept the postulate as true without having to prove it. Now that we're familiar with what a postulate is, let's take a look at theorems! An error occurred trying to load this video. Try refreshing the page, or contact customer support. Coming up next: Interior Angle Theorem: Definition & Formula ### You're on a roll. Keep up the good work! Replay Your next lesson will play in 10 seconds • 0:04 What Is a Postulate? • 1:54 What Is a Theorem? • 4:13 Real-World Applications • 6:25 Lesson Summary Save Save Want to watch this again later? Timeline Autoplay Autoplay Speed Speed ## What Is a Theorem? Now, suppose that one of the side effects of the hair product is that it can cause shortening of the user's right index finger, so another question on the questionnaire asks for the length of the participant's index finger before using the product. George, a member of the group, states that his index finger is the longest in the group. Hmm...George's statement isn't as obvious as Angie's statement because the lengths of everyone's index fingers look pretty similar. In this case, you go ahead and measure the right index fingers of the group, and it turns out that George's index finger is, in fact, the longest in the group. In this case, George's statement was formally proven and then accepted as true. In mathematics, we call this a theorem. A theorem is a statement that's not always obvious but has been proven using mathematical reasoning, along with other theorems and postulates. For instance, a well-known mathematical theorem is the midpoint theorem, and it states the following: • Midpoint Theorem: If M is the midpoint of a line segment AB, then AM = (1/2)AB and MB = (1/2)AB. The proof of this theorem makes use of the segment addition postulate and is shown in the image, but let's quickly move through the different steps: We see that both postulates and theorems are extremely useful in the proofs of theorems! Let's take a look at an application of postulates and theorems in a couple of real-world situations. To unlock this lesson you must be a Study.com Member. ### Register to view this lesson Are you a student or a teacher? Back ### Resources created by teachers for teachers Over 30,000 video lessons & teaching resources‐all in one place. Video lessons Quizzes & Worksheets Classroom Integration Lesson Plans I would definitely recommend Study.com to my colleagues. Itâ€™s like a teacher waved a magic wand and did the work for me. I feel like itâ€™s a lifeline. Jennifer B. Teacher Create an account to start this course today Used by over 30 million students worldwide<\|endoftext\|>	4.65625
351	### Theory: Decimal point position: 1. When a decimal number is multiplied by $$10$$, $$100$$, $$1000$$, $$10000$$, we move the decimal point by $$1$$, $$2$$, $$3$$ and $$4$$ places respectively to the right side of the number. 2. When a decimal number is divided by $$10$$, $$100$$, $$1000$$, $$10000$$, we move the decimal point $$1$$, $$2$$, $$3$$ and $$4$$ places respectively to the left side of the number. Example: 1. Multiply $$125.643$$ by $$10$$. Number of zeroes in $$10$$ is $$1$$. So, move the decimal point by one place to the right. $$125.643 \times 10 = 1256.43$$ 2. Multiply $$125.643$$ by $$100$$. Number of zeroes in $$100$$ is $$2$$. So, move the decimal point by two places to the right. $$125.643 \times 100 = 12564.3$$ 3. Divide $$125.643$$ by $$10$$. Number of zeroes in $$10$$ is $$1$$. So, move the decimal point by one place to the left. $\frac{125.643}{10}$ $$= 12.5643$$ 4. Divide $$125.643$$ by $$1000$$. Number of zeroes in $$1000$$ is $$3$$. So, move the decimal point by three places to the left. $\frac{125.643}{1000}$ $$= 0.125643$$<\|endoftext\|>	4.59375
2,370	Math.Practice.MP3 Common core State Standards • Math: Math • Practice: Mathematical Practice Standards • MP3: Construct viable arguments and critique the reasoning of others. Mathematically proficient students understand and use stated assumptions, definitions, and previously established results in constructing arguments. They make conjectures and build a logical progression of statements to explore the truth of their conjectures. They are able to analyze situations by breaking them into cases, and can recognize and use counterexamples. They justify their conclusions, communicate them to others, and respond to the arguments of others. They reason inductively about data, making plausible arguments that take into account the context from which the data arose. Mathematically proficient students are also able to compare the effectiveness of two plausible arguments, distinguish correct logic or reasoning from that which is flawed, and--if there is a flaw in an argument--explain what it is. Elementary students can construct arguments using concrete referents such as objects, drawings, diagrams, and actions. Such arguments can make sense and be correct, even though they are not generalized or made formal until later grades. Later, students learn to determine domains to which an argument applies. Students at all grades can listen or read the arguments of others, decide whether they make sense, and ask useful questions to clarify or improve the arguments. \| Math.Practice.MP8 Common core State Standards • Math: Math • Practice: Mathematical Practice Standards • MP8: Look for and express regularity in repeated reasoning. Mathematically proficient students notice if calculations are repeated, and look both for general methods and for shortcuts. Upper elementary students might notice when dividing 25 by 11 that they are repeating the same calculations over and over again, and conclude they have a repeating decimal. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1, 2) with slope 3, middle school students might abstract the equation (y – 2)/(x – 1) = 3. Noticing the regularity in the way terms cancel when expanding (x – 1)(x + 1), (x – 1)(x2 + x + 1), and (x – 1)(x3 + x2 + x + 1) might lead them to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of their intermediate results. \| Math.1.OA.B.3 Common core State Standards • Math: Math • OA: Operations & Algebraic Thinking • B: Understand and apply properties of operations and the relationship between addition and subtraction • 3: **Apply properties of operations as strategies to add and subtract. Students need not use formal terms for these properties. Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 • 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = Lesson Objective: Compare and contrast a related set of problems Grade 1 / Math / Argumentation 6 MIN Math.Practice.MP3 \| Math.Practice.MP8 \| Math.1.OA.B.3 ## Discussion and Supporting Materials ### Thought starters 1. What kinds of problems work best for this activity? 2. How are students encouraged to learn from each other throughout the lesson? 3. What do student learn from analyzing related problems? Absolutely love this. Excellent lesson. The "private think time" is a good way to maintain a quiet atmosphere. Reviewed prior days lesson. Presented the day's lesson and allowed "turn and talk" followed with sharing with whole group. Students were able to step forward to explain their reasoning. Presented with a challenge and were able to successfully predict the outcome. Wrap up included a review. Recommended (0) My mentee and I watched this video and we loved the different forms of interaction of the teacher and the students. We also noticed that the students could use the strategies with different yet similar concept problems! Recommended (0) I love that this teacher listens to his students while they have turned and talked. This seems to drive his questioning and instruction! Great formative assessment! Recommended (0) Awesome!!! I will do this at home with my grandson. Recommended (0) Great teacher! Awesome talk time with the first graders. I'm going to implement this lesson next week. Thank you! ? Recommended (0) ### Transcripts Speaker 1: Ladies and gentlemen, who remembers what we did yesterday when we looked Speaker 1: Ladies and gentlemen, who remembers what we did yesterday when we looked at some patterns and numbers? And then we came up with a rule or an idea that we had about the pattern we saw. You folks remember that? Children: Yeah. Speaker 1: Yeah. The mission of today's lesson was to get kids to begin to reason and justify arguments. It's really, really helpful if you do private think time so everybody has a chance to think about it first. They're looking for similarities and differences between the problems sets. Also identifying patterns, reasoning through that. Really the goal is to get them to share, to talk out loud and to explain their thinking. Children: Yes. Speaker 1: Okay. Santiago. Santiago: Eleven. Eleven. [inaudible 00:01:02] eleven! Speaker 1: I have one more for you. Children: Eighty. [crosstalk 00:01:08] Five plus seven equals ... Speaker 1: All right, McKenna? McKenna: Twelve. Speaker 1: All right. Can I just get a thumbs up if you've noticed something about those numbers up there? Taking a look at them? Okay. You turn and talk to the person next to you and share what do you notice up here? What do you notice? Children: [crosstalk 00:01:27] Ten, eleven, twelve and five, six, seven. Cause it's the same in the rolls. One and five are the same. Speaker 1: You notice in the rows they all are the same? Cool. What do you noticing about the ones? Children: I noticed that the next one's going to be the one and the next one's going to be the five. Speaker 1: So you're already thinking about what the next one's going to be. So, in terms of noticing. What are we noticing up here? I heard some good things. Juliete what did you notice? I heard what you shared. Juliete: I noticed there that there is a lot of fives. Speaker 1: A lot of fives. Look at those. All right. Leah, what are you noticing? Leah: I notice a lot of plus and equals. Speaker 1: Plus and equal. Woo. Great noticing everybody. Christi: I notice that starting by ten, eleven, twelve. Speaker 1: Okay. Interesting. So Christi just noticed a ten, eleven, twelve. The students did a really nice job of beginning to notice different things. Noticing the fives, noticing that they were adding. The equal sign. Noticing rows. And so I felt really good about that and then as they began to see the patterns it was really cool to see them begin to make sense of the problem. Children: Five, six, seven. Speaker 1: So what number do you thinks going to go here Kim? Kim: Eight. Speaker 1: Let's try that. If we were to do that, is there anything else we could predict up here? Children: Nine. Speaker 1: Lawson? Children: Nine. Speaker 1: Okay, so I hear nines. Children: Five. Five. Speaker 1: If I was to add this, what would be the next number that goes there? Children: Thirteen. Ten, eleven, twelve, thirteen. Speaker 1: Oh. So you're using the pattern ten, eleven, twelve, five, six and seven. A string of related problems is where the students are looking to see a rule or a pattern that's occurring so that they're able to justify a reason through that and then generalize that as a rule. One of the things that you've noticed, that several of you have noticed, is that this is staying the same. These are all fives. And then what's happening here? We're adding one. Speaker 1: And then what's happening here? Children: Adding one. If it was nine plus five it would be fourteen. Speaker 1: Okay. Children: And then five plus ten would be fifteen. Speaker 1: So if we're thinking about this, if this stays the same and we add one and then our answers changing by one, or the totals changing by one ... If I told you five plus twelve equals seventeen, what is five plus thirteen? Children: Eighteen. Speaker 1: Woah. No way. Children: Yes. Yes way. Yes way. Speaker 1: Christi, how did you do that? Christi: If five plus twelve equals seventeen then five plus thirteen is equal eighteen because thirteen is after twelve. Speaker 1: Thirteen is after twelve. Thirteen is one more than twelve. Can we say that? Thirteen is Children: One more than twelve. Speaker 1: That takes us back to here. One more. If we add one more. What's happening? I think that this type of learning with talking to each other, listening to each other, sharing their thinking transfers to all areas of school. It's not just math. They really feel empowered by the tasks that were in front of them today. That's really what's at the heart is when you have those moments. When that student feels confident enough to get up and share with the world who they are and what they're thinking. And their thinking matters and that you can create a space where their thoughts are validated and important and heard. Really, really just good for the soul. Ladies and gentlemen, thank you so much for your thinking. You did a marvelous job of thinking about this problem and looking for patterns in everything that you noticed. I was really, really impressed how you noticed that they were the same and how things were moving by one. We'll have to figure out a way to write that down so that we can continue to come back and revisit it. Ryan Reilly UNCUT CLASSROOMS \| 52 MIN ### Discovering the Properties of Quadrilaterals (Uncut) Discovering the Properties of Quadrilaterals (Uncut) UNCUT CLASSROOMS \| 29 MIN ### Heterogeneous Literature Circles (Uncut) Heterogeneous Literature Circles (Uncut) TCH Special \| 42 MIN ### Hybrid Best Practices Webinar / Distance Learning / Engagement TCH Special \| 58 MIN ### Creating Community Connections Webinar / Class Culture / Coaching TCHERS' VOICE Class Culture TCHERS' VOICE Lesson Planning TCHERS' VOICE New Teachers TCHERS' VOICE Engagement<\|endoftext\|>	4.5625
262	This week we have been applying our understanding of how to raise awareness by making thought provoking artworks. We created 4 draft ideas before receiving feedback and choosing the most thought provoking idea to focus on for our assessment. We have also been developing our writing skills by revising, editing and giving feedback on our narrative about deforestation. We have been learning how to solve open ended maths challenges and have been focussing on articulating our reasons for using a particular strategy. For example, using manipulatives and drawing out the problem. We have also started exploring the concept of VALUE in maths. How to help at home: Look for examples of numbers at home that have value. Parents can guide with questions like: What is the value of the number? What does the value represent? (shoe size, oven, freezer, dashboard, scales) Ask your child to talk you through the learning process guided by the images below. Ask your child to reflect on their chosen art medium and the process of creating their art piece/performance. Help extend your child’s inquiry by finding out more about their chosen art form and about other artists. Athletics Day 10:10 – Thursday the 21st of September Peace Parade 14:00 – Friday the 22nd of September<\|endoftext\|>	3.875
515	Women and girls are disproportionately affected by the impacts of climate change. In many developing contexts, women’s livelihoods are dependent on climate-sensitive sectors such as subsistence agriculture, forestry and water. Women and girls also typically have less capacity and resources than men and boys to prepare for and adapt to climate change. For example, restrictions on women’s land ownership mean that many women do not have access to productive land to farm, while a lack of financial capital and access to technologies means they cannot easily diversify their livelihoods. At the same time, women and girls who experience the consequences of climate change are often leaders in developing effective coping strategies and building resilience, for example by adapting their farming practices. Both women and men have important insights to contribute to designing and implementing effective climate responses and should be fully included in decision-making relating to climate change at all levels. This brief is a contribution to the 22nd Conference of the Parties to the United Nations Framework Convention on Climate Change (COP22) and a submission of recommendations for the renewal of the Lima Work Programme on Gender. It provides an overview of how members of the OECD Development Assistance Committee (DAC) are integrating gender equality into their bilateral ODA to climate change and makes recommendations to improve the gender-responsiveness of climate action. Climate ODA that also supports the achievement of gender equality accounted for 31% of bilateral ODA to climate change in 2014 – a total of USD 8 billion. Just 3% had gender equality as a principal objective, while 28% integrated gender equality as a secondary objective. Gender equality is better integrated in adaptation than in mitigation activities. In 2014, 46% of bilateral ODA to adaptation only targeted gender equality, compared with 28% to mitigation only. Attention to gender equality is uneven across climate-related sectors. While gender equality is quite well integrated in climate-related aid to agriculture and water, it is poorly addressed in economic infrastructure sectors. In 2013-14, only 8% of climate-related aid to energy targeted gender equality. Donors should improve their support to locally-led action on gender and climate change through multi-year and predictable funding for southern civil society organisations, including women’s rights organisations. Only 2% of all gender-responsive climate aid went to southern civil society organisations in 2014, representing USD 132 million. More needs to be done to improve women’s opportunities to participate in the green economy, notably through ensuring that women benefit equally from development projects focusing on clean technology and renewable energy.<\|endoftext\|>	3.75
1,562	I Came From Where? Imagining the Human Future video Public Event: Religious Audiences and the Topic of Evolution: Here of some of the well-tested methods of dating used in the study of early humans: Potassium-argon dating , Argon-argon dating , Carbon or Radiocarbon , and Uranium series. All of these methods measure the amount of radioactive decay of chemical elements; the decay occurs in a consistent manner, like a clock, over long periods of time. Thermo-luminescence , Optically stimulated luminescence , and Electron spin resonance. Modern phylogenetic trees have no input from stratigraphy, so they can be used in a broad way to make comparisons between tree shape and stratigraphy. Dating Fossils – How Are Fossils Dated? The majority of test cases show good agreement, so the fossil record tells the same story as the molecules enclosed in living organisms. Dating in geology may be relative or absolute. Relative dating is done by observing fossils, as described above, and recording which fossil is younger, which is older. The discovery of means for absolute dating in the early s was a huge advance. The methods are all based on radioactive decay:. The first radiometric dates, generated about , showed that the Earth was hundreds of millions, or billions, of years old. Since then, geologists have made many tens of thousands of radiometric age determinations, and they have refined the earlier estimates. Age estimates can be cross-tested by using different isotope pairs. Results from different techniques, often measured in rival labs, continually confirm each other. Every few years, new geologic time scales are published, providing the latest dates for major time lines. Older dates may change by a few million years up and down, but younger dates are stable. For example, it has been known since the s that the famous Cretaceous-Tertiary boundary, the line marking the end of the dinosaurs, was 65 million years old. Repeated recalibrations and retests, using ever more sophisticated techniques and equipment, cannot shift that date. It is accurate to within a few thousand years. The fossil record is fundamental to an understanding of evolution. Fossils document the order of appearance of groups and they tell us about some of the amazing plants and animals that died out long ago. Fossils can also show us how major crises, such as mass extinctions, happened, and how life recovered after them. If the fossils, or the dating of the fossils, could be shown to be inaccurate, all such information would have to be rejected as unsafe. Geologists and paleontologists are highly self-critical, and they have worried for decades about these issues. Repeated, and tough, regimes of testing have confirmed the broad accuracy of the fossils and their dating, so we can read the history of life from the rocks with confidence. Fossil dating methods website Educators have permission to reprint articles for classroom use; other users, please contact editor actionbioscience. Currently, he is studying certain basal dinosaurs from the Late Triassic and the quality of different segments of the fossil record. He holds the Chair in Vertebrate Paleontology at the University of Bristol, UK, in addition to chairing the Masters program in paleobiology at the university. Your one-stop source for information on evolution. Michael Benton wrote another article, Evidence of Evolutionary Transitions , for this website which explains how fossils support the stages of evolutionary history. Data bases and software for studying the quality of the fossil record. Michael Benton has written over 30 books on dinosaurs and paleobiology. Two suggested readings are provided — the first for adults, the second for children:. An online directory of dinosaur exhibits fro around the world. Many natural history museums and universities worldwide offer public participation programs in dinosaur events, such as fossil hunting or fossil cataloguing. No experience needed in most cases! The list is too long to mention here, so a couple of examples are provided to get you going on your search for programs in your area:. January Fossils provide a record of the history of life. One of the first and most basic scientific dating methods is also one of the easiest to understand. Paleontologists still commonly use biostratigraphy to date fossils, often in combination with paleomagnetism and tephrochronology. A submethod within biostratigraphy is faunal association: Sometimes researchers can determine a rough age for a fossil based on established ages of other fauna from the same layer — especially microfauna, which evolve faster, creating shorter spans in the fossil record for each species. The polarity is recorded by the orientation of magnetic crystals in specific kinds of rock, and researchers have established a timeline of normal and reversed periods of polarity. Paleomagnetism is often used as a rough check of results from another dating method. Within hours or days of a volcanic eruption, tephra — fragments of rock and other material hurled into the atmosphere by the event — is deposited in a single layer with a unique geochemical fingerprint. Researchers can first apply an absolute dating method to the layer. They then use that absolute date to establish a relative age for fossils and artifacts in relation to that layer. Anything below the Taupo tephra is earlier than ; anything above it is later. Generally speaking, the more complex a poem or piece of pottery is, the more advanced it is and the later it falls in the chronology. Egyptologists, for example, created a relative chronology of pre-pharaonic Egypt based on increasing complexity in ceramics found at burial sites. Sometimes called carbon dating, this method works on organic material.Radiometric Dating Both plants and animals exchange carbon with their environment until they die. Afterward, the amount of the radioactive isotope carbon in their remains decreases. Measuring carbon in bones or a piece of wood provides an accurate date, but only within a limited range. It would be like having a watch that told you day and night. Also called single crystal argon or argon-argon Ar-Ar dating, this method is a refinement of an older approach known as potassium-argon K-Ar dating, which is still sometimes used. Both methods date rock instead of organic material. As potassium decays, it turns into argon. But unlike radiocarbon dating, the older the sample, the more accurate the dating — researchers typically use these methods on finds at least , years old. While K-Ar dating requires destroying large samples to measure potassium and argon levels separately, Ar-Ar dating can analyze both at once with a single, smaller sample. The uranium-thorium method is often helpful for dating finds in the 40, to ,year-old range, too old for radiocarbon but too young for K-Ar or Ar-Ar. Silicate rocks, like quartz, are particularly good at trapping electrons. Researchers who work with prehistoric tools made from flint — a hardened form of quartz — often use thermoluminescence TL to tell them not the age of the rock, but of the tool. After shaping flint, toolmakers typically dropped the rocks into a fire. Archaeologists also frequently use TL to date ceramics, which are also exposed to high temperatures during manufacture. Similar to TL, optically stimulated luminescence measures when quartz crystals in certain kinds of rock last saw sunlight. That emitted light, the signal, can be used to calculate when the sample was last exposed to sunlight.<\|endoftext\|>	3.71875
364	A fraction is a part of a whole. A fraction has two components. The number on the top of the line is called the numerator. It tells how many equal parts of the whole are given. The number at the bottom of the fraction line (bar) is called the denominator. The denominator represents the total number of parts that make up a whole. In this lesson you will learn about the concept of a fraction and comparing fractions with unlike denominators. You will practice how to compare fractions with different denominators and why, in order to do that, you need to bring them to the Lowest Common Denominator (LCD). To find the LCD you would need to find a number that is evenly divisible by the denominators of the fractions you are comparing. But remember, when the denominator changes, the numerator changes with it. For example, how could you determine which one is greater, 3/4 or 2/5? In order to compare them, express both fractions in terms of the same number of parts that make up a whole. Watch a Video Lesson The Lowest Common Multiple for 4 and 5 is 20. So, the Lowest Common Denominator is also 20. Now, express each fraction in terms of the denominator 20, preserving the ratio. 3/4 is equivalent to 15/20. And 2/5 is equivalent to 8/20. Now, that the denominators are the same, only compare the numerators. Compare them as natural numbers: 15 is greater than 8. Thus, 15/20 is greater than 8/20. Record it as 15/20 > 8/20. In this lesson you will also discover how to compare fractions without using a number line and how to determine equivalent fractions.<\|endoftext\|>	4.65625
1,491	1. The Mexican Revolution deposed the country’s longest-serving president. Porfirio Díaz first made a name for himself at the 1862 Battle of Puebla. In an event celebrated every Cinco de Mayo, he helped the undermanned Mexican Army defeat invading French troops. Then, after trying and failing to get elected president democratically, Díaz seized power in an 1876 coup. Except for one four-year break, at which time a trusted associate served as president, Díaz would lead Mexico until 1911. Under his reign, foreign capital flooded into the country and extensive infrastructure modernizations took place. But land and power were concentrated in the hands of the elite, and elections were a charade. Following an economic downturn in 1907, even some middle- and upper-class citizens began to turn on him. Pro-democracy advocate Francisco Madero, who came from a wealthy family of landowners and industrialists, decided to challenge Díaz in the 1910 presidential race. Díaz jailed him, however, when it became clear he was gaining momentum. Upon his release Madero fled to Texas, where he issued a call for Mexicans to rise up against their government on November 20, 1910. Despite starting off slowly, revolutionaries soon made gains in the northern state of Chihuahua and elsewhere. By May 1911, Díaz had resigned and gone to France in exile. 2. A new Mexican strongman soon took over. Madero became president in November 1911, but fighting continued throughout large segments of the country, including the south, where Emiliano Zapata’s army of peasants seized lands that had purportedly been stolen by rich hacienda owners. Meanwhile, in February 1913, some counterrevolutionary leaders broke out of prison in Mexico City and marched to the National Palace with their troops in tow. Over the next 10 days hard fighting in the city center produced thousands of civilian casualties. Madero had tasked General Victoriano Huerta with putting down the uprising, but Huerta ended up switching sides and arresting Madero. He then had Madero executed and took hold of the presidency himself. 3. The anti-Huerta forces eventually began fighting each other. Huerta proved to be an even fiercer authoritarian than Díaz, and to this day remains among Mexico’s most despised villains. As president, he continued using political assassination as a tool, and forcibly conscripted the poor into his beefed-up federal army. In order to topple him, Zapata and other revolutionary leaders, such as Francisco “Pancho” Villa, Venustiano Carranza and Álvaro Obregón, united together. But since these men came from different parts of the country and had disparate political views, they turned on each other soon after forcing out Huerta in July 1914. Villa and Zapata briefly occupied Mexico City together, whereas Carranza—who for now had allied with Obregón—headed to the port city of Veracruz. Though Villa and Zapata originally appeared to have the upper hand, the tide turned in 1915 when Obregón won a series of battles against Villa with the help of trenches, barbed wire and other World War I-era defensive tactics. Carranza was elected president in 1917, the same year a new constitution formalized many of the reforms sought by rebel groups. Urban workers received an eight-hour workday, a minimum wage and the right to strike, while peasants gained mechanisms for land redistribution and limiting the size of estates. Another provision restricted foreign investment. Even so, armed struggle did not putter out until at least three years later. 4. The United States intervened numerous times in the conflict. Henry Lane Wilson, the U.S. ambassador to Mexico during the William Howard Taft administration, came to believe the revolution was harming American commercial interests. Wrongly convinced that Huerta would be a stabilizing influence, Wilson personally facilitated the general’s betrayal of Madero and rise to power in February 1913. But when President Woodrow Wilson took office the following month, he recalled Wilson and began materially backing Huerta’s opponents. He even ordered a blockade of Veracruz to prevent European arms from reaching Huerta. When U.S. troops landed there in April 1914, about 90 were killed or wounded in a hail of gunfire. U.S. warships responded by blasting the city with shells, bringing the number of Mexican casualties into the hundreds. A full withdrawal of Veracruz came that November. In March 1916, however, U.S. soldiers went back into Mexico as part of the so-called “punitive expedition.” This time around, the goal was to capture or kill Villa, who, upset over President Wilson’s support for Carranza, had launched a surprise cross-border raid of Columbus, New Mexico. General John J. Pershing and over 10,000 men, including Dwight D. Eisenhower and George S. Patton, searched for nearly a year. But although they found themselves in a number of shootouts, they never got their hands on the famed bandit. 5. The Mexican Revolution was followed by decades of one-party rule. Many historians believe the Mexican Revolution ended by the time Obregón assumed the presidency in December 1920, while others say it lasted all the way until 1940 or later. Part of this confusion lies with continuing periodic uprisings, including a so-called Cristero rebellion from 1926 to 1929 that pitted the anti-clerical government of President Plutarco Elías Calles against Catholic rebels. Calles, nicknamed the “Jefe Máximo” (Big Boss), controlled a series of puppet governments after his term expired in 1928. In order to bring divergent groups under a centralized power apparatus, he founded the National Revolutionary Party, later known as the Institutional Revolutionary Party, or PRI. The PRI would go on to rule Mexico until 2000. Despite its earlier reputation for electoral fraud, authoritarianism and corruption, it remains a major political force. In fact, after 12 years in opposition, a reconstituted PRI will be back in charge this December 1 when President-elect Enrique Peña Nieto takes office. 6. Nearly every major revolutionary leader was assassinated. Madero, Zapata, Carranza, Villa and Obregón—arguably the five most important figures of the Mexican Revolution—all met their ends at the hands of assassins. Madero was done in by Huerta’s treachery in 1913, while Zapata fell victim to an April 1919 ambush while trying to get an army colonel to defect. His body was then publicly displayed for all to see. Less than a year later, Carranza was shot by some of his former bodyguards as he fled toward Veracruz with trainloads full of the national treasury. Villa, meanwhile, had agreed to lay down his arms in July 1920. But after three years of working his farmland, he was murdered as part of a government conspiracy. Obregón, the last of the five to go, was felled by a Cristero rebel’s bullet in 1928.<\|endoftext\|>	4.03125
3,031	# 6.4 Graphs of logarithmic functions (Page 2/8) Page 2 / 8 Given a logarithmic function, identify the domain. 1. Set up an inequality showing the argument greater than zero. 2. Solve for $\text{\hspace{0.17em}}x.$ 3. Write the domain in interval notation. ## Identifying the domain of a logarithmic shift What is the domain of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{2}\left(x+3\right)?$ The logarithmic function is defined only when the input is positive, so this function is defined when $\text{\hspace{0.17em}}x+3>0.\text{\hspace{0.17em}}$ Solving this inequality, The domain of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{2}\left(x+3\right)\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(-3,\infty \right).$ What is the domain of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{5}\left(x-2\right)+1?$ $\left(2,\infty \right)$ ## Identifying the domain of a logarithmic shift and reflection What is the domain of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{log}\left(5-2x\right)?$ The logarithmic function is defined only when the input is positive, so this function is defined when $\text{\hspace{0.17em}}5–2x>0.\text{\hspace{0.17em}}$ Solving this inequality, The domain of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{log}\left(5-2x\right)\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(–\infty ,\frac{5}{2}\right).$ What is the domain of $\text{\hspace{0.17em}}f\left(x\right)=\mathrm{log}\left(x-5\right)+2?$ $\left(5,\infty \right)$ ## Graphing logarithmic functions Now that we have a feel for the set of values for which a logarithmic function is defined, we move on to graphing logarithmic functions. The family of logarithmic functions includes the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ along with all its transformations: shifts, stretches, compressions, and reflections. We begin with the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right).\text{\hspace{0.17em}}$ Because every logarithmic function of this form is the inverse of an exponential function with the form $\text{\hspace{0.17em}}y={b}^{x},$ their graphs will be reflections of each other across the line $\text{\hspace{0.17em}}y=x.\text{\hspace{0.17em}}$ To illustrate this, we can observe the relationship between the input and output values of $\text{\hspace{0.17em}}y={2}^{x}\text{\hspace{0.17em}}$ and its equivalent $\text{\hspace{0.17em}}x={\mathrm{log}}_{2}\left(y\right)\text{\hspace{0.17em}}$ in [link] . $x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ ${2}^{x}=y$ $\frac{1}{8}$ $\frac{1}{4}$ $\frac{1}{2}$ $1$ $2$ $4$ $8$ ${\mathrm{log}}_{2}\left(y\right)=x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ Using the inputs and outputs from [link] , we can build another table to observe the relationship between points on the graphs of the inverse functions $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{log}}_{2}\left(x\right).\text{\hspace{0.17em}}$ See [link] . $f\left(x\right)={2}^{x}$ $\left(-3,\frac{1}{8}\right)$ $\left(-2,\frac{1}{4}\right)$ $\left(-1,\frac{1}{2}\right)$ $\left(0,1\right)$ $\left(1,2\right)$ $\left(2,4\right)$ $\left(3,8\right)$ $g\left(x\right)={\mathrm{log}}_{2}\left(x\right)$ $\left(\frac{1}{8},-3\right)$ $\left(\frac{1}{4},-2\right)$ $\left(\frac{1}{2},-1\right)$ $\left(1,0\right)$ $\left(2,1\right)$ $\left(4,2\right)$ $\left(8,3\right)$ As we’d expect, the x - and y -coordinates are reversed for the inverse functions. [link] shows the graph of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g.$ Observe the following from the graph: • $f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$ has a y -intercept at $\text{\hspace{0.17em}}\left(0,1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{log}}_{2}\left(x\right)\text{\hspace{0.17em}}$ has an x - intercept at $\text{\hspace{0.17em}}\left(1,0\right).$ • The domain of $\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ $\left(-\infty ,\infty \right),$ is the same as the range of $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{log}}_{2}\left(x\right).$ • The range of $\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ $\left(0,\infty \right),$ is the same as the domain of $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{log}}_{2}\left(x\right).$ ## Characteristics of the graph of the parent function, f ( x ) = log b ( x ) For any real number $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and constant $\text{\hspace{0.17em}}b>0,$ $b\ne 1,$ we can see the following characteristics in the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right):$ • one-to-one function • vertical asymptote: $\text{\hspace{0.17em}}x=0$ • domain: $\text{\hspace{0.17em}}\left(0,\infty \right)$ • range: $\text{\hspace{0.17em}}\left(-\infty ,\infty \right)$ • x- intercept: $\text{\hspace{0.17em}}\left(1,0\right)\text{\hspace{0.17em}}$ and key point $\left(b,1\right)$ • y -intercept: none • increasing if $\text{\hspace{0.17em}}b>1$ • decreasing if $\text{\hspace{0.17em}}0 [link] shows how changing the base $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ in $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ can affect the graphs. Observe that the graphs compress vertically as the value of the base increases. ( Note: recall that the function $\text{\hspace{0.17em}}\mathrm{ln}\left(x\right)\text{\hspace{0.17em}}$ has base $\text{\hspace{0.17em}}e\approx \text{2}.\text{718.)}$ bsc F. y algebra and trigonometry pepper 2 given that x= 3/5 find sin 3x 4 DB remove any signs and collect terms of -2(8a-3b-c) -16a+6b+2c Will Joeval (x2-2x+8)-4(x2-3x+5) sorry Miranda x²-2x+9-4x²+12x-20 -3x²+10x+11 Miranda x²-2x+9-4x²+12x-20 -3x²+10x+11 Miranda (X2-2X+8)-4(X2-3X+5)=0 ? master The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer master The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer master Y master master Soo sorry (5±Root11* i)/3 master Mukhtar explain and give four example of hyperbolic function What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y? y/y+10 Mr Find nth derivative of eax sin (bx + c). Find area common to the parabola y2 = 4ax and x2 = 4ay. Anurag A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden to find the length I divide the area by the wide wich means 1125ft/25ft=45 Miranda thanks Jhovie What do you call a relation where each element in the domain is related to only one value in the range by some rules? A banana. Yaona given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither what are you up to? nothing up todat yet Miranda hi jai hello jai Miranda Drice jai aap konsi country se ho jai which language is that Miranda I am living in india jai good Miranda what is the formula for calculating algebraic I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it Miranda state and prove Cayley hamilton therom hello Propessor hi Miranda the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial. Miranda hi jai hi Miranda jai thanks Propessor welcome jai What is algebra algebra is a branch of the mathematics to calculate expressions follow. Miranda Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅 Jeffrey lolll who told you I'm good at it Miranda something seems to wispher me to my ear that u are good at it. lol Jeffrey lolllll if you say so Miranda but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it. Jeffrey which grade are you in though Miranda oh woww I understand Miranda Jeffrey Jeffrey Miranda how come you finished in college and you don't like math though Miranda gotta practice, holmie Steve if you never use it you won't be able to appreciate it Steve I don't know why. But Im trying to like it. Jeffrey yes steve. you're right Jeffrey so you better Miranda what is the solution of the given equation? which equation Miranda I dont know. lol Jeffrey Miranda Jeffrey<\|endoftext\|>	4.71875
1,244	From 11:00PM PDT on Friday, July 1 until 5:00AM PDT on Saturday, July 2, the Shmoop engineering elves will be making tweaks and improvements to the site. That means Shmoop will be unavailable for use during that time. Thanks for your patience! # At a Glance - Inscribed Angle Theorem Our old buddies, central angles, always had their vertices at the center. But an inscribed angle's vertex can be anywhere on the circle. Is the center of the circle even relevant anymore? It turns out the center of the circle is still super relevant to inscribed angles. The center can help us get organized here. Let's classify all possible central angles ∠ABC based on whether the center O of the circle is "inside," "outside," or "on" the angle, and examine these brief cases. ### Case 1 O is on one of the chords which make up ∠ABC. In other words, one of the sides of ∠ABC is a diameter of ⊙O. Let x stand for the measure of the inscribed angle ∠ABC, and we'll see if we can find mAC. We'll draw in line segment OA, because if we can find the measure of the central angle ∠AOC, we'll know mAC. By drawing in OA, we've made ΔAOB, which must be an isosceles triangle since OB and OA are radii of the same circle and are therefore congruent. That means m∠BAO = m∠ABC = x. We've also conveniently made ∠AOC an exterior angle to the triangle, which means that m∠AOC = m∠BAO + m∠ABC = x + x = 2x. So mAC = 2x. Now we know that a Case 1 angle intercepts an arc with twice the measure of the angle. We can use that knowledge to examine Case 2. ### Case 2 O is in the interior of ∠ABC. Let's add a diameter in there. Now we have two Case 1 angles ∠ABD and ∠DBC. From our work on Case 1, we can say that m∠ABD = ½ × mAD and m∠DBC = ½ × mDC Since ∠ABD and ∠DBC are adjacent angles, m∠ABC = m∠ABD + m∠DBC. Substituting, we have m∠ABC = ½ × mAD + ½ × mDC. Factoring out the common ½, we have m∠ABC = ½ × (mAD + mDC). Now we bring in our Arc Addition Postulate. Seemed silly at the time, didn't it? Well, it's here to save us now. Now we substitute that in, and we have m∠ABC = ½ × mAC. In other words, a Case 2 inscribed angle intercepts an arc with twice the measure of the angle, just as a Case 1 angle does. Curious. Now let's look at Case 3. ### Case 3 O is in the exterior of ∠ABC. Diameters have been good to us in the past, so how about we throw one in there, just to see what happens? After all, what's the worst that could happen? It's just math. We can actually use the same strategy we used in Case 2 with a little twist. Here, we have two Case 1 angles: ∠CBD, which intercepts arc CD, and ∠ABD, which intercepts arc AD. From our work in Case 1, we know that m∠ABD = ½ × mAD and m∠CBD = ½ × mCD. We'll find our target angle by subtraction this time, not by addition (although you can't have subtraction without addition, so in a way, we're still using addition). If we subtract, we get m∠ABC = m∠CBD – m∠ABD. Substituting, we have m∠ABC = ½ × mCD – ½ × mAD. Factoring, we have m∠ABC = ½ × (mCD – mAD). Now we bring in our old friend, the Arc Addition Postulate, and we get m∠ABC = ½ × mAC. We've shown that a Case 3 inscribed angle intercepts an arc with twice the measure of the angle—same as a Case 1 angle or a Case 2 angle. Since any inscribed angle falls into one of the three cases, we've proven the Inscribed Angle Theorem: for ∠ABC inscribed in a circle containing points A and C, mAC = 2 × m∠ABC. Translation: the arc is twice the angle. ### Sample Problem Given that ∠ACB is inscribed in a circle containing points A and B, and that m∠ACB = 42°, what is the measure of arc AB? Here we have an inscribed angle intercepting an arc, so we can use the Inscribed Angle Theorem. We can start out with mAB = 2 × m∠ACB, and then substitute 42° for m∠ACB. That'll give us mAB = 2 × 42°, or mAB = 84°. ### Sample Problem Given that ∠ACB is inscribed in circle ⊙O with radius 5 cm containing points A and B, and that m∠ACB = 78°, what is the length of arc AB? Before we calculate arc length, we can calculate the measure of the arc using the Inscribed Angle Theorem: mAB = 2 × m∠ACB = 2 × 78° = 156° Now we can use the arc length formula , with θ = 156° and r = 5 cm. This should give us an arc length of about 13.6 cm.<\|endoftext\|>	4.84375
836	In 1937, Colorado's native greenback cutthroat trout (Oncorhynchus clarki stomias) was believed to be extinct. Many attribute the greenback's decline to pollution, habitat loss, and predation as a result of intensifying front-range settlement during the late-1800s and the turn of the century. Importantly, in 1873, an old homesteader named Joseph C. Jones set out to strike it rich by developing a hotel and restaurant along Bear Creek (a tributary of the Arkansas River) in hopes of alluring tired travelers hiking a newly completed trail to and from the summit of Pike's Peak. Unknowingly, Jones would become an accidental conservationist. Researchers now believe that Jones may have stocked Bear Creek and some nearby ponds with a genetically pure strain of greenback cutthroat trout. From where exactly he obtained the stocks remains unclear. But it's believed that by stocking the creek, Jones simply intended to improve the recreational fishing near the hotel. Little did he know, his investment would eventually save the last known population of pure greenback cutthroat trout. Due to physical barriers within Bear Creek downstream of the Jones homestead, the stocked greenbacks were protected from other species of fish in the lower reaches of Bear Creek, allowing the greenbacks to avoid potential predators and hybridization with non-native trout over the ensuing 130 years. By the 1970s, the greenbacks were listed under the Endangered Species Act: first as endangered, then down-listed as threatened a few years later. Since then, an aggressive federal and state recovery effort followed to protect and restore various "greenback" populations across what was then believed to be their historical range on the eastern slope of the continental divide. However, decades later in 2012, cutting edge aquatic genetic research at the University of Colorado revealed earth-shattering results; we were focused on the wrong fish. As it turns out, the greenbacks identified for protection on the eastern slope were, in fact Colorado river cutthroat that were likely propagated and stocked across the continental divide. The 2012 study (Metcalf et al. 2012) found that between 1885 and 1953, up to 750 million trout (including cutthroat trout) were raised and transplanted from the western slope throughout the state, including to the eastern slope. Old notions that greenbacks and Colorado River cutthroat (which are difficult to distinguish based on physical traits alone) can be identified reliably by geographic range (east vs. west slope, respectively) have been repudiated. The 2012 study also concluded that the Bear Creek greenbacks were the last living vestige of the true greenback trout in Colorado. Local Aquatic Biologist with Colorado Parks and Wildlife (CPW), Kendall Bakich recently discussed these issues and their relevance to the Roaring Fork Valley in an article published by Roaring Fork Lifestyle magazine. According to Kendall, "In the Roaring Fork Valley we have two genetically pure green-lineage populations. . . . They're vulnerable." The two populations reside in Hunter Creek, a tributary of the Roaring Fork, near Aspen and in Cunningham Creek, a tributary of the Fryingpan River. The primary threats facing these fish are potential hybridization and invasion by non-native species, such as brook trout and others. Through Kendall's team, CPW conservation and reclamation efforts are beginning to take shape in these two local streams. Will we learn the true origin of the "green-lineage" cutthroat here in our valley? Did the Roaring Fork Valley have its own accidental conservationist, like Joseph C. Jones? Or are these unique populations of native fish? Time will eventually tell. But for now, please stay informed on this important and unfolding effort. The plan will likely take many years of significant work to accomplish. But hopefully the effort will result with yet another important chapter of the storied history of the greenback cutthroat trout. For more background on the Bear Creek greenback recovery program, take a look this video produced by CPW:<\|endoftext\|>	3.6875
4,708	# Connexions You are here: Home » Content » Collaborative Statistics: Custom Version modified by R. Bloom » Homework (modified R. Bloom) ### Recently Viewed This feature requires Javascript to be enabled. Inside Collection (Textbook): Textbook by: Roberta Bloom. E-mail the author # Homework (modified R. Bloom) Module by: Roberta Bloom. E-mail the author Summary: This module provides homework questions related to lessons on descriptive statistics. The original module by Dr. Barbara Illowsky and Susan Dean has been modified by Roberta Bloom. Some homework questions have been changed and/or added. ## Exercise 1 Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows: Table 1 # of movies Frequency Relative Frequency Cumulative Relative Frequency 0 5 1 9 2 6 3 4 4 1 • a. Find the sample mean x¯ x • b. Find the sample standard deviation, ss • c. Construct a histogram of the data. • d. Complete the columns of the chart. • e. Find the first quartile. • f. Find the median. • g. Find the third quartile. • h. Construct a box plot of the data. • i. What percent of the students saw fewer than three movies? • j. Find the 40th percentile. • k. Find the 90th percentile. • a. 1.48 • b. 1.12 • e. 1 • f. 1 • g. 2 • h. • i. 80% • j. 1 • k. 3 ## Exercise 2 The median age for U.S. blacks currently is 30.1 years; for U.S. whites it is 36.6 years. (Source: U.S. Census) • a. Based upon this information, give two reasons why the black median age could be lower than the white median age. • b. Does the lower median age for blacks necessarily mean that blacks die younger than whites? Why or why not? • c. How might it be possible for blacks and whites to die at approximately the same age, but for the median age for whites to be higher? ## Exercise 3 Forty randomly selected students were asked the number of pairs of sneakers they owned. Let X = the number of pairs of sneakers owned. The results are as follows: Table 2 X Frequency Relative Frequency Cumulative Relative Frequency 1 2 2 5 3 8 4 12 5 12 7 1 • a. Find the sample mean x¯ x • b. Find the sample standard deviation, ss • c. Construct a histogram of the data. • d. Complete the columns of the chart. • e. Find the first quartile. • f. Find the median. • g. Find the third quartile. • h. Construct a box plot of the data. • i. What percent of the students owned at least five pairs? • j. Find the 40th percentile. • k. Find the 90th percentile. • a. 3.78 • b. 1.29 • e. 3 • f. 4 • g. 5 • h. • i. 32.5% • j. 4 • k. 5 ## Exercise 4 600 adult Americans were asked by telephone poll, What do you think constitutes a middle-class income? The results are below. Also, include left endpoint, but not the right endpoint. (Source: Time magazine; survey by Yankelovich Partners, Inc.) ### Note: "Not sure" answers were omitted from the results. Table 3 Salary ($) Relative Frequency < 20,000 0.02 20,000 - 25,000 0.09 25,000 - 30,000 0.19 30,000 - 40,000 0.26 40,000 - 50,000 0.18 50,000 - 75,000 0.17 75,000 - 99,999 0.02 100,000+ 0.01 • a. What percent of the survey answered "not sure" ? • b. What percent think that middle-class is from$25,000 - $50,000 ? • c. Construct a histogram of the data 1. i: Should all bars have the same width, based on the data? Why or why not? 2. ii: How should the <20,000 and the 100,000+ intervals be handled? Why? • d. Find the 40th and 80th percentiles ## Exercise 5 Following are the published weights (in pounds) of all of the team members of the San Francisco 49ers from a previous year (Source: San Jose Mercury News). 177; 205; 210; 210; 232; 205; 185; 185; 178; 210; 206; 212; 184; 174; 185; 242; 188; 212; 215; 247; 241; 223; 220; 260; 245; 259; 278; 270; 280; 295; 275; 285; 290; 272; 273; 280; 285; 286; 200; 215; 185; 230; 250; 241; 190; 260; 250; 302; 265; 290; 276; 228; 265 • a. Organize the data from smallest to largest value. • b. Find the median. • c. Find the first quartile. • d. Find the third quartile. • e. Construct a box plot of the data. • f. The middle 50% of the weights are from _______ to _______. • g. If our population were all professional football players, would the above data be a sample of weights or the population of weights? Why? • h. If our population were the San Francisco 49ers, would the above data be a sample of weights or the population of weights? Why? • i. Assume the population was the San Francisco 49ers. Find: • i. the population mean, μ μ . • ii. the population standard deviation, σ σ . • iii. the weight that is 2 standard deviations below the mean. • iv. When Steve Young, quarterback, played football, he weighed 205 pounds. How many standard deviations above or below the mean was he? • j. That same year, the average weight for the Dallas Cowboys was 240.08 pounds with a standard deviation of 44.38 pounds. Emmit Smith weighed in at 209 pounds. With respect to his team, who was lighter, Smith or Young? How did you determine your answer? • k. Based on the shape of the data, what is the most appropriate measure of center for this data: mean, median, or mode? Explain. • l. Are there any outliers in the data? Use an appropriate numerical test involving the IQR to identify outliers, if any, and clearly state your conclusion. • m. Are any data values further away than 2 standard deviations from the mean? Clearly state your conclusion and show numerical work to justify your answer. ### Solution • b. 241 • c. 205.5 • d. 272.5 • e. • f. 205.5, 272.5 • g. sample • h. population • i. • i. 236.34 • ii. 37.50 • iii. 161.34 • iv. 0.84 std. dev. below the mean • j. Young • k. The mean is most appropriate. From the boxplot the data appear to be relatively symmetric. When the data are symmetric, it is appropriate to use the mean because it incorporates more information from the data. (If the data were skewed, then it would be more appropriate to use the median; but these data are not skewed.) • l. IQR = 272.5 – 202.5 = 67; Q1 – 1.5IQR = 205.5 – 1.5(67) = 105; Q3 + 1.5IQR = 272.5 + 1.5(67) = 373. All weights are between 105 and 373. There are no outliers. • m. Mean – 2(standard deviation) = 240.08 – 2(44.38) = 151.32 ; Mean + 2(standard deviation) = 240.08 + 2(44.38) = 328.84 ; All players' weights are between 2 standard deviations above and below the mean. ## Exercise 6 An elementary school class ran 1 mile in an average of 11 minutes with a standard deviation of 3 minutes. Rachel, a student in the class, ran 1 mile in 8 minutes. A junior high school class ran 1 mile in an average of 9 minutes, with a standard deviation of 2 minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran 1 mile in an average of 7 minutes with a standard deviation of 4 minutes. Nedda, a student in the class, ran 1 mile in 8 minutes. • a. Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he? • b. Who is the fastest runner with respect to his or her class? Explain why. ## Exercise 7 In a survey of 20 year olds in China, Germany and America, people were asked the number of foreign countries they had visited in their lifetime. The following box plots display the results. • a. In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected. • b. Explain how it is possible that more Americans than Germans surveyed have been to over eight foreign countries. • c. Compare the three box plots. What do they imply about the foreign travel of twenty year old residents of the three countries when compared to each other? ## Exercise 8 Twelve teachers attended a seminar on mathematical problem solving. Their attitudes were measured before and after the seminar. A positive number change attitude indicates that a teacher's attitude toward math became more positive. The twelve change scores are as follows: 3; 8; -1; 2; 0; 5; -3; 1; -1; 6; 5; -2 • a. What is the average change score? • b. What is the standard deviation for this population? • c. What is the median change score? • d. Find the change score that is 2.2 standard deviations below the mean. ## Exercise 9 Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best G.P.A. when compared to his school? Explain how you determined your answer. Table 4 Student G.P.A. School Ave. G.P.A. School Standard Deviation Thuy 2.7 3.2 0.8 Vichet 87 75 20 Kamala 8.6 8 0.4 ### Solution Kamala ## Exercise 10 Given the following box plot: • a. Which quarter has the smallest spread of data? What is that spread? • b. Which quarter has the largest spread of data? What is that spread? • c. Find the Inter Quartile Range (IQR). • d. Are there more data in the interval 5 - 10 or in the interval 10 - 13? How do you know this? • e. Which interval has the fewest data in it? How do you know this? • I. 0-2 • II. 2-4 • III. 10-12 • IV. 12-13 ## Exercise 11 Given the following box plot: • a. Think of an example (in words) where the data might fit into the above box plot. In 2-5 sentences, write down the example. • b. What does it mean to have the first and second quartiles so close together, while the second to fourth quartiles are far apart? ## Exercise 12 Santa Clara County, CA, has approximately 27,873 Japanese-Americans. Their ages are as follows. (Source: West magazine) Table 5 Age Group Percent of Community 0-17 18.9 18-24 8.0 25-34 22.8 35-44 15.0 45-54 13.1 55-64 11.9 65+ 10.3 • a. Construct a histogram of the Japanese-American community in Santa Clara County, CA. The bars will not be the same width for this example. Why not? • b. What percent of the community is under age 35? • c. Which box plot most resembles the information above? ## Exercise 13 Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, each asked adult consumers the number of fiction paperbacks they had purchased the previous month. The results are below. Table 6: Publisher A # of books Freq. Rel. Freq. 0 10 1 12 2 16 3 12 4 8 5 6 6 2 8 2 Table 7: Publisher B # of books Freq. Rel. Freq. 0 18 1 24 2 24 3 22 4 15 5 10 7 5 9 1 Table 8: Publisher C # of books Freq. Rel. Freq. 0-1 20 2-3 35 4-5 12 6-7 2 8-9 1 • a. Find the relative frequencies for each survey. Write them in the charts. • b. Using either a graphing calculator, computer, or by hand, use the frequency column to construct a histogram for each publisher's survey. For Publishers A and B, make bar widths of 1. For Publisher C, make bar widths of 2. • c. In complete sentences, give two reasons why the graphs for Publishers A and B are not identical. • d. Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not? • e. Make new histograms for Publisher A and Publisher B. This time, make bar widths of 2. • f. Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer. ## Exercise 14 Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all on-board transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Below is a summary of the bills for each group. Table 9: Singles Amount($) Frequency Rel. Frequency 51-100 5 101-150 10 151-200 15 201-250 15 251-300 10 301-350 5 Table 10: Couples Amount($) Frequency Rel. Frequency 100-150 5 201-250 5 251-300 5 301-350 5 351-400 10 401-450 10 451-500 10 501-550 10 551-600 5 601-650 5 • a. Fill in the relative frequency for each group. • b. Construct a histogram for the Singles group. Scale the x-axis by$50. widths. Use relative frequency on the y-axis. ### Exercise 29 #### Solution For pianos, the cost of the piano is 0.4 standard deviations BELOW average. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE average. For drums, the cost of the drum set is 1.0 standard deviations BELOW average. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar cost the most in comparison to the cost of other instruments of the same type. ### Exercise 32 Suppose that a publisher conducted a survey asking adult consumers the number of fiction paperback books they had purchased in the previous month. The results are summarized in the table below. (Note that this is the data presented for publisher B in homework exercise 13). Table 14: Publisher B # of books Freq. Rel. Freq. 0 18 1 24 2 24 3 22 4 15 5 10 7 5 9 1 1. Are there any outliers in the data? Use an appropriate numerical test involving the IQR to identify outliers, if any, and clearly state your conclusion. 2. If a data value is identified as an outlier, what should be done about it? 3. Are any data values further than 2 standard deviations away from the mean? In some situations, statisticians may use this criteria to identify data values that are unusual, compared to the other data values. (Note that this criteria is most appropriate to use for data that is mound-shaped and symmetric, rather than for skewed data.) 4. Do parts (a) and (c) of this problem this give the same answer? 5. Examine the shape of the data. Which part, (a) or (c), of this question gives a more appropriate result for this data? 6. Based on the shape of the data which is the most appropriate measure of center for this data: mean, median or mode? #### Solution 1. IQR = 4 – 1 = 3 ; Q1 – 1.5IQR = 1 – 1.5(3) = -3.5 ; Q3 + 1.5IQR = 4 + 1.5(3) = 8.5 ;The data value of 9 is larger than 8.5. The purchase of 9 books in one month is an outlier. 2. The outlier should be investigated to see if there is an error or some other problem in the data; then a decision whether to include or exclude it should be made based on the particular situation. If it was a correct value then the data value should remain in the data set. If there is a problem with this data value, then it should be corrected or removed from the data. For example: If the data was recorded incorrectly (perhaps a 9 was miscoded and the correct value was 6) then the data should be corrected. If it was an error but the correct value is not known it should be removed from the data set. 3. xbar – 2s = 2.45 – 21.88 = -1.31 ; xbar + 2s = 2.45 + 21.88 = 6.21 ; Using this method, the five data values of 7 books purchased and the one data value of 9 books purchased would be considered unusual. 4. No: part (a) identifies only the value of 9 to be an outlier but part (c) identifies both 7 and 9. 5. The data is skewed (to the right). It would be more appropriate to use the method involving the IQR in part (a), identifying only the one value of 9 books purchased as an outlier. Note that part (c) remarks that identifying unusual data values by using the criteria of being further than 2 standard deviations away from the mean is most appropriate when the data are mound-shaped and symmetric. 6. The data are skewed to the right. For skewed data it is more appropriate to use the median as a measure of center. ## Content actions PDF \| EPUB (?) ### What is an EPUB file? EPUB is an electronic book format that can be read on a variety of mobile devices. #### Collection to: My Favorites (?) 'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'. \| A lens I own (?) #### Definition of a lens ##### Lenses A lens is a custom view of the content in the repository. 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6,039	HIV and AIDS Acquired immunodeficiency syndrome; Human immunodeficiency virus Acquired immunodeficiency syndrome (AIDS) is a chronic and potentially fatal disease of the immune system. It is caused by the human immunodeficiency virus (HIV), which attacks a specific type of white blood cell known as T-lymphocytes. These cells are measured in the blood as the CD4 count. The lower a person's CD4 count, the weaker the person's immune system. As the immune system grows weaker, people with HIV and AIDS are more vulnerable to infections and cancers that the immune system would ordinarily fight off. More than 50,000 new cases of HIV infection occur in the United States every year. A massive research effort has produced better treatments, resulting in longer survival and improved quality of life for those with access to treatments. However, there is still no vaccine or cure. Signs and Symptoms Symptoms of HIV vary. A flu-like syndrome occurs in 40 to 90% of those who contract HIV within the first 2 to 6 weeks, with a combination of symptoms such as: - Sore throat - Swollen lymph nodes - Joint pain - Muscle aches - Mouth ulcers - Nausea and vomiting After infection with HIV, people may remain relatively symptom-free for years, or the disease may progress more rapidly. In this stage, the CD4 count is below 500/microliter. You may develop infections or chronic symptoms, including: - Swollen lymph nodes - Weight loss - Cough and shortness of breath - Low platelet count, which may manifest as easy bruising, bleeding gums, or nose bleeds - Localized herpes or fungal infections During the last stage of the disease, HIV infection may meet the official criteria for AIDS, which is the presence of an opportunistic infection (such as Pneumocystis carinii pneumonia, or PCP) or a CD4 count below 200/microliter. At this stage, symptoms may include: - Pneumonia, including PCP - Night sweats - Persistent fatigue - Extreme weight loss, exacerbated by diarrhea. More than 90% of HIV patients worldwide experience diarrhea, although that number is lower in the developed world. - Meningitis and other brain infections - Fungal infections - Malignancies, such as lymphoma, cervical cancer, and Kaposi sarcoma, which affects the skin, oral mucosa, and may spread to the lungs. It can occur in earlier stages of HIV, as well. What Causes It? HIV infection causes AIDS. HIV is spread primarily through sexual contact, and also through blood-to-blood contact, needle sharing among intravenous drug users, and, in pregnant women, from mother to child. Worldwide, about 80% of HIV transmission occurs through sexual contact. Blood transfusions and blood products caused many infections in the early years of the epidemic, but screening procedures have nearly eliminated this risk in the United States and other developed countries. A mother can spread the virus to a newborn during delivery and through breastfeeding, although drug therapy available in the developed world can greatly reduce the risk to infants. Risk factors include: - Having unprotected sex and having more than one partner, whether you are heterosexual or homosexual - Having another sexually transmitted disease - Using intravenous drugs and sharing needles What to Expect at Your Doctor's Office If your doctor suspects HIV infection, you may receive a "rapid test," which can provide results in 20 minutes. If the test is positive, your doctor will order a blood test to detect antibodies against the virus. If that test is also positive, the doctor will order a CD4 count (see above) and a viral load (an indication of the amount of virus present). This information, along with your symptoms, helps the doctor see what stage the disease is in and determine the best course of treatment, including the appropriate tests and medications. For example, if you are experiencing shortness of breath, your doctor will order a chest x-ray, particularly if your CD4 count is low. Some symptoms and tests may require evaluation in the hospital. HIV tests may not be accurate immediately after you are infected. It can take up to 12 weeks for your body to develop antibodies against the virus. If you suspect you have been infected and your test is negative, you may need to be retested after a short time to confirm the result. If you do test positive for HIV, you will be asked to tell your sexual partners immediately so they can also be tested. Medication can slow the progression of HIV infection to full blown AIDS. Doctors typically prescribe treatment when the CD4 count falls to a certain level. Generally, physicians use a combination of these medicines, including a type called protease inhibitors. In addition, antibiotics and other therapies are used to prevent or treat specific complications. It is important that a doctor who specializes in HIV direct your care. Your doctor will know the most effective treatment for you, including the most current medical regimen, what alternative treatments are safe, and which combinations may be harmful. Tell your doctor if you are using any alternative therapies to complement your medical regimen. Doctors use combination of drugs to treat HIV very aggressively, with the aim of reducing the amount of virus in your blood to very low or undetectable levels, and to suppress symptoms for as long as possible. Antiretroviral drugs help slow the progression of HIV by inhibiting the reproduction of the virus in your blood. It is important to keep a steady dose of antiretroviral drugs in your body to prevent the virus from developing resistance to the drugs. Antiretroviral medications include: - Protease inhibitors (PIs). PIs stop an HIV enzyme from replicating. This class of drugs includes saquinavir (Invirase), nelfinavir (Viracept), ritonavir (Norvir), tipranavir (Aptivus), indinavir (Crixivan), amprenavir (Agenerase), and atazanavir (Reyataz). For people who have not responded to treatment, doctors may prescribe another medicine, darunavir (Prezista), in combination with other drugs. Ritonavir and lopinavir (Kaletra) is among the most prescribed combination. Protease inhibitors are considered the most powerful HIV drugs and often interact with other medications, so your doctor must monitor them carefully. - Nucleoside analogue reverse transcriptase inhibitors (NRTIs). Also stop a particular HIV enzyme from replicating. These drugs were among the first to be developed and include zidovudine or azidodeoxythymidine (Retrovir or AZT), lamivudine (Epivir), didanosine (Videx), abacavir (Ziagen), stavudine (Zerit), and zalcitabine (Hivid). Emtricitabine (Emtriva) is a newer drug in this class and is taken with at least two other HIV medications. Combinations of several other drugs are also available. All have side effects that your doctor must monitor. - Nucleotide reverse transcriptase inhibitors (NtRTIs). Work similarly to NRTIs but act more quickly. So far there is only one drug in this class, tenofovir (Viread), which seems to be effective in people who develop resistance to NRTIs. - Non-nucleoside reverse transcriptase inhibitors (NNRTIs). Stop the virus from making DNA, so it cannot replicate itself. There are 3 drugs in this class: nevirapine (Viramune), efavirenz (Sustiva), and delavirdine (Rescriptor). They are often used if people cannot tolerate the side effects of protease inhibitors, want to delay protease inhibitor therapy, or if they have taken protease inhibitors but did not experience a drop in levels of the virus. Many of these drugs are cross resistant, meaning that if you develop resistance to one drug in this class it is likely you will be resistant to all. - Fusion inhibitors. Prevent the HIV membrane from fusing with the membrane of healthy cells in your body. Enfuvirtide (Fuzeon) is often used in combination with other drugs in people who have become resistant to other medications. It must be administered by injection. - Combination drug therapies also exist. Epzicom is a combination of abacavir (Ziagen) and lamivudine (Epivir). Truvada is a combination of tenofovir (Viread) and emtricitabine (Emtriva). In addition, doctors treat any opportunistic infections with the appropriate medications, or in some cases they give medications to prevent infections from occurring (prophylaxis). Complementary and Alternative Therapies Many people with HIV turn to complementary and alternative therapies to reduce symptoms of the virus, lessen side effects from medications, improve overall health and well being, and gain a sense of empowerment by being actively involved in their own care. Doctors use different therapies to: - Inhibit the virus - Treat symptoms of the virus or side effects of medication - Treat or prevent opportunistic infections - Improve function of the immune system Since the major impact of HIV is that it leaves patients vulnerable to opportunistic infections, making adjustments to enhance your overall health through minimizing stress, getting regular exercise, building a social support network, and having a spiritual practice can significantly boost immune function. In fact, these actions are some of the most powerful tools a person has to impact the course of the disease. Other changes, such as improving oral and general hygiene and limiting exposure to environmental pollutants, can also bolster your health and vitality. These small steps can add up to a longer and healthier life for many people. However, HIV should never be treated with alternative therapies alone. It is extremely important that you inform your doctor about any complementary and alternative therapies you are considering. Your doctor can help you determine what is safe and appropriate. Some herbs and/or nutrients can interfere with HIV/AIDS medications and new information on herb/drug interactions, both beneficial and detrimental, are being uncovered all the time. It is vital that you work with a knowledgeable provider to determine the proper nutrition and supplement program for your health.Nutrition and Supplements These nutritional tips may help reduce symptoms: - Eliminate suspected food allergens, such as dairy (milk, cheese, and ice cream), wheat (gluten), soy, corn, preservatives, and chemical food additives. Your doctor may want to test you for food allergies. - Eat foods high in B-vitamins, calcium, and iron, such as almonds, beans, whole grains (if no allergy), dark, leafy greens (such as spinach and kale), and sea vegetables. - Eat antioxidant foods, including fruits (such as blueberries, cherries, and tomatoes) and vegetables (such as squash and bell peppers). - Avoid refined foods, such as white breads, pastas, and especially sugar. - Use quality protein sources, such as organic meat and eggs, whey, and vegetable protein shakes, as part of a balanced program aimed at gaining muscle and preventing weight loss that can sometimes be a side effect of therapy. Try to focus on eating more lean meats, such as chicken and fish, tofu (soy, if no allergy), or beans for protein. - Use healthy oils in foods, such as olive or coconut oil. - Reduce or eliminate trans-fatty acids, found in commercially-baked goods, such as cookies, crackers, cakes, French fries, onion rings, donuts, processed foods, and margarine. - Avoid coffee and other stimulants, alcohol, and tobacco. - Drink 6 to 8 glasses of filtered water daily. - Exercise at least 30 minutes daily, 5 days a week. Talk to your doctor about how much exercise you need. You may address nutritional deficiencies with the following supplements: - A daily multivitamin containing the antioxidant vitamins A, C, E, the B-complex vitamins, and trace minerals, such as magnesium, calcium, and selenium. The HIV drug Agenerase already contains large amounts of Vitamin E, so speak to your doctor before taking supplements that contain vitamin E. - Omega-3 fatty acids, such as fish oil, to help reduce inflammation and help with immunity. Cold-water fish, such as salmon or halibut, are good sources but not substitutes for supplementation. Omega-3 supplements can increase the blood-thinning effect of certain medications, such as warfarin (Coumadin) and aspirin; speak with your doctor. - Whey protein, mixed in your favorite beverage, when needed as a protein supplement for support of immunity and weight gain or creatine, 5 to 7 grams daily, when needed for muscle weakness and wasting. Creatine can interact with some medications that treat kidney disease. Talk with your doctor. - N-acetyl cysteine, for antioxidant effects. - Probiotic supplement (containing Lactobacillus acidophilus among other strains), when needed for maintenance of gastrointestinal and immune health. Some probiotic supplements need refrigeration for best results. Check labels carefully. People who are severely immune compromised may not be able to take probiotics; speak with your doctor. - Coenzyme Q10 (CoQ10), for antioxidant and immune activity. CoQ10 may interact with some chemotherapy medications and blood pressure medications, and may lower the effectiveness of blood-thinning medications, such as warfarin (Coumadin). - Vitamin C, 1 to 3 times daily, as an antioxidant and for immune support. Some doctors will use higher doses in HIV and AIDS therapies. Check with your doctor. - L-glutamine, for support of gastrointestinal health and immunity. High doses of glutamine may cause manic symptoms in people with a history of psychiatric illness. It may also interact with certain medications. Speak with your doctor. - Melatonin, 1 hour before bedtime, for sleep and immune protection. Ask your doctor about potential drug interactions with the use of melatonin, particularly psychiatric medications. - Dehydroepiandrosterone (DHEA), for hormonal balance. DHEA is a hormone that is often low in people with HIV. One study found that DHEA supplements improved minor depression with no serious side effects. Because DHEA is a hormone, you should not take it without your doctor's supervision. You and your doctor can determine proper dosages after testing blood or saliva levels of DHEA in your body. Weight loss can be a serious problem for people with HIV. This symptom may begin early in the course of the disease and can increase the risk for developing opportunistic infections. Weight loss is exacerbated by other common symptoms of HIV and AIDS, including lesions in the mouth and esophagus, diarrhea, and poor appetite. Over the last several years, weight loss has become less of a problem due to the new protease inhibitors used for treating HIV. Reduction of muscle mass, though, remains a significant concern. Working with a registered dietitian to develop a meal plan to prevent weight loss and muscle breakdown is extremely helpful. Resistance training (lifting weights) can also protect against muscle breakdown and increase lean body mass. Preventing diarrhea and ensuring the body absorbs enough protein to maintain muscle strength has become a major goal of HIV/AIDS care. One program for combating diarrhea includes using soluble fiber (not insoluble fiber, such as Metamucil and psyllium husks). For some people, soluble fiber can help food stay in the digestive tract for longer periods of time, increasing the amount of nutrients that are absorbed and lessening bowel frequency. Good sources of soluble fiber include apple pectin, oat bran, and flax seed. Because diarrhea can be life threatening, use soluble fiber therapy only under the strict supervision of a trained professional.Herbs Herbs are a way to strengthen and tone the body's systems. As with any therapy, you should work with your doctor before starting any treatment. Work with a knowledgeable health care provider to establish a supplement regimen and make sure you have an up-to-date list of any supplements you are taking. You may use herbs as dried extracts (capsules, powders, or teas), glycerites (glycerine extracts), or tinctures (alcohol extracts). Unless otherwise indicated, make teas with 1 tsp. herb per cup of hot water. Steep covered 5 to 10 minutes for leaf or flowers, and 10 to 20 minutes for roots. Drink 2 to 4 cups per day. - Green tea (Camellia sinensis) standardized extract. For antioxidant and immune effects. Use caffeine-free products. You may also prepare teas from the leaf of this herb. In one study, a component of green tea called epigallocatechin gallate (EGCG) inhibited HIV infectivity. - Fermented wheat germ extract, for immune effects. Ask your health care provider for more information about this supplement. - Bitter Melon (Momordica charantia). For antiviral and immune support. Higher dosages may be needed in HIV and AIDs therapy. A health care provider can help with dosages. - Maitake mushroom (Grifola frondosa). For immune and antiviral effects. You may also take a tincture of this mushroom extract. Maitake may lower blood sugar, so take extra care if you already take blood sugar-lowering medications. - Cat's claw (Uncaria tomentosa) standardized extract. For immune and antiviral activity. Cat's claw may be inappropriate for some people with leukemia or certain autoimmune disorders. It may also interact with some medications. Speak with your doctor. You may use herbs as supportive therapies, but never use them alone to treat HIV or AIDS. Tell all of your providers about any treatments, conventional or alternative, you are taking so they can monitor interactions and side effects, and provide the best care. Always work with a complementary and alternative (CAM) practitioner who is knowledgeable in HIV care. HIV medicine is changing continually and practitioners need to stay current on the latest medication and herb/drug/nutrient interactions. You should stop taking St. John's wort (Hypericum perforatum), which has a negative effect on indinavir and could lead to developing resistance to the drug. You should also avoid echinacea (Echinacea spp.) and astragalus (Astragalus membranaceus). Both show conflicting evidence of enhancing immune function and strengthening replication of the HIV virus in test tubes. Garlic may also interfere with certain HIV medications.Homeopathy No specific scientific research supports the use of homeopathy for HIV or AIDS. A licensed, certified homeopathic doctor would evaluate you individually to assess the value of homeopathy for reduction of symptoms or side effects from medication as an adjunct to standard medical treatment.Physical Medicine Exercise is another way to develop a general sense of well being, improve mental attitude, decrease depression, diminish weight loss, and increase lean body mass. Resistance or weight training is particularly useful to increase strength and enhance lean body mass.Acupuncture People with HIV may use acupuncture to improve general well being, alleviate symptoms such as fatigue, insomnia, and night sweats, and to minimize side effects from medications, such as nausea and diarrhea. Some people also find relief from peripheral neuropathy, caused occasionally by certain medications used for HIV, reporting less pain, increased strength, and improved sensation. In China, acupuncture and moxibustion (a heat treatment performed by the acupuncturist over points where the needles are placed) are the standard treatments for HIV-related diarrhea. Health care professionals may also use acupuncture to treat the neuropathic (nerve) pain associated with certain HIV medications. Inserting needles bilaterally in the hand and foot points known as Baaxie and Bafeng, respectively, can lessen neuropathic pain.Massage Massage can relieve chronic muscle tension and stress, which may help the immune system. If you are HIV positive and pregnant, taking certain antiretroviral medications will reduce the likelihood of you transmitting the virus to your baby. Your doctor will determine which medicine is best for you and safe for your baby. Depending on your own condition, you and your doctor may decide to postpone treatment until after your first trimester to reduce the risk of birth defects. The drug efavirenz (Sustiva) should be avoided throughout pregnancy. If you are HIV-positive, you should not breastfeed because of the risk of transmission to your baby. Studies show that people who are HIV positive have increased arterial inflammation compared to non-infected people with the same risk factors. Preliminary research suggests that spirituality and a positive outlook can help slow disease progression and improve quality of life. Bepe N, Madanhi N, Mudzviti T, et al. The impact of herbal remedies on adverse effects and quality of life in HIV-infected individuals on antiretroviral therapy. J Infect Dev Ctries. 2011;5(1):48-53. Bope ET, Kellerman RD, eds. Conn's Current Therapy 2014. 1st ed. Philadelphia, PA: Elsevier Saunders; 2014. Brown J, Hanson JE, Schmotzer B, et al. Spirituality and optimism: a holistic approach to component-based, self-management treatment for HIV. J Relig Health. 2014;53(5):1317-28. Calvert C, Ronsmans C. The contributions of HIV to pregnancy-related mortality: a systematic review and meta-analysis. AIDS. 2013;27(10):1631-9. Catalfamo M, Le Saout C, Lane HC. The role of cytokines in the pathogenesis and treatment of HIV infections. Cytokine Growth Factor Rev. 2012;23(4-5):207-14. Chang BH, Sommers E. Acupuncture and the relaxation response for treating gastrointestinal symptoms in HIV patients on highly active antiretroviral therapy. Acupunct Med. 2011;29(3):180-7. Chu Y, Liu H. Advances of research on anti-HIV agents from traditional Chinese herbs. Adv Dent Res. 2011;23(1):67-75. Ekwunife OI, Oreh C, Ubaka CM. Concurrent use of complementary and alternative medicine with antiretroviral therapy reduces adherence to HIV medications. Int J Pharm Pract. 2012;20(5):340-3. Faintuch J, Soeters PB, Osmo HG. Nutritional and metabolic abnormalities in pre-AIDS HIV infection. Nutrition. 2006;22(6):683-90. Ferri FF. Ferri's Clinical Advisor 2014. 1st ed. Philadelphia, PA: Elsevier Mosby; 2013. Han H, He W, Wang W, et al. Inhibitory effect of aqueous dandelion extract on HIV-1 replication and reverse transcriptase activity. BMC Complement Altern Med. 2011;11:112. Harris A, Bolus NE. HIV/AIDS: An update. Radiol Technol. 2008;79(3):243-52. Hasan SS, See CK, Choong CL, et al. Reasons, perceived efficacy, and factors associated with complementary and alternative medicine use among Malaysian patients with HIV/AIDS. J Altern Complement Med. 2010;16(11):1171-6. Hendricks MK, Eley B, Bourne LT. Colecraft E. HIV/AIDS: nutritional implications and impact on human development. Proc Nutr Soc. 2008;67(1):109-13. Highleyman L. Nutrition and HIV. BETA. 2006;18(2):18-32. Hillier SL, Louw Q, Morris L, et al. Massage therapy for people with HIV/AIDS. [Review]. Cochrane Database Syst Rev. 2010;(1):CD007502. Hoogbruin A. Complementary and alternative therapy (CAT) use and highly active antiretroviral therapy (HAART): current evidence in the literature, 2000-2009. J Clin Nurs. 2011;20(7-8):925-39. Hoppe C, Andersen GS, Jacobsen S, et al. The use of whey or skimmed milk powder in fortified blended foods for vulnerable groups. J Nutr. 2008;138(1):145S-61S. Irlam JH, Visser MM, Rollins NN, et al. Micronutrient supplementation in children and adults with HIV infection. [Review]. Cochrane Database Syst Rev. 2010;(12):CD003650. Review. Joy T, Keogh HM, Hadigan C, et al. Dietary fat intake and relationship to serum lipid levels in HIV-infected patients with metabolic abnormalities in the HAART era. AIDS. 2007;21(12):1591-600. Liu JP, Manheimer E, Yang M. Herbal medicines for treating HIV infection and AIDS. Cochrane Database Syst Rev. 2005;(3):CD003937. Liu X, Han Y, Peng K, et al. Effect of traditional Chinese medicinal herbs on Candida spp. from patients with HIV/AIDS. [Review]. Adv Dent Res. 2011;23(1):56-60. Louie L, Pathanapornpandh N, Pultajuk U, et al. The Mae On Project: using acupuncture for symptom relief and improved quality of life for people living with HIV and AIDS in rural Thailand. Acupunct Med. 2010;28(1):37-41. Nance CL, Siwak EB, Shearer WT. Preclinical development of the green tea catechin, epigallocatechin gallate, as an HIV-1 therapy. J Allergy Clin Immunol. 2009;123(2):459-65. Oliveira JM, Rondó PH. Omega-3 fatty acids and hypertriglyceridemia in HIV-infected subjects on antiretroviral therapy: systematic review and meta-analysis. [Review]. HIV Clin Trials. 2011;12(5):268-74. Perez EM, Carrara H, Bourne L, et al. Massage therapy improves the development of HIV-exposed infants living in a low socio-economic, peri-urban community of South America. Infant Behav Dev. 2015;38(2):135-46. Romanelli F, Matheny S. HIV Infection: The Role of Primary Care. Am Fam Phys. 2009;80(9). Stone CA, Kawai K, Kupka R, et al. Role of selenium in HIV infection. [Review]. Nutr Rev. 2010;68(11):671-81. Subramanian S, Tawakol A, Burto TH, et al. Arterial inflammation in patients with HIV. JAMA. 2012;308(4):379-86. Suttajit M. Advances in nutrition support for quality of life in HIV+/AIDS. Asia Pac J Clin Nutr. 2007;16 Suppl 1:318-22. Tabi M, Vogel RL. Nutritional counseling: an intervention for HIV-positive patients. J Adv Nurs. 2006;54(6):676-82. Wang J, Zou W. Practices, challenges, and opportunities: HIV/AIDS treatment with traditional Chinese medicine in China. [Review]. Front Med. 2011;5(2):123-6. Yeh SS, Lovitt S, Schuster MW. Pharmacological treatment of geriatric cachexia: evidence and safety in perspective. J Am Med Dir Assoc. 2007;8(6):363-77. Wiysonge CS, Shey M, Kongnyuy EJ, et al. Vitamin A supplementation for reducing the risk of mother-to-child transmission of HIV infection. [Review]. Cochrane Database Syst Rev. 2011;(1):CD003648. Zetola NM, Bernstein KT, Wong E, et al. Exploring the relationship between sexually transmitted diseases and HIV acquisition by using different study designs. J Acquir Immune Defic Syndr. 2009;50(5):546-51. Review Date: 4/1/2016 Reviewed By: Steven D. Ehrlich, NMD, Solutions Acupuncture, a private practice specializing in complementary and alternative medicine, Phoenix, AZ. Review provided by VeriMed Healthcare Network.<\|endoftext\|>	3.84375
516	# Zeno Bouncing If you drop a rigid ball onto a rigid surface, it eventually stops bouncing without ever bouncing the last time! This is called Zeno behavior, after Zeno’s most famous paradox (Achilles and the tortoise). Suppose we drop the ball from a height of 1m, and the coefficient of restitution is 0.9. So every bounce is followed by another bounce of 90% its height, ad-infinitum. Yet after 16.7 seconds the ball comes to a complete rest! Yes, there is always another bounce after every bounce, and yes, there are no more bounces after 16.7 seconds. ## Proof Under the conditions described above, the height attained by the ball after the n’th bounce is $\large{h_0 = 1}$ $\large{h_{n>0} = 0.9h_{n-1}}$ Solving the recursive equation above, we find that $\large{h_n = 0.9^n}$ after the n’th bounce. Falling freely from rest in gravity g, the ball travels a distance of $\large{h = \frac{1}{2} g t^2}$ in time t. Therefore it takes the ball a time of $\large{t = \sqrt{2h/g}}$ to fall to the ground from rest at height h, and the same amount of time to rebound from the ground to rest at height h. Therefore, the period of time between the n’th bounce and its following bounce, is $\large{t_n = 2\sqrt{2h_n/g} = 2\sqrt{2\times 0.9^n/g} = 2\sqrt{2/g}\sqrt{0.9}^n}$ The total time that the ball spends bouncing, starting from its first bounce, is therefore $\large{T = \sum_{n=1}^{\infty}t_n = 2\sqrt{2/g}\sum_{n=1}^{\infty}\sqrt{0.9}^n = 2\sqrt{2/g}(\frac{1}{1-\sqrt{0.9}}-1) \approx 16.7}$ Since every bounce is followed by another bounce (of 90% its height) there is never a last bounce, yet after 16.7 seconds there is no more bouncing whatsoever!<\|endoftext\|>	4.625
2,912	The ACTFL Proficiency Guidelines are a very helpful tool in the Foreign/World Language Classroom. They provide teachers and students with clear guidelines and descriptions to assess proficiency levels. They are also an effective tool for students and teachers to set achievable and concrete goals. The ACTFL Can Do Statements provide detailed examples of what students could/should be able to do at each proficiency level. The challenge I have personally had with the Can Do Statements is using them for various age and developmental levels. There are some Can Do statements that address such things as making reservations and asking questions about particular academic subjects. While these are very applicable to older students, they are not developmentally appropriate for younger students. For this reason I have developed, with the help of a few colleagues, Student-Friendly Can Do Statements. These statements honor the text type (individual words and phrases, discrete sentences, connected sentences, paragraphs) of the ACTFL Proficiency Guidelines, but are more applicable to elementary, middle school and high school students. Language teachers know the importance of target language use in the classroom. Regular exposure and interaction with the language leads to acquisition and a higher proficiency level. In order to promote, expect and respect the use of the target language in the classroom teachers should support students by creating a classroom community that makes students feel safe taking risks with the language and teach the tools needed to communicate. Students should also know their proficiency level and be personally accountable for their commitment to using the target language and striving to raise their proficiency level. In my classroom I have a 20 point rubric that I use to assess students each week on four focus areas: Community, Commitment, Proficiency, and Preparation. Each category is based a five point scale. Students are aware of these criteria and they are posted in the classroom so that they can be references regularly. I typically grade each student myself for the first few weeks of the school year and then students self-assess, but I of course reserve the right to modify the self-assessment grade if necessary. The grade is given holistically for the entire week. Here is the breakdown of each category: - 5 Choices and interactions enhance the classroom community. - 4 Choices and interactions almost always enhance the classroom community. - 3 Choices and interactions sometimes enhance the classroom community. - 2 Choices and interactions often hinder the classroom community. - 1 Choices and interactions regularly hinder the classroom community. For recommendations on classroom community building see my post on Building a Community of Confidence. - 5 Always speaks target language and circumlocutes. - 4 Always speaks target language with some effort to circumlocute. - 3 Makes an effort to speak target language, but need to circumlocute more. - 2 Resorts to native language; no circumlocution. - 1 Little use of target language. For tools and strategies for students to remain in the target language see my posts on circumlocution, functional chunks and language ladders. - 5 Regularly speaks at expected proficiency level and strives to speak above level. - 4 Regularly speaks at expected proficiency level. - 3 Usually speaks at expected proficiency level and below level at times. - 2 Regularly speaks below proficiency level. - 1 Always speaks below proficiency level. See my post on Foreign Language Goal Setting Using ACTFL Proficiency Levels to learn about assessing students’ proficiency levels. - 5 Punctual, has all materials, assignments complete. - 4 Punctual, has most materials, assignments complete. - 3 Punctual, has all materials, assignments incomplete. - 2 Late or missing materials. - 1 Late or missing materials, assignments incomplete. Try out this rubric system and modify to fit the needs and of your individual classroom. I’m sure you will see an increase in student accountability for using the target language and you will feel confident that you are supporting your student in their language proficiency growth. How do we get to 90%+ target language use in the foreign language classroom? We need to rethink how we have been teaching over the past few decades and be willing to leave some of it behind. Traditional teaching practices were not designed to promote a high percentage of target language use. If you want to get to 90%+ target language use in the classroom reflect on your teaching through the lens of these four questions and recommendations. Take some time to contemplate how you can move your teaching in a direction that is more proficiency-based and promotes regular and confident use of the target language in your classroom. - Q1: Are prompts and tasks at the appropriate proficiency level? - R1: Assess the proficiency level of students to make sure that prompts are not above students’ proficiency level. - Q3. Are students held accountable for using the target language? - R3: Include goal setting, consistency, commitment and proficiency in grade. - Q4. Are all the students actively engaged and interested? - R4: Provide choice and opportunity for personal interest, investment and active engagement. I’m always looking for ways to get students up and moving in the classroom while they are practicing their foreign language speaking and writing skills. This is an activity that I call “Hide and Speak (or write)” that accomplishes this goal and students enjoy it and often ask to play. I’m happy to oblige because they speak (or write) so much during this activity. - Begin by hiding 20-30 prompt cards. These can be index cards with vocabulary words, an image, a question about a reading, or proficiency-based questions aligned with ACTFL standards. The possibilities are endless for prompts based on the material that is being covered in class. Memory Cards or Task Cards work very well for this this activity. - Pairs of students set out to find the prompts and when they do they return to the teacher with the card and perform the task: identify the image in the target language, use the word or verb in a sentence, answer a proficiency-based question or complete a Task Card. Lots of possibilities. This can all be through speaking or writing. When writing I give pairs a small white board and marker. - If the pair responds correctly they can get a point for their team or the teacher can make it a point for the entire class with the goal being to get a certain number of points collectively in a specified amount of time. The teacher keeps the prompt card and the pair sets back out. - Be sure to tell pairs that they need to wait in line to check in with the teacher so that that they don’t call crowd in. Check out these task cards these task cards and memory cards that work well in this activity. This a very effective go-to activity that requires very little prep and gets students moving and using the target language immediately. It’s also a great way to use a set of memory/concentration cards that you may have in your classroom. If you need cards you can find them here: Set up desks or tables around room, spaced out enough for students to move around, and put several pictures on each table. Play music and kids move around (maybe dance if they are so inclined), then when the music stops students stand behind a table. Choose a word card, say it out loud and student with that picture identified says they he/she has the corresponding picture card on his/her table (int he target language of course). He/she then uses the word in a sentence and puts a point up by his/her name on the board. Play the music again and continue the same process of stopping the music and students saying a sentence with the word if they had the picture match. Students really enjoyed this activity, review lots of vocabulary, and speak a lot. You can also allow the winner of the round to be the one to start and stop the music the next time, choose a word card and say it to the class. Try this with verb forms as well, with the conjugations 0n the desks. El Camino/Le Chemin is an engaging and interactive speaking activity that students can do in pairs or small groups. Very quick set-up with no prep needed. Just print out the two pages that make up the game board and students are ready to go. Students can do this activity in groups of 2 or 3. Each player needs a game piece to move around the board. They can use a bingo chip, a coin or any object of similar size. One die is also needed for the activity. All players start at “Début” or “Comeinzo.” Taking turns, each player rolls the die and moves the number of spaces rolled. The object is to land on the numbered boxes in the correct order (1-12). They can move in any direction, but they can’t use the same box twice in a turn. They can share a box with another player. The winner is the first player to land on square #12. The game can be made longer by having players return to “Début” or “Comienzo”and work toward #12 a second time. Each time a player rolls the die and moves closer to the next number, he/she must say the verb, number, time, category word, etc. of the square he/she lands on. They can also be required to say a complete sentences. You can download these activities here: This activity gives students a chance to express themselves confidently at their current proficiency level. It is easily adapted by simply prompting students as to how they should speak (text type). Typically I have students work in pairs or in groups of 3. Begin by setting up a sheet with 12 categories on it that are number 1-12. Provide 2 dice along with this paper. Give each pair or group a small bag (not transparent) with small slips of colored paper along with a sheet that has a point value assigned to each color. For fun I also include a “Zut” or “Caramba” color which has no points assigned. You could also put slips of parer with point values in the bag, but I like to keep it more engaging and colorful. You can project the category sheet that the entire class can reference, but again I prefer to keep the activity centered in the group, so I provide an individual reference sheet. The plastic frames that can hold a sheet of paper have come in very handy for me with various activities. Students begin by each individually rolling 1 or both dice and attend to the category of the number. If done correctly (group consensus), the student chooses a colored slip out of the bag and keeps a running total of points. He/She puts the slip back in the bag. After a predetermined amount of play time, the “winner” is the students with the highest points. The teacher can easily adapt the speaking to the proficiency level of the students by using the tasks/functions and text types by ACTFL proficiency level. You can learn more about these asks/functions and text types on the ACTFL OPI website. Be sure to download the OPI Familiarization Manual. If the students are at the novice level, they will give one word answers or short phrases, most likely giving an example of something in that category. If they are at the intermediate level they can speak using a series of sentences or be required to ask a question of another player about the topic. If students are at the advanced level they can speak at length in paragraph form. The categories at this level will need to be more complex in nature, perhaps pertaining to world events or characters and plot in a story. The 100th day of school is a very important day in many elementary schools and there are lots of activities to celebrate, all based on the number 100. Each year, I challenge my 3rd graders to list 100 words and expressions that they know in the target language in 20 minutes. I give pairs of students a card with a category and they brainstorm words and expressions. It’s a great way for them to use category words in preparation for circumlocution. We then write the list. I always hold off on using the words for numbers, unless they are needed to reach 100. We did not need to resort to them this year. It is all about the context. Rather than listing words for fruit, ask students to tell you which fruits are their favorite, or to describe the colors. Instead of asking for examples of verbs, have students tell you what they like to do on the weekends with their friends, and follow it up with when and where. Once they communicate in context the words and expressions keep coming. Once you get a hang of the process, creating QR codes to access student recordings is fairly straight forward and students can quickly learn to do it themselves. There are mays ways to use QR codes in the foreign language classroom. One thing I like to do is make the codes available to parents so that they can listen to their kids speaking the language. For example, I made this bulletin board interactive so that the students voices can be heard reading their writing assignment. All it takes is a QR reading app on a smartphone to quickly and instantly hear the student’s voice. Here are the steps for recording audio and creating a QR Code. There are various apps for recording audio and a number of website to create QR codes. These are simply the ones that I use. Record on Recording Lite or Voice Record apps (both available for free). You can use Google drive to upload the audio files or Dropbox . These directions use Dropbox, but Google drive works the same way. Create a folder for the audio. Drag the audio files from email into the dropbox folder. Click on an audio file in dropbox and select share. Copy the URL. Generate a QR link. Use http://www.qrstuff.com/ Paste the URL into the box and QR code will generate to the right. You can download the image or copy the image from the screen.<\|endoftext\|>	3.921875
462	What you eat has a big impact on your overall health, so it’s no surprise that your diet also affects your dental health. Nutrients found in fruits, vegetables, legumes and nuts improve your body’s ability to fight bacteria and inflammation, helping to protect your teeth and gums. Some foods actually help defend against tooth decay in special ways. Recent studies have shown that fresh cranberries can help prevent bacteria from forming plaque. Calcium-fortified juices, milk and other dairy products are rich in calcium and vitamin D, which promotes healthy teeth and bones, and reduces the risk for tooth loss. Cheese unleashes a burst of calcium that mixes with plaque and sticks to your teeth. This protects your teeth from the acid that causes decay and helps rebuild tooth enamel on the spot. Crisp fruits and raw vegetables, like apples, carrots and celery, help clean plaque from the teeth and freshen breath. You may already know that cavity-causing bacteria feed on the sugars in soda, chocolate milk and candy. Sugar is converted to acid, which attacks tooth enamel and causes tooth decay. Did you also know that acidic foods and drinks, such as carbonated beverages, citrus fruits and juices, wine, pickles and honey, can cause tooth enamel to wear away, making teeth sensitive, cracked and discoloured? A diet that promotes good oral health is not just about the foods you eat or avoid. When and how you eat them is equally important. Foods that take a long time to chew or that you hold in your mouth, such as cough drops, cause damage because they hold sugar against teeth longer than other foods. Instead of snacking on sugary, carbohydrate-rich or acidic foods throughout the day, eat these foods just during meal times to minimize the amount of time your teeth are exposed to acid. Regular brushing and flossing can help keep your teeth healthy. If you don’t have a toothbrush handy, chewing sugarless gum that contains xylitol can help reduce plaque and fight cavities. Chewing stimulates saliva, which helps keep teeth clean, while xylitol prevents the growth of bacteria that cause cavities. Following these good practices will help keep your teeth healthy for a lifetime.<\|endoftext\|>	3.84375
661	fdp-4th # In other words you are trading decimal places in one This preview shows page 1. Sign up to view the full content. This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: Move the decimal point LEFT four places on the 40,000 → 4 0.0003 × 40,000 = 3 × 4 = 12 The reason this technique works is that you are multiplying and then dividing by the same power of ten. In other words, you are trading decimal places in one number for decimal places in another number. This is just like trading decimal places for powers of ten, as we saw earlier. Manhattan GMAT Prep * 18 the new standard DIGITS & DECIMALS STRATEGY Chapter 1 DIVISION If there is a decimal point in the dividend (the inner number) only, you can simply bring the decimal point straight up to the answer and divide normally. Ex. 12.42 ÷ 3 = 4.14 4.14 3 12.42 12 04 3 12 However, if there is a decimal point in the divisor (the outer number), you should shift the decimal point in both the divisor and the dividend to make the divisor a whole number. Then, bring the decimal point up and divide. Ex: 12.42÷ 0.3 → 124.2 ÷ 3 = 41.4 41.4 3 124.2 12 04 3 12 Move the decimal one space to the right to make 0.3 a whole number. Then, move the decimal one space in 12.42 to make it 124.2. Remember, in order to divide decimals, you must make the OUTER number a whole number by shifting the decimal point. Division: Divide by whole numbers! You can always simplify division problems that involve decimals by shifting the decimal point in the same direction in both the divisor and the dividend, even when the division problem is expressed as a fraction: 0.0045 45 ⎯ =⎯ 0.09 900 Move the decimal 4 spaces to the right to make both the numerator and the denominator whole numbers. Note that this is essentially the same process as simplifying a fraction. You are simply multiplying the numerator and denominator of the fraction by a power of ten—in this case, 10 4, or 10,000. Keep track of how you move the decimal point! To simplify multiplication, you can move decimals in opposite directions. But to simplify division, you move decimals in the same direction. Equivalently, by adding zeroes, you can express the numerator and the denominator as the same units, then simplify: 0.0045 5 0.0045 45 ⎯ = ⎯ = 45 ten thousandths ÷ 900 ten–thousandths = ⎯... View Full Document ## This note was uploaded on 02/07/2014 for the course MIS 304 taught by Professor Mejias during the Spring '07 term at Arizona. Ask a homework question - tutors are online<\|endoftext\|>	4.71875
1,735	Edit Article # wikiHow to Find the Degree of a Polynomial Polynomial means "many terms," and it can refer to a variety of expressions that can include constants, variables, and exponents. For example, x - 2 is a polynomial; so is 25. To find the degree of a polynomial, all you have to do is find the largest exponent in the polynomial.[1] If you want to find the degree of a polynomial in a variety of situations, just follow these steps. ### Part 1 a Polynomial with One Variable or Fewer 1. 1 Combine like terms. Combine all of the like terms in the expression so you can simplify it, if they are not combined already. Let's say you're working with the following expression: 3x2 - 3x4 - 5 + 2x + 2x2 - x. Just combine all of the x2, x, and constant terms of the expression to get 5x2 - 3x4 - 5 + x. 2. 2 Drop all of the constants and coefficients. The constant terms are all of the terms that are not attached to a variable, such as 3 or 5. The coefficients are the terms that are attached to the variable. When you're looking for the degree of a polynomial, you can either just actively ignore these terms or cross them off. For instance, the coefficient of the term 5x2 would be 5. The degree is independent of the coefficients, so you don't need them. • Working with the equation 5x2 - 3x4 - 5 + x, you would drop the constants and coefficients to get x2 - x4 + x. 3. 3 Put the terms in decreasing order of their exponents. This is also called putting the polynomial in standard form.[2]. The term with the highest exponent should be first, and the term with the lowest exponent should be last. This will help you see which term has the exponent with the largest value. In the previous example, you would be left with -x4 + x2 + x. 4. 4 Find the power of the largest term. The power is simply number in the exponent. In the example, -x4 + x2 + x, the power of the first term is 4. Since you've arranged the polynomial to put the largest exponent first, that will be where you will find the largest term. 5. 5 Identify this number as the degree of the polynomial. You can just write that the degree of the polynomial = 4, or you can write the answer in a more appropriate form: deg (3x2 - 3x4 - 5 + 2x + 2x2 - x) = 3. You're all done.[3] 6. 6 Know that the degree of a constant is zero. If your polynomial is only a constant, such as 15 or 55, then the degree of that polynomial is really zero. You can think of the constant term as being attached to a variable to the degree of 0, which is really 1. For example, if you have the constant 15, you can think of it as 15x0, which is really 15 x 1, or 15. This proves that the degree of a constant is 0. ### Part 2 a Polynomial with Multiple Variables 1. 1 Write the expression. Finding the degree of a polynomial with multiple variables is only a little bit trickier than finding the degree of a polynomial with one variable. Let's say you're working with the following expression: • x5y3z + 2xy3 + 4x2yz2 2. 2 Add the degree of variables in each term. Just add up the degrees of the variables in each of the terms; it does not matter that they are different variables. Remember that the degree of a variable without a written degree, such as x or y, is just one. Here's how you do it for all three terms:[4] • x5y3z = 5 + 3 + 1 = 9 • 2xy3 = 1 + 3 = 4 • 4x2yz2 = 2 + 1 + 2 = 5 3. 3 Identify the largest degree of these terms. The largest degree of these three terms is 9, the value of the added degree values of the first term. 4. 4 Identify this number as the degree of the polynomial. 9 is the degree of the entire polynomial. You can write the final answer like this: deg (x5y3z + 2xy3 + 4x2yz2) = 9. ### Part 3 a Rational Expression 1. 1 Write down the expression. Let's say you're working with the following expression: (x2 + 1)/(6x -2).[5] 2. 2 Eliminate all coefficients and constants. You won't need the coefficients or constant terms to find the degree of a polynomial with fractions. So, eliminate the 1 from the numerator and the 6 and -2 from the denominator. You're left with x2/x. 3. 3 Subtract the degree of the variable in the denominator from the degree of the variable in the numerator. The degree of the variable in the numerator is 2 and the degree of the variable in the denominator is 1. So, subtract 1 from 2. 2-1 = 1. 4. 4 Write the result as your answer. The degree of this rational expression is 1. You can write it like this: deg [(x2 + 1)/(6x -2)] = 1. ## Community Q&A Search • How do I find the degree of the polynomials and the leading coefficients? The degree is the same as the highest exponent appearing in the polynomial. Coefficients do not have degrees. • How do I find the degree of a polynomial that is (x^2 -2)(x+5)=0? The degree is the same as the highest exponent appearing in the final product, so you just multiply the two factors and you'll wind up with x³ as one of the terms in the product. That's the highest exponent in the product, so 3 is the degree of the polynomial. • What is the degree of a polynomial? The degree is the same as the highest exponent appearing in the polynomial. • How do you solve p(x)=3x7-6x5+4x3-x2-5? • How to find proper and improper fractions? 200 characters left ## Tips • This just shows the steps you would go through in your mind. You don't have to do this on paper, though it might help the first time. If you do it on paper, however, you won't make a mistake. • By convention, the degree of the zero polynomial is generally considered to be negative infinity. • For the third step, linear terms like x can be written as x1 and non-zero constant terms like 7 can be written as 7x0 ## Article Info Categories: Algebra In other languages: Español: encontrar el grado de un polinominal, Português: Encontrar o Ângulo de Um Polinômio, Italiano: Calcolare il Grado di un Polinomio, Français: déterminer le degré d'un polynôme, Deutsch: Den Rang eines Polynoms ermitteln, Русский: найти порядок многочлена, Nederlands: De graad van een polynoom bepalen, Bahasa Indonesia: Menentukan Tingkat Polinomial, 中文: 求多项式的次数, हिन्दी: एक बहुपद की घात (Degree of a Polynomial) पता करें, Tiếng Việt: Tính Bậc của Đa thức, ไทย: หาดีกรีของพหุนาม Thanks to all authors for creating a page that has been read 257,337 times.<\|endoftext\|>	4.84375
562	Horses Have Trotted Through Thousands of Years of Human History by Rose Loos-Austin, Age 12 history of horses is a long and interesting story that started around farming communities sprung up starting in the southwest part of central Asia and the edges of Europe where conditions permitted. Early farmers spent their days growing crops and herding livestock. These early farmers were some of the first to domesticate horses. At first, horses were raised for slaughter. It wasn't until about 500 years later, around 4000 B.C.E., when horses were used as transport for the first time. Soon after the first wheeled vehicles arose. Horses were selectively bred for desirable features, such as strength, so they could pull these vehicles. By the third millennium, horses were strong enough to pull whole war chariots. With the addition of spoked wheels, chariots were more maneuverable; those who had chariots on their side dominated the battlefield. 2500 B.C.E., the Surabaya people invented bits and bridles which allowed them to ride horses. This helped shepherds herd livestock over large areas of land and paved the way for the rise of nomadic pastoralism in the first millennium (1000 B.C.E.) This way of life would engulf the area for thousands of years. The nomads lived in tents that could easily be disassembled and transported. These tents were made out of animal hide and called yurts. Herds of sheep, horses, camels and goats were moved between the summer and winter pastures, greatly increasing the available land The Cimmerian people of Russia were the first known nomadic power. They rose to prominence in the 900's. Around 200 years later they bumped into the Scythians, another nomadic people, who migrated into the Cimmerian territory in the 700's. The Scythians went to war with the Cimmerians for around thirty years before driving them into Anatolia. Here, the Cimmerians defeated the Phrygia, and Sardis the capital of Lydia. In about 626 they were fought off by the Lydians. The Cimmerians ceased to be mentioned in historical sources after possibly settling in The Scythians built a powerful empire stretching from Persian borders to southern Russia. Horse-riding was the key to their military success. It gave them a speed and mobility that amazed their enemies. They were so skilled on horseback that they stopped an invasion by the infamous Persian army in 513 B.C. Short History of the World]<\|endoftext\|>	3.71875
1,258	# Volume of a Cylinder In this section, we will learn the formula of the volume of a cylinder and how to find the volume of a cylinder along with proper examples. A cylinder is a 3D geometrical shape with the two-circular base. It has two circular bases, one at top and the other at the bottom. We can also define a cylinder as an arrangement of circular disks in stacked form. There are two types of cylinders: • Cylinder or Solid Cylinder • Hollow Cylinder ### Definition The number of cubic units that will exactly fill a cylinder is called the volume of the cylinder. In other words, the space covered by a cylinder is called the volume of a cylinder. ### The Volume of Cylinder Formula It is the product of the area of base and height of the cylinder. V=πr2 h ### Volume of a Hollow Cylinder A cylinder that is hollow from inside is called a hollow cylinder. There are two radii in the hollow cylinder. One for the inner cylinder and the other for an outer cylinder formed by the base. Suppose, r1 is the radius of the outer circle, r2 is the radius of the inner circle, and h is the height of the cylinder, then the volume of the hollow cylinder will be: ### How to Find Volume of a Cylinder We can find the volume of a cylinder by multiplying the area of a circle by the height of the cylinder. We know the formula of the area of a circle: The Area of a Circle (A)= πr2 Multiply the area of a circle by the height of the cylinder, we get the volume of the cylinder. Since, The Volume of a Cylinder (V)= πr2 h Where: π: It is a constant whose value is 3.142 or 22/7. r: It is the radius of the cylinder. h: It is the height of the cylinder. ### Unit of Volume The unit of the volume is a cubic unit or unit3. For example, if the radius and height are given in centimeters, the volume will also in centimeters, and the unit will be cubic centimeters or cm3. Let's see some examples. Example 1: The radius of a cylinder is 5 cm, and the height is 12 cm. Calculate the volume of the cylinder. Take π=. Solution: Given, radius (r) = 5 cm Height (h) = 12 cm π= Volume (V) =? We know the formula of volume of a cylinder: V= πr2 h Putting the values in the above formula, we get: Hence, the volume of the cylinder is 942.85 cm3. Example 2: Calculate the volume of a cylinder whose radius is 3 cm and the height is 6 cm. (π=3.14) Solution: Given, radius (r) = 3 cm Height (h) = 6 cm π=3.14 Volume (V) =? We know the formula of volume of a cylinder: V= πr2 h Putting the values in the above formula, we get: V=3.14×(32)×6 V=3.14×9×6 V=3.14×54 V=169.56 Hence, the volume of the cylinder is 169.56 cm3. Example 3: What is the volume of the cylinder given below. Solution: Given, radius (r) = 4.5 cm Height (h) = 8 cm Volume (V) =? We know the formula of volume of a cylinder: V= πr2 h Putting the values in the above formula, we get: V=3.14×(4.52)×8 V=3.14×20.25×8 V=508.68 Hence, the volume of the cylinder is 508.68 cm3. Example 4: The volume of a cylinder is 255 cm3 and the height is 15 cm. Find the radius (r) of the cylinder. Solution: Given, Volume (V) = 255 cm3 Height (h) = 15 cm π=3.14 We know the formula of volume of a cylinder: V= πr2 h Putting the values in the above formula, we get: Hence, the radius of the cylinder is 2.3 cm. Example 5: Find the volume of the hollow cylinder. Solution: Given, radius of outer cylinder (r1) = 2.4 cm radius of inner cylinder (r2) = 2 cm Height (h) = 10 cm π=3.14 Volume (V) =? We know the formula of volume of the hollow cylinder: Putting the values in the above formula, we get: V=3.14×10×(2.42-22) V=3.14×10×(5.76-4) V=3.14×10×(1.76) V=55.264 Hence, the volume of the hollow cylinder is 55.264 cm3. Example 6: The outer and inner radius of a pipe is 8 and 6 cm, respectively. The height of the pipe is 15 cm. Find the volume of the pipe. Take pi=3.14. Solution: Given, radius of outer cylinder (r1) = 8 cm radius of inner cylinder (r2) = 6 cm Height (h) = 15 cm π=3.14 Volume (V) =? We know the formula of volume of the hollow cylinder: Putting the values in the above formula, we get: V=3.14×15×(82-62) V=3.14×15×(64-36) V=3.14×15×2 V=1318.8 Hence, the volume of the hollow cylinder is 1318.8 cm3.<\|endoftext\|>	4.71875
1,247	# Characteristic function The characteristic function (cf) is a complex function that completely characterizes the distribution of a random variable. ## How it is used The use of the characteristic function is almost identical to that of the moment generating function: 1. it can be used to easily derive the moments of a random variable; 2. it uniquely determines its associated probability distribution; it is often used to prove that two distributions are equal. The cf has an important advantage over the moment generating function: while some random variables do not possess the latter, all random variables have a characteristic function. ## Definition We start this lecture with a definition of characteristic function. Definition Let be a random variable. Let be the imaginary unit. The function defined byis called the characteristic function of . The first thing to be noted is that exists for any . This can be proved as follows:and the last two expected values are well-defined, because the sine and cosine functions are bounded in the interval . ## Deriving moments with the characteristic function Like the moment generating function of a random variable, the characteristic function can be used to derive the moments of , as stated in the following proposition. Proposition Let be a random variable and its cf. Let . If the -th moment of , denoted by , exists and is finite, then is times continuously differentiable andwhere is the -th derivative of with respect to , evaluated at the point . Proof The proof of this proposition is quite complex (see, e.g., Resnick 2013) and we give here only a sketch, without taking technical details into consideration. By virtue of the linearity of the expected value and of the derivative operator, the derivative can be brought inside the expected value, as follows:When , the latter becomes In practice, the proposition above is not very useful when one wants to compute a moment because it requires to know in advance whether the moment exists or not. A much more useful proposition is the following. Proposition Let be a random variable and its characteristic function. If is times differentiable at the point , then 1. if is even, the -th moment of exists and is finite for any ; 2. if is odd, the -th moment of exists and is finite for any . In both cases,where is the -th derivative of with respect to , evaluated at the point . Proof See, e.g., Ushakov (1999). The next example shows how this proposition can be used to compute the second moment of an exponential random variable. Example Let be an exponential random variable with parameter . Its support is the set of positive real numbers:and its probability density function isIts cf iswhich is proved in the lecture entitled Exponential distribution. Note that the division above does not pose any division-by-zero problem, because the denominator is different from also when (because ). The first derivative of the cf isThe second derivative of the cf isEvaluating it at , we obtainTherefore, the second moment of exists and is finite. Furthermore, it can be computed as ## Characterization of a distribution via the characteristic function Characteristic functions, like moment generating functions, can also be used to characterize the distribution of a random variable. Proposition Let and be two random variables. Denote by and their distribution functions and by and their cfs. Then, and have the same distribution, i.e., for any , if and only if they have the same cf, i.e., for any . Proof See, e.g., Resnick 2013. In applications, this proposition is often used to prove that two distributions are equal, especially when it is too difficult to directly prove the equality of the two distribution functions and . ## More details The following sections contain more details about the characteristic function. ### Characteristic function of a linear transformation Let be a random variable with cf . Definewhere are two constants and . Then, the cf of is Proof Using the definition of cf, we obtain ### Characteristic function of a sum of mutually independent random variables Let , ..., be mutually independent random variables. Let be their sum: Then, the cf of is the product of the cfs of , ..., : Proof It can be demonstrated as follows: ### Computation of the characteristic function When is a discrete random variable with support and probability mass function , its cf is Thus, the computation of the characteristic function is pretty straightforward: all we need to do is to sum the complex numbers over all values of belonging to the support of . When is a continuous random variable with probability density function , its cf is The right-hand side integral is a contour integral of a complex function along the real axis. As people reading these lecture notes are usually not familiar with contour integration (a topic in complex analysis), we avoid it altogether and instead exploit the fact thatto rewrite the contour integral as the complex sum of two ordinary integrals:and to compute the two integrals separately. ### Multivariate generalization The multivariate generalization of the cf is presented in the lecture on the joint characteristic function. ## Solved exercises Below you can find some exercises with explained solutions. ### Exercise 1 Let be a discrete random variable having supportand probability mass function Derive the characteristic function of . Solution By using the definition of characteristic function, we get ### Exercise 2 Use the characteristic function found in the previous exercise to derive the variance of . Solution We can use the following formula for computing the variance:The expected value of is computed by taking the first derivative of the characteristic function:evaluating it at and dividing it by :The second moment of is computed by taking the second derivative of the characteristic function:evaluating it at and dividing it by :Therefore, ### Exercise 3 Read and try to understand how the characteristic functions of the uniform and of the exponential distributions are derived in the lectures entitled Uniform distribution and Exponential distribution. ## References Resnick, S. I. (2013) A Probability Path, Birkhauser. Ushakov, N. G. (1999) Selected topics in characteristic functions, VSP.<\|endoftext\|>	4.53125
1,824	Congruence Criterion: Side, Side, Side 🏆Practice side, side, side There are 4 criteria to determine that two triangles are congruent. In this article, we will learn to use the third criterion of congruence: Side, Side, Side Definition: 2 triangles in which their three sides are of the same length are congruent triangles. In this article, we will study this criterion and see examples of how to apply it. Test yourself on side, side, side! Look at the parallelogram ABCD below. What can be said about triangles ACD and ABD? Definition: Two triangles in which all three sides are of the same length are congruent triangles. To prove that two triangles are congruent we can use one of the following postulates: Example 1 Given the triangles $Δ ABC$ and $Δ DEF$ such that $AB = DE$ (edge) $BC = EF$ (edge) $AC = DF$ (edge) Therefore, we can deduce that: $Δ ABC$ and $Δ DEF$ are congruent triangles according to the Side, Side, Side congruence criterion. We will write it as follows: $Δ DEF ≅ Δ ABC$ according to the congruence criterion: Side, Side, Side (SSS) From this we can also deduce that: $∠A = ∠D$ $∠B = ∠E$ $∠C = ∠F$ since these are corresponding angles and are equal in congruent triangles Join Over 30,000 Students Excelling in Math! Endless Practice, Expert Guidance - Elevate Your Math Skills Today Example 2: Triangle Congruence Exercise Given the two triangles $Δ ABC$ and $Δ ACD$ such that $AC$ is the common side. We are also informed that: $AB=DA$ $DC=CB$ Prove that the triangles $ΔABC$ and $ΔACD$ are congruent triangles. Proof: We will base our proof on the criterion we just learned. Let's see $AC=DC$ (side) $AB=DB$ (side) We realize that $AC$ (side) is common to both triangles From this, it follows that in both triangles $Δ ABC$ and $Δ ADC$ there are three pairs of equal sides. Consequently, we can deduce that $Δ ADC ≅ Δ ABC$ according to the Side, Side, Side congruence criterion. $QED$ If you're interested in this article, you might also be interested in the following articles: Congruence Criterion: Side, Angle, Side Congruence Criterion: Angle, Side, Angle Side, Side, and the Angle Opposite the Larger of the Two Sides The Method of Writing a Formal Proof in Geometry On the Tutorela blog, you'll find a variety of articles on mathematics. Congruence Criteria Exercises: Side, Side, Side Exercise 1 Assignment In the given figure: $EC=EB$ $AC=AB$ By what theorem are the triangles $ΔABE≅ΔACE$ congruent? Solution Since $EC=EB$ Since $AC=AB$ Common side $AE=AE$ The triangles are congruent by $SSS$ Congruent by $SSS$ Do you know what the answer is? Exercise 2 Assignment In an isosceles triangle $\triangle ABC$ we draw the height $AK$. According to which theorem of congruence do the triangles $ΔABK≅ΔACK$ overlap? Solution $AB=AC$ Since triangle $ABC$ is isosceles $BK=KC$ In an isosceles triangle, the height is also a median, and a median cuts the base into two equal parts. $AK=AK$ Common side The triangles overlap according to $S.S.S$ Overlap according to $S.S.S$ Exercise 3 Assignment The segments $BE$ and $AC$ intersect at point $D$. Which congruence theorem explains why the triangles $ΔABD≅ΔCED$ are congruent? Solution $BE$ and $AC$ Intersect at point $D$ $AD=DC$ $D$ intersects $BE$ $\angle ADB=\angle EDC$ Vertically Opposite Angles Congruent triangles by $S.A.S$ Congruent by $S.A.S$ Exercise 4 Assignment The triangles $ΔABC≅ΔEFG$ In triangle $ΔABC$ we draw the median $AD$ and in triangle $ΔEFG$ we draw the median $EH$. We demonstrate: $ΔADB≅ΔEHF$ Solution $AB=EF$ Given that triangles $ΔABC$ and $ΔEFG$ are congruent $AD=EH$ In congruent triangles, the medians are necessarily equal (coming from the same vertex to the same base) $BD=FH$ The median bisects the base it reaches. Congruent triangles by $S.S.S$ Congruent according to $S.S.S$ Exercise 5 Assignment Given the isosceles trapezoid $ABCD$. Inside it contains the square $ABFE$. According to which theorem are the triangles $ΔADE≅ΔBCF$ congruent? Solution $ABCD$ is an isosceles trapezoid (given) $AD=BC$ Isosceles trapezoid Since $ABFE$ is a square $AE=BF$ Since $ABFE$ is a square and all sides in a square are equal $\sphericalangle D=\sphericalangle C$ The base angles in an isosceles trapezoid are equal $\sphericalangle AED=\sphericalangle BFC=90°$ In a square, all angles are right angles and measure $90°$ degrees $\sphericalangle DAE=\sphericalangle FBC$ if two angles are equal then the third is also equal The triangles are congruent according to $S.A.S$ $S.A.S$ Do you think you will be able to solve it? Review Questions What is the congruence criterion for two triangles? There are four triangle congruence criteria, which allow us to determine if two triangles have the same lengths in their sides and likewise the same degrees in their corresponding angles. In this way, we can say that the two triangles, even when they are in different positions or orientations, will have the same shape and size. What is the SSS congruence criterion? This criterion allows us to deduce if two triangles have the same shape and size. According to this criterion, two triangles are congruent when their three sides are equal. What is the difference between the SSS congruence criterion and the SSS similarity criterion? The SSS congruence criterion tells us that if two triangles have their three sides equal (congruent sides), then the two triangles are identical, meaning they have the same measurements in terms of sides and angles. Whereas the SSS similarity criterion tells us that if two triangles are similar, then their three sides are proportional, meaning they do not have the same measurement but they do have some proportion between them and they have the same shape, but with different measurements in terms of their sides. Which pair of triangles are similar by the SSS criterion? Two triangles will be similar when they have the same shape, regardless of orientation, that is, their corresponding angles are equal but their corresponding sides do not necessarily have the same length, instead, they must have a proportion between them. What are the criteria for similarity and congruence of triangles? Congruence criteria The four triangle congruence criteria are: • SAS - Side, Angle, Side. • ASA - Angle, Side, Angle. • SSS - Side, Side, Side. • SSA - Side, Side, Angle. Similarity criteria Unlike the congruence criteria, there are only three triangle similarity criteria: • SSS - Side, Side, Side. • SAS - Side, Angle, Side. • AAA - Angle, Angle, Angle. examples with solutions for side, side, side Exercise #1 Look at the parallelogram ABCD below. What can be said about triangles ACD and ABD? Step-by-Step Solution According to the side-angle-side theorem, the triangles are similar and coincide with each other: AC = BD (Any pair of opposite sides of a parallelogram are equal) Angle C is equal to angle B. AB = CD (Any pair of opposite sides of the parallelogram are equal) Therefore, all of the answers are correct.<\|endoftext\|>	4.8125
718	Maths Resources GCSE Worksheets # Rearranging Formulae Worksheet • Section 1 of the rearranging formulae worksheet contains 20+ skills-based rearranging formulae questions, in 3 groups to support differentiation • Section 2 contains 3 applied rearranging formulae questions with a mix of worded problems and deeper problem solving questions • Section 3 contains 3 foundation and higher level GCSE exam questions on rearranging formulae • Answers and a mark scheme for all rearranging formulae questions are provided • Questions follow variation theory with plenty of opportunities for students to work independently at their own level • All questions created by fully qualified expert secondary maths teachers • Suitable for GCSE maths revision for AQA, OCR and Edexcel exam boards • This field is for validation purposes and should be left unchanged. You can unsubscribe at any time (each email we send will contain an easy way to unsubscribe). To find out more about how we use your data, see our privacy policy. ### Rearranging formulae at a glance We can rearrange a formula, or change the subject of the formula, so that one unknown is alone on the left-hand side of the formula and is equal to the rest of the formula. This allows us to work out the value of that unknown, when we know the other values. We rearrange formulae by applying inverse functions. For example, if we have the formula for the circumference of a circle, C=π times d, we can rearrange the formula to make d the subject, which will help us to work out the diameter, given the circumference of a circle. At the moment, d is multiplied by π. To move π to the other side of the equation, we apply the inverse of this, which is divide by π. If we divide both sides by π we get C divided by π = d which can be written as d = C divided by π. We have now made π the subject of the formula. It is important to remember that the inverse function to the square of a number is a square root. The square root of a number can be written as the number, raised to the power of a half. The square, and the square root of a number are both powers (indices). This is also true for the cube, and the inverse, the cube root. Common equations used to test knowledge of rearranging formulae are: area of 2D shapes, surface area / volume of 3D shapes, converting units such as celsius to fahrenheit, and equations of a straight line. Looking forward, students can then progress to additional rearranging equations worksheets and other algebra worksheets, for example a sequences worksheet, simultaneous equations worksheet or straight line graphs worksheet. For more teaching and learning support on Algebra our GCSE maths lessons provide step by step support for all GCSE maths concepts. ## Do you have students who need additional support to achieve their target GCSE maths grade? There will be students in your class who require individual attention to help them succeed in their maths GCSEs. In a class of 30, it’s not always easy to provide. Help your students feel confident with exam-style questions and the strategies they’ll need to answer them correctly with personalised online one to one tutoring from Third Space Learning Lessons are selected to provide support where each student needs it most, and specially-trained GCSE maths tutors adapt the pitch and pace of each lesson. This ensures a personalised revision programme that raises grades and boosts confidence. GCSE Revision Programme<\|endoftext\|>	4.71875
612	At middle school, each Inquiry By Design course of study is made up of seven units featuring detailed, but flexible, lesson plans aimed at building critical, standards-aligned, grade-level literacy skills. Each unit is built around challenging, grade-appropriate texts and spiraling cycles of work marked by reading, writing, and collaborative small-group and whole-class discussion. All grades of Inquiry By Design’s middle school course of study feature - Independent reading initiatives focused on selecting books, setting goals, and tracking progress. - A literacy notebook and independent writing routines that nestle alongside inquiries into the craft of writing. - Error journal work that supports and reinforces grade-appropriate language conventions by having students identify, track, and solve grammatical errors in their own writing. - Compelling short fiction as a basis for in-depth orientation to the process of developing and writing text-based arguments. - Engaging informational texts to help students learn to determine and analyze the development of central ideas and arguments. - Challenging literary nonfiction to build the strategies, skills, and tenacity to deal with difficulty head on. - Detective fiction, essays, and reportage that push students beyond superficial comprehension into the realm of inference, analysis, and inquiry. - Multi-text studies that invite students to participate in conversations about important ideas as they take on significant narrative or informative/explanatory writing projects. - Repeated, supported summary writing practice opportunities to ensure growth in this critical form of writing. Grade 6 Units - Introduction to the Reading Life - Introduction to the Writing Life - Introduction to Interpretive Work (Bambara and Baxter) - Story and the Brain (Informational Texts) - Dealing With Difficulty (Doyle and Dickinson) - Reading Nonfiction Like a Detective - Reading and Writing Fairy Tales Optional add-on: How Poems Are Built Grade 7 Units - Reading as Problem Solving - Writers on Writing: Advice and Publication in the Writing Life - Introduction to Interpretive Work (Hughes and Jackson) - Texting and Language Change (Informational Texts) - Dealing With Difficulty (cummings and Dillard) - Investigative Report Writing: Explanations and Arguments - Growing Up: Thinking With Literature Optional add-on: Creating Characters in Poetry: A Study of Two Poets Grade 8 Units - Studying Craft: Apprenticeship and Independent Writing - Vantage Points: A Reading Life Study - Introduction to Interpretive Work (Updike and Rivera) - Superstitions, Patterns, and Control (Informational Texts) - Dealing With Difficulty (Reed and Thomas) - Metaphorically Speaking: Reading Nonfiction Like a Cognitive Linguist - Faces of the Essay: An Orientation to the Form *Optional add-on: Poems as Puzzles: A Pair of Poets Study<\|endoftext\|>	3.828125
292	Class I Navigation Class I navigation refers to any flight being operated in controlled or uncontrolled airspace and typically requires the use of an aeronautical chart. The chart is used as a roadmap for pilots operating under visual flight rules (VFR). An flight that’s en route being operated under VFR by pilotage is conducting class I navigation. Class I navigation also includes instrument flight rules (IFR) or VFR navigational operations on the following: - In the US, published IFR off-airway routes - In the US, published IFR direct routes - Federal airways - Advisory routes, airways, indirect, and direct routes approved or published by a foreign government that fall within the operational service volume (or equivalent) of ICAO standard NAVAIDs. Pilotage in the Skies Pilotage is simply defined as the act of piloting. However, it’s more commonly used to describe the act of navigating an aircraft using landmarks and references. It also includes the use of an aeronautical chart to help obtain its positioning. Pilotage is practiced under visual meteorological conditions (VMC) acceptable for flight. In order to successfully navigate an aircraft when relying on landmarks, the pilot must consider altitude. Flying too high can make spotting landmarks difficult. Flying too low could result in collisions with tall structures. This could result in what is known as a controlled flight into terrain.<\|endoftext\|>	3.890625
305	The rapid growth of the internet and social media has provided a new platform for bullying, although traditional forms continue to exist. Cyber-bullying is the term used to describe any aggressive, intentional act, behaviour or communication undertaken an individual or group, using electronic and digital means against a vulnerable victim, repeatedly and over time. The impact of bullying has been the subject of extensive research over many years in all countries. Bullying occurs throughout the world and can occur at many life stages, from childhood to adolescence and adulthood, in private, educational and work settings. Bullying is a distinctive pattern of harm and humiliation of others, especially seen as smaller, weaker, younger or in any way more vulnerable than the bully. The report produced is the result of the Research and Analysis of School Needs conducted by Universal Learning Systems during the first year of the project and forms the first stage of the Intellectual Output 1 of APPs project. The report summarizes key factors involved and provides a comparative analysis based on research undertaken in four countries (Spain, Italy, Poland and Romania) and in seven schools. In addition, the paper identifies innovative learning strategies, digital resources and detailed findings from surveys of teachers, students and parents that offer techniques and actions for educational establishments to ameliorate this phenomenon. This is designed to identify and pinpoint critical issues involved in developing evidence-based responses to cyber-bullying in European schools and it represents the an analytical framework that will be used in the development of the Vertical Learning Modules (VLM).<\|endoftext\|>	3.796875
758	In Egypt’s Valley of the Kings, British archaeologists Howard Carter and Lord Carnarvon become the first souls to enter King Tutankhamen’s tomb in more than 3,000 years. Tutankhamen’s sealed burial chambers were miraculously intact, and inside was a collection of several thousand priceless objects, including a gold coffin containing the mummy of the teenage king. When Carter first arrived in Egypt, in 1891, most of the ancient Egyptian tombs had been discovered, and the majority of these had been hopelessly plundered by tomb raiders over the millennia. However, Carter was a brilliant excavator, and in the first years of the 20th century he discovered the tombs of Queen Hatshepsut and King Thutmose IV. Around 1907, he became associated with the Earl of Carnarvon, a collector of antiquities who commissioned Carter to supervise excavations in the Valley of the Kings. By 1913, most experts felt there was nothing in the Valley left to be uncovered. Carter, however, persisted in his efforts, convinced that the tomb of the little-known King Tutankhamen might still be found. King Tutankhamen was enthroned in 1333 B.C. when he was still a child. He died a decade later at the age of 18 and thus made only a faint impression on the history of ancient Egypt. In the 13th century B.C., Tutankhamen and the other “Amarna” kings were publicly condemned, and most records of them were destroyed–including the location of Tutankhamen’s tomb. A century later, in the 12th century B.C., workers building a tomb for Ramses VI inadvertently covered Tutankhamen’s tomb with a deep layer of chips, further protecting it from future discovery. After World War I, Carter began an intensive search for Tutankhamen’s tomb and on November 4, 1922, discovered a step leading to its entrance. Lord Carnarvon rushed to Egypt, and on November 23 they broke through a mud-brick door, revealing the passageway that led to Tutankhamen’s tomb. There was evidence that robbers had entered the structure at some point, and the archaeologists feared they had discovered yet another pillaged tomb. However, on November 26 they broke through another door, and Carter leaned in with a candle to take a look. Behind him, Lord Carnarvon asked, “Can you see anything?” Carter replied, “Yes, wonderful things.” It was the antechamber of Tutankhamen’s tomb, and it was gloriously untouched. The dusty floor still showed the footprints of the tomb builders who left the room more than 3,000 years before. Apparently, the robbers who had broken into Tutankhamen’s tomb had done so soon after it was completed and were caught before moving into the interior chambers and causing serious damage. Thus began a monumental excavation process in which Carter carefully explored the four-room tomb over several years, uncovering an incredible collection of several thousand objects. In addition to numerous pieces of jewelry and gold, there was statuary, furniture, clothes, a chariot, weapons, and numerous other objects that shed a brilliant light on the culture and history of ancient Egypt. The most splendid find was a stone sarcophagus containing three coffins nested within each other. Inside the final coffin, made out of solid gold, was the mummified body of the boy-king Tutankhamen, preserved for 3,200 years. Most of these treasures are now housed in the Cairo Museum.<\|endoftext\|>	3.734375
569	Courses Courses for Kids Free study material Offline Centres More Store # A simple harmonic motion has an amplitude $A$ and time period $T$ . Find the time required by it to travel directly from $x = 0$ to $x = \dfrac{A}{2}$ . Last updated date: 20th Jun 2024 Total views: 54.9k Views today: 0.54k Verified 54.9k+ views Hint: Use the formula of the displacement of the simple harmonic motion and substitute the displacement in it. From the simplified relation substitute the angular frequency formula to find the time required to travel the displacement of half the amplitude. Useful formula: (1) The formula of the displacement of the simple harmonic motion is given by $x = A\sin \omega t$ Where $x$ is the displacement of the wave, $A$ is the amplitude of the simple harmonic motion and $\omega$ is the angular frequency of the wave and $t$ is the time taken for the displacement. (2) The formula of the angular frequency of the simple harmonic motion is given by $\omega = \dfrac{{2\pi }}{T}$ Where $\omega$ is the angular frequency of the wave and $T$ is the time period of the wave. Complete step by step solution: It is given that the amplitude and the time period of the simple harmonic motion is $A$ and $T$ respectively. Initial position is $x = 0$ and the final position is $x = \dfrac{A}{2}$ Using the formula of the displacement of the simple harmonic motion, $x = A\sin \omega t$ Substituting the displacement in the above equation, we get $\dfrac{A}{2} = A\sin \omega t \\ \omega t = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\ \omega t = \dfrac{\pi }{6} \\$ Substituting the formula of the angular frequency in the above step, we get $\dfrac{{2\pi }}{T}t = \dfrac{\pi }{6}$ By cancelling the similar terms in the above equation, $t = \dfrac{T}{{12}}$ Hence the time required to travel from one position to the other is $\dfrac{T}{{12}}$ . Note: In the simple harmonic motion, the restoring force is equal to the object magnitude and it is the periodic motion. The sine wave is an example of this type of motion. Time period is the reciprocal of the angular frequency. Remember the formula of the displacement of simple harmonic motion.<\|endoftext\|>	4.625
1,338	To understand that all species have some capacity for communication. The focus of this lesson is threefold. First, to expose students to the fact that all species have a capacity for communication. Second, to enlighten students to the fact that communication abilities range from very simple to extremely complex, depending upon the species. Third, to realize that communication is influenced by a species' genetic makeup, its environment, and the numerous ways by which animals and humans respond to and adapt to their surroundings. At the end of elementary school, students should know that learning means using what one already knows to make sense out of new experiences or information, not just storing new information in one's head. In addition, they should understand that unlike human beings, behavior in insects and other species is determined almost entirely by biological inheritance. This prerequisite knowledge should help middle-school students understand that some animal species are limited to a repertoire of genetically determined behaviors whereas others have more complex brains that enable them to learn a wide variety of behaviors. Further, students should realize that specialized roles within species are genetically programmed, whereas human beings are able to invent and modify a wider range of social behavior. The study of animal communication requires a broader set of perspectives than nearly any other topic in biology because it involves both simple (at the cellular level) and complex (at the neural level) concepts. For example, the importance of basic communication within an organism is seen at the cellular level by means of simple chemical reactions. Conversely, the importance of communication between organisms is seen at the neural level by means of complex speech patterns. Important disciplines surrounding the topic of animal communication include biophysics, chemistry, physiology, pharmacology, neurobiology, cognitive science, evolutionary biology, and behavioral ecology. In this section, students will do two activities that will spark their thinking about animal and human communication. These activities may take an entire class period depending upon the time limitations you set forth. Write the following sentence on the board and ask students to think about what it means as they do the two activities: Although all species have some capacity for communication, there is a range of abilities among species. Using the Animal Communication student esheet, students will go to the Orangutan U Science Update, read the transcript, and then read the research. They should answer the questions about this Science Update on the Animal Communication student sheet. Then, as a class, discuss the questions found in the Science Update: - What is Rob Shumaker having the orangutans do? - What must be in the orangutan's response, in order for it to be correct? - What will Shumaker focus on first as he and the orangutans start working on math? Do you remember the first mathematical thing you learned? - Why would it be important to have an understanding of the similarities orangutans have to humans when it comes to learning or communicating? - What are some other ways humans might communicate with primates? Next, students will play a game that will demonstrate some key elements of good communication. Instructions for the Communication Puzzle game can be found on the Communication Puzzle teacher sheet. Share the instructions with students and make sure that each group is clear on what they are supposed to do. If you like, you can make a copy of the instructions for students, but make sure not to include the puzzle solution in any handouts that you give to the students who are supposed to solve the puzzle. Once students have finished this game, ask them these questions. They will motivate student thinking about the importance of communication regarding humans and other species. Note: there are no right or wrong answers to these questions. - Why was communication not optimal (or ideal) in this exercise? - What are the different ways humans communicate? - Were all ways of human communication being used during this exercise? - What were some of the key elements for better communication? Part I: All Animals Communicate Tell students that specialized roles within species are genetically programmed. And, that depending on the neural abilities of an animal, it may be able to do only simple or very complex tasks. For example, human beings have superior neural abilities compared to other animals and are therefore able to invent and modify a wider range of social behavior, including their own communication abilities. Using the Animal Communication student esheet, students should read All Animals Communicate. As they are reading, they should look for answers to these questions (they can record their answers on the Animal Communication student sheet): - Are humans constantly communicating? How about other animals? - Are there any special areas in human brains that make us distinct from other animals? - Which animals have communication abilities that are closer to humans? - Do scientists believe that chimps have a language like humans? - What are some examples of ways that humans can communicate that are different from other animals? Part II: Factors Affecting Animal Communication In this part of the lesson, students will examine animal communication in terms of how species' behaviors are affected by inheritance, environment, and experience. Using the esheet, students will read these articles: Then the esheet will direct them to answer these questions on their student sheet: - Is the ability to communicate the exclusive possession of human beings? - List some of the ways that animals express themselves. - Who do you think will have greater communication abilities: a dog that lives in a regular home or a dog that lives in a circus, why? Refer students to the "Understanding What You Learned" section of the esheet. Here they will be asked to reconsider the sentence that you shared with them at the beginning of the lesson: Although all species have some capacity for communication, there is a range of abilities among species. Students should reflect on what it means in terms of what they have learned. Student answers should generally reflect an understanding of these three concepts: - Communication is a common link between all known species. - The ability of communication varies from simple to complex among different species, finding its climax in humans, since humans are able to invent and modify a wider range of social behavior, including communication. - Communication styles are influenced by an animal's genetic makeup, environment, and adaptation abilities. The Science NetLinks lesson, Pets: Oh Behave, explores some of the same ideas covered in this lesson. Students may find it fun and interesting to search through these websites to get further information regarding animal communication. These resources can be accessed from the "Going Further" section of the E-Sheet.<\|endoftext\|>	4
170	By Hamid Bloem at May 08 2019 09:24:40 There are many types of writing worksheets. There is the cursive writing worksheets and the kindergarten worksheets. The latter is more on letter writing and number writing. This is typically given to kids of aged four to seven to first teach them how to write. Through these worksheets, they learn muscle control in their fingers and wrist by repeatedly following the strokes of writing each letter. These writing worksheets have traceable patterns of the different strokes of writing letters. By tracing these patterns, kids slowly learn how a letter is structured. Microsoft Excel provides several methods for linking values across worksheets or across workbooks. Which method you choose will depend on your desired end result. This article will review the pros and cons of each method.<\|endoftext\|>	3.921875
1,528	Suppose that the population (in thousands) of a certain kind of insect after t months is given by the following formula. P(t) = 3t + sin(4t) + 100 Determine the minimum and maximum population in the first 4 months. Solution: The question that we’re really asking is to find the absolute extrema of P(t) on the interval [0,4]. Step 1: Find the derivative P'(t) = 3 + 4cos(4t) Step 2: Isolate t 3 + 4cos(4t) = 0 4cos(4t) = -3 cos(4t) = -3/4 4t = arccos(-3/4) 4t = 2.4189 + 2πn, n = 0, ±1, ±2, ... -> I get why it's 2.4189 and the degree is 138.59, but I don't get why it had + 2πn, n - 0, ±1, ±2, ... 4t = 3.8643 + 2πn, n= 0, ±1, ±2, ... -> I don't know where 3.8643 came from and I want to know it's degree too. t = 2.4189/4 + 2πn, n= 0, ±1, ±2, ... t = 3.8643/4 + 2πn, n= 0, ±1, ±2, ... t = 0.6047 + πn/2, n= 0, ±1, ±2, ... t = 0.9661 + πn/2, n= 0, ±1, ±2, ... Step 3: Find the critical points at n = 0 -> t = 0.6047, t = 0.9661 -> i don't get why we had to use positive n=0,1,2 at n = 1 -> t = 0.6047 + π/2 = 2.1755, t = 0.9661 + π/2 = 2.5369 at n = 2 -> t = 0.6047 + π = 3.7463, t = 0.9661 + π = 4.1077 -> because [0,4] we don't need this Step 4: Plug critical points into the original function P(t) = 3t + sin(4t) + 100 P(0) = 100 P(0.6047) = 102.4756 P(0.9661) = 102.2368 P(2.1755) = 107.1880 P(2.5369) = 106.9492 P(3.7463) = 111.9004 P(4) = 111.7121 * remember that P is in thousands Absolute Minimum Population is 100,000 which occurs at t = 0 or (0, 100) Absolute Maximum Population is 111,900 which occurs at t = 3.7463 or (3.7463, 111.9004) A graph may help you. Remember that we are dealing in radians, not degrees for t. Also, it’s not clear whether we are only looking at integer values for t. You can see that there are maxima and minima in between months. The approx values shown on the graph will tell you which values of t are meaningful. So when using calculus you need to keep in mind what the graph looks like so that you know which value of t applies. You can also draw up a table of values for t and P(t). The value of t itself is not the issue: sin(4t) is what matters because it’s that value which is a component of P(t). You already know that t ranges between 0 and 4 (months). The absolute minimum is at the start (t=0, P(0)=100). The absolute maximum is at t=3.746, P(3.746)=111.9. 3.746 is the value of t without having to add 2πn, in other words, n=0 for the whole range of t. You get a better idea of how sin(4t) contributes to P, if you plot Q(t)=3t+100 on the same graph, you will see how the sine wave is superimposed on what is otherwise a linear function. See below: The sine function has a cycle of 2π radians (about 6.263), which corresponds to 4t, so the interval t=π/2 (1.571) indicates how often P(t) will cycle on the line Q(t)=3t+100. sinθ=0 when θ=0, π, 2π, 3π, etc., so 4t=0, π, 2π, 3π, etc., making intervals of π/4=0.785 between values of t where P(t) intersects with Q(t). And when θ=π/2, 5π/2, 9π/2, ... (t=π/8=0.393, 5π/8=1.963, 9π/8=3.534, ...) we have maxima, and when θ=3π/2, 7π/2, ... (t=1.178, 2.749, ...) we have minima. Now, bring in the calculus: cos(4t)=-¾, 4t=2.419, t=0.605 for an extremum. To find out if it’s max or min we take the next derivative: -16sin(4t), which is negative when t=0.605, implying a maximum. The next maximum is ¼arccos(-¾)+π/2=2.1755, the next at ¼arccos(-¾)+π=3.7463, and so on. The P values for these t values can then be calculated. Note that at t=0 we are in the middle of a cycle, and this is the absolute minimum on the interval t in [0,4], because P(0)=100. We would need t<0 in order to drop below 100. When is -16sin(4t)>0? sin(4t) must be negative. So we look for a solution of cos(4t)=-¾ such that sin(4t)<0. Since cos(4t)=cos(-4t)=-¾, t=-0.605 is also a solution, but we need t>0, so we move to the next minimum by adding π/2=0.966. The next minimum is 2.536, and so on. This should explain in detail how to think this one out. by Top Rated User (804k points)<\|endoftext\|>	4.40625
440	1. Class 11 2. Important Question for exams Class 11 3. Chapter 10 Class 11 Straight Lines Transcript Ex10.3, 8 Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3. Let equation of line AB be x – 7y + 5 = 0 Let line CD be perpendicular to line AB and having x-intercept 3 Since Line CD has x-intercept 3 So, line CD passes through the point (3, 0) We have to find equation of line CD, Finding slope of line AB x − 7y + 5 = 0 − 7y = −x − 5 − 7y = −(x + 5) 7y = (x + 5) y = 1/7 (x + 5) y = 𝑥/7 + 5/7 The above equation is of the form y = mx + c where m = slope of line Thus, slope of line AB = 1/7 Now, Given that line AB and line CD are perpendicular We know that, product of slope of perpendicular lines is –1 So, (Slope of line AB) × (Slope of line CD) = –1 1/7 × (Slope of line CD) = –1 Slope of line CD = 7 × –1 Slope of line CD = –7 Equation of a line passing through a point (x0, y0) & having slope m is (y – y0)= m(x – x0) Equation of line CD passing through point P(3, 0) & having slope -7 is (y − 0) = –7(x – 3) y = –7 (x – 3) y = –7x – (–7)3 y = –7x + 21 Thus, the required equation of line is y = –7x + 21 Chapter 10 Class 11 Straight Lines Class 11 Important Question for exams Class 11<\|endoftext\|>	4.53125
228	Python break statement It terminates the current loop and resumes execution at the next statement, just like the traditional break statement in C. The most common use for break is when some external condition is triggered requiring a hasty exit from a loop. The break statement can be used in both while and for loops. If you are using nested loops, the break statement stops the execution of the innermost loop and start executing the next line of code after the block. The syntax for a break statement in Python is as follows − #!/usr/bin/python for letter in 'Python': # First Example if letter == 'h': break print 'Current Letter :', letter var = 10 # Second Example while var > 0: print 'Current variable value :', var var = var -1 if var == 5: break print "Good bye!" When the above code is executed, it produces the following result − Current Letter : P Current Letter : y Current Letter : t Current variable value : 10 Current variable value : 9 Current variable value : 8 Current variable value : 7 Current variable value : 6 Good bye!<\|endoftext\|>	4.3125
681	Knowing where viruses are illuminates where carbon may be sinking A. Deniaud Garcia/Tara Oceans Foundation Arctic waters turn out to be teeming with some of the world’s smallest entities — viruses. Water samples taken during a three-year expedition around the world’s oceans identified around 200,000 virus species, roughly 12 times the number found in a previous smaller survey. And 42 percent of those viruses were found exclusively in the Arctic, researchers report April 25 in Cell. The results come from the Tara Oceans global oceanographic research expedition. From 2009 to 2013, researchers dropped tanks off of an aluminum sailboat called Tara to collect 145 water samples from dozens of sites worldwide, at water depths from 0 to 4,000 meters. Scientists collected everything ranging in size from fish eggs down to viruses. Filtering isolated the viruses, which were then genetically compared. The researchers identified 195,728 virus species parsed into five global regions that are home to distinct viral communities. The most diversity was found in shallow, temperate and tropical waters, followed closely by Arctic waters. Almost all of the viruses were bacteriophages, which attack bacteria — not people. “So you can swim in the ocean and not worry about it,” says Ahmed Zayed, a microbiologist at Ohio State University in Columbus. Bacteriophages and other viruses are credited with killing roughly 20 percent of bacteria in the ocean every day. That process stops carbon in the bacteria from passing up the food chain, and instead releases the carbon back into the ocean for other microorganisms — some of which also consume carbon dioxide. These microbes eventually produce a form of carbon that can’t be recycled and stays stored in the ocean. Viruses may serve an important role in counteracting human-induced climate change by indirectly stashing away carbon in this way, though viruses have rarely been included in climate simulations. Having the global map of virus locations could help scientists know where carbon fallout occurs and increase the accuracy of climate simulations. But the study gives only a limited view of viruses at play. “There’s still swaths of the ocean that haven’t been looked at,” such as the Western Pacific, says Curtis Suttle, an environmental virologist at the University of British Colombia in Vancouver who wasn’t involved in the study. And because microbiologists can only isolate and identify viral species with DNA, viruses with RNA were excluded from the new analysis, despite being thought to make up half of the ocean’s virus diversity. “So we’re still really only scratching the surface of what’s there,” Suttle says. A. Gregory et al. Marine DNA viral macro- and microdiversity from pole to pole. Cell. Vol 177, May 16, 2019, p. 1. doi:10.1016/j.cell.2019.03.040. C. Gramling. Viruses may help phytoplankton make clouds – by tearing the algae apart. Science News. Vol. 194, September 15, 2018, p. 13. S. Milius. Ocean plankton held hostage by pirate viruses. Science News. Vol. 190, July 9, 2016, p. 12.<\|endoftext\|>	3.859375
1,375	1Question # Factors micro-lesson Do you struggle with balancing your kids’ screen time with their education? Say hello to 1Question, the app that solves this problem with a fun twist! Our app lets your child earn screen time minutes in their favourite apps by watching educational videos and correctly answering quiz questions. And the best part? You get to decide which apps to lock behind learning using 1Question. ## Learning time Kids learn by watching short, engaging video lessons. Factors are the numbers you multiply together to find a product. Every number has at least two factors, one and itself. The number 12 has 1 x 12, 2 x 6 and 3 x 4, for a total of six factors. The factor pairs for 32 are 1 and 32, 2 and 16 and 4 and 8. When you want to make a list of your factors from least to greatest, you connect the rainbow on each end with your factors starting with the least and going to the greatest. 3 x 5 is 15 if I put commas in between there's my list from least to greatest. When you look at the number 20, we notice 20 is an even number, so it must have a factor of two. We also know that it ends in zero, so it will have factors of both 5 and 10 so now you know how to find factors of a number Kids earn more screen time by answering fun educational questions. • #### Which number is a factor of 20? 1) All of the above. 2) 2 3) 4 4) 10 1) 8 2) 2 3) 4 4) 6 1) 4 2) 2 3) 3 4) 4 1) 2 2) 3 3) 1 4) 4 1) 2 2) 3 3) 4 4) 5 1) 6 2) 4 3) 2 4) 12 1) 6 2) 4 3) 8 4) 9 1) 5 2) 4 3) 7 4) 3 1) 6 2) 4 3) 1 4) 2 1) 8 2) 4 3) 6 4) 2 1) 6 2) 8 3) 2 4) 16 • #### If a number ends in a zero, it has a factor of _______. 1) all of the above 2) 10 3) 5 4) 2 • #### Which is the correct list of factors for 15? 1) 1, 3, 5, 15 2) 1, 2, 3, 15 3) 1, 3, 9, 15 4) 1, 2, 5, 15 • #### Which is the correct list of factors for 24? 1) 1, 2, 3, 4, 6, 8, 12, 24 2) 1, 2, 4, 6, 8, 9, 24 3) 1, 3, 4, 6, 7, 12, 24 4) 1, 2, 3, 4, 6, 8, 9, 12 1) 36 and 48 2) 12 and 20 3) 18 and 36 4) 28 and 32 1) 3 and 14 2) 2 and 22 3) 4 and 8 4) 13 and 4 1) 4 and 8 2) 3 and 6 3) 2 and 18 4) 4 and 9 1) 4 and 12 2) 6 and 9 3) 7 and 8 4) 8 and 9 1) 6 and 9 2) 7 and 9 3) 5 and 4 4) 7 and 8 • #### Multiply 2 or more factors to find the __________. 1) Product 2) Sum 3) Difference 4) Quotient 1) Yes 2) No 1) Yes 2) No 1) No 2) Yes 1) No 2) Yes 1) No 2) Yes 1) No 2) Yes ## Recreational screen time Kids can use the screen time minutes they earned to unlock the apps selected by their parent e.g. games, social media, streaming, etc. ## Out of time When screen time minutes run out, kids are locked out and need to complete more learning in 1Question to earn more time. Our micro-courses are developed and delivered by qualified educators from around the world. #### Chris Lindsey More lessons from the micro-course ##### Number Patterns Sometimes numbers follow a pattern and we need to figure out what the rule is… ##### Use Area Models to Multiply 2 x 2 Digits Area models are a great way to solve multiplication problems. With larger numbers, you’re going… ##### Skip Counting to find multiples Did you know that you can find multiples by using skip counting? If you skip… ##### Multiply Using Partial Products When you multiply two digits by two digits, you might want to try the partial… ##### Multiply by Multiples of Ten When you multiply with multiples of 10, remember, it’s all about the facts. For example,… ##### Multiplication Word Problems When solving multiplication word problems, here’s a few strategies you can use. Sal earns \$12… ##### Prime and Composite Numbers What’s the difference between prime and composite numbers? Here is a list of the first… ##### Find Area of a Rectangle When we talk about the area of a rectangle, we’re talking about measuring the space… ##### Multi-Step Word Problems Sometimes when you’re doing word problems, they require more than one step. Two classes of… ##### Multiply Area Models If you want to multiply with 2- or 3-digit numbers, you should use area models…. ##### Estimate Products When you multiply 2 numbers, sometimes you don’t want an exact answer. You want an…<\|endoftext\|>	4.59375
572	# Time Shifting and Scaling of Functions We’ll begin with a square function, f(t), that has a an amplitude of 1, a start time of 2 seconds and an end time of 4 seconds. Next, a time shift is demonstrated. Here our function is changed from f(t) to f(t-2). Notice that subtracting 2 from t in the function results in a positive shift of the graph. In the next two graphs t will be scaled. Scaling t is not quite as intuitive as we may have expected. When we multiply t by 2, corresponding points of the function now occur at 1/2 the time they previously had. When we divide t by 2, each corresponding time on the graph occurs at a t that is now multiplied by 2. Notice that each of these factors directly affects the duration of the signal. Scaling the amplitude has more intuitive results. If we multiply f(t) by 2, the amplitude of 1 is changed to 2. Multiplying f(t) by 1/2 results in an amplitude of 1/2. Finally, multiplying t by -1 mirrors our function over the y-axis. Each time now occurs at its negative. Example: Here we will attempt to convert f(t) into 2f(.5t+3). The graph of f(t) is shown below. The easiest way to handle this type of problem without error is to manipulate the function one step at a time. First, I have converted f(t) into 2f(t). Only the peaks are changed here (by a factor of 2). Next, I convert 2f(t) into 2f( $\frac{1}{2}$ t). Notice how the $\frac{1}{2}$ actually expands our graph duration by a factor of 2 (from a 6 sec duration to a 12 sec duration). Finally, we move from 2f( $\frac{1}{2}$ t) to 2f( $\frac{1}{2}$ t + 3). As shown in the discussion above, this is a time shift. Time shifts can be a little confusing because adding results in a negative shift of our graph. Try to think of it as our signal occurring 3 seconds earlier than before, reading from left to right on the graph. The easiest way to do this part is shift each x-intercept by 3 seconds (to the left, of course). ## One Reply to “Time Shifting and Scaling of Functions” 1. You are correct. I performed the time scaling and time shifting in opposite order. The signal should have been shifted by 6 seconds (3/(.5)) to the left and then scaled appropriately for the 1/2 factor. Updates on the way..<\|endoftext\|>	4.6875
721	Probability Distribution Function for a Single Variable One of the most important concepts in Finance is the concept of a random variable. For example this can be viewed as the outcome of throwing a die where the process is fixed by the outcome is not. Form the definition of a random variable is derived the definition of the distribution function which is the probability that the final value of the random variable is less than equal to a given number. The other word for this term is cumulative distribution function. Where $F(X)=p(X\leq x)$ is the cumulative distribution function There are two possibilities in this case. When the variable takes discrete values and when it is continuous in nature. In case the variable is discrete in nature the distribution is obtained by summing the step values less than or equal to the reference number and the function that defines this variable is called the frequency function or the probability density function. When the variable is continuous in nature this is obtained by taking the integral of the probability density function over all possible values until the reference number. The probability density function is then the derivative of the first derivative of the distribution function with respect to the variable. For the discrete case: $F(X)= \sum_{x(i)\leq x} f(x(i))$ For the continuous case: $\int_{-\infty }^{x}f(u)du$ In the continuous case the integral of the probability density function from negative infinity to complete infinity will be 1. $fx=\frac{dF(x)}{dx}$ If the cumulative distribution function is integrated over the complete range of values from -∞ to +∞ then it should be equal to 1. $\int_{-\infty }^{+\infty }f(u)du=1$ Example of Density Functions (for the discrete case) Take a case of rolling two dice at once. Since there are six faces on one side of the dice the total number of outcomes of the total of the addition of the two numbers that can appear on each face of the dice is 6*6=36. The range of final result is from 2-12 with varied frequencies representing the number of times each outcome can happen. The probability of each outcome is the frequency of each outcome divided by the total number of outcomes possible. The addition of all these probabilities should of course be 1. For any outcome the cumulative probability is the sum of the probabilities for each outcome lesser and equal to that outcome. This is very useful in finance because the same approach can be applied to financial information like stock quotes, foreign exchange rates, commodity prices and exchange rates etc. You may find these interesting Single Index Model The Single Index Model (SIM) is an asset pricing model, according to which the returns on a security... Accelerate your finance career with cutting-edge data skills. Join Finance Train Premium for unlimited access to a growing library of ebooks, projects and code examples covering financial modeling, data analysis, data science, machine learning, algorithmic trading strategies, and more applied to real-world finance scenarios. I WANT TO JOIN JOIN 30,000 DATA PROFESSIONALS Free Guides - Getting Started with R and Python Enter your name and email address below and we will email you the guides for R programming and Python. Accelerate your finance career with cutting-edge data skills. Join Finance Train Premium for unlimited access to a growing library of ebooks, projects and code examples covering financial modeling, data analysis, data science, machine learning, algorithmic trading strategies, and more applied to real-world finance scenarios.<\|endoftext\|>	4.5625
1,370	August 20 marks the passing of Fred Hoyle. Hoyle was a well-known British astronomer who was the first to outline the creation of elements within stars. Hoyle’s work on nucleosynthesis showed how elements greater than helium could be formed through fusion reactions within stars. His first paper showed the core temperature of stars could evolve hot enough to fuse elements up to iron. Fusion reactions build heavier elements and those elements would fuse to form even heavier elements. Eventually, the core temperature would reach an equilibrium point where iron would be more abundant than other heavy elements. This process, known as the e Process, explains why iron has such a high natural abundance. His second paper, in cooperation with three other physicists, showed the creation of elements from carbon to iron required special conditions generally found in pre-supernova stars. Each element is created through fusion reactions between concentric shells of elements within the star. The paper also explained the creation of elements greater than iron through neutron capture reactions. This work forms the basis of much of the study of cosmology and stellar chemistry. This paper would be enough to earn one of the authors, William Fowler, part of the 1983 Nobel Prize in Physics. The reason Hoyle’s contribution was ignored and left out of the Prize is still unknown. In spite of making such a large contribution to cosmology, he is probably best known for his outspoken defense of the Steady State Universe theory. Steady State Universe theory holds the universe is constantly expanding and creating new matter in order to maintain a homogenous density. The universe has no beginning and no end. This theory competed with another new theory that suggested the universe was formed from a great explosion of a super-dense state and has been expanding and cooling ever since. This theory adopted a popular name jokingly given by Hoyle during one of his popular BBC radio broadcasts: The Big Bang. He would continue to try to find fault with Big Bang theory even after the discovery of background microwave radiation. This background radiation could be explained by Big Bang, but could not be explained by Steady State. Hoyle would never accept Big Bang and would go to his grave in 2001 trying to discredit it. His reputation would steadily decline as he put forth several unconventional theories. One theory was the concept of panspermia. This was the idea that life began on Earth from cells that arrived from space and evolution is driven by the continual arrival of viruses arriving from comets. He was quoted to say the idea that life formed by chance from some primordial soup was “evidently nonsense of a high order”. Notable Science Events for August 20 2001 – Fred Hoyle died. 1977 – NASA launched the Voyager 2 space probe. NASA launched the Voyager 2 space probe that would go on to explore four of the outer planets of the solar system. Voyager visited the planets Jupiter, Saturn, Uranus, and Neptune during its trek. Voyager 2 left our solar system in 2007 and is expected to be able to continue transmissions until 2025. 1975 – NASA launched the Viking 1 mission to Mars. NASA launched the Viking 1 spacecraft on its mission to Mars. The craft contained an orbiter component and a lander that would land on the surface to conduct experiments. The lander successfully landed on Mars on July 20, 1976, and operated continuously for nearly six and a half years. 1961 – Percy Williams Bridgman died. Bridgman was an American physicist who was awarded the 1917 Nobel Prize in Physics for his work in high-pressure physics. His work started with the modification of an existing high-pressure machine that enabled a pressure in excess of 10 GPa. He used this machine to test physical properties of crystals and liquids at extremely high pressures. Bridgman was suffering from metastatic cancer and committed suicide. His note expressed the sentiment that it wasn’t fair society makes it necessary for a man to kill himself when the end is certain. His case is often brought up in discussions of assisted suicide. 1936 – Hideki Shirakawa was born. Shirakawa is a Japanese chemist who shares the 2000 Nobel Prize in Chemistry with Alan J. Heeger and Alan G. MacDiarmid for the discovery of conductive polymers. He developed polyacetylene and discovered it conducted electricity and enhanced the conductivity with Heeger and MacDiarmid. 1917 – Adolf von Baeyer died. Baeyer was a German organic chemist who was awarded the 1905 Nobel Prize in Chemistry in recognition for his contributions to organic and industrial chemistry through his work on organic dyes and hydroaromatic compounds. He was the first to artificially synthesize the purple dye indigo previously available only from plants and was the first to synthesize synthetic fluorescein, a dye that glows under ultraviolet light. Baeyer also discovered the phthalein dyes and barbituric acid, the basic compound of barbiturates. 1915 – Paul Ehrlich died. Ehrlich was a German biologist who shares the 1908 Nobel Prize in Medicine with Ilya Ilyich Mechnikov for their respective work on immunity. Ehrlich is best known for his side-chain theory which explains the effects of serum and enabled the measurement of antigens. He coined the terms chemotherapy and magic bullet. A magic bullet is a method of selectively targeting a specific bacteria without harming any other organisms. He was also the first to observe the blood-brain barrier that separates blood from cerebrospinal fluid. 1913 – Roger Wolcott Sperry was born. Sperry was an American neurobiologist who was awarded half the 1981 Nobel Prize in Medicine for discovering the function of the different sides of the brain. He found the left lobe of the brain is responsible for analytical and verbal tasks, where the right was responsible for spacial and artistic tasks. 1848 – Jöns Jakob Berzelius was born. Berzelius was a Swedish chemist who created a table of known elements based relative atomic weights where the weight of oxygen was set to 100. He was the first to use the chemical notation system in use today, with symbols for elements and numbers to denote the proportions. The difference between his system and today’s is where the number is placed. Today, a subscript is used to denote the proportion where Berzelius used a superscript. For example, H2O would be written as H2O. Berzelius is credited with the discovery of silicon, selenium, thorium, and cerium. He also coined the terms for catalysis, polymer, isomer, and allotrope.<\|endoftext\|>	4.3125
8,512	# Fundamentals of Complex Analysis Submitted to: Sir MAQSOOD ALI ABBAS Submitted By: Muzamil Hussain Roll No. : 4010 ASSIGNMENT: COMPLEX ANALYSIS CHAPTER # ##### Citation preview Submitted to: Sir MAQSOOD ALI ABBAS Submitted By: Muzamil Hussain Roll No. : 4010 ASSIGNMENT: COMPLEX ANALYSIS CHAPTER # 2 Analytic or Regular Or Holomorphic Functions Level Curves: Example #2: If 𝒘 = 𝟏/𝒛 show that the level curves 𝑼=𝑪𝟏 and 𝑼 = 𝑪𝟐 are orthogonal circles which pass through the origin and have their centres on x-axis and y-axis. Solution: 𝑍̅ 1 𝑥−𝑖𝑦 Given 𝑊 = 𝑍 = \|𝑍\|2 = 𝑥 2 +𝑦 2 Step I 𝑈= When 𝑥 , 𝑥2 + 𝑦2 𝑉=− 𝑦 𝑥2 +𝑦2 𝑈 = 𝐶1 . 𝑥2 𝑥 = 𝐶1 + 𝑦2 i.e.𝐶1 (𝑥 2 + 𝑦 2 ) − 𝑥 = 0 … … … … … . (1) and passes through the origin. Centre:(−𝑓, −𝑔) From eq. 1 we get: 𝑓=− 1 𝑎𝑛𝑑 𝑔 = 0 2𝐶1 1 Hence centre=(2𝐶 , 0) 1 Also . 𝑉 = 𝐶2 ⇒ 𝐶2 (𝑥 2 +𝑦 2 ) + 𝑦 = 0 1 Here 𝑓 = 0 𝑎𝑛𝑑 𝑔 = − 2𝐶 2 1 Hence centre= (0, − 2𝐶 ) and passes through the origin. 2 Step II: 𝜕𝑈 Now 𝜕𝑥 = (𝑥 2 +𝑦 2 ).1−𝑥.2𝑥 (𝑥 2 +𝑦 2 )2 (𝑦 2 −𝑥 2 ) = (𝑥 2 +𝑦 2 )2 𝜕𝑈 −2𝑥𝑦 = 2 2 2 𝜕𝑦 (𝑥 +𝑦 ) 𝜕𝑉 Now 𝜕𝑥 𝜕𝑉 𝜕𝑦 = = (𝑥 2 +𝑦2 ).0−(−𝑦).2𝑥 (𝑥 2 +𝑦 2 )2 (𝑥 2 +𝑦 2 )(−1)−(−𝑦).2𝑦 (𝑥 2 +𝑦 2 )2 2𝑥𝑦 = (𝑥 2 +𝑦 2)2 𝑦 2 −𝑥2 =(𝑥 2 +𝑦 2 )2 For orthogonal system we have 𝜕𝑈 𝜕𝑉 𝜕𝑈 𝜕𝑉 . + . =0 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 i.e. 𝑦 2 −𝑥 2 2𝑥𝑦 −2𝑥𝑦 𝑦2 − 𝑥2 × + × =0 (𝑥 2 +𝑦 2 )2 (𝑥 2 +𝑦 2 )2 (𝑥 2 +𝑦 2 )2 (𝑥 2 +𝑦 2 )2 The condition is satisfied therefore the level curves are orthogonal. Example # 3: If 𝒇(𝒁) = 𝒁+𝟒 𝒁−𝟒 find the level curves 𝑼 = 𝑪𝟏 , 𝑽 = 𝑪𝟐 . Also verify that the level curves 𝑼 = 𝑪𝟏 𝒂𝒏𝒅 𝑽 = 𝑪𝟐 from an orthogonal system of circles. Solution: Step I: (𝑥+4)+𝑖𝑦 Given 𝑓(𝑍) = 𝑈 + 𝑖𝑉 = (𝑥−4)+𝑖𝑦 𝑈 + 𝑖𝑉 = [(𝑥 + 4) + 𝑖𝑦][(𝑥 − 4) − 𝑖𝑦] (𝑥 − 4)2 + 𝑦 2 (𝑥 2 − 16 + 𝑦 2 ) + 𝑖(𝑥𝑦 − 4𝑦 − 𝑥𝑦 − 4𝑦) = (𝑥 − 4)2 + 𝑦 2 𝑥 2 +𝑦 2 −16 𝑈 = (𝑥−4)2 +𝑦2 , −8𝑦 𝑉 = (𝑥−4)2 +𝑦 2 . If 𝑈 = 𝐶1 ⇒ 𝐶1 (𝑥 2 + 𝑦 2 − 8𝑥 + 16) − 𝑥 2 − 𝑦 2 + 16 = 0 (𝐶1 − 1)𝑥 2 + (𝐶1 − 1)𝑦 2 − 8𝐶1 𝑥 + 16𝐶1 + 16 = 0 Which is a circle. . Also when 𝑉 = 𝐶2 ⇒ 𝐶2 (𝑥 2 +𝑦 2 − 8𝑥 + 16) + 8𝑦 = 0 Which is also a circle. Step II: 𝑈= 𝑥 2 + 𝑦 2 − 16 (𝑥 − 4)2 + 𝑦 2 , 𝑉= −8𝑦 (𝑥 − 4)2 + 𝑦 2 𝜕𝑈 ((𝑥 − 4)2 + 𝑦 2 )2𝑥 − 2(𝑥 2 + 𝑦 2 − 16)(𝑥 − 4) = 𝜕𝑥 [(𝑥 − 4)2 + 𝑦 2 ]2 = = 2𝑥 3 − 16𝑥 2 + 32𝑥 + 2𝑥𝑦 2 − 2𝑥 3 − 2𝑥𝑦 2 + 32𝑥 + 8𝑥 2 + 8𝑦 2 − 128 [(𝑥 − 4)2 + 𝑦 2 ]2 8𝑦 2 −8𝑥 2 +64𝑥−128 [(𝑥−4)2 +𝑦 2 ]2 ………………(1) 𝜕𝑈 [(𝑥 − 4)2 + 𝑦 2 ]2𝑦 − 2𝑦(𝑥 2 + 𝑦 2 − 16) = 𝜕𝑦 [(𝑥 − 4)2 + 𝑦 2 ]2 2𝑥 2 𝑦 − 16𝑥𝑦 + 32𝑦 + 2𝑦 3 − 2𝑦𝑥 2 − 2𝑦 3 + 32𝑦 = [(𝑥 − 4)2 + 𝑦 2 ]2 = 16𝑦[4 − 𝑥] … … … … … … … (2) [(𝑥 − 4)2 + 𝑦 2 ]2 𝜕𝑉 8𝑦. 2(𝑥 − 4) = … … … … … … . . (3) 𝜕𝑥 [(𝑥 − 4)2 + 𝑦 2 ]2 𝜕𝑉 [(𝑥 − 4)2 + 𝑦 2 ](−8) + 8𝑦. 2𝑦 = 𝜕𝑦 [(𝑥 − 4)2 + 𝑦 2 ]2 = [8𝑦 2 −8𝑥2 +64𝑥−128 [(𝑥−4)2 +𝑦2 ]2 …………..(4) Now the condition 𝜕𝑈 𝜕𝑉 𝜕𝑈 𝜕𝑉 . + . =0 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 = 8𝑦 2 −8𝑥 2 +64𝑥−128 [(𝑥−4)2 +𝑦 2 ]2 8𝑦.2(𝑥−4) 16𝑦[4−𝑥] × [(𝑥−4)2 +𝑦 2 ]2+[(𝑥−4)2 +𝑦2 ]2 × [8𝑦 2 −8𝑥 2 +64𝑥−128 [(𝑥−4)2 +𝑦 2 ]2 =0 Therefore, it forms an orthogonal system of circles. 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 # 4 If 𝒘 = 𝒇(𝒁) = 𝑼 + 𝒊𝑽 be an analytic function of 𝒁 = 𝒙 + 𝒊𝒚 then show that the curves 𝑼 = 𝑪𝟏 and 𝑽 = 𝑪𝟐 intersect at right angles. SOLUTION: Step I Given 𝑓(𝑍) to be an analytic function 𝜕𝑈 𝜕𝑥 𝜕𝑉 𝜕𝑥 𝜕𝑉 = 𝜕𝑦 𝜕𝑈 = − 𝜕𝑦 … (1) … (2) Which are the Cauchy Riemann equations. Step II Multiplying (1) and (2) we get 𝜕𝑈 𝜕𝑉 𝜕𝑈 𝜕𝑉 =− 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑈 𝜕𝑉 𝜕𝑈 𝜕𝑉 + =0 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 Which is the condition of orthogonality of the two curves 𝑈 = 𝐶1 and 𝑉 = 𝐶2 MISCELLANEOUS PROBLEMS PROBLEM 1 Find an analytic functions in the disk \|𝒁 − 𝟏\| < 1 whose real part is log√𝒙𝟐 + 𝒚𝟐 . Solution: STEP I Given 1 𝑈 = log(𝑥 2 + 𝑦 2 ) 2 𝜕𝑈 1 1.2𝑥 = ∙ 𝜕𝑥 2 𝑥 2 + 𝑦 2 Put𝐱 = 𝐙 and𝒚 =0 𝑈𝑥 (𝑍, 0) = 𝑍 1 = 𝑍2 𝑍 𝜕𝑈 1 2𝑦 = . 2 𝜕𝑦 2 𝑥 + 𝑦 2 Put𝐱 = 𝐙 and𝐲 = 𝐨 𝑈𝑦 (0, 𝑍) = 0 STEP II 𝑓 ′ (𝑍) = 𝑈𝑥 (𝑍, 0) − 𝑖𝑈𝑦 (0, 𝑍) 1 1 𝑍 𝑍 = −0= Integrate w.r.t z 𝑓(𝑍) = 𝑙𝑜𝑔𝑧 + 𝐶 PROBLEM 2 Prove that an analytic function with constant modulus is constant. Solution: Step I Given 𝑓(𝑧) an analytic function i.e. 𝜕𝑈 𝜕𝑥 = 𝜕𝑉 𝜕𝑦 𝜕𝑉 and 𝜕𝑥 Also \|𝑓(𝑧)\| = 𝑐 ′ =− 𝜕𝑈 𝜕𝑦 or 𝑈 2 + 𝑉 2 = 𝐶(1) Differentiating (1) w.r.t. x 2𝑈𝑈𝑥 + 2𝑉𝑉𝑌 = 0 𝑈𝑈𝑥 + 𝑉𝑉𝑌 = 0(2) Likewise differentiating (1) w.r.t y 2𝑈𝑈𝑥 + 2𝑉𝑉𝑌 = 0 𝑈𝑈𝑥 + 𝑉𝑉𝑌 = 0(3) Using CR equation (3) can be written as −𝑈𝑉𝑥 + 𝑉𝑈𝑋 = 0(4) Multiplying (2) by U and (4) by V, then adding. We get 𝑈𝑥 (𝑈 2 + 𝑉 2 ) = 0(5) Using (1)(𝑈 2 + 𝑉 2 ) ≠ 0 ∴ 𝑈𝑥 = 0 By CR equations 𝑈𝑥 = 0 = 𝑉𝑦 (6) From (6) we conclude that U is independent of x and V is independent of y. Again multiplying (2) by v and (4) by U, then subtracting we get 𝑉𝑥 (𝑈 2 + 𝑉 2 ) = 0 ∴ 𝑈 2 + 𝑈 2 ≠ 0 ∴ 𝑉𝑋 = 0 = −𝑈𝑦 (𝒃𝒚 𝐂𝐑 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧𝐬) i.e. 𝑉𝑋 = 0 = 𝑈𝑦 (7) from(7) we infer that V is independent of x and U is independent of y. Thus U and V both are independent of x and y. ∴u is constant and V is also constant. Ultimately𝑓(𝑍) = 𝑈 + 𝔦𝑉 = constant. PROBLEM 3: If 𝒇(𝒛) = 𝑼 + 𝖎𝑽 is analytic and 𝑼 − 𝑽 = 𝒆𝒙 (𝐜𝐨𝐬 𝒚 − 𝖎 𝐬𝐢𝐧 𝒚).find 𝒇(𝒛)in terms of Z. Solution: Step I Given𝑈 − 𝑉 = 𝑒 𝑥 (cos 𝑦 − 𝔦 sin 𝑦)(1) Differentiating (1) w.r.t x 𝜕𝑈 𝜕𝑥 𝜕𝑉 − 𝜕𝑥 = 𝑒 𝑥 (cos 𝑦 − 𝔦 sin 𝑦) (2) Differentiating (1) w.r.t y 𝜕𝑈 𝜕𝑉 − = 𝑒 𝑥 (−sin 𝑦 − cos 𝑦) 𝜕𝑥 𝜕𝑥 𝜕𝑈 𝜕𝑉 − 𝜕𝑥 = −𝑒 𝑥 (sin 𝑦 + cos 𝑦) 𝜕𝑥 ☛ 𝜕𝑈 (3) Step II 𝜕𝑉 Putting 𝜕𝑦 = − 𝜕𝑥 and 𝜕𝑉 𝜕𝑦 = 𝜕𝑈 𝜕𝑥 in (3) we get 𝜕𝑉 𝜕𝑈 + = 𝑒 𝑥 (sin 𝑦 + cos 𝑦) 𝜕𝑥 𝜕𝑥 Adding (2) and (4) we get 2 𝜕𝑈 = 2𝑒 𝑥 cos 𝑦 𝜕𝑥 𝑈(𝑥, 𝑦) = 𝑒 𝑥 cos 𝑦 + 𝑐 Now subtract (4) from(2) we get 2 𝜕𝑉 = 2𝑒 𝑥 sin 𝑦 𝜕𝑥 Integrating w.r.t x 𝑉(𝑥, 𝑦) = 𝑒 𝑥 sin 𝑦 + 𝑐 , 𝑓(𝑍) = 𝑈(𝑥, 𝑦) + 𝔦 𝑉(𝑥, 𝑦) = 𝑒 𝑥 (cos 𝑦 + 𝔦 sin 𝑦) 𝑓(𝑍) = 𝑒 𝑥 𝑒 𝔦𝑦 𝑓(𝑍) = 𝑒 𝑧 + 𝐶1 (𝐶1 = 𝑐 + 𝔦𝑐 / ) Problem #4: 𝑰𝒇 𝑼 = (𝒙 − 𝟏)𝟑 − 𝟑𝒙𝒚𝟐 + 𝟑𝒚𝟐 . 𝑫𝒆𝒕𝒆𝒓𝒎𝒊𝒏𝒆 𝑽 𝒔𝒐 𝒕𝒉𝒂𝒕 𝑼 + 𝒊𝑽 𝒊𝒔 𝒂𝒓𝒆𝒈𝒖𝒍𝒂𝒓 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒙 + 𝒊𝒚 . Solution: Step I: 𝐺𝑖𝑣𝑒𝑛 𝑈(𝑥, 𝑦) = (𝑥 − 1)3 − 3𝑥𝑦 2 + 3𝑦 2 𝜕𝑈 = 3(𝑥 − 1)2 − 3𝑦 2 = 𝑈𝑥 (𝑥, 𝑦) … … … … … . . (1) 𝜕𝑥 𝜕𝑈 = −6𝑥𝑦 + 6𝑦 = 𝑈𝑦 (𝑥, 𝑦) … … … … … … . . (2) 𝜕𝑦 𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑥 = 𝑧 & 𝑦 = 0 𝑖𝑛 𝑒𝑞 (1)&(2)𝑤𝑒𝑔𝑒𝑡 𝑈𝑥 (𝑍, 0) = 3(𝑧 − 1)2 𝑈𝑦 (𝑍, 0) = 0 Step II 𝑓 ′ (𝑧) = 𝑈𝑥 (𝑍, 0) + 𝑖𝑈𝑦 (𝑍, 0) = 3(𝑍 − 1)2 Integrating w .r .t ‘Z’ 𝑓(𝑍) = (𝑧 − 1)3 + 𝐶 Step III Calculation of 𝑣(𝑥, 𝑦) 𝑓(𝑍) = 𝑈 + 𝑖𝑉 = [(𝑥 − 1) + 𝑖𝑦]3 = (𝑥 − 1)3 + 3(𝑥 − 1)2 𝑖𝑦 − 3(𝑥 − 1)𝑦 2 − 𝑖𝑦 3 𝑈𝑥 (𝑥, 𝑦) = (𝑥 − 1)3 − 3(𝑥 − 1)𝑦 2 𝑈𝑦 (𝑥, 𝑦) = 3(𝑥 − 1)2 𝑦 − 𝑦 3 Problem#5: Examine the nature of function 𝒙𝟐 𝒚𝟓 (𝒙 + 𝒊𝒚) 𝒇(𝒛) = 𝒙𝟒 + 𝒚𝟏𝟎 ,𝒁 ≠ 𝟎 =𝟎 ,𝒁 = 𝟎 In a region including the region Solution: Step I: C R equations 𝑥3𝑦5 𝑥2𝑦6 Given 𝑓(𝑧) = 𝑈 + 𝑖𝑉 = 𝑥 4 +𝑦 10 + 𝑖 𝑥 4 +𝑦 10 𝑈𝑥 (𝑥, 𝑦) = 𝑥3𝑦5 𝑥 4 + 𝑦 10 , 𝑉(𝑥, 𝑦) = 𝑥2𝑦6 𝑥 4 + 𝑦10 Derivative at the origin 𝜕𝑈 𝑈(𝑥, 0) − 𝑈(0,0) = lim 𝜕𝑥 𝑥→0 𝑥−0 0−0 =0 𝑥→0 𝑥 − 0 = lim 𝜕𝑈 𝜕𝑉 𝜕𝑉 Likewise 𝜕𝑦 = 0 , 𝜕𝑥 = 0 , 𝜕𝑦 = 0 Therefore Cauchy Riemann Equations are true. Step II Calculation of𝑓 ′ (𝑧) 𝑎𝑡 𝑍 = 0 𝑓 ′ (𝑍) = lim 𝑧→0 (i) [ 𝑥 2 𝑦 5 (𝑥+𝑖𝑦) 𝑥 4 +𝑦 10 − 0] 𝑥 + 𝑖𝑦 … … … … … . (𝐴) Suppose z→ 0 𝑎𝑙𝑜𝑛𝑔 𝑦 = 𝑥 𝑒𝑞 (𝐴)𝑏𝑒𝑐𝑜𝑚𝑒 𝑥7 𝑥→0 𝑥 4 + 𝑥 10 𝑓 ′ (0) = lim ⇒𝑓 ′ (0) 𝑥3 = lim =0 𝑥→0 1 + 𝑥 6 Suppose z→ 0 𝑎𝑙𝑜𝑛𝑔 𝑦 5 = 𝑥 2 𝑒𝑞 (𝐴)𝑏𝑒𝑐𝑜𝑚𝑒 (ii) 𝑓 ′ (0) 𝑓(𝑧) − 𝑓(0) 𝑥2𝑦5 = lim = lim 4 𝑍→0 𝑍→0 𝑥 + 𝑦 10 𝑧−0 𝑝𝑢𝑡 𝑦 5 = 𝑥 2 ⇒𝑓 ′ (0) 𝑥4 1 = lim 4 = 𝑥→0 𝑥 + 𝑥 4 2 Since both the limits are different. So, 𝑓 ′ (𝑧) 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 𝑎𝑡 Z=0. Problem#𝟔 Investigate the value of 𝒁 for which 𝑾 = 𝑼 + 𝒊𝑽is not analytic when 𝒁 = 𝒔𝒊𝒏𝒉𝑼 𝒄𝒐𝒔𝑽 + 𝒊𝒄𝒐𝒔𝒉𝑼𝒔𝒊𝒏𝑽 Solution: Given𝑍 = 𝑠𝑖𝑛ℎ𝑈 𝑐𝑜𝑠𝑉 + 𝑖𝑐𝑜𝑠ℎ𝑈𝑠𝑖𝑛𝑉 To prove 𝑑𝑤 𝑑𝑧 is infinite at some Z. Now 𝑤 = 𝑈 + 𝑖𝑉 if 𝑧 is a function of 𝑤 Then 𝑍is a function of 𝑈 𝑎𝑛𝑑 𝑉 𝑑𝑧 𝑑𝑤 𝜕𝑍 = 𝜕𝑈 = 𝑐𝑜𝑠ℎ𝑈𝑐𝑜𝑠𝑉 + 𝑖𝑠𝑖𝑛ℎ𝑈 𝑠𝑖𝑛𝑉 (1) Also 𝑍 2 = (𝑠𝑖𝑛ℎ𝑈 𝑐𝑜𝑠𝑉 + 𝑖𝑐𝑜𝑠ℎ𝑈𝑠𝑖𝑛𝑉)2 = (sin ℎ2 𝑈𝑐𝑜𝑠 2 𝑉 − cos ℎ2 𝑈𝑠𝑖𝑛2 𝑉 + 2𝑖 𝑠𝑖𝑛ℎ𝑈 𝑐𝑜𝑠𝑉𝑐𝑜𝑠ℎ𝑈𝑠𝑖𝑛𝑉) We know that = cos ℎ2 𝑈 − sin ℎ2 𝑈 = 1 Putting in (2) we have 𝑍 2 = (cos ℎ2 𝑈 − 1)𝑐𝑜𝑠 2 𝑉 − (1 + sin ℎ2 𝑈)𝑠𝑖𝑛2 𝑉) + 2𝑖𝑠𝑛ℎ𝑈 𝑐𝑜𝑠𝑉𝑐𝑜𝑠ℎ𝑈𝑠𝑖𝑛𝑉 cos ℎ2 𝑈𝑐𝑜𝑠 2 𝑉 − sin ℎ2 𝑈𝑠𝑖𝑛2 𝑉 + 2𝑖𝑠𝑛ℎ𝑈 𝑐𝑜𝑠𝑉𝑐𝑜𝑠ℎ𝑈𝑠𝑖𝑛𝑉 − (𝑐𝑜𝑠 2 𝑉 + 𝑠𝑖𝑛2 𝑉) 𝑍 2 = (𝑐𝑜𝑠ℎ𝑈 𝑐𝑜𝑠𝑉 + 𝑖 𝑠𝑖𝑛ℎ𝑈𝑠𝑖𝑛𝑉)2 − 1 Using (1) 𝑍2 = ( =( 𝑑𝑧 2 ) −1 𝑑𝑤 𝑑𝑧 2 ) = 1 + 𝑧2 𝑑𝑤 𝑑𝑤 = ±√1 + 𝑧 2 𝑑𝑧 𝑑𝑤 1 =± 𝑑𝑧 √1 + 𝑍 2 𝑑𝑤 𝑑𝑧 does not exist. When 1 + 𝑍 2 = 0 = 𝑧 = ±𝑖 Therefore , the function 𝑤 = 𝑓(𝑧) is not analytic when 𝑧 = ±𝑖 and 𝑍 is given in the question. Problem#7 If 𝒇(𝒛) = is an analytic function of z in any domain prove that 𝑽 𝟐 \|𝒇(𝒛)\|𝒑 𝝏𝟐 𝝏𝟐 = ( 𝟐 + 𝟐 ) \|𝒇(𝒛)\|𝒑 = 𝒑𝟐 \|𝒇(𝒛)\|𝒑−𝟐 \|𝒇(𝒛)\|𝟐 𝝏𝒙 𝝏𝒚 Deduce the result when 𝒑 = 𝟐 Solution: 𝑝 Suppose 𝑆 = \|𝑓(𝑧)\|𝑝 = (𝑈 2 + 𝑉 2 )2 U and V are function of x and y both 𝜕𝑠 𝑝 𝑝 = 2 (𝑈 2 + 𝑉 2 )2 −1 .(2𝑈𝑈𝑥 + 2𝑉𝑉𝑥 ) 𝜕𝑥 = 𝑝(𝑈 2 + 𝑉 2 ) 𝑝−2 2 (𝑈𝑈𝑥 + 𝑉𝑉𝑥 ) 𝑝−2 𝜕 2𝑠 𝑝−2 2 2 )2 2 2 2) 2 (𝑈 ) = 𝑝 ( ) + 𝑉 . 2(𝑈𝑈 + 𝑉𝑉 + 𝑝(𝑈 + 𝑉 (𝑈𝑈𝑥𝑦 + 𝑈𝑥2 + 𝑉𝑉𝑥 + 𝑉𝑥2 ) 𝑥 𝑥 𝜕𝑥 2 2 Likewise, 𝜕2 𝑠 2 𝑝−2 𝜕𝑦 2 = 𝑝( ) (𝑈 2 + 𝑉 2 )𝑝−4/2 . 2(𝑈𝑈𝑦 + 𝑉𝑉𝑦 ) + 𝑝(𝑈 2 + 𝑉 2 ) 2 𝑝−2 2 (𝑈𝑈𝑦𝑦 + 𝑈𝑦2 + 𝑉𝑉𝑦 + 𝑉𝑦2 ) (2) 𝜕2 𝜕2 Now (𝜕𝑥 2 + 𝜕𝑦 2 ) 𝑠 = 2𝑝 ( 𝑝(𝑈 2 + 𝑉 2 ) 𝑝−2 2 4 𝑝−2 2 ) (𝑈 2 + 𝑉 2 )𝑝−2 [(𝑈𝑈𝑥 + +𝑉𝑉𝑥 ) + (𝑈𝑈𝑦𝑦 + 𝑉𝑉𝑦 )2 ] + [𝑈(𝑈𝑥𝑥 + 𝑈𝑦𝑦 ) + 𝑉(𝑉𝑥𝑥 + 𝑉𝑦𝑦 ) + (𝑈𝑥2 + 𝑐) + (𝑉𝑥2 + 𝑈𝑦2 )] 𝜕2 𝑈 𝜕2 𝑈 In (3) we can use 𝜕𝑥 2 + 𝜕𝑦 2 = 0 i.e𝑈𝑥𝑥 + 𝑈𝑦𝑦 = 0 and 𝑉𝑥𝑥 + 𝑉𝑦𝑦 = 0 𝜕𝑈 𝜕𝑥 𝜕𝑉 𝜕𝑈 𝜕𝑉 = 𝜕𝑥 and𝜕𝑦 = − 𝜕𝑦 𝑖. 𝑒 𝑈𝑥 = −𝑉𝑦 𝑎𝑛𝑑 𝑈𝑦 = −𝑉𝑥 𝜕2 4 𝜕2 We get (𝜕𝑥 2 + 𝜕𝑦 2 ) 𝑠 = 𝑝(𝑝 − 2)(𝑈 2 + 𝑉 2 )𝑝−2 [(𝑈𝑈𝑥 + +𝑉𝑉𝑥 ) + (−𝑈𝑉𝑥 + 𝑉𝑈𝑥 )2 ] + 𝑝(𝑈 2 + 𝑉 2 ) 𝑝−2 2 (𝑈 ∗ 0 + 𝑉 ∗ 0 + 2𝑈𝑥2 + 2𝑉𝑥2 ) 𝑝(𝑝 − 2)(𝑈 2 + 𝑉 2 ) 𝑝−4 2 [𝑉 2 𝑈𝑥2 + 𝑉 2 𝑉𝑥2 + 2𝑈𝑉𝑈𝑥 𝑉𝑥 + 𝑈 2 𝑉𝑥2 + 𝑉𝑥2 − 2𝑈𝑉𝑈𝑥 𝑉𝑥 ] + 2𝑝(𝑈 2 + 𝑉 2 ) 𝑝−4 2 (𝑉𝑥2 + 𝑈𝑦2 ) 4 𝑝−2 𝜕 2𝑠 𝜕 2𝑠 ( 2 + 2 ) = 𝑝(𝑝 − 2)(𝑈 2 + 𝑉 2 )𝑝−2 [(𝑈 2 + 𝑉 2 )(𝑉𝑥2 + 𝑈𝑥2 ) + 2𝑝(𝑈 2 + 𝑉 2 ) 2 (𝑉𝑥2 + 𝑈𝑥2 )] 𝜕𝑥 𝜕𝑦 = 𝑝(𝑝 − 2)(𝑈 2 + 𝑉 2 ) = 𝑝(𝑈 2 − 𝑈 2 ) 𝑝−2 2 = 𝑝2 (𝑈 2 + 𝑉 2 ) 𝑝−2 2 (𝑉𝑥2 + 𝑈𝑥2 ) + 2𝑝(𝑈 2 + 𝑉 2 ) 𝑝−2 2 (𝑉𝑥2 + 𝑈𝑥2 ) (𝑉𝑥2 + 𝑈𝑥2 )(𝑝 − 2 + 2) 𝑝−2 2 (𝑉𝑥2 + 𝑈𝑥2 ) 𝜕 2𝑠 𝜕 2𝑠 ( 2 + 2 ) = 𝑝2 \|𝑓(𝑍)\|𝑝−2 \|𝑓′(𝑍)\|2 𝜕𝑥 𝜕𝑦 When 𝑝 = 2 𝜕2 𝜕2 ( 2 + 2 ) \|𝑓(𝑍)\|2 = 4\|𝑓′(𝑍)\|2 𝜕𝑥 𝜕𝑦<\|endoftext\|>	4.4375
2,567	In Physics, a “Vacuum” is defined as the absence of matter in a control volume. Generally, total vacuum is an ideal extreme condition. Therefore, in reality we experience partial vacuum where ambient pressure is different from zero but much lower than the ambient value. Depending on the pressure we can have different degrees of vacuum, ranging from low vacuum (at 1×105 to 3×103 Pa) to extremely high vacuum (at pressures <10-10 Pa). For the purpose of comparison, space vacuum might present pressures down to ~10-14 Pa in the interstellar regions. Vacuum is needed in research and several industrial sectors for a wide range of different applications and purposes. The main way to create vacuum is by first using primary vacuum pumps -machines that relying on the general principles of viscous fluid dynamics. Though fossil fueled power plants aren’t as commonly used anymore, coal fired power generation is still a major source of global electricity, making up about 25% of the market in total. Compared to other options in fossil fuel power generation, coal is found to be the most economical choice as well as a reliable option. Making demands that are heavily reliant on other fuels, such as oil-fired for example, slowly levers to coal power generation. The global reserve of coal can be found in abundance when compared to other energy sources (such as oil for example) as there is about 3 times more of it. Also, IGCC comes with an economic benefit as the price of coal has remained relatively constant, which results in a higher degree of confidence when relying on coal as an energy source in the future. How Does an IGCC Work? The system uses a high pressure gasifier to turn coal and other carbon based fuels such as high-sulfur coal, heavy petroleum residues and biomass into pressurized clean coal synthesis gas (also known as syngas). The solid coal is gas-fired to produce syngas by gasifying coal in a closed pressurized reactor with a shortage of oxygen to ensure that coal is broken down by the heat and pressure. Before going out of the system, the syngas runs through a pre-combustion separation process to remove impurities, starting with water-gas-shift reaction to increase concentration of hydrogen and efficiency during combustion process, to a physical separation process (through variable methods). After that, a fairly pure syngas is used as a fuel in a combustion turbine that produces electricity. Waste heat contained in a gas turbine’s exhaust is used to produce steam from feed water that further turns a steam turbine to generate additional electricity. In the ever-expanding market for waste-heat recovery methods, different approaches have been established in order to combat the latest environmental restrictions while achieving more attractive power plant efficiencies. As gas turbine cycles continue to expand within the energy market, one particular technology has seen a significant upsurge due to a number of its beneficial contributions. Supercritical CO2 (S-CO2) bottoming cycles have allowed low power units to utilize waste heat recovery economically. For many years, the standard for increasing the efficiency level of a GTU (Gas Turbine Unit) was to set up a steam turbine Rankine cycle to recycle the gas turbine exhaust heat. However, the scalability constraints of the steam system restrict its application to only units above 120MW. Tip leakage is generated inevitably by the clearance between the rotating blades and the stationary casing of a turbine, and is responsible for both the aerodynamic losses in a turbine stage and the high heat-loads in the tip region . To decrease tip leakage and improve component performance, shroud seal structures have been widely applied to modern turbine components, especially to low pressure turbines, because of their advantage on both aerodynamic and structural features. However, due to the complexity of the shroud geometry, the flow structures and thermodynamic process in shroud can be extremely complicated, that is interactions of vortices, separations, jet flow, etc. Thus, because of the complex geometry of shrouds, as well as strong interactions between the tip leakage and main flow, it is not easy to draw a numerical simulation with satisfactory accuracy and time-costing in shrouded turbines. This begs the question of what should the compromise be between using simplified loss models and full 3D CFD analysis for leakage modelling? In the main flow path of a turbine the flow will always be dominated by the blades shape, while for leakage cases the flow will be dominated by the motion and evolution of small eddies. Rosic et al. reviewed the importance of shroud leakage modelling in multistage turbines. The comparison of measurements and 3D calculations shows that the flow in shrouded low aspect ratio turbines is dominated by shroud leakage. This is especially true as regards the loss distribution. The rotor shroud leakage flow greatly increases the secondary flow in the downstream stators and drives low energy fluid towards mid-span. It was pointed out that with very low values of shroud leakage the flow is reasonably well modelled by a simple 1D model of the leakage flow, using sources and sinks on the casing. However, for more representative real clearances, full 3D modelling of the seal and cavity flows is necessary in order to obtain reasonable agreement. Given that developing a simulation method with both high precision and fast solving speed is imperatively demanded for engineers to assess new designs, Zhengping Zou et al. suggested that one of the potential approaches for solving the problem is a method that couples low dimensional models, 1D and 2D models, of the shroud flow with 3D (three-dimensional) simulations of the main flow passage. Specifically, some boundary source and boundary sink is set on the interface between the shroud and the main flow passage, and the source term and sink term are determined by the shroud leakage model. The schematic of this process is given in Fig. 1. The results of his study demonstrate that the proposed models and methods will contribute to pursue deeper understanding and better design methods of shrouded axial turbines. The term, “mixed flow compressor”, refers to a type of compressor that combines axial and radial flow paths. This phenomenon produces a fluid outflow angle somewhere between 0 and 90 degrees with respect to the inlet path. Referred to as the meridional exit angle, the angled outflow of this mixed flow configuration possesses the advantages of both axial and centrifugal compressors. Axial compressors can produce higher order efficiencies for gas engines, but they have relatively low-pressure ratios unless compounded into several stages. Centrifugal compressors can produce high-pressure ratios in a single stage, but they suffer from a drop in efficiency. The geometrical distinction of mixed flow compressors allows for higher efficiencies while maintaining a limited cross-sectional area. The trade-off for a mixed flow compressor when introduced to aero gas turbines is that there is an associated weight increase due to the longer impellers needed to cover this diagonal surface. However, when related to smaller gas turbines, the weight increase becomes less significant to the overall performance of the engine. In recent days, many people find themselves spending time and resources on uncovering the best solution to optimize the power generation cycle. Until recently, 80% of power plants worldwide (whether fossil fuel, nuclear, or clean technology) used steam as its main working fluid and while it is still the most common option, today’s power plants are finding another fluid to use. Although supercritical CO2 study began in the 1940’s, it was disregarded as an alternative fluid option because it was expensive to explore and steam was still perfectly reliable at the time. Nowadays due to increasing quantity and quality demand in power, researchers are looking into the possibility of replacing steam with supercritical carbon dioxide. The discover of this property, increases the incentive of exploring the technology further. This year, the US Department of Energy is awarding up to $80 million towards projects to build and operate a supercritical CO2 plant. A major concern for pump system engineers over the last fifty years has been caviation. Cavitation is defined as the formation of vapor bubbles in low pressure regions within a flow. Generally, this phenomenon occurs when the pressure value within the flow-path of the pump becomes lower than the vapor pressure; which is defined as the pressure exerted by a vapor in thermodynamic equilibrium conditions with its liquid at a specified temperature. Normally, this happens when the pressure at the suction of the pump is insufficient, in formulas NPSHa ≤ NPSHr, where the net positive suction head is the difference between the fluid pressure and the vapor pressure at the pump suction and the “a” and “r” stand respectively for the values available in the system and required by the system to avoid cavitation in the pump. The manifestation of cavitation causes the generation of gas bubbles in zones where the pressure gets below the vapor pressure corresponding to that fluid temperature. When the liquid moves towards the outlet of the pump, the pressure rises and the bubbles implode creating major shock waves and causing vibration and mechanical damage by eroding the metal surfaces. This also causes performance degradation, noise and vibration, which can lead to complete failure. Often a first sign of a problem is vibration, which also has an impact on pump components such as the shaft, bearings and seals. The use of computational fluid dynamics (CFD) in turbomachinery design is getting more and more popular given the increased computational resources. For the design process, however, there is no need for extensive CFD capabilities as the effort is put on minimizing engineering time while obtaining a design which is about 90% optimized. Here we are presenting two cases where CFD is used to derive significant information for pump design. First, the influence of the blade shape on the parameters of the single blade hydrodynamic pump was studied by Knížat et al . The investigation of the pump properties was carried out experimentally with a support of CFD methods. The accuracy of applied steady-state calculations was satisfactory for the process of design of a single blade pump, because of the good agreement between measured and calculated power curves. As with any turbomachinery, pump design requires a lot of effort on finding the right blade profile for the specified application. As there is no right or wrong in the process, engineers have to make some general assumptions as a starting point. Generally, we can say that the focus of this task is to minimize losses. It is obvious that the selected blade shape will affect several important hydrodynamic parameters of the pump and especially the position of optimal flow rate and the shape of the overall pump performance curves. In addition to axial and radial pump design in recent years, we also have seen the development of mixed-flow pumps. A mixed flow pump is a centrifugal pump with a mixed flow impeller (also called diagonal impeller), and their application range covers the transition gap between radial flow pumps and axial flow pumps. Let’s consider a dimensionless coefficient called “specific speed” in order to be able to compare different pumps with various configurations and features. The “specific speed” is obtained as the theoretical rotational speed at which a geometrically-similar impeller would run if it were of such a size as to produce 1 m of head at a 1l/s flow rate. In formulas: where ns is the specific speed, n the rotational speed, Q is the volume flow rate, H is total head and g is gravity acceleration. In an internal combustion engine, combustion of air and fuel takes place inside the engine cylinder and hot gases are generated with temperature of gases around 2300-2500°C which may result in not only burning of oil film between the moving parts, but also in seizing or welding of the stationery and moving components. This temperature must be reduced such that the engine works at top efficienc, promoting high volumetric efficiency and ensuring better combustion without compromising the thermal efficiency due to overcooling. Most importantly, the engine needs to function both in the sense of mechanical operation and reliability. In short, cooling is a matter of equalization of internal temperature to prevent local overheating as well as to remove sufficient heat energy to maintain a practical overall working temperature. It is also important to note that about 20-25% of the total heat generated is used for producing brake power (useful work). The cooling system should be designed to remove 30-35% of total heat and the remaining heat is lost in friction and carried away by exhaust gases.<\|endoftext\|>	4.0625
625	According to a study conducted by CDC and published in The Lancet, between 291,000 and 646,000 people worldwide die from seasonal influenza-related respiratory illnesses each year, higher than a previous estimate of 250,000 to 500,000 and based on a robust, multinational survey.The estimate excludes deaths during pandemics. “These findings remind us of the seriousness of flu and that flu prevention should really be a global priority,” says Joe Bresee, M.D., associate director for global health in CDC’s Influenza Division and a study co-author. The new estimates use more recent data, taken from a larger and more diverse group of countries than previous estimates. Forty-seven countries contributed to this effort. Researchers calculated annual seasonal influenza-associated respiratory deaths for 33 of those countries (57 percent of the world’s population) that had death records and seasonal influenza surveillance information for a minimum of four years between 1999 and 2015. Statistical modeling with those results was used to generate an estimate of the number of flu-associated respiratory deaths for 185 countries across the world. Data from the other 14 countries were used to validate the estimates of seasonal influenza-associated respiratory death from the statistical models. Researchers calculated region-specific estimates and age-specific mortality estimates for people younger than 65 years, people 65-74 years, and people 75 years and older. The greatest flu mortality burden was seen in the world’s poorest regions and among older adults. People age 75 years and older and people living in sub-Saharan African countries experienced the highest rates of flu-associated respiratory deaths. Eastern Mediterranean and Southeast Asian countries had slightly lower but still high rates of flu-associated respiratory deaths. Despite World Health Organization recommendations to use flu vaccination to help protect people in high-risk populations, few developing countries have seasonal flu vaccination programs or the capacity to produce and distribute seasonal or pandemic vaccines. CDC works with global partners to improve worldwide capacity for influenza prevention and control. CDC has helped more than 60 countries build surveillance and laboratory capacity to rapidly detect and respond to influenza threats, including viruses with the potential to cause global pandemics. These efforts, along with technical support, has helped some partners generate estimates of influenza-associated deaths, which contributed to this global effort. Global surveillance also provides the foundation for selecting the viruses used to make seasonal flu vaccines each year. This helps improve the effectiveness of flu vaccines used in the United States. Global surveillance also is crucial to pandemic preparedness by identifying viruses overseas that might pose a human health risk to people in the United States. “This work adds to a growing global understanding of the burden of influenza and populations at highest risk,” says CDC researcher Danielle Iuliano, lead author of The Lancet study. “It builds the evidence base for influenza vaccination programs in other countries.” Iuliano, AD, Roguski, KM, Chang, HH et al. Estimates of global seasonal influenza-associated respiratory mortality: a modelling study. Lancet. 2017; (published online Dec 13.)<\|endoftext\|>	3.828125
1,589	# Question: What Are The 10 Prime Numbers? ## What is the smallest prime number? The smallest prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and 23. The number 2 is the only even prime number. The number 7 has only two factors: 1 and itself.. ## What is the smallest prime number between 1 to 10? The first 1000 prime numbers1101–2022921–407311341–6017922961–8028334914 more rows ## Why 1 is not a prime number? The confusion begins with this definition a person might give of “prime”: a prime number is a positive whole number that is only divisible by 1 and itself. The number 1 is divisible by 1, and it’s divisible by itself. But itself and 1 are not two distinct factors. … Excluding 1 from the primes smooths that out. ## Why is 11 not a prime number? For 11, the answer is: yes, 11 is a prime number because it has only two distinct divisors: 1 and itself (11). As a consequence, 11 is only a multiple of 1 and 11. ## How do you know what a prime number is? To prove whether a number is a prime number, first try dividing it by 2, and see if you get a whole number. If you do, it can’t be a prime number. If you don’t get a whole number, next try dividing it by prime numbers: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below). ## What is the greatest prime numbers between 1 to 10? there are 4 prime numbers between 1 and 10 but the greatest is 7 [prime numbers are those numbers which are only divisible from 1 and itself the prime numbers between 1 and 10 are 2,3,5 and 7] 2 is only even prime number. NOTE- 1 is a composite number. 7 is the greatest prime number between 1 to 10. ## Is 1 a prime number for kids? There are 8 prime numbers under 20: 2, 3, 5, 7, 11, 13, 17 and 19. … 1 is not a prime number, because it only has 1 factor – itself! Prime numbers can continue well past 100. For example, 21,577 is a prime number. ## Why is 2 a prime number? Proof: The definition of a prime number is a positive integer that has exactly two distinct divisors. Since the divisors of 2 are 1 and 2, there are exactly two distinct divisors, so 2 is prime. … In fact, the only reason why most even numbers are composite is that they are divisible by 2 (a prime) by definition. ## What are twin prime numbers? In twin prime conjecture. …that there are infinitely many twin primes, or pairs of primes that differ by 2. For example, 3 and 5, 5 and 7, 11 and 13, and 17 and 19 are twin primes. As numbers get larger, primes become less frequent and twin primes rarer still. ## Is there a pattern in prime numbers? Now, however, Kannan Soundararajan and Robert Lemke Oliver of Stanford University in the US have discovered that when it comes to the last digit of prime numbers, there is a kind of pattern. Apart from 2 and 5, all prime numbers have to end in 1, 3, 7 or 9 so that they can’t be divided by 2 or 5. ## What are 4 examples of prime numbers? Prime numbers are numbers that have only 2 factors: 1 and themselves. For example, the first 5 prime numbers are 2, 3, 5, 7, and 11. By contrast, numbers with more than 2 factors are call composite numbers. ## How much money do you get for finding a prime number? (If someone discovers a prime number of 100 million digits, the prize is \$150,000 from the Electronic Frontiers Foundation. The first 1 billion-digit prime is worth \$250,000.) ## How many prime numbers are there? Originally Answered: How many prime numbers are there totally? There are 25 prime numbers from 1 to 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. ## What is 1 called if it is not a prime? A natural number greater than 1 that is not prime is called a composite number. For example, 5 is prime because the only ways of writing it as a product, 1 × 5 or 5 × 1, involve 5 itself. However, 4 is composite because it is a product (2 × 2) in which both numbers are smaller than 4. ## What’s the opposite of a prime number? composite numbersThe opposite of prime numbers are composite numbers. A composite number is a positive nutural number that has at least one positive divisor other than one or itself. ## Is 2 not a prime number? The first five prime numbers: 2, 3, 5, 7 and 11. A prime number is an integer, or whole number, that has only two factors — 1 and itself. Put another way, a prime number can be divided evenly only by 1 and by itself. ## Is there a highest prime number? The Great Internet Mersenne Prime Search (GIMPS) has discovered the largest known prime number, 277,232,917-1, having 23,249,425 digits. ## How do you find the nth prime number? An easy way to determine if a number is prime is by trial division: divide the number n by all the integers less than n, and if no exact divisors–other than 1–are found, then n is prime. You can see how this becomes time-consuming as the value of n increases. ## What is the smallest whole number? Which is the smallest whole number? Solution. Zero (0) is the smallest whole number. ## What are the prime numbers between 1 and 10? Prime numbers between 1 and 10 are 2, 3, 5, 7. ## What are the prime numbers to 100? When a number has more than two factors it is called a composite number. Here are the first few prime numbers: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, etc.<\|endoftext\|>	4.53125
1,454	The quadriceps muscles and tendons are located at the front of the upper legs and are referred to as the four heads of the femur. The quadriceps can be referred to as the quads, the quadriceps femoris musle or the quadriceps extensors. They comprise of four muscles: the vastus lateralis, vastus medialis, vastus intermedius and the rectus femoris. The rectus femoris is the largest of the four muscles, and is located in the middle of the thigh. The vastus lateralis is situated on the outside of the thigh. The vastus medialis is located on the inside of the thigh and the vastus intermedius is deep to the rectus femoris and is found between the vastus lateralis and the vastus medialis. The quadriceps cross the knee, and are extensors of the knee. The quadriceps muscles help stabilise the knee cap (the patella). They are also flexors of the hip, as the rectus femoris originates from the hip, the ilium. They allow activities such as climbing stairs or lunges. The quadriceps play a vital role in jumping, walking, and particularly running. A quadriceps strain is caused when the muscle or tendon is overstretched and becomes strained. A quadriceps strain can also be referred to as a pulled muscle, and can be a tear or a complete rupture. Injury often occurs as a result of an impact, an extrinsic injury, when the muscle is contracted (the muscle fibres are shortened and tight); for example, when a player is kicked. The most typical area for a tear to occur is at the point where the tendon joins the muscle. A quadriceps strain is a common sports injury in runners. It often occurs in athletes involved in sports that involve sprinting, jumping or kicking, such as football, interval training or hockey. It particularly occurs if the quadriceps has not been warmed up sufficiently. It can also happen as a result of a direct impact (extrinsic injury), for instance a kick. A quadriceps strain is divided into 3 categories, grade 1, 2 and 3, depending on its severity. Pain can be felt when the knee is straightened against resistance, and when the knee is bent. Grade 1 (a partial tear) In a grade 1 strain the pain is moderate, and tightness can be felt in the quadriceps. During exercise pain can disappear and might not resume until after exercise. The area around the muscle tear might be tender to touch, warm and red. Grade 2 (a partial tear) Pain is immediate and can be sudden and sharp. There might be slight swelling and the area will be tender to touch. It might be difficult to bend the affected leg. Pain can be felt when the leg is straightened against resistance. The hamstring muscles might be tight, and tightness tends to occur the day after exercise. Grade 3 (a complete rupture) Pain is severe, and walking is difficult. Pain is worse when the knee is flexed (bent), for example walking upstairs, sitting. Inflammation and bruising is more severe. A lump can be felt where thickening has occurred. This happens as a result of the build-up of scar tissue caused by the rupture of the quadriceps muscles, which then forms scar tissue as it heals (these are the nodules). A gap might be felt where the muscle has ruptured. It is important to seek early treatment to avoid developing a chronic quadriceps strain and weakening the quadriceps. Rest: to prevent further damage, and avoid any exercises that put a strain on the quadriceps muscles until the muscles have completely recovered. In the sub acute (3 days to 3 weeks) and the chronic stage (3 weeks to 2 years) it is important that training should be adapted to avoid jumping or any exercises that put excessive strain on the quadriceps. A physiotherapist or sports therapist or sports massage therapist can advise when exercise should be resumed and what exercise would be appropriate. It is important to always warm up and cool down properly when exercising. Ice treatment: Ice can be applied for 10-15 minutes, every 2-3 hours in the acute and sub- acute stage (frequency can be reduced according to recovery, and can be continued for as long as deemed necessary). An ice stick can be good for massaging the specific area. As the ice is directly massaged onto the skin, and is quite intense, it can be applied for less time 7-12 minutes. In the sub-acute stage (3days – 3 weeks) heat therapy can be applied, usually in the form of a hot bath. Compression: to reduce swelling and restrict movement. NSAIDS (anti-inflammatories) and paracetamol can be taken to aid pain relief. Medical advice should be sought, in case of possible side effects. Steroid injections (under medical guidance) can alleviate pain, but it is recommended exercise should be avoided for 1-2 weeks after an injection. Orthotics can prevent overprontation and make sure there is no undue strain placed on the quadriceps by inadequate footwear. It is therefore, worth consulting a podiatrist, who can perform gait analysis and advice on appropriate foot wear. Kinesology taping aids recovery by assisting with lymphatic drainage, and the repair of damaged tissues. A doctor or physiotherapist might recommend an MRI scan to assess the extent of rupture. In severe cases surgery might be performed. Ultrasound may be prescribed by the physio to help speed up the repair process, by breaking down tissues and stretching them. It can also help alleviate pain. Massage can help aid recovery and be an effective form of treatment, as it helps to stretch the muscle, stimulate repair, by bringing vital nutrients and oxygen to the site of injury, and helps break down the formation of scar tissue. It should not be administered during the acute stage. If there is any underlying medical condition, such as a heart condition, it is important to seek medical advice before receiving massage. Kinesio taping techniques for the quadriceps can sometimes show remarkable improvements in low grade quadriceps strains. A physiotherapist or sports therapist can recommend strengthening, flexibility and proprioceptive exercises in the sub-acute and the chronic stage of recovery. Exercises should focus on strengthening the quadriceps, for example, squats and lunges and quadriceps stretches to improve flexibility. The intensity of the exercises should be increased gradually and in a controlled way. Swimming (front/back crawl, not breaststroke) is a good way of maintaining fitness while the quadriceps recover. Initially a pull buoy can be used to rest the leg. Published: March 2, 2012Author: Sophia Cross, BA (Hons) MA Quadriceps Strain Treatments<\|endoftext\|>	3.828125
1,817	Lecture Note 8 ```ST361: Ch5.4 + Ch1.3 Random Variable and Its Probability Distribution ----------------------------------------------------------------------------------------------------------Topics: Random Variable (§5.4) Probability Distribution of a discrete random variable (§5.4, §1.3) Mean and Variance of a discrete random variable (§5.4) Probability Distribution of a continuous variable (§5.4, §1.3) Mean and Variance of a continuous random variable (§5.4) ---------------------------------------------------------------------------------------------------------- Random Variable (r.v.)  A random variable is a variable whose value is a __________________ _________________________________ of an experiment.  We can think that an r.v. is any rule that associates a _______________ with each ____________ in an experiment. Ex. Consider an experiment of tossing 2 coins. One way to define a r.v. is  For numerical variables, most of the time the values themselves can be used as a r.v. Ex. Define a r.v. to for the exam score of a student as ----------------------------------------------------------------------------------------------------- Discrete r.v. The possible values of the r.v. are isolated points along the number line. Ex. x = # of Heads of tossing 2 coins x, =0,1,2 Ex. x = # of telephone lines in a company that is in use, x =0,1,2, 3,…… (c.f. Continuous r.v.: The possible values forms an interval along the real line) 1  Probability Distribution of a Discrete r.v. 1. The probability distribution of a r.v., denoted as _________, describes (a) ________________________________________________ and (b) _________________________________________________ Ex. X = result of tossing a fair dice. The probability distribution of x is 2. In general, the probability that x gets a value c, P(x=c), is defined as the sum of all corresponding outcomes in S (i.e., the sample space) that are assigned to the value x. Ex. X = # of heads in tossing two fair coins. Then the probability distribution of X is 3. There are 3 ways to display a probability distribution for a discrete r.v.: Ex. Toss a (unfair) coin 3 times, and let x= # of heads. Then the probability distribution of x is given as below: (1) Density plot 2 (2) Table x P(x) 0 0.1 1 0.4 2 0.3 3 0.2 (3) Formula  From the probability distribution, we can calculate P( x = 3 ) = P( x < 2 ) = P( x  2 ) = P( x > 0 ) = 4. For any probability distribution P(x), (recall the axiom of probabilities…) (1) (2) Ex. (1) Find the value of c so that the following function is a probability distribution of a r.v. x: P  x   c  x  2 , where x  0,1,2,3 (2) For this probability distribution, find P(x  2) 3  FYI: Mean and Variance of a discrete r.v. with probability distribution P  x  The mean  x   x  P  x  x (The mean of a r.v. is also called as expected value.) The variances  x2    x   x   P  x  2 x The standard deviation  = Ex. A contractor is required by a county planning department to submit from 1 to 5 different forms, depending on the nature of the project. Let x = # of forms required of the next contractor, and px  kx for x=1,2,3,4,5. (a) What is the value of k? (b) What is the probability that at most 3 forms are required? (c) What is the expected number (i.e., mean) of forms required? (d) What is the SD of the number of forms required? (This calculation won’t be included in the exams) 4  Continuous Random Variable (r.v.) A r.v. is continuous if its possible values forms an interval along the real line Ex. x = exam score, 0  x  100 Ex. x = your height, 0  x    Probability Distribution of a continuous r.v. 1. Every continuous r.v. x has a __________________________, denoted as ________ such that for any 2 numbers a and b (a<b), P( a  x  b ) = ________ under the density curve of f(x) between a and b Ex. P( -1  X  1 ) = Ex. P( X  1 ) = Comment: For continuous r.v. x, (1) Probability is the area encompassed by the density curve, the two vertical bars and the x-axis. _______________________________ ______________________________________________________ (2) Because area under the curve represents probabilities, the total area under the density curve should be equal to __________ 5 (3) Unlike the discrete r.v., the Y-axis is not probability ( The height is determined so that ________________ __________________________________________ Ex. ? (4) P( X = a ) = _________ (Think what is the size of the corresponding area?) (5) For a continuous r.v., (6) Ways to presenting the density function of a continuous r.v.: by a density plot or formula (see next page) 6 Ex. Consider a r.v. X= test score. The probability distribution of X is given below. 1) Density plot 2) Density function if x  40  ________  1 1  y if   30 1200 2  if f x    120 3  if  120   if x  100  ______ P( X=70) = P( 60 < X < 80 ) = P(X > 70 ) = P( X  70 ) = 7 Ex. P X  20  P X  20  P5  X  15  Summary: A density function f  x  of a r.v. has to satisfied the following properties: (a) f  x   0 THINK: do we need 0  f  x   1 ? (b) The total area under the curve is 1, i.e.,    f  x  dx  1 Why?  FYI: Mean and Variance of a continuous r.v. with density function f  x   The mean x   x  f  x  dx  (The mean of a r.v. is also called as expected value.) The variances  2      x  x  The standard deviation   2  f  x  dx   x     x 2  f  x  dx 8 ```<\|endoftext\|>	4.71875
534	1. ## Area of segment. A circle has the center of A and the radius of 5. \|BC\| = 8; \|AC\|=\|AB\|=5 Find the area of the gray bit of the circle. 2. Originally Posted by Kane535 A circle has the center of A and the radius of 5. \|BC\| = 8; \|AC\|=\|AB\|=5 Find the area of the gray bit of the circle. Hi Kane535, welcome to MHF. Altitude from A on BC is 3. So area of ABC = 1/283 = 12 sq. units Angle ABC is given by θ = 2arctan(4/3) Area of the sector ABC = 1/2r^2*θ, where θ is in radians. Now find the area of the shaded portion. 3. Originally Posted by Kane535 A circle has the center of A and the radius of 5. \|BC\| = 8; \|AC\|=\|AB\|=5 Find the area of the gray bit of the circle. Use the cos rule to find angle A (I will use radians because it's easier) $\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{5^2+5^2-8^2}{2 \cdot 5 \cdot 5} = -\frac{7}{25}$ $A = \arccos \left(-\frac{7}{25}\right) \approx 1.855$ The area of a sector is given by $A = \frac{1}{2}r^2 \theta$. In this case $\theta$ is angle A and $r=5$ $A_s = \frac{1}{2} \cdot 5^2 \cdot \arccos \left(-\frac{7}{25}\right)$ Since this includes the white triangle we must remove it's area which is given by $A_t = \frac{1}{2} bc \sin A = \frac{1}{2} \cdot 5^2 = \frac{25}{2}$ $A_g = A_s - A_t = \frac{1}{2} \cdot 5^2 \cdot \arccos \left(-\frac{7}{25}\right) - \frac{25}{2}$ Evaluate for the answer, I get an answer of $10.7 \text{ 3sf }$<\|endoftext\|>	4.375
1,843	# Math for the General Class Ham Radio Operator A Prerequisite Math Refresher For The Math-Phobic Ham. ## Presentation on theme: "Math for the General Class Ham Radio Operator A Prerequisite Math Refresher For The Math-Phobic Ham."— Presentation transcript: Math for the General Class Ham Radio Operator A Prerequisite Math Refresher For The Math-Phobic Ham Why is This Lesson for You? Math Vocabulary What are equations and formulas? What do variables mean? What is an operator? C 2 = A 2 + B 2 Math Vocabulary What is an operator? Math operations: –Add: + –Subtract: − –Multiply: X or –Divide: ∕ or –Exponents: Y X –Roots: or Math Vocabulary What does solving an equation mean? Getting the final answer! Getting the Final Answer: Tricks of the Trade: Opposite math operations: Addition  Subtraction Multiplication  Division Roots  Exponents If you do something to one side of the equation, do exactly the same thing to the other side of the equation to keep everything equal XXXX A number divided by the same number is 1, = 1 A number multiplied by 1 is that number, Y * 1 = Y What does solving an equation mean? Example #1 C 2 = A 2 + B 2 Assume A and B are known Want to solve for C.  C 2 =  A 2 + B 2 Apply same operation to both sides  C 2 =  A 2 + B 2 Opposite operations cancel each other C =  A 2 + B 2 Voila!!! What does solving an equation mean? Example #2 The equation for Ohm’s Law is: E = I * R The variables mean: –E represents voltage –I represents current –R represents resistance The math operator is multiplication. What does solving an equation mean? Example #2 E = I * R –Current is 10 (we will disregard units for now) –Resistance is 50 Therefore: E = 1050 E = 500 (in this case volts) Math Vocabulary What does solving an equation mean? What if we know the voltage and the current and want to find the resistance? E = I RR = E / I Let’s do some math! Simple addition Let’s do some math! Multiply R 1 times R 2 –Write the number down Add R 1 and R 2 –Write the number down Divide the first number by the second to find the answer. R 1 = 50 R 2 = 200 R T = Total Resistance = ? Let’s do some math! R 1 * R 2 = ?  50 * 200 = 10,000 R 1 + R 2 = ?  50 + 200 = 250 R T = 10,000/250 = 40 R 1 = 50 R 2 = 200 R T = ? Let’s do some math! Do each fraction in the denominator in turn 1/R n –Write the number down Add all fraction results together. –Write the number down Divide 1 by the sum of the fractions. Let’s do some math! R 1 = 50 R 2 = 100 R 3 = 200 1/R 1 = ?  1/50 = 0.02 1/R 2 = ?  1/100 = 0.01 1/R 3 = ?  1/200 = 0.005 Sum of fractions = ?  0.02 + 0.01 +0.005 =0.035 1/Sum of fractions = ?  R T = 1/0.035 = 28.6 Let’s do some math! Square the numerator E –Same as E * E –Write the number down Divide the squared number by R. E = 300 R = 450 Let’s do some math! E = 300 R = 450 E 2 = ? (square E)  300 2 = 90,000 90,000/R = ?  P = 90000/450 = 200 Let’s do some math! V Peak = 100 V RMS = ? Solve for V RMS  V RMS = V Peak / 1.414 Plug in value for V Peak  V RMS = 100/1.414  100/1.414 = 70.7 Let’s do some math! Sometimes two formulas need to be used to come to a final answer. V Peak = 300 R = 50 PEP = ? Solve equation 1 for V RMS Plug the value of V RMS into equation 2. Let’s do some math! Solve for V RMS  V RMS = 300 / 1.414  300/1.414 = 212.2  Write the number down Plug the value into V RMS.  V RMS 2 = 45,013.6  Write the number down Divide the square by 50  45,013.6 /50 = 900.3 V Peak = 300 R = 50 PEP = ? Let’s do some math! N S = 300 N P = 2100 E P = 115 E S = ? Solve for E S –Multiply both sides by E P –The E P values on the left cancel Solution is Let’s do some math! N S = 300 N P = 2100 E P = 115 E S = ? N S * E P = ?  300 * 115 = 34,500  Write the number down Result / N P = ?  E S = 34500/2100 = 16.4 Let’s do some math! The right side of this equation is a ratio. Ratios are numbers representing relative size A ratio compares two numbers. –Just a fraction with the two numbers being compared making up the fraction. Let’s do some math! Z P = 1600 Z S = 8 Ratio of N P to N S = ? Z P / Z S = ?  1600/8 = 200  Write the number down 200 1/2 = ?  200 1/2 = 14.1 Ratio of N P to N S = 14.1 / 1  Ratio is 14.1 to 1 Let’s do some math! ← Logarithms –“the log of N is L.” –Or “What power of 10 will give you N?” ← Anti-log: Reverse or opposite of the log. Making Sense of Decibels Examples of Power Ratios commonly expressed in dB: Gain of an amplifier stage Pattern of an antenna Loss of a transmission line Ratio of the Power Out to the Power In Common Decibel Tables 1dB=10 x log 10 1.26 3dB=10 x log 10 2 6dB=10 x log 10 4 7dB=10 x log 10 5 9dB=10 x log 10 8 10dB=10 x log 10 10 13dB=10 x log 10 20 17dB=10 x log 10 50 20dB=10 x log 10 100 -1dB=10 x log 10 1/1.26 -3dB=10 x log 10 1/2 -6dB=10 x log 10 1/4 -7dB=10 x log 10 1/5 -9dB=10 x log 10 1/8 -10dB=10 x log 10 1/10 -13dB=10 x log 10 1/20 -17dB=10 x log 10 1/50 -20dB=10 x log 10 1/100 Let’s do some math! Divide P2 by P1. –Write the number down. Press the log key on your calculator and enter the value of P2/P1. –Write the number down. Multiply the result by 10. P2 = 200 P1 = 50 dB = ? Let’s do some math! P2 = 200 P1 = 50 dB = ? P2/P1 = ?  200/50 = 4  Write the number down. Log 4 = ?  Log (4) = 0.602  Write the number down. 0.602 * 10 = ?  0.602 * 10 = 6.02 Thank goodness it’s over! Similar presentations<\|endoftext\|>	4.5625
394	MASTERS of the universe take note. Only cosmic strings can hasten the demise of a black hole – a revelation that suggests the laws of physics should hold true even in the universe’s most extreme environments. The simplest way to witness a black hole’s death is to wait. That’s because theory says it emits radiation, which eventually depletes the black hole of all its mass. Even for a black hole with the mass of the sun, this natural evaporation process takes 1057 times the age of the universe. In 1983, William Unruh at the University of British Columbia in Vancouver came up with a way to speed up the evaporation, in the form of a thought experiment to explore whether it would be possible to tap some of a black hole’s energy. He imagined lowering a bucket to siphon off some of its radiation, for example, using a space elevator stretching from a spacecraft to the black hole’s radiation zone. A consequence of this would be that you hasten the black hole’s death. But for this to work, Unruh assumed that the bucket was unbreakable and the elevator cable had no mass. Now Adam Brown of Stanford University has investigated what materials you could actually use. Brown found that no conventional material will work – even in principle. A cable made of carbon nanotubes, which have the highest known ratio of tensile strength to weight, would have to be so thick at the far end that it would collapse under gravity. While he showed the job may still be possible using the fundamental strings of string theory, these are still hypothetical (Physical Review Letters, doi.org/p9s). “Pretty much anything we can learn about black holes has a good chance of leading to deep insights about the laws of physics,” says Daniel Harlow of Princeton University. More on these topics:<\|endoftext\|>	4
1,042	If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. ## Integral Calculus ### Course: Integral Calculus>Unit 3 Lesson 10: Volume: disc method (revolving around other axes) # Calculating integral disc around vertical line AP.CALC: CHA‑5 (EU) , CHA‑5.C (LO) , CHA‑5.C.2 (EK) Calculate the integral from the last video. Created by Sal Khan. ## Want to join the conversation? • Why is it that 5y is not integrated to 5y^2/2? • Ok, I'm a little (read: a lot!) confused by these previous answers, but you are right, the integral of 5y would be 5(y^2)/2 — but look again! We weren't taking the integral of 5y, we were taking the integral of 5! • Does this shell has or not an empity core with radius 2. ? (sorry for the stupid question by the way,But this doubt is killing me.) • This is the disc method of integration. The disc method can find the integral of a solid of revolution around an axis. It's finding the volume by pir^2w, w = thickness of disc. When the volume is formed by revolving the equation about the y axis, the r is defined as a function of y. r = R(y) As integration is putting together ever smaller parts of it, pir^2delta w as delta w approaches 0. So V = pi * integral of [R(y)]^2 dy, from a to b • I think this is a washer method problem. in my understanding, we do washer's problem when a hole is created when the function is rotated to create inner and outer radius. By looking at this problem i thought there was an outer radius and inner radius. when you rotate the function around the y axis, the outer radius would be 2+x and the inner radius would be 2 because the distance of the inner radius is 2. so then the volume would be pi(outer radius)^2dy-pi(inner radius)^2dy? Is it wrong? Thanks • That would be true, if he had cut the distance between the y axis and x=2 out of the function, but he chose not too and left that whole cylindrical shape inside. TLDR he didn't cut that part out of the shape. • where did the limits from -1 to 3 come from? • In the last video Sal defined the region that we care about (want to evaluate the volume of) to be from y = -1 up to y = 3. • Is there a generalized equation for this somewhere? • How do I interpret when the question says "enclosed by the x axis"? For example if a region is enclosed by the x axis and the curve y = (x^2) - 2x. What is the volume of the solid rotated around the x axis? What is the actual definition of enclosing? Because on this graph does it necessary mean the area under the x axis enclosed by the curve or the area above the x axis enclosed by the curve and the y axis? I'm just looking for a rigorous definition of enclosed by • Your first interpretation is correct. Note that nowhere in the problem statement it was mentioned that the 𝑦-axis is a boundary. There is only one region that is bounded only by the 𝑥-axis and the curve. In general, only consider the region(s) bounded by the provided curves and/or axes. (1 vote) • Why is it that when you evaluate a definite integral with just u-substitution you have to re-calculate the values used for the bounds of integration, but in this case you can use u-substitution and keep your values the same? (1 vote) • We plugged back in the y+1, that is why the bounds didn't change. If you would have kept the u, then you would have to recalculate the bounds reflecting that. • What if I wanted to calculate the volume of this solid using an x integral and using the shell form? What would my integral be and why? (1 vote) • if we substituted back sqrt(y+1)=x and then evaluated our definite integral wouldn't it be much easier ?? (1 vote) • Not sure I understand the question, but I tried using y = x^2 from 0 to 4 and got the same result with easier integration and evaluation as you suggest. (1 vote) • Can you rotate around a non vertical, non horizontal line? (like: y = x - 3) if so, could someone explain how? (1 vote) • One more method would be to rotate the coordinate axes so that the given line is now parallel to one of them and then rotating about these new axes. (1 vote)<\|endoftext\|>	4.40625
506	# 1.2 Modeling Algebraic Expressions using tiles ## Presentation on theme: "1.2 Modeling Algebraic Expressions using tiles"— Presentation transcript: 1.2 Modeling Algebraic Expressions using tiles Which algebraic expression can be found of the next figure? What does the a represent? and b? and c? What kind of polygon has the area : A = ab + ac? c b Using the formula A= base X height, the area of any rectangle can be calculate base = a, height = b + c Area = a X ( b + c ) The expression (b+ c) indicates that this addition must be solved first and the sum should be multiplied by a. In this case, the multiplication sign can be omitted and writen as: A = a(b+c) a 1. How can the area of one of the rectangles be represented? 2. If you add the areas made by the next three rectangles, how can you write the expression that represents the total area? x Use the rectangle below to explain the next equality: 5(x + 4) = 5x + 20 Now, try to explain the next equality: 11(x – 7) = 11x - 77 Draw the rectangles that represent the next equalities 9( x + 6) = 9x + 54 8(x – 4 ) = 8x - 32 As you noticed, parenthesis are used to indicate which operation must be solved first. Exercises: Sebastian: Raise a number to the second power and then subtract 2, divide the difference by 5. Hortensia: Raise a number to the second power,divide it by 5, and then subract 2. Margarita: Raise a number to the second power and then subtract the quotient of 2 over 5 Can you tell which is Sebastian’s card, Hortensia’s card and Margarita’s card? Write as an algebraic expression Eight times the difference of x and y Nine times the difference of y and z, increased by 3z Three times the sum of a and b, decreased by 7a Match the algebraic expression with its equality 4((x – 2) -8) a) 4x-10 4(x-2) -8 b) 4x-16 (4x-2-8) c) 4x-40<\|endoftext\|>	4.90625
3,129	Hi friends, I hope you all are well. In this post, we will talk about scientific notation. Scientific notation is an essential tool that scientists and engineers can use. Archemdies presented the idea of scientific notation in the 3rd century BC. His work and scientific notation ideas are based on the novel of time, known as place value. Scientific notation ideas have evolved over many centuries but are finally represented by Archimedes. In the 16th or 17th century, mathematicians continued to adopt changes in them and invent many other new notable contributions like Rene Descartes who developed algebraic notation. Scientific notation is a way in which we can express large number values in short form in the form of an exponent or a decimal form. Scientific notation not only simplifies the equation, but it can also help scientists and engineers to do calculations easily. In the United Kingdom scientific notation is also known as standard form or standard index form. Scientific notation is commonly used by scientists and on scientific calculators it is referred to as “SCI” display mode. ## What is scientific notation? Scientific notation in detail is given below: ### Definition: Scientific notation is defined as: "it is the simplest way to express the large number into the small number or into in decimal form." It can used by scientists, mathematicians, and or engineers to simplify the calculation and express the calculation in a very efficient way. In scientific notation, the number is expressed in the product form and they have two factors. • A coefficient • A power of ten Generally, the scientific notation is written in: N=m 10n There: • N is the number • The coefficient is m. It is equal to 1 or less than 10. • The integer exponent is n. ### Example: Some examples of scientific notation are given below: • Covert 89,700 to scientific notation. 1. Firstly move the decimal to get 8.79 2. The decimal moved 3 places to the left, then the exponent is 8. 3. Then the scientific notation is 8.79103 • Convert 0.0000023 into scientific notation: 1. Firstly move the decimal to get 2.3. 2. The decimal moved 6 places, then the exponent is 2. 3. Then the scientific notation is 2.3106. ## Normalized notation: We can write any real number equation in the form of m10n in various ways. In the normalized scientific notation, the value of m is 1 or greater than 1 but it is always less than 10 and the value of n depends upon the larger value or decimal place. thus equation 550 can be written as 5.50102. this scientific notation helps to compare the number easily in an efficient way if the exponent value is large that means that the number is normalized as compared to smaller exponents. The order of magnitude can be assumed when we subtract the exponent by separating the numbers.in the table of logarithms, we can use this form of numbers to solve the log questions. The exponent in the normalized equation is negative if the value ranges between 0 and 1 for example: 00.6 then it can be written in 610-2. the exponents may be equal to 10 if the real number value is large like 0.0000000009 then it can be written as 910-10. In many fields, typically or generally normalized equation is used to express large numbers of values into simple and efficient decimal form. Exponential notation is also referred to as normalized notation. But in different fields, unnormalized or simple notation or equations can be used. Generally, the latter term of notation is more used where the value of m is not suppressed in the range between 1 to 10 and the base of the exponent may be changed or other than the 10. ## E notation: Generally, calculators or computer programs use small or large number values to present scientific notation or in some calculators, all the numbers are present or configured uniformly. The capital letter “E” or the small letter “e: can be used to represent or express the exponent which means “ten raised to the power of”. Typically in all computers or programs scientific notation or normalized notation can be abberived or represented in different styles. But the equation men is the scientific notation like m is the coefficient and the e is base 10 or n is for the exponent and it can be written as men or m10n. For example, 1.000 can be written as 1103 or also written as 1e3. ## Engineering notation: On the scientific calculator, the engineering notation can be written or expressed as “ENG”. The engineering notation is different from normalized notation because the value of exponent n is suppressed in normalized notation. Thus the value of m ranges in 1 ≤ \|m\| < 1000, other than 1 ≤ \|m\| < 10. So that's why sometimes the engineering notation is also known as scientific notation. In engineering notation, the exponent value can be orally read by using prefixes such as nano, pico, Giga, and others. For example, 14.510-9 can as read as fourteen-point-five nanometers and can also be written as 14.5nm. ## Significant figures: The number of digits in the value is used to find the degree of accuracy and precision in a value. Significant figures include all nonzeros or zero digits but the first and the last zero digits are not significant figures if zero is present in between the numbers like 1,2,3,4,5 and so on then it is also called a significant figure. Some examples are given below: • 456709800 In this value 7 significant figures are present and at the last two zero is only a placeholder and not referred to as significant figures. • Then 008809 in these values there are 4 significant figures present and the starting first two zeros are not called significant figures. So when we convert 123459900 values into normalized or scientific notation then all significant figures are present and the nonsignificant figure is removed as 123459900 has 7 significant figures then it can be written in scientific notation as 1.234599108 or also written as 1.23459900108. thus the advantage of scientific notation is to signify the significant figures efficiently. ## Estimated final digit: In scientific measurement, it is essential to record or measure all known digits from the measurement which can be calculated, and then estimate one or more digits if any information about this digit is acquirable. The estimated digits provided more information and became a significant figure because the estimated digits can provide more accurate and precise information about the leading and the ending digits. The information that we can take from the estimated digit can help in the notation. This information also helps to choose the final digits or digits that are present in the value. For example, the estimated mass of a proton can be equal to 1.67262192369(51)10-27 kg. But in this value, the error occurs when we estimate the digit inefficiently and can be written as 5.110-37 and this estimated digit can increase the standard error or some other possible errors. ## Rules for scientific notation: The rules that are used to convert the number into scientific notation are given below: • Identify the coefficient • Determine the exponent • Base • Non-zero integer • Positive or negative integer • Add or subtract the coefficient • Arithmetic operation with scientific notation ## Identify the coefficient: When we move the decimal point in the value then the original number which becomes the coefficient a is taken like 230000 then the decimal moves in it and becomes 2.3 and the 2 becomes the coefficient in the scientific notation. ## Determine the exponent: Count the number of decimals that we moved in the value and this count becomes the exponent. For example: We have a value of 46000 Then it becomes, 4.6 The total no of decimals that we moved in the value is 4 then the exponent is 4 the value becomes 4.6x104. ## Base: The base of the exponent in the scientific notation is always 10. ## Non-zero integer: The exponent is always a value it is always a non-zero integer like 104 or 108. ## Positive or negative integer: When we move the decimal from right to left then the exponent integer is positive. Example: 46000 then 4.610+4 When we move from left to right then the exponent integer is negative. Example 0.00677 then 6.7710-3 ## Add or subtract the coefficient: If the exponent is the same then we simply add or subtract them. Because the same exponent can be added or subtracted and also we can multiply or divide with them. For example: • 3.4106 + 9.7106 • The exponent is the same then we add them and then rewrite • 3.4+9.7=10.04 • Then rewrite them into scientific notation • 10.04106 then the new scientific notation is 1.004107 ## Arithmetic operation with notation: ### Division: In the division method, we divide the coefficient and subtract the exponent. An example is given below: 81062104= 8106-42 =4102 To add or subtract the scientific notation, the same exponent must be necessary. And adjust the equation if needed. An example is given below: = 2.6103+ 4.0104 Then firstly we adjust the equation to same the exponent of both equations, 2.6103= 0.26104 =0.26104+ 4.0104 =4.26104 ### Multiplication: When we multiply the scientific notation then we add the exponent and multiply the coefficient. Such as, =(2104) (6102) Then, = (26) 104+2 So, = 12108 ## Application of scientific notation in the real world: Real-world applications of scientific notation in different fields are given below: ### Chemistry: Chemists use scientific notation to measure or calculate the atomic size or the atomic level. Such as Avagardo’s number which can be written scientific notation 6.0221023 which can be approximately equal to the one mole in an atom or a molecule. ### Astronomy: In space, astronomers used scientific notation to calculate the vast distance. Because in space the distance between the moon, the sun, and other planets is vast so that's why to calculate this distance in a short or efficient way astronomers used scientific notation. For example, the distance between the Earth and the nearest star is approximately equal to 4.241013 kilometers. ### Engineers: Engineers work on a large scale and calculate the calculation in a very precise and efficient way so that is why they use scientific notation to express large number calculations in short form. For example, the circumference of the earth which can calculated by engineers is 4.007107. ## Physics: In physics, scientists can deal with very large and small quantities of measurements so that’s why to write these measurements in an efficient way they can use the scientific notation method. For example, the speed of light is 3.00108. ## In education the role of scientific notation: To become a scientist or for a scientific education understanding the scientific notation is essential. Scientific notation education starts in middle school but is explained in detail at higher levels. To become a scientist and, if we choose the engineering and technology career then the education of scientific notation is compulsory. because it helps them to calculate or, measure the very large or small measurements in a very precise or accurate form. ## Challenges and misconceptions: With the advantages and utility of scientific notation, it also becomes difficult or challenging for beginners. The common misconceptions that can be faced by beginners are given below: • Arithmetic operations rules can't be understood or can't be used precisely. • Zero can be confused with exponent. • The decimal point can't be placed properly. • Rules which are used for scientific notation can be misunderstood. Strategies that can be used to manage or overcome these misconceptions or challenges include: • Understand the rules of scientific notation step by step and follow instructions. • Try to solve numerous examples without help. • Understand the rules and solve the arithmetic operations. ## Future perspectives: With the development of modern science and technology, the use of scientific notation is also increased and it can also be improved from time to time. In modern science and technology, very large and small measurements can be calculated so that’s why the scientific notation method is used to express calculation effectively and accurately because scientific notation also helps to overcome or analyze errors in calculations. ## Examples: Examples of scientific notation are given below: • The mass of the electron is approximately equal to 0.000000000000000000000000000000910938356 kg But we can write in scientific notation as, = 9.10938356×10−31 • The circumference of the earth is approximately equal to 40000000m Then we can write in scientific notation as = 4107 • The distance between the sun and earth is approximately equal to 149,600,000 kilometers In scientific notation, it can written as: =1.496108 • The mass of hydrogen is approximately equal to 0.00000000000000000000000167kg In scientific notation, it can be written as, =1.6710-27 • The value 564300 can be converted into scientific notation and written as =5.643105 ## Problems: Some problems are given below: ### Conversion: Convert the following problems into scientific notation: • 456800 • 0.005544 • 98076 • 5544320 • 0.00000000009 • 9888800000 #### Solutions: Solutions to these problems are given below: • 4.568105 • 5.54410-3 • 9.8076104 • 5.544320106 • 910-11 • 9.8888109 ### Multiplication: Multiplication problems are given below: • (2108 ) (4104) • (2.0104) (2102) ### Solution: Solutions to these problems are given below: • (24) 108+4 =81012 • (2.02) 104+2 =4.0106 ### Division: Division problems are given below: • 5.61097.0103 • 8.41052.0102 ### Solution: A solution to these problems is given below: • 8.0105 • 4.2103 Perform the following operations with numbers in scientific notation: • (5.0104)+(2.5104) • (6103) + (4.0103) ### Solution: A solution to these problems is given below: • 7.5104 • 6103<\|endoftext\|>	4.53125
1,428	What's Happening in Room 414 The students will be working on I-station,Think Through Math and Xtra math for homework this week. Please download these programs at home. Your child's account will be checked for completion of homework. Math Unit Summary: This unit reviews and extends students’ thinking about place value, multi-digit addition and subtraction, and problem solving. In the first module, students are introduced to the idea of rounding 2- and 3-digit numbers to the nearest ten and the nearest hundred. This skill is extended into the realm of computation, as students use rounding as a way to estimate and check the results of adding and subtracting multi-digit numbers. Students will continue to deepen their understandings while using the standard algorithms for adding and subtracting multi-digit numbers. November Number Corner Summary: This month's focus will be on multiplication. Students explore area and arrays as they look for patterns and relationships on the Calendar Grid markers and play a game called Array Race in the Computational Fluency workout. Students learn about and practice rounding with a game on the Number Line and they begin exploring fractions as they collect halves, fourths, and eighths in the Calendar Collector’s Unit Fraction Race. The Problem Strings workout switches from string work to problem solving this month, as students work on strategies and skills including writing equations with letters standing for unknown quantities. (Pace Math is the red font in addition to the black font) Students will be able to... • compose and decompose numbers up to 100,000 as a sum of so many ten thousands, so many thousands, so many hundreds, so many tens, and so many ones using objects, pictorial models, and numbers, including expanded notation as appropriate; • interpret the value of each place-value position as 10 times the position to the right and as one-tenth of the value of the place to its left; (whole number only) • represent the value of the digit in whole numbers through 1,000,000,000 using expanded notation and numerals; • compare and order whole numbers to 1,000,000 and represent comparisons using the symbols >, <, or =; • round whole numbers to a given place value through the hundred thousands place; • describe the mathematical relationships found in the base-10 place value system through the hundred thousands place • represent a number on a number line as being between two consecutive multiples of 10; 100; 1,000; or 10,000 and use words to describe relative size of numbers in order to round whole numbers • compare and order whole numbers up to 100,000 and represent comparisons using the symbols >, <, or = • solve with fluency one-step and two-step problems involving addition and subtraction within 1,000 using strategies based on place value, properties of operations, and the relationship between addition and subtraction • add and subtract whole numbers • determine the value of a collection of coins and bills • round to the nearest 10 or 100 or use compatible numbers to estimate solutions to addition and subtraction problems • round to the nearest 10, 100, or 1,000 or use compatible numbers to estimate solutions involving whole numbers • recall facts to multiply up to 10 by 10 with automaticity and recall the corresponding division facts • represent one- and two-step problems involving addition and subtraction of whole numbers to 1,000 using pictorial models, number lines, and equations • represent multi-step problems involving the four operations with whole numbers using strip diagrams and equations with a letter standing for the unknown quantity Language Arts Unit Summary: Children learn to read and write by having opportunities to read and write stories to entertain, to foster artistic expression, to stimulate imagination, to clarify thinking, and to search for identity. In this unit on fiction, teachers model reading strategies as they “think aloud” during interactive reading to raise student awareness of the thought processes during reading to allow deeper understanding of texts. Teachers immerse students in a variety of types of fictional texts with increasingly more complex structures while explicitly teaching comprehension skills. Students read texts with expression and appropriate phrasing while making inferences and drawing conclusions about the structure and elements of fiction including the main events in a plot and their influence on future events, setting, and character interactions. Imaginative writing provides opportunities for writers to invent a situation or story based on their imagination. Students write imaginative stories to express their ideas and feelings about real or imagined people, events, and ideas. Students begin exploring ideas that could possibly become fiction stories by storytelling their ideas which helps a student bring a storyteller’s voice into his or her own story plans. The writer creates an intriguing reading experience, so the reader does not want to stop reading. The writer uses effective word choice that lingers in the reader’s mind and makes the reader want to read aloud and with expression. An imaginative story is organized to keep the reading flowing. In order to keep the reader hooked, the writer builds the plot to a climax and contains details about the characters and setting. Teachers model organizing ideas into paragraphs, and students notice how authors use paragraphs. Students begin experimenting with paragraphs to organize information in their stories. Students continue to practice cursive writing and teachers continue to model writing in cursive for students to increase their ability to read cursive writing. Poetry Integration: Humorous and Lyrical* Poems. Poetry can be used to support expression and fluency practice in reading. In writing, poetry can be used to study word choice as a vehicle to create imagery. (Lyrical Poetry-songlike poetry that has rhythm and sometimes rhyme and is memorable for sensory images and description.) Science/Social Studies Students will build on their knowledge of how people adapt, modify and use the physical environment in which they live and how that physical environment may change, including natural resources. They will investigate the impact of people, organisms and events on the physical environment. Students will explore, investigate, and compare landforms found on Earth and the rapid changes landforms undergo. They will explore changes and interactions between humans and their environment including how people adapt, modify and use the physical environment. Students will explore and investigate natural resources and the impact of rapid changes to natural resources. *Daily expectations for homework Read for 25 minutes Reading Response Monday-Thursday 3 minute math fact practice Spelling Planner signed each day Math fact tests are given in the following order: 50 addition facts in 3 minutes50 subtraction facts in 3 minutes50 multiplication facts in 4 minutes 50 division facts in 4 minutes 50 addition facts in 3 minutes50 subtraction facts in 3 minutes50 multiplication facts in 4 minutes 50 division facts in 4 minutes Multiples #1- 5 minutesMultiples #2 - 5 minutesMixed Multiples #1- 5 minutesMixed Multiples #2- 5 minutes<\|endoftext\|>	4.5625
1,458	Mathematics is a view to the logical structure of the world. The mathematical principles are a priori propositions whose truth-values are independent of empirical evidence, and hence the a priori. Mathematics involves two types of things: Sets and Relations. Sets are the collections of objects, the collection being defined provisionally. For instance, a mathematician may define his set to be the set of all triangles, while another one may work with the set of all imaginary arrows in space or the set of all even numbers. One can make up any collection of objects and consider them to be members of a set. Objects in a set are usually called members of that set. Relations are the relationships that exist between the objects of the set. These relations too are defined provisionally; we define a relation between the objects of a set. For instance, if we define our set to be the set of all triangles and number them, then we can impose the relation that the odd-numbered triangles are twice as big as the even-numbered triangles. This relation created a relationship between every two triangles: They are either the same size or one is twice as the other, or vice versa. So relations are such kinds of things. Notice that how relations can be used as a way of creating sets or collections. We can use a relation as a criterion for creating a set. For instance, I can say that I want a collection whose members are numbers such that they are all divisible by 2. Using this condition I collect all such numbers and define the set; obviously this is the set of all even numbers, and each two members are related to one another such that they are both divisible by 2. An important point is that relations themselves can be treated as objects so that we may create collections whose members are relations. For instance, I can define a collection, a set, whose members are all algebraic operations, say plus, minus, multiplications, etc. These objects are relations; addition is something that exists between two or more numbers. If we define a set whose members are pairs of numbers, any numbers we wish, like (a,b), then we can say addition is a relation between any pair of numbers and a third number c such that (a,b) is related to c in the form a+b equals c or a+b=c. Addition is an instance of a relation between some objects. Therefore, the most general structures in mathematics are sets and relations. Everything else is a specializations and variations of these two concepts. We defined relations in its most general sense. We can specialize it further and refer to all relations such that each object of the set is related to exactly one object and not two. For instance, if a is related to be, then a is not related to c unless b=c. All such specialized relations are known as functions. The another kinds of relations that I mentioned above as operations, the operations between a pair of numbers and a third number, are known as binary operations. Binary operations such as +, -, , / are special kinds of relations. We can play more and define new objects: The new object is a set of numbers equipped with a binary operation. Consider the set S with members a, b, and c. We write it as: S={a,b,c}. The name of the set and its members. Call our binary operation this symbol . We define the set A equipped with * and write it as <S , > which is a set and relation together. We define the relation to be the following: ab=c ac=b aa=a ba=c bc=a bb=b ca=b cb=a cc=c See that ever two members of S are related to one another by the binary operation . Notice that I have not mentioned anything about the nature of a,b, and c and also the nature of the binary operation which can be addition, multiplication, and anything else. But I have defined a structure which can exist regardless of how we fill in the unknown symbols. The object we defined above, which is a set equipped with an operation <S,>, has a special property: Notice that all members are related to one another; there is no member that is not related to another; and no matter how many times we apply the operation between members we never go out of S; the operation between any two members of S always gives another elements which is still a member of S. We never go outside S in order to perform the operation . This property is called closure; we say that the set S is closed under the operation , that is: Operation of * on the members of S always gives another member of S. Now the new object that we define to be a set equipped with an operation under which S is closed has a special name in algebra: It is called a group. To have a group we need some more restrictions: The operation must be associative; it must permit an identity and an inverse. We need not go into these now. The point is to say that everything in mathematics is composed of a set and a relation. In fact, everything in the world can be viewed in the same fashion. Now we can play more and extend our definitions: We define a group to be a set equipped with a binary operation which relates the members of the set with one another. Now we can define a set equipped with two binary operations. We can call these two operations multiplication and addition. In fact the set of real numbers of which we are familiar and use on a daily basis is such a mathematical object; we multiply and add them and get another object which is still a real number. This new object which is a set equipped with two binary operations, under some group constraints, is called a Ring in mathematics. If we add more relations to this ring we can make a more interesting object called a Field. In fact all of us know a field and work with it everyday: The set of real numbers. Field is a special kind of ring. If the operation * that relates two elements a and b is such that ab=ba we say that * is commutative. The real numbers that we know is a commutative ring because 23=32=6. We can now say: Field is a commutative division ring. All of the above are created by playing with set and relations: Set is a collection of object. Relations are the relationships existing between members of a set. Function is a monogamous relation. Binary operation is a function. Group is a set with a binary operation. Ring is a group with a binary operation. Field is a commutative division ring. Division ring is a ring whose non-zero elements are all units. A unit is an element of a set that has an inverse. Inverse of an element a is another element b such that ab is the identity element. We write: ab=e. Identity element e is an element of a set such that ae=ea=a for all elements a of the set. [The purpose of this post was to review the subject for myself and prepare for a lecture on the concepts of mathematics.]<\|endoftext\|>	4.4375
530	The laws that governed the various Maya states were issued by the halach uinic and his council, or by the council alone if the state did not have an halach uinic. The batabs were responsible for carrying out these laws and serving as administrators to smaller towns and cities. Batabs also served as judges for their towns and adjudicated civil and criminal cases. Court cases were generally handled swiftly in public meeting houses known as popilna. Judicial proceedings were conducted orally and written records were not maintained. Witnesses were required to testify under oath and there is evidence to suggest that the parties were represented by individuals who functioned as attorneys. Batabs would review the evidence, evaluate the circumstances of the case, consider whether the criminal act in question was deliberate or accidental, and would order an appropriate punishment. Decisions made by the batabs were final and could not be appealed, though the victims could pardon the accused, thus reducing their punishment. If the accused parties were found guilty, their sentences were carried out immediately by the tupiles. The Maya did not have prisons, but may have had wooden cages that were used as holding cells for individuals who were awaiting capital punishment. If a crime occurred that affected an individual in another town, the batabs in the two towns would work together to ensure that issue was resolved. The batab generally acted independently, but would consult with the halach uinic on serious cases before passing judgement. Because the ancient Maya civilization had peaked before the Spanish Conquest, the amount of primary material on the Maya legal system is limited. The majority of Maya manuscripts and codices were destroyed by Spanish priests, and the surviving codices tend to focus on Maya astronomy, mathematics, history, calendars, and religious rituals. These include the Dresden, Paris, and Madrid Codices. Following the conquest, Maya scribes wrote various books, including the Popul Vuh, and the Books of Chilam Balam (Books of the Jaguar Shaman). Both of these resources contain information about Maya history, myths, and religious traditions. The conquistadores and Spanish missionaries additionally documented their observations of the Maya. Bishop Diego de Landa wrote a detailed chronicle of the Maya, entitled Relación de las Cosas de Yucatan. This manuscript contained information about Maya history, culture, and hieroglyphics. Finally, researchers have relied on Maya monuments, pottery and paintings, hieroglyphic texts, and anthropological studies of modern day Maya to learn more about this civilization. Sources: Foster (2002), Herrera (2001), Salcedo Flores (2009), and Sharer (1996).<\|endoftext\|>	4.03125
2,956	Breast Cancer Screening (PDQ®) Last modified: 2018-10-26 Last downloaded: 2019-05-18 What is screening? Screening is looking for signs of disease, such as breast cancer, before a person has symptoms. The goal of screening tests is to find cancer at an early stage when it can be treated and may be cured. Sometimes a screening test finds cancer that is very small or very slow growing. These cancers are unlikely to cause death or illness during the person's lifetime. Scientists are trying to better understand which people are more likely to get certain types of cancer. For example, they look at the person's age, their family history, and certain exposures during their lifetime. This information helps doctors recommend who should be screened for cancer, which screening tests should be used, and how often the tests should be done. It is important to remember that your doctor does not necessarily think you have cancer if he or she suggests a screening test. Screening tests are done when you have no cancer symptoms. Women who have a strong family history or a personal history of cancer or other risk factors may also be offered genetic testing. See the following PDQ summary for more information about cancer screening: Back to Top General Information About Breast Cancer Breast cancer is a disease in which malignant (cancer) cells form in the tissues of the breast. The breast is made up of lobes and ducts. Each breast has 15 to 20 sections called lobes, which have many smaller sections called lobules. Lobules end in dozens of tiny bulbs that can produce milk. The lobes, lobules, and bulbs are linked by thin tubes called ducts.Anatomy of the female breast. The nipple and areola are shown on the outside of the breast. The lymph nodes, lobes, lobules, ducts, and other parts of the inside of the breast are also shown. Each breast also has blood vessels and lymph vessels. The lymph vessels carry an almost colorless, watery fluid called lymph. Lymph vessels carry lymph between lymph nodes. Lymph nodes are small, bean-shaped structures that filter lymph and store white blood cells that help fight infection and disease. Groups of lymph nodes are found near the breast in the axilla (under the arm), above the collarbone, and in the chest. - Breast Cancer Prevention - Breast Cancer Treatment - Male Breast Cancer Treatment - Genetics of Breast and Gynecologic Cancers Breast cancer is the second leading cause of death from cancer in American women. Breast cancer is more likely to occur as a woman ages. It occurs more often in white women than in black women, but black women die from breast cancer more often than white women. Breast cancer rarely occurs in men. Because men with breast cancer usually have a lump that can be felt, screening tests are not likely to be helpful. Different factors increase or decrease the risk of breast cancer. Back to Top Breast Cancer Screening - Tests are used to screen for different types of cancer when a person does not have symptoms. - Mammography is the most common screening test for breast cancer. - Magnetic resonance imaging (MRI) may be used to screen women who have a high risk of breast cancer. - Whether a woman should be screened for breast cancer and the screening test to use depends on certain factors. - Other screening tests have been or are being studied in clinical trials. - Screening tests for breast cancer are being studied in clinical trials. Tests are used to screen for different types of cancer when a person does not have symptoms. Scientists study screening tests to find those with the fewest harms and most benefits. Cancer screening trials also are meant to show whether early detection (finding cancer before it causes symptoms) helps a person live longer or decreases a person’s chance of dying from the disease. For some types of cancer, the chance of recovery is better if the disease is found and treated at an early stage. Mammography is the most common screening test for breast cancer. A mammogram is an x-ray picture of the breast. Mammography may find tumors that are too small to feel. It may also find ductal carcinoma in situ (DCIS). In DCIS, abnormalcells line the breast duct, and in some women may become invasive cancer. Mammography is less likely to find breast tumors in women with dense breast tissue. Because both tumors and dense breast tissue appear white on a mammogram, it can be harder to find a tumor when there is dense breast tissue. Younger women are more likely to have dense breast tissue.Mammography. The breast is pressed between two plates. X-rays are used to take pictures of breast tissue. Many factors affect whether mammography is able to detect (find) breast cancer: - The age and weight of the patient. - The size and type of tumor. - Where the tumor has formed in the breast. - How sensitive the breast tissue is to hormones. - How dense the breast tissue is. - The timing of the mammography within the woman's menstrual cycle. - The quality of the mammogram picture. - The skill of the radiologist in reading the mammogram. Women aged 50 to 69 years who have screening mammograms have a lower chance of dying from breast cancer than women who do not have screening mammograms. Fewer women are dying of breast cancer in the United States, but it is not known whether the lower risk of dying is because the cancer was found early by screening or whether the treatments were better. Magnetic resonance imaging (MRI) may be used to screen women who have a high risk of breast cancer. MRI is a procedure that uses a magnet, radio waves, and a computer to make a series of detailed pictures of areas inside the body. This procedure is also called nuclear magnetic resonance imaging (NMRI). MRI does not use any x-rays and the woman is not exposed to radiation. MRI may be used as a screening test for women who have a high risk of breast cancer. Factors that put women at high risk include the following: - Certain gene changes, such as changes in the BRCA1 or BRCA2 genes. - A family history (first degree relative, such as a mother, daughter or sister) with breast cancer. - Certain geneticsyndromes, such as Li-Fraumeni or Cowden syndrome. An MRI is more likely than mammography to find a breast mass that is not cancer. Whether a woman should be screened for breast cancer and the screening test to use depends on certain factors. Women with risk factors for breast cancer, such as certain changes in the BRCA1 or BRCA2 gene or certain genetic syndromes may be screened at a younger age and more often. Women who have had radiation treatment to the chest, especially at a young age, may start routine breast cancer screening at an earlier age. The benefits and risks of mammograms and MRIs for these women have not been studied. Breast cancer screening has not been shown to benefit the following women: - Elderly women who, if diagnosed with breast cancer through screening, will usually die of other causes. - In women with an average risk of developing breast cancer, screening mammography before age 40 has not shown any benefit. - In women who are not expected to live for a long time and have other diseases or conditions, finding and treating early stage breast cancer may reduce their quality of life without helping them live longer. Other screening tests have been or are being studied in clinical trials. Studies have been done to find out if the following breast cancer screening tests are useful in finding breast cancer or helping women with breast cancer live longer. A clinical breast exam is an exam of the breast by a doctor or other health professional. He or she will carefully feel the breasts and under the arms for lumps or anything else that seems unusual. It is not known if having clinical breast exams decreases the chance of dying from breast cancer. Breast self-exams may be done by women or men to check their breasts for lumps or other changes. If you feel any lumps or notice any other changes in your breasts, talk to your doctor. Doing regular breast self-exams has not been shown to decrease the chance of dying from breast cancer. Thermography is a procedure in which a special camera that senses heat is used to record the temperature of the skin that covers the breasts. Tumors can cause temperature changes that may show up on the thermogram. There have been no randomized clinical trials of thermography to find out how well it detects breast cancer or the harms of the procedure. Breast tissue sampling is taking cells from breast tissue to check under a microscope. Breast tissue sampling as a screening test has not been shown to decrease the risk of dying from breast cancer. Screening tests for breast cancer are being studied in clinical trials. Information about clinical trials supported by NCI can be found on NCI’s clinical trials search webpage. Clinical trials supported by other organizations can be found on the ClinicalTrials.gov website. Back to Top Harms of Breast Cancer Screening - Screening tests can have harms. The harms of mammography include the - False-positive test results can occur. - False-positive results can lead to extra testing and cause anxiety. - False-negative test results can delay diagnosis and treatment. - Finding breast cancer may lead to breast cancer treatment and side effects, but it may not improve a woman's health or help her live longer. - Mammography exposes the breast to low doses of radiation. - There may be pain or discomfort during a mammogram. - Talk to your doctor about your risk of breast cancer and your need for screening tests. Screening tests can have harms. Not all breast cancers will cause death or illness in a woman's lifetime, so they may not need to be found or treated. Decisions about screening tests can be difficult. Not all screening tests are helpful and most have harms. Before having any screening test, you may want to discuss the test with your doctor. It is important to know the harms of the test and whether it has been proven to reduce the risk of dying from cancer. The harms of mammography include the following: Screening test results may appear to be abnormal even though no cancer is present. A false-positive test result (one that shows there is cancer when there really isn’t) is usually followed by more tests (such as biopsy), which also have risks. Most abnormal test results turn out not to be cancer. False-positive results are more common in the following: - Younger women (under age 50). - Women who have had previous breast biopsies. - Women with a family history of breast cancer. - Women who take hormones for menopause. False-positive results are more likely the first time screening mammography is done than with later screenings. For every ten women who have a single mammogram, one will have a false-positive result. The chance of having a false-positive result goes up the more mammograms a woman has. Comparing a current mammogram with a past mammogram lowers the risk of a false-positive result. The skill of the radiologist also can affect the chance of a false-positive result. If a mammogram is abnormal, more tests may be done to diagnose cancer. Women can become anxious during the diagnostic testing. Even if it is a false-positive test and cancer is not diagnosed, the result can lead to anxiety anywhere from a few days to years later. Several studies show that women who feel anxiety after false-positive test results are more likely to schedule regular breast screening exams in the future. Screening test results may appear to be normal even though breast cancer is present. This is called a false-negative test result. A woman who has a false-negative test result may delay seeking medical care even if she has symptoms. About one in 5 cancers are missed by mammography. The chance of a false-negative test result is more common in women who: - Are younger. - Have dense breast tissue. - Have cancer that is not dependent on hormones (estrogen and progesterone). - Have cancer that is fast growing. Finding breast cancer may lead to breast cancer treatment and side effects, but it may not improve a woman's health or help her live longer. Some breast cancers found by screening mammography may never cause health problems or become life-threatening. Finding these cancers is called overdiagnosis. When these cancers are found, having treatment may cause serious side effects and may not lead to a longer, healthier life. Being exposed to high radiationdoses is a risk factor for breast cancer. The radiation dose with a mammogram is very low. Women who start getting mammograms after age 50 have very little risk that the overall exposure to radiation from mammograms throughout their lives will cause harm. Women with large breasts or with breast implants may be exposed to slightly higher radiation doses during screening mammography. During a mammogram, the breast is placed between two plates that are pressed together. Pressing the breast helps to get a better x-ray of the breast. Some women have pain or discomfort during a mammogram. The amount of pain may also depend on the following: - The phase of the woman's menstrual cycle. - The woman's anxiety level. - How much pain the woman expected. Talk to your doctor about your risk of breast cancer and your need for screening tests. Talk to your doctor or other health care provider about your risk of breast cancer, whether a screening test is right for you, and the benefits and harms of the screening test. You should take part in the decision about whether you want to have a screening test, based on what is best for you. (See the PDQ summary on Cancer Screening Overview for more information.) Back to Top Source: The National Cancer Institute's Physician Data Query (PDQ®) Cancer Information Summaries ( http://www.cancer.gov/cancertopics/pdq)<\|endoftext\|>	3.75
653	# Difference between revisions of "2021 AMC 12B Problems/Problem 15" ## Problem The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written is $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$ $[asy] /* Made by samrocksnature / pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-2,2); pair G=(-3,2); draw(A--B--C--D--E--A); draw(A--F--A--G); draw(B--F--C); draw(E--G--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E^^F^^G); [/asy]$ $\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$ ## Solution Let $M$ be the midpoint of $CD$. Noting that $AED$ and $ABC$ are $120-30-30$ triangles because of the equilateral triangles, $AM=\sqrt{AD^2-MD^2}=\sqrt{12-1}=\sqrt{11} \implies [ACD]=\sqrt{11}$. Also, $[AED]=22\frac{1}{2}\sin{120^o}=\sqrt{3}$ and so $[ABCDE]=[ACD]+2[AED]=\sqrt{11}+2\sqrt{3}=\sqrt{11}+\sqrt{12} \implies \boxed{(\textbf{D})23}$. ~Lcz ~ pi_is_3.14 ## See Also 2021 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. Invalid username Login to AoPS<\|endoftext\|>	4.53125
598	Special education teachers are responsible for assisting each child in achieving his or her goals for the academic year. Setting goals through the development of an Individualized Education Program (IEP) is an important part of working with special needs children. An IEP is a legally binding document developed by a team of teachers, school district representatives, members of a student’s family and sometimes even the student. While literacy and other academic goals are part of an IEP, the team also addresses nonacademic and extracurricular goals. While earning a Master of Science Degree in Education for Special Education, master’s degree candidates gain knowledge and skills that assist in the development and implementation of IEPs for a wide range of special needs students. Development of IEPs A special education teacher is an essential part of an IEP team. In addition to remaining compassionate and respectful towards special needs students and their families, special education teachers assist in the development of IEPs in the following ways: - Bringing a comprehensive knowledge of learning challenges and disabilities - Ability to suggest appropriate technologies or modifications to ensure an effective learning environment for children with different disabilities - Possessing a complete understanding of legal and ethical parameters for providing education to special needs students - Offering experience in providing positive behavior modifications for a wide range of behavioral challenges - Advocating for "least restrictive" learning environments, with an emphasis on special needs children learning alongside peers without disabilities whenever possible Implementation of IEPs Once an IEP is in place, a special education teacher helps to ensure that the classroom adheres to the IEP’s requirements. Because IEPs contain very clear goals and modifications for each student, classroom teachers can follow the IEP as though it were a roadmap. It is important for special education teachers to spend time reading and understanding each student’s IEP. Depending on the IEP, a teacher may need to provide assistive technologies, such as computer software, visual aids or mobility equipment. The IEP may specify that a child receive special modifications, such as extra time for testing, oral instructions or recorded lessons. It is important for the special education teacher to ensure that the classroom adheres to the IEP and that any issues come to the IEP team’s attention. Special education teachers become expert participants in the design and implementation of IEPs. Other members of the IEP team depend on special education teachers’ assessments of disabilities, data analysis or strategic development to help special needs children overcome learning barriers. As such, special education teachers provide essential resources, information and perspective during the IEP process. Working with special needs children as a special education teacher requires a deep level of compassion for students and families, a commitment to collaborating with other entities and a student-focused approach to creating dynamic, effective IEPs. Learn more about the A-State MSE Special Ed Instructional Specialist online program.<\|endoftext\|>	3.828125
1,095	# Complex rectangles 2 Suppose we have four complex numbers $$A, B, C, D$$ on the unit circle such that $$A + B + C + D = 0$$. We want to show that these points must form a rectangle. How can we do this? An intuition I had last time was that we might want to draw the vector-loop picture of these points. Let's do that now. Figure 1 shows the situation. Here, $$P = 0, Q = A, R = A + B$$, and $$S = A+B+C$$. We draw each complex number as a vector joining this four points, and we also draw the diameter $$A + B$$ connecting $$P$$ with $$R$$. The key idea here is to relate these as two isoceles triangles, which forces the angles to be equal. Because $$A, B, C$$, and $$D$$ lie on the unit circle, they are all equal in magnitude, so that the outer sides are all equal. But the diameter $$A + B$$ is common, so the triangles $$PQR$$ and $$PSR$$ are side-side-side congruent. This means the corresponding angles must be equal. The angle of $$A$$ with $$B$$ and the angle of $$C$$ with $$D$$ are equal, to start with. But because these are isoceles triangles, the base angles are equal to each other, so the angles of $$B$$ with $$C$$ and the angle of $$A$$ with $$D$$ are equal. This is true of the end-to-end angles, but also the base-to-base angles because the base-to-base angles are the corresponding complementary angles. So, the angles work out equal. In fact, we can go further: The angle between $$A$$ and $$B$$ is complementary with that between $$B$$ and $$C$$. We know this because the two base angles of each isocoles triangle taken together must be complementary to the third. Here, the third angles are $$\angle PQR$$ and $$\angle PSR$$, also known as the angles between $$A$$ and $$B$$ and between $$C$$ and $$D$$. But the two base angles together are equal to the angle between$$B$$ and $$C$$, or equivalently between $$A$$ and $$D$$. So, the angle between $$A$$ and $$B$$ is complementary with that between $$C$$ and $$D$$. For the same reason the angle between $$C$$ and $$D$$ is complementary with the angle between $$D$$ and $$A$$. So, the angles work out as complementary. This shows almost exactly what we want. Clearly, the angles being complementary means we should have $$A = -C$$ and $$B = -D$$. That in turn implies that $$A + B = -C - D$$ hence $$\|A + C\| = \|C + D\|$$ so that the four points $$A, B, C, D$$ form a rectangle. But this connection between the geometry and the complex numbers is a little tenuous for my taste. I would like to spell it out in more detail. This seems tricky. Can we relate the geometry back to statements about complex numbers? Actually, I think we can. I think we can restate the complementary angles claim as saying that the products of certain complex numbers must be -1, the complex number with angle $$\pi$$. That's because complex numbers adds their angles. The question is what complex number expresses the right angles. I think the answer is $$(B/A)(C/B)$$. Inverting a complex number negates the angle, and we're interested in the angles between complex numbers. Let's make this a bit more formal. Say that we write $$A, B, C, D$$ as $$e^{i \alpha}, e^{i \beta}, e^{i \gamma}, e^{i \delta}$$ where $$0 \leq \alpha < \beta < \gamma < \delta < 2\pi$$. I think we can state the complementary angles Saying that the angle between $$A$$ and $$B$$ are complementary with the angle between $$B$$ and $$C$$ is equivalent to saying that $$(\beta - \alpha) + (\gamma - \beta) = \gamma - \alpha = \pi$$. But $$B/A = e^{i (\beta - \alpha)}$$ and similarly for the other configurations. So as statements about complex numbers this is equivalent to saying $$(B/A)(C/B) = C/A = -1$$, or $$A = -C$$. But we've already proven this. The geometric approach we've just taken shows exactly that $$(\beta - \alpha) + (\gamma - \beta) = \pi$$ as desired. This feels a little disorienting. It feels like I'm still missing a step. I think I'm missing the step where we explicitly relate the angles in the diagram to the complex numbers $$A, B, C$$, and $$D$$. I think we need another diagram. This pretty clearly shows that the relationship just described should hold. Can we make this same kind of argument algrebraically? Tune in next time.<\|endoftext\|>	4.8125
2,048	# 4.11: Simplify Variable Expressions Involving Integer Multiplication Difficulty Level: At Grade Created by: CK-12 Estimated3 minsto complete % Progress Practice Simplify Variable Expressions Involving Integer Multiplication MEMORY METER This indicates how strong in your memory this concept is Progress Estimated3 minsto complete % Estimated3 minsto complete % MEMORY METER This indicates how strong in your memory this concept is Have you ever been skiing? In Alaska, Molly and her family went on a skiing trip. There are four people in Molly's family. When they got there, they saw that four other families of four were also there. Molly isn't sure how much a lift ticket costs. This is an unknown variable in this situation. Can you write a variable expression involving integer multiplication to figure out how many people will need lift tickets? Well, if you aren't sure how to do this, don't worry, you will learn all about it in this Concept. ### Guidance Do you remember what a variable expression is? A variable expression is a math phrase using numbers, operations and variables. A variable expression can also contain like terms. A like term is a term that is common between one or more terms in the equation. When you have like terms, we can combine them using addition and subtraction. Pay attention! Here is a change! How does this happen? Find the value of this expression 3z(2)\begin{align}3z \cdot (-2)\end{align} The Commutative Property of Multiplication states that the order in which factors are multiplied does not matter. 3z(2)=23z\begin{align}3z \cdot (-2) = -2 \cdot 3z\end{align} The Associative Property of Multiplication states that you can group the factors being multiplied in any order. 23z=23z=(23)z\begin{align}-2 \cdot 3z = -2 \cdot 3 \cdot z = (-2 \cdot 3) \cdot z\end{align} Now we can multiply the integers. Since -2 and 3z\begin{align}3z\end{align} have different signs, the product will be negative. 23=6\begin{align}2 \cdot 3=6\end{align}, so (2)3=6\begin{align}(-2)\cdot 3=-6\end{align} and (23)z=6z\begin{align}(-2 \cdot 3)\cdot z = -6z\end{align} The value of the expression is 6z\begin{align}-6z\end{align}. Find the value of this expression (5)(2m)(n)\begin{align}(-5)(-2m)(n)\end{align} The Associative Property of Multiplication states that you can group the factors being multiplied in any order. (5)(2m)(n)=(5)(2)mn\begin{align}(-5)(-2m)(n) = (-5)\cdot (-2) \cdot m \cdot n\end{align} Now we can multiply the integers. Since -5 and -2 are both negative, the product will be positive. 52=52=10\begin{align}-5 \cdot -2=5 \cdot 2 = 10\end{align}, so (5)(2)mn=10mn=10mn\begin{align}(-5) \cdot (-2) \cdot m \cdot n = 10 \cdot mn=10mn\end{align} The value of the expression is 10mn\begin{align}10mn\end{align}. Multiply the following variable expressions. #### Example A 3x(4y)\begin{align}3x(4y)\end{align} Solution:12xy\begin{align}12xy\end{align} #### Example B 6a(4b)\begin{align}-6a(-4b)\end{align} Solution:24ab\begin{align}24ab\end{align} #### Example C 4z(10)\begin{align}-4z(10)\end{align} Solution:40z\begin{align}-40z\end{align} Here is the original problem once again. In Alaska, Molly and her family went on a skiing trip. There are four people in Molly's family. When they got there, they saw that four other families of four were also there. Molly isn't sure how much a lift ticket costs. This is an unknown variable in this situation. Can you write a variable expression involving integer multiplication to show how many people will need lift tickets? To write this variable expression, let's write down what we know first. We know that there are four people in Molly's family who will all need lift tickets. 4x\begin{align}4x\end{align} This is the first part of the variable expression because we don't know the price of the lift ticket. Then there are four groups like Molly's family. We can multiply the variable expression by 4. 4(4x)\begin{align}4(4x)\end{align} Now we can simplify. 16x\begin{align}16x\end{align} This is the number of people who will need lift tickets. ### Vocabulary Here are the vocabulary words in this Concept. Integer the set of whole numbers and their opposites. Product the answer in a multiplication problem Factors the numbers being multiplied Variable Expression a number phrase using numbers, operations and variables. Commutative Property of Multiplication states that the order that we multiply terms does not change the product. Associative Property of Multiplication states that changing the grouping of factors does not change the product. ### Guided Practice Here is one for you to try on your own. The temperature outside Fred's house is dropping at a rate of 2F\begin{align}2^\circ F\end{align} each hour. Represent the total change in the temperature over the next 5 hours as an integer. First, let's write an expressing to represent the situation. (2)(5)=10\begin{align}(-2)(5) = -10\end{align} The temperature change was 10\begin{align}-10^\circ\end{align}. ### Video Review Here is a video for review. ### Practice Directions: Multiply each variable expression. 1. (7k)(6)\begin{align}(-7k)(-6)\end{align} 2. (8)(3a)(b)\begin{align}(-8)(3a)(b)\end{align} 3. 6a(b)(c)\begin{align}-6a(b)(c)\end{align} 4. 8a(6b)\begin{align}-8a(6b)\end{align} 5. (12y)(3x)(1)\begin{align}(12y)(-3x)(-1)\end{align} 6. 8x(4)\begin{align}-8x(4)\end{align} 7. a(5)(4b)\begin{align}-a(5)(-4b)\end{align} 8. 2ab(12c)\begin{align}-2ab(12c)\end{align} 9. 12ab(12c)\begin{align}-12ab(12c)\end{align} 10. 8x(12z)\begin{align}8x(12z)\end{align} 11. 2a(14c)\begin{align}-2a(-14c)\end{align} 12. 12ab(11c)\begin{align}-12ab(11c)\end{align} 13. 22ab(2c)\begin{align}-22ab(-2c)\end{align} 14. 18ab(12)\begin{align}18ab(12)\end{align} 15. 21a(3b)\begin{align}-21a(-3b)\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English TermDefinition Associative property The associative property states that the order in which three or more values are grouped for multiplication or addition will not affect the product or sum. For example: $(a+b) + c = a + (b+c) \text{ and\,} (ab)c = a(bc)$. Commutative Property The commutative property states that the order in which two numbers are added or multiplied does not affect the sum or product. For example $a+b=b+a \text{ and\,} (a)(b)=(b)(a)$. Factors Factors are numbers or values multiplied to equal a product. Integer The integers consist of all natural numbers, their opposites, and zero. Integers are numbers in the list ..., -3, -2, -1, 0, 1, 2, 3... Product The product is the result after two amounts have been multiplied. Variable Expression A variable expression is a mathematical phrase that contains at least one variable or unknown quantity. Show Hide Details Description Difficulty Level: Authors: Tags: Subjects:<\|endoftext\|>	4.96875
993	# How To Solve Problems Number Of Golf Balls School Bus? ## How many golf balls fit in a school bus? So, when we divide 1.6 cubic inches by 2.5 cubic inches we get 660,000 balls. Since a school bus is not empty and has some space occupied by the engine and seats, it is best that we round off this figure to 500,000. This means that a school bus can accommodate 500,000 golf balls. ## How many golf balls fit in a school bus interview question? Divide that 2.5 cubic inches into 1.6 million and you come up with 660,000 golf balls. However, since there are seats and crap in there taking up space and also since the spherical shape of a golf ball means there will be considerable empty space between them when stacked, I’ll round down to 500,000 golf balls. You might be interested: Question: How Wide Is A School Bus Empty? ## How many footballs can fit in a bus? 74*495,000) = 366,000 Balls. Suppose a standard school bus is about 8 feet wide by 6 feet high by 20 feet long – An assumption. ## How many golf balls fit in a 747 answer? Battison: But if we have an airliner, we do have those seats that can’t hold the golf balls. Josaphat: 56,784 cubic feet divided by 12 cubic feet. That total answer would be 4,732 golf balls. ## How many tennis balls actually fit in a limo? About 125,000 tennis balls should fit into a limo. ## How many golf balls are on the moon? There are two golf balls on the Moon. It is TRUE. Astronaut Alan Shepard is the fifth man to walk on the Moon and the first (and only) to have played golf there. He hit two balls during the Apollo 14 mission. ## How many golf balls can you put in a 5 gallon bucket? The math suggests, a spherically sized bucket can contain around 340 balls, although, usually golfers use a 5-gallon bucket to carry an average of fifty to eighty balls in or out of the golf pitch. ## How many golf balls could you fit in to an empty Volkswagen? However, since there are seats and crap in there taking up space and also since the spherical shape of a golf ball means there will be considerable empty space between them when stacked, I’ll round down to 500,000 golf balls. ## How many balls can fit in a room interview questions? Suggested answer: The average mens’ basketball has a diameter of 25cm. There are approximately 30cm in a foot. Therefore you could fit one inflated basketball in a 1 foot cubed space. Therefore you could fit 1000 inflated basketballs inside a room with a volume of 1000 ft cubed. You might be interested: Readers ask: Which Is The Best Estimate For The Length Of A School Bus? ## How many tennis balls can you fit inside a BMW? The tennis ball fits in a regular car 28,0000/4 = 70,000. So, the total number of a tennis ball is 70,000, which easily fit in your car. How many tennis balls fit in a BMW? Some of our honorable visitors ask this question. ## How many ping pong balls are allowed on a plane? The question is fairly satisfactorily answered: “The 747-400 has a passenger volume of 876 cubic meters, plus a cargo volume of 159 cubic meters, for a total of 1035; the balls have a diameter of 40mm; this gives about 22,870,000 ping pong balls.” The answerer notes that the weight of this many balls would prevent the ## How many balls does it take to fill a kiddie pool? Bottom Line So, how many ball pit balls does it take to fill a kiddie pool? At the end of the day, you can fill your kiddie pool with as many ball pit balls as you would like, but we recommend roughly 240-480 ball pit balls for a great time. ## How many balls fit in a box? After finding right packing, you can calculate exact value. Well, assuming each ball occupies ~1 cm 3 of space, you have fit in 1,000,000 balls just by having them in a square lattice. ## How many tennis balls can you fit in a Boeing 747? The volume of the tennis ball would be 47.916πcm^3. Finally, after dividing the volume of the aeroplane by that to the tennis ball we arrive at 688,705. Therefore, we can fit approximately 688,705 tennis balls in a Boeing 747.<\|endoftext\|>	4.46875
2,493	An airborne disorder is any disease that is caused by a microorganism that is transmitted through the air. There are many airborne diseases that are of clinical importance and include bacteria, viruses, and fungi. These organisms may be spread through sneezing, coughing spraying of liquids, the spread of dust or any activity that results in the generation of aerosolized particles. It is important to be aware that airborne diseases, in general, do not include disorders caused by air pollution such as poisons, smog, and dust. The microorganisms transmitted airborne may be spread via fine mist, dust, aerosols, or liquids. The aerosolized particles may be generated from a source of infection such as body secretions of an infected patient or even an animal. In addition, aerosols may be generated from biological waste products that accumulate in garbage, caves and dry arid containers. During aerosolization, the microorganisms that are less than 5 microns in size float in the air. Sometimes, the microorganisms may be contained in dust particles that are present in the air. Once the droplets that contains microorganisms have been formed, they are then dispersed via air currents to varying distances and can be inhaled by susceptible hosts. The infected aerosolized particles often remain suspended in the air currents and may travel considerable distances, although many particles will drop off within the vicinity. As the distance traveled of the aerosol particle increases, the risk of infection starts to drop. Airborne precautions necessitate the prevention of infections and use of available interventions in healthcare facilities to prevent the transmission of airborne particles. The airborne particles often remain suspended in the facility air environment and with air currents move along to different parts of the institution where there is a potential of them being inhaled by other patients. The airborne particles may remain localized to the room or move depending on the airflow. In some cases where there is inadequate ventilation, the airborne particle may remain in the hospital room and be inhaled by a newly admitted patient. The control and prevention of airborne transmission of infections are not simple as it requires the control of airflow with the use of specially designed ventilator systems, the practice of antiseptic techniques, wearing personalized protection equipment (PPE) and performing basic infection prevention measures like hand washing. In almost all cases, airborne pathogens cause an inflammatory reaction of the upper airways affecting the nose, sinuses, throat, and lungs. The involvement of these structures may result in nasal or sinus congestion, and sore throat. Any coughing or sneezing activity may then generate aerosolized particles. Some of the common infections that can spread via airborne transmission include the following: Airborne disease is not exclusive to humans but can also affect non-humans. For example, many types of poultry are often affected by an avian disorder (Newcastle disease), that is also transmitted via an airborne route. However, it is important to understand that exposure to an animal or a patient with an airborne disease does not automatically ensure disease transmission. The infection also depends on the host immunity, how much exposure the individual has suffered and the duration of exposure to the infected patient. Besides patients, there are also several medical and surgical procedures that can generate aerosolized particles. In most cases, these airborne particles are generated during manipulation of the large airways. Examples include: Airborne transmission of microorganisms depends on several physical variables endemic to the infectious particle. Factors that influence the spread of airborne infections includes the following: When patients are seen in the emergency room or admitted to the hospital, it is impossible to know immediately if there have an airborne infection. Hence, healthcare workers need to maintain a high degree of suspicion in patients who present with signs and symptoms that are compatible with an airborne infection. The earlier the airborne prevention methods are implemented, the lower the risk to other patients and staff. Airborne Isolation Room This is a single patient room equipped with special air handling (negative pressure) and ventilation capacity. The negative pressure room is also known as an Airborne Isolation Room. This negative pressure room is usually a single-occupancy patient-care room frequently used to isolated individuals with confirmed or suspected airborne infection. This is relatively clean and is a frequently used area to transition patients/healthcare workers in and out of the airborne isolation room when it is under negative pressure. An anteroom is frequently used as a transitional space between the airborne isolation room and the hallway. It is in this transition area where healthcare workers store their PPE and put on their PPE before entering the airborne isolation room. All healthcare workers who enter the negative pressure room should wear an appropriately fit-tested N95 respirator. The N95 Respirators is individually fitted and can filter particles one micron in size. Studies show that the mask has a 95% filter efficiency and provides a tight facial seal with less than a 10% leak. To use the respirator appropriately, one should do the following: In hospitals, patients admitted to the negative room often have to be transported to other departments like radiology, physical therapy, the pulmonary laboratory or even another hospital. Before sending a patient with an airborne infection, one must always communicate with the relevant department first. This communication should include the following: Patient and Family Education Duration of Precautions In general, airborne precautions can be discontinued once the patient shows no signs or symptoms of an infection. However, one also has to be aware of the period of incubation and if unsure, always consult with an infectious disease expert What if there are many infected patients? Managing Deceased Patients When a patient is deceased, airborne precautions are necessary when handling and preparing the body during transfer to the morgue and an autopsy. Managing specimens and rooms of patients and with airborne infections Ways to prevent airborne diseases include the following: The management of airborne infections is with a multidisciplinary team aided by hospital guidelines and rules. All healthcare workers including the nurse have a vital role in preventing airborne infections. An interdisciplinary team approach involving clinicians and nurses will produce the best results in preventing the spread of infections. The following principles should apply: \|\|\|Gammon J,Hunt J, A review of isolation practices and procedures in healthcare settings. British journal of nursing (Mark Allen Publishing). 2018 Feb 8 [PubMed PMID: 29412028]\| \|\|\|Broussard IM,Bhimji SS, Precautions, Universal null. 2018 Jan [PubMed PMID: 29262198]\| \|\|\|Brouqui P,Boudjema S,Soto Aladro A,Chabrière E,Florea O,Nguyen H,Dufour JC, New Approaches to Prevent Healthcare-Associated Infection. Clinical infectious diseases : an official publication of the Infectious Diseases Society of America. 2017 Aug 15 [PubMed PMID: 28859352]\| \|\|\|MacIntyre CR,Chughtai AA,Rahman B,Peng Y,Zhang Y,Seale H,Wang X,Wang Q, The efficacy of medical masks and respirators against respiratory infection in healthcare workers. Influenza and other respiratory viruses. 2017 Nov [PubMed PMID: 28799710]\| \|\|\|Moriceau G,Gagneux-Brunon A,Gagnaire J,Mariat C,Lucht F,Berthelot P,Botelho-Nevers E, Preventing healthcare-associated infections: Residents and attending physicians need better training in advanced isolation precautions. Medecine et maladies infectieuses. 2016 Feb [PubMed PMID: 26654322]\| \|\|\|Baseer MA,Ansari SH,AlShamrani SS,Alakras AR,Mahrous R,Alenazi AM, Awareness of droplet and airborne isolation precautions among dental health professionals during the outbreak of corona virus infection in Riyadh city, Saudi Arabia. Journal of clinical and experimental dentistry. 2016 Oct [PubMed PMID: 27703605]\| \|\|\|Ito Y,Nagao M,Iinuma Y,Matsumura Y,Yamamoto M,Takakura S,Igawa J,Yamanaka H,Hashimoto A,Hirai T,Niimi A,Ichiyama S,Mishima M, Risk factors for nosocomial tuberculosis transmission among health care workers. American journal of infection control. 2016 May 1 [PubMed PMID: 26777287]\| \|\|\|Lindsley WG,Noti JD,Blachere FM,Szalajda JV,Beezhold DH, Efficacy of face shields against cough aerosol droplets from a cough simulator. Journal of occupational and environmental hygiene. 2014 [PubMed PMID: 24467190]\| \|\|\|Beam E,Gibbs SG,Hewlett AL,Iwen PC,Nuss SL,Smith PW, Clinical challenges in isolation care. The American journal of nursing. 2015 Apr [PubMed PMID: 25811520]\| \|\|\|Seto WH, Airborne transmission and precautions: facts and myths. The Journal of hospital infection. 2015 Apr [PubMed PMID: 25578684]\| \|\|\|Paton R,Tolhurst N,Perisa M,Dempsey K,Tallon J, What mask to use? Australian nursing [PubMed PMID: 29235820]\| \|\|\|Peterson K,Novak D,Stradtman L,Wilson D,Couzens L, Hospital respiratory protection practices in 6 U.S. states: a public health evaluation study. American journal of infection control. 2015 Jan [PubMed PMID: 25564126]\| \|\|\|Katz LM,Tobian AA, Ebola virus disease, transmission risk to laboratory personnel, and pretransfusion testing. Transfusion. 2014 Dec [PubMed PMID: 25403825]\| \|\|\|Tajima K,Nishimura H,Hongo S,Hazawa M,Saotome-Nakamura AI,Tomiyama K,Obara C,Kato T, Estimation of secondary measles transmission from a healthcare worker in a hospital setting. International journal of infectious diseases : IJID : official publication of the International Society for Infectious Diseases. 2014 Jul [PubMed PMID: 24780918]\| \|\|\|Patrick MR,Hicks RW, Implementing AORN recommended practices for prevention of transmissible infections. AORN journal. 2013 Dec [PubMed PMID: 24266933]\| \|\|\|Hines L,Rees E,Pavelchak N, Respiratory protection policies and practices among the health care workforce exposed to influenza in New York State: evaluating emergency preparedness for the next pandemic. American journal of infection control. 2014 Mar [PubMed PMID: 24457143]\| \|\|\|Seto WH,Conly JM,Pessoa-Silva CL,Malik M,Eremin S, Infection prevention and control measures for acute respiratory infections in healthcare settings: an update. Eastern Mediterranean health journal = La revue de sante de la Mediterranee orientale = al-Majallah al-sihhiyah li-sharq al-mutawassit. 2013 [PubMed PMID: 23888794]\| \|\|\|Benson SM,Novak DA,Ogg MJ, Proper use of surgical n95 respirators and surgical masks in the OR. AORN journal. 2013 Apr [PubMed PMID: 23531312]\| \|\|\|Silkaitis C,Bardowski L,Coomer C,Trakas K,Lavin MA,Reddy S,Bolon MK,Zembower TR, Use of acid-fast bacilli staining to determine the need for airborne infection isolation precautions: a comparison of respiratory specimens. American journal of infection control. 2014 Nov [PubMed PMID: 25248484]\|<\|endoftext\|>	3.875
252	It’s never to early to introduce left brain learning and right brain learning into a children’s life. Reggio inspired and Montessori based activities can be presented to children as young as 12 months. In what Mari Montessori called sensitive periods, babies grow and develop rapidly needing as much freedom as possible for movement and becoming increasingly Independent and self-confident kids. In this lesson, we use the infant program initiates hands-on learning for our babies in order to precipitate movement language order and development of senses in the right brain environment. During an activity called marble painting, the children are learning how to follow directions and distinguish their belongings from others. As they grasp the containers and shake them to create movement from the marbles inside, they are practicing their fine motor skills. The end result is an array of unique artwork that the children made themselves. Fine motor skills are also being practiced in the left – brain environment here at the sensory table. As the children explore the materials in front of them, they are experiencing and comparing cold and warm temperatures and identifying different colours these simple activities provide many learning possibilities and help to facilitate whole child development well before the preschool age and most importantly the babies are happy and having fun.<\|endoftext\|>	3.671875
1,965	Chapter 10 Nuclear Chemistry Post on 06-Feb-2016 DESCRIPTIONChapter 10 Nuclear Chemistry. Standards Addressed in this Chapter. SPS3. Students will distinguish the characteristics and components of radioactivity. Differentiate among alpha and beta particles and gamma radiation. Differentiate between fission and fusion. - PowerPoint PPT Presentation Ch 10 Nuclear Chemistry Chapter 10Nuclear Chemistry Standards Addressed in this ChapterSPS3. Students will distinguish the characteristics and components of radioactivity.Differentiate among alpha and beta particles and gamma radiation.Differentiate between fission and fusion.Explain the process half-life as related to radioactive decay.Describe nuclear energy, its practical application as an alternative energy source, and its potential problems. SPS5. Students will compare and contrast the phases of matter as they relate to atomic and molecular motion.Compare and contrast the atomic/molecular motion of solids, liquids, gases and plasmas. Radioactivity is the process in which an unstable atomic nucleus emits charged particles and energy.Radioisotope is short for radioactive isotopes, which is any atom containing an unstable nucleus.Radioisotopes spontaneously change into other isotopes over time and is said to undergo nuclear decay.During nuclear decay, atoms of one element can change into atoms of a different element altogether. Types of Nuclear RadiationNuclear radiation is charged particles and energy that are emitted from the nuclei of radioisotopesCommon types of nuclear radiation include alpha particles, beta particles and gamma rays 1. Alpha DecayAlpha particle is a positively charged particle made up of two protons and two neutrons (the same as helium nucleus)Alpha particles are the least penetrating type of nuclear radiation.They can be stopped by a sheet of paper of by clothing.The alpha particle has no electrons so it has a 2+ charge. 42He is the symbol for an alpha particle Alpha DecayAlpha decay is expressed as an equation 2. Beta DecayBeta particle is an electron emitted by an unstable nucleusBeta particles are abbreviated or 0-1eBeta particles are more penetrating than alpha particles.Beta particles pass through paper but can be stopped by a thin sheet of metal. 2. Beta DecayThe beta particle has no mass During beta decay a neutron decomposes into a proton and an electronThe proton stays trapped in the nucleus while the electron is releasedBeta DecayBeta decay is expressed as an equation 3. Gamma DecayGamma ray is a penetrating ray of energy emitted by an unstable nucleus.The symbol for a gamma ray is The gamma radiation has no mass and no chargeDuring gamma decay the atomic number and mass number of the atom remain the same but the energy of the nucleus decreases Gamma DecayGamma decay often accompanies alpha or beta decay.Gamma rays have the most energy of the three, gamma rays can pass through paper and aluminum but is stopped by thick concrete or leadGamma DecayGamma decay Comparing Strength of Nuclear radiation Nuclear Ration SummaryAlpha Particles Symbol 42He2 protons & 2 neutrons Has a charge +2 and mass of 4 atmWeakestStopped by paper Symbol or 0-1eAn electronHas no massStronger than AlphaStopped by sheet of metalGamma Ray Symbol Only energyNo mass, No chargeStrongestStopped by thick lead or thick concrete Types of RadiationAlpha ()helium nucleuspaper2+Beta-minus (-)electron 1-leadGamma ()high-energy photon0concrete10.2 Rates of Nuclear DecayHalf-life is the time required for one half of a sample of radioisotope to decayAfter one half-life, half of the atoms in a sample have decayed, while the other half remains unchanged. Half-lives can vary from fractions of a second to billions of yearsTime in which of the original isotopes decayA. Half-LifeFirst Half-life original isotopes remain decayedSecond Half-life original isotopes remain decayedThird Half-life 1/8 original isotopes remain7/8 decayedUnlike chemical reaction rates, which vary with the conditions of a reaction, nuclear decay rates are constant. Half-Life progression of Iodine-131100 gram sample with 8.1 day lifeFirst lifeEtc.8.1 days 50 g remains16.2 days25 g remains32.4 days6.25g remains40.5 days3.125 g remains24.3 days12.5 g remains0 days100 gSecond lifeThird lifeFourth lifeFifth lifeHalf-life graph http://einstein.byu.edu/~masong/htmstuff/Radioactive2.html life calculationsAmount of sample divide by two for each life that passedAmount of time = (# of lives) X ( length of one life)A. Half-Life PracticeIf we start with 800 atoms of a radioactive substance, how many would remain after one half-life?_________ after two half-lives? _________ after three half-lives? _______2. If we start with 48 g of a radioactive substance with a 2 hour life , how much is left after two half-lives? _____ after four half-lives?___ how much time has passed for 4 lives? ______If we start with 16 grams of a radioactive substance that has a 6 day life, How much will remain after three half-lives?________ How much time would have passed?_______ 400 atoms200 atoms100 atoms12 g3 g2 grams18 days 8 hours6. If a sample originally had 150 grams of carbon-14, how many atoms will remain after 16,110 years? _______ 4. How much of the sample has decayed after zero years?________5. If only 25% of the carbon-14 remains, how old is the material containing the carbon-14? ___________zero10740 years old grams10.4 Fission and FusionStrong nuclear force is the attractive force that binds protons and neutrons together in the nucleus. Over very short distances the strong nuclear force is much great than the electric forces among protons. 1. The effect of size on Nuclear ForcesThe greater the number of protons in a nucleus the greater is the electric force that repels those protons.In larger nuclei, the repulsive electric force is stronger than in smaller nucleiLarger numbers of electric forces make larger nucleus less stable 2. Unstable NucleiA nucleus becomes unstable (radioactive) when the strong nuclear force can no longer overcome the repulsive electric forces among protons.All nuclei with more than 83 protons are radioactive FissionFission is the splitting of an atomic nucleus into two smaller parts.In nuclear fission, tremendous amounts of energy can be produced from very small amounts of mass. Chain ReactionA chain reaction refers to a process in which neutrons released in fission produce an additional fission in at least one further nucleus. This nucleus in turn produces neutrons, and the process repeats. The process may be controlled (nuclear power) or uncontrolled (nuclear weapons). Critical MassThe minimum amount of a substance that can sustain a chain reaction. It takes very little Uranium-235 to reach critical mass. FusionFusion is a process in which the nuclei of two atoms combine to form a larger nucleus.During fusion a small fraction of the reactant mass is converted into energy.Inside the sun an estimated 600 millions tons of hydrogen undergo fusion each secondFusion requires extremely high temperatures (10,000,000C).At these temperature matter can exist as plasma C. FusionPlasma is a state of matter in which atoms have been stripped of their electrons.Fusion reactions produce much more energy per gram of fuel and produce less radioactive waste than fission.Two main problems in designing a fusion rector 1st they need to achieve high temperatures required to start the reaction It requires a heat of about 10 million degrees Celsius. Scientist have to find a way of producing and containing that much heat.2nd they must contain the plasmaFusion can occur only in the plasma state of matter (super-heated gas). FissionSplitting a larger atom into smaller atomsReleases two or three neutronsReleases large amounts of energyUsed as a source for electricity FusionCombining small atoms into a larger atomRequires very high temperaturesReleases large amounts of energy 3. Nuclear Energy from FissionNuclear power plants generate about 20% of the electricity in the USNuclear power plant do not emit air pollutantsBut workers are made to wear protective clothing to recue their exposure to nuclear radiation. Nuclear power plants produce radioactive waste that must be isolated and stored so that it does not harm people or the environment.If the reactors cooling systems failed a meltdown might occurDuring a meltdown the core of the reactor melts and radioactive material may be released. Nuclear PowerFission Reactors Nuclear PowerFusion Reactors (not yet sustainable) Tokamak Fusion Test Reactor Princeton UniversityNational Spherical Torus ExperimentNuclear Power235U is limiteddanger of meltdowntoxic wastethermal pollutionHydrogen is abundantno danger of meltdownno toxic wastenot yet sustainable FISSIONFUSIONvs.Dangers Nuclear Decay nuclear wasteNuclear radiationBenefits MedicalCancer TreatmentRadioactive tracersNuclear Power Other Uses of RadiationIrradiated Food (p.676)Radioactive Dating (p.683)Nuclear Medicine (p.692-693)<\|endoftext\|>	4.0625
1,623	Understanding the political setting of the Language Movement and 21st February requires an examination of how religion played very different roles before and after the 1947 Partition in East Bengal. In British India, colonial subjects, both Hindus and Muslims, fought a common foe: the British colonial power. Yet, since the 19th century, the political and social status of the Hindus and Muslims differed significantly. Within the colonial bureaucracy, the Hindus were in a much more advantageous position than the Muslims. Among other reasons for the decline of Muslims in colonial India were the repressive agrarian systems since the Permanent Settlement of 1793 that negatively affected Muslim petty landowners and peasants and the replacement of Persian with English as the court language in 1837. Governor-General of India William Bentinck's act “displaced most of the Muslim professionals from traditional judicial service.” The Muslims either shunned or were slow to embrace the new system of education, while the Hindus zealously learned English, solidifying their position within the colonial administration. The immediate outcome of these differing paths was the “monopolisation of new employment opportunities” by the Hindus, while Muslims lagged far behind. In the context of the growing material success of the Hindus, the Muslims became self-conscious of their backwardness and, consequently, began to emphasise the separateness of their economic and political interests. A communal schism between the Hindus and Muslims began to calcify since, at least, the mid-19th century. In 1905, Viceroy Curzon's inchoate Partition of Bengal was designed to create two Bengals, West Bengal for Hindu Bengalis and East Bengal for Muslim Bengalis. Even though the Partition of Bengal was annulled in 1911 because of widespread Hindu protests, the brewing Hindu-Muslim antagonism eventually created the ideological foundation for the Partition of 1947. But, after the Partition, the political mood changed rapidly in East Pakistan, from Islamic nationalism or the pursuit of a separate homeland for Muslim Bengalis to a path of secular Bengali nationalism. “Whereas the pre-1947 nationalism was cloaked under the religious and/or communal surplice, the post-1947 nationalism was entirely secular.” Disillusionment with the idea of Pakistan as a country in which all Muslims—irrespective of their ethnicities—would enjoy equal access to political power and economic opportunities gradually set in. The Bengali Muslims' excitement about economic emancipation under the statehood of Pakistan soon proved to be a pipedream. Despite representing 54 percent of Pakistan's total population, East Pakistan had very little influence on the political decision-making process. The vacuum created by departing Hindus in East Pakistan's administration was soon filled by the Punjabi- and Urdu-speaking Muslim elites from Uttar Pradesh and other parts of India. For many Bengali observers, “the creation of Pakistan symbolised for the Bengali Muslims a change of masters only and not the mode of domination and exploitation.” Peter Bertocci elaborated on the background of this state of affairs and noted that “differences of language and culture between Bengalis and the ethnic groups of West Pakistan became accentuated in the context of the growing regional inequities characteristic of Pakistan's political economy.” The language factor pushed the thorniest wedge between the Bengalis and the ruling class of West Pakistan. The vast majority of East Pakistanis spoke Bengali, while only a fraction of West Pakistanis spoke Urdu. However, the Urdu-speaking political oligarchy dominated the politics of West Pakistan. Muhammad Ali Jinnah, the founder and Governor-General of Pakistan, announced the following, while addressing the Dhaka University Special Convocation on March 24, 1948: “There can be only one state language. If the component parts of this state are to march forward in unison, that language, in my opinion, can only be Urdu.” Such flagrant denial of the complexity of Pakistan's intricate ethnic and linguistic composition predictably ignited a nationalist passion among the Bengalis. Realising that the prevailing economic and political disparities between the two wings of Pakistan would not go away anytime soon, many Bengalis called into question the religion-based political vision that drove the creation of Pakistan. Many felt compelled to search for cultural, social, and linguistic roots that made them Bengali. Acclaimed Bengali educator and philologist Muhammad Shahidullah sought to celebrate, albeit with a tinge of essentialism, his concept of Bengaliness at the first Bengali literary conference of East Bengal, held in Dhaka, on December 31, 1948: “It is true that there are Hindus and Muslims. But what is transcending is that they are in essence Bengali. This is a reality. Nature with her own hand has stamped the indelible mark of Bengali in such a manner on our appearance and language that it is no longer possible to conceal it.” Bengali activist-historian Badruddin Umar called the prevailing Bengali sentiment a “homecoming of Bengali Muslims.” What Umar meant was that the Bengali Muslims finally realised the political fallacy of creating a pan-Muslim country, ignoring the robust complexity of the roles that religion, culture, tradition, and language play in forming ethnic identities. While not rejecting the importance of the Islamic faith in the Bengali Muslim character, some intellectuals endeavoured to explain the syncretic nature of Bengali Islam, influenced by many organic factors of pre-Islamic folk values. Many secular-minded Bengali Muslims disputed the argument that a faith that originated in the Arabian Peninsula could remain pure as it travelled across geographical and cultural boundaries. They deemed that a pan-Islamic ideology—buttressed by the imposition of Urdu as the state language—able to cohere the ethnically divergent peoples of Pakistan was a doomed political position from the beginning. Their point of contention was to understand how faith blends with local cultural forms, values, and practices. In The Islamic Syncretistic Tradition in Bengal (1983), Asim Roy cogently demonstrated how medieval Muslim mystics converted Bengali Muslims by fusing the Islamic faith with native myths and cosmologies. Far from being a corruption of the faith, Bengal's syncretic Islam provided a rich tapestry of cultural amalgamation. As Rafiuddin Ahmed stated: “As in many other Muslim societies, Islam in Bengal has taken many forms and has assimilated values and symbols not always in conformity with Qur'anic ideals and precepts. The cultural idioms of Islam underwent rapid transformations here, giving birth to a set of popular beliefs and practices, which, in essence, represented the popular culture of rural Bengal rooted in the pre-Islamic past….Islam in Bengal has not been able to escape the influences of local culture: The Bengali Muslims have remained Bengalis.” The ordinary Bengali Muslims' psychological predicament of opposing allegiances—to local culture and to an uncompromised interpretation of faith—would magnify in favour of indigenous roots of Bengali Islam after the Partition, when their disillusionment with Pakistan's promise of a safe haven for all Muslims became widespread. The coexistence of contradictory values explains “why the same region could support movements first for Pakistan, and then for Bangladesh.” It is in this post-Partition shifting context that the Language Movement had sown the seeds of wider Bengali agitation for political and economic emancipation. Adnan Zillur Morshed, PhD, is an architect, architectural historian, urbanist, and columnist. He teaches at the Catholic University of America in Washington, DC, and serves as Executive Director of the Centre for Inclusive Architecture and Urbanism at BRAC University. This essay has been excerpted from his article, “Modernism as Post-Nationalist Politics: Muzharul Islam's Faculty of Fine Arts (1953–56),” published in the Journal of the Society of Architectural Historians (2017).<\|endoftext\|>	4.28125
689	Asking young girls to “do science” leads them to show greater persistence in science activities than does asking them to “be scientists,” researchers at New York University and Princeton University find. The study results are published in Psychological Science, a journal of the Association for Psychological Science. “Describing science as actions, by saying ‘let’s do science,’ leads to more science engagement than does describing science in terms of identities, by asking them to ‘be scientists,’” explains Marjorie Rhodes, an associate professor in NYU’s Department of Psychology and the senior author of the study. “These effects particularly hold for children who are the target of stereotypes suggesting that they might not be the kind of person who succeeds in science—in this case, girls,” she adds. These findings suggest that efforts encouraging girls to enter science—a field in which they are underrepresented—might benefit from focusing on describing the activity of doing science rather than on encouraging children to adopt scientist identities, at least in early childhood. “The roots of gender disparities in science achievement take hold in early childhood,” Rhodes observes. “This research identifies an element of children’s environments that could be targeted to reduce early gender differences in science behavior among young children.” Rhodes and her coauthors, who include Princeton’s Sarah-Jane Leslie, note that the messaging children often receive through television shows centers on identity rather than action when it comes to science. In the newly published Psychological Science work, the researchers conducted four studies with children aged 4 to 9 years old. Here, the children received an introduction to science that described science as an identity (“Let’s be scientists! Scientists explore the world and discover new things!”) or as action (“Let’s do science! Doing science means exploring the world and discovering new things!”). Children were then asked to complete a new science game designed to illustrate the scientific method. Persistence was measured by how long they continued to play this game. Notably, girls who were initially asked to “do science” showed more persistence on the subsequent science game than did girls who had been asked to “be scientists.” By contrast, the effects of language for boys were more variable. For instance, one of their studies found that boys younger than 5 years old showed greater persistence when language was action-oriented while those older than 5 revealed higher levels of persistence when language was identity-oriented. Overall, these findings suggest that identity-focused language can undermine persistence in some children as they acquire new skills, particularly when cultural stereotypes lead children to question if they hold the relevant identity. Additional coauthors on the work include Kathryn Yee, a researcher in NYU’s Department of Psychology at the time of the study, and Katya Saunders, a postdoctoral researcher in NYU’s Department of Psychology. This research was supported, in part, by a grant from the Eunice Kennedy Shriver National Institute of Child Health and Human Development of the National Institutes of Health (R01HD087672). All data, materials, and analytic code are publicly available via the Open Science Framework. This article has received the badges for Open Data and Open Materials.<\|endoftext\|>	3.734375
593	One of the newest exoplanet discoveries is also one of the rarest. KELT-4Ab sits in a triple-star system. It takes three days to orbit the closest star, KELT-A. That star is orbited by two nearby stars, KELT-B and KELT-C. According to Phys.org, KELT-B and C are a lot further away and take nearly 30 years to orbit each other. The pair takes even longer to circle KELT-A. About four thousand years according to new research. What would the sky look like from KELT-4Ab? It would be blinding. Remember, the exoplanet takes just three days to orbit the closest star, KELT-A. That also means the planet is tidally locked. One side always faces the sun, while the other is plunged into eternal darkness. Here’s how the researchers describe it via Phys.org. The view from KELT-4Ab would likely be one where its sun, KELT-A, would appear roughly forty times as big as our sun does to us due to its proximity. The two other orbiting stars, on the other hand, would appear much dimmer due to their great distance, shining no brighter than our moon. A hot Jupiter Image of HD 80606b based on data from NASA’s Spitzer Space Telescope Triple star systems are cool to look at, but researchers hope to use this exoplanet to answer another mystery. ‘Hot Jupiters.’ When we began our search for planets outside our solar system, we assumed we would find things much like our own. Rocky planets near the front and gas giants in the back. Instead, we started finding gas planets like Jupiter, but orbits that take them much closer to their stars. Scientists were shocked when they discovered the first ‘hot Jupiter.’ But, the more exoplanet discoveries scientists made, the more our solar system seemed like the odd one out. We don’t even understand the basics of these planets. How do they form? How do they end up in such close proximity to their stars? One theory suggests a ‘hot Jupiter’ forms just like your typical gas giant. Then, the gravity from nearby planets or stars pushes them into closer orbits. Over hundreds of millions of years, these gas giants go from eccentric orbits to close, circular orbits. Scientists are using NASA’s Spitzer telescope to study what they believe is a ‘hot Jupiter’ in the making. Other theories suggest ‘hot Jupiters’ form in their current orbits. Scientists still don’t have the answers, but planets like HD 80606b and KELT-4Ab present perfect observation opportunities to figure it out.<\|endoftext\|>	4.0625
1,446	Types of Stroke According to the CDC, every year about 610,000 people in the United States have a new stroke. Almost 130,000 of the 800,000 Americans who die of cardiovascular disease each year are killed by strokes —that’s 1 in every 19 deaths from all causes. Anyone, including children, can have a stroke, and there are different kinds of stroke that can occur. Read on to learn about the most common types and what the signs are that your or someone else may be having a stroke. Most strokes (87%) that occur fall into the category of ischemic strokes. An ischemic stroke happens when blood flow through the artery that supplies blood to the brain becomes blocked. Ischemic strokes are primarily caused by blood clots and atherosclerosis, which is fatty deposits lining the vessel walls. According to the American Stroke Association, atherosclerosis can cause two types of obstruction: - Cerebral thrombosis is a thrombus (blood clot) that develops at the fatty plaque within the blood vessel. - Cerebral embolism is a blood clot that forms at another location in the circulatory system, usually the heart and large arteries of the upper chest and neck. Part of the blood clot breaks loose, enters the bloodstream and travels through the brain’s blood vessels until it reaches vessels too small to let it pass. A main cause of embolism is an irregular heartbeat called atrial fibrillation. This condition can cause clots to form in the heart, dislodge and travel to the brain. Ischemic stroke can also occur in other scenarios, one of which is during a procedure involving a catheter, such as TAVR. During a TAVR procedure, a long flexible tube (catheter) is advanced through an artery (often through a small incision in the leg) to reach the aortic valve. Once in position, an artificial replacement valve at the tip of the catheter is expanded, pushing the diseased natural valve aside. In the process of advancing the catheter and positioning the artificial replacement valve, embolic debris – calcium, tissue or other organic or foreign matter – may break loose and travel through the bloodstream toward the brain. If this matter sticks, it could potentially limit the blood and oxygen supply to the brain, which can lead to long-term damage. This can lead to a stroke, silent cerebral infarctions or other brain injury. Silent cerebral infarction (SCI), or “silent stroke,” is a brain injury without the typical symptoms and sign of a stroke and is likely caused by a blood clot that interrupts blood flow in the brain. It’s a risk factor for future strokes and a sign of progressive brain damage. Hemorrhagic stroke accounts for around 13% of all stroke and happens when a weakened vessel in the brain leaks blood or breaks open and bleeds into the surrounding brain. The leaked blood puts too much pressure on brain cells, which damages them. There are two types of hemorrhagic strokes: Intracerebral hemorrhage, the most common type, occurs when an artery in the brain bursts, flooding the surrounding tissue with blood. Two types of weakened blood vessels usually cause hemorrhagic stroke: aneurysms and arteriovenous malformations (AVMs). The less common type of hemorrhagic stroke is called a subarachnoid hemorrhage. It refers to bleeding in the area between the brain and the thin tissues that cover it. High blood pressure and aneurysms are examples of conditions that can cause a hemorrhagic stroke. Transient Ischemic Attack (TIA) A transient ischemic attack (TIA) is sometimes called a “mini-stroke.” It is a temporary blockage of blood flow, which is different than an ischemic or hemorrhagic stroke. In TIA, the blockage occurs for a short time, usually less than 5 minutes, and per definition symptoms and signs disappear within 24 hours. Similar to an ischemic stroke, TIAs can often be caused by a blood clot. There isn’t a way to distinguish between a TIA and a major stroke in the beginning, since the symptoms and signs are identical. Although the blockage does not last as long and it does not typically cause permanent damage, it is important to treat it just as seriously and call 911. Having a TIA may signal a major stroke may happen in the future. According to the CDC, more than a third of people who have a TIA and don’t get treatment have a major stroke within 1 year. As many as 10% to 15% of people will have a major stroke within 3 months of a TIA. Recognizing and treating TIAs can lower the risk of a major stroke. If you have a TIA, your health care team can find the cause and take steps to prevent a major stroke. The signs of a stroke The American Stroke Association outlines the best method to determining stroke: Use the letters in “F.A.S.T.” to spot stroke signs and know when to call 9-1-1 Face Drooping – Does one side of the face droop or is it numb? Ask the person to smile. Is the person’s smile uneven or lopsided? Arm Weakness – Is one arm weak or numb? Ask the person to raise both arms. Does one arm drift downward? Speech – Is speech slurred? Is the person unable to speak or hard to understand? Ask the person to repeat a simple sentence Time to Call 9-1-1 – If the person shows any of these symptoms, even if the symptoms go away, call 9-1-1 and get them to the hospital immediately. Additional Symptoms of Stroke: - Sudden numbness or weakness of face, arm, or leg, especially on one side of the body - Sudden confusion – trouble speaking or understanding speech - Sudden trouble seeing in one or both eyes - Sudden trouble walking, dizziness, loss of balance or coordination - Sudden severe headache with no known cause - Know the Facts about Stroke. Available at: https://www.cdc.gov/stroke/docs/ConsumerEd_Stroke.pdf. Accessed Feb. 22, 2019. - Types of Strokes. Available at: https://www.strokeassociation.org/en/about-stroke/types-of-stroke. Accessed Feb. 22, 2019. - Stroke Symptoms. Available at: https://www.strokeassociation.org/en/about-stroke/stroke-symptoms. Accessed Feb. 22, 2019. - Type of Strokes. Available at: https://www.cdc.gov/stroke/types_of_stroke.htm. Accessed Feb. 22, 2019. This educational blog was provided by Boston Scientific.<\|endoftext\|>	3.765625
627	In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2 - Mathematics Sum In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2 [see figure]. Find the length and breadth of the pond. Solution Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions 50 mx 40 m So, the distance between pond and lawn would be same around the pond. Say x m. Now, length of rectangular lawn (l1) = 50 m and breadth of rectangular lawn (b1) = 40 m Length of rectangular pond (l2)= 50 – (x + x) = 50 – 2x And breadth of rectangular pond (b2) = 40 – (x + x)= 40 – 2x Also, area of the grass surrounding the pond = 1184 m2 Area of rectangular lawn – Area of rectangular pond = Area of grass surrounding the pond l_1 xx b_1 - l_2 xx b_2 = 1184 ......[∵ Area of rectangle = length × breadth] ⇒ 50 xx 40 - (50 - 2x)(40 - 2x) = 1184 ⇒ 2000 - (2000 - 80x - 100x + 4x^2) = 1184 ⇒ 80x + 100x - 4x^2 = 1184 ⇒ 4x^2 - 180x + 1184 = 0 ⇒ x^2 - 45x + 296 = 0 ....[By splitting the middle term] ⇒ x^2 - 37x - 8(x - 37) = 0 ⇒ x(x - 37) - 8(x - 37) = 0 ⇒ (x - 37)(x - 8) = 0 ∴ x = 8 At x = 37, Length and Breadth of pond are – 24 and – 34, respectively but length and So, x = 37 cannot be possible ∴ Length of pond = 50 – 2x = 50 – 2(8) = 50 – 16 = 34 m And breadth of pond = 40 – 2x = 40 – 2(8) = 40 – 16 = 24 m Hence, required length and breadth of pond are 34 m and 24 m, respectively. Concept: Nature of Roots Is there an error in this question or solution? APPEARS IN NCERT Mathematics Exemplar Class 10<\|endoftext\|>	4.625
1,346	0 Q: # A card is drawn from a pack of 52 cards. The card is drawn at random. What is the probability that it is neither a spade nor a Jack? A) 9/13 B) 4/13 C) 10/13 D) 8/13 Explanation: There are 13 spade and 3 more jack Probability of getting spade or a jack: =13+352=1652=413 So probability of getting neither spade nor a jack: =1−413 = 9/13 Q: In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color? A) 2/9 B) 5/9 C) 4/9 D) 0 Explanation: Number of white marbles = 4 Number of Black marbles = 5 Total number of marbles = 9 Number of ways, two marbles picked randomly = 9C2 Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2 = 1/6 + 5/18 = 4/9. 0 182 Q: A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red? A) 2/3 B) 1/8 C) 3/8 D) 3/4 Explanation: Given number of balls = 3 + 5 + 7 = 15 One ball is drawn randomly = 15C1 probability that it is either pink or red = 5 216 Q: Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M? A) 1/4 B) 1/6 C) 1/8 D) 4 Explanation: Required probability is given by P(E) = 10 419 Q: 14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together. A) 11/379 B) 21/628 C) 24/625 D) 26/247 Explanation: Total no of ways = (14 – 1)! = 13! Number of favorable ways = (12 – 1)! = 11! So, required probability = $\left(\frac{\left(\mathbf{11}\mathbf{!}\mathbf{×}\mathbf{3}\mathbf{!}\right)}{\mathbf{13}\mathbf{!}}\right)$ = $\frac{39916800×6}{6227020800}$ = $\frac{\mathbf{24}}{\mathbf{625}}$ 10 630 Q: Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the on the two faces divisible by 3 or 4? A) 3/7 B) 7/11 C) 5/9 D) 6/13 Explanation: Here n(S) = 6 x 6 = 36 E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)} => n(E)=20 Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9. 24 775 Q: A person starting with 64 rupees and making 6 bets, wins three times and loses three times, the wins and loses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is: A) A gain of Rs. 27 B) A loss of Rs. 37 C) A loss of Rs. 27 D) A gain of Rs. 37 Explanation: As the win leads to multiplying the amount by 1.5 and loss leads to multiplying the amount by 0.5, we will multiply the initial amount by 1.5 thrice and by 0.5 thrice (in any order). The overall resultant will remain same. So final amount with the person will be (in all cases): 64(1.5)(1.5)(1.5)(0.5)(0.5)(0.5)= Rs. 27 Hence the final result is: 64 − 27 37 A loss of Rs.37 12 877 Q: A card is drawn from a pack of 52 cards. The probability of getting a queen of the club or a king of the heart is? A) 1/26 B) 1/13 C) 2/13 D) 1/52 Explanation: Here in this pack of cards, n(S) = 52 Let E = event of getting a queen of the club or a king of the heart Then, n(E) = 2 P(E) = n(E)/n(S) = 2/52 = 1/26 8 703 Q: A room contains 3 brown, 5 black and 4 white chairs. Two chairs are picked and are put in the lawn. What is the probability that none of the chairs picked is white ? A) 14/33 B) 14/55 C) 12/55 D) 13/33 Explanation: Total number of chairs = (3 + 5 + 4) = 12. Let S be the sample space. Then, n(s)= Number of ways of picking 2 chairs out of 12 12×11/2×66 Let n(E) = number of events of selecting 2 chairs for selecting no white chairs. => 8C8×7/2×28 Therefore required probability = 28/66 = 14/33.<\|endoftext\|>	4.4375
531	Mars Global Surveyor NASA's Mars Global Surveyor was launched on November 7, 1996 from Cape Canaveral Air Force Station (Florida) on a McDonnell Douglas-built Delta II-7925 rocket. Mars Global Surveyor was the first successful US mission launched to Mars in 20 years since the Viking mission in 1976. On November 2, 2006 it was lost due to loss of power (through incorrect alignment of the solar panels). On arriving into Mars orbit, the Mars Reconnaissance Orbiter made attempts to image the lost Mars Global Surveyor so mission control could diagnose the problem, but the spacecraft could not be found. Observations from the Mars Global Surveyor orbiter are used on the interactive Google Mars website. In its highly successful decade of operations, the Mars Global Surveyor orbiter became the longest-lasting Mars mission collecting a massive quantity of data about the Martian environment. The Mars Global Surveyor completed its primary mission to: - Characterize the surface features and geological processes on Mars. - Determine the composition, distribution and physical properties of surface minerals, rocks and ice. - Determine the global topography, planet shape, and gravitational field. - Establish the nature of the magnetic field and map the crustal remnant field (a crustal remnant field is evidence of magnetism within the planet's crust or rocks, produced by the planet's own magnetic field at the time of formation). - Monitor global weather and the thermal structure of the atmosphere. - Study interactions between Mars' surface and the atmosphere by monitoring surface features, polar caps that expand and recede, the polar energy balance, and dust and clouds as they migrate over a seasonal cycle. To achieve this primary objectives, the mission consisted of five principal instruments: - Mars Orbiter Camera (MOC) - to produce daily wide-angle and narrow-angle images of Mars. - Mars Orbiter Laser Altimeter (MOLA) - to measure the height of Martian surface features such as mountains and depth of valleys. - Thermal Emission Spectrometer (TES) - analysis of infrared radiation by scanning the heat emitted from the surface of Mars to gain insight into the mineral composition of the surface. - Electron Reflectometer (magnetometer) - to understand the magnetic properties of Mars to probe the interior of the planet. - Gravity Field Experiment (Radio Science) - mapping the gravity variations in Mars' gravitation. Pictures from Mars Orbital Camera Evidence from the Mars Global Surveyor MOC instrument that spurts of liquid water may sporadically flow on the Martian surface]]<\|endoftext\|>	3.921875
597	Published at Saturday, 04 May 2019. Worksheet. By Natalie Riou. I go into a lot of detail about meaningful experiences in this post. Basically, a child needs to have a reason for learning the concept. Completing a worksheet is not a good enough reason for a child. Providing activities that connect to real life gives children a reason to learn it. If you present a worksheet to a child and say “Read this so you can answer these questions.” Are they going to be motivated? Most likely not! But if a child is trying to learn how to build a sturdy fort, but must read the directions to learn how to do so, then that gives them a reason to learn. I see parents all the time in different Facebook groups mention something like this… I’m at a lost. My 2 year old is frustrating me with learning her letters. I have tried everything, we do a worksheet a day, but I feel like I’m beating a dead horse. People often wonder about the effectiveness of analogies. What do they teach? How do they work? Why are they so useful? What makes analogies so effective is their ability to get students to think critically. In order to answer an analogy question correctly, the student has to form a logical relationship, or ”bridge” between two words. They must think about how the words are related. Since words represent particulars (not universals), there is a nearly infinite number of ways they might be related. It is the student’s job to narrow this number, and focus on the most essential relation — the most basic aspect of the word’s function or definition. This page contains analogies worksheets. In these worksheets, students must be able to recognize the relationship between the words in a word pair and to recognize when two word pairs display parallel relationships. To answer an analogy question, you must formulate the relationship between the words in the given word pair and then select the answer containing words related to one another in most nearly the same way. Each question has five answer choices, and 12 questions total. Worksheets typically have a ”right answer.” Jamaica is expected to circle the rhyming words or match the pictures of things that start with the letter ”G.” She may learn quickly that putting down a wrong answer is emotionally costly. Worksheet activities may make her feel ignorant and incompetent, so that she learns to stop taking risks by guessing. Any content, trademark’s, or other material that might be found on the Kiches website that is not Kiches’s property remains the copyright of its respective owner/s. In no way does Kiches claim ownership or responsibility for such items, and you should seek legal consent for any use of such materials from its owner. Copyright © 2019 Kiches. All Rights Reserved.<\|endoftext\|>	3.859375
636	- Rhythmic instruments, such as drums, tambourines, shakers, and rhythm sticks - Parade music - Masking tape Objective: Children will use gross-motor skills and explore a variety of rhythms and movements as they parade around the classroom. To Prepare: Adjust furniture, if necessary, and use masking tape to create an open path for marching around the room. Warm-Up: Invite children to sit around you on the floor. Sing "A-Marching We Will Go" (to the tune of "The Farmer in the Dell"). Sing the song again, this time clapping your hands to the beat. Encourage children to sing and clap their hands along with you. Step 1: Pass out rhythm instruments to use while singing "A-Marching We Will Go." Encourage children to make sounds with the instruments by shaking, rattling, and hitting them. Discuss what kinds of sounds they hear. Use words such as loud and soft to describe the sounds. Step 2: Explain to children that they will march like soldiers around the room. Ask them to stand up. Have children stomp their feet in place to the beat of the song. Invite them to stomp to the beat you create as you clap your hands. Encourage them to march, clap, or play their instrument as you sing. Step 3: Divide children into two groups. Have one group sit down on the floor while members of the other group march along the path. Start the parade music. Encourage marchers to shake, rattle, and play their instruments. Invite members of the other group to clap as the marchers go by. Step 4: Switch groups a few times. As children march, ask them to change direction or to jump, hop, and so on. Remember: The main purposes of this activity are to practice rhythm and have fun marching to music. It isn't necessary for twos and threes to chant the words just right or follow the exact foot pattern (left, right, left). Avoid accidents by making sure that children have plenty of room between each other, especially when carrying instruments. - On a nice day, take this activity outdoors. Invite children to think of different ways they can march. can they march with high, bent legs? Strong, stomping legs? On tiptoes? - Let children create and decorate their own instruments. Encourage them to pretend-play their instruments, creating their own sounds as they march along to music. - Put out real and toy instruments in the dramatic-play area, along with other musical props, such as a microphone. Books - Children will want to march and sing as you share these musical inspirations! I Make Music by Eloise Greenfield (Writers and Readers Publishing) Shake My Sillies Out by Raffi (Crown Books) We're Going on a Bear Hunt by Michael Rosen (Simon & Schuster Children's) This activity originally appeared in the October, 1998 issue of Early Childhood Today.<\|endoftext\|>	3.921875
1,348	Regression Inferential Methods Presentation on theme: "Regression Inferential Methods"— Presentation transcript: Regression Inferential Methods Height Weight Suppose you took many samples of the same size from this population & calculated the LSRL for each. Using the slope from each of these LSRLs – we can create a sampling distribution for the slope of the true LSRL. What is the standard deviation of the sampling distribution? What is the mean of the sampling distribution equal? What shape will this distribution have? b b b b b b b mb = b What would you expect for other heights? Weight What would you expect for other heights? How much would an adult female weigh if she were 5 feet tall? This distribution is normally distributed. (we hope) She could weigh varying amounts – in other words, there is a distribution of weights for adult females who are 5 feet tall. What about the standard deviations of all these normal distributions? We want the standard deviations of all these normal distributions to be the same. Where would you expect the TRUE LSRL to be? Regression Model The mean response my has a straight-line relationship with x: Where: slope b and intercept a are unknown parameters For any fixed value of x, the response y varies according to a normal distribution. Repeated responses of y are independent of each other. The standard deviation of y (sy) is the same for all values of x. (sy is also an unknown parameter) What distribution does their weight have? Person # Ht Wt 1 64 130 10 175 15 150 19 125 21 145 40 186 47 121 60 137 63 143 68 120 70 112 78 108 83 160 Suppose we look at part of a population of adult women. These women are all 64 inches tall. What distribution does their weight have? The slope b of the LSRL is an unbiased estimator of the true slope b. We use to estimate The slope b of the LSRL is an unbiased estimator of the true slope b. The intercept a of the LSRL is an unbiased estimator of the true intercept a. The standard error s is an unbiased estimator of the true standard deviation of y (sy). Note: df = n-2 Let’s review the regression model! x & y have a linear relationship with the true LSRL going through the my sy is the same for each x-value. For a given x-value, the responses (y) are normally distributed What is the slope of a horizontal line? Height Weight Suppose the LSRL has a horizontal line –would height be useful in predicting weight? A slope of zero – means that there is NO relationship between x & y! Assumptions for inference on slope The true relationship is Linear Check the scatter plot & residual plot The observations are Independent and random Check that you have an SRS For any fixed value of x, the response y varies Normally about the true regression line. Check a histogram or boxplot of residuals Equal variance about regression line. The standard deviation of the response is constant. L I N E Hypotheses Be sure to define b! H0: b = 0 1 Ha: b > 0 Ha: b < 0 This implies that there is no relationship between x & y Or that x should not be used to predict y What would the slope equal if there were a perfect relationship between x & y? H0: b = 0 Ha: b > 0 Ha: b < 0 Ha: b ≠ 0 1 Be sure to define b! Because there are two unknowns a & b Formulas: Confidence Interval: Hypothesis test: df = n -2 Because there are two unknowns a & b Body fat = -27.376 + 0.250 weight r = 0.697 r2 = 0.485 Example: It is difficult to accurately determine a person’s body fat percentage without immersing him or her in water. Researchers hoping to find ways to make a good estimate immersed 20 male subjects, and then measured their weights. Find the LSRL, correlation coefficient, and coefficient of determination. Body fat = weight r = 0.697 r2 = 0.485 b) Explain the meaning of slope in the context of the problem. For each increase of 1 pound in weight, there is an approximate increase in .25 percent body fat. c) Explain the meaning of the coefficient of determination in context. Approximately 48.5% of the variation in body fat can be explained by the regression of body fat on weight. a = -27.376 b = 0.25 s = 7.049 d) Estimate a, b, and s. e) Create a scatter plot and residual plot for the data. Weight Body fat Weight Residuals f) Is there sufficient evidence that weight can be used to predict body fat? Assumptions: Scatterplot and residual plot shows Linear association. Have an Independent SRS of male subjects Since the boxplot of residual is approximately symmetrical, the responses are approximately Normally distributed. Since the points are evenly spaced across the LSRL on the scatterplot, sy is approximately Equal for all values of weight H0: b = 0 Where b is the true slope of the LSRL of weight Ha: b ≠ 0 & body fat Since the p-value < a, I reject H0. There is sufficient evidence to suggest that weight can be used to predict body fat. Be sure to show all graphs! g) Give a 95% confidence interval for the true slope of the LSRL. Assumptions: Scatter plot and residual plot show LINEAR association Have an INDEPENDENT SRS of male subjects Since the boxplot of residualS is approximately symmetrical, the responses are approximately NORMALLY distributed. Since the points are evenly spaced across the LSRL on the scatterplot, sy is approximately EQUAL for all values of weight We are 95% confident that the true slope of the LSRL of weight & body fat is between 0.12 and 0.38. Be sure to show all graphs! What does “s” represent (in context)? h) Here is the computer-generated result from the data: Sample size: 20 R-square = 48.5% s = df? What does “s” represent (in context)? Parameter Estimate Std. Err. Intercept Weight Correlation coeficient? Be sure to write as decimal first! What does this number represent? What do these numbers represent?<\|endoftext\|>	4.40625
182	The Trust has carried out work on fish since the 1970s when work began on fish populations and other wildlife in lakes created after the extraction of gravel. Declining wild brown trout populations were researched in the early 1990s to try and halt the declines and successes through habitat management were followed by the River Monnow Project. This project aimed to rejuvinate a neglected and damaged river and benefit the local community. The habitat restoration was carried out on a landscape scale and was the largest ever river habitat restoration project. The Trust took over the East Stoke Salmon & Trout Research Centre on the Frome in Dorset and continued the monitoring of salmon started with the installation of a fish counter at the site in 1973. This series of data makes it one of the most comprehensive records of salmon movements in Europe. The research into the causes of the decline of salmon and how to reverse this decline is of global importance.<\|endoftext\|>	3.671875
274	Definition - What does True Leaf mean? True leaves are the leaves of a seed plant that contain vascular tissue. Unlike seed leaves, true leaves tend to be quite low on the stem and are normally produced first, after the seed leaves (cotyledons). True leaves look distinctively different than the seed leaves. They tend to be hairier than their seed counterparts and tend to boast a more decorative shape. The true leaves appear above the cotyledons on the seedling, and appear as smaller versions of the plant’s adult foliage. MaximumYield explains True Leaf Prior to the appearance of true leaves, the plant produces cotyledons, which typically emerge from the soil upon germination. These provide enough nutrients and minerals to the plants until the true leaves start to emerge. All of the leaves that emerge after the true leaves have appeared will look the same as the true leaves, rather than match the look of the seed leaves. Most plants don’t begin the process of photosynthesis until the true leaves start to appear. While some growers prefer to pinch off the cotyledons once the true leaves emerge, some botanists recommend that you leave them intact to avoid accidentally damaging the stem. It is also advisable to wait until the appearance of true leaves before transplanting a seedling.<\|endoftext\|>	3.96875
703	What Is 22/27 as a Decimal + Solution With Free Steps The fraction 22/27 as a decimal is equal to 0.814. The fraction 22/27 is a non-terminating repeating decimal fraction. Here, 22 is the dividend, and 27 is the divisor. When division is performed, it will give us a decimal number having a group of terms that repeat endlessly. Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers. Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 22/27. Solution First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively. This can be done as follows: Dividend = 22 Divisor = 27 Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents: Quotient = Dividend $\div$ Divisor = 22 $\div$ 27 This is when we go through the Long Division solution to our problem. The following figure shows the solution for fraction 22/27. Figure 1 22/27 Long Division Method We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 22 and 27, we can see how 22 is Smaller than 27, and to solve this division, we require that 22 be Bigger than 27. This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later. Now, we begin solving for our dividend 22, which after getting multiplied by 10 becomes 220. We take this x1 and divide it by y; this can be done as follows: 220 $\div$ 27 $\approx$ 8 Where: 27 x 8 = 216 This will lead to the generation of a Remainder equal to 220 – 216 = 4. Now this means we have to repeat the process by Converting the 4 into 40 and solving for that: 40 $\div$ 27 $\approx$ 1 Where: 27 x 1 = 27 This, therefore, produces another Remainder which is equal to 40 – 27 = 13. Now this means we have to repeat the process by Converting the 13 into 130 and solving for that: 130 $\div$ 27 $\approx$ 4 Where: 27 x 4 = 108 Finally, we have a Quotient generated after combining the three pieces of it as 0.814, with a Remainder equal to 22. Images/mathematical drawings are created with GeoGebra.<\|endoftext\|>	4.84375
838	Sixty-six million years ago, an asteroid the size of a small city smashed into Earth. This impact, the one that would lead to the end of the dinosaurs, left a scar several miles underground and more than 115 miles wide. Chicxulub, which lies underneath the Yucatán Peninsula of Mexico, is the best-preserved large impact crater on Earth, although it’s buried underneath a half mile of rocks. It’s also the only crater on the planet with a mountainous ring of smashed rocks inside its outer rim, called a peak ring. How these features form has long been debated, but a new study in Nature shows they’re a product of extremely strong vibrations that let rock flow like liquid for a crucial few minutes after the impact. When an asteroid crashes into Earth, it leaves a bowl-shaped pit, just like you’d expect. But it doesn’t just leave a dent. If the asteroid is big enough, the resulting crater can be more than 20 miles deep, at which point it becomes unstable and collapses. “For a while, the broken rock behaves as a fluid,” said Jay Melosh, a professor of earth, atmospheric and planetary sciences at Purdue University. “There have been a lot of theories proposed about what mechanism allows this fluidization to happen, and now we know it’s really strong vibrations shaking the rock constantly enough to allow it to flow.” This mechanism, known as “acoustic fluidization,” is the process that allows the ring of mountains in the crater’s center to rise within minutes of the asteroid’s strike. (This idea was first proposed by Melosh in 1979). Craters are essentially the same on all the terrestrial planets (Earth, Mercury, Venus, Mars and our moon), but they’re hard to study in space for obvious reasons: We can’t look at them with the same detail we can on Earth. The Chicxulub crater isn’t easily accessible by traditional standards either; it’s been buried throughout the last 66 million years. So the International Ocean Discovery Program (a group within the International Continental Scientific Drilling Program), did the only thing they could — they dug. The team drilled a core roughly six inches in diameter and a mile into Earth, collecting rock that was shattered and partly melted by the impact that wiped out the dinosaurs. In examining fracture zones and patterns in the core, the international research team found an evolution in the vibration sequence that would allow debris to flow. “These findings help us understand how impact craters collapse and how large masses of rock behave in a fluid-like manner in other circumstances, such as landslides and earthquakes,” Melosh said. “Towns have been wiped out by enormous landslides, where people thought they were safe but then discovered that rock will flow like liquid when some disturbance sets a big enough mass in motion.” The extinction of the dinosaurs itself was probably not directly affected by the crater’s internal collapse — other, external effects of the impact did them in, Melosh said. Regardless, it’s important to understand the consequences of a large asteroid strike on Earth. Because cratering is the same on all the terrestrial planets, these findings also validate the mechanics of impacts everywhere in the solar system. Ulrich Riller, Michael H. Poelchau, Auriol S. P. Rae, Felix M. Schulte, Gareth S. Collins, H. Jay Melosh, Richard A. F. Grieve, Joanna V. Morgan, Sean P. S. Gulick, Johanna Lofi, Abdoulaye Diaw, Naoma McCall, David A. Kring. Rock fluidization during peak-ring formation of large impact structures. Nature, 2018; 562 (7728): 511 DOI: 10.1038/s41586-018-0607-z Note: The above post is reprinted from materials provided by Purdue University.<\|endoftext\|>	4.375
795	Courses Courses for Kids Free study material Offline Centres More Store # What is the value of $0.\overline {57}$A. $\dfrac{{57}}{{10}}$B. $\dfrac{{77}}{{99}}$C. $\dfrac{{19}}{{33}}$D. $\dfrac{{52}}{9}$ Last updated date: 06th Sep 2024 Total views: 365.1k Views today: 9.65k Verified 365.1k+ views Hint:In order to find the given value, we should first know what this bar is over $57$. This bar is known as the bar notation, which says that the numbers having a bar over it is being repeated over and over again. For example: $0.\overline x = 0.xxxxxxxxxxx........$ and it continues. Naming the given value as $x$; so, it becomes $x = 0.\overline {57}$. Since, it is a bar notation, so the value can be written as $x = 0.57575757575757.....$………..(1) Since the bar is on two consecutive digits and we need the same value after decimal, we are moving the decimal point towards right by two digits. For that we need to multiply the value (1) by $100$. So, multiplying the equation (1) by $100$, and we get: $100x = 100 \times \left( {0.57575757575757.....} \right)$ $\Rightarrow 100x = 57.575757575757......$………(2) Since, we need to simplify the terms in order to get the value of $x$. So, subtracting (1) from (2), and we get: $100x - x = \left( {57.575757575757......} \right) - \left( {0.575757575757......} \right)$ Solving the left-hand side and the right side separately, and we can see that the values after decimals are the same and can be cancelled out from simple subtraction. So, we get: $100x - x = \left( {57.575757575757......} \right) - \left( {0.575757575757......} \right) \\ \Rightarrow 99x = 57 \\$ Dividing both the sides by $99$ in order to get the value of $x$ , and we get: $\Rightarrow \dfrac{{99x}}{{99}} = \dfrac{{57}}{{99}} \\ \Rightarrow x = \dfrac{{57}}{{99}} \\$ Since, we can see that the value on the right-side can be further simplified, by dividing both the numerator and denominator by $3$ as it’s the highest factor that can divide both of them. So, dividing and multiplying the right-side by$3$, we get: $\Rightarrow x = \dfrac{{\dfrac{{57}}{3}}}{{\dfrac{{99}}{3}}} \\ \Rightarrow x = \dfrac{{19}}{{33}} \\ \therefore x = 0.\overline {57}$ Therefore, the value of $0.\overline {57}$ and the correct answer is option C. Note:Do not commit a mistake by repeating only $5$ or $7$ from $0.\overline {57}$, as because the bar is upon both $5$ and $7$, so must be repeated in the same order. Since, the bar was on two digits that’s why we multiplied the value by $100$, in order to take the two digits out. If the bar is on one digit then we can multiply by $10$, then subtract.<\|endoftext\|>	4.65625
372	How do video games and apps lead to learning? At the most basic level, play equals learning. When kids play, they learn. Play is helpful in teaching about problem solving, testing hypotheses, and developing social skills. Play also practices skills such as planning, prioritization, and self-management. Play that involves technology, what we refer to as digital play, provides the same opportunities for learning as does traditional play. Digital play can be even more powerful because technologies can be customized to an individual’s need, provide immediate feedback, and adaptively increase the level of challenge to meet a child’s mastery level. The incredible engagement of many video games and apps enhances the attention and motivation of children for learning. We have only begun to understand how to use these powerful technologies to help children learn academic, social, emotional, and executive-functioning skills. The following articles and videos can be shared with your patients or used by you to learn more about the science of play and how video games and apps can enhance learning. Can Video Games Make you Smarter? (Really fun four-minute video that covers much of the research on games and learning.) Gamification – How the Principles of Play Apply to Real Life (Daniel Floyd introduces Gamification, a concept that takes the principles of play and uses them to make activities in reality more engaging. However gamification can also have some risks, and this video talks about how we can find a balance and safely use gaming as a positive method of education.) The Science of Play (How play leads to learning and how digital play needs to be considered a major learning opportunity in the 21st century.) Cognitive Benefits of Video Games (Article by noted expert on play, Peter Gray, where he cites how game play improves visual, attention, executive- functioning, and job skills.)<\|endoftext\|>	4.21875
275	Earwigs are in the Class Insecta and are of the Order Dermaptera. Dermaptera means “skin wing” due to the leathery appearance of the wings. Earwigs have four wings and the hind wings typically fold under the front wings. There are over 20 species of earwigs in North America; but there are about 1,800 species worldwide. Most earwigs encountered in North America are up to 25 mm long; however, there are species in other parts of the world which can reach almost 80 mm in length, although the largest known earwig is extinct. Three common species in the U.S. are European earwigs, red-legged earwigs, and striped earwigs. European earwig adults are approximately 16 mm long and are rust-brown colored with light-colored wing covers. They are one of the most commonly encountered species and found in many parts of North America. Red-legged earwig adults are 12 to 15 mm long and are brown to black in color. They are most common in the southern half of the U.S. Striped earwigs, sometimes called the shore earwig, are dark brown, bordering on a reddish brown. They are 20 to 25 mm long and are most likely found in the southern half of the U.S.<\|endoftext\|>	3.953125
1,071	We have an intuitive understanding of how the world works. Our intuition works well in our everyday lives, but it doesn’t quite match reality. One of the basic assumptions we make about objects is that they are objectively real. For example, when you leave your apartment in the morning, you imagine your house is right where you left it as you go about your day. Barring some calamity, you expect to find your apartment just as you left it when you return in the evening. This idea is known as macrorealism, in that a macroscopic object such as your house is still there even when you’re not looking at it. In scientific terms we would say the state of an object exists independent of an observer. While this seems obviously true, we can test the idea experimentally. To show this, let’s consider an experiment. Suppose we have a box that contains a coin. At any time we can open the box and see whether it’s heads or tails. The catch is that the box has to follow a certain rule. Whatever you observe initially, you must observe the same thing ten seconds later. If you open the box and see heads, then ten seconds later when you open the box you must also see heads. For the experiment we’ll open the box, then open it five seconds later, then finally open it again at ten seconds. We know the first and last observations have to agree, but what about the observation in the middle? It depends upon what the box does with the coin. One possibility is that the box does nothing, so if you see tails the first time, you will see tails both other times. Another possibility is that each time you open the box it flips the coin when you close it. So the first time you see heads, the second time tails, and the third time heads again. But we could also make a tricky box that flips the coin at seemingly random times, but does so in such a way that every ten seconds it repeats the pattern. In that case, we would always see the same thing the first and last time, but the middle observation would seem random. If macrorealism is true, then middle observations being always the same as the first and last observation, always opposite or always random are the only possibilities for the experiment. This reality condition can be expressed in a simple inequality known as the Leggett-Garg Inequality. This is done by looking at the correlation between each observation. Since the first and last observations must always agree, then we can say the correlation C13 = 1. If the middle observation is also the same, then we can say C12 = C23 = 1. If the middle observation is always opposite, then we can sayC12 = C23 = – 1. If the middle observation is random thenC12 = C23 = 0. The Leggett-Garg inequality simply states that C12 + C23 – C13 must be no bigger than 1. This makes sense because the three cases give sums of 1, -3, or -1. If any experiment violates this inequality, we can say it violates the assumption of microrealism. What’s interesting is that quantum systems violate this inequality all the time. Take, for example, a recent experiment on neutrinos. Neutrinos have a strange property known as oscillation, where the “flavor” of a neutrino changes. Each flavor interacts with other matter in a unique way, so when we observe neutrinos we also observe their flavor. It’s tempting to imagine the three flavors of a neutrino as being like the state of a traffic light. It’s always red, green, or yellow, but changes over time. If that were the case, then the Leggett-Garg inequality should hold for neutrino oscillation. But it doesn’t. Recently a team measured the flavor of neutrinos beamed from Fermilab to the MINOS far detector near Soudan, MN. They found a clear violation of the Leggett-Garg inequality, which means that neutrino flavor violates microrealism. In other words neutrino flavor is a truly quantum state. A particular neutrino doesn’t have a specific flavor that changes over time, because flavor isn’t macroscopically real. It’s one more example of how our intuition fails when it comes to quantum theory. Paper: A. J. Leggett and Anupam Garg. Quantum mechanics versus macroscopic realism: Is the flux there when nobody looks? Phys. Rev. Lett. 54, 857 (1985) DOI:http://dx.doi.org/10.1103/PhysRevLett.54.857 Paper: J. A. Formaggio, et al. Violation of the Leggett-Garg Inequality in Neutrino Oscillations. Phys. Rev. Lett. 117, 050402 (2016) DOI:http://dx.doi.org/10.1103/PhysRevLett.117.050402<\|endoftext\|>	3.75
520	The Nettle - Urtica dioica The nettle is a highly successful plant found all over the temperate areas of the world. It spreads by means of seeds and underground rhizomes that creep around just under the surface of the soil. The jagged leaves held in pairs along the square stems are easily recognisable particularly after having experienced the sting. The plant itself is variable growing from 0.6 to 2 metres plus in height and can be found in a variety of habitats and soil types. It prefers rich soils and therefore does well around human settlements benefiting from the waste we produce - often indicating where old settlements have long since disappeared from the countryside. How did the nettle get its name? The latin name of the plant dioica means 'two houses' - this refers to the fact that the male and female flowers are normally carried on separate plants. It is possible that the 'nettle' is derived from Noedl meaning a needle - referring to the stinging mechanism in the nettle leaves. Others suggest that it comes from the Latin nere and other similar old European verbs meaning to sew. What's in the sting? The stinging structure of the nettle is very similar to the hypodermic needle although it predates that man-made invention by millions of years! Each sting is actually a hollow hair stiffened by silica with a swollen base that contains the venom. The tip of this hair is very brittle and when brushed against, no matter how lightly, it breaks off exposing a sharp point that penetrates the skin and delivers its stinging payload. It used to be thought that the main constituent of the sting was formic acid - the same chemical used by ants, giving that never forgotten burning sensation that demands to be scratched. Although formic acid is present in the sting, recent research has shown that the main chemicals are histamine, acetylcholine and 5-hydroxytryptamine (serotonin). A fourth ingredient has yet to be identified. Remember when stung a natural remedy will often be found close at hand. The leaves of the dock contain chemicals that neutralise the sting and also cool the skin. A real sting in the tail The sting of our native nettle is nothing compared to some of its tropical cousins! One species in Timor causes a burning sensation and symptoms like lockjaw which can last for days or weeks. The effects of another species from Java last for months and have frequently caused the death of some of its unfortunate victims.<\|endoftext\|>	3.65625
4,509	Dots per inch The lead section of this article may need to be rewritten. (September 2016) (Learn how and when to remove this template message) Dots per inch (DPI, or dpi) is a measure of spatial printing, video or image scanner dot density, in particular the number of individual dots that can be placed in a line within the span of 1 inch (2.54 cm). Similarly, the more newly introduced dots per centimeter (d/cm or dpcm) refers to the number of individual dots that can be placed within a line of 1 centimeter (≈ 0.393 in). Monitors do not have dots, but do have pixels; the closely related concept for monitors and images is pixels per inch or PPI. Many resources, including the Android developer guide, use the terms DPI and PPI interchangeably. DPI measurement in printing DPI is used to describe the resolution number of dots per inch in a digital print and the printing resolution of a hard copy print dot gain, which is the increase in the size of the halftone dots during printing. This is caused by the spreading of ink on the surface of the media. Up to a point, printers with higher DPI produce clearer and more detailed output. A printer does not necessarily have a single DPI measurement; it is dependent on print mode, which is usually influenced by driver settings. The range of DPI supported by a printer is most dependent on the print head technology it uses. A dot matrix printer, for example, applies ink via tiny rods striking an ink ribbon, and has a relatively low resolution, typically in the range of 60 to 90 DPI (420 to 280 μm). An inkjet printer sprays ink through tiny nozzles, and is typically capable of 300–720 DPI. A laser printer applies toner through a controlled electrostatic charge, and may be in the range of 600 to 2,400 DPI. The DPI measurement of a printer often needs to be considerably higher than the pixels per inch (PPI) measurement of a video display in order to produce similar-quality output. This is due to the limited range of colors for each dot typically available on a printer. At each dot position, the simplest type of color printer can either print no dot, or print a dot consisting of a fixed volume of ink in each of four color channels (typically CMYK with cyan, magenta, yellow and black ink) or 24 = 16 colors on laser, wax and most inkjet printers, of which only 14 or 15 (or as few as 8 or 9) may be actually discernible depending on the strength of the black component, the strategy used for overlaying and combining it with the other colors, and whether it is in "color" mode. Higher-end inkjet printers can offer 5, 6 or 7 ink colors giving 32, 64 or 128 possible tones per dot location (and again, it can be that not all combinations will produce a unique result). Contrast this to a standard sRGB monitor where each pixel produces 256 intensities of light in each of three channels (RGB). While some color printers can produce variable drop volumes at each dot position, and may use additional ink-color channels, the number of colors is still typically less than on a monitor. Most printers must therefore produce additional colors through a halftone or dithering process, and rely on their base resolution being high enough to "fool" the human observer's eye into perceiving a patch of a single smooth color. The exception to this rule is dye-sublimation printers, which can apply a much more variable amount of dye—close to or exceeding the number of the 256 levels per channel available on a typical monitor—to each "pixel" on the page without dithering, but with other limitations: - lower spatial resolution (typically 200 to 300 dpi), which can make text and lines look somewhat rough - lower output speed (a single page requiring three or four complete passes, one for each dye color, each of which may take more than fifteen seconds—generally quicker, however, than most inkjet printers' "photo" modes) - a wasteful (and, for confidential documents, insecure) dye-film roll cartridge system - occasional color registration errors (mainly along the long axis of the page), which necessitate recalibrating the printer to account for slippage and drift in the paper feed system. These disadvantages mean that, despite their marked superiority in producing good photographic and non-linear diagrammatic output, dye-sublimation printers remain niche products, and devices using higher resolution, lower color depth, and dither patterns remain the norm. This dithered printing process could require a region of four to six dots (measured across each side) in order to faithfully reproduce the color in a single pixel. An image that is 100 pixels wide may need to be 400 to 600 dots in width in the printed output; if a 100×100-pixel image is to be printed in a one-inch square, the printer must be capable of 400 to 600 dots per inch to reproduce the image. Fittingly, 600 dpi (sometimes 720) is now the typical output resolution of entry-level laser printers and some utility inkjet printers, with 1200/1440 and 2400/2880 being common "high" resolutions. This contrasts with the 300/360 (or 240) dpi of early models, and the approximate 200 dpi of dot-matrix printers and fax machines, which gave faxed and computer-printed documents—especially those that made heavy use of graphics or colored block text—a characteristic "digitized" appearance, because of their coarse, obvious dither patterns, inaccurate colors, loss of clarity in photographs, and jagged ("aliased") edges on some text and line art. DPI or PPI in digital image files This section does not cite any sources. (January 2010) (Learn how and when to remove this template message) In printing, DPI (dots per inch) refers to the output resolution of a printer or imagesetter, and PPI (pixels per inch) refers to the input resolution of a photograph or image. DPI refers to the physical dot density of an image when it is reproduced as a real physical entity, for example printed onto paper. A digitally stored image has no inherent physical dimensions, measured in inches or centimeters. Some digital file formats record a DPI value, or more commonly a PPI (pixels per inch) value, which is to be used when printing the image. This number lets the printer or software know the intended size of the image, or in the case of scanned images, the size of the original scanned object. For example, a bitmap image may measure 1,000 × 1,000 pixels, a resolution of 1 megapixel. If it is labeled as 250 PPI, that is an instruction to the printer to print it at a size of 4 × 4 inches. Changing the PPI to 100 in an image editing program would tell the printer to print it at a size of 10 × 10 inches. However, changing the PPI value would not change the size of the image in pixels which would still be 1,000 × 1,000. An image may also be resampled to change the number of pixels and therefore the size or resolution of the image, but this is quite different from simply setting a new PPI for the file. For vector images, there is no equivalent of resampling an image when it is resized, and there is no PPI in the file because it is resolution independent (prints equally well at all sizes). However, there is still a target printing size. Some image formats, such as Photoshop format, can contain both bitmap and vector data in the same file. Adjusting the PPI in a Photoshop file will change the intended printing size of the bitmap portion of the data and also change the intended printing size of the vector data to match. This way the vector and bitmap data maintain a consistent size relationship when the target printing size is changed. Text stored as outline fonts in bitmap image formats is handled in the same way. Other formats, such as PDF, are primarily vector formats which can contain images, potentially at a mixture of resolutions. In these formats the target PPI of the bitmaps is adjusted to match when the target print size of the file is changed. This is the converse of how it works in a primarily bitmap format like Photoshop, but has exactly the same result of maintaining the relationship between the vector and bitmap portions of the data. Computer monitor DPI standards Since the 1980s, the Microsoft Windows operating system has set the default display "DPI" to 96 PPI, while Apple/Macintosh computers have used a default of 72 PPI. These default specifications arose out of the problems rendering standard fonts in the early display systems of the 1980s, including the IBM-based CGA, EGA, VGA and 8514 displays as well as the Macintosh displays featured in the 128K computer and its successors. The choice of 72 PPI by Macintosh for their displays arose from existing convention: the official 72 points per inch mirrored the 72 pixels per inch that appeared on their display screens. (Points are a physical unit of measure in typography, dating from the days of printing presses, where 1 point by the modern definition is 1/72 of the international inch (25.4 mm), which therefore makes 1 point approximately 0.0139 in or 352.8 µm). Thus, the 72 pixels per inch seen on the display had exactly the same physical dimensions as the 72 points per inch later seen on a printout, with 1 pt in printed text equal to 1 px on the display screen. As it is, the Macintosh 128K featured a screen measuring 512 pixels in width by 342 pixels in height, and this corresponded to the width of standard office paper (512 px ÷ 72 px/in ≈ 7.1 in, with a 0.7 in margin down each side when assuming 8.5 in × 11 in North American paper size (in Europe, it's 21cm x 30cm - called "A4". B5 is 176 millimeters x 250 millimeters)). A consequence of Apple's decision was that the widely used 10-point fonts from the typewriter era had to be allotted 10 display pixels in em height, and 5 display pixels in x-height. This is technically described as 10 pixels per em (PPEm). This made 10-point fonts be rendered crudely and made them difficult to read on the display screen, particularly the lowercase characters. Furthermore, there was the consideration that computer screens are typically viewed (at a desk) at a distance 30% greater than printed materials, causing a mismatch between the perceived sizes seen on the computer screen and those on the printouts. Microsoft tried to solve both problems with a hack that has had long-term consequences for the understanding of what DPI and PPI mean. Microsoft began writing its software to treat the screen as though it provided a PPI characteristic that is of what the screen actually displayed. Because most screens at the time provided around 72 PPI, Microsoft essentially wrote its software to assume that every screen provides 96 PPI (because ). The short-term gain of this trickery was twofold: - It would seem to the software that more pixels were available for rendering an image, thereby allowing for bitmap fonts to be created with greater detail. - On every screen that actually provided 72 PPI, each graphical element (such as a character of text) would be rendered at a size larger than it "should" be, thereby allowing a person to sit a comfortable distance from the screen. However, larger graphical elements meant less screen space was available for programs to draw; indeed, although the default 720-pixel wide mode of a Hercules mono graphics adaptor (the one-time gold standard for high resolution PC graphics) – or a "tweaked" VGA adaptor – provided an apparent 7.5-inch page width at this resolution, the more common and color-capable display adaptors of the time all provided a 640-pixel wide image in their high resolution modes, enough for a bare 6.67 inches at 100% zoom (and barely any greater visible page height – a maximum of 5 inches, versus 4.75). Consequently, the default margins in Microsoft Word were set, and still remain at 1 full inch on all sides of the page, keeping the "text width" for standard size printer paper within visible limits; despite most computer monitors now being both larger and finer-pitched, and printer paper transports having become more sophisticated, the Mac-standard half-inch borders remain listed in Word 2010's page layout presets as the "narrow" option (versus the 1-inch default). - Without using supplemental, software-provided zoom levels, the 1:1 relationship between display and print size was (deliberately) lost; the availability of different-sized, user-adjustable monitors and display adaptors with varying output resolutions exacerbated this, as it was not possible to rely on a properly-adjusted "standard" monitor and adaptor having a known PPI. For example, a 12" Hercules monitor and adaptor with a thick bezel and a little underscan may offer 90 "physical" PPI, with the displayed image appearing nearly identical to hardcopy (assuming the H-scan density was properly adjusted to give square pixels) but a thin-bezel 14" VGA monitor adjusted to give a borderless display may be closer to 60, with the same bitmap image thus appearing 50% larger; yet, someone with an 8514 ("XGA") adaptor and the same monitor could achieve 100 DPI using its 1024-pixel wide mode and adjusting the image to be underscanned. A user who wanted to directly compare on-screen elements against those on an existing printed page by holding it up against the monitor would therefore first need to determine the correct zoom level to use, largely by trial and error, and often not be able to obtain an exact match in programs that only allowed integer percent settings, or even fixed pre-programmed zoom levels. For the examples above, they may need to use respectively 94% (precisely, 93.75) – or 90/96, 63% (62.5) – or 60/96; and 104% (104.167) – or 100/96, with the more commonly accessible 110% actually being a less precise match. Thus, for example, a 10-point font on a Macintosh (at 72 PPI) was represented with 10 pixels (i.e., 10 PPEm), whereas a 10-point font on a Windows platform (at 96 PPI) at the same zoom level is represented with 13 pixels (i.e., Microsoft rounded 13.3333 to 13 pixels, or 13 PPEm) – and, on a typical consumer grade monitor, would have physically appeared around 15/72 to 16/72 of an inch high instead of 10/72. Likewise, a 12-point font was represented with 12 pixels on a Macintosh, and 16 pixels (or a physical display height of maybe 19/72 of an inch) on a Windows platform at the same zoom, and so on. The negative consequence of this standard is that with 96 PPI displays, there is no longer a 1-to-1 relationship between the font size in pixels and the printout size in points. This difference is accentuated on more recent displays that feature higher pixel densities. This has been less of a problem with the advent of vector graphics and fonts being used in place of bitmap graphics and fonts. Moreover, many Windows software programs have been written since the 1980s which assume that the screen provides 96 PPI. Accordingly, these programs do not display properly at common alternative resolutions such as 72 PPI or 120 PPI. The solution has been to introduce two concepts: - logical PPI: The PPI that software claims a screen provides. This can be thought of as the PPI provided by a virtual screen created by the operating system. - physical PPI: The PPI that a physical screen actually provides. Software programs render images to the virtual screen and then the operating system renders the virtual screen onto the physical screen. With a logical PPI of 96 PPI, older programs can still run properly regardless of the actual physical PPI of the display screen, although they may exhibit some visual distortion thanks to the effective 133.3% pixel zoom level (requiring either that every third pixel be doubled in width/height, or heavy-handed smoothing be employed). How Microsoft Windows handles DPI scaling Displays with high pixel densities were not common up to the Windows XP era. High DPI displays became mainstream around the time Windows 8 was released. Display scaling by entering a custom DPI irrespective of the display resolution has been a feature of Microsoft Windows since Windows 95. Windows XP introduced the GDI+ library which allows resolution-independent text scaling. Windows Vista introduced support for programs to declare themselves to the OS that they are high-DPI aware via a manifest file or using an API. For programs that do not declare themselves as DPI-aware, Windows Vista supports a compatibility feature called DPI virtualization so system metrics and UI elements are presented to applications as if they are running at 96 DPI and the Desktop Window Manager then scales the resulting application window to match the DPI setting. Windows Vista retains the Windows XP style scaling option which when enabled turns off DPI virtualization for all applications globally. DPI virtualization is a compatibility option as application developers are all expected to update their apps to support high DPI without relying on DPI virtualization. Windows Vista also introduces Windows Presentation Foundation. WPF .NET applications are vector-based, not pixel-based and are designed to be resolution-independent. Developers using the old GDI API and Windows Forms on .NET Framework runtime need to update their apps to be DPI aware and flag their applications as DPI-aware. Windows 7 adds the ability to change the DPI by doing only a log off, not a full reboot and makes it a per-user setting. Additionally, Windows 7 reads the monitor DPI from the EDID and automatically sets the system DPI value to match the monitor's physical pixel density, unless the effective resolution is less than 1024 x 768. In Windows 8, only the DPI scaling percentage is shown in the DPI changing dialog and the display of the raw DPI value has been removed. In Windows 8.1, the global setting to disable DPI virtualization (only use XP-style scaling) is removed and a per-app setting added for the user to disable DPI virtualization from the Compatibility tab. When the DPI scaling setting is set to be higher than 120 PPI (125%), DPI virtualization is enabled for all applications unless the application opts out of it by specifying a DPI aware flag (manifest) as "true" inside the EXE. Windows 8.1 retains a per-application option to disable DPI virtualization of an app. Windows 8.1 also adds the ability for different displays to use independent DPI scaling factors, although it calculates this automatically for each display and turns on DPI virtualization for all monitors at any scaling level. Windows 10 adds manual control over DPI scaling for individual monitors. There are some ongoing efforts to abandon the DPI Image resolution unit in favor of a metric unit, giving the inter-dot spacing in dots per centimeter (px/cm or dpcm), as used in CSS3 media queries or micrometres (µm) between dots. A resolution of 72 DPI, for example, equals a resolution of about 28 dpcm or an inter-dot spacing of about 353 µm. - Pixel density - Samples per inch – a related concept for image scanners - Lines per inch - Metric typographic units - Display resolution - Mouse DPI - The acronym appears in sources as either "DPI" or lowercase "dpi". See: "Print Resolution Understanding 4-bit depth – Xerox" (PDF). Xerox.com. September 2012. - CSS3 media queries draft - "OKI's Technology Guide to Inkjet Printing". www.askoki.co.uk. Archived from the original on 2009-08-15. - Hitchcock, Greg (2005-10-08). "Where does 96 DPI come from in Windows?". Microsoft Developer Network Blog. Microsoft. Retrieved 2009-11-07. - Hitchcock, Greg (2005-09-08). "Where does 96 DPI come from in Windows?". blogs.msdn.com. Retrieved 2010-05-09. - Connare, Vincent (1998-04-06). "Microsoft Typography – Making TrueType bitmap fonts". Microsoft. Retrieved 2009-11-07. - fbcontrb (2005-11-08). "Where does 96 DPI come from in Windows?". Blogs.msdn.com. Retrieved 2018-04-03. - "Why text appears different when drawn with GDIPlus versus GDI". Support.microsoft.com. 2018-02-04. Retrieved 2018-04-03. - "Win32 SetProcessDPIAware Function". - "Windows Vista DPI scaling: my Vista is bigger than your Vista". December 11, 2006. - Christoph Nahr /. "High DPI Settings in Windows". Kynosarges.org. Retrieved 2018-04-03. - "Media Queries". - "Class ResolutionSyntax". Sun Microsystems. Retrieved 2007-10-12.<\|endoftext\|>	3.8125
1,000	What are vitamin B12 deficiency and folic acid deficiency? — Vitamin B12 and folic acid (also called “folate” or “vitamin B9”) are 2 different vitamins that your body needs to work normally. A “deficiency” means that your body does not have as much of something as it needs. Why are vitamin B12 and folic acid important? — Your body needs vitamin B12 for many things. For example, you need vitamin B12 to make new blood cells, such as red blood cells, and for your nervous system to work normally. Your body needs folic acid, too, to make new cells. Folic acid is especially important for women who are planning to get pregnant (or might get pregnant). That’s because folic acid is needed for a baby to develop normally, especially during the first few weeks of pregnancy. Folic acid is needed for the spinal cord to develop correctly. ●They don’t eat enough foods that have the vitamin in them – Folic acid is added to many grains and cereals, so most people get enough. People who eat a vegetarian diet, especially strict vegans, need to take extra vitamin B12. (Vitamin B12 is found in meats and animal products like eggs.) ●They eat foods with vitamin B12 and folic acid, but their body can’t absorb or use the vitamins normally – These things can happen with certain medical problems, after weight loss surgery, or as a side effect of certain medicines. This is more common in the case of vitamin B12. ●They have a condition that makes them need extra folic acid – Examples include certain types of anemia and skin diseases. What are the symptoms of vitamin B12 deficiency and folic acid deficiency? — A deficiency of vitamin B12, folic acid, or both can cause low blood counts. Your doctor can check this by doing a blood test. A low blood count can affect 1 or more of the following: ●Red blood cells –Doctors use the term “anemia” when a person has too few red blood cells. Anemia can cause headaches or trouble breathing, especially with exercise. It can also make you feel tired or weak. ●White blood cells – A low white blood cell count can increase your risk of infection. ●Platelets – A low platelet count might increase your risk of bleeding. Vitamin B12 deficiency can also cause other symptoms related to the brain and nerves. They include: ●Tingling or numbness in the hands or feet ●Memory problems or trouble thinking clearly Should I see a doctor or nurse? — Yes. You should see your doctor or nurse if you have any of the symptoms listed above. Is there a test for vitamin B12 deficiency and folic acid deficiency? — Yes. Your doctor or nurse can do blood tests to check for deficiencies of these vitamins. How are vitamin B12 deficiency and folic acid deficiency treated? — Doctors treat vitamin B12 deficiency by giving people vitamin B12. It comes in the form of a shot or pills. You and your doctor will decide which form is better for your situation. Doctors treat folic acid deficiency by giving people folic acid. This treatment comes as a pill. Women who are pregnant or planning to get pregnant should also make sure they are getting enough folic acid. Your doctor or nurse will also treat the cause of your deficiency, if it can be treated. Can vitamin B12 deficiency and folic acid deficiency be prevented? — Yes. You can lower your chances of getting vitamin B12 deficiency and folic acid deficiency by eating foods that have the vitamins in them. Foods with vitamin B12 include any foods from animals, such as meat, fish, eggs, and milk. Foods with folic acid include fruits and green leafy vegetables. Many grains, like bread and cereals, have folic acid added to them. If you don’t eat these foods, your doctor or nurse might recommend that you take a vitamin supplement. Vitamin supplements are pills, capsules, or liquids that have nutrients in them. If you are a strict vegetarian or vegan and do not eat any meats, dairy foods, or eggs, you will not get enough vitamin B12 from your diet alone. To avoid getting vitamin B12 deficiency, you should take a daily vitamin B12 supplement. What if I want to get pregnant? — If you want to get pregnant, you should start taking a multivitamin at least 1 month before you start trying to get pregnant. Choose a multivitamin that has at least 400 micrograms of folic acid. Show your doctor or nurse the vitamin you plan to take to make sure the doses are right for you and your baby. Too much of some vitamins can be harmful.<\|endoftext\|>	3.75
9,305	a model species of roundworm The nematodes (UK: //, US: //) or roundworms constitute the phylum Nematoda (also called Nemathelminthes). They are a diverse animal phylum inhabiting a broad range of environments. Taxonomically, they are classified along with insects and other moulting animals in the clade Ecdysozoa, and unlike flatworms, have tubular digestive systems with openings at both ends. Nematode species can be difficult to distinguish from one another. Consequently, estimates of the number of nematode species described to date vary by author and may change rapidly over time. A 2013 survey of animal biodiversity published in the mega journal Zootaxa puts this figure at over 25,000. Estimates of the total number of extant species are subject to even greater variation. A widely referenced article published in 1993 estimated there may be over 1 million species of nematode, a claim which has since been repeated in numerous publications, without additional investigation, in an attempt to accentuate the importance and ubiquity of nematodes in the global ecosystem (rather than as a sign of agreement with the estimated taxonomic figure). Many other publications have since vigorously refuted this claim on the grounds that it is unsupported by fact, and is the result of speculation and sensationalism. More recent, fact-based estimates have placed the true figure closer to 40,000 species worldwide. Nematodes have successfully adapted to nearly every ecosystem: from marine (salt) to fresh water, soils, from the polar regions to the tropics, as well as the highest to the lowest of elevations. They are ubiquitous in freshwater, marine, and terrestrial environments, where they often outnumber other animals in both individual and species counts, and are found in locations as diverse as mountains, deserts, and oceanic trenches. They are found in every part of the earth's lithosphere, even at great depths, 0.9–3.6 km (3,000–12,000 ft) below the surface of the Earth in gold mines in South Africa. They represent 90% of all animals on the ocean floor. Their numerical dominance, often exceeding a million individuals per square meter and accounting for about 80% of all individual animals on earth, their diversity of lifecycles, and their presence at various trophic levels point to an important role in many ecosystems. They have been shown to play crucial roles in polar ecosystem. The roughly 2,271 genera are placed in 256 families. The many parasitic forms include pathogens in most plants and animals. A third of the genera occur as parasites of vertebrates; about 35 nematode species occur in humans. In short, if all the matter in the universe except the nematodes were swept away, our world would still be dimly recognizable, and if, as disembodied spirits, we could then investigate it, we should find its mountains, hills, vales, rivers, lakes, and oceans represented by a film of nematodes. The location of towns would be decipherable, since for every massing of human beings, there would be a corresponding massing of certain nematodes. Trees would still stand in ghostly rows representing our streets and highways. The location of the various plants and animals would still be decipherable, and, had we sufficient knowledge, in many cases even their species could be determined by an examination of their erstwhile nematode parasites. - 1 Etymology - 2 Taxonomy and systematics - 3 Anatomy - 4 Reproduction - 5 Free-living species - 6 Parasitic species - 7 Epidemiology - 8 Soil ecosystems - 9 Society and culture - 10 See also - 11 Notes - 12 References - 13 Further reading - 14 External links The word nematode comes from the Modern Latin compound of nemat- "thread" (from Greek nema, genitive nematos "thread," from stem of nein "to spin"; see needle) + -odes "like, of the nature of" (see -oid). Taxonomy and systematics The name of the group Nematoda, informally called "nematodes", came from Nematoidea, originally defined by Karl Rudolphi (1808), from Ancient Greek νῆμα (nêma, nêmatos, 'thread') and -eiδἠς (-eidēs, 'species'). It was treated as family Nematodes by Burmeister (1837). At its origin, the "Nematoidea" erroneously included Nematodes and Nematomorpha, attributed by von Siebold (1843). Along with Acanthocephala, Trematoda, and Cestoidea, it formed the obsolete group Entozoa, created by Rudolphi (1808). They were also classed along with Acanthocephala in the obsolete phylum Nemathelminthes by Gegenbaur (1859). In 1861, K. M. Diesing treated the group as order Nematoda. In 1877, the taxon Nematoidea, including the family Gordiidae (horsehair worms), was promoted to the rank of phylum by Ray Lankester. The first clear distinction between the nemas and gordiids was realized by Vejdovsky when he named a group to contain the horsehair worms the order Nematomorpha. In 1919, Nathan Cobb proposed that nematodes should be recognized alone as a phylum. He argued they should be called "nema" in English rather than "nematodes" and defined the taxon Nemates (later emended as Nemata, Latin plural of nema), listing Nematoidea sensu restricto as a synonym. However, in 1910, Grobben proposed the phylum Aschelminthes and the nematodes were included in as class Nematoda along with class Rotifera, class Gastrotricha, class Kinorhyncha, class Priapulida, and class Nematomorpha (The phylum was later revived and modified by Libbie Henrietta Hyman in 1951 as Pseudoceolomata, but remained similar). In 1932, Potts elevated the class Nematoda to the level of phylum, leaving the name the same. Despite Potts' classification being equivalent to Cobbs', both names have been used (and are still used today) and Nematode became a popular term in zoological science. Since Cobb was the first to include nematodes in a particular phylum separated from Nematomorpha, some researchers consider the valid taxon name to be Nemates or Nemata, rather than Nematoda, because of the zoological rule that gives priority to the first used term in case of synonyms. The phylogenetic relationships of the nematodes and their close relatives among the protostomian Metazoa are unresolved. Traditionally, they were held to be a lineage of their own, but in the 1990s, they were proposed to form the group Ecdysozoa together with moulting animals, such as arthropods. The identity of the closest living relatives of the Nematoda has always been considered to be well resolved. Morphological characters and molecular phylogenies agree with placement of the roundworms as a sister taxon to the parasitic Nematomorpha; together, they make up the Nematoida. Along with the Scalidophora (formerly Cephalorhyncha), the Nematoida form the clade Cycloneuralia, but much disagreement occurs both between and among the available morphological and molecular data. The Cycloneuralia or the Introverta—depending on the validity of the former—are often ranked as a superphylum. Due to the lack of knowledge regarding many nematodes, their systematics is contentious. An earliest and influential classification was proposed by Chitwood and Chitwood — later revised by Chitwood — who divided the phylum into two — the Aphasmidia and the Phasmidia. These were later renamed Adenophorea (gland bearers) and Secernentea (secretors), respectively. The Secernentea share several characteristics, including the presence of phasmids, a pair of sensory organs located in the lateral posterior region, and this was used as the basis for this division. This scheme was adhered to in many later classifications, though the Adenophorea were not in a uniform group. As it seems, the Secernentea are indeed a natural group of closest relatives, but the "Adenophorea" appear to be a paraphyletic assemblage of roundworms simply retaining a good number of ancestral traits. The old Enoplia do not seem to be monophyletic, either, but to contain two distinct lineages. The old group "Chromadoria" seems to be another paraphyletic assemblage, with the Monhysterida representing a very ancient minor group of nematodes. Among the Secernentea, the Diplogasteria may need to be united with the Rhabditia, while the Tylenchia might be paraphyletic with the Rhabditia. The understanding of roundworm systematics and phylogeny as of 2002 is summarised below: - Basal order Monhysterida - Class Dorylaimida - Class Enoplea - Class Secernentea - "Chromadorea" assemblage Later work has suggested the presence of 12 clades. The Secernentea—a group that includes virtually all major animal and plant 'nematode' parasites—apparently arose from within the Adenophorea. A major effort to improve the systematics of this phylum is in progress and being organised by the 959 Nematode Genomes. A complete checklist of the world's nematode species can be found in the World Species Index: Nematoda. An analysis of the mitochondrial DNA suggests that the following groupings are valid - subclass Dorylaimia - orders Rhabditida, Trichinellida and Mermithida - suborder Rhabditina - infraorders Spiruromorpha and Oxyuridomorpha The monophyly of the Ascaridomorph is uncertain. Nematodes are very small, slender worms: typically about 5 to 100 µm thick, and 0.1 to 2.5 mm long. The smallest nematodes are microscopic, while free-living species can reach as much as 5 cm (2 in), and some parasitic species are larger still, reaching over 1 m (3 ft) in length.:271 The body is often ornamented with ridges, rings, bristles, or other distinctive structures. The head of a nematode is relatively distinct. Whereas the rest of the body is bilaterally symmetrical, the head is radially symmetrical, with sensory bristles and, in many cases, solid 'head-shields' radiating outwards around the mouth. The mouth has either three or six lips, which often bear a series of teeth on their inner edges. An adhesive 'caudal gland' is often found at the tip of the tail. The epidermis is either a syncytium or a single layer of cells, and is covered by a thick collagenous cuticle. The cuticle is often of complex structure, and may have two or three distinct layers. Underneath the epidermis lies a layer of longitudinal muscle cells. The relatively rigid cuticle works with the muscles to create a hydroskeleton, as nematodes lack circumferential muscles. Projections run from the inner surface of muscle cells towards the nerve cords; this is a unique arrangement in the animal kingdom, in which nerve cells normally extend fibres into the muscles rather than vice versa. The oral cavity is lined with cuticle, which is often strengthened with structures, such as ridges, and especially in carnivorous species, may bear a number of teeth. The mouth often includes a sharp stylet, which the animal can thrust into its prey. In some species, the stylet is hollow, and can be used to suck liquids from plants or animals. The oral cavity opens into a muscular, sucking pharynx, also lined with cuticle. Digestive glands are found in this region of the gut, producing enzymes that start to break down the food. In stylet-bearing species, these may even be injected into the prey. No stomach is present, with the pharynx connecting directly to a muscleless intestine that forms the main length of the gut. This produces further enzymes, and also absorbs nutrients through its single-cell-thick lining. The last portion of the intestine is lined by cuticle, forming a rectum, which expels waste through the anus just below and in front of the tip of the tail. Movement of food through the digestive system is the result of body movements of the worm. The intestine has valves or sphincters at either end to help control the movement of food through the body. Nitrogenous waste is excreted in the form of ammonia through the body wall, and is not associated with any specific organs. However, the structures for excreting salt to maintain osmoregulation are typically more complex. In many marine nematodes, one or two unicellular 'renette glands' excrete salt through a pore on the underside of the animal, close to the pharynx. In most other nematodes, these specialised cells have been replaced by an organ consisting of two parallel ducts connected by a single transverse duct. This transverse duct opens into a common canal that runs to the excretory pore. Four peripheral nerves run the length of the body on the dorsal, ventral, and lateral surfaces. Each nerve lies within a cord of connective tissue lying beneath the cuticle and between the muscle cells. The ventral nerve is the largest, and has a double structure forward of the excretory pore. The dorsal nerve is responsible for motor control, while the lateral nerves are sensory, and the ventral combines both functions. At the anterior end of the animal, the nerves branch from a dense, circular nerve (nerve ring) round surrounding the pharynx, and serving as the brain. Smaller nerves run forward from the ring to supply the sensory organs of the head. The bodies of nematodes are covered in numerous sensory bristles and papillae that together provide a sense of touch. Behind the sensory bristles on the head lie two small pits, or 'amphids'. These are well supplied with nerve cells, and are probably chemoreception organs. A few aquatic nematodes possess what appear to be pigmented eye-spots, but whether or not these are actually sensory in nature is unclear. Most nematode species are dioecious, with separate male and female individuals, though some, such as Caenorhabditis elegans, are androdioecious, consisting of hermaphrodites and rare males. Both sexes possess one or two tubular gonads. In males, the sperm are produced at the end of the gonad and migrate along its length as they mature. The testis opens into a relatively wide seminal vesicle and then during intercourse into a glandular and muscular ejaculatory duct associated with the vas deferens and cloaca. In females, the ovaries each open into an oviduct (in hermaphrodites, the eggs enter a spermatheca first) and then a glandular uterus. The uteri both open into a common vulva/vagina, usually located in the middle of the morphologically ventral surface. Reproduction is usually sexual, though hermaphrodites are capable of self-fertilization. Males are usually smaller than females or hermaphrodites (often much smaller) and often have a characteristically bent or fan-shaped tail. During copulation, one or more chitinized spicules move out of the cloaca and are inserted into the genital pore of the female. Amoeboid sperm crawl along the spicule into the female worm. Nematode sperm is thought to be the only eukaryotic cell without the globular protein G-actin. Eggs may be embryonated or unembryonated when passed by the female, meaning their fertilized eggs may not yet be developed. A few species are known to be ovoviviparous. The eggs are protected by an outer shell, secreted by the uterus. In free-living roundworms, the eggs hatch into larvae, which appear essentially identical to the adults, except for an underdeveloped reproductive system; in parasitic roundworms, the lifecycle is often much more complicated. Nematodes as a whole possess a wide range of modes of reproduction. Some nematodes, such as Heterorhabditis spp., undergo a process called endotokia matricida: intrauterine birth causing maternal death. Some nematodes are hermaphroditic, and keep their self-fertilized eggs inside the uterus until they hatch. The juvenile nematodes then ingest the parent nematode. This process is significantly promoted in environments with a low food supply. The nematode model species C. elegans and C. briggsae exhibit androdioecy, which is very rare among animals. The single genus Meloidogyne (root-knot nematodes) exhibits a range of reproductive modes, including sexual reproduction, facultative sexuality (in which most, but not all, generations reproduce asexually), and both meiotic and mitotic parthenogenesis. The genus Mesorhabditis exhibits an unusual form of parthenogenesis, in which sperm-producing males copulate with females, but the sperm do not fuse with the ovum. Contact with the sperm is essential for the ovum to begin dividing, but because no fusion of the cells occurs, the male contributes no genetic material to the offspring, which are essentially clones of the female. Different free-living species feed on materials as varied as algae, fungi, small animals, fecal matter, dead organisms, and living tissues. Free-living marine nematodes are important and abundant members of the meiobenthos. They play an important role in the decomposition process, aid in recycling of nutrients in marine environments, and are sensitive to changes in the environment caused by pollution. One roundworm of note, C. elegans, lives in the soil and has found much use as a model organism. C. elegans has had its entire genome sequenced, the developmental fate of every cell determined, and every neuron mapped. Nematodes commonly parasitic on humans include ascarids (Ascaris), filarias, hookworms, pinworms (Enterobius), and whipworms (Trichuris trichiura). The species Trichinella spiralis, commonly known as the 'trichina worm', occurs in rats, pigs, bears, and humans, and is responsible for the disease trichinosis. Baylisascaris usually infests wild animals, but can be deadly to humans, as well. Dirofilaria immitis is known for causing heartworm disease by inhabiting the hearts, arteries, and lungs of dogs and some cats. Haemonchus contortus is one of the most abundant infectious agents in sheep around the world, causing great economic damage to sheep. In contrast, entomopathogenic nematodes parasitize insects and are mostly considered beneficial by humans, but some attack beneficial insects. One form of nematode is entirely dependent upon fig wasps, which are the sole source of fig fertilization. They prey upon the wasps, riding them from the ripe fig of the wasp's birth to the fig flower of its death, where they kill the wasp, and their offspring await the birth of the next generation of wasps as the fig ripens. A newly discovered parasitic tetradonematid nematode, Myrmeconema neotropicum, apparently induces fruit mimicry in the tropical ant Cephalotes atratus. Infected ants develop bright red gasters (abdomens), tend to be more sluggish, and walk with their gasters in a conspicuous elevated position. These changes likely cause frugivorous birds to confuse the infected ants for berries, and eat them. Parasite eggs passed in the bird's feces are subsequently collected by foraging C. atratus and are fed to their larvae, thus completing the lifecycle of M. neotropicum. Similarly, multiple varieties of nematodes have been found in the abdominal cavities of the primitively social sweat bee, Lasioglossum zephyrum. Inside the female body, the nematode hinders ovarian development and renders the bee less active, thus less effective in pollen collection. Plant-parasitic nematodes include several groups causing severe crop losses. The most common genera are Aphelenchoides (foliar nematodes), Ditylenchus, Globodera (potato cyst nematodes), Heterodera (soybean cyst nematodes), Longidorus, Meloidogyne (root-knot nematodes), Nacobbus, Pratylenchus (lesion nematodes), Trichodorus, and Xiphinema (dagger nematodes). Several phytoparasitic nematode species cause histological damages to roots, including the formation of visible galls (e.g. by root-knot nematodes), which are useful characters for their diagnostic in the field. Some nematode species transmit plant viruses through their feeding activity on roots. One of them is Xiphinema index, vector of grapevine fanleaf virus, an important disease of grapes, another one is Xiphinema diversicaudatum, vector of arabis mosaic virus. Other nematodes attack bark and forest trees. The most important representative of this group is Bursaphelenchus xylophilus, the pine wood nematode, present in Asia and America and recently discovered in Europe. Agriculture and horticulture Depending on the species, a nematode may be beneficial or detrimental to plant health. From agricultural and horticulture perspectives, the two categories of nematodes are the predatory ones, which kill garden pests such as cutworms and corn earworm moths, and the pest nematodes, such as the root-knot nematode, which attack plants, and those that act as vectors spreading plant viruses between crop plants. Predatory nematodes can be bred by soaking a specific recipe of leaves and other detritus in water, in a dark, cool place, and can even be purchased as an organic form of pest control. Rotations of plants with nematode-resistant species or varieties is one means of managing parasitic nematode infestations. For example, marigolds, grown over one or more seasons (the effect is cumulative), can be used to control nematodes. Another is treatment with natural antagonists such as the fungus Gliocladium roseum. Chitosan, a natural biocontrol, elicits plant defense responses to destroy parasitic cyst nematodes on roots of soybean, corn, sugar beet, potato, and tomato crops without harming beneficial nematodes in the soil. Soil steaming is an efficient method to kill nematodes before planting a crop, but indiscriminately eliminates both harmful and beneficial soil fauna. The golden nematode Globodera rostochiensis is a particularly harmful variety of nematode pest that has resulted in quarantines and crop failures worldwide. CSIRO has found a 13- to 14-fold reduction of nematode population densities in plots having Indian mustard Brassica juncea green manure or seed meal in the soil. About 90% of nematodes reside in the top 15 cm of soil. Nematodes do not decompose organic matter, but, instead, are parasitic and free-living organisms that feed on living material. Nematodes can effectively regulate bacterial population and community composition — they may eat up to 5,000 bacteria per minute. Also, nematodes can play an important role in the nitrogen cycle by way of nitrogen mineralization. Society and culture Nematode worms (C. elegans), part of an ongoing research project conducted on the Space Shuttle Columbia mission STS-107, survived the re-entry breakup. It is believed to be the first known life form to survive a virtually unprotected atmospheric descent to Earth's surface. - Biological pest control - List of organic gardening and farming topics - List of parasites of humans - Toxocariasis: A helminth infection of humans caused by the dog or cat roundworm, Toxocara canis or Toxocara cati - "Nematode Fossils — Nematoda". The Virtual Fossil Museum.[permanent dead link] - Classification of Animal Parasites - Garcia, Lynne (29 October 1999). "Classification of Human Parasites, Vectors, and Similar Organisms" (PDF). Los Angeles, California: Department of Pathology and Laboratory Medicine, UCLA Medical Center. Retrieved 21 July 2017. - Hodda, M (2011). "Phylum Nematoda Cobb, 1932. In: Zhang, Z.-Q. (Ed.) Animal biodiversity: An outline of higher-level classification and survey of taxonomic richness". Zootaxa. 3148: 63–95. - Zhang, Z (2013). "Animal biodiversity: An update of classification and diversity in 2013. In: Zhang, Z.-Q. (Ed.) Animal Biodiversity: An Outline of Higher-level Classification and Survey of Taxonomic Richness (Addenda 2013)". Zootaxa. 3703 (1): 5–11. doi:10.11646/zootaxa.3703.1.3. - "Recent developments in marine benthic biodiversity research". ResearchGate. Retrieved 5 November 2018. - Lambshead, PJD (1993). "Recent developments in marine benthic biodiversity research". Oceanis. 19 (6): 5–24. Anderson, Roy C. (8 February 2000). Nematode Parasites of Vertebrates: Their Development and Transmission. CABI. pp. 1–2. ISBN 9780851994215. Estimates of 500,000 to a million species have no basis in fact. - Borgonie G, García-Moyano A, Litthauer D, Bert W, Bester A, van Heerden E, Möller C, Erasmus M, Onstott TC (June 2011). "Nematoda from the terrestrial deep subsurface of South Africa". Nature. 474 (7349): 79–82. Bibcode:2011Natur.474...79B. doi:10.1038/nature09974. PMID 21637257. - Lemonick MD (8 June 2011). "Could 'worms from Hell' mean there's life in space?". Time. ISSN 0040-781X. Retrieved 8 June 2011. - Bhanoo SN (1 June 2011). "Nematode found in mine is first subsurface multicellular organism". The New York Times. ISSN 0362-4331. Retrieved 13 June 2011. - "Gold mine". Nature. 474 (7349): 6. June 2011. doi:10.1038/474006b. PMID 21637213. - Drake N (1 June 2011). "Subterranean worms from hell: Nature News". Nature News. doi:10.1038/news.2011.342. Retrieved 13 June 2011. - Borgonie G, García-Moyano A, Litthauer D, Bert W, Bester A, van Heerden E, Möller C, Erasmus M, Onstott TC (2 June 2011). "Nematoda from the terrestrial deep subsurface of South Africa". Nature. 474 (7349): 79–82. Bibcode:2011Natur.474...79B. doi:10.1038/nature09974. ISSN 0028-0836. PMID 21637257. - Danovaro R, Gambi C, Dell'Anno A, Corinaldesi C, Fraschetti S, Vanreusel A, Vincx M, Gooday AJ (January 2008). "Exponential decline of deep-sea ecosystem functioning linked to benthic biodiversity loss". Curr. Biol. 18 (1): 1–8. doi:10.1016/j.cub.2007.11.056. PMID 18164201. Lay summary – EurekAlert!. - Platt HM (1994). "foreword". In Lorenzen S, Lorenzen SA (eds.). The phylogenetic systematics of freeliving nematodes. London, UK: The Ray Society. ISBN 978-0-903874-22-9. - Cary, S. Craig; Green, T. G. Allan; Storey, Bryan C.; Sparrow, Ashley D.; Hogg, Ian D.; Katurji, Marwan; Zawar-Reza, Peyman; Jones, Irfon; Stichbury, Glen A. (2019-02-15). "Biotic interactions are an unexpected yet critical control on the complexity of an abiotically driven polar ecosystem". Communications Biology. 2 (1): 62. doi:10.1038/s42003-018-0274-5. ISSN 2399-3642. - Adams, Byron J.; Wall, Diana H.; Storey, Bryan C.; Green, T. G. Allan; Barrett, John E.; S. Craig Cary; Hopkins, David W.; Lee, Charles K.; Bottos, Eric M. (2019-02-15). "Nematodes in a polar desert reveal the relative role of biotic interactions in the coexistence of soil animals". Communications Biology. 2 (1): 63. doi:10.1038/s42003-018-0260-y. ISSN 2399-3642. - Roy C. Anderson (8 February 2000). Nematode Parasites of Vertebrates: Their development and transmission. CABI. p. 1. ISBN 978-0-85199-786-5. - Cobb, Nathan (1914). "Nematodes and their relationships". Yearbook. United States Department of Agriculture. pp. 472, 457–490. Archived from the original on 9 June 2016. Retrieved 25 September 2012. Quote on p. 472. - Chitwood BG (1957). "The English word "Nema" revised". Systematic Zoology in Nematology Newsletter. 4 (45): 1619. doi:10.2307/sysbio/6.4.184. - Siddiqi MR (2000). Tylenchida: parasites of plants and insects. Wallingford, Oxon, UK: CABI Pub. ISBN 978-0-85199-202-0. - Schmidt-Rhaesa A (2014). "Gastrotricha, Cycloneuralia and Gnathifera: General History and Phylogeny". In Schmidt-Rhaesa A (ed.). Handbook of Zoology (founded by W. Kükenthal). 1, Nematomorpha, Priapulida, Kinorhyncha, Loricifera. Berlin, Boston: de Gruyter. - Cobb NA (1919). "The orders and classes of nemas". Contrib. Sci. Nematol. 8: 213–216. - Wilson, E. O. "Phylum Nemata". Plant and insect parasitic nematodes. Retrieved 29 April 2018. - "ITIS report: Nematoda". Itis.gov. Retrieved 12 June 2012. - "Bilateria". Tree of Life Web Project. Tree of Life Web Project. 2002. Retrieved 2 November 2008. - Chitwood BG, Chitwood MB (1933). "The characters of a protonematode". J Parasitol. 20: 130. - Chitwood BG (1937). "A revised classification of the Nematoda". Papers on Helminthology published in commemoration of the 30 year Jubileum of ... K.J. Skrjabin ... Moscow: All-Union Lenin Academy of Agricultural Sciences. pp. 67–79. - Chitwood BG (1958). "The designation of official names for higher taxa of invertebrates". Bull Zool Nomencl. 15: 860–895. doi:10.5962/bhl.part.19410. - Coghlan, A. (7 Sep 2005). "Nematode genome evolution" (PDF). WormBook: 1–15. doi:10.1895/wormbook.1.15.1. PMC 4781476. PMID 18050393. Retrieved 13 January 2016. - Blaxter ML, De Ley P, Garey JR, Liu LX, Scheldeman P, Vierstraete A, Vanfleteren JR, Mackey LY, Dorris M, Frisse LM, Vida JT, Thomas WK (March 1998). "A molecular evolutionary framework for the phylum Nematoda". Nature. 392 (6671): 71–75. Bibcode:1998Natur.392...71B. doi:10.1038/32160. PMID 9510248. - "Nematoda". Tree of Life Web Project. Tree of Life Web Project. 2002. Retrieved 2 November 2008. - Holterman M, van der Wurff A, van den Elsen S, van Megen H, Bongers T, Holovachov O, Bakker J, Helder J (2006). "Phylum-wide analysis of SSU rDNA reveals deep phylogenetic relationships among nematodes and accelerated evolution toward crown Clades". Mol Biol Evol. 23 (9): 1792–1800. doi:10.1093/molbev/msl044. PMID 16790472. - "959 Nematode Genomes – NematodeGenomes". Nematodes.org. 11 November 2011. Retrieved 12 June 2012. - World Species Index: Nematoda. 2012. - Liu GH, Shao R, Li JY, Zhou DH, Li H, Zhu XQ (2013). "The complete mitochondrial genomes of three parasitic nematodes of birds: a unique gene order and insights into nematode phylogeny". BMC Genomics. 14 (1): 414. doi:10.1186/1471-2164-14-414. PMC 3693896. PMID 23800363. - Nyle C. Brady & Ray R. Weil (2009). Elements of the Nature and Properties of Soils (3rd ed.). Prentice Hall. ISBN 9780135014332. - Ruppert EE, Fox RS, Barnes RD (2004). Invertebrate Zoology: A Functional Evolutionary Approach (7th ed.). Belmont, California: Brooks/Cole. ISBN 978-0-03-025982-1. - Weischer B, Brown DJ (2000). An Introduction to Nematodes: General Nematology. Sofia, Bulgaria: Pensoft. pp. 75–76. ISBN 978-954-642-087-9. - Barnes RG (1980). Invertebrate zoology. Philadelphia: Sanders College. ISBN 978-0-03-056747-6. - "The sensory cilia of Caenorhabditis elegans". www.wormbook.org. - Kavlie, RG; Kernan, MJ; Eberl, DF (May 2010). "Hearing in Drosophila requires TilB, a conserved protein associated with ciliary motility". Genetics. 185 (1): 177–88. doi:10.1534/genetics.110.114009. PMC 2870953. PMID 20215474. - Lalošević, V.; Lalošević, D.; Capo, I.; Simin, V.; Galfi, A.; Traversa, D. (2013). "High infection rate of zoonotic Eucoleus aerophilus infection in foxes from Serbia". Parasite. 20: 3. doi:10.1051/parasite/2012003. PMC 3718516. PMID 23340229. - Bell G (1982). The masterpiece of nature: the evolution and genetics of sexuality. Berkeley: University of California Press. ISBN 978-0-520-04583-5. - Johnigk SA, Ehlers RU (1999). "Endotokia matricida in hermaphrodites of Heterorhabditis spp. and the effect of the food supply". Nematology. 1 (7–8): 717–726. doi:10.1163/156854199508748. ISSN 1388-5545. - Yanoviak SP, Kaspari M, Dudley R, Poinar G (April 2008). "Parasite-induced fruit mimicry in a tropical canopy ant". Am. Nat. 171 (4): 536–44. doi:10.1086/528968. PMID 18279076. - Batra, Suzanne W. T. (1965-10-01). "Organisms Associated with Lasioglossum zephyrum (Hymenoptera: Halictidae)". Journal of the Kansas Entomological Society. 38 (4): 367–389. JSTOR 25083474. - Purcell M, Johnson MW, Lebeck LM, Hara AH (1992). "Biological Control of Helicoverpa zea (Lepidoptera: Noctuidae) with Steinernema carpocapsae (Rhabditida: Steinernematidae) in Corn Used as a Trap Crop". Environmental Entomology. 21 (6): 1441–1447. doi:10.1093/ee/21.6.1441. - Riotte L (1975). Secrets of companion planting for successful gardening. p. 7. - US application 2008072494, Stoner RJ, Linden JC, "Micronutrient elicitor for treating nematodes in field crops", published 2008-03-27 - Loothfar R, Tony S (22 March 2005). "Suppression of root knot nematode (Meloidogyne javanica) after incorporation of Indian mustard cv. Nemfix as green manure and seed meal in vineyards". Australasian Plant Pathology. 34 (1): 77–83. doi:10.1071/AP04081. Retrieved 14 June 2010. - Pramer C (1964). "Nematode-trapping fungi". Science. 144 (3617): 382–388. Bibcode:1964Sci...144..382P. doi:10.1126/science.144.3617.382. PMID 14169325. - Hauser JT (December 1985). "Nematode-trapping fungi" (PDF). Carnivorous Plant Newsletter. 14 (1): 8–11. - Ahrén D, Ursing BM, Tunlid A (1998). "Phylogeny of nematode-trapping fungi based on 18S rDNA sequences". FEMS Microbiology Letters. 158 (2): 179–184. doi:10.1016/s0378-1097(97)00519-3. PMID 9465391. - "Columbia Survivors". Astrobiology Magazine. Jan 1, 2006. - Szewczyk, Nathaniel J.; Mancinelli, Rocco L.; McLamb, William; Reed, David; Blumberg, Baruch S.; Conley, Catharine A. (December 2005). "Caenorhabditis elegans Survives Atmospheric Breakup of STS–107, Space Shuttle Columbia". Astrobiology. 5 (6): 690–705. Bibcode:2005AsBio...5..690S. doi:10.1089/ast.2005.5.690. PMID 16379525. - Atkinson, H.J. (1973). "The respiratory physiology of the marine nematodes Enoplus brevis (Bastian) and E. communis (Bastian): I. The influence of oxygen tension and body size" (PDF). J. Exp. Biol. 59 (1): 255–266. - "Worms survived Columbia disaster". BBC News. 1 May 2003. Retrieved 4 Nov 2008. - Gubanov, N.M. (1951). "Giant nematoda from the placenta of Cetacea; Placentonema gigantissima nov. gen., nov. sp". Proc. USSR Acad. Sci. 77 (6): 1123–1125. [in Russian]. - Kaya, Harry K.; et al. (1993). "An Overview of Insect-Parasitic and Entomopathogenic Nematodes". In Bedding, R.A. (ed.). Nematodes and the Biological Control of Insect Pests. Csiro Publishing. ISBN 9780643105911. - "Giant kidney worm infection in mink and dogs". Merck Veterinary Manual (MVM). 2006. Archived from the original on 3 March 2016. Retrieved 10 February 2007. - White JG, Southgate E, Thomson JN, Brenner S (August 1976). "The structure of the ventral nerve cord of Caenorhabditis elegans". Philos. Trans. R. Soc. Lond. B Biol. Sci. 275 (938): 327–348. Bibcode:1976RSPTB.275..327W. doi:10.1098/rstb.1976.0086. PMID 8806. - Lee, Donald L, ed. (2010). The biology of nematodes. London: Taylor & Francis. ISBN 978-0415272117. Retrieved 16 December 2014. - De Ley, P & Blaxter, M (2004). "A new system for Nematoda: combining morphological characters with molecular trees, and translating clades into ranks and taxa". In R Cook & DJ Hunt (eds.). Nematology Monographs and Perspectives. 2. E.J. Brill, Leiden. pp. 633–653.CS1 maint: Uses authors parameter (link) CS1 maint: Uses editors parameter (link) \|Wikimedia Commons has media related to Nematoda.\| \|Wikisource has the text of the 1911 Encyclopædia Britannica article Nematoda.\| - Harper Adams University College Nematology Research - Nematodes/roundworms of man - European Society of Nematologists - Nematode.net: Repository of parasitic nematode sequences. - NeMys World free-living Marine Nematodes database - Nematode Virtual Library - International Federation of Nematology Societies - Society of Nematologists - Australasian Association of Nematologists - Research on nematodes and longevity - Nematode on BBC - Nematode worms in an aquarium - Phylum Nematoda – nematodes on the UF / *IFAS Featured Creatures Web site<\|endoftext\|>	4.1875
649	Gaussian Elimination Example By Rcantua considered the following equations: x - 3y + z = 4 2x - 8y + 8z = -2 -6x + 3y - 15z = 9 We can convert the equations into an augmented matrix: (1) \begin{align} \left(\begin{array}{cc} 1 & -3 & 1 & 4 \\ 2 & -8 & 8 & -2 \\-6 & 3 & -15 & 9\end{array}\right) \end{align} When we look at the start of the left column, we find 1 which is a non-zero number, which means that this works as a pivot. Next we try to get entry (2,1) to equal zero. To do this we can take one-third times the 3rd row and add it to the 2nd row. (1/3)R3+ R2 = R2. (*we could have also done -2R1+R2=R2) Now our matrix is: (2) \begin{align} \left(\begin{array}{cc} 1 & -3 & 1 & 4 \\ 0 & -7 & 3 & 1 \\-6 & 3 & -15 & 9\end{array}\right) \end{align} Now we can try to get a zero in entry (3,1). We can do (6)R1+R3 = R3 Now our matrix is: (3) \begin{align} \left(\begin{array}{cc} 1 & -3 & 1 & 4 \\ 0 & -7 & 3 & 1 \\0 & -15 & -9 & 33\end{array}\right) \end{align} Now we can try to get a zero in entry (3,2). We can do (-15/7)R2+R3 = R3 Now our matrix is: (4) \begin{align} \left(\begin{array}{cc} 1 & -3 & 1 & 4 \\ 0 & -7 & 3 & 1 \\0 & 0 & (-108/7) & (216/7)\end{array}\right) \end{align} From this we can figure out that z = -2 as the third row means 0x + 0y + (-108/7)z = 216/7. Now we can substitute Z back into the 2nd equation. 0x + -7y + 3(-2) = 1 y = -1 And finally we can substitute y and z into the 1st equation 1x + -3(-1) + 1(-2) = 4 x = 3 so we have x = 3 y = -1 z = -2 page revision: 3, last edited: 01 Feb 2012 18:50<\|endoftext\|>	4.9375
1,719	The Holocaust was a genocide in which the Nazi rule in Germany and its collaborators executed about six million Jews. Those who died comprised of 1.5 million children and 4 million women, moreover, the figure was about two-thirds of all the Jews who were residing in Europe at the time. The killings began in 1933 when the Nazis assumed leadership in Germany. The term holocaust is of Greek origin which means death by fire, the Greece was a society that was deemed religious, and as a result, the Nazi felt that by carrying out the Holocaust, they had some religious affiliation in killing the Jews. The Nazi believed that the Germans were racially superior. Therefore, the increasing population of Jews in Europe indicated a threat to their development as an elite and superior society. In response to this threat, the Nazi rule under the leadership of Adolf Hitler implemented a plan to execute at least two out of three Jews. This is because the Nazi regime felt that the Jews were racially inferior and were just serving as a threat to their development. Therefore, the Nazi authorities began executing the Jews who were initially living in Germany and then those that were residing in the then German occupied territories. Due to the tyranny that surrounded the leadership in Germany, their search for Jews for execution spread to Europe, everywhere in Europe the Nazi authorities were searching for Jews and killing them. However, this actions did not affect only the Jews, other communities such as the Soviet citizens, the Roma, and the Polish were affected too. The nature in which the killings were conducted has aroused many questions, for instance, was the plan to execute the Jews prepared earlier and was just waiting for execution or was the act instantaneous? Moreover, who bears the responsibility for the actions that occurred during the war? In addressing these particular concerns, this paper reviews the whole Holocaust experience, the effects of the killings, and most important, who bears the responsibility of holocaust. Events of Holocaust: A review The Germans were a super power nation and had occupied many regions in Europe, however, they were perturbed by the spread and increase in the population of the Jews. Therefore, they decided to reduce the number of Jews in Europe. The number of Jews in the year 1933 stood at around nine million in Europe only. The Germanys thought that a day could come when the Jews would decide to unite, in so doing they could achieve many feats, one of them being to subdue or outdoor the reigning superpowers which were the Germans. Therefore, when the Nazi ascended to power in 1933 with Hitler as its head, all the beurocratic and logistic factors aligned with Hitler in commissioning mass execution if the Jews. The Jews had spared all over Europe and had no defences. Moreover, few nations could match the strength and the mighty of the Germanys by then. Through a series of killings that followed, the Nazi authorities in assistance with the elite police and battalions squads of the elite Germany Force murdered about six million Jews while the number if those captured exceeded the dead ones. This called for setting up of various camps to accommodate the captured Jews; these were called concentration camps. The killing by the Nazi occurred in different stages. However, all the steps were leading to a final objective which was obtaining a final solution to the Jews question. This referred to the attainment of an agenda to exterminate all the Jews in Europe. Initially, the Nazi regime had enacted laws to exclude the Jews from civil society, especially the Nuremberg Laws of 1935. After the outbreak of world war two, the Nazis created a network of concentration camps in various places which were used to hold those who were considered to be their opponents, however, it was just a facade because these camps were filled with Jews who had been captured in the Nazi-controlled regions and were awaiting transport to ghettos, holding camps, and forced labour camps. In the year 1942, Germany became a worry of the high numbers of their prisoners, who comprised majorly of the Jews, the Polish, and the Soviet citizens. They feared that keeping them together could make them form an alliance with each other which could be detrimental. Moreover, it could show or exploit a weakness in their leadership. Therefore, to solve this, the Germans began moving the prisoners either by foot or on forced march to extermination camps. In the extermination camps, the prisoners were killed using a range of means such as starvation and use of special gassing facilities. The continued movement of prisoners across Europe coupled with the engagements in war with the aired forces were proving to be difficult for the Nazis, often the moving trains came in contact with sections of the allied forces who battled the Nazis and rescued the prisoners. In other places, what the German leadership had feared most began happening, rescuing of a vast number of prisoners by the allied forces. With the end of world war two nearing, Germans grasp on more prisoners began to weaken; they now resorted to defensive strategies as opposed to prisoner protection. However, they still continued with the execution of the detainees and the Jews in their extermination camps though on a dwindling scale. The end of the second world war marked the end of Holocaust, Germanys Nazi forces had been defeated and their collaborators had been defeated too. Holocaust brought many consequences both to the direct victims and those who were not affected directly. The effects were not just felt in Europe but in the whole world. Common effects to the people included psychological conditions that victims still have, as they age, visions of murder and inhuman deaths plunge into their minds. It shows how prejudiced a society can become towards one another. In societal and moral consideration, Holocaust lead to the eradication of society, the initial ethical and human thoughts of a model society were eliminated. The Jews as a community was scattered all over the planet, those who survived lived in fear whereas others still have health and mental condition. Indeed, Holocaust was a destruction of the entire human nature. Germany too felt this, since the end of the holocaust and world war, it has struggled to reshape its way as a nation so that it can for once be viewed as a palatable society. The effects on the larger scale involved formation and loss of various institutions. One development after the war was the formation if Israel as a state, this was as a result of the settlement of the Jews who lacked places to go and whose earlier society had been eradicated. Moreover, most Jews felt that they could not go back to the places they were living previously. Therefore, the settlement by the allied nations chose an area near Parkston where they settled the Jews, and the area came to be known as Israel. The aftermath of the war also led to a resurgence of antisemitism; this was witnessed mostly in Poland after the return of Jew remnants from the Soviet Union and Germany. This later resulted in moving of the Jews from Poland as they no longer felt safe there. Another significant effect of Holocaust was the decline of Yiddish language and culture. This is because the people who were previously practising the culture were dispersed and others were killed. Responsibility for Holocaust Holocaust was applied systematically within a short time and over a large area. This posed a question of whether the plan had just been hatched or the plan was in existence awaiting execution. Despite the presence of various suggestions, what is clear is that Adolf Hitler is the one who commissioned it and encouraged the German battalion soldiers to carry out the execution. However, the total responsibility of the war was not with Hitler; his entire capital was responsible too. However, the ordinary German citizens should not be blamed. This is because they were fed with wrong information and were meant to believe and obey the sentiments posted by Hitter. In conclusion, Holocaust represents one of the darkest points in history, not only to Germany but the whole world. It leaves us asking to what extent prejudice can go. However, what is important is that it provides us with a point to make, the choice of leaders should be carefully made to avoid having leaders who will not hesitate to sanction such moves again. Chang, Iris. The rape of Nanking: The forgotten holocaust of World War II. Basic Books, 2012. Horowitz, Sara R. Voicing the void: muteness and memory in Holocaust fiction. SUNY Press, 2012. Lipstadt, Deborah E. Denying the Holocaust: The growing assault on truth and memory. Simon and Schuster, 2012. Wardi, Dina. Memorial candles: Children of the Holocaust. Routledge, 2014.<\|endoftext\|>	3.6875
1,199	Understanding the weather on alien planets just got a little less complicated. For the first time, astronomers have detected helium in the atmosphere of an alien world, a new study reports. The discovery shows that it's possible to probe the air of at least some exoplanets without launching a new space telescope dedicated to this endeavor, the researchers said. "This is a new method to probe the upper parts of an exoplanet atmosphere, where high-energy radiation is observed," study lead author Jessica Spake, an exoplanet hunter at the University of Exeter in England, told Space.com via email. "Hopefully, we'll be able to study many more upper planetary atmospheres this way." [Gallery: The Strangest Alien Planets] Revealing the unseen To study the atmosphere of a faraway planet, scientists must wait for it to pass between its host star and Earth. Just before and after the bulk of the planet blocks some of the light streaming from the star, some of that light passes through the world's atmosphere. By studying how the light from the star changes as it passes through that alien air, scientists are able to characterize the atmosphere's composition. The process is difficult, and scientists have probed only a handful of worlds' atmospheres in this manner to date. So far, the study of exoplanet air has focused on hydrogen, the most abundant element in the universe and one of the main constituents of the solar system's gas giants. Early theoretical models predicted that helium should be among the most easily spotted elements in exoplanet atmospheres, but until now, it has remained unseen. Enter WASP-107b, a super-Neptune world that lies 200 light-years from Earth in the constellation of Virgo. Discovered in 2017, the gas giant is comparable in diameter to Jupiter but has just one-eighth of Jupiter's mass, making it one of the lowest-density worlds known. WASP-107b lies about eight times closer to its star than Mercury does to the sun, completing one orbit every 5.7 Earth days. At a sweltering 932 degrees Fahrenheit (500 degrees Celsius), this planet's atmosphere is one of the hottest among any known exoplanet. Spake and her colleagues turned NASA's Hubble Space Telescope toward the overheated world to study it in infrared wavelengths. By analyzing the light passing through WASP-107b's atmosphere, the researchers were able to identify helium in its excited state. This was possible thanks to a quirk of quantum mechanics that kept the element's electrons at a higher-energy state for more than 2 hours instead of a handful of nanoseconds. This focus on infrared light was key, the researchers said. Previous measurements of exoplanet atmospheres have relied on ultraviolet (UV) light, which Earth's atmosphere largely blocks. But infrared radiation passes through our air and clouds, meaning ground-based scopes can pick it up. Hubble isn't the only tool that astronomers can use for such infrared work. Spake said that NASA's $8.8 billion James Webb Space Telescope, which is scheduled to launch in May 2020, will likely be able to make similar observations, as will ground-based instruments like the European Southern Observatory's Very Large Telescopein Chile, the Telescopio Nazionale Galileoin the Canary Islands and the Keck telescopesin Hawaii. [Photos: Hubble's Successor, the James Webb Space Telescope] In contrast, only a single space-based telescope has been capable of probing exoplanet atmospheres in the ultraviolet. "UV observations are … technically difficult, and only Hubble can do them right now," Spake said. "That's the main reason only a handful of upper atmospheres have been studied so far." The research was published online today (May 2) in the journal Nature. WASP-107b is a great world to study with the new method, because a large quantity of the exoplanet's atmosphere is being lost to space. The researchers determined that ultraviolet radiation from WASP-107b's star is stripping away the planet's air, resulting in a long, comet-like tail on the world. "The helium we detected extends far out to space as a tenuous cloud surrounding the planet," study co-author Tom Evans, also of the University of Exeter, said in a statement. The abundance of helium doesn't mean that WASP-107b lacks hydrogen, team members added. "I think we'd expect hydrogen to be dominant in the atmosphere of WASP-107b," Spake said. "We just haven't directly detected it on this planet yet." The new method will also allow researchers to study planets farther away from our sun than could be probed in the ultraviolet, Spake said. Clouds of hydrogen in space can absorb UV light across longer distances, so UV studies generally must look in Earth's cosmic backyard. Infrared wavelengths, on the other hand, easily pass through hydrogen clouds, allowing for the examination of planets at greater distances. So far, the study of exoplanet atmospheres has focused on "hot Jupiters," the bloated gas giants that orbit their suns in mere days or even hours. Spake said that's because larger atmospheres are easier to study. WASP-107b's atmosphere has a height of about 620 miles (1,000 kilometers); for comparison, about 99 percent of Earth's air is found within 20 miles (32 km) of the ground. But a cloud of escaping helium could make it easier to study the atmospheres of smaller, rockier worlds, study team members said. "If smaller, Earth-sized planets have similar helium clouds, this new technique offers an exciting means to study their upper atmospheres in the very near future," Evans said.<\|endoftext\|>	3.6875
527	Chloride is one of the most common anions found in tap water. It generally combines with calcium, magnesium, or sodium to form various salts: for example, sodium chloride (NaCl) is formed when chloride and sodium combine. Chloride occurs naturally in groundwater but is found in greater concentrations where seawater and run-off from road salts (salts used to de-ice icy roads) can make their way into water sources. As such, well owners near snowy roads or road salting storage facilities are especially at risk for high levels of sodium chloride. Although chlorides are harmless at low levels, well water high in sodium chloride can damage plants if used for gardening or irrigation, and give drinking water an unpleasant taste. Over time, sodium chloride’s high corrosivity will also damage plumbing, appliances, and water heaters, causing toxic metals to leach into your water. Interestingly, there is no federally enforceable standard for chlorides in drinking water, though the EPA recommends levels no higher than 250 mg/L to avoid salty tastes and undesirable odors. At levels greater than this, sodium chloride can complicate existing heart problems and contribute to high blood pressure when ingested in excess. The good news is that chlorides can easily be removed from water with either a reverse osmosis system or a distiller. Reverse osmosis works by passing water through a semi-permeable membrane that separates pure water into one stream and salt water into another stream. In regular osmosis water flows from a lower concentration of salts to higher concentrations; in reverse osmosis the application of pressure greater than the osmotic pressure reverses the water flow from higher concentrations to much lower concentrations, producing pure water. With this method, about 50% of water can be recovered as pure water, while about 50% becomes salty wastewater. (If you’d like, you can read the reverse osmosis FAQ on our website for more info.) The prolonged boiling process kills virtually all types of microorganisms, including bacteria, viruses, and parasites. Microorganisms are not evaporated into the product water but remain in the boiling chamber as part of the residue. Distillers work very well, but use a lot of electricity. If you looking to remove chloride from your water, see our web store for a wide selection of both point-of-use and point-of-entry RO systems, as well as distillers. We love to hear from you and will get back to you usually within one business day.<\|endoftext\|>	3.6875
1,267	# Find the prime factors of $3^{32}-2^{32}$ I'm having a go at BMO 2006/7 Q1 which states: "Find four prime numbers less than 100 which are factors of $3^{32}-2^{32}$." My working is as follows (basically just follows difference of two squares loads of times): $$3^{32}-2^{32}$$ $$=(3^{16}+2^{16})(3^{16}-2^{16})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{8}-2^{8})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{4}-2^{4})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3^{2}-2^{2})$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)(3-2)$$ $$=(3^{16}+2^{16})(3^{8}+2^{8})(3^{4}+2^{4})(3^{2}+2^{2})(3+2)$$ Now it is simple to get $3$ of the primes here: $$3+2=5$$ $$3^2+2^2=13$$ $$3^4+2^4=97$$ Now I was having some trouble with finding the fourth prime. It's clear that the fourth prime must either be a prime factor of $3^8+2^8$ or $3^{16}+2^{16}$. I started off with $3^8+2^8$: $$3^8+2^8$$ $$=3^{4^2}+256$$ $$=81^2+256$$ $$=6561+256$$ $$=6817$$ Here is where I am stuck because google tells me that $6817=17401$ and so the fourth prime is $17$. But this is a non-calculator paper so is there a way of working out this answer without working out $\frac{6817}{3},\frac{6817}{7},\frac{6817}{11}$ etc until one of them is an integer solution? Finally, before anyone says this is a duplicate, there is a similar question here (prime factors of $3^{32}-2^{32}$) that I didn't know existed until writing this question. However, none of those answers give a way of finding $17$ without using any computational methods. So are they just expecting you to essentially do trial and error throughout all of the primes under $100$ until one works? • Well, you have only three primes to actually try ($3$ is obviously not a divisor). Of course, you don't know that beforehand. And you have only to try up to $79$, since $\sqrt{6817}<83$. – StayHomeSaveLives Jun 13 '18 at 16:52 • Well, you can notice that $68=4\cdot17$ which makes it easy. – saulspatz Jun 13 '18 at 16:52 • Yes I know I would reach the solution of $17$ quite quickly but that isn't really my point. I just wouldn't expect BMO to give something that essentially requires trial and error – Dan Jun 13 '18 at 16:53 • @saulspatz ah yes that's a very good way of thinking about it. Is that what they would expect from you though or is there some kind of method they would want you to use? – Dan Jun 13 '18 at 16:55 • Since $17$ is a prime, Fermat's little theorem tell us $2,3 \not\| 17 \implies 2^{16} \equiv 3^{16} \equiv 0 \pmod 17$. One thing one can try is looking for prime of the form $2^k + 1$ for $k \le 5$, you immediate get $5$ and $17$. – achille hui Jun 13 '18 at 17:02 Apply Fermat's little theorem: $17$ is prime, so $$3^{16}\equiv1\pmod{17}$$ $$2^{16}\equiv1\pmod{17}$$ Note that $\forall a \perp p, a^{p-1\over2} \equiv \pm 1 \pmod p$ 2 and 3 are already coprime to the other primes you're looking for. $82+1=17$ is a strong candidate. In fact, it is the only strong candidate for this method of attack. I checked this in Python and there are only 4 such primes, $5$, $13$, $17$ and $97$. As has been noted, FLT implies $17$ is such a prime, and we can get $5$ the same way. We also get $13$ as a prime factor of $3^4-2^4$, and $97$ as a prime factor of $3^8-2^8$. In the timed conditions of an olympiad, it helps to first seek prime factors of $3^2-2^2$, then the remaining prime factors of $3^4-2^4$ (i.e. those of $3^2+2^2$) etc. In particular $5=3^2-2^2,\,13=3^2+2^2,\,97=3^4+2^4,\,17\times 401=3^8+2^8$.<\|endoftext\|>	4.4375
887	What is El Niño? The El Niño Southern Oscillation (ENSO) is a large-scale climatic phenomenon that originates in the tropical Pacific but affects global climate patterns. The warm phase is known as El Niño and the cold phase is La Niña. El Niño occurs irregularly every two to seven years and peaks around in winter. What causes an El Niño event? In a normal year, when ENSO is inactive, the equatorial Pacific trade winds blow from east to west. The winds push the warmer water towards the west, and colder water rises up from deeper in the ocean to replace it. This creates an east-west difference in sea surface temperature and hence an east-west difference in sea level pressure that maintains the trade winds and so drives a positive feedback loop. During an El Niño year, the east-west SST difference weakens, the pressure difference weakens and the trade winds and their effects on the ocean weaken, so the eastern Pacific warms further. During a La Nina year, the opposite happens, the east west difference in temperature strengthens, the pressure difference strengthens and the trade winds and their effects on the ocean strengthen, so the east Pacific cools further. Such changes in sea surface temperatures affect the atmosphere over vast areas, with both local and global repercussions. Locally, the associated atmospheric circulation changes drive increased atmospheric convection and precipitation over the central and eastern Pacific while rainfall is reduced over the western Pacific. These changes have remote impacts throughout the tropics but also at higher latitudes via the atmosphere, especially when ENSO is at its strongest in winter. How do we measure El Niño? A network of ocean buoys measure water temperature, currents and the wind. Satellites also provide surface temperature and current data, providing the ability to monitor ENSO in real-time. We are also able to predict El Niño months in advance as we can measure the heat content of the upper-ocean, one of the precursors for the onset of El Niño and use our climate models to predict its evolution. Although various criteria exist, we normally say an El Niño event is underway when sea surface temperatures in the equatorial Pacific (officially called Niño region 3.4), rise 0.5°C above the historical average for at least three months in a row. Impacts of El Niño and La Niña The strongest impacts are experienced by those countries near the tropical Pacific origin of ENSO. Changes in surface temperatures, winds and moisture affect rainfall intensity and patterns which can lead to extreme events such as flooding and drought. During an El Niño event, Peru, Ecuador and south-eastern parts of South America receive heavy rainfall. In northern Brazil, drier conditions or even drought results. Indonesia, South Asia and parts of Australia are also more likely to experience drought during El Niño. The change in weather patterns associated with El Niño can greatly impact the economy, in particular agriculture, water resources, fisheries and public health. ENSO also has a strong influence on the occurrence and intensity of tropical cyclones and Atlantic hurricane activity weakens during El Niño but strengthens during La Niña. In the UK, we experience the socio-economic impacts of an El Niño event, partly through increased food prices and there are effects on the jet stream and European weather, especially in late winter when El Nino increases the chances of cold snaps and La Nina increases the chances of wet and stormy conditions. How does El Niño affect global temperature? El Niño releases heat to the atmosphere and increases subsequent global temperatures. For example, global average temperatures for 2016 were around 1.1°C above preindustrial values and the strong El Niño episode of 2015/2016 partly contributed. However, researchers have concluded that the warming from El Niño is only accountable for about 0.2°C of this overall figure. El Niño and climate change Because of the large event to event variations of El Niño, we don’t have enough past years of observations to show a clear impact of climate change on its properties. There is now some evidence that the effects of El Niño on rainfall may increase in the future and that we may even see more very strong ENSO events, but these remain active research questions. Listen to the Society's podcast about El Niño and La Niña with Professor Adam Scaife.<\|endoftext\|>	4.375
601	## What is proportion in confidence interval? More specifically, the confidence interval is calculated as the sample proportion ± z* times the standard deviation of the sample proportion, where z* is the critical value of z that has (1-C)/2 of the normal distribution to the right of the value, and the standard deviation is . ### How do you find the 95% CI of a proportion? Suppose you take a random sample of 100 different trips through this intersection and you find that a red light was hit 53 times. Because you want a 95 percent confidence interval, your z-value is 1.96. The red light was hit 53 out of 100 times. So ρ = 53/100 = 0.53. What is the formula for confidence interval for proportions? The result is the following formula for a confidence interval for a population proportion: p̂ +/- z (p̂(1 – p̂)/n)0.5. Here the value of z* is determined by our level of confidence C. For the standard normal distribution, exactly C percent of the standard normal distribution is between -z* and z*. What is the 95% conservative confidence interval for the population proportion? The 95% confidence interval for the true binomial population proportion is ( p′ – EBP, p′ + EBP) = (0.810, 0.874). ## What does 95 confidence interval represent? The 95% confidence interval defines a range of values that you can be 95% certain contains the population mean. With large samples, you know that mean with much more precision than you do with a small sample, so the confidence interval is quite narrow when computed from a large sample. ### How do you find P value from confidence interval? (a) CI for a difference 1. 1 calculate the test statistic for a normal distribution test, z, from P3: z = −0.862 + √[0.743 − 2.404×log(P)] 2. 2 calculate the standard error: SE = Est/z (ignoring minus signs) 3. 3 calculate the 95% CI: Est –1.96×SE to Est + 1.96×SE. When solving the sample size is required a 95% confidence interval for a population proportion P having a given error bound E we choose a value? When solving for the sample size needed to compute a 95 percent confidence interval for a population proportion p, having a given error bound E, we choose a value of p^ that makes pˆ (1−pˆ)p^⁢ (1−p^) as large as reasonably possible. Which z score is used in a 90% confidence interval for a population proportion? Confidence Intervals Desired Confidence Interval Z Score 90% 95% 99% 1.645 1.96 2.576<\|endoftext\|>	4.59375
485	# Calculator search results Formula Expand the expression $2x-3+6x-x+5$ $7 x + 2$ Organize polynomials $\color{#FF6800}{ 2 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 5 }$ Organize the similar terms $\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ x } \color{#FF6800}{ + } \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right )$ $\left ( \color{#FF6800}{ 2 } \color{#FF6800}{ + } \color{#FF6800}{ 6 } \color{#FF6800}{ - } \color{#FF6800}{ 1 } \right ) \color{#FF6800}{ x } + \left ( - 3 + 5 \right )$ Arrange the constant term $\color{#FF6800}{ 7 } \color{#FF6800}{ x } + \left ( - 3 + 5 \right )$ $7 x + \left ( \color{#FF6800}{ - } \color{#FF6800}{ 3 } \color{#FF6800}{ + } \color{#FF6800}{ 5 } \right )$ Arrange the constant term $7 x + \color{#FF6800}{ 2 }$ Solution search results<\|endoftext\|>	4.40625
1,209	New research suggests that the geological staying power of continents comes partly from their losing battle with the Earth's oceans over magnesium. The research finds continents lose more than 20 percent of their initial mass via chemical reactions involving the Earth's crust, water and atmosphere. Because much of the lost mass is dominated by magnesium and calcium, continents ultimately gain because the lighter, silicon-rich rock that's left behind is buoyed up by denser rock beneath the Earth's crust. The Earth's continents seem like fixtures, having changed little throughout recorded human history. But geologists know that continents have come and gone during the Earth's 4.5 billion years. However, there are more theories than hard data about some of the key processes that govern continents' lives. "Continents are built by new rock that wells up from volcanoes in island arcs like Japan," said lead author Cin-Ty Lee, assistant professor of Earth science at Rice University. "In addition to chemical weathering at the Earth’s surface, we know that some magnesium is also lost due to destabilizing convective forces beneath these arcs." Lee's research, which appeared in the March 24 issue of the Proceedings of the National Academy of Science, marks the first attempt to precisely nail down how much magnesium is lost through two markedly different routes -- destabilizing convective forces deep inside the Earth and chemical weathering reactions on its surface. Lee said the project might not have happened at all if it weren't for some laboratory serendipity. "I'd acquired some tourmaline samples in San Diego with my childhood mentor, Doug Morton," Lee said. "We were adding to our rock collections, like kids, but when I got back to the lab, I was curious where the lithium, a major element in tourmaline, needed to make the tourmalines came from. I decided to measure the lithium content in the granitic rocks from the same area, and that's where this started." In examining the lithium content in a variety of rocks, Lee realized that lithium tended to behave like the magnesium that was missing from continents. In fact, the correlation was so close, he realized that lithium could be used as a proxy to find out how much magnesium continents had lost due to chemical weathering. Continents ride higher than oceans, partly because the Earth's crust is thicker beneath continents than it is beneath the oceans. In addition, the rock beneath continents is made primarily of silicon-rich minerals like granite and quartz, which are less dense than the magnesium-rich basalt beneath the oceans. Lee said he always assumed that processes deep in the Earth, beneath the volcanoes that feed continents, accounted for far more magnesium loss than weathering. In particular, a process called "delamination" occurs in subduction zones, places where one piece of the Earth's crust slides beneath another and gets recycled into the Earth's magma. As magma wells up beneath continent-feeding volcanoes, it often leaves behind a dense, magnesium-rich layer that ultimately founders back into the Earth's interior. In previous research, Lee found that about 40 percent of the magnesium in basaltic magma was lost to delamination. He said he was thus surprised to find that chemical weathering alone accounted for another 20 percent. "Weathering occurs in just the top few meters or so of the Earth's crust, and it's driven by the hydrosphere, the water that moves between the air, land and oceans," Lee said. "It appears that our planet has continents because we have an active hydrosphere, so if we want to find a hydrosphere on distant planets, perhaps we should look for continents." Jade Boyd \| EurekAlert! Cause for variability in Arctic sea ice clarified 14.05.2019 \| Max-Planck-Institut für Meteorologie Arctic rivers provide fingerprint of carbon release from thawing permafrost 08.05.2019 \| Stockholm University Engineers at the University of Tokyo continually pioneer new ways to improve battery technology. Professor Atsuo Yamada and his team recently developed a... With a quantum coprocessor in the cloud, physicists from Innsbruck, Austria, open the door to the simulation of previously unsolvable problems in chemistry, materials research or high-energy physics. The research groups led by Rainer Blatt and Peter Zoller report in the journal Nature how they simulated particle physics phenomena on 20 quantum bits and how the quantum simulator self-verified the result for the first time. Many scientists are currently working on investigating how quantum advantage can be exploited on hardware already available today. Three years ago, physicists... 'Quantum technologies' utilise the unique phenomena of quantum superposition and entanglement to encode and process information, with potentially profound benefits to a wide range of information technologies from communications to sensing and computing. However a major challenge in developing these technologies is that the quantum phenomena are very fragile, and only a handful of physical systems have been... Working group led by physicist Professor Ulrich Nowak at the University of Konstanz, in collaboration with a team of physicists from Johannes Gutenberg University Mainz, demonstrates how skyrmions can be used for the computer concepts of the future When it comes to performing a calculation destined to arrive at an exact result, humans are hopelessly inferior to the computer. In other areas, humans are... Scientists develop a molecular recording tool that enables in vivo lineage tracing of embryonic cells The beginning of new life starts with a fascinating process: A single cell gives rise to progenitor cells that eventually differentiate into the three germ... 29.04.2019 \| Event News 17.04.2019 \| Event News 15.04.2019 \| Event News 17.05.2019 \| Materials Sciences 17.05.2019 \| Physics and Astronomy 17.05.2019 \| Materials Sciences<\|endoftext\|>	4.03125
737	Dysphagia is a medical term used to cover several conditions involving chewing and swallowing disorders. People who are suffering from swallowing difficulties can get effective medical help. This is a condition that can happen at any age but happens more as people age. It is important to go to the family physician for diagnosis and treatment. What Is Dysphagia and What Causes It? Dysphagia is actually a collection of symptoms and conditions grouped under one name by the medical profession because they are closely related. There are three types of dysphagia. The first is oral dysphasia in the mouth, the second is oropharyngeal dysphagia involving the throat, and the third type is esophageal dysphagia that involves the esophagus. The causes and treatments for these conditions depend on which areas are affected and what caused the problem. Some common causes are aging, stokes, Parkinson’s disease, other neurological problems, cancer, multiple sclerosis, radiation, cleft palate or lip, and other conditions. Symptoms That signal Trouble and Call For A Visit To The Doctor The symptoms that tell a person they should visit their doctor include trouble swallowing, sudden weight loss, choking, coughing, and vomiting when trying to swallow food, painful swallowing or being unable to swallow, drooling, stomach acid backing up into the throat, and having frequent heartburn. When it feels like food is getting stuck in the throat or chest, there may be a problem. When a patient visits their family physician, the doctor will perform certain tests to determine which type of dysphagia is causing problems and, if possible, determine what the underlying cause is. Then a treatment plan can be designed. There Are Four Types of Treatments The promising treatment options for dysphagia fall into four categories: - Lifestyle and eating habit changes. A speech-language pathologist can help patients with exercises and therapies to improve muscle and nerve function. There may be strategies to position the head while swallowing. Modifying the feeding environment to eliminate distractions may help. There are also modified utensils plates, cups, and straws to help. - Changes to the diet. There are foods that trigger or worsen dysphagia and foods and drinks that are easier to eat with this condition. The diet should be altered to include a well-balanced collection of more easily swallowed foods. Foods and drinks can have their texture modified with thickening liquids such as Simply Thick. This allows liquids and other foods to have an easier texture to swallow. Foods can also be pureed in a blender to make them easier to swallow. - There are medications that provide relief from symptoms including Parkinson’s disease medication, Botox, GERD medications, H2 blockers, and proton pump inhibitors. - Some causes and types of dysphagia require surgery. Surgical procedures can include the insertion of feeding tubes through the nasal passage or directly into the stomach or inserting a plastic or metal tubes or stents to hold the esophagus open. Surgery may involve partial removal of the esophagus to remove damaged parts or cancerous growths. The remaining esophagus is then attached to the stomach. Esophageal dilation is a surgery that involves inserting a small balloon attached to an endoscope into the esophagus to stretch it. The balloon is removed. Good medical treatment allows people who suffer from dysphagia to live better lives. Their symptoms are reduced, eliminated, or effectively controlled. People who receive treatment avoid complications such as malnutrition, dehydration, excessive weight loss, choking, and aspiration pneumonia.<\|endoftext\|>	3.6875
4,459	$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 10.1: Add and Subtract Polynomials [ "article:topic", "authorname:openstax", "showtoc:no" ] $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ Skills to Develop • Identify polynomials, monomials, binomials, and trinomials • Determine the degree of polynomials • Evaluate a polynomial for a given value be prepared! Before you get started, take this readiness quiz. 1. Simplify: 8x + 3x. If you missed this problem, review Example 2.22. 2. Subtract: (5n + 8) − (2n − 1). If you missed this problem, review Example 7.29. 3. Evaluate: 4y2 when y = 5 If you missed this problem, review Example 2.18. ### Identify Polynomials, Monomials, Binomials, and Trinomials In Evaluate, Simplify, and Translate Expressions, you learned that a term is a constant or the product of a constant and one or more variables. When it is of the form axm, where a is a constant and m is a whole number, it is called a monomial. A monomial, or a sum and/or difference of monomials, is called a polynomial. Definition: Polynomials polynomial — A monomial, or two or more monomials, combined by addition or subtraction monomial — A polynomial with exactly one term binomial — A polynomial with exactly two terms trinomial — A polynomial with exactly three terms Notice the roots: • poly- means many • mono- means one • bi- means two • tri- means three Here are some examples of polynomials: Polynomial Monomial Binomial b + 1 4y2 − 7y + 2 5x5 − 4x4 + x3 + 8x2 − 9x + 1 5 4b2 -9x3 3a - 7 y2 - 9 17x3 + 14x2 x2 - 5x + 6 4y2 - 7y + 2 5a4 - 3a3 + a Notice that every monomial, binomial, and trinomial is also a polynomial. They are special members of the family of polynomials and so they have special names. We use the words ‘monomial’, ‘binomial’, and ‘trinomial’ when referring to these special polynomials and just call all the rest ‘polynomials’. Example 10.1: Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial: (a) 8x2 − 7x − 9 (b) −5a4 (c) x4 − 7x3 − 6x2 + 5x + 2 (d) 11 − 4y3 (e) n ##### Solution Polynomial Number of terms Type (a) 8x2 − 7x − 9 3 Trinomial (b) −5a4 1 Monomial (c) x4 − 7x3 − 6x2 + 5x + 2 5 Polynomial (d) 11 − 4y3 2 Binomial (e) n 1 Monomial Exercise 10.1: Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. (a) z (b) 2x3 − 4x2 − x − 8 (c) 6x2 − 4x + 1 (d) 9 − 4y2 (e) 3x7 Exercise 10.2: Determine whether each polynomial is a monomial, binomial, trinomial, or other polynomial. (a) y3 − 8 (b) 9x3 − 5x2 − x (c) x4 − 3x2 − 4x − 7 (d) −y4 (e) w ### Determine the Degree of Polynomials In this section, we will work with polynomials that have only one variable in each term. The degree of a polynomial and the degree of its terms are determined by the exponents of the variable. A monomial that has no variable, just a constant, is a special case. The degree of a constant is 0 —it has no variable. Definition: Degree of a Polynomial The degree of a term is the exponent of its variable. The degree of a constant is 0. The degree of a polynomial is the highest degree of all its terms. Let's see how this works by looking at several polynomials. We'll take it step by step, starting with monomials, and then progressing to polynomials with more terms. Remember: Any base written without an exponent has an implied exponent of 1. Example 10.2: Find the degree of the following polynomials: (a) 4x (b) 3x3 − 5x + 7 (c) −11 (d) −6x2 + 9x − 3 (e) 8x + 2 ##### Solution (a) 4x The exponent of x is one. x = x1 The degree is 1. (b) 3x3 − 5x + 7 The highest degree of all the terms is 3. The degree is 3. (c) −11 The degree of a constant is 0. The degree is 0. (d) −6x2 + 9x − 3 The highest degree of all the terms is 2. The degree is 2. (e) 8x + 2 The highest degree of all the terms is 1. The degree is 1. Exercise 10.3: Find the degree of the following polynomials: (a) −6y (b) 4x − 1 (c) 3x4 + 4x2 − 8 (d) 2y2 + 3y + 9 (e) −18 Exercise 10.4: Find the degree of the following polynomials: (a) 47 (b) 2x2 − 8x + 2 (c) x4 − 16 (d) y5 − 5y3 + y (e) 9a3 Working with polynomials is easier when you list the terms in descending order of degrees. When a polynomial is written this way, it is said to be in standard form. Look back at the polynomials in Example 10.2. Notice that they are all written in standard form. Get in the habit of writing the term with the highest degree first. In The Language of Algebra, you simplified expressions by combining like terms. Adding and subtracting monomials is the same as combining like terms. Like terms must have the same variable with the same exponent. Recall that when combining like terms only the coefficients are combined, never the exponents. Example 10.3: ##### Solution Combine like terms. 23x2 Exercise 10.5: Exercise 10.6: Example 10.4: Subtract: 11n − (−8n). ##### Solution Combine like terms. 19n Exercise 10.7: Subtract: 9n − (−5n). Exercise 10.8: Subtract: −7a3 − (−5a3). Example 10.5: Simplify: a2 + 4b2 − 7a2. ##### Solution Combine like terms. −6a2 + 4b2 Remember, −6a2 and 4b2 are not like terms. The variables are not the same. Exercise 10.9: Add: 3x2 + 3y2 − 5x2. Exercise 10.10: Add: 2a2 + b2 − 4a2. Adding and subtracting polynomials can be thought of as just adding and subtracting like terms. Look for like terms—those with the same variables with the same exponent. The Commutative Property allows us to rearrange the terms to put like terms together. It may also be helpful to underline, circle, or box like terms. Example 10.6: Find the sum: (4x2 − 5x + 1) + (3x2 − 8x − 9) ##### Solution Identify like terms. Rearrange to get the like terms together. Combine like terms. Exercise 10.11: Find the sum: (3x2 − 2x + 8) + (x2 − 6x + 2). Exercise 10.12: Find the sum: (7y2 + 4y − 6) + (4y2 + 5y + 1) Parentheses are grouping symbols. When we add polynomials as we did in Example 10.6, we can rewrite the expression without parentheses and then combine like terms. But when we subtract polynomials, we must be very careful with the signs. Example 10.7: Find the difference: (7u2 − 5u + 3) − (4u2 − 2). ##### Solution Distribute and identify like terms. Rearrange the terms. Combine like terms. Exercise 10.13: Find the difference: (6y2 + 3y − 1) − (3y2 − 4). Exercise 10.14: Find the difference: (8u2 − 7u − 2) − (5u2 − 6u − 4). Example 10.8: Subtract: (m2 − 3m + 8) from (9m2 − 7m + 4). ##### Solution Distribute and identify like terms. Rearrange the terms. Combine like terms. Exercise 10.15: Subtract: (4n2 − 7n − 3) from (8n2 + 5n − 3). Exercise 10.16: Subtract: (a2 − 4a − 9) from (6a2 + 4a − 1). ### Evaluate a Polynomial for a Given Value In The Language of Algebra we evaluated expressions. Since polynomials are expressions, we'll follow the same procedures to evaluate polynomials—substitute the given value for the variable into the polynomial, and then simplify. Example 10.9: Evaluate 3x 2 − 9x + 7 when (a) x = 3 (b) x = −1 ##### Solution (a) x = 3 Substitute 3 for x. 3(3)2 − 9(3) + 7 Simplify the expression with the exponent. 3 • 9 − 9(3) + 7 Multiply. 27 − 27 + 7 Simplify. 7 (b) x = −1 Substitute -1 for x. 3(-1)2 − 9(-1) + 7 Simplify the expression with the exponent. 3 • 1 − 9(-1) + 7 Multiply. 3 + 9 + 7 Simplify. 19 Exercise 10.17: Evaluate: 2x2 + 4x − 3 when (a) x = 2 (b) x = −3 Exercise 10.18: Evaluate: 7y2 − y − 2 when (a) y = −4 (b) y = 0 Example 10.10: The polynomial −16t 2 + 300 gives the height of an object t seconds after it is dropped from a 300 foot tall bridge. Find the height after t = 3 seconds. ##### Solution Substitute 3 for t. -16(3)2 + 300 Simplify the expression with the exponent. -16 • 9 + 300 Multiply. -144 + 300 Simplify. 156 Exercise 10.19: The polynomial −8t2 + 24t + 4 gives the height, in feet, of a ball t seconds after it is tossed into the air, from an initial height of 4 feet. Find the height after t = 3 seconds. Exercise 10.20: The polynomial −8t2 + 24t + 4 gives the height, in feet, of a ball x seconds after it is tossed into the air, from an initial height of 4 feet. Find the height after t = 2 seconds. Subtracting Polynomials ### Practice Makes Perfect #### Identify Polynomials, Monomials, Binomials and Trinomials In the following exercises, determine if each of the polynomials is a monomial, binomial, trinomial, or other polynomial. 1. 5x + 2 2. z2 − 5z − 6 3. a2 + 9a + 18 4. −12p4 5. y3 − 8y2 + 2y − 16 6. 10 − 9x 7. 23y2 8. m4 + 4m3 + 6m2 + 4m + 1 #### Determine the Degree of Polynomials In the following exercises, determine the degree of each polynomial. 1. 8a5 − 2a3 + 1 2. 5c3 + 11c2 − c − 8 3. 3x − 12 4. 4y + 17 5. −13 6. −22 In the following exercises, add or subtract the monomials. 1. 6x2 + 9x2 2. 4y3 + 6y3 3. −12u + 4u 4. −3m + 9m 5. 5a + 7b 6. 8y + 6z 7. Add: 4a, − 3b, − 8a 8. Add: 4x, 3y, − 3x 9. 18x − 2x 10. 13a − 3a 11. Subtract 5x6 from − 12x6 12. Subtract 2p4 from − 7p4 In the following exercises, add or subtract the polynomials. 1. (4y2 + 10y + 3) + (8y2 − 6y + 5) 2. (7x2 − 9x + 2) + (6x2 − 4x + 3) 3. (x2 + 6x + 8) + (−4x2 + 11x − 9) 4. (y2 + 9y + 4) + (−2y2 − 5y − 1) 5. (3a2 + 7) + (a2 − 7a − 18) 6. (p2 − 5p − 11) + (3p2 + 9) 7. (6m2 − 9m − 3) − (2m2 + m − 5) 8. (3n2 − 4n + 1) − (4n2 − n − 2) 9. (z2 + 8z + 9) − (z2 − 3z + 1) 10. (z2 − 7z + 5) − (z2 − 8z + 6) 11. (12s2 − 15s) − (s − 9) 12. (10r2 − 20r) − (r − 8) 13. Find the sum of (2p3 − 8) and (p2 + 9p + 18) 14. Find the sum of (q2 + 4q + 13) and (7q3 − 3) 15. Subtract (7x2 − 4x + 2) from (8x2 − x + 6) 16. Subtract (5x2 − x + 12) from (9x2 − 6x − 20) 17. Find the difference of (w2 + w − 42) and (w2 − 10w + 24) 18. Find the difference of (z2 − 3z − 18) and (z2 + 5z − 20) #### Evaluate a Polynomial for a Given Value In the following exercises, evaluate each polynomial for the given value. 1. Evaluate 8y2 − 3y + 2 1. y = 5 2. y = −2 3. y = 0 2. Evaluate 5y2 − y − 7 when: 1. y = −4 2. y = 1 3. y = 0 3. Evaluate 4 − 36x when: 1. x = 3 2. x = 0 3. x = −1 4. Evaluate 16 − 36x2 when: 1. x = −1 2. x = 0 3. x = 2 5. A window washer drops a squeegee from a platform 275 feet high. The polynomial −16t2 + 275 gives the height of the squeegee t seconds after it was dropped. Find the height after t = 4 seconds. 6. A manufacturer of microwave ovens has found that the revenue received from selling microwaves at a cost of p dollars each is given by the polynomial −5p2 + 350p. Find the revenue received when p = 50 dollars. ### Everyday Math 1. Fuel Efficiency The fuel efficiency (in miles per gallon) of a bus going at a speed of x miles per hour is given by the polynomial $$− \frac{1}{160} x^{2} + \frac{1}{2} x$$. Find the fuel efficiency when x = 40 mph. 2. Stopping Distance The number of feet it takes for a car traveling at x miles per hour to stop on dry, level concrete is given by the polynomial 0.06x2 + 1.1x. Find the stopping distance when x = 60 mph. ### Writing Exercises 1. Using your own words, explain the difference between a monomial, a binomial, and a trinomial. 2. Eloise thinks the sum 5x2 + 3x4 is 8x6. What is wrong with her reasoning? ### Self Check (a) After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. (b) If most of your checks were: …confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific. …with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved? …no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.<\|endoftext\|>	4.8125
1,430	Imagine trying to figure out your car’s fuel economy by driving only 20 miles. Sure, the number might look pretty good, but it wouldn’t be a very accurate picture of how your vehicle burns fuel over the long term. Predicting how the climate will change is a bit more complicated than calculating miles per gallon, but scientists who estimate Earth’s future temperatures face a similar challenge—having enough data to see the big picture. The big goal is to gauge how the atmosphere responds to changes, and to fully understand the long-term trends, you’d better understand the short-term trends really well. “The big goal is to gauge how the atmosphere responds to changes, and to fully understand the long-term trends, you’d better understand the short-term trends really well,” said Eric Fetzer, a project scientist at NASA’s Jet Propulsion Laboratory. In 2002, NASA launched the Atmospheric Infrared Sounder (AIRS) aboard its Aqua satellite, designed to make precise measurements of global temperatures, greenhouse gases and clouds. Now that the instrument has amassed 12 years of data, researchers are using its cutting-edge observations to better understand how climate feedbacks will impact warming rates. Water vapor: The hot and cold cycles of El Niño When it comes to global climate, one of the strongest triggers of short-term variability is the El Niño Southern Oscillation (ENSO). Normally, most of the Earth’s warm water is concentrated in a deep pool in the western Pacific. But every three to five years, El Niño makes its appearance, bringing warm water to the surface and making the tropical Pacific Ocean warmer than average. (Conversely, during La Niña, sea-surface temperatures become colder than normal.) As a potent greenhouse gas, water vapor plays a key role in global climate: As temperatures rise, water vapor increases, and because water vapor is itself a greenhouse gas, this produces even more warming. This vicious cycle is known as a feedback. By looking at the AIRS water vapor data during both the cold and warm cycles of the ENSO (El Niño and La Niña), Dessler can see how water vapor levels are responding to temperature changes and calculate the strength of the water vapor feedback. “If the globe warmed seven-tenths of a degree during the last El Niño, I can see in the AIRS data how water vapor changed in response to that. This tells me a lot about how water vapor is going to interact with climate change,” he said. Now that AIRS has 12 years of data, Dessler is able to measure water vapor at all altitudes of the lower atmosphere during several El Niño cycles. “Water vapor doubles the amount of warming you get from CO2 alone, so it’s really this very significant effect,” he said. “If you have only three years of data, you really don’t know if what you’re seeing in the atmosphere is representative of the longer-term picture.” Clouds: Using the present to predict the future While water vapor has a powerful effect on the climate system, the role that clouds play in affecting climate change is not well understood. “Models all predict that the water vapor effect doubles the amount of warming, but we’re still pretty uncertain about how clouds amplify warming,” said Hui Su, a research scientist at NASA’s Jet Propulsion Laboratory. Low clouds can shelter the Earth like an umbrella, reflecting solar radiation back to space and producing a net cooling effect. But high clouds can act like a blanket over the planet, triggering a greenhouse (warming) effect. The big conundrum for climate modelers is, how much cloud cover will there be in the future? The answer to that question is directly related to humidity. “More moisture in the atmosphere means more cloud cover,” said Su. “Based on the AIRS data, we know what the current humidity should be, and that should predict the amount of clouds,” said Su. “So, we’re really using current observations to infer the realism of model predictions.” It’s kind of an ‘inconvenient truth’—the models that are realistic all happen to be the ones that are predicting a very strong warming in the future. In a recent study, Su and her research team used AIRS observations of humidity to analyze the performance of 15 leading global climate models. What they found was surprising: Only five of the models reproduced humidity levels that matched the current range observed by AIRS and other instruments. Even more startling, those models that did accurately forecast current humidity were the ones that predicted a much warmer future climate—about .7 degrees C (1.3 degrees F) higher than the average warming predicted by all 15 models combined. “It’s kind of an ‘inconvenient truth’—the models that are realistic all happen to be the ones that are predicting a very strong warming in the future,” said Su. According to Fetzer, Su’s findings beg two important questions: “First, why are we looking at models that don’t accurately produce what we’re experiencing today? And second, if the models that are accurately forecasting humidity say that future warming is going to be closer to 4 degrees C than 1.5 degrees C, shouldn’t we be concerned?” CO2 is the big control knob just like the guy playing guitar is making the music, but it’s the amplifier that makes it loud. The guitar player by himself can’t blow your eardrums out. There is indeed reason for vigilance, if not concern. According to Dessler, only about one-third of the warming predicted for the future will come directly from CO2; the rest will result from feedback effects—like water vapor and clouds. “CO2 is the big control knob,” Dessler said, “just like the guy playing guitar is making the music, but it’s the amplifier that makes it loud. The guitar player by himself can’t blow your eardrums out.” Despite their complexity, climate models are the most robust tools we have for estimating Earth’s future temperatures. But can we be confident that their estimates of the “amplifier” effects are accurate? The best way to assess that, says Fetzer, is to evaluate their ability to forecast current conditions. “We need to come up with a set of stringent observational requirements that models must meet, and the basic prerequisite is that they are able to reproduce current reality—that’s just logic,” said Fetzer. “That’s really why we need to look at 10-plus years of data.”<\|endoftext\|>	4.1875
244	Copy of introduction First grade (primary school) level To provide speaking and listening practice for young learners To practice counting to 10 while using the target language Procedure (30-45 minutes) 1. Play Hello, Hello song nonstop and let them listen and watch you do the actions 2. Introduce yourself and ask what's your name. (probably you get no answers here) 3. Help kids write their names on name cards and put the name cards on the desks. 4. Play What's your name song? 1. PPT page 4. Show the picture. Talk about the picture to expose them to the language, it does not matter that they cannot follow at this stage. The goal is exposure to the target language by picture discription. 2. Play the CD. 3. Explain the difficult parts if necessary. 4. Repeat the sentences (chain drill my favorite) 5. Ask 2 or 4 students to role play the converstaion. 1. Play what's your name song again? 2. When the song is finished, ask a random student the question :"what's your name?", still you must say it like the song to get better results.<\|endoftext\|>	4
1,177	# Subtract the numbers: 868 + 1,862 - 416 + 224 = ? Calculate the numbers difference and learn how to do the subtraction, column subtracting method, from right to left ## The operation to perform: 868 + 1,862 + 224 ### Stack the numbers on top of each other. #### And so on... 8 6 8 1 8 6 2 + 2 2 4 ? ## Add column by column; start from the column on the right. ### Add the digits in the ones column: #### 1 is the tens digit. Carry it over to the tens column. Write the digit above that column. Add it with the rest of the digits in that column. 1 8 6 8 1 8 6 2 + 2 2 4 4 ### Add the digits in the tens column: #### 1 is the hundreds digit. Carry it over to the hundreds column. Write the digit above that column. Add it with the rest of the digits in that column. 1 1 8 6 8 1 8 6 2 + 2 2 4 5 4 ### Add the digits in the hundreds column: #### 1 is the thousands digit. Carry it over to the thousands column. Write the digit above that column. Add it with the rest of the digits in that column. 1 1 1 8 6 8 1 8 6 2 + 2 2 4 9 5 4 ### Add the digits in the thousands column: #### 2 is the thousands digit. Write it down at the base of the thousands column. 1 1 1 8 6 8 1 8 6 2 + 2 2 4 2 9 5 4 ## Now subtract the final numbers above: 2,954 - 416 = ? ### Stack the numbers on top of each other. #### And so on... 2 9 5 4 - 4 1 6 ? ## Subtract column by column; start from the column on the right ### Subtract the digits in the ones column: #### When borrowing, 1 ten = 10 ones. Add 10 to the top digit in the column of the ones: 10 + 4 = 14. 4 2 9 5 14 - 4 1 6 #### After borrowing, the subtraction has become: 14 - 6 = 10 + 4 - 6 = 10 + 4 - 6 = 4 + (10 - 6) = 4 + 4 = 8. 8 is the ones digit. Write it down at the base of the ones column. 4 2 9 5 14 - 4 1 6 8 ### Subtract the digits in the tens column: #### 54 - 1 = 3. 3 is the tens digit. Write it down at the base of the tens column. 4 2 9 5 14 - 4 1 6 3 8 ### Subtract the digits in the hundreds column: #### 9 - 4 = 5. 5 is the hundreds digit. Write it down at the base of the hundreds column. 4 2 9 5 14 - 4 1 6 5 3 8 ### Subtract the digits in the thousands column: #### There is a single digit in this column: 2. 2 is the thousands digit. Write it down at the base of the thousands column. 4 2 9 5 14 - 4 1 6 2 5 3 8 ## The latest numbers that were subtracted 868 + 1,862 - 416 + 224 = ? Jan 18 11:00 UTC (GMT) 559 - 212 - 288 - 262 + 266 - 256 - 332 = ? Jan 18 11:00 UTC (GMT) - 3,376 + 1,460 = ? Jan 18 11:00 UTC (GMT) 580 + 530 + 655 + 521 - 1,094 = ? Jan 18 11:00 UTC (GMT) 454 + 523 - 448 + 691 + 588 = ? Jan 18 11:00 UTC (GMT) - 4,054 + 9,560 = ? Jan 18 11:00 UTC (GMT) 3,953 - 9,474 - 10,113 = ? Jan 18 11:00 UTC (GMT) 7,054 - 5,547 = ? Jan 18 11:00 UTC (GMT) 451 - 4,168 - 674 - 2,066 + 561 - 576 = ? Jan 18 11:00 UTC (GMT) - 905 + 1,887 + 457 + 249 = ? Jan 18 11:00 UTC (GMT) 965 + 1,985 - 521 + 324 = ? Jan 18 11:00 UTC (GMT) 3,864 - 482 + 507 + 2,054 + 136 = ? Jan 18 11:00 UTC (GMT) 377 - 4,083 - 570 - 1,975 - 484 + 521 = ? Jan 18 11:00 UTC (GMT) All the numbers that were subtracted by users<\|endoftext\|>	4.53125

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