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672	The 12 ribs on each side of the chest are an important part of the chest wall that protect underlying internal organs. Rib fractures commonly occur in the middle ribs. Fracture of the first two ribs is rare as they are somewhat protected by the clavicle (collarbone), whilst the lowest two ribs are also rarely fractured because they do not insert into the sternum (breastbone) so they are able to move more freely. Fractures usually occur from direct blows or from indirect crushing injuries. Rib fractures can be painful and may lead to severe injury inside the abdomen and chest.With any rib trauma, x-ray is suggested to rule out a fracture. Also a chest examination is recommended to be certain that there is no injury to the lungs as a result of the fracture. Rib fractures may be sustained during sports or recreational activities or they may be the result of a motor vehicle accident or from a fall. However, ribs can be fractured by means other than direct trauma. Prolonged coughing, cancer, osteoporosis or weakened bones from bone infections can all lead to fractured ribs. Stress fractures of the ribs may occur with overuse, usually related to repetitive movements in sports such as rowing or baseball pitching in which the muscles that insert into the ribs work excessively hard leading to the stress fracture. The most common indicator of a fractured rib is sharp localised pain during breathing, especially with inspiration (breathing in). There may be a grating sound with breathing or movement of the fractured rib that causes the chest wall to flail. A flail chest is a potentially fatal condition in which a portion of the chest wall moves separately from the rest of the chest. The flail segment moves in while the rest of the chest moves out. If this occurs, immediate medical attention is necessary as it may lead to a pneumothorax (a punctured lung). Pain from rib fractures can make any movement, especially changes in position such as getting out of bed, very difficult due to pain.Also, because breathing is painful, the person may try to suppress deep breathing or coughing, which can lead to a chest infection. Early mobilisation after a rib fracture is important to prevent infection, pneumonia, constipation and deep vein thrombosis (DVT). Physiotherapy can help patients with fractured ribs in a number of ways: - Assist you to walk and show you the best ways to change positions, while heeding necessary precautions - Apply padding or strapping to stabilise and support the fracture site and ease pain - Provide pain relief in the form of ice, heat and ultrasound - Strengthening and/or flexibility exercises to help return to sport If you have sustained a trauma to the ribs, come and see how we can help you. We do not warrant or represent that the information in this site is free from errors or omissions or is suitable for your intended use. We recommend that you seek individual advice before acting on any information in this site. We have made every effort to ensure that the information on our website is correct at the time of publication but recommend that you exercise your own skill and care with respect to its use. If you wish to purchase our services, please do not rely solely on the information in this website.<\|endoftext\|>	3.65625
1,357	We cover every section of the GMAT with in-depth lessons, 5000+ practice questions and realistic practice tests. ## Up to 90+ points GMAT score improvement guarantee ### The best guarantee you’ll find Our Premium and Ultimate plans guarantee up to 90+ points score increase or your money back. ## Master each section of the test ### Comprehensive GMAT prep We cover every section of the GMAT with in-depth lessons, 5000+ practice questions and realistic practice tests. ## Schedule-free studying ### Learn on the go Study whenever and wherever you want with our iOS and Android mobile apps. # Integers: Prime Numbers Is $$\sqrt[3]{{x}}$$ non-negative? >(1) The product of $$x$$ and positive integer $$y$$ is not $$x$$. >(2) $$x$$ is a prime number. Incorrect. [[snippet]] According to Stat. (1), the product of $$x$$ and positive integer $$y$$ is not $$x$$. Notice that we know nothing of $$y$$, but we don't really care. Actually, $$y$$ is of no great importance here; it's only utility is in telling us something about $$x$$. The only number that would be equal to itself when multiplied by any other number is 0. Thus, what this sentence actually means is that $$x$$ is not 0. Since the only thing this statement tells us is that $$x$$ isn't 0, $$x$$ can still be any other number. Try plugging different values for $$x$$. Remember to check both positive and negative numbers, just to be safe. Also, plugging in numbers which actually have an integer third root will make things easier: * If $$x=8$$, then $$\sqrt[3]{x}=2$$, which is non-negative, so the answer is "Yes." * If $$x=-8$$, then $$\sqrt[3]{x}=\sqrt[3]{-8}=-2$$, which is negative, so the answer is "No." Therefore, the answer is "Maybe," and Stat.(1) → IS → BCE. Incorrect. [[snippet]] According to Stat. (1), the product of $$x$$ and positive integer $$y$$ is not $$x$$. Notice that we know nothing of $$y$$, but we don't really care. Actually, $$y$$ is of no great importance here; it's only utility is in telling us something about $$x$$. The only number that would be equal to itself when multiplied by any other number is 0. Thus, what this sentence actually means is that $$x$$ is not 0. Since the only thing this statement tells us is that $$x$$ isn't 0, $$x$$ can still be any other number. Try plugging different values for $$x$$. Remember to check both positive and negative numbers, just to be safe. Also, plugging in numbers which actually have an integer third root will make things easier: * If $$x=8$$, then $$\sqrt[3]{x}=2$$, which is non-negative, so the answer is "Yes." * If $$x=-8$$, then $$\sqrt[3]{x}=\sqrt[3]{-8}=-2$$, which is negative, so the answer is "No." Therefore, the answer is "Maybe," and Stat.(1) → IS → BCE. Incorrect. [[snippet]] According to Stat. (2), $$x$$ is a prime number. Remember, prime numbers are positive by definition, so according to this statement, $$x$$ must be positive. As long as $$x$$ is positive, its third root is also positive. This means $$\sqrt[3]{x}$$ is always non-negative, and the answer is "Yes," so Stat.(2) → S → BD. Incorrect. [[snippet]] According to Stat. (2), $$x$$ is a prime number. Remember, prime numbers are positive by definition, so according to this statement, $$x$$ must be positive. As long as $$x$$ is positive, its third root is also positive. This means $$\sqrt[3]{x}$$ is always non-negative, and the answer is "Yes," so Stat.(2) → S → BD. Correct. [[snippet]] According to Stat. (1), the product of $$x$$ and positive integer $$y$$ is not $$x$$. Notice that we know nothing of $$y$$, but we don't really care. Actually, $$y$$ is of no great importance here; it's only utility is in telling us something about $$x$$. The only number that would be equal to itself when multiplied by any other number is 0. Thus, what this sentence actually means is that $$x$$ is not 0. Since the only thing this statement tells us is that $$x$$ isn't 0, $$x$$ can still be any other number. Try plugging different values for $$x$$. Remember to check both positive and negative numbers, just to be safe. Also, plugging in numbers which actually have an integer third root will make things easier: * If $$x=8$$, then $$\sqrt[3]{x}=2$$, which is non-negative, so the answer is "Yes." * If $$x=-8$$, then $$\sqrt[3]{x}=\sqrt[3]{-8}=-2$$, which is negative, so the answer is "No." Therefore, the answer is "Maybe," and Stat.(1) → IS → BCE. According to Stat. (2), $$x$$ is a prime number. Remember, prime numbers are positive by definition, so according to this statement, $$x$$ must be positive. As long as $$x$$ is positive, its third root is also positive. This means $$\sqrt[3]{x}$$ is always non-negative, and the answer is "Yes," so Stat.(2) → S → B. Statement (1) ALONE is sufficient, but Statement (2) alone is not sufficient to answer the question asked. Statement (2) ALONE is sufficient, but Statement (1) alone is not sufficient to answer the question asked. BOTH Statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.<\|endoftext\|>	4.46875
642	Diabetes is a disease where the body cannot properly produce or use insulin. Insulin is a hormone that turns the foods you eat into energy. If your body cannot turn food into energy, not only will your cells be starved for energy, you will also build up glucose (sugar) in your blood. This will lead you to have "high blood glucose levels." Over years, the high blood glucose level can damage major organs like your heart, eyes, and kidneys. Types of Diabetes: Type 1 Diabetes is caused by a total lack of insulin that, in turn, produces high blood glucose levels. Type 1 is most often is seen in children, but can develop in adults. If you have Type 1, your health care provider might recommend scheduled, nutritious meals, exercise, medication, and frequent blood sugar level tests. Type 2 Diabetes occurs when the body does not produce enough insulin or cannot properly use insulin. This is the most common type. The treatment may be similar to Type 1. Pre Diabetes or Borderline Diabetes may occur before a Type 2 diagnosis. Blood glucose levels will be higher than normal. Good nutrition and exercise may be recommended by your health care provider as treatment for pre diabetes. Even a slightly high blood sugar level is insidious and could affect major organs over time. Gestational Diabetes occurs in pregnant women that have high blood glucose levels. This type of Diabetes can harm both Mother and baby. If you have Gestational Diabetes, your health care provider may prescribe meal plans, exercise, daily testing and medicine. The exact causes of Diabetes are still unknown. However, heredity, obesity and lack of exercise may play a role. Here are some general risk factors: - Your siblings or parents have diabetes. - You are more than 20% overweight. - You do not exercise. - You have had gestational diabetes or you have had a baby over 9 lbs. - You have high blood pressure. - Your cholesterol level is not normal. How to care for you Diabetes: According to the American Diabetes Association, people with Diabetes have the same nutritional needs as everyone else. In addition to prescribed medications, well-balanced meals may help you keep your blood glucose level as normal as possible. Also, just like everyone else, exercise is an important part of staying healthy. Exercising with diabetes does require a few extra safety steps that your health care professional can make you aware of. Nutritious meals, an exercise routine, and using your NutriCounter along with the help of your doctor may aid you in controlling your Diabetes. Diabetes will never truly go away, but with proper nutrition, exercise and prescribed medications, it can be controlled. For more information on specific exercise and eating advice for diabetics, see the following web sites: American Diabetes Association Canadian Diabetes Association Online Risk Test Excellent article about Borderline Diabetes Am I at risk for Type 2 Diabetes? Safety Tips for Exercising with Diabetes Come and visit the NutriCounter web site for more information on how nutrition influences weight loss, diabetes, pregnancy, heart disease and more! healthfitcounter.com<\|endoftext\|>	3.875
656	# An object's two dimensional velocity is given by v(t) = ( e^t-2t , t-4e^2 ). What is the object's rate and direction of acceleration at t=5 ? Mar 7, 2018 The rate of acceleration is $= 146.4 m {s}^{-} 1$ in the direction of ${0.4}^{\circ}$ anticlockwise from the x-axis #### Explanation: The acceleration is the derivative of the velocity $v \left(t\right) = \left({e}^{t} - 2 t , t - 4 {e}^{2}\right)$ $a \left(t\right) = v ' \left(t\right) = \left({e}^{t} - 2 , 1\right)$ When $t = 5$ $a \left(5\right) = \left({e}^{5} - 2 , 1\right)$ $a \left(5\right) = \left(146.4 , 1\right)$ The rate of acceleration is $\| \| a \left(5\right) \| \| = \sqrt{{146.4}^{2} + {1}^{2}} = 146.4 m {s}^{-} 1$ The angle is $\theta = \arctan \left(\frac{1}{146.4}\right) = {0.4}^{\circ}$ anticlockwise from the x-axis Mar 7, 2018 #### Explanation: As velocity is given by $v \left(t\right) = \left({e}^{t} - 2 t , t - 4 {e}^{2}\right)$ accelaration is $\frac{\mathrm{dv}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left({e}^{t} - 2 t\right) \hat{i} + \frac{d}{\mathrm{dt}} \left(t - 4 {e}^{2}\right) \hat{j}$ or $\left({e}^{t} - 2\right) \hat{i} + \hat{j}$ Note that $- 4 {e}^{2}$ is a constant hence its differential is $0$. and at $t = 5$, accelaration is $\left({e}^{5} - 2\right) \hat{i} + \hat{j}$ Its magnitude is $\sqrt{{\left({e}^{5} - 2\right)}^{2} + 1} = \sqrt{{\left(148.4132 - 2\right)}^{2} + 1}$ = $146.417$ and direction is ${\tan}^{- 1} \left(\frac{1}{{e}^{5} - 2}\right) = {\tan}^{- 1} \left(\frac{1}{146.4132}\right)$ = ${0.39}^{\circ}$<\|endoftext\|>	4.53125
657	Researchers at Liverpool John Moore’s University have reconstructed the face of a man who lived in Dublin some 500 years ago. Incredibly accurate reconstructions like this are providing archaeologists with new way of studying the past—while also allowing them to visualize some of the most forgotten figures in history. Known only as SK2, the remains of this man were found in 2014 by archaeologists from Rubicon Heritage and Transport Infrastructure Ireland. He was one of five skeletons pulled from a construction site near the entrance of Trinity College in Dublin, Ireland. At first, the archaeologists figured the remains were either Vikings or Norsemen, but further analysis suggests the skeletons date back to Tudor Ireland, a period that ran from 1485 to 1603. The skull of SK2 was in excellent shape, prompting a 3D digital facial reconstruction. The work was carried out by Caroline Wilkinson and her team from Liverpool John Moore’s University’s Face Lab. This is the same team that has performed facial reconstructions of Richard III, Robert the Bruce, and several other historical figures. But unlike these historical bigshots, SK2 was a relative nobody—an individual who came from the lower end of the socioeconomic spectrum. Indeed, this latest facial reconstruction, like the one done of an ordinary man who lived in England 700 years ago, is providing archaeologists with a novel way of studying a portion of the population that rarely gets mentioned in historical documentation or through art—the very poor. Typically, it’s only the rich and famous that get to be remembered. To build the face of SK2, Wilkinson’s team first created a 3D scan of the skull, which formed the basis of the reconstruction. Using well-established marker points from forensic science and specialized software, the researchers were able to lay down the critical facial components, namely facial muscles, soft tissue, and skin. The rest was guesswork—but informed guesswork. Skeletal analysis suggests the man was between 25 to 35 years of age when he died, he measured 5'6" in height, he suffered from childhood malnutrition, and he performed heavy manual labor during his life. Isotopic analysis of his teeth indicate he hailed from Dublin, where he was subsequently buried. His remains, along with those of four adolescents, were located in place that used to be known as Hoggen Green—an open area that was controlled by the municipality of Dublin. Their graves were shallow, and none of the bodies were placed in a coffin. Based on these findings, and a review of illustrations made of 16th century Irish people, Wilkinson’s team adjusted SK2's face to match. They endowed him with long hair, a beard, and typical Irish features such as fair skin, blue eyes, and brown hair. They even gave him bland, simple clothing to match his social status. This may not be a perfect representation of SK2's mug, but perfection isn’t the point. By pooling all of the scientific and historical evidence together, and by using cutting edge computer technology, the researchers have done something that, until recently, we’ve never been able to do before—create a photo-like impression of a common person who lived centuries ago.<\|endoftext\|>	3.828125
1,125	# Derivative of Fourth Root of x: Proof by First Principle The derivative of fourth foot of x is equal to 1/(4x3/4). Fourth root of x is denoted by ∜x = x1/4, so its derivative formula is given by $\dfrac{d}{dx}$(∜x) = $\dfrac{1}{4x^{3/4}}.$ In this article, we will learn how to differentiate fourth root of x with respect to x by the following methods: • Power rule of derivatives • Substitution method • First principle of differentiation • Logarithmic differentiation. ## By Power Rule As the fourth root of x is expressed as ∜x = x1/4, its derivative can be easily computed by the power rule of differentiation. This rule says that the derivative of xn is equal to nxn-1, that is, d/dx (xn) = nxn-1. Put n=1/4. So we obtain that So the derivative of fourth root of x by power rule is equal to 1/(4x3/4). ALTERNATIVE METHOD: To find the derivative of fourth root of x, let us substitute $y=\sqrt[4]{x}.$. This implies that y4=x. Differentiate both sides with respect to x. So we get that $4y^3 \dfrac{dy}{dx}=1$ ⇒ $\dfrac{dy}{dx}=\dfrac{1}{4y^3}$ ⇒ $\dfrac{dy}{dx}=\dfrac{1}{4(\sqrt[4]{x})^3}$ as we have $[\because y=\sqrt[4]{x}]$ ⇒ $\dfrac{dy}{dx}=\dfrac{1}{4x^{3/4}}$ This proves that the derivative of fourth root of x is 1/(4x3/4), and we get this by the substitution method. Also Read: Derivative of cube root of x ## Derivative of Fourth Root of x by First Principle By first principle, the derivative of f(x) is given by the following limit formula: d/dx (f(x)) = limh→0 $\dfrac{f(x+h)-f(x)}{h}$. Let f(x) = ∜x. Note that f(x) = x1/4 and f(x+h) = (x+h)1/4. Then by first principle, Put x+h = z. So z→x as h→0. Also, h=z-x. Therefore, Hence, $\dfrac{d}{dx}$(∜x) = $\dfrac{1}{4x^{3/4}}$. Therefore, the derivative of fourth root of x is equal to 1/(4x3/4) and this is obtained by the first principle of differentiation. Related Topics: ## By Logarithmic Differentiation Let y = ∜x = x1/4. Taking natural logarithms on both sides, we get that ln y = 1/4 ln x Differentiating with respect to x, $\dfrac{1}{y} \dfrac{dy}{dx}=\dfrac{1}{4} \cdot \dfrac{1}{x}$ ⇒ $\dfrac{dy}{dx}=\dfrac{y}{4x} = \dfrac{x^{1/4}}{4x} = = \dfrac{1}{4x^{3/4}}$. So by logarithmic differentiation, the derivative of fourth root of x is equal to 1/(4x3/4). ## Solved Problems Question 1: Find the derivative of fourth root of a where a is a constant. That is, find d/dx(∜a). Note that the fourth root of $a$ is a constant. We know that the derivative of a constant is zero. So we obtain that d/dx(∜a) = 0. ⇒ d/dx(a1/4) = 0. Remark: Put a=1. So the derivative of fourth root of 1 with respect to x is equal to 0, that is, d/dx(∜1) = 0. Question 2: Find the derivative of fourth root of x+1, i.e, find d/dx(∜(x+1)). Put t=x+2. So dt/dx = 1. Now by the chain rule of derivatives, $\dfrac{d}{dx}$ (∜(x+1)) = $\dfrac{d}{dt}$ (∜t) × $\dfrac{dt}{dx}$ = 1/(4t3/4) × 1 = 1/{4(x+1)3/4}. So the derivative of (x+1)1/4, that is, fourth root of x+1 is equal to 1/{4(x+1)3/4}. ## FAQs Q1: What is the derivative of x1/4 (fourth root of x)? Answer: The derivative of x1/4 (fourth root of x) is equal to 1/(4x3/4). Q2: What is d/dx(∜x)?<\|endoftext\|>	4.875
289	In this historical photo from the U.S. space agency, another first for NASA, an all-female crew of scientific experimenters began a five-day exercise on December 16, 1974, to test the feasibility of experiments that were later tested on the Space Shuttle/Spacelab missions. The experimenters, Dr. Mary H. Johnston (seated, left), Ann F. Whitaker and Carolyn S. Griner (standing, left to right), and the crew chief, Doris Chandler, spent spend eight hours each day of the mission in the Marshall Space Flight Centers General Purpose Laboratory (GPL). They conducted 11 selected experiments in materials science to determine their practical application for Spacelab missions and to identify integration and operational problems that might occur on actual missions. Inside the GPL, the four women worked under conditions simulating, as nearly as practical, those that would exist in a space station in Earth orbit, excepting, of course, weightlessness. Air circulation, temperature, humidity and other factors were carefully controlled. The test was conducted at NASAs Marshall Space Flight Center, Huntsville, AL, where the GPL is part of the centers Concept Verification Test (CVT), a project oriented to reducing future costs of experimentation in space by involving potential experimenters early in the development cycle of their hardware. Each weekday, SPACE.com looks back at the history of spaceflight through photos (archive).<\|endoftext\|>	3.671875
601	Electron beams boost nanotechnology. (Miniaturization). Operated in the Microdevices Laboratory at JPL, it provides a tool for delving into the realm of nanotechnology, where the individual molecules become accessible to electronic probing. "The E-Beam lithography system will allow researchers to work at the equivalent level of nature's biological building blocks, by allowing them to create and research technologies at the cellular and subcellular level," notes Paul Maker, manager of JPL's Electron Beam Lithography Laboratory. (Lithography is the process of printing a pattern onto a surface, such as a silicon chip or a high-resolution film.) "The E-Beam lithography system is like a very fast, very high-resolution camera, but instead of exposing photo-sensitive film to light, a thin layer of electron-sensitive material is exposed to electrons. Instead of using a shutter that imprints the whole image at once, an intense electron beam focused to a tiny spot is rastered [scanned] over the chip like the beam that creates the image on a television screen." Just as with photographic film, subsequent processing steps develop the image that was imprinted on the film, in this case, the device structure. The system allows users to "write" 10 times faster with a spot that is two times smaller than can be done with the system currently in place, installed 12 years ago. "The faster `writing' speed means we can fabricate many more of these experimental chips, thereby reducing the time it takes to perfect a new chip design. The higher resolution translates into device designs with much-finer detail, leading to smaller, more capable chips," Maker explains. NASA faces the challenge of miniaturizing all aspects of its space systems, with the ultimate goal of reducing the size and mass of instruments by orders of magnitude without sacrificing performance--like creating an entire laboratory on a chip with the same sensitivity as the room-size version. "Since this machine is capable of producing patterns with feature sizes on the scale of molecules," Maker indicates, "we can now develop miniature devices that allow us to manipulate and characterize these minute building blocks of nature, and create tools that can be used to search for the signatures of life in a controlled manner." \|Printer friendly Cite/link Email Feedback\| \|Publication:\|\|USA Today (Magazine)\| \|Article Type:\|\|Brief Article\| \|Date:\|\|Jun 1, 2002\| \|Previous Article:\|\|Protecting nanotubes from static electricity. (Circuitry).\| \|Next Article:\|\|Self-assembled tubes for electronic devices. (Nanotechnology).\| \|Carbon nanotubes beam electrons. (Materials Science).\| \|Atomic uniformity key to miniaturization.\| \|Microscopy techniques focus on nanotech.\| \|Carl Zeiss and Seiko partner for Nanotech market.\|<\|endoftext\|>	3.875
1,921	Methods of Jesus.02 Lesson 2: Jesus and Motivation Theory Upon completion of this module the learner should be equipped to motivate learners by means of: - directing needs assessment. - allowing reflection on activity. - encouraging self-reinforcement. Motivation is essential at the beginning, the middle, and the end of spiritual learning. It is eye opening to see how many ways Jesus motivated learners and that modern educators use similar means. Needs: physical, mental, spiritual. Boundless enthusiasm is still driven by perceived basic human needs. Motivation for learning is first activated by the extent to which learners perceive that the new information, attitude, or modeled experience directly satisfies a strong physical, mental, emotional, or spiritual need in harmony with their beliefs and values. Class members are also motivated to learn by a need to be competent and successful. Passion to study the Bible kicks in when class members get insight into how the passage meets their needs. Thus it is the task of the facilitator to help learners see how the Bible does this. Learners must constantly lead the discussion in such a manner as to deal with members’ needs and to address them with Bible content. In order to do this the facilitator must develop skills to discern the needs underlying verbal expressions, and for this a conscious presence of the Holy Spirit is mandatory. A daily baptism of the Holy Spirit helped Jesus to recognize the needs of the weary and oppressed. “His words were given Him fresh from the heavenly courts, words that He might speak in season to the weary and oppressed” (Christ’s Object Lessons, p. 139). Weariness and oppression most often lie beneath the surface in the experiences expressed by class members. The class facilitator must become expert at listening with the “third ear.” Example. A member of the class says: “I agree with Solomon when he said: ‘So I hated life, for the work which had been done under the sun was grievous to me’ ” (Eccl. 2:17, NASB). The facilitator may perceive that the reason this person hated his work is because he did not see himself as successfully doing important work. The class facilitator might encourage the man to talk about his pain. By leading the discussion in a competent manner, and through discussion of the biblical portion, the facilitator can help other members to support this person. Together they can help the man to recognize that daily labor for the necessities of life has value in itself. The needy man should experience healing by realizing that life itself is his portion given to him by a loving God (Eccl. 3:11-14). In this way the class facilitator may relate the experiences of the Bible openly to the daily needs of class members. Money, treasure, pearls, building houses, sowing seed, investments, food, clothes, vineyards, riches, and other tangibles fill the teachings of Jesus. These all deal with basic physical needs. Jesus also appealed to psychological needs such as a desire for greatness, security, success, competence, love, nobility, courage, and virtue. Narratives about thugs attacking travelers on the highway, thieves breaking into a house and fighting with the owner, people being cut to pieces and their houses being turned into a pile of dung, floods washing houses away, and treasures hidden in a field kept His listeners spellbound. All these were dealing with life’s most basic requirements. Jesus went beyond dealing with physical and psychological needs, however. He knew the great longing of the human heart to be reconciled to God. He opened the door to spiritual needs by means of physical and emotional needs. His encounter with the woman at the well was classic in this regard. He created thirst for the Water of Life through the metaphor of living water. He also addressed higher needs through parables such as the prodigal son and through demonstrations of healing. Effective class facilitators may lead class members to appreciate God in new ways. Example. Ask the question: “Does it serve a practical purpose for us to visit an art museum or attend a concert?” The discussion that follows may focus on the fact that art and music, along with daily necessities, can be instruments used by God to help one appreciate His loving character. It may be as important to receive His gifts with thanksgiving as it is to accomplish some great thing. Dynamics of Changing Needs Life stages, gender differences, circumstances, world conditions, changes in relationships, and the dynamics of human experience all bring changes in perceived needs. Jesus shared things when the time was right (John 16:12; Christ’s Object Lessons, p. 139). The following teachable moments in adult life have been identified by modern educators Habermas and Issle, K., Teaching for Reconciliation, p. 162: - establishing independent living arrangements - entering the work world - deciding whether or not to marry - deciding whether or not to be parents - selecting the right friends - being a community leader - living with teenagers - living within one’s income - choosing leisure activities - living with a spouse - accepting physiological changes - adjusting physiological changes - coping with personal aging - adjusting to reduced income in retirement - adjusting to the loss of your spouse - living with peers - being a good friend, good citizen - relocation of living arrangements Next to intercessory prayer, needs analysis or assessment should be a primary step in motivating learners. There are several ways to do this: Example 1. A facilitator might hand out a list of life situations such as the “teachable moments” list, and ask class members to prioritize the list in reference to their personal situation. The class facilitator should always be alert to pick up expressed or hidden needs that come out of class discussion. Example 2. Pass out a list of questions about the current lesson and ask member to select the question they wish to discuss first. Example 3. Ask class members to complete sentences such as: - The thing that concerns me most right now is . . . - Prepare a sentence completion list dealing with issues in the current Adult Sabbath School Bible Study Guide. Ask members to share their answers with the class. - The most helpful thing to me in today’s lesson is . . . Stephen Brookfield says that the way to facilitate adult motivated education is for learners and facilitators to be involved in a continual ongoing process of activity, reflection upon activity, and collaborative analysis of activity. He says that “activity” is not limited to “doing” but may include such things as verbally exploring a whole new way of doing one’s work (Brookfield, S. D., Understanding and Facilitating, p. 10). Class facilitators should lead in discussion and reflection on activities of individuals in the current Bible lesson that relate to needs of members. By sharing personal experiences, members bring real-life activities into focus. Example. Adjusting to aging parents will surface as an immediate need for several class members. The discussion may center on the expressions of Solomon that are similar to those of a depressed person. Elderly people may be subject to various forms of depression. This may develop into understanding and action solutions supported by prayer on the part of other members of the class. This can be one of the most effective means of bringing about changes in activities. Competency and Success Adult learners are also motivated by a need for success or competence not only with information but also in life’s actions. They want to improve skills, become better Christians, husbands, wives, and parents. They want to become more joyous, more satisfied with life.They want to successfully glorify God. Jesus appealed to this need when He told the parable of the man with 10 talents who productively invested them and earned more for His master. The facilitator who can help members experience success or perceive their activities as successful will have motivated learners. Self-motivation for Success It is much better for the learner to be self-motivated for competence than for facilitators or members to impose external rewards, punishment, test, or other means of extrinsic reinforcement. This is where testimonies of successful application of biblical principles in the lesson not only provide intrinsic reinforcement but model success to other members. The class facilitator may simply ask: “Has anyone experienced a positive change in facing a life problem by applying the principles we have studied today?” Of course it is the power of the Holy Spirit, through prayer and contemplation of the Scriptures and experiences of class members that bring a converting effect that most motivates learners (Testimonies, vol. 5, pp. 82, 83; Education, p. 80). Prepare a needs assessment by means of a sentence completion list, dealing with issues in the current Adult Sabbath School Bible Study Guide. Ask members to share their answers with the class. - I most identify with Solomon in the following ways . . . - The part of today’s lesson that I am most curious about is . . . - The most helpful thing to me in today’s lesson is . . . © 2014 General Conference of Seventh-day Adventists<\|endoftext\|>	3.859375
1,964	# Integration by Parts Let $f(x)$ and $g(x)$ be differentiable functions. Then the product rule $$(f(x)g(x))’=f’(x)g(x)+f(x)g’(x)$$ \label{eq:intpart} \int f(x)g’(x)dx=f(x)g(x)-\int f’(x)g(x)dx. The formula \eqref{eq:intpart} is called integration by parts. If we set $u=f(x)$ and $v=g(x)$, then \eqref{eq:intpart} can be also written as \label{eq:intpart2} \int udv=uv-\int vdu. Example. Evaluate $\int x\cos xdx$. Solution. Let $u=x$ and $dv=\cos xdx$. Then $du=dx$ and $v=\sin x$. So, \begin{align} \int x\cos xdx&=x\sin x-\int\sin xdx\\ &=x\sin x+\cos x+C, \end{align} where $C$ is a constant. Example. Evaluate $\int\ln xdx$. Solution. Let $u=\ln x$ and $dv=dx$. Then $du=\frac{1}{x}dx$ and $v=x$. So, \begin{align} \int\ln xdx&=x\ln x-\int x\cdot\frac{1}{x}dx\\ &=x\ln x-x+C, \end{align} where $C$ is a constant. Often it is required to apply integration by parts more than once to evaluate a given integral. In that case, it is convenient to use a table as shown in the following example. Example. Evaluate $\int x^2e^xdx$ Solution. In the following table, the first column represents $x^2$ and its derivatives, and the second column represents $e^x$ and its integrals. $$\begin{array}{ccc} x^2 & & e^x\\ &\stackrel{+}{\searrow}&\\ 2x & & e^x\\ &\stackrel{-}{\searrow}&\\ 2 & & e^x\\ &\stackrel{+}{\searrow}&\\ 0 & & e^x. \end{array}$$ This table shows the repeated application of integration by parts. Following the table, the final answer is given by $$\int x^2e^xdx=x^2e^x-2xe^x+2e^x+C,$$ where $C$ is a constant. Example. Evaluate $\int x^3\sin xdx$. Solution. In the following table, the first column represents $x^3$ and its derivatives, and the second column represents $\sin x$ and its integrals. $$\begin{array}{ccc} x^3 & & \sin x\\ &\stackrel{+}{\searrow}&\\ 3x^2 & & -\cos x\\ &\stackrel{-}{\searrow}&\\ 6x & & -\sin x\\ &\stackrel{+}{\searrow}&\\ 6 & & \cos x\\ &\stackrel{-}{\searrow}&\\ 0 & & \sin x. \end{array}$$ Following the table, the final answer is given by $$\int x^3\sin xdx=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+C,$$ where $C$ is a constant. Example. Evaluate $\int e^x\cos xdx$. Solution. In the following table, the first column represents $e^x$ and its derivatives, and the second column represents $\cos x$ and its integrals. $$\begin{array}{ccc} e^x & & \cos x\\ &\stackrel{+}{\searrow}&\\ e^x & & \sin x\\ &\stackrel{-}{\searrow}&\\ e^x & & -\cos x. \end{array}$$ Now, this is different from the previous two examples. While the first column repeats the same function $e^x$, the functions second column changes from $\cos x$ to $\sin x$ and to $\cos x$ again up to sign. In this case, we stop there and write the answer as we have done in the previous two examples and add to it $\int e^x(-\cos x)dx$. (Notice that the integrand is the product of functions in the last row.) That is, $$\int e^x\cos xdx=e^x\sin x-e^x\cos x-\int e^x\cos xdx.$$ For now we do not worry about the constant of integration. Solving this for $\int e^x\cos xdx$, we obtain the final answer $$\int e^x\cos xdx=\frac{1}{2}e^x\sin x-\frac{1}{2}e^x\cos x+C,$$ where $C$ is a constant. Example. Evaluate $\int e^x\sin xdx$. Solution. In the following table, the first column represents $e^x$ and its derivatives, and the second column represents $\sin x$ and its integrals. $$\begin{array}{ccc} e^x & & \sin x\\ &\stackrel{+}{\searrow}&\\ e^x & & -\cos x\\ &\stackrel{-}{\searrow}&\\ e^x & & -\sin x. \end{array}$$ This is similar to the above example. The first columns repeats the same function $e^x$, and the functions in the second column changes from $\sin x$ to $\cos x$ and to $\sin x$ again up to sign. So we stop there and write $$\int e^x\sin xdx=-e^x\cos x+e^x\sin x-\int e^x\sin xdx.$$ Solving this for $\int e^x\sin xdx$, we obtain $$\int e^x\sin xdx=-\frac{1}{2}e^x\cos x+\frac{1}{2}e^x\sin x+C,$$ where $C$ is a constant. Example. Evaluate $\int e^{5x}\cos 8xdx$. Solution. In the following table, the first column represents $e^{5x}$ and its derivatives, and the second column represents $\cos 8x$ and its integrals. $$\begin{array}{ccc} e^{5x} & & \cos 8x\\ &\stackrel{+}{\searrow}&\\ 5e^{5x} & & \frac{1}{8}\sin 8x\\ &\stackrel{-}{\searrow}&\\ 25e^{5x} & & -\frac{1}{64}\cos 8x. \end{array}$$ The first columns repeats the same function $e^{5x}$ up to constant multiple, and the functions in the second column changes from $\cos 8x$ to $\sin 8x$ and to $\cos 8x$ again to constant multiple. This case also we do the same. $$\int e^{5x}\cos 8xdx=\frac{1}{8}e^{5x}\sin 8x+\frac{5}{64}e^{5x}\cos 8x-\frac{25}{64}\int e^{5x}\cos 8xdx.$$ Solving this for $\int e^{5x}\cos 8xdx$, we obtain $$\int e^{5x}\cos 8xdx=\frac{8}{89}e^{5x}\sin 8x+\frac{5}{89}e^{5x}\cos 8x+C,$$ where $C$ is a constant. The evaluation of a definite integral by parts can be done as \label{eq:intpart3} \int_a^b f(x)g’(x)dx=[f(x)g(x)]_a^b-\int_a^b f’(x)g(x)dx. Example. Find the area of the region bounded by $y=xe^{-x}$ and the x-axis from $x=0$ to $x=4$. The graph of y=xexp(-x), x=0..4 Solution. Let $u=x$ and $dv=e^{-x}dx$. Then $du=dx$ and $v=-e^{-x}$. Hence, \begin{align} A&=\int_0^4 xe^{-x}dx\\ &=[-xe^{-x}]0^4+\int_0^4 e^{-x}dx\\ &=-4e^{-4}+[-e^{-x}]_0^4\\ &=1-5e^{-4}. \end{align}<\|endoftext\|>	4.6875
687	# Proof 12: Circle equation There is a round amphitheatre 1000 metres East and 3500 metres North from your hotel (let’s call those numbers $c_1$ and $c_2$). You go there for a visit. Wow, it is pretty, when standing in the middle! You wonder what size it has… Your GPS watch can tell this! It can tell you your Eastwards and Northwards coordinates. And Pythagoras’ theorem can then give the radius. So, you walk due East and then due North until you reach the edge. You read off the coordinates but realise that you forgot to reset the GPS. It was last reset at the hotel, which is now the origin, the reference. So, the GPS tells the total distance East – let’s call it $x$ – and North – let’s call it $y$ from the hotel and not from the theatre centre. The distances $c_1$ and $c_2$ shouldn’t be included. The actual distances are thus $x-c_1$ and $y-c_2$. Pythagoras’ theorem ($a^2+b^2=c^2$ for a right-angled triangle) now gives us the circle equation: $$(x-c_1)^2+(y-c_2)^2=r^2$$ Pick any possible $x$ and there is a matching $y$ that makes this equation hold true; meaning, a coordinate set $(x,y)$ on the circle. Run through all possible $x$ values and you get the full circle. We just must only use $x$ values that are inside the circle; any other values are of course impossible. $\quad_\blacksquare$ ## Proof 13: Ellipsis equation The circle equation (see Proof 12) can be rearranged: $$(x-c_1)^2+(y-c_2)^2=r^2\\\frac{(x-c_1)^2}{r^2}+\frac{(y-c_2)^2}{r^2}=1\\\left(\frac{x-c_1}{r}\right)^2+\left(\frac{y-c_2}{r} \right)^2 =1$$ The $x-c_1$ and $y-c_2$ distances are being ‘squeezed’ by $r$, so to say.… ## Proof 21: Sine and cosine values We here derive the sine and cosine values of a few chosen angles. Half ($\pi$) and quarter turns ($\frac \pi 2$, $-\frac \pi 2$): Cosine… ## Proof 5: Ellipsis area Imagine a circle with a radius of exactly 1, $r=1$ (called a unit circle). Circle area is $A_\text{circle}=\pi r^2$ (see Proof 4), so such a… ## Proof 6: Prism and cylinder volume In a room with 3 metres to the ceiling, there are 3 cubic metres above each square metre floor regardless of the shape. The 3… ## Proof 4: Circle and circle sector area A pizza slice’s edge is the pizza’s radius $r$. The slice is almost a triangle if it wasn’t for the rounding. That rounding has a…<\|endoftext\|>	4.78125
605	Skip the main content if you do not want to read it as the next section. An electrocardiogram (ECG) is a test which measures the electrical activity of your heart to show whether or not it is working normally. An ECG records the heart’s rhythm and activity on a moving strip of paper or a line on a screen. Your doctor can read and interpret the peaks and dips on paper or screen to see if there is any abnormal or unusual activity. What can an ECG (electrocardiogram) show? An electrocardiogram can be a useful way to find out whether your high blood pressure has caused any damage to your heart or blood vessels. Because of this, you may be asked to have an ECG when you are first diagnosed with high blood pressure. Some of the things an ECG reading can detect are: - cholesterol clogging up your heart’s blood supply - a heart attack in the past - enlargement of one side of the heart - abnormal heart rhythms How is an ECG carried out? An ECG (electrocardiogram) is a safe and painless test which normally only takes a few minutes. Leads from an electrocardiograph machine are attached to the skin on your arms, legs and chest using sticky patches. These leads read signals from your heart and send this information to the electrocardiograph. The machine then prints the reading on a paper strip or on a screen. There are three main types of ECG: - Resting ECG – if your doctor is interests in how your heart is working while you are at rest, you will be asked to lie down and relax while the heartbeat is being recorded. - Exercise ECG – your doctor may be interested in how your heart reacts to activity and you will be asked to walk or run on a treadmill or cycle on an exercise bike while your heartbeat is recorded. - 24-hour ECG – sometimes it can be helpful to monitor your heartbeat throughout the day, in which case you will be asked to wear a small electrocardiograph machine. The recordings from the machine are then read by your doctor when you return the machine. Who will carry out my electrocardiogram? You may be able to have your ECG in your GP surgery. However not all surgeries have this equipment so you may be referred to a local hospital to carry out the test. Is there anything I need to do before an ECG (electrocardiogram)? If you are having a resting ECG, there is no need to do anything different before having the test. If you are having an exercise ECG, make sure that you are wearing comfortable clothes and shoes that you can exercise in. It is also helpful to avoid eating a meal or drinking a caffeinated drink for two hours before the test because these can affect your heart’s rhythm.<\|endoftext\|>	3.921875
1,321	# Origami Silver Rectangle Alignments to Content Standards: 8.G.A.1 G-CO.D.12 This task examines the mathematics behind an origami construction of a rectangle whose sides have the ratio $(\sqrt{2}:1)$. Such a rectangle is called a silver rectangle. Beginning with a square piece of paper, first fold and unfold it leaving the diagonal crease as shown here: Next fold the bottom right corner up to the diagonal: After unfolding then fold the left hand side of the rectangle over to the crease from the previous fold: Here is a picture, after the last step has been unfolded, with all folds shown and some important points marked. In the picture $T$ is the reflection of $S$ about $\ell$. 1. Suppose $s$ is the side length of our square. Show that $\|PT\| = s$. 2. Show that $\triangle PQT$ is a 45-45-90 isosceles triangle. 3. Calculate $\|PQ\|$ and conclude that $PQRS$ is a silver rectangle. ## IM Commentary The purpose of this task is to apply geometry in order analyze the shape of a rectangle obtained by folding paper. The central geometric ideas involved are reflections (used to model the paper folds), analysis of angles in triangles, and the Pythagorean Theorem. The task is appropriate either at the 8th grade level or in high school: the only difference would be the level of rigor expected in the explanation. The solution given is appropriate for either level. The silver rectangle is one of three rectangles identified in ancient times as having important properties: the bronze rectangle has a side ratio of $(\sqrt{3},1)$ and the golden rectangle has a side ratio of $\left(1+\sqrt{5}:2 \right)$. Each of these three rectangles can be constructed by folding paper. Further properties of silver rectangles are examined in www.illustrativemathematics.org/illustrations/1489. Much more interesting information about the silver rectangle can be found here: http://en.wikipedia.org/wiki/Silver_ratio This task could be given in a much more open ended form, particularly if done in high school. Teachers could go through the different steps of the construction and then prompt students to examine the ratio of side lengths in the final rectangle. There are many possible solution paths revolving around the different triangles in the picture that are similar: half of the original square, $\triangle TQP$, and the small triangle in the lower left of the last picture. These are all 45-45-90 triangles and the dimensions of the rectangle can be found by calculating the scale factor between any pair of these triangles. The Standards for Mathematical Practice focus on the nature of the learning experiences by attending to the thinking processes and habits of mind that students need to develop in order to attain a deep and flexible understanding of mathematics. Certain tasks lend themselves to the demonstration of specific practices by students. The practices that are observable during exploration of a task depend on how instruction unfolds in the classroom. While it is possible that tasks may be connected to several practices, the commentary will spotlight one practice connection in depth. Possible secondary practice connections may be discussed but not in the same degree of detail. Students engage in Mathematical Practice Standard 5, Use appropriate tools strategically,” by considering a tool’s usefulness, its strengths and limitations, as well as knowing how to use it appropriately. This task examines the mathematics of the paper folding of a square piece of paper resulting in the creation of a silver rectangle. The tools, including paper, directions, straight edges, etc. allow students to analyze the shape of the paper as it is folded and unfolded revealing different triangular and rectangular shapes. This experimentation with paper folding allows students to explore and discuss observations before forming conclusions about the final rectangle. Students will also ''Reason abstractly and quantitatively'' (MP.2) as they need to represent the paper folding geometrically in order to calculate angles and side lengths. ## Solution 1. Since reflection about $\ell$ maps $P$ to itself and maps $S$ to $T$ this means that segments $\overline{PS}$ and $\overline{PT}$ are interchanged by reflection about $\ell$. Hence $$\|PS\| = \|PT\|$$ because reflections preserve lengths of segments. We are given that $\|PS\| = s$ since it is one side of the square so $\|PT\| = s$. 2. We know that $m(\angle QPT) = 45$ because the diagonal of a square bisects the 90 degree angles. Alternatively, reflection over the diagonal $\overleftrightarrow{PT}$ is a symmetry of the square: this reflection interchanges angles $QPT$ and $SPT$ and so these must each measure 45 degrees. Angle $TQP$ is a right angle because $\overline{QR}$ is a crease resulting from a horizontal fold of the paper. Since $$m(\angle QTP) + m(\angle QPT) + m(\angle TQP) = 180$$ this means that $m(\angle QTP) = 45$. Thus $\triangle PQT$ is a 45-45-90 triangle: it is isosceles because $m(\angle QTP) = m(\angle QPT)$. 3. To find $\|PQ\|$ we use the fact that $\triangle PQT$ is a right isosceles triangle. Thus we know that $\|TQ\| = \|PQ\|$ and, from the Pythagorean Theorem, $$\|PQ\|^2 + \|TQ\|^2 = \|PT\|^2.$$ Substituting $\|PQ\|$ for $\|TQ\|$ this is equivalent to $$2\|PQ\|^2 = \|PT\|^2.$$ From part (a) this means that $$\|PQ\|^2 = \frac{s^2}{2}.$$ So $$\|PQ\| = \frac{s}{\sqrt{2}}.$$ We can now check that $PQRS$ is a silver rectangle. Our calculations show that $\|PS\|:\|PQ\|$ is $s:\frac{s}{\sqrt{2}}$. But the ratio $s:\frac{s}{\sqrt{2}}$ is equivalent to the ratio $\sqrt{2}:1$ as we can see using a scaling factor of $\frac{\sqrt{2}}{s}$.<\|endoftext\|>	4.65625
25,051	An Overview of Religion in Los Angeles from 1850 to 1930 by Clifton L. Holland When California was admitted to the Union in September 1850, Southern California had experienced few changes through American control and settlement. Only three small towns existed in all of Southern California: San Diego, Los Angeles and Santa Barbara. However, smaller settlements were to be found around the old Spanish missions and on some of the large ranchos that dominated the economy of Southern California, a region that had a total population of only 6,367 in 1851 (McWilliams 1946:64). religion in Los Angeles was Roman Catholicism, which had been established throughout California by Franciscan friars who came from Mexico, beginning in the 1770s, to establish a chain of missions for the purpose of evangelizing the Native Americans and of developing agricultural colonies using forced Indigenous labor. The pueblo of "Our Lady the Queen of the Angels" had been established in 1781 on the banks of the Rio Porciuncula, now known as the Los Angeles River. As the Anglo American population of Southern California began to grow after 1850, small Protestant denominational churches grew out of union services in small towns. During the late 1860s and the decade of the 1870s, as more churches were planted in new settlements and the size of denominational groups increased along with the rapidly growing population, regional and state associations of Protestant churches were formed. Whereas the Baptist and Congregational preachers simply came with the people as part of the western migration, the Methodist preachers were usually sent west to form new churches, and the Presbyterian ministers were called to serve a church by a previously 1869 and 1909, the construction of several railroad lines within the Southland spurred a series of real estate booms that brought a flood of Anglo American settlers from Northern California, the Midwest and East Coast, as well as from many foreign countries, to begin a new life in "sunny Southern California." Between 1880 and 1900, hundreds of towns and thousands of orchards and farms emerged in the region, and the population of the City of Los Angeles grew from about 50,000 in 1890 to more than 100,000 It is worth noting that the majority of Anglo Americans in Southern California during this period were strongly biased and discriminatory against Indians, Mexicans, Asians and Roman Catholics (see McWilliams, 1968; Bean, 1968; and Wollenburg, 1970). The Spanish-speaking population (mostly Roman Catholic) of Los Angeles totalled about 12,000 in 1887, or less than 10% of the Anglo- American population; also, there were a few hundred Chinese in Los Angeles, most of whom lived in an area north of the Mexican Plaza, near the present-day Chinatown (Pitt, 1970). Excerpts from Clifton L Holland, The Religious Dimension in Hispanic Los Angeles: A Protestant Case Study. South Pasadena, CA: William Carey Library, 1974. The Religious Situation in Los Angeles from 1850 to 1900: the beginnings of ethnic and religious pluralism historic presence of the Roman Catholic Church in Los Angeles and environs accentuates the region's distinct spiritual heritage. Under the charge of a resident bishop, by 1859, this denomination benefited from the ministrations of clergy and nuns, outside funding and capable administration. Roman Catholics also remained a numerical plurality in the Los Angeles area through the early years of U.S. statehood, and their clergymen spoke the Spanish of the Californio residents, who were the descendants of Spanish and Mexican settlers and the local Indigenous population (see Leonard Pitt, The Decline of the Californios: A Social History of Spanish-Speaking Californians, 1846-1890). The institutions of local Catholicism therefore endured the tumultuous 1850s and 1860s with a stable church organization, staffed by clerics conversant in the language of the majority of the citizenry: Spanish. The remnants of the Native American Indian population in the Los Angeles area lived near the Spanish Franciscan Missions established in San Fernando, San Gabriel and San Juan Capistrano, as well as in the poorer sections of the growing town of Los Angeles, and were ministered to by clergy and lay workers of the Roman Catholic Church. establishment of lasting congregations in the town of Los Angeles, Protestant settlers were initially far less fortunate than Roman Catholics, Jews and Chinese. The greatest obstacle faced by pioneers of the Protestant tradition was the division of a small number of people into separate denominations. These "godly" folk struggled with a scarcity of clergy, a paucity of funds, local violence, and the distance from fellow believers. Lack of familiarity with Spanish also precluded outreach to the broader local community. Increased Ango American immigration was necessary for the survival of the churches so long identified with western frontier Christianity. Neither the circuit-riding Methodist nor the Baptist farmer-preacher could succeed initially in this isolated pueblo of Los Angeles. - The Rev. John Brier, a Methodist minister, was the first Protestant clergyman to preach in Los Angeles, in the spring of - The Rev. John W. Douglas, a New-School Presbyterian minister, arrived in Los Angeles a few months after the departure of Brier in 1850. Douglas departed for San Francisco in August 1851. - The Roman Catholic Diocese of Monterey-Los Angeles was established in 1853, which included the southern California counties, under Bishop Thaddeus Amat (1810-1878, a Spanish-born theologian of the Vincentian Order who had previously served in Missouri and Pennsylvania), who arrived in Los Angeles in late 1855. February 1853, the Northern Methodist Episcopal Church in San Francisco sent the Rev. Adam Bland to Los Angeles to begin Methodist work in the southland; he has replaced by the Rev. J. MacHenry Caldwell in 1854. Occasional Methodist worship services were held in private homes and rented halls, as well as a series of camp meetings for the Ango American settlers. - A group of Texas Baptists settled in the town of Lexington (now named El Monte) on the banks of the San Gabriel River in the 1850s. The first Baptist church was organized there in November 1853 with four members. During 1854, the Methodists and Baptists held joint services in Lexington. Episcopal Church services were first held in Los Angeles in October 1855 in a rented hall, where the Methodists also held their services. However, it was not until October 1857 that St. Luke's Parish was Hebrew Benevolent Society was organized in Los Angeles on 6 July 1854, with 30 members, and a Jewish Cemetery was established on 9 April 1855. preachers only occasionally supplied the Southern California District of the Methodist Episcopal Church, North between 1854 and 1869, when first permanent Methodist congregation was established: Fort Street Methodist Episcopal Church. In 1856, this district reported only 42 worshippers. - As early as 1854 a congregation of the African Methodist Episcopal Church existed in Los Angeles. The colored people of this denomination first held services at the house of Robert Owen (" Uncle Bob") in 1854. In 1869 a church was organized and a building was erected in April 1871, known as "Little Church on the Hill." The first members of this congregation were Mrs. Winnie Owen ("Aunt Winnie "), Mrs. Biddy Mason and Miss Alice Coleman. The Wesley Chapel (colored) was organized August 24, 1888, with twenty-three members and eighteen probationers; now there are fifty-six members and seventeen probationers. Services are held in a hall on Los Angeles street; Rev. F. H. Tubbs (white) has been the pastor of this body from its beginning. - On 18 March 1855, the Rev. James Wood, an Old-School Presbyterian minister, organized the first Protestant church in Los Angeles, but it lasted only about six months, due to the departure of Wood for greener pastures. - In May 1859, the "First Protestant Society of the City of Los Angeles" was organized as an interdenominational effort by the Rev. William E. Boardman, an Old-School Presbyterian minister. A modest church building was constructed in 1862 that was passed over to the Episcopalians in 1866, who later sold it to the County of Los Angeles in 1883. - On 2 November 1860, Boardman organized the First Presbyterian Church of Los Angeles, but only sporadic services were held until this congregation was reestablished in 1874 under the Rev. Fraser. - In 1864, the Rev. Elias Birdsall, a "missionary pastor" from Indiana, arrived to reestablish the Protestant Episcopal Church in Los Angeles. In March 1865, Birdsall established St. Athanasius Episcopal Church, which was renamed St. Paul's Episcopal Church in 1883. - In 1866, the American Home Missionary Society of the Congregational Church (related to New England Puritans) in San Francisco sent its first two missionaries to southern California: the Rev. R. A. Johnson to San Bernardino (originally a Mormon settlement) and the Rev. Alexander Parker to Los Angeles. - In 1867, the Rev. J. C. Miller of the Methodist Episcopal Church, South arrived in Los Angeles to establish the Los Angeles Circuit; Miller was succeeded by the Rev. Abram Adams in 1869. The center of their work was in the rural community of Los Nietos-Downey, but by 1869 Trinity Methodist Church was established in Los Angeles. - The First Congregational Church of Los Angeles was organized on 21 July 1867; the first church building was completed and dedicated on 28 August 1869, the Los Angeles Baptist Association was organized in El Monte (formerly known as Lexington), composed of five rural churches in the Los Angeles basin; by 1876, the Baptists had 16 organized congregations with 633 members in five counties of southern California. - In 1870, the Rev. William C. Harding arrived in Los Angeles, where he organized a short-lived Presbyterian Church with only eight members; however, Harding retreated from Los Angeles and reestablished his ministry in the community of Wilmington, near the port of San Pedro. By 1872, six Presbyterian congregations had been organized in southern California under the Presbytery of Los Angeles. After several years of frustrating ministry in Los Angeles by the Rev. Thomas Fraser, he was successful in reestablishing the First Presbyterian Church of Los Angeles on 11 January 1874. However, a permanent pastor for this church did not arrive until 1875, when the Rev. A. F. White (from Carson City, Nevada) was appointed. By 1870 there were only five Protestant churches in Los Angeles that were fully organized: Fort Street Methodist Church (Methodist Episcopal Church, North) with 40 members; Trinity Methodist Church (Methodist Episcopal Church, South); First Congregational Church with 36 members; St. Athanasius Episcopal Church (later known as St. Paul's Episcopal Church); and the African Methodist Episcopal Church (organized in 1869). In 1870 the town of Los Angeles only had 5,728 inhabitants. - In 1872, Carl Zahn, a German dentist, established an independent "deutsch-evangelisch" (German Evangelical) church in Los Angeles; this congregation later joined the Methodist Episcopal Church, North and part of its German Mission, in 1876. - On 8 September 1874, the First Baptist Church of Los Angeles was organized with eight members, under the leadership of the Rev. William Hobbs. He was followed by the Rev. Winfield Scott, who served as pastor first Chinese Temple ("joss house") was established in Los Angeles in 1875 among an estimated 170 Chinese immigrants, some of whom had lived there since the early 1850s as "house servants" for Anglo American settlers. Most of the Chinese population lived together in an ethnic barrio (the first "Chinatown") near the town Plaza. It was here that the famous "Chinese massacre" occurred on 7 October 1871, when a mob of Anglo Americans killed 19 Chinese in the neighborhood known as "Nigger Alley." Chinese religion at that time was a polytheistic blend of Buddhist, Confucianist and Taoist teachings and traditions, known as the "Three Teachings." Joss house or Miu is a place for worshiping a variety of indigenous Chinese deities, saints and supernatural beings from Taoist, Buddhism, Confucianism, heroes and folklore. Joss house is usually translated as temple, although it was in common use in English in western North America during frontier times, when joss houses were a common feature of places with Chinatowns. Joss houses are distinct from Taoist temples and Buddhist monasteries in that they are established by nearby villagers or fishermen to pray for good luck; only few or none of monks, nuns or priests study religion or stay in joss houses. Joss houses are usually small houses decorated with traditional figures on their roofs although some evolve into significant structures. The name "joss house" describes the environment of worship. Joss sticks, a kind of incense, are burned inside and outside of house. The Chinese character Miu means "ancestor hall," a place to worship ancestors. It is later extended to places for worshipping others. also see http://en.wikipedia.org/wiki/Chinese_language) - The First Christian Church/Disciples of Christ was founded in Los Angeles in 1874; previous congregations of this denomination had been founded in San Bernardino, Los Nietos-Downey and El Monte. - The Unitarian Church (Unitarian-Universalist) was informally established in Los Angeles under the leadership of Caroline M. Seymour Severance in May 1877; the congregation was inactive between 1880 and 1884 when Mr. and Mrs. Severance returned to Boston, but it was formally established in 1884 with the Rev. J. H. Allen as pastor and with the support of the Severances. - Trinity German Lutheran Church (Missouri Synod) was founded in Los Angeles in 1882. Later, other Lutheran churches were established that used English and Swedish for worship. - The Reorganized Church of Latter-Day Saints was organized in Los Angeles in the autumn of 1882, with about a dozen members; it had about eighty members in 1890. Worship was held in a hall rented by the society. 1862, the Jewish community of Los Angeles (about 200 persons) established the first Jewish synagogue with 32 charter members, Congregation B'nai B'rith, led by Rabbi Abraham Wolf Edelman and lay leader Joseph Newman. Mission was established in Los Angeles by T. P. and Manie Ferguson in - In 1887, the Roman Catholic Cathedral of St. Vibiana was dedicated at the corner of Second and Main Streets in Los Angeles; whereas most of the other Roman Catholic churches and Missions were constructed of wood, rock and adobe materials, the new Cathedral (see picture below) was made of steel and cement, and designed by Los Angeles architect Ezra F. Kysor. Several Theosophy groups (founded in New York City in 1875 by Madame H. P. Blavatsky, William Q. Judge and others) arrived in Southern California during the 1880s. A branch of The Theosophical Society existed in Los Angeles as early as 1886. Jerome A. Anderson was active on the Pacific Coast and was an author of elementary books on Reincarnation and Karma, Immortality, and Septenary Man. He was a frequent contributor to the pages of the New Californian, a Theosophical monthly founded in Los Angeles in 1891. The editor of this magazine, Miss Louise A. Off, was among the most active members on the Pacific Coast, writing on Theosophical subjects for the California newspapers as well as in the New Californian. She also conducted in her home well-attended weekly meetings for the discussion of Theosophy. The United Lodge of Theosophists was formed in 1909 in Los Angeles under the inspiration and guidance of Robert Crosbie, who was a Boston Theosophist during the time of William Q. Judge. He worked very closely with Judge, enjoying his confidence. When, after Judges death in 1896, the members most active at the New York headquarters raised Mrs. Katherine Tingley to the position of Judges successor, Crosbie gave her his loyalty and support. About 1900 he went to Point Loma, near San Diego, to be of assistance in the work of The Theosophical Society there, founded in 1900 by Mrs. Tingley. However, in the course of a few years, he came to feel that nothing constructive was to be accomplished by remaining at Point Lomathat, in fact, the teachings and philosophy of Theosophy had suffered an almost complete eclipse by the methods and sensational program instituted by Mrs. Tingleyand he quietly left the Point Loma Society in 1904 and came to Los Angeles. He secured work in Los Angeles and gradually began to gather around him a few studentsmost of them entirely new to Theosophyto undertake once more the task of promulgating Theosophy in the same form as originally presented by the founders of the movement. When, in 1909, he had been joined by a small nucleus of persons who shared this ideal, The United Lodge of Theosophists was formed to carry out the purposes in view. The Theosophy Company began in 1925 as a non-profit corporation devoted to the dissemination of theosophical literature and the publication of the quarterly journal, Theosophy (a synthesis of science, religion and philosophy), first published in 1912. German Evangelical Friedenskirche ("Church of Peace") was founded in Los Angeles in 1887 and had about 50 families by 1890. This church was named after the Peace of Westphalia of 1648, which ended both the Thirty Years' War in Germany and the Eighty Years' War between Spain and The Netherlands. This permitted the Lutherans in the Roman Catholic parts of the Roman Empire to construct their own churches, to be built outside the city walls and made of wood with no nails. the Seventh-Day Adventist Church had about eighty members in Los Angeles, and it also had churches in Pasadena, Norwalk and Santa Ana. arrived in southern California during the 1880s and established an agricultural colony at Loma Linda, near Redlands. In the fall of 1893, Plymouth Brethren ("Open Brethren" tradition) evangelist W. J. Mc Clure held tent meetings in Los Angeles, which led to the establishment of one of the earliest Brethren Assemblies in California at 806 Temple Street in Los Angeles. This assembly later moved to 1231 West Jefferson Blvd. and built the West Jefferson Gospel Hall, which remained at that location for many years. This assembly held Saturday night street meetings and sponsored tent meetings of one to two month's duration from time to time. For more information about their history, see: http://www.emmaus.edu/files/Documents/History%20of%20Brethren%20Mvt/history_3.htm 1870 the growth of religious groups in Los Angeles can be attributed to four specific ways: increased membership, the construction of new church buildings, the formation of new ecclesiastical jurisdictions, and the appearance of additional denominations. population of Los Angeles doubled from 15,309 in 1870 to 33,381 in 1880 as more Ango-American Protestant families arrived from the Midwest and East, thanks to the new intercontinental railway connections via San Francisco that reached Los Angeles in September 1876 and to major real estate promotional efforts to attract newcomers to "sunny southern California." Southern Pacific Railroad reached Los Angeles in 1876, followed by the Santa Fe Railroad nine years later. The two rival companies conducted a rate war that eventually drove the price of a ticket from the eastern United States down to five dollars. This price slashing brought thousands of settlers to the area, sending real estate prices to unrealistically high levels. By 1887, lots around the central plaza sold for up to one thousand dollars a foot, but the market collapsed in that same year, making millionaires destitute overnight. People in vast numbers abandoned Los Angeles, sometimes as many as three thousand a day. This flight prompted the creation of the Chamber of Commerce, which began a worldwide advertising campaign to attract new citizens. By 1890, the population had climbed back up to fifty thousand residents. the 1890s, oil was discovered in the City of Los Angeles, and soon another boom took hold. By the turn of the century almost fifteen hundred oil wells operated throughout Los Angeles. In the early 1900s, agriculture became an important part of the economy, and a massive aqueduct project was completed. The city's growth necessitated the annexation of the large San Fernando Valley, and the port at San Pedro was also added to give Los Angeles a position in the international trade market. to Engh, by 1889, 17 religious congregations had been established in Los Angeles by the denominations previously reported above, and four of these denominations had begun missionary activities among the Chinese, Hispanic, German, Swedish and French population: the Presbyterians, Northern Methodists, Congregationalists and Northern Baptists. Between 1880 and 1885, the Seventh-day Adventists, the United Presbyterians, the German Lutherans (Missouri Synod), various Holiness groups, the Church of Jesus Christ of Latter-day Saints (Mormons), the Reorganized Church of Jesus Christ of Latter-day Saints, Spiritualists, Theosophy and "a society of American Buddhists" established themselves in Los Angeles. Other religious groups created new settlements in nearby areas, such as the Quakers in Whittier, the Presbyterians in Westminster, and the German Brethren (now known as the Brethren Church) in Lordsburg, now known as La Verne). source states that, in 1890, Los Angeles contained forty-four church organizations, of twelve different denominations, besides a few representatives of other faiths, such as Spiritualism, Mohammedanism (Islam), Buddhism, Parseeism (a religious community of India, practicing Zoroastrianism), Confucianism, etc., and also an organization auxiliary to the National Secular Union (SOURCE: http://www.calarchives4u.com/history/losangeles/socal1890-770.htm). Protestants finally began to exercise an influence in the community of Los Angeles similar to what they had known in settlements extending from the Old Northwest to the Midwest. Gregory H. Singleton, in Religion in the City of the Angels: American Public Culture and Urbanization, Los Angeles, 1850-1930, presents an interesting discussion of the nature of the "voluntaristic" denominations that came to denominate the religious scene in Los Angeles at the turn of from Chapter 1 of Michael E. Engh's, Frontier Faiths: Church, Temple and Synagogue in Los Angeles, 1846-1888; for Plymouth Brethren sources, see: http://www.emmaus.edu/files/Documents/History%20of%20Brethren%20Mvt/history_3.htm) a more detailed history of the religious development of the City of Los Angeles, see: Churches in Los Angeles in 1890. Pentecostal Movement Arrives in Los Angeles in 1905-1906 has surely come and with it the Bible evidences are following," writes the editor of The Apostolic Faith: many are being converted and sanctified and filled with the Holy Spirit, speaking in tongues as they did on the day of Pentecost." The modern Pentecostal Movement began on January 1, 1901, with Agnes N. Ozman at Charles Fox Parham's Bethel Bible College in Topeka, Kansas. Then on April 9, 1906, at 214 Bonnie Brae Street, "the first Pentecostal effusion came" to Los Angeles. Although those events delineate the beginning of the modern Pentecostal Movement, their foundations are clearly found in the nineteenth-century Holiness Movement. Los Angeles, Joseph Smale, pastor of the First Baptist Church, after visiting Wales (then in the midst of a great revival) in 1905, began prayer meetings in his church modeled after what he had seen in Wales, and healings began to occur there. However, the things Smale was doing caused him trouble with his board, and he left to found a "New Testament Assembly," which met in a house on Bonnie Brae Street. Meanwhile, in April, 1906, at the instance of Neeley Terry, who had just visited Houston, the small black [Church of the] Nazarene congregation she attended invited [black Holiness preacher William J.] Seymour to preach in their church. He accepted the invitation, and preached his first sermon out of Acts 2:4 on the baptism of the Holy Spirit. Many in that Nazarene church believed Seymour to have preached false doctrine, and he returned that evening to find the door padlocked. Those who followed Seymour out of the Nazarene church started to meet in the home of some Baptists (from Smale's flock) on Bonnie Brae Street. On April 9, 1996, "the Spirit fell upon this small group of African-American believers." The group soon moved to a former Methodist church building at 312 Azusa St., where it met 3 times a day, 7 days a week for the next is noteworthy that California had by far the most diverse population of any state in the U. S. and had no apartheid laws requiring racial segregation of public meetings. What started at Azusa Street was entirely inclusive; under Seymour's leadership, the Azusa Street congregation would tolerate no racial or ethnic divisions in the Body of Christ. Although it started among a group of African-Americans, the Azusa Street meetings were completely interracial, and many whites became involved. Many people of all races, and from various countries, came to Azusa Street to observe or to receive the baptism of the Holy Spirit. Though Parham continued to preach in Houston and elsewhere, and his students also spread around the U.S., after April 1906 the focus of activity was in Los Angeles, not in Parham's The practice of glossolalia at Azusa Street was important for more reasons than its controversial nature. "The great significance of the Azusa Street revival," writes William Menzies, "is its role in transforming the embryo Pentecostal outpouring into a worldwide movement." Mills goes a step further: Azusa Street was not only the spark that ignited the Pentecostal revival, but in its earliest days, most of the doctrinal issues surfaced there that would later become determinative for the formation of the major Pentecostal groups. Those issues were: 1) doctrine of sanctification; 2) Jesus only doctrine; 3) latter rain covenant; and 4) race as a basis for denominational division. After 1909 the influence of Azusa Street began to fade. One reason for this was the success of many of the children of Azusa Street. Pentecostal missions sprang up all across Los Angeles. Many of these missions replicated the success of Azusa. Added to this were the many thriving centers of Pentecostalism across both the United States and Canada. It became increasingly unnecessary to travel to Azusa Street to Almost one hundred years later, the Azusa Street revival remains an important touchstone in the history of modern Pentecostalism. The building has long since been torn down. Today the Japanese-American Cultural and Community Center sits on the site where the Azusa Street mission was once located. However, Pentecostals fondly look on the site and the revival housed there as the cradle of Pentecostalism. famous Azusa Street Revival (1906 - 1913) in Los Angeles was a key milestone in the history of Christianity, and it helped to place the City of the Angeles on the world map as thousands of people came from far and wide to witness the new phenomena of "speaking in tongues," prophesy and divine healing. Los Angeles, with a population of about 300,000 in 1910, became known as the modern birthplace of the Pentecostal movement, which has had a significant worldwide impact as one of the fastest-growing religious movements on the planet. See the following websites for more information: http://www.ag.org/enrichmentjournal/199904/026_azusa.cfm Also, the interracial Azusa Street Revival touched the lives of hundreds of Afro-Americans and Hispanics in Los Angeles, and led to the establishment of several new Pentecostal denominations: the Pentecostal Assemblies of the World (originally interracial but later composed mostly of Afro-Americans) in 1907 and the Apostolic Assembly of Faith in Jesus Christ (Asamblea Apostólica de Fe en Cristo Jesús) in 1914 among Mexican immigrants and Mexican-Americans. Groups in Los Angeles & Environs in 1914 Most of the Protestant churches and denominations that existed in Southern California by 1914 were mainstream groups: Baptist (Northern Baptist Convention: 1853 in El Monte, 1874 in Los Angeles; Swedish Baptist Church in Los Angeles, 1887; Second Baptist Church in East Los Angeles, an Afro-American congregation, prior to 1890), Congregational (1865-67, American Home Missionary Society; the first Congregational Church in Los Angeles was organized in 1867 by the Rev. Alexander Parker), Protestant Episcopal (1857), Methodist Episcopal (1853), Presbyterian (1855) and the Christian Church/Disciples of Christ 1913, a Comity Agreement had been established in Los Angeles between the major Protestant denominations, whereby they agreed not to enter neighborhoods occupied by another denomination. By 1914, some of these denominations had begun to minister to the minority population of Mexican, Asian and Portuguese immigrants. Ten of thousands of Mexicans began arriving in the Los Angeles area after the beginning of the Mexican Revolution in 1910; this created an emergency refugee situation in Los Angeles, which was responded to by most of the Protestant denominations by organizing settlement houses, welfare services and a variety of Spanish-speaking ministries. Also, several other Protestant denominations had arrived in Los Angeles by 1914: the African Methodist Episcopal Church (1854), Lutheran (Trinity German Lutheran Church in Los Angeles, 1882), the German Evangelical Friedenskirche ("Peace Church," 1887), the Free Methodists in 1903, and various Holiness groups in between 1886 and 1906, including the nondenominational Peniel Mission (established by T. P. and Manie Ferguson in 1886); the Burning Bush Holiness Church; the First New Testament Church (Joseph Smale, formerly the pastor of the First Baptist Church, established this congregation in Burbank Hall at 542 South Main Street, Los Angeles, in early 1906); and the Household of God Church (1904-05, W. F. Manley, possibly linked to the Free Methodists). * * * Union Rescue Mission (URM) was established in 1891, dedicated to serving the poor and homeless in downtown Los Angeles. Today, the URM is one of the largest rescue missions of its kind in the United States and the oldest in Los Angeles. The mission provides a comprehensive array of emergency and long-term services to their guests, including: food, shelter, clothing, medical and dental care, recovery programs, transitional housing, legal assistance, education, counselling and job training to needy men, women, children and In 1902 Captain Charles Farr approached the Executive Committee of the Los Angeles City Christian Endeavor Union and asked it to support a Sailors Mission to be conducted by him in San Pedro, CA. After an investigation by the committee, it took over the work. Meetings were held aboard the Warrior, an abandoned tugboat moored on the east side of the San Pedro Bay. After about two years, a site was selected at 331 S. Beacon Street. A corporation was formed in 1905 under the name of Southern California Floating Christian Endeavor Association and the mission was named The Sailors Rest Mission. In 1945 the name was changed to Beacon Light Mission and it became a regular rescue mission. * * * first Jewish synagogue (B'nai B'rith) was organized in Los Angeles in 1862 by Rabbi A. W. Edelman, the Unitarians began to hold services in Los Angeles in 1877, and the Reorganized Church of Latter-day Saints in 1882. By 1890, the City of Los Angeles contained more than forty-four Christian churches, of at least twelve different denominations, besides a few representatives of other faiths, such as Spiritualism, Metaphysical-New Thought (including Christian Science and Theosophy), Islam, Buddhism, Parseeism (a Zoroastrian religious sect from India), Confucianism, Shinto, etc., and also an organization auxiliary to the American Secular Union and Freethought Federation (formed in Albany, NY, in 1885 by Colonel Robert Green Ingersoll and his associates). * * * Claremont School of Theology traces its history back to 1885 with the founding of the Maclay College of Theology in San Fernando, California. In 1900, Maclay College moved to the campus of the then Methodist-affiliated University of Southern California in Los Angeles. In 1956, the School withdrew from the University and became an independent corporation, related to the Southern California-Arizona Annual Conference of The United Methodist Church. The School moved to its present Claremont site in 1957. distinguished past presidents have provided extraordinary leadership and have set a strong foundation for Claremont School of Theology: Ernest Cadman Colwell (1957-1968), Gordon Elliott Michalson (1968-1977), Richard Wilson Cain (1977-1990), and Robert W. Edgar (1990 - 2000). Dr. Philip A. Amerson became the fifth president of the School on February 24, 2001. Dr. Jerry D. Campbell began as Claremont's sixth president in June 2006. SOURCE: * * * Society of Friends established the town of Whittier in 1887 and the Whittier Academy the same year. Whittier College grew from the academy and was chartered by the State of California in 1901 with a student body of 25. The college began construction on its first building, Founders Hall, in 1893. Initially, Founders Hall housed all classes, dormitories and the library. Both the town and the college were named in honor of John Greenleaf Whittier, prominent Quaker, poet, and leader in the abolitionist movement. the college is no longer affiliated with the Society of Friends, the college is proud of its Quaker heritage, which is evidenced in many ways, including respect for the individual, commitment to a diverse student body and faculty, freedom of conscience, and respect for human differences. * * * Los Angeles Pacific College was founded in 1903 as a four-year liberal-arts college by a group of ministers and laymen of the Free Methodist Church. The college ceased to exist as an independent entity in 1965 and was merged with another college to eventually form Azusa Pacific University. The founders of the college were the original founders of the community of Hermon, situated in a half square-mile valley bordered by the Arroyo Seco and the historic 110 freeway to the west, Monterey Hills to the south, and South Pasadena to the north and east. In 1903 a group of Free Methodists obtained the isolated valley from owner Ralph Rogers to establish a school. The school grew to become Los Angeles Pacific College in 1934, then merged with Azusa Pacific University in the 1960s. called Los Angeles Free Methodist Seminary, it was not a seminary for the education of ministers, but a school for young children of the community who wanted to raise their children in a Christian atmosphere. The Seminary (grades 1-12) opened in the fall of 1904 with 70 students. In 1911 the seminary added a junior college to its school, the first junior college in the state of California. As the community of Hermon continued to expand, a four-year college course was added in 1934 and the school came to be called Los Angeles Pacific College (LAPC). origins of Azusa Pacific University reside in 1899, when a group of spiritual leaders from various denominations met in Whittier, California, and established a Bible college geared to training students for service and missionary endeavors. This was the first Bible college founded on the West Coast. The initial class of students met on March 3, 1900, with Mary A. Hill serving as the earliest president. institution, named the Training School for Christian Workers, moved three times before settling in Huntington Park in 1907. In 1939, the Training School became Pacific Bible College, and four-year degrees were offered. Cornelius P. Haggard, Th.D., was appointed president and served for 36 years, until his death in 1975. the mid 1940s, Pacific Bible College had outgrown its Huntington Park campus. The Board of Trustees decided then to purchase a 12-acre school for girls in Azusa. Classes began on the new campus in 1947, and in 1956, the name was changed to Azusa College. College merged first in 1965 with Los Angeles Pacific College, a four-year liberal arts institution founded in 1903 by the Free Methodists, acquiring the name Azusa Pacific College, and again three years later with Arlington College, which had been founded in 1954. * * * The Adventists also arrived in southern California during the 1880s and established an agricultural colony at Loma Linda, near Redlands. The Seventh-Day Adventist Church had about eighty members in Los Angeles in 1890, along with churches at Pasadena, Norwalk and Santa Ana. September 29, 1913, the Adventist College of Medical Evangelists in Loma Linda (near Redlands, in San Bernardino County) opened a small storefront clinic at 941 East First Street, in the heart of Los Angeles. It was from these humble beginnings that White Memorial Medical Center was born. Three years later, the influx of patients was so great that there was a need to expand the clinic. The fundraising campaign began for a hospital to be built at a nearby site on Boyle Avenue and named in honor of Adventist prophetess Ellen G. White. Thanks to the Herculean efforts of 50 Adventist women, the Adventist Church purchased property on Boyle Avenue in 1916. In 1917, a new dispensary opened on the site. Meanwhile, construction began on cottage-style buildings that were to become a permanent hospital. On April 21, 1918, a crowd of 2,500 people gathered to dedicate White Memorial Hospital, built at a cost of $61,000. When the first patient entered White Memorial "Cottage" Hospital in 1918, its 11 one- and two-story buildings could accommodate up to 200 patients. It quickly emerged as the largest facility of its kind west of Chicago. By the mid-1930s, the initial jolt of the Depression had passed, and White Memorial Hospital began looking again to the future. Responding to an ever-growing demand on its original facilities, the hospital built a 180-bed, five-story concrete and steel structure at a cost of $330,000. Dedicated in 1937, the building was the first earthquake-resistant hospital in California. The Boyle Heights section of East Los Angeles was part of an area that eventually became the largest Hispanic community in the U.S. by 1950. Situated just east of the Los Angeles River, Boyle Heights has long been a gateway for newcomers to the city. From the 1920s to the 1950s it was Los Angeles' most heterogeneous neighborhood, serving as home to large concentrations of Jews, Mexicans and Japanese Americans, as well as Russian Molokans, African Americans, and people of Armenian, Italian, and Chinese descent. Today the neighborhood is primarily Latino, and it continues to serve as a port-of-entry for a number of the city's immigrant groups. * * * The University of La Verne was founded in 1891 as Lordsburg College by members of the Church of the Brethren. Both the surrounding agricultural community and the College were renamed La Verne in 1917. The College reorganized in 1977 as the University of La Verne. At present, the structure of the University consists of the College of Arts and Sciences, the College of Business and Public Management, the College of Education, the College of Law, and Regional Campuses. The school conferred its first master's degree in 1965 and began an adult education program in 1969. ULV awarded its first doctorate in 1979. In 1981, the University founded a campus in Orange County and has since opened campuses throughout Southern California. Today, the University of La Verne is an independent, nonsectarian and non-profit institution. SOURCE: East Los Angeles Church of the Brethren 3231 N. Broadway and Gates St., East Los Angeles, CA Photos and text below courtesy of Jeanie L. Woo In 1906, my grandmother, Caroline Dierdorff (Deardorff), moved with her family from Illinois to Los Angeles. They lived on Manitou Ave. I am sure you know there were many members of the Church of the Brethren of German descent who lived in Lincoln Heights at the turn of the century. They built the East Los Angeles Church of the Brethren on Hancock St. (234 S. Hancock St. now renumbered as 2218 N. Hancock St.) in 1896. I believe that church is now the First American Indian church. The Brethren also built the Berean Bible School, a three-story brick structure, at 3231 N. Broadway and Gates St. in 1911. That building no longer stands which surprises me since there are so many other vintage structures in the community. No one seems to know when, why, or how the building was removed from the site. are photos of the students, teachers, and leaders of the Berean Bible School ca. 1912. Berean Bible School, Chinese Sunday School, 1908-1951 A ministry of the East Los Angeles Church of the Brethren in Los Angeles, California row: L. to R., (third) Tom Yee Woo (student); (sixth) Clarence Lehmer (Superintendent); unidentified leader and students. row: far right, Elder Solomon G. Lehmer (Trustee); unidentified students and little girl. Berean Bible School, Minister and Missionary Training Branch, 1908-1916 A ministry of the East Los Angeles Church of the Brethren in Los Angeles, California Back row: L. to R., (fourth) Bessie Deardorff Lehmer (teacher); (sixth) (Caroline) Carrie Deardorff (teacher); unidentified teachers. row: L. to R., (fifth) Clarence Lehmer (Superintendent) with unidentified students. Berean Bible School, Minister and Missionary Training Branch, 1908-1916 A ministry of the East Los Angeles Church of the Brethren in Los Angeles, California Back row: L. to R., (second) Bessie Deardorff Lehmer (teacher); (third) Clarence Lehmer (Superintendent); (Evangeline) Rhea Deardorff (teacher); (thirteenth) (Caroline) Carrie Deardorff (teacher); unidentified teachers and leaders. Center row: unidentified students; Front row: unidentified students * * * Los Angeles Examiner, February 7, 1909 COSTLIEST CHURCH ON THE PACIFIC COAST NEARS COMPLETION When the Second Church of Christ, Scientist, on West Adams Street, near Hoover Street, is finished, Los Angeles can boast of having the largest and most magnificent church west of Chicago. Of Roman Corinthian architecture, it will be an imposing structure and will cost nearly a quarter of a million dollars. Church of Christ, Scientist Rosenheim, Architect, 1907-1910 at 948 West Adams Boulevard in the Historic West Adams District, the Second Church of Christ, Scientist is a most imposing edifice in the Beaux-Arts Classical Style. Authors David Gephard & Robert Winter in their authoritative work An Architectural Guidebook to Los Angeles claim the church was inspired by the Mother Church of Christian Science located in Boston, Massachusetts. The church was declared a Historic-Cultural Monument in the City of Los Angeles in 1968 (No. 57). Six massive Corinthian columns and a copper-clad dome are its most striking First Church of Christ, Scientist, was established by M. Paul Martin in Los Angeles in 1902; under Elmer Grey in 1911, First Church of Christ, Scientist, of Los Angeles moved to its second building (address unknown). Pacific Historical Review Vol. 72, No. 2, Pages 229-263 at the Golden Gate: Christian Scientists on the Pacific Coast, 18801915 There has never been a social history of Christian Science, a distinctive and controversial new religious group that emphasized metaphysical healing. The group appeared in the United States in the 1870s and 1880s under the leadership of Mary Baker Eddy. This article examines the early rapid growth of Christian Science on the Pacific Coast, for the religion flourished to a greater degree in this health-conscious and socially fluid region than in any other section of the world. Analysis of the occupations of more than 1,000 members and spouses of six Christian Science churches in California, Oregon, and Washington for the years 1905-1907 provides detailed conclusions at variance with previous conjecture. The new evidence shows that Christian Scientists on the Pacific Coast were an ethnically homogeneous, uprooted, and energetic lot from all social levels, with a surprisingly large contingent from the working classes. Science is ''a religion and a system of healing founded by Mary Baker Eddy c. 1866, based on an interpretation of the Scriptures asserting that disease, sin, and death may be overcome by understanding and applying the divine principles of Christian teachings." -- Webster Dictionary publication in 1875 of Science and Health, Eddys primary work on spirituality and healing, readers began meeting to discuss the ideas and share their healing results. Then, in 1879, Eddy established what became The First Church of Christ, Scientist (The Mother Church). Church is designed "to commemorate the word and works of our Master, which should reinstate primitive Christianity and its lost element of healing." (Church Manual, page 17). Eddy had a lifelong reverence for the life and teachings of Jesus Christ and a deep desire that his healing works be universally practiced. consists of The First Church of Christ, Scientist, in Boston, Massachusetts, and around 2,000 branch churches and societies of Christ, Scientist, worldwide. Church has no ordained clergy. In 1895, Eddy named the Bible and Science and Health with Key to the Scriptures as Pastor for worldwide Churches of Christ, Scientist. Holiness Movement in Southern California of the Holiness Movement Originating in the U. S. in the 1840s and 50s, this was an endeavor to preserve and propagate John Wesley's teaching on entire sanctification and Christian perfection. Wesley held that the road from sin to salvation is one from wilful rebellion against divine and human law to perfect love for God and man. Following Wesley, Holiness preachers emphasized that the process of salvation involves two crises. the first, conversion or justification, one is freed from the sins he has committed. In the second, entire sanctification or full salvation, one is liberated from the flaw in his moral nature that causes him to sin. Man is capable of this perfection even though he dwells in a corruptible body marked by a thousand defects arising from ignorance, infirmities, and other creaturely limitations. It is a process of loving the Lord God with all one's heart, soul, and mind, and it results in the ability to live without conscious or deliberate sin. However, to achieve and then remain in this blessed state requires intense, sustained effort, and one's life must be marked by constant self renunciation, careful observance of the divine ordinances, a humble, steadfast reliance on God's forgiving grace in the atonement, the intention to look for God's glory in all things, and an increasing exercise of the love which itself fulfils the whole law and is the end of the commandments. the mid-nineteenth century several factors converged that contributed to the renewal of the Holiness emphasis, among them the camp meeting revivals that were a common feature in rural America, the Christian perfectionism of Charles Finney and Asa Mahan (the Oberlin theology), the "Tuesday Meeting" of Phoebe Palmer in New York, the urban revival of 1857 - 58, and protests within the Methodist churches about the decline of discipline which resulted in the Wesleyan Methodist secession in 1843 and Free Methodist withdrawal in 1860. These two became the first denominations formally committed to Holiness. After the Civil War a full fledged Holiness revival broke out within the ranks of Methodism, and in 1867 the National Camp Meeting Association for the Promotion of Holiness was formed. From 1893 it was known as the National Holiness Association (NHA) and in 1971 was renamed the Christian Holiness Association. Until the 1890s Methodists dominated the movement and channeled its enthusiasm into their The increasing number of Holiness evangelists, many of whom were unsanctioned by their superiors, a flourishing independent press, and the growth of nondenominational associations gradually weakened the position of mainline Methodism in the movement. By the 1880s the first independent Holiness denominations had begun to appear, and tensions between Methodism and the Holiness associations escalated. The gap between the two widened as Methodist practice drifted steadily toward a sedate, middle-class American Protestantism, while the Holiness groups insisted they were practicing primitive Wesleyanism and were the true successors of Wesley in America. The small schismatic bodies gradually coalesced into formal denominations, the largest of which were the Church of God, Anderson, Indiana (1880), Church of the Nazarene (1908), and Pilgrim Holiness Church (1897, merged with the Wesleyan Methodists in 1968 to form the Wesleyan Church). polity of these bodies was a modified Methodism in that there was generally somewhat more congregational autonomy, and the "second blessing" of entire sanctification was an integral part of their theology. Most operated with a strict perfectionist code of personal morality and demanded from their adherents plain dress and abstinence from "worldly" pleasures and amusements. Also, nearly all of them allowed women to be ordained to the ministry and occupy leadership positions. Holiness Movement on the Pacific Coast By the late 1870s some rural holiness preachers were organizing their converts into holiness "bands" independent of the regular denominations, and more local and regional "associations" were sprouting. By the 1880s the first independent holiness churches had begun to form. In California a radical Methodist named Hardin Wallace, who had already evangelized throughout Texas, began to preach in Los Angeles and elsewhere, often together with evangelist Harry Ashcraft and gospel singer James Jayns. Out of their work the Southern California and Arizona Holiness Association was formed, led by James and Josephine Washburn. This organization was very strict: all members had to experience sanctification; all had to dress plainly, abstain from tobacco and the use of gold ornaments, and abjure membership in any secret society. They erected plain buildings and forbade musical instruments in church; ordination was by the "baptism of fire," with no preacher designated before the service. The Southern California and Arizona Holiness Association remained quite small, establishing churches in only a handful of southern California towns; a more moderate organization, the Pacific Coast Holiness Association, appeared in 1885. Nevertheless the radicals caused concern in the Methodist church when in 1885 one of their leaders, B. A. Washburn, proposed that all holiness groups separate from the mainstream churches. The California Methodist establishment was not at that time antagonistic to the holiness movementquite the contrary. The radicals, however, felt alienated from the regular Methodists. Eventually (1896) they organized into the Holiness Church, which continued to be very The urban sector of the movement, more intellectual and interdenominational, less concerned about regulating details of outer behavior, had so far stayed within the Methodist church. Nevertheless tensions were building. Holiness Christians were inclined to ally with those in other denominations, at least for revivals and general meetings. They wanted more evangelism focusing specifically on sanctification, whereas the Methodist bishops believed all church activities were already designed to promote holiness, and no special means should be instituted. Meanwhile many churches supported activities such as fairs, plays, and concertsnot to mention higher biblical criticismwhich, to holiness people, were tangential to the Christian life. Those seeking to help the poor through missions to urban families and neighborhoods were not getting much support for their efforts. The stage was set for a split in the was rapidly being urbanized as Los Angeles grew, and developments there were similar to those in other large cities. The city mission approach was vigorously represented by the work of T. P. and Manie Ferguson. T. P. Ferguson, born in Ohio in 1853 and converted at Oberlin in 1875, came to Santa Barbara in 1879 and soon thereafter was sanctified at a holiness revival. He became an itinerant preacher and settled in Los Angeles during the boom of 188586. Late in 1886 he set up the Los Angeles Peniel Mission, the first in what would become a chain of Peniel ("Face of God") Missions dotting the Pacific Coast and mountain states. Together with his wife Manie, he offered street-corner meetings in the afternoons and evangelistic services nightly, with a meal afterwards. Their entire work, like that of most of the city holiness missions, was oriented toward soul saving and the promotion of holiness. The mission was not a church, however; converts were supposed to join one of the regular denominations. It was, rather, a holiness revival station spreading the message of Christian perfection. development in Southern California came when Phineas F. Bresee arrived on the scene. Born in 1838 in Delaware County, New York, he had gone to Iowa in 1855 as a circuit pastor, and was highly successful for a time. In the early 1880s, however, he went bankrupt due to the failure of some Mexican iron mines in which he had invested, and he left Iowa for California, arriving in 1883. Soon he won fine appointments in the Methodist church, notably as pastor of First Methodist Church in Los Angeles and as one of the editorial committee of the Southern California Christian Advocate himself with the holiness movement and experienced sanctification himself in 1884 or 1885. In his church he emphasized revivals, gospel singing, and spontaneous congregational responses. Some ministers opposed his outright holiness stance, but he was supported by his general popularity and the approvalor at least the neutralityof the bishops until 1892. In that year an anti-holiness clergyman, John Vincent, became Bishop, and he assigned Bresee to churches that could not offer adequate financial support. In 1894 Bresee sought a supernumerary relation so that he could do mission work instead of a regular pastorate, but Vincent refused permission. At that point Bresee withdrew from the Methodist ministry. At first Bresee joined with the Fergusons at the Peniel Mission in Los Angeles, where he tried to persuade them to open a school and organize to receive members like a church. They refused, however, and other difficulties led to his parting with them after one year. In the fall of 1895 he, together with Joseph P. Widney, began holding independent services in a rented hall. Their ministry was so popular that three and a half weeks later they organized as a church, the Church of the Nazarene. Bresee and Widney were appointed to life tenure as pastors and superintendents. * * * Church of the Nazarene The Church of the Nazarene, a holiness body, was founded in 1895 in the Los Angeles area by Dr. Phineas F. Bresee (1838-1915) and Dr. Joseph P. Widney (1841-1938). Their primary purpose was to bring the Gospel to the poor and underprivileged. Widney came up with the name for the new church. He explained the choice of the name had come to him one morning after spending the whole night in prayer. He said that the word "Nazarene" symbolized "the toiling, lowly mission of Christ. It was the name that Christ used of Himself, the name which was used in derision of Him by His enemies, the name which above all others linked Him to the great toiling, struggling, sorrowing heart of the world. It is Jesus, Jesus of Nazareth to whom the world in its misery and despair turns, that it may have hope" (Called Unto Holiness, Volume I). after moving to Iowa from New York State in 1856, was granted a district preacher's license by the Methodist Episcopal Church, North. Soon he was given his own church to pastor. He had a difficult time there, which was pleasantly interrupted by a trip back to New York in 1860, where Phineas married Maria Hibbard, the sister of a close friend of his. Shortly after the Civil War broke out, Bresee was ordained by the Methodist Episcopal Conference, which meant that he was now a full minister in the Methodist church. years that followed were rather difficult for the Bresee family. Numerous different preaching assignments and other occupations were assigned to Mr. Bresee, and the family lived in poverty most of the time. In 1883, they decided to move to California. With their six children they made the eight-day trip in a train wagon, into which they had stuffed most of their belongings. Pasadena, near Los Angeles, Phineas became the minister of a Methodist church, which grew strongly under his leadership. At the same time he was heavily involved in addressing social issues, such as the liquor business, which brought him many threats, but also contributed to his preaching, as more and more people became Christians and members of his church. Yet, Dr. Bresee (by this time he had received an honorary degree from the University of Southern California, established in Los Angeles in 1880) felt a calling for a new ministry - reaching out to the poor, the needy. He helped organize (along with T. P. and Manie Ferguson) a nondenominational project, which they called "Peniel Mission." The leaders of the Methodist church did not like this project. They feared it might hurt the image of the church. So Dr. Bresee was forced to either give up the mission or leave the church. After a night of struggle, he decided to leave the Methodist Church. October 20, 1895, the first Church of the Nazarene was organized in Pasadena, with 135 charter members who pledged to commit their lives to the work of Jesus the Nazarene. By the end of the first year, 350 people had joined the church. After five years the membership had increased to almost 1,000. New churches were started, and other groups joined the fledging Church of the Nazarene. sought to return to John Wesley's original goals of preaching to the poor and needy. The original name of the denomination was the Pentecostal Church of the Nazarene, however the term "Pentecostal" soon proved to be problematic. In the Wesleyan-holiness movement, the word was used widely as a synonym simply for "holiness." But the rise of 20th century Pentecostalism, especially after 1906, gave new meanings and associations to the term--meanings that the Nazarenes rejected. In 1919, the name was shortened to avoid any confusion in the public mind about the church's place on the theological spectrum. Widney was a medical doctor and the second President (1892-1895) of the University of Southern California in Los Angeles. Prior to that, he was the first dean of the USC College of Medicine. He was the brother of Robert Maclay Widney, one of the founders of USC. * * * Fundamentalism, which was decidedly anti-Liberal and anti-Pentecostal, received its name and crucial promotion in Los Angeles. In 1909, Lyman Stewart and his brother Milton (co-owners of the Union Oil Company of California, currently known as Unocal) anonymously funded the publication of a twelve-volume series of articles called The Fundamentals, published between 1910 and 1915, and distributed free of charge to a wide range of Christian teachers and leaders, "Compliments of Two Christian Laymen." These volumes were intended as a restatement of conservative Protestant theological teachings, primarily in response to the growing influence of modernist theology in the Protestant churches. In 1917 these articles were republished in a revised, four volume set by the nondenominational Bible Institute of Los Angeles (BIOLA). The term "Fundamentalism" is in part derived from these volumes. founded in 1908 by Lyman Stewart and T. C. Horton, a well-known preacher and Christian writer. By 1912, the school had grown sufficiently in its outreach and constituency to call Dr. Reuben A. Torrey (1856-1928), a well-known leader in the field of Christian education, as the first dean. Dr. Torrey previously had been president of Moody Bible Institute in Chicago and had conducted many well-publicized evangelistic crusades across America and in Great Britain. Between 1912 and 1928, BIOLA was an established leader in conservative Protestant Christianity in North America, publishing The King's Business (a magazine similar to Christianity Today), operating one of the largest Christian radio stations in the U.S. (KTBI), and running the BIOLA Press, which sold and distributed Christian literature worldwide, including material for the Los Angeles- based Pentecostal preacher, Amy Semple McPherson. After Stewart's death and Torrey and Horton's retirements, William P. White, a well-known Christian leader and speaker, became BIOLA's first president in Torrey also helped to organize and served as the first pastor of the non-denominational Church of the Open Door (1915-1924). There he preached to great throngs and thousands were trained at the school, including Charles E. Fuller (1887-1968), famed radio preacher of the next generation. For decades, the Church of the Open Door was the largest Protestant church in Los Angeles, located adjacent to BIOLA at Fifth and Hope streets. Fuller was the radio pastor of "The Old-Fashioned Revival Hour" (1937-1968); for nearly 17 years (1941-1958), beginning with World War II, the program was broadcast each Sunday afternoon from the Municipal Auditorium in Long Beach, where it drew huge audiences. At the time of Dr. Fuller's death, the broadcast was heard on more than 500 stations around the world. Charles E. Fuller, a graduate of BIOLA, became chairman of the board and he later founded the nondenominational Fuller Theological Seminary in Pasadena, which later became one of the largest Protestant seminaries in the world. Meanwhile, BIOLA fell into hard times during the Great Depression and was forced to sell its publishing company and radio station. The 13-story downtown building that housed the school was also under threat of loss. It was during this time that Dr. Louis T. Talbot became BIOLA's second president in 1932. Talbot also served as the pastor of the Church of the Open Door, which held services in the school's downtown building, with its famous red neon "Jesus Saves" sign on the roof. In 1935, Paul W. Rood became BIOLA's third president. He was instrumental in establishing the Torrey Memorial Bible Conference, which is one of the longest standing Bible conferences today. He resigned in 1938. During Rood's presidency, Talbot was instrumental in helping to save the school from financial ruin caused by the Great Depression. entered a second term as BIOLA's president from 1938 to 1952. During this time, the Institute program became a four-year course, leading to degrees in theology, Christian education and sacred music. The School of Missionary Medicine came into being in 1945, laying the foundation for BIOLA's current baccalaureate nursing program. In 1946, Talbot also established the Biola Institute Hour, a national radio program, that was later called the BIOLA Hour. The Institute was renamed BIOLA College in 1949. Under the leadership of Samuel H. Sutherland, president from 1952 to 1970, BIOLA moved its campus to its current location in La Mirada in the summer of 1959, where it later became an accredited four-year evangelical university. * * * Groups and Activities in Los Angeles after 1914 Midnight Mission established in 1914 Mission is one of the oldest continuously operating human services organizations in the Los Angeles region. Centered in the Skid Row area of downtown Los Angeles, the Mission runs one of the most efficient direct service operations in the country. With only four executive managers through out its ninety-year history, the Mission has been a consistent beacon of light for those with no where else to turn. The story begins with the founder of The Midnight Mission - Tom Liddecoat. In 1914 this kind man, nicknamed "father of the poor," opened the doors of the Mission as a refuge to the men of Skid Row. A successful business man and lay minister, Liddecoat would serve a meal at midnight (hence the origin of the name) after church services were completed. Realizing the need to offer more than a meal and that additional resources were necessary Liddecoat sought help from the local community. as a non-profit in 1922, The Midnight Mission named a Board of Directors and continued to expand their services with showers, shaves and haircuts. Religious services were no longer a requisite and the organization began to focus on the rehabilitation of men and boys. Billy Sunday Comes to Town Fundamentalist evangelist William "Billy" Ashley Sunday (1862-1935) held an evangelistic campaign in Los Angeles during September and October of 1917. He returned to the Los Angeles area for a series of meetings in 1931 and 1934. to evangelical Christianity in the 1880s, Sunday left his [major league] baseball career for the Christian ministry. He gradually developed his skills as a pulpit evangelist in the Midwest and then, during the early 20th century, he became the nation's most famous evangelist with his colloquial sermons and frenetic delivery. He became an ordained Presbyterian minister in 1903. 1910, Sunday began to conduct meetings (usually longer than a month) in small cities like Youngstown, Bend, and Denver, and then finally, between 1915 and 1917, in the major cities of Philadelphia, Syracuse, City, Detroit, Boston, and New York City. During the 1910s, Sunday was front page news in the cities where he held campaigns. Newspapers often printed his sermons in full, and during World War I, local coverage of his campaigns often surpassed that of the war. Sunday was the subject of over sixty articles in major periodicals, and he was a staple of the religious press regardless of denomination. was welcomed into the circle of the social, economic, and political elite. He counted among his neighbors and acquaintances several prominent businessmen. Sunday dined with numerous politicians, including Presidents Theodore Roosevelt and Woodrow Wilson, and counted both Herbert Hoover and John D. Rockefeller, Jr. as friends. During and after the 1917 Los Angeles campaign, the Sundays visited stars, and members of Sunday's organization played a charity baseball game against a team of show business personalities that included Douglas held heavily reported campaigns in America's largest cities, made a great deal of money, and was welcomed into the homes of the wealthy and influential. Perhaps more than a million people came forward at his invitations, and he may have personally preached the gospel of Jesus Christ to more people than any other person in history up to that time. Sunday was a strong supporter of Prohibition, and his preaching almost certainly played a significant role in the adoption of the Eighteenth Amendment in SOURCES: adapted from See also http://billysunday.org/ http://www.wheaton.edu/bgc/archives/GUIDES/061.htm Aimee and the Foursquare Gospel In the 1920s, the flamboyant Pentecostal evangelist Aimee Semple McPherson (1890-1944) established a thriving ministry at the $1.5 million Angelus Temple in the Echo Park district of Los Angeles. This church created notoriety by allowing both Blacks and Whites to become members, as well as people of many nationalities, including Mexican immigrants. She also developed an international radio ministry under the auspices of the International Church of the Foursquare Gospel, which today has affiliated churches in 83 countries and claims more than two million "Sister Aimee" went she was an immediate success. The novelty of a woman preacher brought out the crowds, but McPherson's power as a speaker and her reputation as a formidable "soul-saver" and healer built her reputation. In 1913, she embarked upon a preaching career in Canada and the US. In keeping with a promise she made to God during a serious illness, she began evangelizing and holding tent revivals, first by traveling up and down the eastern part of the US, then expanding to other parts of the country. Finally, in 1919, McPherson found her home base in the rapidly expanding City of Los Angeles, where the movie business was booming. She frequently recalled that she arrived there with "ten dollars and a tambourine" and her ministry quickly grew from a simple storefront to large auditoriums. "Sister Aimee" did not promote herself as a healer, but the crowds came in hope of miracles. She herself said, "Jesus is the healer. I am only the office girl who opens the door and says, 'Come In."' Sister Aimee, ca. music, and she is credited with bringing popular music into the churchjazz in particular. She later composed operas, a natural outgrowth of her performances in the pulpit, which were elaborate spectacles featuring "Sister Aimee" in costume, props (which included animals) and a supporting cast of followers. In just four years she opened the 5,300 seat Angelus Temple in 1923, built by the contributions of her faithful followers, "entirely debt-free" as she proudly asserted. McPherson allegedly became the first woman in history to preach a radio sermon; and, with the opening of Foursquare-owned radio station KFSG on February 6, 1924, she became one of the first women in the USA to be granted a broadcast license by the Federal Radio Commission (which became the Federal Communications Commission in 1934). the Great Depression, McPherson was active in creating "soup kitchens," free clinics and other charitable activities; with the outbreak of World War II, she became involved in war bond rallies. On September 27, 1944, shortly after giving a sermon, she was found dead in her hotel room in Oakland, California, of an overdose of prescription barbiturates. Once again, rumors flew, this time conjecturing suicide. However, it is generally agreed that the overdose was accidental, as stated on the more information, see: http://www.pbs.org/wgbh/amex/sister/filmmore/fd.html * * * Fellowship is a religious organization founded by Paramahansa Yogananda (1893-1952) in 1920 and based in Los Angeles, California. The group carries on Yogananda's teachings, including Kriya Yoga, a form of yoga the group claims originated millennia ago in India. of Trinity Methodist Church in Los Angeles Pierce Shuler (1880-1965): A short biography of an out-spoken Methodist preacher Shuler was born August 4, 1880, in the foothills of the Blue Ridge Mountains of Virginia. At the age of nine, kneeling between his mother and his preacher-uncle in "the meetin' house" at Comer's Rocks, he received Christ to be his Lord and Saviour. His primary education consisted of a three- month school, where he mastered the McGuffey's Readers. In 1897 he entered Emory and Henry College as a sub-freshman, and was graduated in 1903. Two years later he married Nelle Revees, and the same year entered the Holston Conference of the Methodist Church. Endowed with a good mind and an even better wit, he was an excellent extemporaneous speaker. In addition to this, his great courage, coupled with his conservative theology and evangelistic fervor, prompted him to ever preach with the altar call in view. In 1920 he became pastor of the Trinity Methodist Church in downtown Los Angeles (organized in 1869), a position he occupied until his death. He began with a depleted congregation and saw it grow to 5,000 in the 1930s. In 1929, he was given a radio station that was housed in the tower of his church. It became a strong voice against crime and corruption in Southern California. His life was threatened many times, his church was bombed, he was sued and put in jail. He ran for United States Senator on the Prohibition ticket in 1932 and lost by only 50,000 votes. writings included The Methodist Challenge, What New Doctrine Is This?, Some Dogs I Have Known, and I Met Them on the Trail. Three of his sons followed him in The growth of Shuler's church paralleled the growth of the population on the West Coast with its "rootless" people from all parts of America. These masses found in him a "champion of the common man," for Shuler's cry against corruption was the complaint of the masses. The politicians hated Shuler and tried every means to silence his preaching. His life was threatened, his church was bombed, he was sued and finally put in Wikipedia, the free encyclopedia Party candidate who received the highest vote in any election in U.S. history was Rev. Robert P. Shuler. In the 1932 California election for the US Senate" US Senate, he received 560,088 votes (25.8%) and carried Orange and Riverside counties. Following his defeat, Shuler placed an awful curse on Southern California. Bob Shuler owned radio station KGEF, which existed from 1926 to 1932. He said that KGEF stood for Keep God Ever First and your Kind Gentle Emphatic Friend. The temperance movement leader lost the broadcasting license for his station in 1932 after his controversial broadcasts attacking Catholics, Jews, African Americans, and the Hollywood elite for their consumption of alcoholic beverages and their alleged dishonesty, corruption, and immorality (1). Shuler was pastor of the Trinity Methodist Church, South in Los Angeles, California. He is unrelated to the pastor of the Crystal Cathedral in Garden Grove, California. Prohibition Party candidate who received the highest vote total in a single election was Rev. Robert P. Shuler in a 1932 California race for the US Senate. He garnered 560,088 votes (25.8%) and carried Orange and Riverside counties. He had previously played a key role in exposing corruption in other states. He was one of those involved in the investigation which led to the ouster of Gov. Ferguson in Texas. thought I knew all the meanings of the early Los Angeles radio stations, but recently, I did a websearch on KGEF in Los Angeles (1926 to 1932), and found a website that had the slogan for KGEF radio. KGEF was owned by Rev. Robert P. Shuler of Trinity Methodist Church in downtown L.A. He was also known as "Fightin" Bob Shuler. story on the website says KGEF stood for: K)eep G)od E)ver F)irst That may be entirely possible, though I have never seen it in print in any of the Los Angeles radio magazines of the day....Unknown if it was heard on the air on KGEF, but that is likely. These slogans were mostly created after the call letters were assigned, and KGEF got their call letters in December 1926, assigned in sequential order. Shuler lost his license for KGEF in 1932, due to his controversial broadcasts attacking Jews, Catholics, Blacks, and going after the sinners in the L.A. Hollywood community who he deemed to be corrupt, dishonest, immoral and such. found another meaning for KGEF from an individual who did a thesis on the station in 1975 for journalism class. He found KGEF not only stood for K)eep G)od E)ver F)irst, but they also made up this slogan: K)ind, G)entle, E)mphatic F)riend. I'm not sure if they mean the radio station or the pastor of the church, Bob Shuler, who owned the station and was pastor of Trinity Methodist Church where the KGEF studios were. Rev. Robert P. The Prohibition Party candidate who received the highest vote in any election in U.S. history was Rev. Robert P. Shuler. In the 1932 California election for the US Senate he received 560,088 votes (25.8%) and carried Orange and Riverside counties. Following his defeat, Shuler placed an awful curse on Southern California and some people attributed a later earthquake in that region to his curse. Bob Shuler owned radio station KGEF, which existed from 1926 to 1932. He said that KGEF stood for Keep God Forever First. The temperance movement leader lost the license for his station after his controversial broadcasts attacking Catholics, Jews, African Americans, and the Hollywood elite for their consumption of alcoholic beverages and their alleged dishonesty, corruption, and immorality. However, there is no evidence that he was a member of the Ku Klux Klan, which also strongly supported Prohibition. Shuler was pastor of the Trinity Methodist Church in Los Angeles, California. In the 1940s, Charlotta A. Bass (1880?-1969), the African American editor of The California Eagle, discovered that the powerful pastor, Robert Shuler had aided the Klan from his bully pulpit at his Los Angeles Trinity Methodist Church With his fire and brimstone Air Raids from the Pulpit radio shows, Shuler delivered vivid scriptural revelations aimed at civil rights leaders and minorities, including Roberto Galvan [1911-1958, a labor union organizer and tireless worker for human rights], calling them criminals who spoil paradise. Historian Kevin Starr has labeled Shuler the Methodist Savonarola of Los Angeles, referring to the Dominican priest who preached against the moral corruption of the clergy in the early and his close friend, John Clinton Porter, mayor of Los Angeles from 1929 to 1933, insisted that civil rights leaders would bring about Armageddon. They also fought against relief programs to aid those in poverty. Material Dreams: Southern California Through the 1920s is the foremost chronicler of the California dream and indeed one of the finest narrative historians writing today on any subject. The first two installments of his monumental cultural history, "Americans and the California Dream," have been hailed as "mature, well-proportioned and marvelously diverse (and diverting)" (The New York Times Book Review) and "rich in details and alive with interesting, and sometimes incredible people" (Los Angeles Times). Now, in Material Dreams [published by Oxford University Press in 1999], Starr turns to one of the most vibrant decades in the Golden State's history, the 1920s, when some two million Americans migrated to California, the vast majority settling in or around Los Angeles. In a lively and eminently readable narrative, Starr reveals how Los Angeles arose almost defiantly on a site lacking many of the advantages required for urban development, creating itself out of sheer will, the Great Gatsby of American cities. He describes how William Ellsworth Smyth, the Peter the Hermit of the Irrigation Crusade, the self-educated, Irish engineer William Mulholland (who built the main aqueducts to Los Angeles), and George Chaffey (who diverted the Colorado River, transforming desert into the lush Imperial Valley) brought life-supporting water to the arid South. He examines the discovery of oil, the boosters and land developers, the evangelists (such as Bob Shuler, the Methodist Savanarola of Los Angeles, and Aimee Semple McPherson), and countless other colorful figures of the period. There are also fascinating sections on the city's architecture the impact of the automobile on city planning, the Hollywood film community, the L.A. literati, and much more. By the end of the decade, Los Angeles had tripled in population and become the fifth largest city in the nation. In Material Dreams, Starr captures this explosive growth in a narrative tour de force that combines wide-ranging scholarship with captivating prose. Strange Case of Maurice M. Johnson A religious group known as The Church Which is Christ's Body was founded in 1925 in Los Angeles by Maurice McArdle Johnson [1893-1979], a former minister of the Methodist Episcopal Church, South [MECS]. He was licensed to preach by the MECS in Texas in 1912 and moved to California in 1921, where he served as an assistant to the Rev. Robert [Fighting Bob] Pierce Shuler [1880-1965] at Trinity Methodist Church in downtown Los Angeles, from 1921 to 1923. He was known as a gifted singer and preacher and served as a MECS Conference Evangelist and pastor during part of 1923-1925. the Fall of 1925, Johnson left the MEPS with about 75 followers and established an independent Fundamentalist church, Maranatha Tabernacle, in nearby Glendale. Then, in 1927, he renounced all formal denominational structures with their salaried pastors and began to form house churches, which he called The Church which is Christs body, led by laymen who were called to preach and teach a New Testament message along the lines of the Exclusive Plymouth Brethren Assemblies. Johnson and his associates are known as undenominational Christians, today they have affiliated assemblies in California, Texas, Oklahoma, Missouri, Maryland, Virginia, Mexico and Central America; their mission work in El Salvador is known as "Christian's who meet in the Name of the Lord" -- "Cristianos congregados en el Nombre del Señor." There are no formal headquarters and each affiliated group is an autonomous assembly. At the time of Johnsons official retirement in 1972 at age 79, he and his wife were living in Orangevale, CA, in Sacramento County, but until 1969 his ministry was centered in the Los Angeles metro area. In 1972, he turned over his radio ministry to his associates Berl Chisum, Jack Langford and James Cox. The current presiding elder is alleged to be Robert A. Grove (known as RAG), President and Chairman of "Robert A. Grove Ministries, Inc." (A Virginia Corporation), 149 Edgemoor Street, San Leandro, CA 94579-1414. Adapted from A Classification System of Religious Groups in the Americas by Clifton L. Holland, pages 55-56; see: http://www.prolades.com/clas-eng.pdf information, see the following websites: http://www.churchgrowth.cc/Holy%20Love.htm http://www.churchgrowth.cc/April%201927.htm http://www.bibletruths.org/ http://www.bibletruths.org/contacts.html familiar with the "non-denominational" no-name church assemblies that go by "The Church Which is Christ's Body" or "Christ's True Church"? There are groups in California, Texas, Virginia, Maryland, Canada, Mexico and Peru. Maybe a couple thousand members, all led by Robert A. Grove (California, was in Virginia), and his sons Scott Grove (Virginia) and Jeff Grove (Texas). This church was founded by Maurice Johnson in the 1920's in California, and has been under the leadership of Robert Grove since the 1970's. purports to be a manifestation of the church from Christ's time, and requires of its members utmost subservience to the leaders and complete conformity to a rigid, hyper-orthodox dress code, speech code and behavior code. There is no room for any individual thought or life choices. Members deemed threatening are marked and shunned, and this fear of ex-communication from family serves to keep members in line. OF THE PAST AND PRESENT ("Fighting Bob") Shuler, D.D., LL.D. appeared in the December 1945 issue of "The Methodist Challenge" and reprinted in 1955 in "Bob Shuler - Met These On the Trail" I passed him on Spring Street. He was looking into a window where old books were being displayed. Something about his sagging shoulders and rather frayed clothing shook me. For a moment I felt compelled to speak to him. But a voice within seemed to forbid. I walked on. He did not see me. I had first seen Maurice Johnson in Eastland, Texas. He was my song leader in a tabernacle revival, and God was certainly with us. It seemed as though all Eastland County came the way of that revival. had been a taxicab driver in Fort Worth, had been wonderfully converted, and God was using him as few young fellows I have ever known. He could sing like a lark. He could win the toughest to Christ. Nothing was too hard for the Christ of Maurice Johnson. too, was a bit spectacular in those days! So, one Saturday night I preached on "Some Dogs I Have Known." Maurice walked out just before my sermon and sang, "You've Got to Quit Kicking My Dog Around." No Negro in the Old South could sing the Negro songs with more telling effect, and he could mimic a "Hard-shell Baptist" preacher to perfection. persuaded him to go to Los Angeles with me. He became my young people's leader and directed the music of the church. It was a departure from all the rules of the game. He had never been trained for any of this work. He was simply God's man. That was enough. the leadership of Trinity Methodist Church is largely made up of men and women led to Christ and typed in their Christian lives by the influence of this remarkable fellow. No man ever came the way of Trinity who had a more vitalizing and invigorating effect upon the people, young and old. But Maurice Johnson had what we call a "kink". Most of us have a dozen. His was fatal. He was impulsive and stubbornly insistent when he thought that he was right, but he lacked that something which the really great men of history have had -- a balance! He came just that near to being great as God's prophet. He was so effective that his faults seemed trivial. I recall that once I had an invitation to go to a camp as speaker with a Y.M.C.A. group of picked high school students. At the last moment, I discovered that I could not keep the engagement. I persuaded the Y.M.C.A. secretary, at first very reluctant, to take Maurice. That secretary came back tremendously enthused. It seemed the whole camp had been led to Christ through the personal touch and magnetic messages and singing of young Johnson. on earth did you get that fellow?" asked my Y.M.C.A. friend. "Off a Fort Worth taxicab," I said. where was he trained?" he insisted. "He got his training, equipment, and everything else from Heaven," I replied. Had Maurice Johnson been able to hold his balance, it is my honest opinion that he would have become a nationally-known power in the evangelization of this nation. But suddenly he became obsessed with the idea that he must brand all the scribes, Pharisees, and hypocrites, castigate the false teachers and anathematize, as the agents of Hell, every man who happened to disagree with his thinking. almost every position that he took, he was right. He discovered Unitarianism and other poison in the Sunday School literature. Immediately he published a pamphlet denouncing the Sunday School board of the church and quoting Scripture to show that the Sunday School editors of Methodism were the direct agents of the Devil. Time has pretty well vindicated his statements. ecclesiastical heads of our church in California would not stand for such an attack. They brought charges of insubordination against him and moved his location, forcing him out. I fought for him to the last breath. I begged the brethren to give me a chance to talk with him and see if I could calm him down. But just as I had about succeeded in my defense, Maurice got the floor and reiterated all that he had said, and then said a little more. I confess I admired him for it. But his speech in his own defense finished it. They put him out. organized an independent church in Glendale, which failed. He tried out an auto, touring about, preaching to anybody who would listen. He got on the radio. He turned his guns on all organized churches. He fired cannonballs at men whom he had held up in his earlier ministry as his idols. He became a spiritual isolationist. So, there he stood on Spring Street, dressed shabbily and looking haggard. His face was the gray of ashes. His hungry eyes read the titles of the moldy volumes in the window of that secondhand book store. I said to myself as I walked down the street, "Why did it have to happen!" could have packed the Church of the Open Door, preached to crowds that would have overflowed the great Moody Church in Chicago. He could have done what Appelman did in the great tent in Los Angeles. He could have thrilled the crowd I saw in Hollywood Bowl on the Saturday night of October 6th. Why did it have to happen? Especially when he was so nearly right practically all of the time! don't feel good about it. I wonder if somehow I failed God and this fine young fellow who left his taxicab and followed Jesus! Every now and then I tune in on Maurice [Johnson's radio programs], when he can get hold of enough money to buy some time over the air. He is still 90 per cent of the time right! I wonder if I am! can hear him now singing, "There's honey in the rock!" . . . . . . . . . Robert (Bob) P. Shuler 1201 S. Flower Street Los Angeles, California Dear Brother Shuler: said: "Faithful are the wounds of a friend; but the kisses of an enemy are deceitful" (Prov. 27:6). He also said: "The legs of the lame are not equal ..." (Proverbs 26:7). It was, therefore, not at all surprising to me to observe the "long and short" of your article concerning me. believe no Spirit-taught child of God can doubt that a preacher is "lame" when his "talking-leg" keeps fair step with the "fundamentalists" while his "walking-leg" maintains his place in what he condemns as often as Lot must have verbally condemned Sodom. For instance, in your Bob Shuler's Magazine, Sept. 1925, you thus daringly declared: need not attempt to deceive ourselves longer ... Modernism is massing ... How can we live together if we be not agreed? How can a house divided stand? ... Therefore, I say the clash is certain ... I cannot see how there can come anything else less than a division ..." Then in December: "We Southerners are a loud-mouthed set. If Modernism begins to creep into our Sunday School literature, we talk. If our Mission Board begins to foster a liberal movement in China, we discuss. If the Methodist Review begins to look like one of Bob Ingersoll's books, we remark upon it ... We are hard to silence." "We reserve the right as a Methodist preacher, personally supporting with our money our Methodist schools, and taking collections every year from our people for this cause, to voice our protest when our educational leaders come out boldly and go on record on the side of materialistic philosophies that are wrecking the religious faith and zeal of our young people to right and left ... NOR WILL OUR PEOPLE AROUSE. Many of them recognize the futility of protest. There will never be another formidable stand against modernism in Methodism. The tide is set in against us."(March 1927) "We can make, my brethren, by turning against the whole church and fighting everything and everybody." inch by inch we retreat before the advance of that sure and steady gain of modernism that few men will longer deny ... It is needless that we seek to comfort ourselves with the idea that such retreat will not in the ultimate prove fatal ... Methodism is being steadily, surely, purposely liberalized and modernized ... We are loyal to the church. We will not desert the banner of the fathers. We expect to stand and fight to the end." (February 1928) pastor of the leading church within my denomination (So. Meth.), I wish to say that I have never failed to send up to the proper authorities every cent of the assessments laid against my charge." (May 1928) Bro. Shuler, I am rather ashamed that I "shook" you by permitting you to discover me in front of that Spring Street secondhand book store as my "hungry eyes read the titles of the moldy volumes in the window" when my "face was the gray of ashes, looking haggard" with "sagging shoulders and rather frayed clothing ... dressed shabbily ..." a moment I felt impelled to speak to him. But a voice within me seemed to forbid ... I walked on," you wrote. Was that the voice, Bro. Shuler, that says in I John 3:17: "Whoso hath this world's good, and seeth his brother have need, and shutteth up his bowels of compassion from him, how dwelleth the love of God in him?" But I was no longer in "the church" to which you are so "loyal" and generous. It is quite true that I am now merely your brother I honestly wasn't in actual need of anything at the time unless it was that I was in need of a little more consideration as to my personal appearance in such a public place. You see, just about nineteen years ago, Bro. Shuler, Phil. 4:79 canceled all my earthly insurance policies; 2 Tim. 3:16,17 dynamited all my sectarian straight-jackets; Col. 2:6 and I John 1:5-7 provided me with the most intimate fellowship with the Triune God and all Spirit-led saints; and 2 Cor. 12:9,10 with Phil. 4:11-13 have been showing me the wicked extravagance and the obvious poverty of covetously "window-shopping" at Babylon's windows, so, for the life of me, I don't know how I happened to present such an inaccurate spectacle as that which you so minutely and tenderly described. were right when you said that I speak over the radio "when he can get hold of enough money to buy some time." Frankly I think that's all the radio speaking a Christian should do - only what he can pay for. By-the-way, Bro. Shuler, did you ever hear me asking for money over the radio? Nor has anyone else! But how in the world do you suppose I "get hold of enough to buy" radio time for six weekly programs now in four states, Ohio, Kansas, Washington, and California? To be exact, the answer isn't "in the world" but in the WORD -- the Word of God. Remember that you said: "He is still 90% of the time right." (Of course, you couldn't KNOW that I was 90% of the time right unless you were likewise right so as to know right from wrong. It is possible, I admit, for one to be a hypocrite, knowing better than he is doing.) the grace of God, I believe that "no good thing will He withhold from them that walk uprightly" (Psa. 84:11). Incidentally, every penny of the money that I invest in radio time, renting halls, etc, is voluntarily placed in my hands without any stipulation as to what I do with it, so, you see, Bro. Shuler, I COULD buy better clothes and PINK powder for my face (now "ashen gray"). Indeed, my "hungry eyes" could have read more than "the titles of the moldy volumes in the window" of that secondhand book store! let your conscience bother you too much, Bro. Shuler, for having obeyed that "voice within" that forbade you to speak to me (or even slip me 15 cents to satisfy my "hungry eyes" for I think I later went in and bought one or more rapidly molding books that were Methodist "best sellers" a couple or more years ago. see, there are others besides "we Southerners" who are "loud-mouthed", therefore, their yesterday's writings are quickly discarded to make room for their today's mouthings. I am thus enabled to keep just "a second-hand-book-store" behind you "men of the cloth" who buy the first-hand modernist mouthings before they are obviously "moldy". I only read those by BIG men "of the cloth", however. And there is some value, at least, in the light Bible prophecy, to see what the "black shirts", the black coats, the black vests, the "silver shirts", the silver tonguers, the "brown shirts", the Popish lock-steppers, the modernist high-steppers and the fundamentalist side-steppers are doing. (I'm speaking of the systems rather than the individuals.) "He became a spiritual isolationist", you wrote of me. I pray that you are right for that would mean that I am separated ONLY to the spiritual. I can now sing: "Blest be the tie that binds our hearts in Christian love" and sing it with understanding and satisfaction that I never knew nor could I know before I became "spiritual isolationist." "Awake thou that sleepest, and arise from out from among the dead and Christ shall give thee light" (Eph. 5:14). "And the Lord said unto Abraham, after that Lot was separated from him, Lift up now thine eyes, and look from the place where thou art, northward, and southward, and eastward, and westward: for all the land which thou seest, to thee will I give it ..." (Gen. 14:14,15). I would to God, Bro. Shuler, that you saw the inverted pyramid that your whole man-made Methodism is. You apparently don't dream that TIME and SENSE are the greedy Mortgage Co., from which you have so largely borrowed in building yourself up as a BIG preacher Fighting Bob Shuler. And when that heartless pair, TIME and SENSE, foreclose on you it will be the most sickening and frightful awakening you have ever known. Your fight has NOT been "the good fight of faith", but of sight. It has been the confused and confusing bluffs and wild swingings of a blood-bought, fearless and fervent soul that somewhere back in his life became an orthodox "Front" for that immense "holding company", TIME and SENSE, the silent but controlling partners. believe the day will come when your two usually talented boys, Bob Jr. and Jack, will all but curse you for not pleading with them to forsake their father's religio-political appeasement policy; your "magnanimous-back-slapping" with your denomination modernists and your "Good Friday" orthodoxy with inter-denominational fundamentalists. I say these severe things ONLY because I love your soul and believe they may be used of God to arrest you before you bluff away the rest of your earthly Later, I expect to go more into the subject of just what is a "great man." Suffice it now for me to say that I firmly believe you would rather have had your son, Jack, appear on the platform, as he did, at that "Youth For Christ" mass meeting in Hollywood Bowl than to have had him be alone with John the Beloved on the Isle of Patmos or caught up to the third heaven with Paul, to have him, or yourself, DIRECTLY called by an angel of the Lord, as was Philip, to LEAVE the city and "go to Gaza, which is desert" (Acts 8:26) would suggest a "kink" or spiritual "isolationism" to you, I cannot doubt. A servant of Christ, /signed/ Maurice Johnson Letter from Maurice M. Johnson about his retirement Orangevale, Ca. 95662 May 8, 1972 my brothers and sisters in the family of God and fellow-soldiers in the good fight of As most of you know, in November of 1969 shortly before my seventy-sixth birthday, my wife and I moved permanently from Los Angeles County which had been our home base, so to speak, since our marriage in June, 1923. At that time I was soloist, choir director and director of young peoples work at Trinity Methodist Church in Los Angeles. In the fall of 1923 I resigned these positions and was appointed one of the Conference Evangelists at the Annual Methodist Conference in November. first meeting I held, preaching and singing with my wife assisting at the piano, was in the Southern Methodist Church building in Sacramento, Ca. The pastor and his wife were dear Christians, "fundamental" in their Bible beliefs, but completely discouraged by the coldness and indifference of most of the members. I preached my heart out, with no visible results except a profession of faith on the part of a young hitch-hiker who had stopped in the church vestibule late one evening for a drink of water. meeting followed sponsored by the young people of the Central Methodist Church in Phoenix, Ariz. I had known many of these young people in Epworth League work. They were enthusiastic and worked hard and there were several who professed faith in Christ for the first time, but the pastor soon revealed himself as a modernistic infidel. I left Phoenix with a heavy burden for those young "lambs" with a wolf in sheeps clothing for a pastor and so-called shepherd. third meeting in San Diego was likewise a disappointment. The pastor and his wife were earnest Christians, but the Sunday School literature was almost completely modernistic in content; and again the few converts were left to feed on infidelity except for the pastor's Sunday sermons. I left that meeting convicted that I could not with a good conscience continue to help add to the membership of such a congregation. There were some troubled Christians in all three of these places; but they could do no more than "just Lot" vexing his righteous soul (2 Pet. 2:7). The next nine months were spent in Chicago as choir director and assistant pastor to J. C. O'Hair at the North Shore Congregational Church; but the burden to preach the Word myself was growing on me. I went back to California to the annual Methodist Conference in November and was assigned to the pastorate of the Broadway Methodist Church in Glendale, Ca. During the year [1924-1925] we doubled the membership of the local church and many of the new members were admitted on profession of faith in Christ. Many of you have heard me say to my shame that I "opened the door of the Methodist church" and shut these babes in Christ in, because when I led in saying every Sunday morning, "I believe in the holy, catholic church" I had never really believed that there was only one church and that Christ was the only Door. At the end of the conference year I was put out of the Methodist ministry on the grounds of general unacceptability, mainly because we had refused to use the official Sunday School literature and were teaching the Bible in All the Sunday School classes. Over seventy-five people came out of the Methodist denomination with me and we straightway organized a "fundamentalist church" [Maranatha Tabernacle in Glendale] modeled after three of the "best" in the nation. I thank the Lord, the Head of His church, that He did not let me succeed. A little over a year later we dissolved our corporation, I gave up my salary and began the walk of faith with El Shaddai, the All-Sufficient One. The small handful of Christians that had stood the storm with me also came out unto the Lord, free from all sectarian ties. We still have the joy of fellowship with many of them, and in some cases with their children and their grand-children. November of this year I will be seventy-nine years old; my wife will be seventy-four. It has been a grand and glorious battle, and we have worked together joyously and in unity, but we are both tired---not tired of the work, but tired in the work. I had hoped to continue in public ministry in the Orangevale area, but the Lord saw fit to "retire" me by means of major surgery in 1970 and a stroke in 1971 that left me with a severe form of aphasia---a condition in which I know WHAT I want to say, but the right words do not come out. My poor health has placed a heavier burden of responsibility on my wife than she should have to bear and her health has suffered also. After much prayer and waiting on the Lord, I have decided to give up the radio broadcasts, and since most of the money received by me is designated for the M. M. Johnson Gospel Fund I shall close that out, too. We have set the time for this closing out of funds and broadcasts for the end of May---the end of forty-nine years together in the work of the Lord. We do not intend to stop serving the Lord, of course, but will seek His mind for further direction. broadcasts I now am responsible for will be turned over to Bros. Berl Chisum, Jack Langford and James Cox, to be carried on in whatever way they may decide. The continuation of the local broadcast over K-POP will be left to the decision of the men in the Orangevale assembly and the financial responsibility will be theirs. explanation may be necessary for some of you who do not know how the gospel fund is used. Until a problem with the Internal Revenue Service arose as to whether contributions to Maurice Johnson were tax deductible, our personal funds and the funds for the Lord's work were in one account. Twice before I had set up a gospel fund when we were working in Texas and the mid-west, but they were used up and closed out before 1953. 1957 I set up the Maurice Johnson Gospel Fund in South Pasadena so gifts to it would be tax deductible, and opened a separate joint account for my wife and me in Arroyo Grande, where we moved temporarily on account of her poor health. I was still active and she would have been alone at times and perhaps need immediately accessible funds. 1957 until now, the gospel fund has been used ONLY for the work of the Lord---hall rentals, radio broadcasts, advertising, office supplies and machines, traveling expenses, etc. I have taken for our personal needs only the cash in the offering box and checks made out to me personally without any designation as to where they were to be used. In many cases where we did not need the cash or the checks they were also deposited in the gospel fund. If it had been necessary to draw on the gospel fund for any of our living expenses I'm sure I would have felt free to do so, but it was never necessary. When we left Los Angeles in November of 1969 Russell Ross made an examination of all accounts handled by us from 1942 to 1969. A copy of these accounts can be made for anyone who needs to verify Orangevale in December, 1969 I set up a new M. M. Johnson Gospel Fund in which were deposited checks from here and other areas. As before, we used the cash and undesignated checks for our personal expenses when needed. The expenses of the Orangevale assembly, office expenses, tapes, recording supplies and other miscellaneous expenses have been paid from the gospel fund as well as a total of $11.996.00 to the Berl Chisum Gospel Fund up to date, as a part of the radio broadcasting expense to stations on which I shared time with Bro. Chisum. A total of $2,900.00 has been sent to Bro. Jack Langford to help with his printing and radio expenses, and $100.00 to Tom Murley for general expenses. Bro. Bill Hagan is preparing a summary of this account from December 1969 to May 31st of 1972 which can also be inspected upon request. At no time has our son, Jim Johnson, had knowledge of the amount in the fund, nor any access to it, nor profited from it personally. of you who have contributed regularly to the Lord's work through my ministry are asked to send no more to me, but to seek the mind of the Lord through the leading of the Holy Spirit as to the distribution of your gifts to the other ministering brethren. second fund should be mentioned, called by my wife the "Postage Fund". It consists of all cash received from the radio audience and is used for mailing tapes, literature, issues of Sound Words, yearly calendars, etc. It is put in a desk drawer and used when needed, generally used up after each big mailing. All checks from the radio audience are put in the gospel fund. A shorter, but similar letter will be sent to our regular radio contributors as soon as it can be done. lengthy letter may seem unnecessary to some, but my wife and son and I are in full agreement that we want to be crystal clear in this matter, so that there will be no need for explanations or interpretations from other sources. If there remain any questions in the minds of any of you, will you please for Christ's sake, ask us FIRST? We will be happy to answer them. ask your prayers for further direction in our lives. For many years our children have begged their mother to write the story of our "adventures" in our long and eventful and happy Christian service, especially the years they were too young to remember. If the Lord allows us that much time we may do it, and have already begun to assemble material for that purpose. Whatever He allows us to do will be worthy of Him as long as we have a single eye and seek only the honor and glory of our Savior and Lord If present plans work out we also hope to be able soon to offer free tapes of previous sermons and radio broadcasts, and to mail out literature by request as long as our supply lasts. If you should be interested in the radio tapes, please let us know. We hope those of you passing through the Sacramento area will stop by often for a little visit. We do not expect to do much traveling now. EBENEZER -- I Sam. 7:12. the God of peace that brought again from the dead our Lord Jesus, that great shepherd of the sheep, through the blood of the everlasting covenant make you perfect in every good work to do his will, working in you that which is well-pleasing in his sight, through Jesus Christ, to whom be glory forever and ever, Amen." servant and soldier of the Lord Jesus Christ, the parallel history of religion among the major race-ethnic groups in Los Angeles Race, Ethnic, Ancestry and Religious Info & Histories The Hispanic American Community The Asian American & Pacific Islander Communities (an overview of all subgroups) City of Los Angeles County of Los Angeles County of Orange African American Community The Arab Community The Native American Indian Community<\|endoftext\|>	3.75
5,051	Practical Geometry Mind Maps Class VI - Maths: Practical Geometry Q) Definition of Dimension? Q) Definition of Points? Q) Definition of Plane or Plane Surface? Q) Definition of Solid Geometry? Q) Definition of Line? Q) Draw a triangle ABC such that BC = 5 cm, AB = 6.5 cm, AC = 7.5 cm? Q) (a)Draw a line segment of length 3.9 cm using a ruler? (b)Construct a copy of a given segment? Q) Construct a ?ABC, in which ?B = 700, AB = 4.8 cm and BC = 5.2 cm. Q) Construct an isosceles triangle PQR such that PQ = PR = 7cm and P = 1400? Q) Construct a triangle ABC in which BC = 6 cm, ∠B = 35° and ∠C = 100°? Q) Theorem on parallel lines and plane are explained step-by-step along with the converse of the theorem? Q) If two straight line are both perpendicular to a plane then they are parallel? Q) (a) Construct a right triangle PQR in which ?Q = 105° , PR = 8 cm and QR = 5 cm. (b)Construct a triangle ABC in which BC = 5 cm, ∠B = 65° and ∠C = 90° ? Q) Given a line l and a point not on it, We used of some ruler and compasses construction. transversal diagram to draw a line parallel to l. We could also have used the idea of 'equal corresponding angles' to do the construction? Q) (a) Construct a right-angle triangle whose hypotenuse is 8 cm long and one of the legs is 5 cm long. (b) Construct an isosceles right-angled triangle ABC, where m∠ACB = 9° and AC = 5 cm. • Construction of parallel lines • Theorem on Parallel Lines and Plane • Converse of the theorem on parallel lines and plane • Solid Geometry • Definitions on solid geometry terms • Angle Between Two Skew Lines • Orthogonal Projection • Angle Between a Straight Line and a Plane • Dihedral Angle • Construction of Triangles • Construction of Line Segment • SSS Construction • SSS Construction of Δ • SAS Construction • SAS Triangle Construction • ASA Triangle Construction • RHS Construction • RHS Triangle Construction • In this Chapter, we looked into the methods of some ruler and compasses constructions. 1. Given a line l and a point not on it, we used the idea of ‘equal alternate angles’ in a transversal diagram to draw a line parallel to l. We could also have used the idea of ‘equal corresponding angles’ to do the construction. • We studied the method of drawing a triangle, using indirectly the concept of congruence of triangles. The following cases were discussed: (i) SSS: Given the three side lengths of a triangle. (ii) SAS: Given the lengths of any two sides and the measure of the angle between these sides. (iii) ASA: Given the measures of two angles and the length of side included between them. (iv) RHS: Given the length of hypotenuse of a right-angled triangle and the length of one of its legs. Construction of parallel lines Theorem on Parallel Lines and Plane Theorem on parallel lines and plane are explained step-by-step along with the converse of the theorem. Theorem: If two straight lines are parallel and if one of them is perpendicular to a plane, then the other is also perpendicular to the same plane. Let PQ and RS be two parallel straight lines of which PQ is perpendicular to the plane XY. We are to prove that the straight line RS is also perpendicular to the plane XY. Construction: Let us assume straight line PQ and RS intersect the plane XY at Q and S respectively. Join QS. Evidently, QS lies in the XY plane. Now, through S draw ST perpendicular to QS in the XY plane. Then, join QT, PT and PS. Proof: By construction, ST is perpendicular to QS. Therefore, from the right-angled triangle QST we get, QT2 = QS2 + ST2 ..............(1) • Since PQ is perpendicular to the plane XY at Q and the straight lines QS and QT lie in the same plane, therefore PQ is perpendicular to both the lines QS and QT.Therefore, from the right-angle PQS We get, PS2 = PQ2 + QS2 ..............(2) And from the right-angle PQT we get, PT2= PQ2 + QT2= PQ2+ QS2 + ST2 [using (1)] or, PT2 = PS2 + ST2 [using (2)] • Therefore, ∠PST = 1 right angle. i.e., ST is perpendicular to PS. But by construction, ST is perpendicular to QT. Thus, ST is perpendicular to both PS and QS at S.Therefore, ST is perpendicular to the plane PQS, containing the lines PS and QS. Now, S lies in the plane PQS and RS is parallel to PQ; hence, RS lies in the plane of PQ and PS i.e., in the plane PQS. Since ST is perpendicular to the plane PQS at S and RS lies in this plane, hence ST is perpendicular to RS i.e., RS is perpendicular to ST. Again, PQ and RS are parallel and ∠PQS = 1 right angle. Therefore, ∠RSQ = 1 right angle i.e., RS is perpendicular to QS. Therefore, RS is perpendicular to both QS and ST at S; hence, RS is perpendicular to the plane containing QS and ST i.e., perpendicular to the XY. Converse of the theorem on parallel lines and plane If two straight lines are both perpendicular to a plane then they are parallel. Let two straight lines PQ and RS be both perpendicular to the plane XY. We are to prove that the lines PQ and RS are parallel. Following the same construction as in theorem on parallel lines and plane • it can be proved that ST is perpendicular to PS. • Since, RS is perpendicular to the plane XY, hence RS is perpendicular to TS, a line through S in the plane XY i.e., • TS is perpendicular to RS. Again, by construction, TS is perpendicular QS. • Therefore, TS is perpendicular to each of the straight lines QS, PS and RS at S. • Hence, QS, PS and RS are co-planar (by theorem on co-planar). • Again, PQ, QS and PS are co-planar (Since they lie in the plane of the triangle PQS). • Thus, PQ and RS both lie in the plane of PS and QS i.e., • PQ and RS are co-planar. Again, by hypothesis, ∠PQS = 1 right angle and ∠RSQ = 1 right angle. Therefore, ∠PQS + ∠RSQ = 1 right angle + 1 right angle = 2 right angles. Solid Geometry Definitions on solid geometry terms (i) Dimension: Each of length, breadth and thickness of any body is called a dimension of the body. (ii) Point: A point has no dimension, that is, it has neither length nor breadth nor thickness ; it has position only. (iii) Line: A line has length only but no breadth and thickness. Therefore, a line has one dimension, that is, it is one dimensional. (iv) Surface: A surface has length and breadth but no thickness. Therefore, a surface has two dimensions, that is, it is two dimensional. (v) Solid: A solid has length, breadth and thickness. Therefore, a solid has three dimensions, that is, it is three dimensional. The book is a solid, each of its six faces is a surface, each of its edges is a line and each of its corners is a point. A line is bounded by points, a surface is bounded by lines and a solid is bounded by surfaces. In other words, a line is generated by the motion of a point, a surface is generated by the motion of a line and a solid is generated by the motion of a surface. (vi) Solid Geometry: The branch of geometry which deals with the properties of points, lines, surfaces and solids in three dimensional space is called solid geometry. (vii) Plane or Plane Surface: If the straight line joining two points on a surface lies wholly on the surface then the surface is called a plane surface or a plane. A straight line may be extended indefinitely in either direction, that is, straight lines are supposed to be of infinite length. Similarly, planes are also assumed to be of infinite extent, unless otherwise stated. The statement that a straight line lies wholly on a surface signifies that every point on the line (however produced in both directions) lies on the surface. A surface is called curved surface when it is not a plane surface. (i) Lines or points are said to be co-planar if they lie on the same plane; in other words, lines or points are co-planar if a plane can be made to pass through them. (ii) Two co-planar straight lines are either parallel or they intersect at a point. Two straight lines are said to be parallel when they are co-planar and they do not meet however indefinitely they are produced in both directions. (iii) Two straight lines are said to be skew (or non-coplanar) if a plane cannot be made to pass through them. In other words, two straight lines are said to be skew when they do not meet at a point and they are not parallel. (iv) Two planes are said to be parallel if they do not meet when extended infinitely in all directions. (v) A straight line is said to be parallel to a plane if they do not meet when both are produced infinitely. In the given picture we observe that, the lines LM, MN, NO and OL lie in the plane LMNO, that is, they are co-planar. The lines LM and LO meet at L and the lines LM and ON are parallel. LP and MN are skew lines and the line QR is parallel to the plane LPSO. The planes ABFE and DCGH are parallel. A straight line is said to be perpendicular to a plane if it is perpendicular to every straight lines drawn in the plane through the point where the line meets the plane. A straight line perpendicular to a plane is called a normal to the plane. In the given figure, the straight line OP meets the plane XY at O ; if OP is perpendicular to every straight lines OI, OJ, OK, OD etc. drawn through O in the XY plane then OP is perpendicular (or a normal) to the plane XY. A straight line parallel to the direction of a plumb-line hanging freely at rest is called a vertical line. A plane which is perpendicular to a vertical line is called the horizontal plane. A straight line drawn in a horizontal plane is called a horizontal line. Angle Between Two Skew Lines: The angle between two skew lines (i.e., two non-co-planar straight lines) is measured by the angle between one of them and a straight line drawn parallel to the other through a point on the first line. In the given figure, let MN and QR be two skew straight lines. Take any point O on the line MN and draw the straight line OP parallel to QR through O. Then ∠NOP gives the measure of the angle between the skew straight lines MN and QR. A triangle is a plane figure since all its three sides lie in one plane. Similarly a parallelogram is also a plane figure. But a quadrilateral may or may not be plane figures since all its four sides always do not lie in one plane. A quadrilateral whose two adjacent sides lie in one plane and other two adjacent sides lie in a different plane is called a skew quadrilateral. Orthogonal Projection: (a) If a perpendicular be drawn from an external point on a given line then the foot of the perpendicular is called the orthogonal projection (or simply the projection) of the external point on the given line. In the above left side figure, Pp is perpendicular from the external point P on the straight line AB. Since the foot of the perpendicular is p hence, p is the projection of P on the on the line AB. Again, we can observe that in the above right side figure, 5 the point P lies on the line AB ; hence, in this case the projection of P on AB is the point P itself. (b) The locus of the feet of the perpendiculars drawn from all points of a line (straight or curved) on a given straight line is called the projection of the line on the given straight line. In the above left side figure, Pp and Qq are perpendiculars from P and Q respectively on the straight line AB; p and q are the respective feet of perpendiculars. Then, pq is the projection of the straight line PQ on the straight line AB. Again, previous left hand side figure, the projection of the straight line PQ on the straight line AB is Pq. Again, similarly we can observe that in the above right side figure, pq is the projection of the curved line PQ on the straight line AB. Again, suppose the straight line PQ intersects the straight line AB at R; in this case, the projections of QR and RP on AB are qR and Rp respectively. (c) The locus of the feet of the perpendiculars drawn from all points of a line (straight or curved) on a given plane is called the projection of the line on the plane. In this figure, the locus of the feet of the perpendiculars drawn from points of the line MN on the plane XY is the line mn; hence, the projection of the line MN on the plane XY is the line mn. #### Note: 1. The projection of a straight line on a plane is a straight line ; but the projection of a curved line on a plane may be a straight line as well as a curved line. If the curved line MN lies in a plane which is perpendicular to the plane XY then the projection of MN on the plane XY is a straight line. 2. A straight line and its projection on a plane are co-planar. Angle Between a Straight Line and a Plane The angle between a straight line and a plane is measured by the angle between the given straight line and its projection on the given plane. Let mn be the projection of the straight line MN on the plane XY. Suppose, in the given figure, straight lines MN and mn (when produced) meet at the point R in the plane XY. Then the angle between the straight line MN and the plane XY is measured by ∠MRm. Dihedral Angle The plane angle between two intersecting planes is called a dihedral angle between the planes. The angle between two intersecting planes (i.e., a dihedral angle) is measured as follows: Take any point on the line of intersection of the two planes. From this point draw two straight lines, one in each plane, at right angles to the line of intersection. Then the plane angle between the drawn two lines gives the measure of the dihedral angle between the planes. Let XY and LM be two intersecting planes and XL be their line of intersection. From any point A on XL draw the straight line AB perpendicular to XL in the XY plane and the straight line AC perpendicular to XL in the LM plane. Then the plane ∠BAC is the measure of the dihedral angle between the two intersecting planes XY and LM. If the dihedral angle between two intersecting planes is a right angle then one plane is said to be perpendicular to the other. Construction of Triangles #### To Construct a Triangle whose Three Sides are given: To construct a triangle whose three sides are given; let us follow the examples. 1. Draw a triangle ABC such that BC = 4 cm, AB = 3.5 cm, AC = 4.5 cm. [Let us draw a rough sketch of the triangle ABC. Name the vertices and show the given measures.] Steps of Construction: Now let’s follow these steps one by one (i) Draw a line segment BC = 4 cm. (ii) With B as Centre and radius 3.5 cm, draw an arc. (iii) With C as centre and radius 4.5 cm, draw another arc cutting the previous arc at A. (iv) Join AB and AC. Therefore, Δ ABC is the required triangle. 2. Draw a triangle ABC such that BC = 4.5 cm, AB = 3.5 cm, AC = 5 cm. [Let us draw a rough sketch of the triangle ABC. Name the vertices and show the given measures.] Steps of Construction: Now let’s follow these steps one by one (i) Draw a line segment BC = 4.5 cm. (ii) With B as Centre and radius 3.5 cm, draw an arc. (iii) With C as centre and radius 5 cm, draw another arc cutting the previous arc at A. (iv) Join AB and AC. Therefore, Δ ABC is the required triangle. Construction of Line Segment 1). Draw a line segment of length 3.9 cm using a ruler. Solution : Step 1: Draw a line l. Mark a point on the line l. Step 2: Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point up to 3.9 cm. Step 3: Taking caution that the opening of the compasses has not changed, place the pointer on A and swing an arc to cut l at B. Step 4: segment AB is a line segment of required length. 2). Construct a copy of a given segment. Step 1: Draw line segment CD of any length. Step 2: Fix the compasses pointer on C and the pencil end on D. Now this opening of the compasses gives the length of segment CD. Step 3: Draw a line l. Mark a point A on the line l. Without changing the compasses setting, place the pointer on A. Step 4: Cut an arc that cuts l at a point B. Now, AB is a copy of CD. SSS Construction Construction Of Triangle Note that it is not sufficient to know only three angles. In order to be able to construct a Δ from three segments, one of the segments must be shorter than the sum of the other two: a < b+c, or b < a+c, or c < a+b SSS Construction of Δ Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6cm. Step 1. Draw a line YZ of length 5 cm. Step 2. From Y, point X is at a distance of 4.5 cm. So, with Y as center, draw an arc of radius 4.5 cm. Step 3. From Z, point X is at a distance of 6 cm. So, with Z as center, draw an arc of radius 6 cm. Step 4. X has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark the point of interaction of arcs as X. Join XY and XZ. ΔXYZ is the required triangle as shown in the figure. SAS Construction Triangle given two sides and included angle (SAS) Multiple triangles possible It is possible to draw more than one triangle has the side lengths and angle measure as given. Depending on which line you start with, which end of the line you draw the angles, and whether they are above or below the line, the four triangles below are possible. All four are correct in that they satisfy the requirements, and are congruent to each other. #### Printable step-by-step instructions The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available. #### Proof The image below is the final drawing above with the red items added. Argument Reason 1 Line Segment MN Is Congruent To AB. Drawn With The Same Compass Width.For Proof See Copying A Line Segment 2 Line Segment ML Is Congruent To AC. Drawn With The Same Compass Width. 3 The Angle LMN Is Congruent To The Angle A Copied Using The Procedure Shown In Copying An Angle.See That Page For The Proof 4 Triangle LNM satisfies the side lengths and angle measure given. SAS Triangle Construction #### Δ Construction 1.Construct a ΔABC, in which ∠B = 70°, AB = 4.8 cm and BC = 5.2 cm. Before doing construction, draw a rough sketch, so that you will get an idea which side is taken as a base.Here after drawing a rough sketch, we are sure that the base is AB = 4.8 cm. Step 1 : Draw AB = 4.8 cm. Step 2 : Using protractor, draw ∠ABK = 70° . Step 3 : On the line segment BK, cut off BC = 5.2 cm. Step 4 : Join A and C. ΔABC is the required triangle. 2.Construct an isosceles triangle PQR such that PQ = PR = 5cm and P = 110°. Before doing construction, draw a rough sketch, so that you will get an idea which side is taken as a base.Here after drawing a rough sketch, we are sure that the base is AB = 4.8 cm. Step 1 : Draw PR = 5 cm. Step 2 : Using protractor, draw ∠QPM = 110°. Step 3 : On the line segment PM, cut off PQ = 5 cm. Step 4 : Join Q and R. ΔPQR is the required triangle. ASA Triangle Construction To construct a triangle when two of its angles,say B and C, and the included side BC are given,we proceed as follows : Step 1 : Draw a line segment BC. Step 2 : Draw ∠CBX of measure equal to that of ∠B. Step 3 : Draw ∠BCY with Y on the same side of BC as X, such that its measure is equal to that of ∠C. Let BX and CY intersect at A. Then, ΔABC is the required triangle. ΔPQR is the required triangle. #### Example Construct a ΔABC in which BC = 6 cm, ∠B = 35° and ∠C = 100°. Step 1 : Draw a line segment BC = 6 cm. Step 2 :Draw ∠CBX, such that ∠CBX = 35° Step 3 : Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 100° Step 4 : Let BX and CY intersect at A. ΔABC is the required triangle. RHS Construction #### HL Triangle Construction To construct a right triangle ABC right angled at C when its hypotenuse is AC and one side is BC is given, we follow the following steps : Step 1 : Draw a line segment BC of given length. Step 2: Draw ∠BCX of measure 90°. Step 3 : With center B and radius equal to the hypotenuse AB, draw an arc of the circle to intersect ray CX at A. Step 4 : Join BA to obtain the required triangle ABC #### Example RHS Triangle Construction Construct a right triangle PQR in which ∠Q = 90° , PR = 6 cm and QR = 4 cm. Step 1: Draw QR = 4 cm. Step 2 : Using compass, at Q draw ∠RQK = 90° . Step 3 : Cut off RP = 6 cm. Step 4 : Join P and R. Now we get the required ΔPQR. Meritpath...Way to Success! Meritpath provides well organized smart e-learning study material with balanced passive and participatory teaching methodology. Meritpath is on-line e-learning education portal with dynamic interactive hands on sessions and worksheets.<\|endoftext\|>	4.59375
944	CLICK HERE for The Top-Rated & Best-Selling Health and Personal Care Products What is Cerebral Palsy? In the United States and across continental Europe, it is estimated that cerebral palsy is occurring in two to four newly born babies out of 1,000. All around the globe, the inborn condition is affecting thousands of infants annually. An analysis of the etymology of the word would lead one to easily infer that cerebral palsy is affecting the brain in general. The word ‘cerebral’ is referring to cerebrum, the affected brain area, while ‘palsy’ connotes movement disorder. Many studies have tried to explain the root cause of the condition. It was initially found that cerebral palsy is brought about by damages to the developing brain’s motor control centers. These damages may have been incurred during pregnancy or during actual childbirth. In some instances, cerebral palsy develops long after birth or up until the infant reaches the age of three. Contrary to what many people believe, cerebral palsy is not in any way contagious. Thus, the condition is non-communicable and could not be passed on from a patient to another person. It is just okay to be dealing with and interacting with children with cerebral palsy. The sad news is that the condition has always been curable, though modern medicine is continuously looking for possible cures and treatments. However, in many instances, an appropriate physical therapy into a child could spell a really huge difference. Motor disorders related to cerebral palsy are accompanied more often by disturbances in perception, sensation, communication, behavior, and cognition. As mentioned, there is no cure, but there are usual medical interventions, which are limited to prevention and treatment of possible complications. There are also findings or studies that show that improvements and progress in neonatal nursing could possibly help lower the number of infants who are born with the condition. There are three major categories or classifications of cerebral palsy, namely, spastic, ataxic, and athetoid or dyskinetic. Spastic is the most common type of cerebral palsy. It occurs in about 80% of reported cases. Patients exhibit neuromuscular conditions that stem from damages to the motor cortex, which in turn influences the brain’s ability to cognate. Ataxic cerebral palsy is the type that is linked to possible damages to the cerebellum. It occurs less frequently, accounting for only about 10% of reported cerebral palsy cases. Patients exhibit tremors and hypotonia. Their skills in typing, writing, and using scissors are affected, along with physical balance while walking. Athetoid or dyskinetic type of cerebral palsy is rare. Patients often show signs of involuntary movements. They find it hard to keep still in a position and hold objects. Once an infant is diagnosed to have cerebral palsy, there is an automatic need for further optional diagnostic tests. It is also interesting to note that male infants are more likely to have cerebral palsy than females. Specific advances and improvements in care of expecting mothers have been found not to help decrease or eliminate occurrence of cerebral palsy. Selected ArticlesCerebral Palsy And Education Risk Factors That Increase Risks Of Cerebral Palsy Facts About Cerebral Palsy Cerebral Palsy And Its Known Risk Factors Hand In Hand: Cerebral Palsy And Family Support What Is Cerebral Palsy? Cerebral Palsy And Dealing With Malnutrition Maternal Illnesses Linked To Cerebral Palsy The Importance Of Understanding Cerebral Palsy Symptoms Of Cerebral Palsy How To Cope With Cerebral Palsy How To Help Boost The Self Esteem Of People With Cerebral Palsy Understanding Cerebral Palsy Don’t Panic, After Your Child Has Been Diagnosed With Cerebral Palsy The Power Of Encouragement To Help People With Cerebral Palsy Coping And Support For Cerebral Palsy Patients Cerebral Palsy, The Signs To Look For Treatments And Drugs For Cerebral Palsy Common Misconceptions About Cerebral Palsy Cerebral Palsy Risks While Pregnant Types Of Cerebral Palsy Cerebral Palsy, Do My Kids Have It? Risks To Avoid Cerebral Palsy A Promising Modern Device For Treating Cerebral Palsy<\|endoftext\|>	3.703125
570	Let’s Have a Luau! Project Title: Hawaiian themed art projects Description: leis and parrot puppets Author: Mary Hager Grade level or Target Age Range: Preschool Vocabulary: lei, luau, Hawaii, hula Materials: For leis: string, cutouts of paper flowers with holes in the middle and small cut pieces of plastic straws. Plan to prepare the flowers and straws before class. For parrot puppets: paper cups, red, yellow, and orange construction paper, glue, crayons, scissors, and fake feathers. Before we made the leis, we learned about Hawaii. For instance: - We talked about how Hawaii is an island. - We learned the different animals we might find on Hawaii, such as parrots - Last week we made paper volcanoes and the students learned that volcanoes can be found on Hawaii - We also talked about leis: how they are usually made with real flowers and how they are important to the Hawaiian culture (people are now greeted with leis when they arrive in Hawaii). - Take one bead and string it to the bottom of your necklace and tie a knot around it. This will ensure that the students’ beads will not fall off of their leis. - Demonstrate how to make a lei by stringing a paper flower followed by a small plastic bead (the cut straws). - Explain that there is a pattern to making a lei (bead, flower, bead, flower, bead). When the students completed their leis, I taught them how to hula dance and then let them dance to Hawaiian music. They absolutely loved this!! *** Dancing video <– watch the students learn how to hula dance! - Cover each paper cup with red or orange paper. - Pre-cut 3 wings (in small, medium and large sizes) from different colored paper. - Glue these wings glued in layers on top of each other to make one colorful wing. - Once the wings are complete, the students will glue the wings on either side of the cup. - Next, cut out one large circle for the head (out of red or orange paper). - Cut a large beak out of paper and glue this in the middle of the circle. - Have students draw eyes on either side of the beak. - Add feathers to the top of the head with glue. - Glue the head to the cup and allow time for the parrot to dry - Once everything is dry, students can use the parrot as a puppet by placing their hand inside the cup. “Transforming Lives Through Art” ~ Mary Hager<\|endoftext\|>	3.90625
274	Learning sight words takes a lot of practice. There are many different sight word activities that parents can use to help their child learn. These sight word activity sheets include four very basic activities for all 40 of the traditional Dolch pre-primer words. *This post includes Amazon affiliate links. Please see my disclosure policy for more information.* Dolch Sight Word Activity Sheets This printable pack includes one sheet of simple activities for each of the 40 Dolch pre-primer words. The activities are: - Read the word - Color the word - Trace the word - Write the word What You Will Need How to Use the Sight Word Activity Sheets - Print out the page or pages that you want to use with your child. - Make sure your child has any tools that they will need (pencils, crayons, etc.). - Decide how you want your child to use the activities. Your child can do all four activities for a word. Or you can choose one activity for your child to do with several words. Just cut along the dotted lines. For example, your child can color a few words of your choosing. I hope that you and your child find these sight word activities helpful! To download this printable, simply click on the link below.<\|endoftext\|>	3.8125
560	# GMAT Quant \| Linear Algebra Question 2 #### GMAT Sample Questions \| Linear Equations Questions The gmat practice question given below is an Algebra word problem and tests your ability to frame a couple of linear equations in two variables and solve the system of equations to determine the answer to the age of the son 3 years from now. This GMAT sample question in quant is a 600 level problem solving question. A relatively easy question. Question 2: Three years back, a father was 24 years older than his son. At present the father is 5 times as old as the son. How old will the son be three years from now? 1. 12 years 2. 6 years 3. 3 years 4. 9 years 5. 27 years ## GMAT Live Online Classes #### Step 1 of solving this GMAT Linear Equations Question: From words to mathematical equations Let the age of the son 3 years back be 'x' years. Information 1: Three years back, a father was 24 years older than his son. Therefore, the age of the father 3 years back was (x + 24). If the age of the son 3 years back was 'x' years, the present age of the son is x + 3. Present age of father = x + 24 + 3 Information 2: The father is at present 5 times as old as the son. i.e., (x + 24 + 3) = 5(x + 3) Or x + 27 = 5x + 15 Or 4x = 12 or x = 3. #### Step 2 of solving this GMAT Algebra Word Problem: From unknown to the answer x was the age of the son 3 years back. Therefore, the son was 3 years old 3 years back. The question is "How old will the son be three years from now?" If the son was 3 years old, 3 years back, the son is 6 years old now. Hence, he will be 9 years old three years from now. #### Choice D is the correct answer. As a good practice when solving GMAT Math questions, after solving equations and finding values for the unknown, check whether further action has to be taken to get the answer. In this example, the value of x gave the age of the son 3 years back; the question was to find his age 3 years from now. (x + 6) was what was to be found. #### GMAT Online CourseTry it free! Register in 2 easy steps and Start learning in 5 minutes!<\|endoftext\|>	4.53125
7,739	Explicitly, it is used to define the degree of a polynomial and the notion of homogeneous polynomial, as well as for graded monomial orderings used in formulating and computing Gröbner bases. If you do not have two perfect square terms, then this trinomial is not a perfect square trinomial. A trinomial is a perfect square trinomial if it can be factored into a binomial multiplied to itself. n It may contain on both positive and negative values. ( , n ) The result is a perfect square trinomial. Both uses of this notion can be found, and in many cases the distinction is simply ignored, see for instance examples for the first[2] and second[3] meaning. In algebraic geometry the varieties defined by monomial equations To see why, remember how you rationalize a binomial denominator; or just check what happens when you multiply those two factors. The square roots of two of the terms multiplied by two will equal either the negative or positive version of the third term. Find the GCF of all the terms of the polynomial. 1 Using the square root property on both sides of the equation yields a linear on one side and a positive/negative number on the other making it much easier to solve. for some set of α have special properties of homogeneity. The difference of squares, the sum of cubes, and the difference of cubes are other polynomials that fall into the special products category. The square root of x2 is x, the square root of 36 is 6, and 2 times x (which is the same as 1) times 6 equals 12x/-12x, which does equal the other term. Check by multiplying the factors. For example, take the binomial (x + 2) and multiply it by itself (x + 2). When studying the structure of polynomials however, one often definitely needs a notion with the first meaning. If the variables being used form an indexed family like + For this: Step Three: Multiply 2 by a by 'b2(2x)(-3y) = -12xy, Step Four: Add a2, b2, and 2ab4x2 - 12xy + 9y2. x Definition of Like Terms. All other trademarks and copyrights are the property of their respective owners. x squared times x squared equals x to the fourth, so x to the fourth is a perfect square. Plus, get practice tests, quizzes, and personalized coaching to help you Perfect square trinomials are a vital component of the completing the square algorithm. How to find c to make a perfect square trinomial? This expression can also be given in the form of a binomial coefficient, as a polynomial expression in d, or using a rising factorial power of d + 1: The latter forms are particularly useful when one fixes the number of variables and lets the degree vary. , ..., then multi-index notation is helpful: if we write. In informal discussions the distinction is seldom important, and tendency is towards the broader second meaning. d You should be able to take the binomials and find the perfect square trinomial and you should be able to take the perfect square trinomials and create the binomials from which it came. So, this trinomial factors as, 2. The number to be added to both sides of the equation to create a perfect square trinomial is the value of (b / 2a)2. Use the Distributive Property âin reverseâ to factor the expression. ; these numbers form the sequence 1, 3, 6, 10, 15, ... of triangular numbers. The trinomial is a perfect square trinomial. {\displaystyle x_{3}} d 0 Before we can get to defining a perfect square trinomial, we need to review some vocabulary. In mathematics, a monomial is, roughly speaking, a polynomial which has only one term.Two definitions of a monomial may be encountered: A monomial, also called power product, is a product of powers of variables with nonnegative integer exponents, or, in other words, a product of variables, possibly with repetitions. Create your account. Try refreshing the page, or contact customer support. We'll now progress beyond the world of purely linear expressions and equations and enter the world of quadratics (and more generally polynomials). Log in here for access. 2 Notation for monomials is constantly required in fields like partial differential equations. They will factor into (a + b)(a + b) or (a - b)(a - b) where a and b are the square root of the perfect square terms. y flashcard set{{course.flashcardSetCoun > 1 ? It is called a fifth degree polynomial. All rights reserved. Multiplication : Multiplication is the repeated addition of the same number denoted with the symbol x. Learn to factor expressions that have powers of 2 in them and solve quadratic equations. A monomial will never have an addition or a subtraction sign. 1 Any time you take a binomial and multiply it to itself, you end up with a perfect square trinomial. {\displaystyle x_{2}} = A trinomial is a perfect square trinomial if it can be factored into a binomial multiplied to itself. The degree of a monomial is defined as the sum of all the exponents of the variables, including the implicit exponents of 1 for the variables which appear without exponent; e.g., in the example of the previous section, the degree is Already registered? For example, 2 × x × y × z is a monomial. ( ( d Once you've finished, you should be able to: To unlock this lesson you must be a Study.com Member. Multiple : The multiple of a number is the product of that number and any other whole number. {\displaystyle x_{1}} Examples: $$6x$$, $$7x^3$$, $$2ab$$ Binomial . 2 x The constant 1 is a monomialâ¦ n A perfect square trinomial is a special kind of polynomial consisting of three terms. ) How to find c in a perfect square trinomial? With perfect square trinomials, you will need to be able to move forwards and backwards. This video covers common terminology like terms, degree, standard form, monomial, binomial and trinomial. In mathematics, an irreducible polynomial is, roughly speaking, a polynomial that cannot be factored into the product of two non-constant polynomials.The property of irreducibility depends on the nature of the coefficients that are accepted for the possible factors, that is, the field or ring to which the coefficients of the polynomial and its possible factors are supposed to belong. n = This area is studied under the name of torus embeddings. 2 3. For those that are polynomials, state whether the polynomial is a monomial, a binomial, or a trinomial. (This is the part where you are moving the other way). ) The number of monomials of degree d in n variables is the number of multicombinations of d elements chosen among the n variables (a variable can be chosen more than once, but order does not matter), which is given by the multiset coefficient In the process of removing parentheses we have already noted that all terms in the parentheses are affected by the sign or number preceding the parentheses. Factor the greatest common factor from a polynomial. Determine if the following trinomials are perfect square trinomials. Monomial; Binomial; Trinomial; Monomial. + Now you should find the square root of both perfect square terms. What is an example of a perfect square trinomial? This is for instance the case when considering a monomial basis of a polynomial ring, or a monomial ordering of that basis. She has over 10 years of teaching experience at high school and university level. succeed. 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We now extend this idea to multiply a monomial by a polynomial. 's' : ''}}. Perfect square trinomials are used to solve equations, primarily quadratics by completing the square. 3x 3: This is a one-term algebraic expression that is actually referred to as a monomial. 1 Introduction to polynomials. {{courseNav.course.topics.length}} chapters \| We'll also learn to manipulate more general polynomial expressions. How Long is the School Day in Homeschool Programs? Monomial degree is fundamental to the theory of univariate and multivariate polynomials. For example, 3x+2x-5 is a polynomial. ) Polynomials may also contains on decimal values. x Polynomials are sums of terms of the form kâxâ¿, where k is any number and n is a positive integer. Enrolling in a course lets you earn progress by passing quizzes and exams. 2 It consists of only three variables. {\textstyle \left(\!\! d = + 7 times 7 is 49, so 49 is a perfect square. ( x2 - 12x + 36 can be factored into (x - 6)(x- 6), also written as (x - 6)2. A binomial is a polynomial that consists of two terms. . From these expressions one sees that for fixed n, the number of monomials of degree d is a polynomial expression in d of degree A monomial consists of only one term with a condition that this term should be non-zero. If the third term is negative, you will have (a - b)2, and if the third term is positive, it will be (a + b)2. The Hilbert series is a compact way to express the number of monomials of a given degree: the number of monomials of degree d in n variables is the coefficient of degree d of the formal power series expansion of. (This is the part where you are moving the other way). In order to solve a quadratic equation, it is possible to add the same number to both sides of the equation; thus creating a perfect square trinomial on one side and a number on the other side of the equal sign. Example: (3x + 2y)2 = 9x2 + 12xy + 4y2. 9x 5 - 2x 3x 4 - 2: This 4 term polynomial has a leading term to the fifth degree and a term to the fourth degree. Once again, if this is not the case, you do not have a perfect square trinomial. A binomial can be considered as a sum or difference between two or more monomials. â Definition of Trinomial â Factoring Quadratics â Solve literal equations for a given variable ... â Real World Examples of Quadratic Equations â Solving Word Questions. 3 Sociology 110: Cultural Studies & Diversity in the U.S. Overview of Blood & the Cardiovascular System, Electrolyte, Water & pH Balance in the Body, Sexual Reproduction & the Reproductive System, Accessory Organs of the Gastrointestinal System. For example, = is a monomial. − A trinomial is an expression which is composed of exactly three terms. For example, in the trinomial x2 - 12x + 36, both x2 and 36 are perfect squares. To find the perfect square trinomial from the binomial, you will follow four steps: Let's add some numbers now and find the perfect square trinomial for 2x - 3y. With either definition, the set of monomials is a subset of all polynomials that is closed under multiplication. If (xâ2+â3) is a factor of a polynomial with rational coefficients, then (xâ2ââ3) must also be a factor. 2.Polynomial Equation: Polynomial Equation can be expressed in terms of monomial, binomial, trinomial and higher order polynomials. d The unique pattern with perfect square trinomials is that their factors consist of the repetition of one binomial. What is a perfect square trinomial example? A trinomial is a polynomial or algebraic expression, which has a maximum of three non-zero terms. The degree of a monomial is sometimes called order, mainly in the context of series. {\displaystyle x^{\alpha }=0} ) Even though the first and last terms are perfect squares, the middle term is not equal to 2 times the product of the square roots of the first and last terms. These polynomials are grouped this way because they have a unique pattern to factoring them. The trinomial can then be written as the square of a binomial. Sciences, Culinary Arts and Personal Monomials A monomial is a number, a variable, or the product of a number and one or more variables. d What is the perfect square trinomial formula? . What are the factors of a perfect square trinomial? lessons in math, English, science, history, and more. + Perfect square trinomials are algebraic expressions with three terms that are created by multiplying a binomial to itself. If a polynomial has two terms it is called a binomial. If you start with the standard form of a quadratic equation and complete the square on it, the result would be the quadratic formula. d Since the word "monomial", as well as the word "polynomial", comes from the late Latin word "binomium" (binomial), by changing the prefix "bi" (two in Latin), a monomial should theoretically be called a "mononomial". It is also called total degree when it is needed to distinguish it from the degree in one of the variables. Monomial: An algebraic expression made up of one term. David holds a Master of Arts in Education. ¯ In the following practice problems, students will identify and factor perfect square trinomials, solve a quadratic equation by completing the square, and derive the quadratic formula by completing the square. − The trinomial is not a perfect square trinomial. In mathematics, a monomial is, roughly speaking, a polynomial which has only one term. 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Constants are monomials that contain no variables. Perfect square trinomials are a special group of polynomials that can be factored into a very convenient pattern, making them very useful in solving equations. 2 ( x "Monomial" is a syncope by haplology of "mononomial".[1]. , Perfect Square Trinomial: Definition, Formula & Examples, Perfect Square Binomial: Definition & Explanation, Solving Quadratic Trinomials by Factoring, What is Factoring in Algebra? Perfect square trinomials are often introduced in algebra courses in a section that would be entitled 'Special Products.' 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You should get the positive or negative version of the other term. 2 n This follows from the one-to-one correspondence between the monomials of degree d in n+1 variables and the monomials of degree at most d in n variables, which consists in substituting by 1 the extra variable. 4 x 3 is equal to 3 + 3 + 3 + 3. ! Log in or sign up to add this lesson to a Custom Course. 2. If they are, factor the perfect square trinomial as a binomial squared. + {\displaystyle n-1} - Definition & Example, Factoring By Grouping: Steps, Verification & Examples, How to Multiply and Divide Rational Expressions, Solving Problems Using Rational Equations, How to Add and Subtract Rational Expressions, How to Solve Quadratics That Are Not in Standard Form, How to Factor a Perfect Cube: Formula & Examples, Solving Quadratic Inequalities in One Variable, Practice Adding and Subtracting Rational Expressions, What is an Equation in Math? c n . Example: 3, A perfect square trinomial is a special polynomial consisting of three terms, A perfect square trinomial is created by multiplying a binomial to itself, Two of the terms in a perfect trinomial are perfect squares, They can be used to solve quadratics by completing the square, Describe what constitutes a perfect square trinomial, Explain how to use perfect square trinomials to solve quadratics. ( ( + Kathryn earned her Ph.D. in Mathematics from UW-Milwaukee in 2019. It is called a second-degree polynomial and often referred to as a trinomial. Binomials are algrebraic expressions containing only two terms. = It is written as the sum or difference of two or more monomials. Solve by using the perfect squares method: x^2 - 12x + 36 = 0, Working Scholars® Bringing Tuition-Free College to the Community. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons For example, the number of monomials in three variables ( . Completing the square using perfect square trinomials is also helpful when manipulating the terms in the equation of a circle so that the center and radius of the circle can be easily read from the equation. To solve by completing the square, we first move the constant term to the other side of the equation. If a polynomial has three terms it is called a trinomial. Every quadratic equation can be written as ax2 + bx + c = 0, which is called the standard form. A few examples of binomials are: â 5x+3, 6a 4 + 17x; xy 2 +xy; Trinomial. 3 © copyright 2003-2021 Study.com. The degree of a nonzero constant is 0. ) of degree d is ... â Definition of Monomial â Polynomial Definition â Like Terms â Divide a polynomial by a monomial or binomial, where the quotient has no remainder. ) + A binomial is a polynomial expression which contains exactly two terms. n Recognizing when you have these perfect square trinomials will make factoring them much simpler. Perfect squares are numbers or expressions that are the product of a number or expression multiplied to itself. {\displaystyle n=3} Zero polynomial (degree undefined or −1 or −∞), https://en.wikipedia.org/w/index.php?title=Monomial&oldid=1008073202, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, A monomial is a monomial in the first sense multiplied by a nonzero constant, called the, This page was last edited on 21 February 2021, at 11:59. ) {\textstyle {\frac {1}{(n-1)!}}} a For example, the degree of −7 is 0. This can be phrased in the language of algebraic groups, in terms of the existence of a group action of an algebraic torus (equivalently by a multiplicative group of diagonal matrices). 2, 4, 6, and 8 are multiples of 2. They are also very helpful when solving and graphing certain kinds of equations. {\displaystyle xyz^{2}} The square root of the first term is 2x and the square root of the last term is 5 and 22x5 = 20x which is the opposite of the middle term. 1.a. b. ( The most obvious fact about monomials (first meaning) is that any polynomial is a linear combination of them, so they form a basis of the vector space of all polynomials, called the monomial basis - a fact of constant implicit use in mathematics. Get unlimited access to over 83,000 lessons. Solve the following quadratic equation by completing the square. The degree of b The number of monomials of degree at most d in n variables is Start by moving the constant to the right side and dividing everything by a. Example: Trinomials are algebraic expressions that contain three terms. {\textstyle {\binom {n+d}{n}}={\binom {n+d}{d}}} {\textstyle {\frac {1}{2}}(d+1)^{\overline {2}}={\frac {1}{2}}(d+1)(d+2)} multiply the two square roots together and then by two. Two definitions of a monomial may be encountered: In the context of Laurent polynomials and Laurent series, the exponents of a monomial may be negative, and in the context of Puiseux series, the exponents may be rational numbers. {{courseNav.course.mDynamicIntFields.lessonCount}}, Exponential Notation: Definition & Examples, Composite Function: Definition & Examples, Horizontal Line Test: Definition & Overview, Polynomial Inequalities: Definition & Examples, Biological and Biomedical Is x^2 + 4x + 4 a perfect square trinomial? Rewrite each term as a product using the GCF. z ) As a member, you'll also get unlimited access to over 83,000 1 x is 1+1+2=4. \| {{course.flashcardSetCount}} 3. α 1 with leading coefficient As Jeff Beckman pointed out (20 June 2006), this is emphatically not true for odd roots. Example are: 2x 2 + y + z, r + 10p + 7q 2, a + b + c, 2x 2 y 2 + 9 + z, are all trinomials having three variables. Negative Exponent a nand a for any real number a 0 and any integer n. When you simplify an expression, you rewrite it without parentheses or negative exponents. An error occurred trying to load this video. In a perfect square trinomial, two of your terms will be perfect squares. {\displaystyle a+b+c} The remainder of this article assumes the first meaning of "monomial". An argument in favor of the first meaning is also that no obvious other notion is available to designate these values (the term power product is in use, in particular when monomial is used with the first meaning, but it does not make the absence of constants clear either), while the notion term of a polynomial unambiguously coincides with the second meaning of monomial. Implicitly, it is used in grouping the terms of a Taylor series in several variables. Trinomial. Solve the standard form of a quadratic equation. The square root of the first term is x and the square root of the last term is 2, but 2*2x = 4x which is not equal to the middle term - 8x. 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487	Nearly all plants go dormant in winter—whether they’re growing indoors or out in the garden. This period of rest is crucial to their survival in order to regrow each year. While plant dormancy during cold conditions is important, it may be equally important during times of stress. For instance, during periods of extreme heat or drought, many plants (especially trees) will go into a dormancy-like state, shedding their leaves early in order to conserve what little moisture may be available to ensure their survival. Making a Plant Go Dormant Normally, you don’t need to do anything to get a plant to go dormant. This usually happens on its own, though some indoor plants may need to be coaxed. Most plants can detect the shorter days towards the end of summer or early fall. As cooler temperatures begin to approach soon after, plant growth will start to decline as they enter into dormancy. With houseplants, it may help to move them to a darker and cooler area of the home in order to allow them to go dormant. Once a plant is dormant, foliage growth may be limited and even drop off, but the roots will continue to grow and thrive. This is why fall is often an ideal and preferable time for transplanting. Outdoor plants that are in the ground won’t need any help, though outdoor potted plants may need to be moved, depending on the climate and type of plant. Most potted plants can be moved indoors or for hardier types, an unheated garage will be sufficient over winter. For a fully dormant plant (one that loses its leaves), monthly watering during winter dormancy can also be given, though no more than this. Revive a Dormant Plant Depending on your location, it can take weeks for plants to come out of dormancy in spring. To revive a dormant plant indoors, bring it back into indirect light. Give it a thorough watering and a boost of fertilizer (diluted at half strength) to encourage new growth. Do not move any potted plants back outdoors until all threat of frost or freezing temps has passed. Most outdoor plants require little maintenance other than trimming back to allow for new growth to come through. A dose of fertilizer in spring can also help encourage the regrowth of foliage, though it will oftentimes occur naturally whenever the plant is ready.<\|endoftext\|>	3.78125
710	# If f(x) = (x^2-4)/ (x-1), what is f(0), f(1/2), f(-2), f(x-2), f(r^2) and f(1/t)? Dec 3, 2017 $f \left(0\right) = 4$ $f \left(\frac{1}{2}\right) = \frac{15}{2}$ $f \left(- 2\right) = 0$ $f \left(x - 2\right) = \frac{x \left(x - 4\right)}{x - 3}$ $f \left({r}^{2}\right) = \frac{{r}^{4} - 4}{{r}^{2} - 1}$ $f \left(\frac{1}{t}\right) = \frac{\left(\frac{1}{t} ^ 2\right) - 4}{\left(\frac{1}{t}\right) - 1}$ #### Explanation: To find the expressions that replaces the x in f(x), we have to replace the x in the equation with whatever is in the bracket, so if $f \left(x\right) = \frac{{x}^{2} - 4}{x - 1}$ $f \left(0\right) = \frac{{0}^{2} - 4}{0 - 1} = \frac{- 4}{- 1} = 4$ $f \left(\frac{1}{2}\right) = \frac{{\left(\frac{1}{2}\right)}^{2} - 4}{\left(\frac{1}{2}\right) - 1} = \frac{\left(\frac{1}{4}\right) - 4}{\left(\frac{1}{2}\right) - 1} = \frac{15}{2}$ $f \left(- 2\right) = \frac{{\left(- 2\right)}^{2} - 4}{\left(- 2\right) - 1} = \frac{0}{-} 3 = 0$ $f \left(x - 2\right) = \frac{{\left(x - 2\right)}^{2} - 4}{\left(x - 2\right) - 1} = \frac{{x}^{2} - 4 x + 4 - 4}{x - 3} = \frac{x \left(x - 4\right)}{x - 3}$ $f \left({r}^{2}\right) = \frac{{\left({r}^{2}\right)}^{2} - 4}{\left({r}^{2}\right) - 1} = \frac{{r}^{4} - 4}{{r}^{2} - 1}$ $f \left(\frac{1}{t}\right) = \frac{{\left(\frac{1}{t}\right)}^{2} - 4}{\left(\frac{1}{t}\right) - 1} = \frac{\left(\frac{1}{t} ^ 2\right) - 4}{\left(\frac{1}{t}\right) - 1}$<\|endoftext\|>	4.53125
2,161	## Activity: Vector Differential--Curvilinear Vector Calculus II 2022 (7 years) In this small group activity, students are given a picture as a guide. They then write down an algebraic expression for the vector differential in different coordinate systems (cartesian, cylindrical, spherical). Use Vector Differential--Rectangular as an introduction. This activity can be done simultaneously with Pineapples and Pumpkins where students or the instructor cut volume elements out of pineapples and/or pumpkins to show the geometry. • This activity is used in the following sequences • Media Cylindrical Coordinates: Find the general form for $d\vec{r}$ in cylindrical coordinates by determining $d\vec{r}$ along the specific paths below. • Path 1 from $(s,\phi,z)$ to $(s+ds,\phi,z)$: $d\vec{r}=\hspace{35em}$ • Path 2 from $(s,\phi,z)$ to $(s,\phi+d\phi,z)$: $d\vec{r}=\hspace{35em}$ • Path 3 from $(s,\phi,z)$ to $(s,\phi,z+dz)$: $d\vec{r}=\hspace{35em}$ If all three coordinates are allowed to change simultaneously, by an infinitesimal amount, we could write this $d\vec{r}$ for any path as: $d\vec{r}=\hspace{35em}$ This is the general line element in cylindrical coordinates. Spherical Coordinates: Find the general form for $d\vec{r}$ in spherical coordinates by determining $d\vec{r}$ along the specific paths below. • Path 1 from $(r,\theta,\phi)$ to $(r+dr,\theta,\phi)$: $d\vec{r}=\hspace{35em}$ • Path 2 from $(r,\theta,\phi)$ to $(r,\theta+d\theta,\phi)$: $d\vec{r}=\hspace{35em}$ • Path 3 from $(r,\theta,\phi)$ to $(r,\theta,\phi+d\phi)$: (Be careful, this is a tricky one!) $d\vec{r}=\hspace{35em}$ If all three coordinates are allowed to change simultaneously, by an infinitesimal amount, we could write this $d\vec{r}$ for any path as: $d\vec{r}=\hspace{35em}$ This is the general line element in spherical coordinates. ## Instructor's Guide ### Main Ideas This activity allows students to derive formulas for $d\vec{r}$ in cylindrical, and spherical coordinates, using purely geometric reasoning. These formulas form the basis of our unified view of all of vector calculus, so this activity is essential. For more information on this unified view, see our publications, especially: Using differentials to bridge the vector calculus gap Using a picture as a guide, students write down an algebraic expression for the vector differential in different coordinate systems (cylindrical, spherical). ### Introduction Begin by drawing a curve (like a particle trajectory, but avoid "time" in the language) and an origin on the board. Show the position vector $\vec{r}$ that points from the origin to a point on the curve and the position vector $\vec{r}+d\vec{r}$ to a nearby point. Show the vector $d\vec{r}$ and explain that it is tangent to the curve. It may help to do activity Vector Differential--Rectangular as an introduction. ### Student Conversations For the case of cylindrical coordinates, students who are pattern-matching will write $d\vec{r} = dr\, \hat{r} + d\phi\, \hat{\phi} + dz\, \hat{z}$. Point out that $\phi$ is dimensionless and that path two is an arc with arclength $r\, d\phi$. Some students will remember the formula for arclength, but many will not. The following sequence of prompts can be helpful. • What is the circumference of a circle? • What is the arclength for a half circle? • What is the arclength for the angle $\pi\over 2$? • What is the arclength for the angle $\phi$? • What is the arclength for the angle $d\phi$? For the spherical case, students who are pattern matching will now write $d\vec{r} = dr\, \hat{r} + d\phi\, \hat{\phi} + d\theta\, \hat{\theta}$. It helps to draw a picture in cross-section so that they can see that the circle whose arclength gives the coefficient of $\hat{\theta}$ has radius $r\sin\theta$. It can also help to carry around a basketball to write on to talk about the three dimensional geometry of this problem. ### Wrap-up The only wrap-up needed is to make sure that all students have (and understand the geometry of!) the correct formulas for $d\vec{r}$. • assignment_ind Vector Differential--Rectangular assignment_ind Small White Board Question 10 min. ##### Vector Differential--Rectangular Static Fields 2022 (7 years) Integration Sequence In this introductory lecture/SWBQ, students are given a picture as a guide. They then write down an algebraic expression for the vector differential in rectangular coordinates for coordinate equals constant paths. This activity can be done as a mini-lecture/SWBQ as an introduction to Vector Differential--Curvilinear where students find the vector differential in cylindrical and spherical coordinates.. • group Vector Integrals (Contour Map) group Small Group Activity 30 min. ##### Vector Integrals (Contour Map) • group Number of Paths group Small Group Activity 30 min. ##### Number of Paths Student discuss how many paths can be found on a map of the vector fields $\vec{F}$ for which the integral $\int \vec{F}\cdot d\vec{r}$ is positive, negative, or zero. $\vec{F}$ is conservative. They do a similar activity for the vector field $\vec{G}$ which is not conservative. • group Work By An Electric Field (Contour Map) group Small Group Activity 30 min. ##### Work By An Electric Field (Contour Map) Students will estimate the work done by a given electric field. They will connect the work done to the height of a plastic surface graph of the electric potential. • group Covariation in Thermal Systems group Small Group Activity 30 min. ##### Covariation in Thermal Systems Students consider how changing the volume of a system changes the internal energy of the system. Students use plastic graph models to explore these functions. • group Quantifying Change (Remote) group Small Group Activity 30 min. ##### Quantifying Change (Remote) In this activity, students will explore how to calculate a derivative from measured data. Students should have prior exposure to differential calculus. At the start of the activity, orient the students to the contour plot - it's busy. • accessibility_new Time Dilation Light Clock Skit accessibility_new Kinesthetic 5 min. ##### Time Dilation Light Clock Skit Students act out the classic light clock scenario for deriving time dilation. • assignment Paramagnet (multiple solutions) assignment Homework ##### Paramagnet (multiple solutions) Energy and Entropy 2021 (2 years) We have the following equations of state for the total magnetization $M$, and the entropy $S$ of a paramagnetic system: \begin{align} M&=N\mu\, \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}}\\ S&=Nk_B\left\{\ln 2 + \ln \left(e^{\frac{\mu B}{k_B T}}+e^{-\frac{\mu B}{k_B T}}\right) +\frac{\mu B}{k_B T} \frac{e^{\frac{\mu B}{k_B T}} - e^{-\frac{\mu B}{k_B T}}} {e^{\frac{\mu B}{k_B T}} + e^{-\frac{\mu B}{k_B T}}} \right\} \end{align} 1. List variables in their proper positions in the middle columns of the charts below. 2. Solve for the magnetic susceptibility, which is defined as: $\chi_B=\left(\frac{\partial M}{\partial B}\right)_T$ 3. Using both the differentials (zapping with d) and chain rule diagram methods, find a chain rule for: $\left(\frac{\partial M}{\partial B}\right)_S$ 4. Evaluate your chain rule. Sense-making: Why does this come out to zero? • group Paramagnet (multiple solutions) group Small Group Activity 30 min. ##### Paramagnet (multiple solutions) • Students evaluate two given partial derivatives from a system of equations. • Students learn/review generalized Leibniz notation. • Students may find it helpful to use a chain rule diagram. • assignment Free Expansion assignment Homework ##### Free Expansion Energy and Entropy 2021 (2 years) The internal energy is of any ideal gas can be written as \begin{align} U &= U(T,N) \end{align} meaning that the internal energy depends only on the number of particles and the temperature, but not the volume.* The ideal gas law \begin{align} pV &= Nk_BT \end{align} defines the relationship between $p$, $V$ and $T$. You may take the number of molecules $N$ to be constant. Consider the free adiabatic expansion of an ideal gas to twice its volume. “Free expansion” means that no work is done, but also that the process is also neither quasistatic nor reversible. 1. What is the change in entropy of the gas? How do you know this? 2. What is the change in temperature of the gas? Author Information Corinne Manogue, Tevian Dray, & Katherine Meyer Learning Outcomes<\|endoftext\|>	4.71875
1,129	Show and Tell with Powerpoint® If you speak it and don’t show it; the auditory learners will get it, the kinetic learners might get it, but the visual learners probably won’t. If you speak it and show it – the positive impact on understanding and retention is dramatic. The Dual Coding Theory of memory was initially proposed by Paivio (1971). The core idea is the human mind operates with two distinct classes of mental “codes”: verbal representations and mental images. Human memory thus comprises two functionally independent (although interacting) systems: verbal memory and image memory. Imagery potentiates recall of verbal material because when a word evokes an associated image (either spontaneously, or through deliberate effort) two separate but linked memory traces are laid down, one in each of the memory stores. Obviously the chances that a memory will be retained and retrieved are much greater if it is stored in two distinct functional locations rather than in just one. See, Nigel J.T. Thomas in the Stanford Encyclopdia of Philosophy. http://bit.ly/9e3HCo. To use less scientific terminology: seeing is believing. Consider what types of visual aids you will be using and why they will help. Make sure they match to the presentation. They should not be too cutsie. They should add rather than detract from your message. Do not show visuals filled with words. Do not show visuals filled with words. If you show a visual filled with words, then Paivio’s dual processing channels cross wires. While you are speaking, the audience will be reading. They will have to tune you out to read. Or they will have to ignore the printing to listen to what you are saying. Comprehension and retention will significantly decrease. It is actually better to show no visual, than to show one filled with words. Cliff Atkinson is the author of Beyond Bullet Points: Using Microsoft PowerPoint to Create Presentations that Inform, Motivate, and Inspire. © 2005 Cliff Atkinson. Microsoft Press He suggests ways to structure slides so that they are not filled with bullet points. He recommends dispensing with templates and using blank slide formats. He has a website filled with free materials and helpful advice. www.beyondbulletpoints.com. Here are some suggestions for using Powerpoint® visuals during your presentation: - Check out the courtroom ahead of time and figure out where the screen and projector should go. - Get advance permission from the bailiff for the set up. - Use a projector that's bright enough so you don't have to dim the lights (jurors will fall asleep if you do that) - Use a remote mouse - The mouse should have a "black screen" button - Having the screen black focuses the attention on you and should be used periodically to drive points home. - Situate the laptop screen in front of you off to the side so you can glance at it w/o having to turn your head back to look at the screen - Always have a backup plan. - If it takes longer than two minutes to fix a glitch and a recess isn’t available, go to that other plan. - Don’t talk to the screen. - Some people stand in front of the screen (not obstructing it but in front of it newscaster style) and never look at the screen. Other people interact with the screen. There is no rule governing this other than – don’t talk to the screen. - There is a “notes” view that allows you to view your notes on the computer monitor so you don’t need to talk to the screen or shuffler paper notes. - The slides should match up with what you are talking about without using the same words - If you show a document, blow up and highlight the important part of it - Words must be big enough for the audience to read - Don't use all capitals. - The general rule is no more than three to four chunks of data per slide - Avoid more than one photo per slide unless you are comparing pictures - Crop photos - Don’t use templates - Headlines should be short complete sentences - Rarely use headlines - Don't interact with every slide - Interact with some of the slides - Contrast should be bright and clear - If you are trying to color code slides - remember, men in particular may be colorblind - Special effects should rarely be used - Same for clip art - Get photos and images of real things from the case file and from the internet - If it is too cute or basic, it will be seen as condescending - If you modify a photo to make a point, type out a disclaimer at the bottom of the slide, i.e. "This photo has been altered" - Pause before/after presenting a dramatic/humorous slide; or risk the audience forgetting what else you say - Try out the slide show on others before you show it to a judge/jury - Use more than a few, but not too many slides. A couple per minute. - If the judge requires you to show the other side your slides, then make sure you do so. There won't be a problem if you are simply showing evidence or things you disclosed prior to trial. Other images may require a bit of discussion, negotiation, or motion practice.<\|endoftext\|>	3.84375
1,064	home PDF (letter size) PDF (legal size) ## Finding roots of unity using Euler and De Moivreś June 14,2006 Compiled on May 22, 2020 at 4:22am To ﬁnd the roots of $f(x) = x^{n}-1$ Solving for $$x$$ from \begin{align} 0 & =x^{n}-1\nonumber \\ x^{n} & =1\nonumber \\ x & =1^{\frac{1}{n}}\tag{1} \end{align} Now $$1^{\frac{1}{n}}$$ is evaluated. Since $1=e^{i\left ( 2\pi \right ) }$ Substituting (2) in the RHS of (1) gives \begin{align} x & =(e^{i2\pi })^{\frac{1}{n}}\nonumber \\ & =\left ( \cos 2\pi +i\sin 2\pi \right ) ^{\frac{1}{n}}\tag{3} \end{align} Using De Moivre’s formula $\left ( \cos \alpha +i\sin \alpha \right ) ^{\frac{1}{n}}=\cos \left ( \frac{\alpha }{n}+k\frac{2\pi }{n}\right ) +i\sin \left ( \frac{\alpha }{n}+k\frac{2\pi }{n}\right ) \ \ \ \ \ \ \ k=0,1,\cdots n-1$ Therefore (3) is rewritten as$x=\cos \left ( \frac{2\pi }{n}+k\frac{2\pi }{n}\right ) +i\sin \left ( \frac{2\pi }{n}+k\frac{2\pi }{n}\right ) \ \ \ \ \ \ \ k=0,1,\cdots n-1$ The above gives the roots of $$f(x)=x^{n}-1$$. The following examples illustrate the use of the above. 1. Solve $$f(x)=x^{2}-1$$. Here $$n=2$$, therefore $$k=0,1$$. For $$k=0$$ \begin{align} x & =\cos \left ( \frac{2\pi }{2}\right ) +i\sin \left ( \frac{2\pi }{2}\right ) \\ & =-1 \end{align} And for $$k=1$$\begin{align} x & =\cos \left ( \frac{2\pi }{2}+\frac{2\pi }{2}\right ) +i\sin \left ( \frac{2\pi }{2}+\frac{2\pi }{2}\right ) \\ & =1 \end{align} Hence the two roots are $$\{1,-1\}$$ 2. Solve $$f(x)=x^{3}-1$$. Here $$n=3$$, hence for $$k=0$$\begin{align} x & =\cos \left ( \frac{2\pi }{3}\right ) +i\sin \left ( \frac{2\pi }{3}\right ) \\ & =-\frac{1}{2}+i\frac{\sqrt{3}}{2} \end{align} And for $$k=1$$ \begin{align} x & =\cos \left ( \frac{2\pi }{3}+\frac{2\pi }{3}\right ) +i\sin \left ( \frac{2\pi }{3}+\frac{2\pi }{3}\right ) \\ & =\cos \left ( \frac{4\pi }{3}\right ) +i\sin \left ( \frac{4\pi }{3}\right ) \\ & =-\frac{1}{2}-i\frac{\sqrt{3}}{2} \end{align} And for $$k=2$$\begin{align} x & =\cos \left ( \frac{2\pi }{3}+2\frac{2\pi }{3}\right ) +i\sin \left ( \frac{2\pi }{3}+2\frac{2\pi }{3}\right ) \\ & =\cos \left ( \frac{6\pi }{3}\right ) +i\sin \left ( \frac{6\pi }{3}\right ) \\ & =1 \end{align} Therefore the roots are $$\{1,$$ $$-\frac{1}{2}+i\frac{\sqrt{3}}{2},-\frac{1}{2}-i\frac{\sqrt{3}}{2}\}$$<\|endoftext\|>	4.5
2,481	Lesson Video: The Median of a Data Set \| Nagwa Lesson Video: The Median of a Data Set \| Nagwa # Lesson Video: The Median of a Data Set Mathematics • First Year of Preparatory School ## Join Nagwa Classes In this video, we will learn how to find and interpret the median of a data set. 14:59 ### Video Transcript In this video, we will learn how to find and interpret the median of a data set. The median is a type of average. And we will begin by defining it. The median is an example of a measure of center or a measure of central tendency. The mode and mean are also measures of center, although in this video we will only discuss the median. The median is a single number which gives us some information about the typical values in a data set. Specifically, the median of a set of data represents the middle value. Half of the data is above the median and half of the data is below the median. We will begin by looking at some examples where we calculate the median of a small data set. Find the median of the values six, eight, 16, six, and 19. The median is the middle value of a set of data. In order to calculate the median of a small set of data, we follow two simple steps. Firstly, we put the numbers in ascending order. In this question, the order will be six, six, eight, 16, and 19. Our second step is to find the middle number or value. With a small data set, we can do this by crossing off numbers from either end. We begin by crossing off six and 19, the smallest and largest numbers. We then cross off six and 16. We are left with one number in the middle. Therefore, the median of the set of values is eight. An alternative way of finding the median after putting the numbers in ascending order is to use a little rule or formula. The median position can be calculated using the formula ๐ plus one divided by two, where ๐ is the number of values. In this question, we have five numbers. Therefore, the median position can be calculated by adding five to one and then dividing by two. This is equal to three. The median number will therefore be the third number in the list. Counting the numbers from left to right in ascending order, we see that the third number is eight. Therefore, the median is eight. Find the median of the values 13, five, nine, 10, two, and 15. We can calculate the median of any data set by following two steps. Firstly, we put the numbers in ascending order. In this question, our six numbers in ascending order are two, five, nine, 10, 13, and 15. The median is the middle number. Therefore, we need to find the middle value from our list. One way to do this is to cross off a number from each end of the list. We cross off the highest number, 15, and the lowest number, two. We then cross off the next highest and next lowest, 13 and five. We are now left with two middle values, nine and 10. To find the median in this case, we find the number that is halfway between the middle values. This can be calculated by adding the two middle values and then dividing by two. Nine plus 10 is equal to 19, and dividing this by two gives us 9.5. The median of the set of six values is therefore equal to 9.5. Half of our values must be above this, in this case, 10, 13, and 15. And half of the values must be below 9.5, nine, five, and two. An alternative method here wouldโve been to have found the median position. We do this using the formula ๐ plus one divided by two, where ๐ is the number of values. In this question, we had six values. We need to add six to one and divide by two. This is equal to 3.5. The median will therefore lie between the third and fourth value. As the third value was equal to nine and the fourth value 10, once again, we have proved that the median was 9.5. What is the median of the following numbers: 11, 11, eight, eight, nine, and nine? The median of any data set is the middle number. We can calculate this using two steps. Firstly, we put the numbers in ascending order. In this question, we have two eights, followed by two nines, followed by two 11s. For a small data set like this, we can then find the middle number by crossing off one number from either end, firstly, the eight and the 11. We can then cross off the second eight and the second 11. We are left with two middle numbers. Normally, when there are two middle values, we have to calculate the number that is halfway between them. In this case, since both the middle numbers are the same, this value is the median. The median of the six numbers 11, 11, eight, eight, nine, and nine is nine. The next question weโll look at is more complicated as there is a missing number, but we are given the median. Jennifer has the following data: 10, eight, seven, nine, and ๐. If the median is eight, which number could ๐ be? Is it (A) seven, (B) 8.5, (C) nine, (D) 9.5, or (E) 10? The median represents the middle of the data. This means that half of the data values are above it and half are below it. We can start by putting the data in order and thinking about how we can work out the number ๐. Writing the four values that we know in order gives us seven, eight, nine, and 10. We are told that the median is eight. As the median is the middle of five values, it must be the third number when listed in order from lowest to highest. There must be two values above eight and two values below eight. For the number to be written to the left of eight, it must be less than or equal to eight. If ๐ was any integer value less than seven, it would be the first number in the list. Positive integers here could be one, two, three, four, five, and six. None of these are options in the question, though. The missing number could also be seven as both the sevens would be written to the left-hand side of eight as they are less than eight. Out of our five options, the correct answer is option (A), seven. ๐ could not be 8.5, nine, 9.5, or 10 as all of these are greater than eight and would be on the right-hand side of eight. It is possible that the missing number couldโve been eight. If this was the case, seven would be the first number in our list. We would then have two eights. The third number would still be eight, which is the median. Whilst the value of ๐ could be any number less than or equal to eight, the correct answer in this case is seven. The final question that weโll look at is more complicated as it involves a large data set. Find the median of the set of data represented in this line plot. The number line here goes from two to 14. And the number of crosses represents how many of each value we have. There are four crosses above the number four. Therefore, we have four fours. There are six crosses above the number five, so we have six fives. We have two sixes and two sevens. Continuing this, we have six eights, three nines, two 10s, three 11s, three 12s, six 13s, and three 14s. We know that the median is the middle value when the numbers are in ascending order. One way to answer this question would be to write all the numbers out from smallest to largest. We would write the number four four times. We would write the number five six times. There would be two sixes and two sevens. The list would continue as shown all the way up to three 14s. There are a total here of 40 values. We could find the middle number by crossing off one from each end, firstly, the highest number and the lowest number. We repeat this by crossing another four and another 14. Crossing off the next 10 smallest numbers and next 10 largest numbers would leave us with the numbers from seven to 11. We could continue this process until weโre left with two middle values, eight and nine. When there are an even number of values in total, there will always be two middle values. We can then find the median by finding the number that is halfway between them. We do this in this case by adding eight and nine and then dividing the answer by two. Eight plus nine is equal to 17, and half of this is 8.5. Clearly, 8.5 is halfway between eight and nine. Therefore, this is the median of the set of data. An alternative method here would be to calculate the median position first. The median position can be calculated using the formula ๐ plus one divided by two, where ๐ is the total number of values. 40 plus one is equal to 41. And dividing this by two gives us 20.5. As 20.5 lies between the integers 20 and 21, we know that the median will be halfway between the 20th and 21st value. To find the 20th and 21st values, we can work out the running total or cumulative frequency. We do this by adding the number of values we have. Four plus six is equal to 10. Adding another two gives us 12. This means that 12 values are six or lower. Adding another two gives us 14, and adding six gives us 20. This means that there are 20 values that are eight or less. As there are 40 values in total, there must therefore be 20 values that are nine or greater. The 20th value is equal to eight, and the 21st value is equal to nine. Once again, finding the midpoint of these two values gives us a median of 8.5. This method is useful when we have a large number of values as it saves writing out the whole data set. We will now summarize the key points from this video on the median of a set of data. In order to calculate the median from a data set, our first step is to write the data in ascending order from least to greatest. Next, we need to find the middle value or values. If there are an odd number of data values, there will be exactly one value in the middle. However, if there are an even number of data values, there will be two values in the middle. Our third and final step is to find the median. If there are an odd number of data values, the median is the middle number. If there are an even number of data values, the median is halfway between the two middle numbers. Finally, if we have a large data set, we can calculate the median position using the formula ๐ plus one divided by two, where ๐ is the total number of data values. ## Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! • Interactive Sessions • Chat & Messaging • Realistic Exam Questions<\|endoftext\|>	4.96875
1,787	S k i l l i n A R I T H M E T I C Lesson 29 # The Method of Proportions In this Lesson, we will answer the following: 1. Any number is what percent of 100? 2. How do we find the Percent by the method of proportions? Section 2 3. What is a general method for finding the Percent? 1. Any number is what percent of 100? 5 is ? % of 100. Any number is that same percent of 100. 5 is 5% of 100. 12 is 12% of 100. 250 is 250% of 100. For, percent is how many for each 100, which is to say, percents are hundredths (Lesson 14). 5 is 5 hundredths of 100. That is the ratio of 5 to 100. 5 100 = 5%, 12 100 = 12%, 250100 = 250%. When the percent is less than or equal to 100%, then we can say "out of" 100. 5% is 5 out of 100. 12% is 12 out of 100. But 250% cannot mean 250 out of 100. That makes no sense. It means 250 for each 100, which is two and a half times (Lesson 15). Example 1. \$42.10 is what percent of \$42.10? Answer. 100% 100% is the whole thing. The method of proportions Example 2. 24 out of 100 is 24%. But what percent is 24 out of 200? If there are 24 for each 200, then for each 100 there are 12. (12 + 12 = 24.) 24 is 12% of 200. Again, percent is how much for each 100. 24 200 = ? 100 "24 out of 200 is how many out of 100?" To change 200 into 100, we must take half. Therefore the missing term is half of 24, which is 12. Example 3. 8 out of 25 is what percent? That is, 8 is what percent of 25? Solution. Now, percent is not out of 25, it is out of 100. But since there are 8 out of each 25, and in 100 there are four 25's, then in 100 there are four 8's: 32. 8 is 32% of 25. Again, if we had to do this problem formally, then we could write the proportion, 8 25 = ? 100 "8 out of 25 is how many out of 100?" Since four times 25 is 100, then the missing term is four times 8, which is 32. 2. How do we find the Percent by the method of proportions? Solve the proportion in which 100 is the fourth term, and the Amount and Base are the first and second. Amount Base = ? 100 "The Amount is to the Base as what number is to 100?" See the vocabulary of Lesson 27: "Amount," "Base." Example 4. 7 out of 28 students got A. What percent got A? Solution. Form this proportion, 7 28 = ? 100 Now, 7 is one quarter of 28. And 25 is one quarter of 100. Therefore, 7 28 = 25 100 = 25%. 7 is 25% of 28. Example 5. 11 out of 50 students studied French. What percent studied French? Solution. Form the proportion, 1150 = ? 100 "11 out of 50 is how many out of 100?" In this case, looking directly at the ratio of 11 to 50 does not help. We must look alternately (across). To make 50 into 100, we have to multiply by 2. Therefore, we must multiply 11 by 2, also: 1150 = 22 100 = 22%. Example 6. 11 out of 200 studied French. What percent studied French? Solution. In this case, to make 200 into 100, we must divide by 2, or take half. Therefore we must also take half of 11: 11 200 = 5½100 = 5½%. We see that to deal with fractions (ratios): We must divide both terms by the same number, or we must multiply both terms by the same number. Example 7. 3 is what percent of 25? Solution. This question means the same as, 3 out of 25 is what percent? In every case, the Base follows "of." Proportionally, 3 25 = ? 100 25 has been multiplied by 4. Therefore we must multiply 3 by 4, also: 3 25 = 12 4 × 3 100 4 × 25 3 25 = 12%. 3 is 12% of 25. With practice, this need not be a written calculation. By simply looking at the numbers, and realizing that the Base must be 100, the student can know the Percent. Example 8. 9 out of 20 students were able to stop smoking. What percent were able to stop smoking? Answer. 45%. For, to make 20 into 100, we must multiply by 5. Therefore we must multiply 9 by 5, also: 9 out of 20 is equal to 45 out of 100. To find the Percent, the Base must be 100. Example 9. What percent of 400 is 33? Solution. 400 is the Base; it follows "of." 33 400 = ? 100 To go from 400 to 100, we must divide by 4. Therefore we must divide 33 by 4, also. 33 400 = 8¼ 100 = 8¼%. "4 goes into 33 eight (8 ) times (32) with 1 left over." Example 10. 1000 people voted in the recent election, and 763 voted for Jones. What percent voted for Jones? Solution. 763 out of 1000 voted for Jones. 763 1000 = ? 100 1000 has been divided by 10. Therefore 763 also must be divided by 10. We will separate one decimal place: (Lesson 3, Question 5) 763 1000 = 76.31 76.3% voted for Jones. Example 11. In a class of 45 students, there were 9 A's. What percent got A? Solution. 9 out of 45 got A. 9 45 = ? 100 Here, we must look directly (down). 9 is a fifth of 45. And 20 is a fifth of 100. 9 45 = 20 100 . 20% got A. Example 12. 21 is what percent of 75? Solution. 2175 = ? 100 Now, 100 is a third more than 75, because the difference between them is 25, and 25 is a third of 75. Therefore, the missing term is a third more than 21: 21 + 7 = 28. 2175 = 28 100 . 21 is 28% of 75. But it will not alway be clear how to make the 4th term 100. We will consider that in the next Section. At this point, please "turn" the page and do some Problems. or Continue on to the next Section. 1st Lesson on Percent www.proyectosalonhogar.com<\|endoftext\|>	4.625
402	A check for short circuits is one of the most basic tests you can perform with a multimeter. On the simplest meters, you use the resistance setting; sophisticated models have a continuity setting that flashes a light or beeps a tone to let you know a connection is a short circuit. Turn Off Power Turn off all power to the circuit or device under test. Unplug the equipment from the AC outlet. Probing an electrical circuit with a multimeter may pose a dangerous shock hazard if the circuit's power is on. Set Multimeter to Resistance or Continuity Switch the multimeter on and turn its selector knob to the resistance setting. Use the continuity setting if your meter has that function. Some multimeters may have several resistance settings; choose the lowest resistance scale on the meter. Touch Probe Tips Together Touch the test probes together and observe that the resistance reading goes to nearly zero. For continuity, the light flashes or a tone sounds. Locate Circuit Component Locate the component or portion of the circuit you want to check for a short. The tested part should not normally have zero electrical resistance; for example, the input of an audio amplifier should have a resistance of at least several hundred ohms. Touch Probe Tips to Circuit Touch the metal tip of the black probe to the circuit’s chassis or electrical ground, and touch the tip of the red probe to the parts of the circuit you suspect may have a short. The tips of the probes must touch metal parts of the circuit, such as a component lead, circuit board foil or wire. Observe Meter Display Observe what the meter does when you touch the probes to the circuit. A high resistance signifies an open circuit. Very low resistance -- about 2 ohms or less -- indicates a short circuit. A meter with a continuity setting flashes or beeps only if it detects a short circuit. Some meters may indicate an open circuit as "overload" or infinite resistance.<\|endoftext\|>	3.78125
1,562	# Factors of 93: Prime Factorization, Methods, Tree, and Examples Factors of 93 are the numbers that can be divided by 93 without leaving any remainder. For the factors, the condition is that they must be exactly divisible by the given number or must have zero as a remainder when divided. Factors are also known as divisors of the given number. Figure 1 – All possible factors of 93 In this article, we will be finding the factors of 93. There are several ways to find factors of any number. We are going to learn how to find factors by the division methodAfter reading this article, you will have a clear understanding of prime factorization, prime numbers, and factor pairs by using positive and negative factors and a factor tree. In the end, there are some examples for better understanding and your practice. ## What Are the Factors of 93? The factors of 93 are 1, 3, 31, and 93, as all of them are exactly divisible by 93.The numbers that can completely divide 93 are included in the list of its factors. In other words, the remainder should always be zero. The given number 93 is not a prime number so it has more than 2 factors. It has both positive and negative factors although negative factors are not often considered. 93 has four factors in total. A number that has more than 2 factors is known as a composite number. ## How To Calculate the Factors of 93? To calculate the factors of 93, divide it by the smallest natural number which is 1.1 is a factor of all the whole numbers because it divides every number completely which means the remainder is zero.$\dfrac{93}{1} = 93,\ r = 0$As a result, 1 will be included in the factors list of 93.93 is an odd number, so it cannot be divided by 2. So, we will determine its smallest prime factor which is 3. Now, divide 93 by 3.$\dfrac{93}{3} = 31$This means 3 and 31 both are factors of 93 because both divide 93 completely and the remainder is zero in both cases.Check for the other natural numbers as well.Dividing 93 by 6 gives:$\dfrac{93}{6} =15.5$The remainder is 3, which is a non-zero number so 6 is not a factor of 93.Now divide 93 by 9:$\dfrac{93}{9}=10.33$The remainder is 3, which is also a non-zero number so 9 is also not a factor of 93.The last factor will be the number itself because every number divides itself fully.Following are the numbers that entirely divide the number 93 without leaving any remainder.$\dfrac{93}{1} = 93$$\dfrac{93}{3} = 31$$\dfrac{93}{31} = 3$$\dfrac{93}{93} = 1$The positive and negative factors of 93 are listed below:Positive factors are 1, 3, 31, and 93.Negative factors are -1, -3, -31, and -93. ### Properties of Factors of 93 Following are some important properties of factors of 93: 1. 93 is an odd number therefore, it has no even prime factor. 2. The factor of 93 can never be in the form of a decimal or fraction. 3. 93 is a semiprime. Semiprime is the natural number that is the product of two prime numbers. 4. 93 is also the first natural number in the third triples of successive semiprime numbers. The triplet is 93, 94, and 95. 5. The additive inverse of every factor of 93 is also its factor which is called a negative factor. ## Factors of 93 by Prime Factorization Prime numbers are the numbers that have only 2 factors. Those two factors are 1 and the other is the number itself. For example: 2,3,5,7,11….31 etc.(NOTE: 0 and 1 are not prime numbers)Prime Factorization means representing numbers by the product of their prime factors. The list of prime factors contains the factors which are prime numbers. This is an important topic. As mentioned above in the article factors of 93 are 1, 3, 31, & 93. The numbers 3 and 31 are prime numbers because they are not divisible on any number completely except for 1 and itself. So the prime factorization of 93 is 3 x 31. It can be expressed as: 93 = 3 x 31 This means when we multiply prime factors of the number the product will be the number itself. In simple words, prime factoring means writing the factors of a number that are prime only. ## Factor Tree of 93 The factor tree of 93 is shown below in figure 1: Figure 2 – Factor tree of 93 This diagram is known as a factor tree. The factor tree consists of factors of the number. At the top of the factor tree, each branch will contain its factors. It is a pictorial representation of factors of the given number.By looking at the factor tree, one could easily understand that by multiplying 3 and 31 we will get the original number which is 93. ## Factors of 93 in Pairs Pairing the factors of a number means writing them in such pairs that the product must be equal to the number itself. 3× 31=93 1× 93=93 The factor pairs for 93 will be (3, 31) and (1, 93).We can also find factor pairs with negative factors of 93 -3×- 31=93 -1× -93=93 The negative factor pairs of 93 are (-1, -93), and (-3, -31).When a negative sign is multiplied by a negative sign their product is always positive. ## Factors of 93 Solved Examples Following are some solved examples related to factors of 93. ### Example 1 Find the sum of all factors of 93. ### Solution Factors of 93 are 1, 3, 31, and 93.Add up all the factors to find the sum.Sum of all factors of 93 is given as: Sum = 1 + 3 + 31 + 93 Sum = 128 ### Example 2 Find the common factors of 93 and 3. ### Solution Factors of 93 are 1, 3, 31, and 93.As we know 3 is a prime number so it will have only 2 factors 1 and the number itself Factors of 3 are 1 and 3.Common factors mean factors that are part of both lists.Common factors of 3 and 93 are 1 and 3. ### Example 3 Find the negative factor pair of 93. ### Solution Negative factors of 93 are -1, -3, -31and -93.The first Factor pair will be (-1, -3).The second Factor pair will be (-31, -93).Negative factor pair of 93 are (-1, -3) and (-31, -93)Images/mathematical drawings are created with GeoGebra.<\|endoftext\|>	4.90625
1,493	# Thread: For which numbers a,b,c,d will the function f(x) satisfy... 1. ## For which numbers a,b,c,d will the function f(x) satisfy... Problem: For which numbers $\displaystyle a,b,c,d$ will the function $\displaystyle f(x)=\frac{ax+b}{cx+d}$ satisfy $\displaystyle f(f(x))=x$ for all $\displaystyle x$? ------------------------------------------------------------------------ Attempt: $\displaystyle f(f(x))=\frac{a\left[\frac{ax+b}{cx+d}\right]+b}{c\left[\frac{ax+b}{cx+d}\right]+d}\\ = \frac{\left[\frac{a^2x+ab}{cx+d}\right]+b}{\left[\frac{acx+bc}{cx+d}\right]+d}\\ =\frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)}$ By using polynomial division I get the following result: $\displaystyle \left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd) x+(bc+d^2)}$ By doing some hopefully correct algebra I get: $\displaystyle (a^2+bc)x+(ab+bd)=x$ 1. $\displaystyle (ab+bd)=0$ So, $\displaystyle a=-d$ 2. $\displaystyle (a^2+bc)=1$ So, $\displaystyle a=\sqrt{1-bc}$ If $\displaystyle b$ and $\displaystyle c$ have opposite signs, their product is negative, and they can be any number. If on the other hand, their signs are the same, their product is positive and: $\displaystyle -1 \leq b \leq 1$ $\displaystyle -1 \leq c \leq 1$ ----------------------------------------------------------------------- Am I making this more complicated than it is? How can I better express my results regarding $\displaystyle b$, and $\displaystyle c$? Thanks! 2. Originally Posted by Mollier Problem: For which numbers $\displaystyle a,b,c,d$ will the function $\displaystyle f(x)=\frac{ax+b}{cx+d}$ satisfy $\displaystyle f(f(x))=x$ for all $\displaystyle x$? if $\displaystyle f[f(x)] = x$ , then $\displaystyle f(x)$ is its own inverse, and as such, its graph is symmetrical to the line $\displaystyle y = x$ $\displaystyle \frac{a \cdot f(x) + b}{c \cdot f(x) + d} = x$ $\displaystyle a \cdot f(x) + b = cx \cdot f(x) + dx$ $\displaystyle a \cdot f(x) - cx \cdot f(x) = dx - b$ $\displaystyle f(x)[a - cx] = dx - b$ $\displaystyle f(x) = \frac{dx-b}{a-cx} = \frac{ax+b}{cx+d}$ $\displaystyle \frac{-dx+b}{cx-a} = \frac{ax+b}{cx+d}$ the above equation only requires that $\displaystyle a = -d$. 3. Originally Posted by Mollier Problem: For which numbers $\displaystyle a,b,c,d$ will the function $\displaystyle f(x)=\frac{ax+b}{cx+d}$ satisfy $\displaystyle f(f(x))=x$ for all $\displaystyle x$? ------------------------------------------------------------------------ Attempt: $\displaystyle f(f(x))=\frac{a\left[\frac{ax+b}{cx+d}\right]+b}{c\left[\frac{ax+b}{cx+d}\right]+d}\\ = \frac{\left[\frac{a^2x+ab}{cx+d}\right]+b}{\left[\frac{acx+bc}{cx+d}\right]+d}\\ =\frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)}$ Correct up to here. By using polynomial division I get the following result: $\displaystyle \left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd) x+(bc+d^2)}$ The trouble with this is that you are about to find that a = –d, so the denominator of the fraction in parentheses is zero. By doing some hopefully correct algebra I get: $\displaystyle (a^2+bc)x+(ab+bd)=x$ 1. $\displaystyle (ab+bd)=0$ So, $\displaystyle a=-d$ 2. $\displaystyle (a^2+bc)=1$ So, $\displaystyle a=\sqrt{1-bc}$ If $\displaystyle b$ and $\displaystyle c$ have opposite signs, their product is negative, and they can be any number. If on the other hand, their signs are the same, their product is positive and: $\displaystyle -1 \leq b \leq 1$ $\displaystyle -1 \leq c \leq 1$ ----------------------------------------------------------------------- Am I making this more complicated than it is? How can I better express my results regarding $\displaystyle b$, and $\displaystyle c$? The condition f(f(x)) tells you that $\displaystyle \frac{(a^2+bc)x+(ab+bd)}{(ac+cd)x+(bc+d^2)} = x$. Cross-multiply to get $\displaystyle (a^2+bc)x+(ab+bd) = x\bigl((ac+cd)x+(bc+d^2)\bigr)$. If that quadratic equation holds for all x then you can equate each coefficient to zero. You should find that there are two sets of solutions: 1. $\displaystyle a+d=0$ (b and c can be anything, so long as you don't have all four of a, b, c, d equal to 0); 2. $\displaystyle a = d \ne0,\;b=c=0$. 4. Originally Posted by skeeter if [tex] the above equation only requires that $\displaystyle a = -d$. The way you come to that conclusion is very understandable, thank you. Originally Posted by Opalg $\displaystyle \left(\frac{a^2+bc}{ac+cd}\right)+\frac{(ab+bd)-(bc+d^2)\left(\frac{a^2+bc}{ac+cd}\right)}{(ac+cd) x+(bc+d^2)}$ The trouble with this is that you are about to find that a = –d, so the denominator of the fraction in parentheses is zero. Ah, I did not see that! Thank you! Merry Christmas<\|endoftext\|>	4.4375
1,919	# Intersection of a line and a curve. Given that the line $y = 2x + 3$ intersects the curve $y = x^2 + 3x + 1$ at two separate points, I have to find these two points. Here is what I did: $$2x + 3 = x^2 + 3x + 1$$ $$0 = x^2 + 1x - 2$$ Using factorisation: $$x = -2 \text{ or } 1$$ Substituting each values of $x$ obtained into the equation of the straight line gives two points of intersections at $(-2, -1)$ and $(1, 5)$ Here is my issue: Why does this work? Equating the curve and the straight line means they share a single similar value of $y$ while they clearly share two. • Equating the curve and straight line in this case is to find the values of $x$ where the equations are equal––in this case, there are two $x$ where they are equal. Commented Jun 7, 2017 at 6:27 • But is the value of y no supposed to be equal as well otherwise this would not make much sense now, would it. Commented Jun 7, 2017 at 6:31 • You're mistaken. It doesn't mean they share a single similar value of $y$. Equating the two expressions means that IF $y$ is any shared value, THEN the resulting equation must be satisfied by the corresponding $x$ values. – MPW Commented Jun 7, 2017 at 12:02 • Note that it it's an equal sign, not an identical sign. – SOFe Commented Jun 7, 2017 at 19:10 Equating the two equations doesn't mean the curve and the line share a single value of $y$; it means that you're assuming they share a value of $y$, and then getting an equation for the corresponding shared value of $x$. This says nothing about how many shared pairs $(x, y)$ there might be. What is an intersection? An intersection is where both $y$ for each function are equal for the same $x$. Consider $f(x)=g(x)$. Your intersection(s) are ALL the points where you can plug (the same) $x$ into both $f(x)$ and $g(x)$ and get an equality. Because $f(x)=g(x)$, $y=y$, if that makes sense. And this can happen say $0,1,2...\infty$ times. • Nice explanation! Commented Jun 7, 2017 at 7:23 Define two systems of equations to be equivalent if they have precisely the same sets of solutions. Abstracting, "solving a system of equations" is the process of successively replacing a given system by equivalent systems until one reaches a "tautological" system whose solutions can be read off by inspection. For example, the system \left. \begin{aligned} y &= 2x + 3 \\ y &= x^{2} + 3x + 1 \end{aligned} \right\} \tag{1} is equivalent, by subtracting the first equation from the second, to the system \left. \begin{aligned} y &= 2x + 3 \\ 0 &= x^{2} + x - 2 \end{aligned} \right\} \tag{2} in which $y$ has been eliminated from the second equation. You solved the second equation using the quadratic formula, obtaining \left. \begin{aligned} y &= 2x + 3 \\ x &= -2\quad\text{or}\quad 1 \end{aligned} \right\} \tag{3} then implicitly used the first equation to deduce the corresponding value(s) of $y$. "Why this works" should be apparent. From this perspective, it should be clear that the reasoning Equating the curve and the straight line means they share a single similar value of $y$ while they clearly share two. would point to a logical gap only if the equation $y = y$ had a unique solution. But the opposite is true: $y = y$ is a tautology; it has no non-solutions. In general, any "reversible operation" on a system of equations yields an equivalent system. The following operations (non-exhaustive list!) are reversible is this sense: • Adding a (constant) multiple of one equation to another equation. • Multiplying an equation by a non-zero constant, or by a non-vanishing expression. • Exchanging two equations. • Replacing an equation $a = b$ with $f(a) = f(b)$ for some injective function $f$. (For equations involving real variables, this includes cubing or exponentiating both sides, squaring both sides when both sides are known to be non-negative, and so forth.) • If $f$ and $g$ are functions, replacing $f(y) = g(y)$ with $f(\phi(x)) = g(\phi(x))$ for some injective function $\phi$. Compare the first three with the Gaussian elimination algorithm for systems of linear equations in several variables. In this case, you are actually not assuming that they share the same value for y. Here, you are equating at the point where both points intersect each other. It is easier to visualise this when you plot a graph online. And in this case, you are only able to equate the two together BECAUSE they intersect which allows you to find the points where they meet, and in this case, at two different intersections. Let $y=f(x)$ and $y=g(x)$ define two curves in the x-y plane. When you set the two equations, $y=f(x)$ and $y=g(x)$ equal to one another (i.e. $f(x)=g(x)$ ), you are asking to find all values of $x$ for which the points $(x,f(x))$ and $(x,g(x))$ are the same points. In other words, you are asking "Which point of the curve $y=f(x)$ is also a point on the curve $y=g(x)$?" Depending on the functions $f$ and $g$, there may be zero, 1 or many points in common to both curves. When you solve $f(x)=g(x)$, you get a collection of $x$ values that you can then plug into either of the equations $y=f(x)$ or $y=g(x)$ to determine the $y$-value for a specific $x$-value. For two different values of $x$, the resulting values of $y$ may be different, but for any given value of $x$ in the solution set, $f(x)$ and $g(x)$ will both yield the same $y$-value. Here's how you can think about it. You've got two curves with equations $y = 2x + 3$ and $y = x^2 + 3x + 1$. You're looking for a point of intersection, that is, a point $(x_0, y_0)$, for which both of these equations hold. Then for this point you have: $y_0 = 2x_0 + 3$ $y_0 = x_0^2 + 3x_0 + 1$ Therefore: $2x_0 + 3 = x_0^2 + 3x_0 + 1$ If I subtract a number from itself, I get zero. Conversely, if I subtract a number from any number, and I get a non-zero number, then the first number isn't equal to the number from which I subtracted it. Therefore, for an x I select, if $x^2 + x - 2$ isn't zero, then your first equation does not hold at that x value. The term intersection of a line and a curve indicates set operations. In fact when intersecting a line $\mathcal{l}: y=2x+3$ and a curve $\mathcal{C}: y=x^2+3x+1$ we consider the graphs of these functions \begin{align} G(\mathcal{{l}})&=\{(x,y)\in\mathbb{R}\times\mathbb{R}\,\big\|\,y=2x+3\}\\ G(\mathcal{C})&=\{(x,y)\in\mathbb{R}\times\mathbb{R}\,\big\|\,y=x^2+3x+1\}\\ \end{align} and the intersection of these two sets \begin{align} G(\mathcal{l})\cap G(\mathcal{C})=\{(x,y)\in\mathbb{R}\times\mathbb{R}\,\big\|\,2x+3=x^2+3x+1\} \end{align}<\|endoftext\|>	4.5625
2,195	• Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month • Level: GCSE • Subject: Maths • Word count: 1344 # Open box problem Extracts from this document... Introduction Junaid Iqbal 11B Mathematics Coursework The Open Box Problem Aim:To determine the size of the square cut which makes the volume of the box as large as possible for any given rectangular sheet of card. Part 1 I will be investigating the size of the cut out square which makes an open box of the largest volume for any sized square sheet of card. The volume for any sized square sheet of card will be gained by using the following formula. Volume = height × area of base V = x × ((L - 2x) × (W - 2x)) X = height of cut out square L= length of square W = width of square I am going to use the following formula to work out the volume of an 8cm by 8cm square sheet of card. I am going to keep the size of card the same while changing the height of the cut out square. V = x × ((8 - 2x) × (8 - 2x)) SQUARE Length (L)cm Width (W)cm Height (X)cm Volume cm3 (2dp) 8 8 0 0.00000 8 8 0.25 14.06250 8 8 0.5 24.50000 8 8 0.75 31.68750 8 8 1 36.00000 8 8 1.25 37.81250 8 8 1.5 37.50000 8 8 1.75 35.43750 8 8 2 32.00000 8 8 2.25 27.56250 8 8 2.5 22.50000 8 8 2.75 17.18750 8 8 3 12.00000 8 8 3.25 7.31250 8 8 3.5 3.50000 8 8 3.75 0.93750 8 8 4 0.00000 Middle 8 8 1.4 37.85600 8 8 1.45 37.71450 As it can clearly be seen from the table above that biggest volume has found to be 37.92cm3 not 37.81cm3 as shown in the previous table. This isn’t the highest possible volume; therefore I am now going to use calculus to find the maximum possible volume. V = x × ((8 - 2x) × (8 - 2x)) V = x × (64 - 16x - 16x + 4x2) V = x × (64 - 32x + 4x2) V = 64x - 32x2 + 4x3 If we rearrange this, it gives us: 4x3 - 32x2 + 64x Now, to find the maximum volume I will differentiate the following to find the slope at where it’s at its maximum. dv = 12x2 - 64x + 64 dx = 12x2 - 64x + 64 3 = 3x2 – 16x + 16 The equation above tells us the slope of a curve at any given point, so now I am going to find out where the maximum volume lies on the curve. Using the quadratic formula -b±√b2 – 4ac : 2a = 16±√162 - 4(3) (16) 2(3) = 16±√256 - 192 6 = 16±√64 6 = 16 + 8OR = 16 - 8 6 6 = 4 = 4 3 The first value is out of range because it’s greater than half the width of the box. Conclusion V = x × ((10 - 2x) × (8 - 2x)) V = x × (80 - 20x - 16x + 4x2) V = x × (80 - 36x + 4x2) V = 80x - 36x2 + 4x3 If we rearrange this, it gives us: 4x3 - 36x2 + 80x Now, to find the maximum volume I will differentiate the following to find the slope at where it’s at its maximum. dv = 12x2 - 72x + 80 dx = 12x2 - 72x + 80 4 = 3x2 – 18x + 20 The equation above tells us the slope of a curve at any given point, so now I am going to find out where the maximum volume lies on the curve. Using the quadratic formula -b±√b2 - 4ac : 2a = 18±√182 - 4(3) (20) 2(3) = 18±√324 - 240 6 = 18±√80 6 = 18 +√80OR = 18 -√80 6 6 = 4.49 (2dp) = 1.51 (2dp) The first value is out of range because it’s greater than half the width of the box. So the second value must give the maximum height for the volume. Now I am going to substitute it back into the equation. 1.51 × ((10 - 2 × (1.51) × (8 - 2 × (1.51))) = 1.51 × ((6.98) × (4.98)) = 1.51 × 34.77 = 52.50 (2dp) This is the highest possible volume for a 10 by 8cm rectangular sheet of card. This value could have also been found using the first method, but it would have taken lot more time. This student written piece of work is one of many that can be found in our GCSE Number Stairs, Grids and Sequences section. ## Found what you're looking for? • Start learning 29% faster today • 150,000+ documents available • Just £6.99 a month Not the one? Search for your essay title... • Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month # Related GCSE Number Stairs, Grids and Sequences essays 1. ## Mathematics Coursework: problem solving tasks 3 star(s) Formulas I have inserted a column labeled terms. 1 represents the first arrangement of tiles, 2 represents the second arrangement of tiles, and so on. n represents the nth term. The common difference is the 2nd term - 1st term. 2. ## Open Box Problem. The scatter diagram on the next page also shows us that the maximum value is 6. Notice that the shape of the scatter diagram is the same as the two scatter diagrams done for the other two squares. 1. ## Investigate the size of the cut out square, from any square sheet of card, ... Now, I shall do some calculations to check for any other connections. 864 cm cubed divided by 18cm = 48 - No connection 18cm divided by 864cm cubed = 0.02083333333 - No connection 18cm 2. ## Mathematical Coursework: 3-step stairs term 20. Thus making it's time consuming. After analysing the grids, the table of results and my prediction I have summed up an algebraic equation which would allow me to find out the total of any 3-step stair shape in a matter of minutes. The criterion of why I haven't explained what B stands for is because I haven't found it yet. 1. ## The Open Box Problem � = V Is also relevant for the rectangle but A is split into W and L so the formula becomes: X (W - 2X) (L - 2X) = V As no immediate connection between X and V is obvious other means of calculation are necessary. 2. ## the Open Box Problem My aim is to find the maximum volume. To solve x, I have to use the quadratic formula, since this quadratic equation cannot be factorised. a =12 b =-160 c=400 x = -b +- Vb�-4ac 2a x = 160+- V160�-412400 2*12 x = 160+- V25600-19200 24 x = 160+- V6400 1. ## First Problem,The Open Box Problem 0.2 0.2 0.056 1.5 4cm by 4cm, piece of square card Length of the section (cm) Height of the section (cm) Depth of the section (cm) Width of the section (cm) Volume of the cube (cm3) 0.1 0.1 3.8 3.8 1.444 0.2 0.2 3.6 3.6 2.592 0.3 0.3 3.4 3.4 2. ## Investigate Borders - a fencing problem. prediction was 28 which is the correct answer, for the number of squares needed for border number 6, and which also proves that the formula is in working order. Diagram of Borders of square: 4x1 Table of results for Borders of square: 4x1 Formula to find the number of squares • Over 160,000 pieces of student written work • Annotated by experienced teachers • Ideas and feedback to<\|endoftext\|>	4.5
616	Middle course of a river The middle course of a river has more energy and volume then in the upper course. The gradient is more gentle and lateral (sideways) erosion has widened the channel. The river channel has also become deeper. Meanders are typical landforms found in this stage of the river. A meander is a winding curve or bend in a river. They are typical of the middle and lower course of a river. This is because vertical erosion is replaced by a sideways form of erosion called LATERAL erosion, plus deposition within the floodplain. The image below shows a series of meanders. Notice the deposition on the inside of the meanders (pale material) and the river cliffs or bluffs (indicated by dark shadows) on the outside of the meanders. Again, the image below shows a series of meanders. The river shown in the photograph is swollen due to recent rainfall. Once again we can see deposition on the inside curves of the meander. The video above shows a meander on the River Derwent, North Yorkshire. Erosion, transportation and deposition are all processes that create the characteristic features of meanders shown in the images above. There are several stages involved in the creation of meanders. These are discussed below. In low flow conditions straight river channels have bars of sediment on their beds. Flowing water weavers around these bars of sediment. This creates deeper pathways where most of the water flows called pools and shallow areas where less water flows called riffles. This causes the river flow to swing from side to side. The map below shows bars of sediment exposed due to the low level of water in the channel. Notice how the flow of water weaves around the sediment bars. View Larger Map Where the river swings towards the bank lateral (sidewayes) erosion causes undercutting. On the opposite side of the channel where the velocity (speed of the flow of water) is lower material is deposited. Therefore the river does not get any wider. The image below shows evidence of undercutting on the outer bank and deposition on the inner bank of the meander. Continued erosion along the outer bank, as the result of hydraulic action and abrasion, creates a river cliff or bluff. A point bar forms on the inner bank. This is a gently sloping deposit of sand, gravel and pebbles. The image below shows a point bar. Meanders are perpetuated through a process called helicoidal flow. As the surface flow of water hits the outer bank it corkscrews, flows along the river bed then deposits eroded material on the inner bank. Eventually the neck of the meander will be breached by the river creating an ox-bow lake. The aerial photograph below shows a meander in the River Derwent, North Yorkshire which will soon be breached. The photograph below shows how close the nexk of the meander is to being breached. Once this occurs an ox-bow lake will form.<\|endoftext\|>	4.4375
1,551	Human Memory and Covenant Faithful memory is covenantal memory. In His good pleasure, God chooses to relate to humanity by means of the special binding arrangement known as covenant. Though the idea of covenant is developed throughout the Bible, elements of this relationship are found as early as the story of Adam and Eve in the garden. Even at this early point, memory becomes an issue of utmost importance. When the serpent tempts Eve to eat of the fruit, he does so by calling her memory of God’s commandment into question: “Did God actually say . . . ?” (Gen. 3:1). Some commentators argue that Eve’s memory is awed, since she adds “touching” to God’s actual prohibition of only “eating” (v. 3). In any case, even if she correctly remembers God’s command, she is not convinced of it. In other words, memory, or the lack thereof, was operative in the fall of humanity, and memory will also play an important role in the redemption of humanity from sin. God remembers Noah in the flood (Gen. 8:1), He remembers Abraham and saves Lot (19:29), and He remembers Rachel and gives her a child (30:22). The preeminent act of redemption in the Old Testament, the exodus, is predicated on God’s remembering His promises to Abraham, Isaac, and Jacob (Ex. 2:24). After their deliverance, the Israelites are called to remember God’s creative work by modeling their week after creation—work for six days, rest for one (Ex. 20:8–10). In Deuteronomy, the rationale for the same command is expanded to explain why the servants of the Israelites should also enjoy weekly rest. The sons and daughters of Israel must show mercy to those who serve in their households just as they themselves were delivered from oppressive work conditions while they were enslaved in Egypt (Deut. 5:15). Such redemptive memory is actually encoded in the biblical covenants themselves. Covenantal documents throughout the ancient Near East typically begin with an account of the benevolence shown by the greater party to the lesser party of the covenant. The same is true of biblical covenants. Deuteronomy—itself an exemplary covenant document—begins with a retelling of God’s faithfulness to Israel during its wandering in the wilderness, a benevolence made all the more extraordinary by Israel’s nearly unmitigated rebellion (1:1–4:43). The implication is hard to miss: because of God’s past faithfulness, Israel should respond with love and obedience (6:4–9). In the best of cases, covenantal memory would become a righteous habit of the mind. The young David showed that he was fit to serve as Israel’s covenantally faithful king when his memory of the Lord’s past blessings inspired him to confront Goliath near the Valley of Elah. Unlike the failed potentate Saul, who huddled in his tent away from the battle, the young shepherd boy sprinted out to the conflict, emboldened by his recollection of God’s past faithfulness to him when he confronted lions and bears that had threatened his flock (1 Sam. 17:34–37). David’s covenantal logic was unassailable, and the victory was sure. Later, as the precipice of the national exile drew near, the role of the biblical prophets evolved into one of accountability, calling the leadership and people to remember the promises and demands of the Lord, promises to show mercy and demands to show justice. Prophets often served as covenant reminders, drawing the people’s attention back to the identity of the Lord and the claims He had made on their lives. Isaiah is exemplary in this way: “Have you not known? Have you not heard? The LORD is the everlasting God, the Creator of the ends of the earth. He does not faint or grow weary; his understanding is unsearchable” (Isa. 40:28). Of course, they had “known” this to be true of the Lord, but they had chosen to forget it as a formative truth in their lives; they had opted for selective amnesia, and exile would prove to be a powerful reminder. Remembering the Apostolic Faith The New Testament is often presented as the Christian part of the Bible, a new piece of revelation that supplants the teaching of the Old Testament. However, this conception misses the organic relationship between the two testaments. In reality, the New Testament is a collection of writings about memory. At its core, the Apostolic teaching in the New Testament is an inspired remembrance and understanding of the teaching of the Old Testament in light of the death and resurrection of Jesus Christ. Every time we find a statement about the Old Testament being fulfilled in Jesus or a statement that gives some explanation of an Old Testament text in light of Jesus, we are experiencing an example of memory that is at once covenantal and Apostolic. Because we are the recipients of the Word of God, we are called by the Apostles to remember God’s mighty works of creation and redemption as revealed to us throughout the course of redemptive history, culminating in the person and work of Jesus Christ. Every time we proclaim Jesus as Messiah, High Priest, true vine (a reference to His identity as true Israel), atoning sacrifice, Immanuel, Creator, Redeemer, and so on, we are committing the sacred act of Apostolic covenant remembering. To be sure, we have a multitude of distractions that compete for our attention, distractions that would confuse our knowledge of the Lord of our salvation. That is why we must nurture our knowledge of the Lord through corporate and individual remembering. That is why it is so crucial for us to remember what we believe about God and His designs for us when we gather together as a worshiping church. One way that we remember is by reciting the Apostles’ Creed or a similar confession of faith. While any corporate proclamation of Christianity involves remembering, the recital of a historic creed of the church improves upon our confession by reminding us of our solidarity with the communion of saints. Similarly, as a covenant people, we receive the Word from the pulpit with hearts inclined to the way in which God has revealed Himself. The faithful preacher prioritizes the clear exposition of God’s Word in a way that fortifies the congregation’s knowledge and love of the Lord. Covenant memory also finds expression in the sacraments of baptism and the Lord’s Supper. The former admonishes parents and new believers to remember the promises made to members of the covenant community, while the latter explicitly marks our collective memory of Jesus Christ’s giving Himself on the cross and of His blood shed as the blood of the new covenant. Lest we miss the point, Jesus Himself explicitly commands us to participate in such regular remembering during Christian worship: “Do this in remembrance of me” (Luke 22:19; 1 Cor. 11:24). Finally, let us not forget the critical covenant remembering that occu rs in the personal, day-to-day practice of meditating upon God’s Word and applying it to our lives as believers. In this way, human consciousness becomes shaped by the story of salvation so that individual believers begin to understand their own biographies as part of the history of the covenant people of God.<\|endoftext\|>	3.875
128	# How do you solve x + 3/8 = 1/4? Jul 16, 2015 color(blue)(x = -1/8 #### Explanation: $x + \frac{3}{8} = \frac{1}{4}$ $\frac{8 x + 3}{\cancel{8}} = \frac{1}{\cancel{4}}$ $\frac{8 x + 3}{2} = 1$ $8 x + 3 = 2$ $8 x = - 1$ color(blue)(x = -1/8<\|endoftext\|>	4.40625
245	Spirograph Nebula, IC-418. Glowing like a multi-faceted jewel, the planetary nebula IC 418 lies about 2,000 light-years from Earth in the constellation Lepus. In this picture, the Hubble telescope reveals some remarkable textures weaving through the nebula. Their origin, however, is still uncertain. A planetary nebula represents the final stage in the evolution of a star similar to our Sun. The star at the center of IC 418 was a red giant a few thousand years ago, but then ejected its outer layers into space to form the nebula, which has now expanded to a diameter of about 0.1 light-year. The stellar remnant at the center is the hot core of the red giant, from which ultraviolet radiation floods out into the surrounding gas, causing it to fluoresce. Over the next several thousand years, the nebula will gradually disperse into space, and then the star will cool and fade away for billions of years as a white dwarf. Our own Sun is expected to undergo a similar fate, but fortunately this will not occur until some 5 billion years from now. Image Credit : NASA The Hubble Heritage Team STScI AURA 2002<\|endoftext\|>	3.9375
1,635	# 09-08 SC Views: Category: Education ## Presentation Description No description available. ## Presentation Transcript ### 5-Minute Check 1: 5-Minute Check 1 A B C D The graph of y = 4 x is shown. State the y -intercept. Then use the graph to approximate value of 4 0.6 ? Determine whether the data in the table display exponential behavior. ### Splash Screen: Splash Screen Homework: Page 547 WM – 3 problems 33 – 69 odd CW – 7 problems HW – 18 problems Lesson Menu Five-Minute Check (over Lesson 9–7) Then/Now New Vocabulary Example 1: Identify Geometric Sequences Example 2: Find Terms of Geometric Sequences Key Concept: n th term of a Geometric Sequence Example 3: Find the n th Term of a Geometric Sequence Example 4: Real-World Example: Graph a Geometric Sequence ### Then/Now: Then/Now You related arithmetic sequences to linear functions. (Lesson 3–5) Identify and generate geometric sequences. Relate geometric sequences to exponential functions. ### Vocabulary: Vocabulary geometric sequence common ratio ### Example 1: Example 1 Identify Geometric Sequences A. Determine whether each sequence is arithmetic , geometric , or neither . Explain. 0, 8, 16, 24, 32, ... 0 8 16 24 32 8 – 0 = 8 Answer: The common difference is 8. So the sequence is arithmetic. 16 – 8 = 8 24 – 16 = 8 32 – 24 = 8 ### Example 1: Example 1 Identify Geometric Sequences B. Determine whether each sequence is arithmetic , geometric , or neither . Explain. 64, 48, 36, 27, ... 64 48 36 27 Answer: The common ratio is , so the sequence is geometric. __ 3 4 __ 3 4 ___ 48 64 = __ 3 4 ___ 36 48 = __ 3 4 ___ 27 36 = ### Example 2: Example 2 Find Terms of Geometric Sequences A. Find the next three terms in the geometric sequence. 1, –8, 64, –512, ... 1 –8 64 –512 The common ratio is –8. = –8 __ 1 –8 ___ 64 –8 = –8 = –8 ______ –512 64 Step 1 Find the common ratio. ### Example 2: Example 2 Find Terms of Geometric Sequences Step 2 Multiply each term by the common ratio to find the next three terms. 262,144 × (–8) × (–8) × (–8) Answer: The next 3 terms in the sequence are 4096; –32,768; and 262,144. –32,768 4096 –512 ### Example 2: Example 2 Find Terms of Geometric Sequences B. Find the next three terms in the geometric sequence. 40, 20, 10, 5, .... 40 20 10 5 Step 1 Find the common ratio. = __ 1 2 ___ 40 20 = __ 1 2 ___ 10 20 = __ 1 2 ___ 5 10 The common ratio is . __ 1 2 ### Example 2: Example 2 Find Terms of Geometric Sequences Step 2 Multiply each term by the common ratio to find the next three terms. 5 __ 5 2 __ 5 4 __ 5 8 × __ 1 2 × __ 1 2 × __ 1 2 Answer: The next 3 terms in the sequence are , __ 5 2 __ 5 4 , and . __ 5 8 Concept ### Example 3: Example 3 Find the n th Term of a Geometric Sequence A. Write an equation for the n th term of the geometric sequence 1, –2, 4, –8, ... . The first term of the sequence is 1. So, a 1 = 1. Now find the common ratio. 1 –2 4 –8 = –2 ___ –2 1 = –2 ___ 4 –2 = –2 ___ –8 4 a n = a 1 r n – 1 Formula for the n th term a n = 1 ( –2 ) n – 1 a 1 = 1 and r = –2 The common ration is –2. Answer: a n = 1(–2) n – 1 ### Example 3: Example 3 Find the n th Term of a Geometric Sequence B. Find the 12 th term of the sequence. 1, –2, 4, –8, ... . a n = a 1 r n – 1 Formula for the n th term a 12 = 1 ( –2 ) 12 – 1 For the n th term, n = 12. = 1(–2) 11 Simplify. = 1(–2048) (–2) 11 = –2048 = –2048 Multiply. Answer: The 12 th term of the sequence is –2048. ### Example 4: Example 4 Graph a Geometric Sequence ART A 50-pound ice sculpture is melting at a rate in which 80% of its weight remains each hour. Draw a graph to represent how many pounds of the sculpture is left at each hour. Compared to each previous hour, 80% of the weight remains. So, r = 0.80. Therefore, the geometric sequence that models this situation is 50, 40, 32, 25.6, 20.48,… So after 1 hour, the sculpture weighs 40 pounds, 32 pounds after 2 hours, 25.6 pounds after 3 hours, and so forth. Use this information to draw a graph. ### Example 4: Example 4 Graph a Geometric Sequence Answer: ### Example 1: A B C Example 1 A. Determine whether the sequence is arithmetic , geometric , or neither . 1, 7, 49, 343, ... ### Example 1: A B C Example 1 B. Determine whether the sequence is arithmetic , geometric , or neither . 1, 2, 4, 14, 54, ... ### Example 2: A B C D Example 2 A. Find the next three terms in the geometric sequence. 1, –5, 25, –125, .... ### Example 2: A B C D Example 2 B. Find the next three terms in the geometric sequence. 800, 200, 50, , .... __ 2 25 ### Example 3: A B C D Example 3 A. Write an equation for the n th term of the geometric sequence 3, –12, 48, –192, .... ### Example 3: A B C D Example 3 B. Find the 7 th term of this sequence using the equation a n = 3(–4) n – 1 . ### Example 4: A B C D Example 4 Soccer A soccer tournament begins with 32 teams in the first round. In each of the following rounds, on half of the teams are left to compete, until only one team remains. Draw a graph to represent how many teams are left to compete in each round. ### End of the Lesson: End of the Lesson Homework: Page 547 WM – 3 problems 33 – 69 odd CW – 7 problems HW – 18 problems<\|endoftext\|>	4.71875
299	# Over a 6 month period, a bakery sold an average of 29 pies per day. The number of apple pies they sold was four less than twice the number of blueberry pies they sold. How many blueberry pies did the bakery average selling per day during that period? Jun 13, 2016 Let $x$ be the average number of apple pies sold and $y$ be the average number of blueberry pies sold per day at the bakery. $x + y = 29$ $x = 2 y - 4$ $2 y - 4 + y = 29$ $3 y = 33$ $y = 11$ The bakery sold an average of $11$ blueberry pies per day. Hopefully this helps! Jun 13, 2016 11 blueberry pies. #### Explanation: It is also possible to work this out using only one variable, as long as we write an expression for each type of pie. There were fewer blueberry pies sold than apple pies, so: Let the Number of blueberry pies be $x$ The number of apple pies is $2 x - 4 \text{ (4 less than double x)}$ There are 29 pies sold every day. Blueberry pies + apple pies = 29 $x + 2 x - 4 = 29$ $3 x = 33$ $x = 11$ $x$ is the number of blueberry pies.<\|endoftext\|>	4.5
321	about Favorite Author A commemorative essay is a type of writing that celebrates or simply analyzes an important figure or event in the past. People often write commemorative essays as tributes on the anniversary or centennial of a person’s date of birth, date of death or the date that an event occurred. For this assignment, based on the works read in this course, choose an author whose work you admire or appreciate. First review the commemorative essay by Toni Morrison written about James Baldwin. See this website: “James Baldwin: His Voice Remembered; Life in His Language“. Next, read or review at least one critical essay on the work of your chosen author. Decide whether or support or refute this critical essay. Then, read at least one biographical sketch about this author. Draft a sentence that states what and why you admire or appreciate about the writing of your chosen author. Modeling the pattern of Morrison’s essay, write a 3-4 page commemorative essay based one author we’ve read during this course. Consider the qualities of person and prose that Morrison mentioned in her essay. Include quotations from (1) three different short works or three different sections of a full length work (2) quotation(s) from critical essay, and from the (3) biographical sketch. Include appropriate citations in body of essay and attach a bibliography of sources consulted. MINIMUM REQUIREMENTS and GRADING RUBRIC (See attached copy of full assignment for Commemorative Essay)<\|endoftext\|>	3.75
844	# How to do this question 10? Then teach the underlying concepts Don't copy without citing sources preview ? #### Explanation Explain in detail... #### Explanation: I want someone to double check my answer 1 Feb 12, 2018 $\textcolor{b l u e}{y = - x - 12}$ #### Explanation: All points on the line $y = - 2 x + 8$ will be transformed by the matrix $\boldsymbol{A}$ Any points on this line will have coordinates of the form $\left(k , - 2 k + 8\right)$ $\therefore$ $\left(\begin{matrix}x ' \\ y '\end{matrix}\right) = \left(\begin{matrix}0 & - 2 \\ - 4 & 0\end{matrix}\right) \left[\left(\begin{matrix}k \\ - 2 k + 8\end{matrix}\right) + \left(\begin{matrix}- 2 \\ 2\end{matrix}\right)\right]$ $\textcolor{w h i t e}{888888} = \left(\begin{matrix}0 & - 2 \\ - 4 & 0\end{matrix}\right) \left(\begin{matrix}k - 2 \\ - 2 k + 10\end{matrix}\right)$ $\textcolor{w h i t e}{888888} = \left(\begin{matrix}4 k - 20 \\ - 4 k + 8\end{matrix}\right)$ i.e. $x ' = 4 k - 20$ and $y ' = - 4 k + 8$ Eliminating $k$: $k = \frac{x + 20}{4}$ $y = - 4 \left(\frac{x + 20}{4}\right) + 8$ $\textcolor{b l u e}{y = - x - 12}$ So all images of $\left(x ' , y '\right)$ lie on the line $y = - x - 12$. The line $y = - x - 12$ is the image of $y = - 2 x + 8$ under the transformation. Note: We could have found this line and alternate way. If we had generated two pairs of coordinates using $\boldsymbol{y = - 2 x + 8}$, and used them in the transformation i.e for $\boldsymbol{X}$, we would have 2 pairs of coordinates of the image. This would enable us to find the equation of the line. Example: From $y = - 2 x + 8$ For $x = 3$ and $x = 4$ $y = 2$ and $y = 0 \textcolor{w h i t e}{88}$ respectively. Under the transformation we have: $x = - 8$ and $y = - 4$ $x = - 4$ and $y = - 8$ $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} = \frac{- 8 - \left(- 4\right)}{- 4 - \left(- 8\right)} = - 1$ $y - \left(- 4\right) = - \left(x - \left(- 8\right)\right)$ $y + 4 = - x - 8$ $y = - x - 12 \textcolor{w h i t e}{888}$ as expected. • 7 minutes ago • 8 minutes ago • 10 minutes ago • 12 minutes ago • A minute ago • 2 minutes ago • 3 minutes ago • 3 minutes ago • 6 minutes ago • 6 minutes ago • 7 minutes ago • 8 minutes ago • 10 minutes ago • 12 minutes ago<\|endoftext\|>	4.59375
1,608	Problem of the Week Problem E and Solution Wipe Away 3 Problem Tyra writes consecutive positive integers on a whiteboard starting with the integer $$1$$. However, when she writes a number that is a multiple of $$9$$, or contains the digit $$9$$, Juliana immediately erases it. If they continue this for a long time, what is the 400th number that Juliana will erase? Note: In solving this problem, it may be helpful to use the fact that a number is divisible by $$9$$ exactly when the sum of its digits is divisible by $$9$$. For example, the number $$214\,578$$ is divisible by $$9$$ since $$2+1+4+5+7+8 = 27$$, which is divisible by $$9$$. In fact, $$214\,578=9 \times 23\,842.$$ Solution We first consider the integers between $$1$$ and $$999$$, inclusive. Since $$999 = 111 \times 9$$, there are $$111$$ multiples of $$9$$ between $$1$$ and $$999$$. Now letâ€™s figure out how many of the integers from $$1$$ to $$999$$ contain the digit $$9$$. The integers from $$1$$ to $$99$$ that contain the digit $$9$$ are $$9,~19, \ldots, 79,~ 89$$ as well as $$90,~ 91, \ldots, 97~ , 98,~ 99$$. Thus, there are $$19$$ positive integers from $$1$$ to $$99$$ that contain the digit $$9$$. Since there are $$19$$ integers from $$1$$ to $$99$$ that contain the digit $$9$$, it follows that there are $$19 \times 9 = 171$$ integers from $$1$$ to $$899$$ that contain the digit $$9$$. Between $$900$$ and $$999$$, every integer contains the digit $$9$$. Thus, there are $$100$$ numbers that contain the digit $$9$$. Thus, in total, $$171+100 = 271$$ of the integers from $$1$$ to $$999$$ contain the digit $$9$$. However, some of the integers that contain the digit $$9$$ are also multiples of $$9$$, so were counted twice. To determine how many of these such numbers there are, we use the fact that a number is divisible by $$9$$ exactly when the sum of its digits is divisible by $$9$$. • The only one-digit number that contains the digit $$9$$ and is also a multiple of $$9$$ is $$9$$ itself. • The only two-digit numbers that contain the digit $$9$$ and are also multiples of $$9$$ are $$90$$ and $$99$$. • To find the three-digit numbers that contain the digit $$9$$ and are also multiples of $$9$$, we will look at their digit sum. • Case 1: Three digit-numbers with a digit sum of $$9$$: The only possibility is $$900$$. Thus, there is $$1$$ number. • Case 2: Three digit-numbers with a digit sum of $$18$$: • If two of the digits are $$9$$, then the other digit must be $$0$$. The only possibilities are $$909$$ and $$990$$. Thus, there are $$2$$ numbers. • If only one of the digits is $$9$$, then the other two digits must add to $$9$$. The possible digits are $$9,~4,~5$$, or $$9,~3,~6$$, or $$9,~2,~7$$, or $$9,~8,~1$$. For each of these sets of digits, there are $$3$$ choices for the hundreds digit. Once the hundreds digit is chosen, there are $$2$$ choices for the tens digit, and then the remaining digit must be the ones digit. Thus, there are $$3 \times 2=6$$ possible three-digit numbers for each set of digits. Since there are $$4$$ sets of digits, then there are $$4 \times 6 = 24$$ possible numbers. • Case 3: Three digit-numbers with a digit sum of $$27$$: The only possibility is $$999$$. Thus, there is $$1$$ number. Therefore, there are $$1+2+24+1=28$$ three-digit numbers from $$1$$ to $$999$$ that contain the digit $$9$$, and are also multiples of $$9$$. Thus, there are $$1+2+28=31$$ integers from $$1$$ to $$999$$ that contain the digit $$9$$, and are also multiples of $$9$$. It follows that Juliana erases $$111+271-31=351$$ of the numbers from $$1$$ to $$999$$ from the whiteboard. Since we are looking for the 400th number that Juliana erases, we need to keep going. Next, we consider the integers between $$1000$$ and $$1099$$, inclusive. Since $$1099=(122 \times 9)+1$$, there are $$122$$ multiples of $$9$$ between $$1$$ and $$1099$$. Since there are $$111$$ multiples of $$9$$ between $$1$$ and $$999$$, it follows that there are $$122-111=11$$ multiples of $$9$$ between $$1000$$ and $$1099$$. The integers between $$1000$$ and $$1099$$ that contain the digit $$9$$ are $$1009,~1019, \ldots, 1079,~ 1089$$ as well as $$1090,~ 1091, \ldots, 1097~ , 1098,~ 1099$$. Thus, there are $$19$$ integers from $$1000$$ to $$1099$$ that contain the digit $$9$$. Of these, the only integers that are also multiples of $$9$$ are $$1089$$ and $$1098$$. Thus, Juliana erases $$11+19-2=28$$ of the numbers from $$1000$$ to $$1099$$ from the whiteboard. In total, she has now erased $$351+28=379$$ numbers. Next, we consider the integers between $$1100$$ and $$1189$$, inclusive. Since $$1189=(132 \times 9) +1$$, there are $$132$$ multiples of $$9$$ between $$1$$ and $$1189$$. Since there are $$122$$ multiples of $$9$$ between $$1$$ and $$1099$$, it follows that there are $$132-122=10$$ multiples of $$9$$ between $$1100$$ and $$1189$$. The integers between $$1100$$ and $$1189$$ that contain the digit $$9$$ are $$1109,~1119,\dots,1179,~1189$$. Thus there are $$9$$ integers from $$1100$$ to $$1189$$ that contain the digit $$9$$. The only one of these that is also a multiple of $$9$$ is $$1179$$. Thus, Juliana erases $$10+9-1=18$$ of the numbers from $$1100$$ to $$1189$$ from the whiteboard. In total, she has now erased $$379+18=397$$ numbers. The next three numbers that Juliana will erase are $$1190,~1191,$$ and $$1192$$. Thus, the 400th number that Juliana erases is $$1192$$.<\|endoftext\|>	4.65625
178	How do you recognise a racist incident? What should you do if one of your pupils uses a racist term? How can you promote better race relations in your school? The new Race Equality resource helps teachers answer these questions and tackle the problem of racism in schools. The website clarifies what racism means in Scottish society today and identifies a wide range of racist behaviour. By looking at the effect racism has on victims and the broader community, it explores the role that schools can play in discouraging racist behaviour and dealing with it appropriately when it happens. A selection of anti-racist teaching materials have been developed to encourage children to understand the history of Scotland’s diverse society, the difficulties faced by immigrants today, the causes and devastating effects of racism, and the importance of racial tolerance. The new site supports the Scottish Government’s One Scotland campaign to tackle racist attitudes in Scotland.<\|endoftext\|>	4.40625
807	# Infinite Series $\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m!\:n!}{(m+n+2)!}$ Evaluating $$\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!}{(m+n+2)!}$$ involving binomial coefficients. My attempt: $$\frac{1}{(m+1)(n+1)}\sum_{m=0}^\infty \sum_{n=0}^\infty\frac{(m+1)!(n+1)!}{(m+n+2)!}=\frac{1}{(m+1)(n+1)} \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{1}{\binom{m+n+2}{m+1}}=?$$ Is there any closed form of this expression? • How can you take the factor out of the summation ? Jan 3, 2016 at 1:57 • @Shailesh. Also,Olivier Oloa. Thanks for pointing out Jan 3, 2016 at 2:27 One may observe that, $$\frac{m!}{(m+n+2)!}=\frac{m!}{(n+1)(m+n+1)!}-\frac{(m+1)!}{(n+1)(m+n+2)!}$$ giving, by telescoping terms, $$\sum_{m=0}^N\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}-\frac{(N+1)!}{(n+1)(N+n+2)!}$$ thus, as $N \to \infty$, $$\sum_{m=0}^\infty\frac{m!}{(m+n+2)!}=\frac1{(n+1)(n+1)!}.$$ Then the initial series reduces to $$\sum_{n=0}^\infty \sum_{m=0}^\infty\frac{n!\:m!}{(m+n+2)!}=\sum_{n=0}^\infty\frac{n!}{(n+1)(n+1)!}=\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6.$$ • Sneaky... :-$)$ Jan 3, 2016 at 5:27 $$S=\sum_{m\geq 0}\sum_{n\geq 0}\frac{\Gamma(m+1)\,\Gamma(n+1)}{(m+n+2)\,\Gamma(m+n+2)}=\sum_{m,n\geq 0}\iint_{(0,1)^2} x^m(1-x)^n y^{m+n+1}\,dx\,dy \tag{1}$$ hence: $$S = \iint_{(0,1)^2}\frac{y\,dx\,dy}{(1-xy)(1-y+xy)}=2\int_{0}^{1}\frac{-\log(1-y)}{2-y}\,dy=2\int_{0}^{1}\frac{-\log(t)}{1+t}\,dt\tag{2}$$ and by expanding $\frac{1}{1+t}$ as a geometric series, $$S = 2\sum_{n\geq 0}\frac{(-1)^n}{(n+1)^2} = \color{red}{\zeta(2)} = \frac{\pi^2}{6}.\tag{3}$$<\|endoftext\|>	4.4375
738	1. Governments can raise taxes. Governments can compel a large swathe of the population to give it unrequited transfers known as taxes. Households cannot. 2. Governments can print money (at least those that enjoy monetary sovereignty). Governments can print pieces of paper that will be gladly accepted as payment for goods and services. Households cannot. Note though that this power is not a widow's cruse. The limiting factor, constraint, or trade-off is inflation. See this discussion: Smith (2014). 3. People die. There is thus a (literal) deadline at which a person's assets and debts must be resolved. A person cannot remain in debt forever. In contrast, there is no obvious deadline for governments. A government can be in debt for pretty much forever: The US has been in debt every year since its founding (TreasuryDirect.gov — the debt came close to zero in 1835–36). The UK has been in debt every year since at least 1694 (Ellison & Scott, 2017, Fig 1). Yes, a high and rising debt is undesirable. However, if say a government consistently maintains a 40% debt-to-GDP ratio for 200 years, then most economists would consider this perfectly healthy and sustainable. In contrast, it is not generally possible for an individual to be in debt to the tune of 40% of her annual income for 200 consecutive years. 4. Governments can borrow at much lower interest rates (thanks to the above factors). On 2019-01-31, the US government's average interest rate was 2.574%. This is lower than the rate at which most Americans would be able to borrow and certainly much lower than their credit card interest rates. 5. Government budgets directly influence economic growth. When an individual decides not to buy a new fridge, this does not affect her income. But when government cuts spending, this likely reduces national income, which, by the way, also tends to reduce the government's own income (i.e. tax revenue). Conversely, when an individual buys a new fridge, this expenditure does not increase her income. One individual's expenditure adds to others' income, but not to her own. But when government increases spending (on anything other than imports), this does result in increased national income. A country's expenditure is its own income. When an individual suffers a fall in income, it may be prudent for her to cut back on her expenses. In contrast, when a country suffers a recession, it is not usually prudent for its government to reduce spending. Every family in America has to balance their budget. Every small business. Should we expect anything less from a great nation? (Jeb Hensarling, 2011.) The above is a common piece of rhetoric used by pro-balanced-budget politicians. There are two mistakes. First, as already alluded to above, we have the fallacy of composition: What is true of the parts need not be true of the whole. Second, the premise is not even true. Many families and businesses do not have balanced budgets and are in debt. This though is not necessarily a bad thing. An individual in law school, a family that's just bought a home, and a firm that's just built a factory may all be in debt, but we do not ipso facto disapprove of them. Likewise, if a government spends its money on productive and legitimate uses (and not mostly on fireworks and bridges to nowhere), we should not condemn it simply because it is running a deficit.<\|endoftext\|>	3.875
753	An EPFL doctoral student has come up with methods to map out forests more effectively using aerial remote sensing, in support of on-the-ground forest inventories. Forests are an essential component of the world's ecosystems and a key indicator of our planet's health. They provide valuable resources – like wood for construction and heating – and they filter rainwater, protect against erosion and avalanches, and can be used for numerous leisure pursuits. For these and other reasons, it's important to monitor their development through regular forest inventories. On-the-ground inventories, in addition to being subject to the observers' subjectivity, are costly and laborious and can only be done in easy-to-reach regions. As a result, they are not carried out very often, and only in limited areas. In Switzerland, for example, the national inventory has been updated only every ten years or so since 1985. Aerial remote sensing can be a good complement to on-the-ground monitoring. It is more objective and less expensive, and it can cover a larger area. Two techniques are currently used: airborne laser scanning, which determines the three-dimensional structure of the forest, and hyperspectral imaging, which identifies the precise color of the tree canopy, even beyond the visible light spectrum. Scientists know how to collect these two types of data, but extracting the information needed to monitor and manage forests is more complicated. "Chopping down" 5,000 trees by hand For his Ph.D. thesis, Matthew Parkan, from EPFL's Laboratory of Geographic Information Systems, developed a number of algorithms capable of automatically determining certain inventory parameters – such as trunk location, estimated diameter and species – over large areas. These algorithms can be used, for example, to create a detailed map of an area in preparation for tree marking (prior to cutting), to closely monitor the development of individual trees and to identify habitats more suited to certain animal species. To calibrate and validate his algorithms, Parkan had to build a reference dataset by manually extracting more than 5,000 trees from a 3-D point cloud. For this, he created a digital forestry toolbox to facilitate the manual extraction of trees and the visual identification of tree species. This allowed him to verify that the algorithms could reliably detect the location and shape of trees, and to calibrate his classification models for nine tree species commonly found in Swiss forests. A complement, not a substitute "My aim was to develop methods and tools that can supplement on-the-ground inventories rather than replace them," says Parkan. On-the-ground inventories are still essential for calibrating models, validating results and identifying subtle characteristics – like the dead wood on the ground, habitat trees and the detailed health of trees – that cannot be detected by most of the aerial remote sensing techniques currently available. Since trees are complex organisms whose shape and spatial structure vary enormously within a forest, it is very difficult to automatically detect all of their characteristics. "For the time being, no algorithm can provide a totally reliable set of results," says Parkan. "That said, huge progress will be made in the coming years as more and more very-high-resolution data becomes available and we develop algorithms that work almost as well as the human brain." Citation: Algorithms to enhance forest inventories (2019, April 15) retrieved 15 April 2019 from https://phys.org/news/2019-04-algorithms-forest.html This document is subject to copyright. Apart from any fair dealing for the purpose of private study or research, no part may be reproduced without the written permission. The content is provided for information purposes only.<\|endoftext\|>	3.8125
797	Click on sites below of The Creative Process Art?? Why Art in School Art Graduation and Achievement Projects Art Skills Map forTeachers Skills for Art Proficiency of The Creative Process 7, 8 Art Class Assignments Links to some Great Art on Net Artsmart Know things Artwork by Students Ideas -Concepts to Aid Creativity Basics of Composition Organization in Art Beginning A Picture, Checklist 7, design, " breakup" Additional Clay Designs slab and coil Cropping, finding a Picture Inside a Picture Little Things Make Interesting Pictures & Invention things which enhance creativity & Imagination, Exercises General Info for New Students How to Transform 2d to 3d How to look at Art Ideas for Art Ideas that follow Themes Development in Art Index of Design Lessons to Look at Art, Part one Learning to Look at Art, Part Two Painting and Color Assignments Students at Work or - - RECENT STUDENT SAMPLES Lesson for 7th Grade Beginning Art BREAKING UP SPACE A Basic Skill and Understanding INTRODUCTION TO DESIGN & ARRANGEMENT Break Up Number 1The first design skill in art is to think about the whole space and make an effort to divide that space into interesting areas. become more interesting when there is a variety of shapes and sizes. In the first design series. You will make small trial sketches (thumbnail sketches) to break up the space. You will use only strait line segments which can change directions 2 or 3 times. Selected the best design from your thumbnails. You will reproduce that design on a larger sheet of paper using first a pencil to copy the sketch then painting in the spaces (you should have at least 10 areas in your breakup design). In the first set. You will use only black and white and as many shades of gray that you can mix. The water soluble poster paint is mixed using a flat palette and blending colors together with your Sample of Breakup Design #1 to Break Up Design 2 In the second design series you will break up the space again using 3 line movements which go from edge to edge. These will follow movement pattern of: Strait, Curve, Strait. On your thumbnail sketches try to achieve ten to fourteen areas inside of your How to Paint This Design. Paint these spaces by mixing combinations of black, white and one Hue. A hue would be the name of a color, like red, blue, green, etc. Example of Breakup Design #2 \|Directions to Break Up Number the third design series you will break up the space again using 3 line movements that go from edge to edge. This one will break up space using the following movements: Strait, Curve, Curve, Strait. Again try to achieve about ten or more shapes or areas inside of your design. to Paint Number 3 this design using 2 hues plus white. Do NOT use any black in your color mixtures. Example of Breakup Design #3 \|How to Break Up Number 4 will be an extra choice if time allows. Use 3 line movements which go from edge to edge. Use only compound curves, (S curve movements.) These can get scribbley, so don't use too many S curves. Limit your compound curves to 2 per line. Try to achieve about ten or more shapes or areas inside of your design . to Paint Number 4 the color scheme for this design use only bright primary colors, blue, yellow, red, and black. Do not mix the colors<\|endoftext\|>	3.828125
3,900	# Understanding Tail Recursion April 03, 2022 In this post, I try to explain how tail call optimization can be used to make recursive functions (of certain kind) more efficient. ### Understanding Recursion A recursive process can be divided in two parts: 1. A base case(s), which defines a simple case (such as the first item in a sequence) 2. A recursive step, where new cases are defined in terms of previous cases. When a recursive process is called with the base case, it simply returns the result. If the process is called with a more complex case, it would break down the problem into two parts: a part it knows how to evaluate and a part which it does not. The latter part will resemble the original case, except that it will be a slightly smaller or simpler version of it. It would then call a fresh instance of itself on this latter part to work on the simpler case. What is essential for a proper use of recursion is that the objects can be expressed in terms of simpler objects, where “simpler” means closer to the basis of the recursion. [Source] To put it in another way, a recursive algorithm can be defined as solving a problem by solving a smaller version of the same problem (a sub-problem) and then using that solution to solve the original problem. To solve a sub-problem, we need to solve a still smaller version of that sub-problem and so on. To prevent an infinite series of sub-problems, we need the series of sub-problems to eventually terminate at a base case. ## Examples of recursive functions We can use recursion to find the factorial of a positive natural number: ``````function factorial(n){ if(n <= 1)//.... (1) return 1 return n * factorial(n - 1)// ... (2) } `````` In the above example, line (1) represents the base case and line (2) represents the recursive call. ## Keeping Track of Recursive Calls Other than the base case, a recursive function cannot compute the final result (of a given input) without first knowing the result of the same function with a smaller input. For example, in the case of the factorial function, for any value of `n` (where `n != 1`), we need to first know the value of `factorial(n - 1)`, and to calculate the value of `factorial(n - 1)` we need to compute `factorial(n - 2)` and so on, till `n == 1` is true. This means that for every step of this recursive ladder, we must wait for return value of the recursive call. Till this is done, we must pause our execution. Once we have this result, we can complete our evaluation by carrying out our operation(s) on it and return the result so computed. So, to find the factorial of 4, we need to suspend our execution till we know the result of the factorial of 3 (and so on). Also, note that at each step of the recursive process, we do not carry out the same operation with the result we get from the recursive call. Operations at each recursive step involves variables (not constants) relevant to that recursive step. In the case of `factorial`, for each step, we need to multiply the current value of `n` with the value we get from `factorial(n - 1)`. The answer to the new factorial subproblem is not the answer to the original problem. The value obtained for (n - 1)! must be multiplied by n to get the final answer. [Source] From the above discussion we know that for a recursive operation- • we need a series of invocations of the same function with different values starting from a given value to the base-case value; • to compute the value at a given recursive step, we need to wait till all the function calls from the base case leading up to it are completed; • while waiting, we need to store the state of the values at that recursive step. To illustrate this, let’s take the example of `factorial(n)`, where `n = 4`: ``````factorial(4) = 4 * factorial(3) factorial(3) = 3 * factorial(2)// ... factorial(4) waiting factorial(2) = 2 * factorial(1)// ... factorial(4), factorial(3) waiting factorial(1) = 1// ... factorial(4), factorial(3), factorial(2) waiting factorial(2) = 2 * 1// ... factorial(3), factorial(4) waiting factorial(3) = 3 * 2// ... factorial(4) waiting factorial(4) = 4 * 6 `````` Since there is no ex ante way of knowing the number of recursive calls required to reach the base case for a given value, we need to be able to store the states of a dynamic number of recursive steps. Also, as we can see above, the order in which the states of each recursive step is accessed is one of Last-In-First-Out, i.e., the values of each recursive step are accessed in the reverse order of their insertion. Because of this property of recursive function calls, we can use stacks to store the state of each recursive step. ## Understanding Call Stack Whenever a function is called, a memory block (known as ‘stack frame’ or ‘activation record’) representing its state (mainly, its parameters, local variables, its return address) is pushed to a call stack that is maintained during the run-time of the program. When the execution of the function is complete, it’s corresponding memory block is popped from the call stack. At any point in time, the call stack only contains memory blocks representing functions that have been called, but not yet been executed. (As an aside, most programming languages have the functionality to display the call stack at any particular point of time during run-time. This is called stack trace and it is a useful tool for debugging run-time errors) In the case of recursive functions, a function that is already part of the call stack, makes another call of the same function (with a smaller sub-problem). The original function cannot be executed till this newly called function returns its value. So, a new memory block representing the newly called function gets pushed on to the call stack. The process repeats itself till we reach the base case. Once we have reached the base case, we will have reached the top of the recursive call-stack. As soon as the base-case function call returns its value, it’s associated memory block gets popped from the stack. The function that made the base-case function call, gets popped next which returns its value to the next function that had called it. This goes on till we reach the bottom of the stack. The bottom-most block on the stack represents the original function call. By the time we reach there, we will have all the values needed to complete this evaluation. Here’s a visualization of a recursive call stack for the factorial of 5: ## Computing Space Complexity for Recursive Calls The space complexity for a recursive function depends on the maximum size of the call stack at any point of time during its execution. For a linear recursive function, i.e., a function that makes a single call to itself during each recursive step, the maximum size of the call stack will be the sum of the number of recursive calls made by each recursive step. Thus, for the `factorial` function, the maximum size of the call stack will be `n` (as shown above). For non-linear recursive functions, if we visualise the pattern of recursive calls as a recursive tree, where each node represents a recursive call, the maximum size of the call stack will be the height of the tree. Thus, for the following function generating `n`th Fibonacci number, the maximum size of the call stack will be `n` ``````function fibo(n){ if (n == 0) return 0 if(n==1) return 1 return fibo(n - 1) + fibo(n - 2) } `````` In the above example, even though we are making two recursive calls at each step, we will still only require a call stack of size `n`. This is because at each node of the recursive tree, we need to wait for it’s children nodes to return their values, and the same goes for their children nodes as well. We essentially do a depth-first traversal from each node in the recursive tree, thereby requiring only a max stack size of the height (or depth) of the recursive tree. ## Stack Overflow The size of call stacks are finite and limited (their exact size depend on multiple factors, including the programming language and the run time environment). This means that there are only so many recursive calls we can make for a given call stack. When we exceed this size, we get a ‘stack overflow’ error. Let’s take the following function that recursively finds the sum of the first `n` natural numbers: ``````function sumOfNaturalNumbers(n){ if(n == 1) return 1 return n + sumOfNaturalNumbers(n - 1) } `````` If we pass a relatively large number to the above function, say `100000`, we get the following error ``````RangeError: Maximum call stack size exceeded at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:13:5) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) at sumOfNaturalNumbers (/home/otee/projects/recursive_factorial.js:15:16) `````` However, if we write the same function using a `for` loop and pass the same parameter (`100000`), we do not get any errors. ``````function sumOfNaturalNumbersIteration(n){ sum = 0; for(let i = 0; i <= n; i++){ sum += i; } return sum; } `````` The reason why we the above prograat ism does not break is because it takes constant space to run a `for` loop, irrespective of the number of iterations being executed. To run the above loop, all we need to keep track of are the current values of `n`, `i` and `sum`. ## Tail recursion Tail recursion is a special kind of recursion where the recursive call is the last operation carried out in the recursive case and the result of the recursive call is not manipulated by the caller. A call from procedure f ( ) to procedure g( ) is a tail call if the only thing f ( ) does, after g( ) returns to it, is itself return. The call is tail-recursive if f ( ) and g( ) are the same procedure (Steven S. Muchnick, Advanced Compiler Design and Implementation, p 461) The most important aspect of a tail call, is that a new frame does not need to be added to the call stack for every function call. For tail calls, there is no need to store the state of the function making the tail call (as all the operations inside that function are executed by the time the tail call is made and all that is left to do is to pass on the value so returned, to its original caller). Instead, the tail-called function returns the value directly to the original caller. As a result, tail recursion takes constant space to be executed. In fact, a tail-call is essentially a goto statement: In this way, if the last thing a procedure does is call another (external) procedure, that call can be compiled as a GOTO. Such a call is called tail-recursive, because the call appears to be recursive, but is not, since it appears at the tail end of the caller. [Source] The above implementation of the sum of natural numbers,`sumOfNaturalNumbers`, is not written in a tail-recursive style, as we do one operation after receiving the return value from the recursive call (namely, adding the current value of `n` to the returned value). Here’s a tail-recursive implementation of that function: ``````function sumOfNaturalNumbersTailRecursive(n, acc = 0){ if(n <= 0) return acc return sumOfNaturalNumbersTailRecursive(n - 1, acc + n) } `````` However, this implementation, too, fails when we pass `100000` to it: ``````RangeError: Maximum call stack size exceeded at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:34:42) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) at sumOfNaturalNumbersTailRecursive (/home/otee/projects/recursive_factorial.js:37:12) `````` The reason why this implementation also results in stack overflow is because JavaScript (in the NodeJs runtime environment) does not support tail call optimization. In fact, many other programming languages, including Java, do not support tail call optimization: Not all programming languages require tail-call elimination. However, in functional programming languages, tail-call elimination is often guaranteed by the language standard, allowing tail recursion to use a similar amount of memory as an equivalent loop. [Wikipedia] Clojure, being a functional programming language, supports tail call optimization. ### Tail Recursion In Clojure Unlike compilers of some other functional languages, Clojure’s compiler will not automatically detect a recursive tail-call and optimise it accordingly. We need to make the tail call using a special form, `recur`, to explicitly utilise tail recursion optimisation. Many such languages guarantee that function calls made in tail position do not consume stack space, and thus recursive loops utilize constant space. Since Clojure uses the Java calling conventions, it cannot, and does not, make the same tail call optimization guarantees. Instead, it provides the recur special operator, which does constant-space recursive looping by rebinding and jumping to the nearest enclosing loop or function frame. While not as general as tail-call-optimization, it allows most of the same elegant constructs, and offers the advantage of checking that calls to recur can only happen in a tail position.[Source] Here’s a non-tail recursive implementation of the sum of first `n` natural numbers: ``````(defn sum-of-natural-numbers [n] (if (<= n 0) 0 (+ n (sum-of-natural-numbers (dec n))))) `````` If we pass `100000` to this function, we get the following output on the REPL: ``````(sum-of-natural-numbers 100000) ; Execution error (StackOverflowError) at flash.tail-recursive/sum-of-natural-numbers (form-init869174823328265034.clj:6). ; null `````` Here’s a tail recursive implementation of the above function: ``````(defn sum-of-natural-numbers-tail-recursive ([n] (if (<= n 0) 0 (sum-of-natural-numbers-tail-recursive n 0))) ([n acc] (if (<= n 0) acc (recur (dec n) (+ acc n))))) `````` This function successfully completes evaluation when we pass `100000` ``````(sum-of-natural-numbers-tail-recursive 100000) ;; => 5000050000 `````` Here’s a tail-recursive implementation of finding the `n`th fibonacci number: ``````(defn fibo-tail-recursive ([n] (if (<= n 0) 0 (fibo-tail-recursive n 0 1))) ([n prev curr] (if (= n 1) curr (recur (dec n) curr (+ prev curr))))) `````` The above tail recursion examples involve a common pattern: • For each recursive call, we pass an accumulator parameter, in addition to the main argument. • The initial value of this accumulator represents the base case • At each recursive step, we modify the main argument as per the definition of the problem (in the case of `sum-of-natural-numbers-tail-recursive`, decrementing the value of `n`) and update the current value of the accumulator. We pass both these parameters along with the recursive call. • The operations we were doing after the return of the non-tail recursive function, are now done before calling the tail recursive function and this is passed as the accumulator • The value of the accumulator in the base-case, is the final result, which is returned by the base case. ## Takeaways 1. Representing a problem in a recursive manner, involves breaking down the problem into a series of simpler sub-problems and providing a result for the simplest version of the problem 2. Although recursion is intuitive and easy to reason about while solving computational problems, they take significantly more space than iterative processes 3. A special kind of recursion, namely tail recursion, solves the problem of stack overflow, when we can express our recursive calls as tail calls. 4. We can transform (ordinary) recursive functions to tail recursive functions, with the help of an additional accumulator parameter.<\|endoftext\|>	4.40625
898	Venus Express searching for life – on Earth Scientists using ESA’s Venus Express are trying to observe whether Earth is habitable. Silly, you might think, when we know that Earth is richly stocked with life. In fact, far from being a pointless exercise, Venus Express is paving the way for an exciting new era in astronomy. Venus Express took its first image of Earth with its Visible and Infrared Thermal Imaging Spectrometer (VIRTIS) soon after its launch in November 2005. About a year after the spacecraft established itself in Venus’s orbit, David Grinspoon, a Venus Express Interdisciplinary Scientist from the Denver Museum of Nature & Science, Colorado, suggested a programme of sustained Earth observation. “When the Earth is in a good position, we observe it two or three times per month,” says Giuseppe Piccioni, Venus Express VIRTIS Co-Principal Investigator, at IASF-INAF, Rome, Italy. The instrument has now amassed approximately 40 images of Earth over the last two years. The images of Earth cover both visible and near-infrared regions of the spectrum and can be split into spectra, in order to search for the signature of molecules in the Earth’s atmosphere. The value of the images lies in the fact that Earth spans less than a pixel in Venus Express’s cameras. In other words, it appears as a single dot with no visible surface details. This situation is something that astronomers expect to soon face in their quest for Earth-sized worlds around other stars. “We want to know what can we discern about the Earth’s habitability based on such observations. Whatever we learn about Earth, we can then apply to the study of other worlds,” says Grinspoon. Since 1995, astronomers have been discovering these extrasolar planets and now know of more than three hundred. As observational techniques have been refined and the data continuously taken, so smaller and smaller planets have been discovered. Now, with CNES–ESA’s COROT and NASA’s Kepler missions, the prospect of discovering Earth-sized worlds in Earth-like orbits around other stars is better than ever. “We are now on the verge of finding Earth-like planets,” says Grinspoon. As has been proved with the discovery of gas giant planets, as soon as astronomers know that they are there, they invent all sorts of innovative methods to separate the planet’s feeble light from the overwhelming glare of the star. One thing has become obvious from the study of Earth using Venus Express: determining whether a planet is habitable is not going to be easy. “We see water and molecular oxygen in Earth’s atmosphere, but Venus also shows these signatures. So looking at these molecules is not enough,” says Piccioni. Instead, astronomers are going to have to search for more subtle signals, perhaps the so-called red edge caused by photosynthetic life. “Green plants are bright in the near infrared,” says Grinspoon. The analysis to see whether this red edge is visible is just beginning. The team will also compare spectra of the Earth’s oceans with those taken when the continents are facing Venus Express. “We have initiated the first sustained programme of Earth observation from a distant platform,” says Grinspoon. Although the observations may not tell us anything new about the Earth, they will allow us to unveil far-off worlds, making them seem more real than simply dots of light. For more information: David Grinspoon, Venus Express interdisciplinary scientist, Denver Museum of Nature & Science, Colorado, USA Email: David.Grinspoon @ dmns.org Giuseppe Piccioni, VIRTIS co-Principal Investigator, IASF-INAF, Rome, Italy Email: Giuseppe.piccioni @ iasf-roma.inaf.it Pierre Drossart, VIRTIS co-Principal Investigator, Observatoire de Paris-LESIA, France Email: Pierre.Drossart @ obspm.fr Håkan Svedhem, ESA Venus Express Project Scientist Email: Hakan.Svedhem @ esa.int<\|endoftext\|>	3.8125
411	# How do you find dy/dx for the function; cos(x^2+2Y)+xe^Y^2=1? Apr 11, 2017 $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x \sin \left({x}^{2} + 2 y\right) - {e}^{{y}^{2}}}{- 2 \sin \left({x}^{2} + 2 y\right) + 2 x y {e}^{{y}^{2}}}$ #### Explanation: Using the implicit function theorem: $f \left(x , y\right) = \cos \left({x}^{2} + 2 y\right) + x {e}^{{y}^{2}} = 1$ $\mathrm{df} = 0 = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}}$ So: ${f}_{x} = - 2 x \sin \left({x}^{2} + 2 y\right) + {e}^{{y}^{2}}$ ${f}_{y} = - 2 \sin \left({x}^{2} + 2 y\right) + 2 x y {e}^{{y}^{2}}$ $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}} = \frac{2 x \sin \left({x}^{2} + 2 y\right) - {e}^{{y}^{2}}}{- 2 \sin \left({x}^{2} + 2 y\right) + 2 x y {e}^{{y}^{2}}}$<\|endoftext\|>	4.5625
348	(1852–1931). Because Marshal Joffre halted the German invasion of 1914 on the Marne River he has been called the “victor of the Marne.” He was born on Jan. 12, 1852, in southern France. His father made wine casks. The boy, however, had little interest in this work so he was sent to prepare for a military career at the École Polytechnique in Paris. Before he completed the course, Joffre was called into service in the Franco-Prussian War of 1870 and 1871. He saw the victorious Germans march into Paris. In the next 40 years he prepared himself and France for the next Prussian blow. He spent some of those years in the French colonies in Africa and Asia, and superintended the building of many important defenses. In 1914 at the outbreak of World War I he was the French chief of staff. Subsequently, he was made supreme commander of the French forces on the Western front. Before the powerful German thrust, Joffre retreated from Belgium into France. Joffre, however, was choosing his own time and his own ground for battle. On Sept. 6, 1914, after five weeks of retreating, he gave the command for attack. The result was the victory of the Marne. Joffre was acclaimed as the savior of France. His country made him marshal of France and decorated him with the grand cross of the Legion of Honor. Joffre’s removal from supreme command came in December 1916, after losses at Verdun. Later he served on the French High Commission as technical adviser. He died in Paris, on Jan. 3, 1931.<\|endoftext\|>	3.703125
662	Math + You = 1 # How to approximate the perimeter of an ellipse? It is easy to calculate the perimeter of the well-known special case of ellipse: the circle. For the circle, its perimeter is P=2\pi R where R is the radius of the circle. In the rest of this article, we will consider an ellipse with a Cartesian equation: \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1 Any translation or rotation does not change the value of the perimeter, so we have chosen to center this ellipse at the origin of the reference, aligned with the axes. ## The classical approximation of the perimeter of the ellipse The classic formula, which is found almost everywhere around the perimeter of the ellipse, is the following P \approx 2\pi \sqrt{\dfrac{a^2+b^2}{2}} ## Ramanujan's formulas for the perimeter of the ellipse ### Ramanujan's first formula Ramanujan has accustomed us to formulas, each one more incredible than the other, especially those on the number pi. Here is his first approximation of the perimeter of the ellipse. P \approx \pi\left(3(a+b) - \sqrt{(3a+b)(a+3b)}\right) ### Ramanujan's second formula And here is his second formula, there we arrive on the incredible! We pose h = \dfrac{(ab)^2}{(a+b)^2} The approximation is then: P \approx \pi (a+b) \left( 1 + \dfrac{3h}{1 + \sqrt{4-3h}}\right) Nice huh? Fortunately we are not asked in class to learn such formulas! ## Using an integral We are going to establish an absolute formula but which does not admit a directly calculable value. The ellipse verifies the parametric equation \left\{\begin{array}{lcl} x(t) & = & a \cos(t)\\ y(t) & = & b \sin(t) \end{array} \right., 0 \leq t \leq 2 \pi We then have, according to a formula on the parametrized curves: P = \int_0 ^{2\pi}\sqrt{x'(t)^2 +y'(t)^2 }dt Which give P = \int_0 ^{2\pi}\sqrt{a^2\sin^2(t) +b^2 \cos^2 (t)}dt And thanks to various symmetries, we obtain: P = 4\int_0 ^{\frac{\pi}{2}}\sqrt{a^2\sin^2(t)+b^2 \cos^2 (t)}dt This integral has no explicit solution. One can then use methods of approximation of integrals to obtain again an approximation of this perimeter: • Packages rectangle • Trapezoid Formula • Simpson's Formula • Monte Carlo method And to find out why this perimeter cannot be calculated, I recommend this video which will allow you to learn more!<\|endoftext\|>	4.59375
1,599	# Basic concepts of percentage Basic concepts of percentage are essential for everyone. Application of concepts of percentage can be seen everywhere in arithmetic section of mathematics. With the intention of gaining knowledge of percentage, let us explore the topic. Percentage can be defined as “a part out of each hundred”or in other words “for every hundred”. For example, 20 out of 100 can be written as, \frac { 20 }{ 100 } or 20%. Similarly, 1%=\frac { 1 }{ 100 } 50%=\frac { 50 }{ 100 } 99%=\frac { 99 }{ 100 } We can express percentage in many ways: • Express percentage as fraction: “x%” can be expressed as fraction which= \frac { x }{ 100 } to explain, 40%= \frac { 40 }{ 100 } = \frac { 2 }{ 5 } • Express percentage as decimal: We know, 2%= \frac { 2 }{ 100 } to get it decimal form we have to divide 2 by 100, i.e ⇒ 2%= \frac { 2 }{ 100 } =0.02 • Convert fraction into percentage: Let \frac { a }{ b } be a fraction, to convert it into percentage we multiply it by 100 . =( \frac { a }{ b }×100 )% example, percentage value of \frac { 1 }{ 4 } will be ⇒(\frac { 1 }{ 4 } ×100)%=25% • Convert decimal into percentage: In order to convert a decimal number into percentage, we multiply the decimal number with 100. →1.5 can be in percentage as, = 1.5× 100 = \frac { 15 }{ 10 } × 100 = 150% Before moving further, let us solve some questions. Question 1. Convert the following into fraction: i) 75% ii) 2% Solution: i) 75%=\frac { 75 }{ 100 } =\frac { 3 }{ 4 } ii) 2%=\frac { 2 }{ 100 } =\frac { 1 }{ 50 } Question 2: Convert the following into decimals: i) 25% ii) 20% Solution: i) 25%=\frac { 25 }{ 100 } =0.25 ii) 20% =\frac { 20 }{ 100 } =0.2 Question 3. What is 20% of 300 ? Solution : This can be written as, = \frac { 20 }{ 100 } × 300 = 60 Question 4. Solve 2% of 50 + 5% of 10 . Solution : This is, =\frac { 2 }{ 100 } ×50 + \frac { 5 }{ 100 } × 10 =1+\frac { 1 }{ 2 } = \frac { 3 }{ 2 } =1.5 Question 5. 3 is what percent of 50 ? Solution: Required percentage = \frac { 3 }{ 50 } × 100 =6% ### Percentage and their fractional values Following are some percentages and their fractional values that are often used in calculations.In fact, these values will be useful for solving questions swiftly. We can find other fractions from these values. To enumerate, ⇒If 33\frac { 1 }{ 3 } %= \frac { 1 }{ 3 } then, by multiplying both sides by 2 we will get, {(33\frac { 1 }{ 3 }) × 2 }%=\frac { 1 }{ 3 } ×2 Thus 66\frac { 2 }{ 3 } % = \frac { 2 }{ 3 } . See its quite simple. Similarly let us look another example. ⇒if 20% =\frac { 1 }{ 5 } then 60% will be 3 times 20% ie. 20% ×3=\frac { 1 }{ 5 } × 3 60%=\frac { 3 }{ 5 } ### Some important tricks for percentage: 1. If a is increased to a’, then percentage increase #### percentage increase%= \frac { a-a' }{ a } ×100 2. When a is decreased to { a }_{ 0 } then, #### percentage decrease%= \frac { a-{ a }_{ 0 } }{ a } × 100 3. If amount is first increase by x% and then decreased by x%.The percentage change will be #### Decrease of \frac { { a }^{ 2 } }{ 100 } % 4. If amount is first increase by x% and then decreased by y%.The percentage change will be , #### = (x-y-\frac { xy }{ 100 } )% [ – sign for decrease and +sign for increase] 5. If amount is first decreased by x% and then increased by y%.The percentage change will be , #### = (-x+y-\frac { xy }{ 100 } )% 6. If amount is first increased by x% and then decreased by y%.Then net percentage decrease will be , ##### =(-x-y+\frac { xy }{ 100 } )% 7. If amount is first increased by x% and then increased by y%.Then net percentage increase will be , ### Comparision between two quantities: 1. If A is x% more than B, Then B is less than A by #### (\frac { x }{ 100+x } ×100)% 2. If A is x% less than B, Then B is more than A by ### Price of commodity (for no change in expenditure) • If the price of commodity is increase by x% then reduction in comsumption as not to inrease the expenditure is #### (\frac { x }{ 100+x } ×100)% • If the price of commodity is decrease by x% then increase in comsumption as not to decrease the expenditure is ### Population based questions If the population of a town is P & it increases at a rate of R% per annum, Then ### DEPRECIATION Let the present value of an object be P, if it depreciates at the rate of R% per annum , then • The value of object after n years #### =P({ 1-\frac { R }{ 100 } })^{ n } • The value of object n years ago ### =\frac { P }{ ({ 1-\frac { R }{ 100 } ) }^{ n } } These trick will be a boon in your performance and will increase your speed. Not only it will save your time but also will be beneficial in other sections of mathematics. Do not try to memorize these formulas.The only way to learn them, is by using them in problems. Solve as many questions as you can.You can solve questions from previous years of your exam. Concepts of percentage will be beneficial in sections like profit-loss,discount and Simple interest-compound interest. Make sure you practice a lot, because maths is nothing without practice.Once the basic concepts of percentage are clear there is no looking back. At the same time if you want to learn other basic topics of mathematics,you can have a look. BASIC TOPICS OF MATHEMATICS<\|endoftext\|>	4.84375
244	In 1928, Paul Kollsman forever changed the way pilots would fly. By introducing the first accurate barometric altimeter, an instrument used to measure the altitude by calculating barometric pressure, "flying on the gauges" became possible. In 1929, the first blind flight was made using the altimeter as a guide through the sky. Kollsman's altimeter with a barometric setting display, which is commonly called the Kollsman Window, is still integral to aircraft flight. Paul Kollsman was born and educated in Germany, studying science and technology before he immigrated to America in 1923 in hopes of selling a radical new automotive engine. He was introduced to altimeters during employment with the Pioneer Instrument Company. Working from his attic in 1928 he developed an altimeter with unprecedented accuracy. The Navy purchased 300, launching Kollsman Instrument Company. Kollsman acquired more than a hundred patents, and the altimeter is considered one of the major milestones in aviation. A man of many interests, Kollsman was active in the mining industry and was granted many patents for his work in converting seawater to fresh water and a patent for a slip-resistant surface for bathtubs and showers.<\|endoftext\|>	3.890625
456	Pyramids are three-dimensional (3-D), unlike triangles, which are flat, or two-dimensional (2D). This means they have depth as well as length and width. Pyramids have a polygon at their base (a polygon is a plane or flat shape with 3 or more sides). The name of a pyramid depends upon the shape of its base. For example, pyramids with a square at their base are known as 'square based' pyramids, such as the one in the picture and those in Egypt. A pyramid with a triangular base is called a 'tetrahedron'. The other faces are triangles, which meet at a vertex at the pyramids top. © Islington Artefact Library Pyramids are members of the polyhedron family. A polyhedron is a solid shape which is made from lots of polygons. The word polyhedron comes from a Greek word meaning many faces. Polyhedra have flat faces, straight edges and vertices. There is a formula for connecting the number of faces, edges and vertices on a solid, made famous by the Swiss mathematician Euler (1707-1783). The formula is: Vertices + Faces - Edges = 2 The height of a pyramid is the perpendicular distance from the top (vertex) to the plane of the base. The volume of a pyramid is one-third of the volume of a prism that has the same base and altitude. The equation for the volume of a pyramid is hK/3, in which h is the height of the pyramid and K is the area of the base. The most famous pyramids of all were built in ancient Egypt between about 3200 - 2500 BC. They were the monuments and burial places for the royal families of Egypt - in particular their kings who were called Pharaohs. We may never know why the Egyptians chose the pyramid shape. It may have developed from earlier burial mounds, or been a symbol of the sun's rays or a stairway to heaven. Pyramids were also built in Central America by the Mayan people between AD 300 and 900, and later by the Aztecs. They were used to worship their sun, moon and rain gods.<\|endoftext\|>	4.28125
907	Ecologists are embracing the potential to increase the storage of carbon in terrestrial ecosystems. This could reduce the build-up of carbon dioxide in Earth’s atmosphere. At least a portion of their enthusiasm derives from the observation that wetlands are very efficient in soil carbon storage, because the rate of decomposition is slow in anoxic (without oxygen) conditions. Most wetlands store carbon at rates about 10X greater than the rate of carbon storage in upland soils. Now, at least several investigations suggest some tempering of enthusiasm for carbon storage in restored or constructed wetlands, inasmuch as wetlands yield a greater flux of methane to the atmosphere, and methane has a much greater effect on global warming than carbon dioxide—the normal pathway returning carbon to the atmosphere from upland soils. The enhanced carbon storage in wetland soils must be discounted by their greater flux of methane to the atmosphere. Flooding can convert soils from a methane sink to a methane source to the atmosphere—even in uplands as shown in a new paper by Ni and Groffman. One of the earliest investigations of the comparative flux of methane and carbon dioxide from peatland soils found that the ratio of carbon dioxide to methane was >2500 in relatively dry soils, but decreased to values of 4 to 173 when the soils were flooded. Any ratio less than about 80 means that the impact of flooding soils with the goal of enhanced carbon storage is completely negated by the higher flux of methane from flooded soils to the atmosphere. Many restored wetlands in California are net sources of global warming potential, despite storing more carbon in their soils. Reservoirs created by hydroelectric dams in the Amazon basin are also large sources of greenhouse gases to the atmosphere. The “discount” to the role of wetlands as a carbon sink is less in mangrove forests, salt marshes, and seagrass meadows inasmuch as the methane flux from saline habitats is less than that from flooded freshwater soils. Mangroves are seldom constructed to provide a new sink for atmospheric carbon dioxide. Nevertheless about 20% of the benefit of carbon storage in mangrove soils is lost due to methane flux to the atmosphere. Wetlands are great for wildlife, improving water quality, and for tempering the flow of water to the sea. But, as tools to mitigate global warming, we need to look more carefully at their net gas flux. Barros, N. and 7 others. 2011. Carbon emission from hydroelectric reservoirs linked to reservoir age and latitude. Nature Geoscience 4: 593-596. Bortolotti, L.E., V.L. St Louis, R.D. Vinebrooke, and A.P. Wolfe. 2016. Net ecosystem production and carbon greenhouse gas fluxes in three prairie wetlands. Ecosystems 19: 411-425. Griscom, B. and 31 others. 2017. Natural climate solutions. Proceedings of the National Academy of Science doi: 10.1073/pnas.1710465114 Hemes, K.S., S.D. Chamberlain, E. Eichelmann, S.H. Knox and D.D. Baldocchi. 2018. A biogeochemical compromise: the high methane cost of sequestering carbon in restored wetlands. Geophysical Research Letters doi.org/10.1029/2018GL077747 Moore, T.R. and R. Knowles. 1989. The influence of water table levels on methane and carbon dioxide emissions from peatland soils. Canadian Journal of Soil Science 69: 33-38. Neubauer, S.C and J.P. Megonigal. 2015. Moving Beyond Global Warming Potentials to Quantify the Climatic Role of Ecosystems. Ecosystems 18:1000-1013 Ni, X. and P.M. Groffman. 2018. Declines in methane uptake in forest soils. Proceedings of the National Academy of Sciences doi:10.1073/pnas.1807377115. Rosentreter, J.A., D.T. Maher, D.V. Erler, R.H. Murray and B.D. Eyre. 2018. Methane emissions partially offset “blue carbon” burial in mangroves. Science Advances 4: DOI: 10.1126/sciadv.aao4985<\|endoftext\|>	3.921875
1,848	The myriad faces of the coasts A million kilometres of coastsThe coasts of the world are highly diverse. The northern coast of Brittany in France is characterized by granite cliffs interspersed with numerous bays. In Namibia the high dunes of the Namib Desert extend to the Atlantic shore, where the coast runs nearly parallel to the dunes. In Siberia, by contrast, the flat coastal region is dominated by permafrost, a metres-thick layer of frozen soil whose surface thaws out for a few weeks each year during the short Arctic summer, when it is especially susceptible to wave action. During storm-flood events, several metres of the saturated banks can break off, creating a constantly changing shore face. - 1.19 > This satellite photo shows the Lena River delta in Siberia with all its fine structures, extending around 150 kilometres into the Laptev Sea. A large proportion of the sea ice that eventually drifts out into the Arctic Ocean originates in this marine region. - What all of these coasts have in common is that they are narrow strips of land exposed to the forces of the sea. Depending on the context they can be classified in different ways. Coasts can be distinguished based on whether they are strongly or weakly washed by the surf and currents. They can, alternatively, be classified according to the materials they comprise or by the rate that the material is eroded away by the sea. Coasts can furthermore be characterized by their ability to capture sediments that are delivered by rivers or currents. The ultimate form exhibited by a coast also depends significantly on the interplay between the materials that make up the substrate or that rivers transport to the coast, and the physical forces of wind and wave action that impact those materials. Geologists estimate the total global length of coastline to be around one million kilometres. This projection, of course, depends on how fine a scale is applied. When considering the entire globe, any differentiation of the coasts is only practical at a relatively coarse scale. For this categorization the continental margins can be traced in their present forms, which are in part a result of plate tectonics. Researchers created such a classification system in the 1970s, under which six different categories of coasts were distinguished. - Coastal plain: an area where the land gently flattens toward the sea. An example is the coast of the West African country of Mauritania, where the land merges into the sea through a broad strip of coastal marshes and low dunes. - Major delta: a large river mouth where sediments from the river are deposited because the ocean currents or tides are not strong enough to transport the material away. This is the case with the delta of the Lena River in Russia, which flows into the Laptev Sea in the Arctic Ocean. - Tropical coral reef: a structure composed of carbonate produced by sessile corals (Cnidarians). It develops as a fringe along the coasts in near-surface waters penetrated by abundant light. Reef-building corals occur in tropical and subtropical waters at temperatures consistently greater than 20 degrees Celsius. A spectacular tropical coral reef is situated along the Central American Caribbean coast between Honduras and Belize. It is around 250 kilometres long and is among the most popular diving areas in the world. - Rocky coast and fjord: a coast of solid rock. Fjords, like those found abundantly on the west coast of Norway, represent a special kind of rocky coast. They were formed during glacial periods, when the motion of the glaciers scoured deep valleys into the bedrock. - Permafrost coast: a deeply frozen soil covering large areas of the Arctic land masses in the northern hemisphere since the last glacial period. Permafrost is found over many thousands of kilometres along the coasts of North America, Siberia and Scandinavia. - Headland-bay coast: a coast where rocky headlands extend into the sea. The headlands act as barriers to obstruct the surf and currents. Slow eddy currents form in the sheltered areas between headlands, gradually eroding the shore and forming bays. An example of this is Half Moon Bay on the Pacific Coast of the United States near San Francisco. There, over thousands of years, a half-moon shaped bay has formed behind a prominent headland. Wind and waves shape the coastsThe physical forces of the sea – the waves, currents and winds – have a substantial effect on the shape of the coasts. The intensity of these forces is used to distinguish between low-energy and high-energy coasts. The kind of material that makes up the substrate of a coastal area is also a key factor influencing the formation of the coasts. Tidal flats comprising relatively fine sediments can be reworked fairly quickly because these materials are easily transported by the currents. Fine sands can also be easily transported, as illustrated by the East Frisian Islands off the German North Sea coast. Because the prevailing winds there blow from the west, wave action carries sand away from the northwest side of the islands and redeposits it on the east side. In the past this has caused the islands to slowly migrate eastward. To impede this motion, rock jetties and breakwaters were built as early as the 19th century to fortify the islands. This greatly helped in preventing further migration. - 1.22 > The village of Porthleven in the English county of Cornwall is located on an extremely high-energy rocky coast. Accordingly, the shoreline fortifications, including massive walls, are very substantial. Under conditions of very high seas, however, they are hardly noticeable. The intertidal zone The intertidal zone is the area of the coast that is flooded and then exposed again by the rhythm of daily tidal cycles. The surface can be mudflats, sandy beaches or rocky cliffs. Rocky shores are exceptional because they occur on steep coastlines, while most other intertidal zones are found on flat coasts. Large-scale flat intertidal areas which include the salt marshes on the shore are called mudflats. - While changes in the shape of sandy coasts are often visible with the naked eye, they can be more difficult to recognize when other material is involved. But even high-energy rocky coasts do change their appearance over time. The rate of change, however, depends largely on the properties of the rocks. Coasts composed of compacted but not yet lithified ash, generated over time by ash falls from volcanic eruptions, are especially easily eroded. Examples of this kind of coast are found in New Zealand. Up to ten metres of coast can be lost there within a single year at some locations. Chalk cliffs, like the White Cliffs of Dover in the extreme southeast of England are also relatively soft. When exposed to strong currents they can be eroded by several centimetres per year. By contrast, hard granitic rock is depleted at most by only a few millimetres in the same space of time. Harder still, black volcanic basalt is only destroyed by water each year by a maximum of a few hundred billionths of a metre. - 1.23 > Depending on the material making up the coasts, they can be eroded slowly or more rapidly. Some can be depleted by several metres in a single year. A question of particle sizeUnderstanding the nature of the substrate in coastal areas is especially crucial for coastal protection, coastal management, and the planning of waterways and port installations. In particular, the size and density of the particles that make up the material play an important role. These can be factors, for example, in determining whether the shore of a populated island is in danger of erosion or whether shipping channels might shift their positions causing ships to run aground. With respect to the size of particles, the following coastal types are differentiated: - muddy coasts, - sandy coasts, - pebble coasts, - cobble coasts, - rocky or boulder coasts. The filtering function of the coastsIn many areas the character of the coasts is strongly shaped by rivers – through both their current strength and the material loads that they transport. They carry many minerals and nutrients that are incorporated to some extent into the sediments. Coasts that are rich in such sediments are also highly productive. A good example are the Sundarbans in Bangladesh and India, which, with a total area of around 10,000 square kilometres, comprise the largest block of mangrove forest in the world. The Sundarbans formed in the estuarine areas of the Ganges and Brahmaputra Rivers, which deliver immense amounts of material into the Gulf of Bengal. The Sundarbans are a vital unspoiled natural region. They are home to abundant birds, fish, crocodiles, pythons, deer and wild boar. Furthermore, rare animals such as the axis deer and Bengal tiger may also find refuge here. Depending on the ability of a particular coast to filter and store the material transported by rivers, it can be designated as having an active or inactive filtering function. ><\|endoftext\|>	4.21875
574	Today’s Wonder of the Day was inspired by Cohen. Cohen Wonders, “How does touch screen work?” Thanks for WONDERing with us, Cohen! From mall kiosks to smartphones to tablet computers, touch screens are everywhere you look these days. As technology advances, keyboards and mice are quickly becoming a thing of the past. Why be burdened with cords when you can have what you want with just a touch? Touch screens are electronic visual displays that allow a user to interact directly with what is displayed on the screen, rather than using a pointing device, such as a mouse. Touch screens are designed to respond to the touch of a finger, although an object — like a stylus — can also be used. Touch screens are used in all sorts of modern electronic devices, including personal digital assistants (PDAs), satellite navigation systems and video games. Their popularity has surged recently, but the idea for the touch screen goes back several decades. Given the many different types of devices that use touch screens, it's no surprise that there are several different types of touch screens. Each type of touch screen works a little differently from the others. Resistive touch screen systems use two thin layers separated by spacers. An electrical current runs through the two layers. When the screen is touched, the two layers make contact in the exact spot where the screen is touched. This contact creates a change in the electrical field, which a device's computer operating system can understand. Capacitive touch screen systems feature a special layer that stores an electrical charge. When the screen is touched, some of the electrical charge is transferred to the user. This decreases the charge on the capacitive layer. The device's computer operating system can determine from this change in electrical charge where the screen was touched. For a capacitive system to work, some of the electrical charge must be able to be transmitted to the user. This is why capacitive touch screens may not work properly if you wear gloves that block the transmission of the electrical charge. Capacitive systems are newer and tend to be more popular than resistive systems, because they transmit more light and provide a clearer picture. Of course, capacitive systems also tend to be more expensive than resistive systems, too. Surface acoustic wave touch screen systems use transducers and reflectors to measure changes in the reflection of ultrasonic waves caused when the screen is touched. These systems are the most advanced and offer the clearest picture possible. Unfortunately, they're also extremely expensive. When touch screens first became popular, they could only sense one point of contact at a time. Technology has advanced greatly in recent years, though. Today, many touch screen devices feature multi-touch technology. This technology allows a touch screen device to interpret multiple points of contact simultaneously.<\|endoftext\|>	4.03125
213	Otitis externa is the inflammation and infection of the external ear canal. The skin in the ear canal can become red and swollen due to a bacterial or fungal infection. Symptoms include pain in the ear canal, itchiness, foul smelling yellow or green pus in the ear canal, fever, hearing loss and tinnitus. Water can enter the ear canal and not drain out, which can lead the skin to become soft and an ideal environment for bacteria and fungus to grow, particularly after swimming. Inspection (otoscopy) of the ear canal is the first stage of diagnosis. A medical professional may also take a swab of the infection to determine if it is bacterial or fungal. Treatment and prevention Drops or ointment may be prescribed to treat the infection and the ear may be suctioned. To prevent otitis externa, keep the ear dry while swimming and showering by using ear plugs. Avoid swimming in polluted water and do not poke your fingers or other objects into your ear.<\|endoftext\|>	3.796875
1,161	##### Objectives 1. Learn to replace a system of linear equations by an augmented matrix. 2. Learn how the elimination method corresponds to performing row operations on an augmented matrix. 3. Understand when a matrix is in (reduced) row echelon form. 4. Learn which row reduced matrices come from inconsistent linear systems. 5. Recipe: the row reduction algorithm. 6. Vocabulary words: row operation, row equivalence, matrix, augmented matrix, pivot, (reduced) row echelon form. In this section, we will present an algorithm for “solving” a system of linear equations. We will solve systems of linear equations algebraically using the elimination method. In other words, we will combine the equations in various ways to try to eliminate as many variables as possible from each equation. There are three valid operations we can perform on our system of equations: • Scaling: we can multiply both sides of an equation by a nonzero number. • Replacement: we can add a multiple of one equation to another, replacing the second equation with the result. • Swap: we can swap two equations. ##### Augmented Matrices and Row Operations Solving equations by elimination requires writing the variables and the equals sign over and over again, merely as placeholders: all that is changing in the equations is the coefficient numbers. We can make our life easier by extracting only the numbers, and putting them in a box: This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belongs; a matrix is a grid of numbers without the vertical line. In this notation, our three valid ways of manipulating our equations become row operations: • Scaling: multiply all entries in a row by a nonzero number. Here the notation simply means “the first row”, and likewise for etc. • Replacement: add a multiple of one row to another, replacing the second row with the result. • Swap: interchange two rows. The process of doing row operations to a matrix does not change the solution set of the corresponding linear equations! Indeed, the whole point of doing these operations is to solve the equations using the elimination method. ##### Definition Two matrices are called row equivalent if one can be obtained from the other by doing some number of row operations. So the linear equations of row-equivalent matrices have the same solution set. In the previous subsection we saw how to translate a system of linear equations into an augmented matrix. We want to find an algorithm for “solving” such an augmented matrix. First we must decide what it means for an augmented matrix to be “solved”. ##### Definition A matrix is in row echelon form if: 1. All zero rows are at the bottom. 2. The first nonzero entry of a row is to the right of the first nonzero entry of the row above. 3. Below the first nonzero entry of a row, all entries are zero. Here is a picture of a matrix in row echelon form: ##### Definition A pivot is the first nonzero entry of a row of a matrix in row echelon form. A matrix in row-echelon form is generally easy to solve using back-substitution. For example, We immediately see that which implies and See this example. ##### Definition A matrix is in reduced row echelon form if it is in row echelon form, and in addition: 1. Each pivot is equal to 1. 2. Each pivot is the only nonzero entry in its column. Here is a picture of a matrix in reduced row echelon form: A matrix in reduced row echelon form is in some sense completely solved. For example, When deciding if an augmented matrix is in (reduced) row echelon form, there is nothing special about the augmented column(s). Just ignore the vertical line. If an augmented matrix is in reduced row echelon form, the corresponding linear system is viewed as solved. We will see below why this is the case, and we will show that any matrix can be put into reduced row echelon form using only row operations. # Subsection1.2.3The Row Reduction Algorithm We will give an algorithm, called row reduction or Gaussian elimination, which demonstrates that every matrix is row equivalent to at least one matrix in reduced row echelon form. The uniqueness statement is interesting—it means that, no matter how you row reduce, you always get the same matrix in reduced row echelon form. This assumes, of course, that you only do the three legal row operations, and you don’t make any arithmetic errors. We will not prove uniqueness, but maybe you can! Here is the row reduction algorithm, summarized in pictures. It will be very important to know where are the pivots of a matrix after row reducing; this is the reason for the following piece of terminology. ##### Definition A pivot position of a matrix is an entry that is a pivot of a row echelon form of that matrix. A pivot column of a matrix is a column that contains a pivot position. In the above example, we saw how to recognize the reduced row echelon form of an inconsistent system. In other words, the row reduced matrix of an inconsistent system looks like this: We have discussed two classes of matrices so far: 1. When the reduced row echelon form of a matrix has a pivot in every non-augmented column, then it corresponds to a system with a unique solution: 2. When the reduced row echelon form of a matrix has a pivot in the last (augmented) column, then it corresponds to a system with a no solutions: What happens when one of the non-augmented columns lacks a pivot? This is the subject of Section 1.3.<\|endoftext\|>	4.875
489	This Trippy Image Shows How Supercomputers Helped Build Better Jet Engines It's not a psychedelic desktop wallpaper from the future; it’s a compressor blade simulation from the past, produced by a Cray supercomputer. Re-Exposure is an occasional Motherboard feature where we look back on delightful old tech photos from wire service archives. Most people have likely never seen a supercomputer in person, let alone sat on one, as the vintage Cray supercomputers memorably allowed people to do. But for those unfamiliar with these multi-million-dollar supercomputers, perhaps one of the most awesome and interesting use cases for them is perhaps jet engine design. Above, computer scientist John Aldag (a Cray employee) is shown in 1983, using a Cray machine to virtually test the compressor blades on the front of a jet for any inconsistencies that might show themselves. (Aldag had deep familiarity with graphics related to Cray machines, by the way, once helping to produce a paper titled "Requirements, Status and Plans for Graphics on Cray Computers" that he presented to the Cray user group in 1986.) So why a use an extremely expensive Cray supercomputer for this? Why not an IBM PC? A 1997 press release from Stanford University, announcing the launch of Department of Energy-funded supercomputer simulations for aircraft engine design, breaks down the benefits of testing compressor blades using high-powered supercomputing technology such as Cray's: The first component in an aircraft jet engine is the compressor, which raises the pressure of the air flowing through the engine. The efficiency of the engine is strongly dependent on the quality of the air flow through the compressor. Improper flow patterns can cause compressor-blade vibrations that can cause premature engine failure. A good engine design requires large investments in testing because the flows that occur in compressors are not now adequately predictable. A goal of the center is to be able to predict these complex flows, predict the compressor performance under all possible operating conditions, and investigate flow-driven blade vibration problems, thereby dramatically reducing the need for costly testing. In other words, jet-engine design is such an important thing not to get wrong that aerospace companies are willing to spend a whole lot of money to get it right. As a result, this use case remains an important part of Cray's business, even today.<\|endoftext\|>	3.78125
907	In a transformation the shape does not change, only position, size and direction can change. That means, all the angles of the transformed shape do not change. A combination of transformations just means doing more than one transformation to the same shape. For example, an enlargement and reflection means you enlarge the shape, then you reflect the enlarged shape. ### How do you do transformations? There are four transformations, which TERRy can help us remember: Translation Enlargement Rotation Reflection y ### Translation The position of the shape is moved up/down/sideways, but it does not change in size or direction. ### Enlargement This means increasing the size of a shape by a 'scale factor' from a particular point, which is called the centre of enlargement. Enlargement changes size and position but not direction. If the shape is complicated to enlarge, take it step by step by taking the corners of the shape and 'enlarging' it one by one. Enlarging a shape We want to enlarge a shape with scale factor 3, with centre of enlargement (1,1) Suppose we have an 'L' shape. Let's start with Corner A, which has co-ordinates (2,3). [picture missing] To get to A from the centre of enlargement, we have to go right 1 step and up 2 steps. A scale factor of 3 means we multiply these distances by 3. So now, to get to A' from the centre of enlargement, we have to go right 3 steps and up 6 steps. These co-ordinates (4,7), because (1,1) is the centre of enlargement. Finding the centre of enlargement You can use a ruler to draw lines connecting each corner from the original shape to same corner in the enalrged shape. where all these points meet is the centre of enlargement. ### Rotation Every point on a rotated shape is at the same distance from the centre of rotation. It stays the same size, but changes in direction. There are no tricks to finding the centre of rotation, it is done by observing, but remember every point must be at the same distance. To rotate an object, again take each corner one by one and rotate it using a protractor. If the rotation is by 90o,180o,270o then we you can use the grid on the paper. For example, if you want to rotate the point (2,3) about the origin 90o clockwise, insteading of going right 2 steps and up 3 steps, we go down 2 steps and right 3 steps, giving you the co-ordinates (3,2) (remember up and down give you y-coordinate, and left and right give you x co-ordinates, it doesn't matter what order it is in!) ### Reflection Reflection means we take themirror image of a shape, where the mirror would be, we call it the line of symmetry. It stays the same, but changes in direction. You can use tracing paper to help you get an idea of how to reflect a shape. Nothing in this section yet. Why not help us get started? ## Related Topics Requires a knowledge of… ## Related Questions • 1 Vote 4 ### What is a coordinate grid? By Ryan McGuire on the 10th of January, 2013 • 1 Vote 3 ### What is the general form for a quadratic equation? By Filsan on the 10th of January, 2013 • 1 Vote 2 ### What does the graph of Y = x Linear look like? By Filsan on the 10th of January, 2013 • 1 Vote 3 ### What is the equation of a straight line graph? By Filsan on the 10th of January, 2013 • 0 4 ### How do I find 15% of a value? By Verity Painter on the 8th of January, 2013 • 1 Vote 2 ### What is an index law? By Filsan on the 6th of December, 2012 • 0 1 ### What does write the terms of a sequence mean? By Filsan on the 6th of December, 2012 • 0 1<\|endoftext\|>	4.5
1,822	# What is 9/156 as a decimal? ## Solution and how to convert 9 / 156 into a decimal 9 / 156 = 0.058 Fraction conversions explained: • 9 divided by 156 • Numerator: 9 • Denominator: 156 • Decimal: 0.058 • Percentage: 0.058% Convert 9/156 to 0.058 decimal form by understanding when to use each form of the number. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. If we need to convert a fraction quickly, let's find out how and when we should. 9 / 156 as a percentage 9 / 156 as a fraction 9 / 156 as a decimal 0.058% - Convert percentages 9 / 156 9 / 156 = 0.058 ## 9/156 is 9 divided by 156 Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 9 divided by 156. To solve the equation, we must divide the numerator (9) by the denominator (156). This is our equation: ### Numerator: 9 • Numerators sit at the top of the fraction, representing the parts of the whole. Comparatively, 9 is a small number meaning you will have less parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Ultimately, having a small value may not make your fraction easier to convert. So how does our denominator stack up? ### Denominator: 156 • Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. 156 is a large number which means you should probably use a calculator. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. Now it's time to learn how to convert 9/156 to a decimal. ## How to convert 9/156 to 0.058 ### Step 1: Set your long division bracket: denominator / numerator $$\require{enclose} 156 \enclose{longdiv}{ 9 }$$ To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well. ### Step 2: Extend your division problem $$\require{enclose} 00. \\ 156 \enclose{longdiv}{ 9.0 }$$ We've hit our first challenge. 9 cannot be divided into 156! So we will have to extend our division problem. Add a decimal point to 9, your numerator, and add an additional zero. Now 156 will be able to divide into 90. ### Step 3: Solve for how many whole groups you can divide 156 into 90 $$\require{enclose} 00.0 \\ 156 \enclose{longdiv}{ 9.0 }$$ We can now pull 0 whole groups from the equation. Multiply this number by 156, the denominator to get the first part of your answer! ### Step 4: Subtract the remainder $$\require{enclose} 00.0 \\ 156 \enclose{longdiv}{ 9.0 } \\ \underline{ 0 \phantom{00} } \\ 90 \phantom{0}$$ If your remainder is zero, that's it! If you have a remainder over 156, go back. Your solution will need a bit of adjustment. If you have a number less than 156, continue! ### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit. Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable. ### Why should you convert between fractions, decimals, and percentages? Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. This is also true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But 9/156 and 0.058 bring clarity and value to numbers in every day life. Without them, we’re stuck rounding and guessing. Here are real life examples: ### When you should convert 9/156 into a decimal Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 9/156 MPH. The radar will read: 90.5 MPH. This simplifies the value. ### When to convert 0.058 to 9/156 as a fraction Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'. ### Practice Decimal Conversion with your Classroom • If 9/156 = 0.058 what would it be as a percentage? • What is 1 + 9/156 in decimal form? • What is 1 - 9/156 in decimal form? • If we switched the numerator and denominator, what would be our new fraction? • What is 0.058 + 1/2? ### Convert more fractions to decimals From 9 Numerator From 156 Denominator What is 9/157 as a decimal? What is 10/156 as a decimal? What is 9/158 as a decimal? What is 11/156 as a decimal? What is 9/159 as a decimal? What is 12/156 as a decimal? What is 9/160 as a decimal? What is 13/156 as a decimal? What is 9/161 as a decimal? What is 14/156 as a decimal? What is 9/162 as a decimal? What is 15/156 as a decimal? What is 9/163 as a decimal? What is 16/156 as a decimal? What is 9/164 as a decimal? What is 17/156 as a decimal? What is 9/165 as a decimal? What is 18/156 as a decimal? What is 9/166 as a decimal? What is 19/156 as a decimal? What is 9/167 as a decimal? What is 20/156 as a decimal? What is 9/168 as a decimal? What is 21/156 as a decimal? What is 9/169 as a decimal? What is 22/156 as a decimal? What is 9/170 as a decimal? What is 23/156 as a decimal? What is 9/171 as a decimal? What is 24/156 as a decimal? What is 9/172 as a decimal? What is 25/156 as a decimal? What is 9/173 as a decimal? What is 26/156 as a decimal? What is 9/174 as a decimal? What is 27/156 as a decimal? What is 9/175 as a decimal? What is 28/156 as a decimal? What is 9/176 as a decimal? What is 29/156 as a decimal? ### Convert similar fractions to percentages From 9 Numerator From 156 Denominator 10/156 as a percentage 9/157 as a percentage 11/156 as a percentage 9/158 as a percentage 12/156 as a percentage 9/159 as a percentage 13/156 as a percentage 9/160 as a percentage 14/156 as a percentage 9/161 as a percentage 15/156 as a percentage 9/162 as a percentage 16/156 as a percentage 9/163 as a percentage 17/156 as a percentage 9/164 as a percentage 18/156 as a percentage 9/165 as a percentage 19/156 as a percentage 9/166 as a percentage<\|endoftext\|>	4.59375
830	Choosing a Type of Resistor: Fixed or Variable Resistors come in two basic flavors: fixed and variable. Both types are commonly used in electronic circuits. Here’s the lowdown on each type and why you would choose one or the other: A fixed resistor supplies a constant, factory-determined resistance. You use it when you want to restrict current to within a certain range or divide voltage in a particular way. Circuits with LEDs use fixed resistors to limit the current, thus protecting the LED from damage. A variable resistor, commonly called a potentiometer (pot for short), allows you to adjust the resistance from virtually zero ohms to a factory-determined maximum value. You use a potentiometer when you want to vary the amount of current or voltage you’re supplying to part of your circuit. A few examples of where you might find potentiometers are light-dimmer switches, volume controls for audio systems, and position sensors, although digital controls have largely replaced potentiometers in consumer electronics. Take a closer look at fixed and variable resistors. The following figure shows the circuit symbols that are commonly used to represent fixed resistors, potentiometers, and another type of variable resistor called a rheostat. The zigzag pattern should remind you that resistors make it more difficult for current to pass through, just as a kink in a hose makes it more difficult for water to pass through. Fixed resistors are designed to have a specific resistance, but the actual resistance of any given resistor may vary (up or down) from its nominal value by some percentage, known as the resistor’s tolerance. Say you choose a that has a 5% tolerance. The actual resistance it provides could be anywhere from (because 5% of 1,000 is 50). You might say that the resistance is give or take 5%. There are two categories of fixed resistors: Standard-precision resistors can vary anywhere from 2% to (gulp) 20% of their nominal values. Markings on the resistor package will tell you just how far off the actual resistance may be (for instance, +/–2%, +/–5%, +/–10%, or +/–20%). You use standard-precision resistors in most hobby projects because (more often than not) you’re using resistors to limit current or divide voltages to within an acceptable range. Resistors with 5% or 10% tolerance are commonly used in electronic circuits. High-precision resistors come within just 1% of their nominal value. You use these in circuits where you need extreme accuracy, as in a precision timing or voltage reference circuit. Fixed resistors often come in a cylindrical package with two leads sticking out so you can connect them to other circuit elements. Feel free to insert fixed resistors either way in your circuits — there’s no left or right, up or down, or to or from when it comes to these little two-terminal devices. Most fixed resistors are color-coded with their nominal value and tolerance, but some resistors have their values stamped right onto the tiny package, along with a bunch of other letters and numbers guaranteed to cause confusion. If you aren’t sure of the value of a specific resistor, pull out your multimeter, set it to measure resistance in ohms, and place its probes across the resistor (either way), as shown. Make sure your resistor is not wired into a circuit when you measure its resistance; otherwise, you won’t get an accurate reading. Circuit designs usually tell you the safe resistor tolerance to use, whether for each individual resistor or for all the resistors in the circuit. Look for a notation in the parts list or as a footnote in the circuit diagram. If the schematic doesn’t state a tolerance, you can assume it’s okay to use standard-tolerance resistors (+/–5% or +/–10%).<\|endoftext\|>	3.859375
2,198	Special Education: Meeting Student Needs in School and Beyond by Caryn Darwin Can you recall your school experience? You probably do not consider it much until it is time for your class reunion. Most people probably recollect a variety of academic, social, and cultural events, which were a blend of positive and negative experiences with a variety of peers that shaped one’s outlook on achievement, friendships, conflict resolution, and societal functioning that is often carried into adulthood. Attending school should prepare students to become productive citizens by teaching content knowledge, social/emotional, and prerequisite career attainment skills by learning to formulate goals, perseverance, and experiencing feelings of accomplishment. What about the students who have deficits in learning, social skills, emotional functioning, or physical capabilities? Some students with disabilities may possess deficits in gathering and comprehending information, whereas others may misunderstand or misinterpret social cues, become overwhelmed by their environments, become anxious or depressed, lack friendships, and possess attention difficulties, which can often result in severe learning deficits, decreased self-esteem, and negative self-concepts. Not to mention those students who have difficult home lives and fail to get their basic needs met, such as adequate food, shelter, safety, and consistency. Currently, IDEA identifies 13 eligibility categories in special education to identify students with disabilities. - Emotional Disturbance - Hearing Impairment - Intellectual Disability - Multiple Disabilities - Orthopedic Impairment - Other Health Impairment - Specific Learning Disability - Speech or Language Impairment - Traumatic Brain Injury - Visual Impairment including Blindness Whether the deficits are mild, moderate, or severe, any difference in learning can be debilitating and lead to self-fulfilling prophecies of failure. In reality, it is the perceived limitation of a disability and the stigma associated with possessing an educational diagnosis which may be more debilitating than the disability itself. For instance, too often a student might comment “I’m having a bad day because I did not take my medicine this morning.” Once students become mature enough to understand that they are treated differently than their peers or they go to “special classes,” often times they feel ostracized and disempowered. It is unrealistic to believe that all the needs of students in special education can be met within general education classrooms. However, depending upon the severity of the disability and the adverse educational impact, how a disability is manifested in the school setting can vary substantially. So how can students identified as disabled in one of the aforementioned areas learn how to function as productive citizens in society without developing self-fulfilling prophecies of failure? What does it truly mean to be “disabled”? And finally, does every student who has a disability demonstrate the same characteristics? Merriam-Webster dictionary defines disabled as incapacitated by illness or injury or physically or mentally impaired in a way that substantially limits activity especially in relation to employment or education. What disability eligibility does not entail is individual student motivation, ambition, and supportive environments with enriched curriculum. Furthermore, special education services are akin to any other remedy, it has to be tailored to meet the student’s needs. How long would a physician be in practice if his remedy for every illness from a headache to a hepatoma was to take two aspirin every four hours? IDEA 2004 defines least restrictive environment as to the maximum extent appropriate, children with disabilities are educated with non-disabled peers. When collaboration and encouragement from professionals such as general and special education teachers, school principals, school counselors, school psychologists, and parents occur and multiple discussions take place about student needs, strengths, and deficits, this increases the maximum benefit appropriate for students within educational and social experiences from their general education peers as well as receive specialized curriculum to remedy learning and emotional gaps that may be present. Further it is equally if not more important to recognize students for their educational capabilities and their contribution as it is to recognize their disabilities. Through communication and advocacy efforts, interventions in the general education setting can work toward addressing the needs of all students. Further, when professionals and parents advocate for the least restrictive environment for students receiving special education services, students can learn from the most authentic source: themselves. As a result, this fosters heterogeneous environments that prepare our youth for diverse educational, employment, and extracurricular activities that mirror our culturally and sociodemographic diverse society, which will give students more tools and better potential to become productive, happy, and thriving citizens with fulfilling lives upon graduation to make meaningful contributions to society. Isn’t that what education is all about? Caryn Darwin, Ed.S. LSSP (Licensed Specialist in School Psychology) Cypress-Fairbanks Independent School District, Houston, TX National Psychologist Trainee Register Scholarship Recipient, Fall 2015 Pre-doctoral Psychology Intern, Rockdale Regional Juvenile Justice Center in Rockdale, TX School Psychology Doctoral Candidate, Tennessee State University, Nashville, TN School Psychologist, Metropolitan Nashville Public Schools, Nashville, TN Explanation of the Eligibility Categories in Special Education - Autism: a developmental disability significantly affecting verbal and nonverbal communication and social interaction, generally evident before age three that adversely affects a child’s educational performance. Other characteristics often associated with autism are engaging in repetitive activities and stereotyped movements, resistance to environmental change or change in daily routines, and unusual responses to sensory experiences. - Deaf-Blindness: concomitant [simultaneous] hearing and visual impairments, the combination of which causes such severe communication and other developmental and educational needs that they cannot be accommodated in special education programs solely for children with deafness or children with blindness. - Deafness: means a hearing impairment so severe that a child is impaired in processing linguistic information through hearing, with or without amplification that adversely affects a child's educational performance. - Emotional Disturbance: a condition exhibiting one or more of the following characteristics over a long period of time and to a marked degree that adversely affects a child’s educational performance: (a) An inability to learn that cannot be explained by intellectual, sensory, or health factors. (b) An inability to build or maintain satisfactory interpersonal relationships with peers and teachers. (c) Inappropriate types of behavior or feelings under normal circumstances. (d) A general pervasive mood of unhappiness or depression. (e) A tendency to develop physical symptoms or fears associated with personal or school problems. The term includes schizophrenia. The term does not apply to children who are socially maladjusted, unless it is determined that they have an emotional disturbance. - Hearing Impairment: means an impairment in hearing, whether permanent or fluctuating, that adversely affects a child’s educational performance but is not included under the definition of “deafness.” - Intellectual Disability: significantly sub average general intellectual functioning, existing concurrently with deficits in adaptive behavior and manifested during the developmental period that adversely affects a child’s educational performance. - Multiple Disabilities: concomitant impairments (such as intellectual disability-blindness, intellectual disability-orthopedic impairment), the combination of which causes such severe educational needs that they cannot be accommodated in special education programs solely for one of the impairments. The term does not include deaf-blindness. - Orthopedic Impairment: a severe orthopedic impairment that adversely affects a child’s educational performance. The term includes impairments caused by a congenital anomaly, impairments caused by disease (e.g., poliomyelitis, bone tuberculosis), and impairments from other causes (e.g., cerebral palsy, amputations, and fractures or burns that cause contractures). - Other Health Impairment: having limited strength, vitality, or alertness, including a heightened alertness to environmental stimuli, that results in limited alertness with respect to the educational environment, that— (a) is due to chronic or acute health problems such as asthma, attention deficit disorder or attention deficit hyperactivity disorder, diabetes, epilepsy, a heart condition, hemophilia, lead poisoning, leukemia, nephritis, rheumatic fever, sickle cell anemia, and Tourette syndrome; and (b) adversely affects a child’s educational performance. - Specific Learning Disability: a disorder in one or more of the basic psychological processes involved in understanding or in using language, spoken or written, that may manifest itself in the imperfect ability to listen, think, speak, read, write, spell, or to do mathematical calculations. The term includes such conditions as perceptual disabilities, brain injury minimal brain dysfunction, dyslexia, and developmental aphasia. The term does not include learning problems that are primarily the result of visual, hearing, or motor disabilities; of intellectual disability; of emotional disturbance; or of environmental, cultural, or economic disadvantage. - Speech or Language Impairment: a communication disorder such as stuttering, impaired articulation, a language impairment, or a voice impairment that adversely affects a child’s educational performance. - Traumatic Brain Injury: an acquired injury to the brain caused by an external physical force, resulting in total or partial functional disability or psychosocial impairment, or both, that adversely affects a child's educational performance. The term applies to open or closed head injuries resulting in impairments in one or more areas, such as cognition; language; memory; attention; reasoning; abstract thinking; judgment; problem solving; sensory, perceptual, and motor abilities; psychosocial behavior; physical functions; information processing; and speech. The term does not apply to brain injuries that are congenital or degenerative, or to brain injuries induced by birth trauma. - Visual Impairment including Blindness: an impairment in vision that, even with correction, adversely affects a child’s educational performance. The term includes both partial sight and blindness. Merriam-Webster, (2015). Merriam-Webster Collegiate Dictionary (11th Eds.). Springfield, MA. Retrieved from http://www.merriam-webster.com/dictionary/disabled National Dissemination Center for Children with Disabilities, (2012). Categories of disability under IDEA. Retrieved from http://www.parentcenterhub.org/wp-content/uploads/repo_items/gr3.pdf US Department of Education, (2015). Building the legacy IDEA 2004. Retrieved from http://idea.ed.gov/explore/view/p/%2Croot%2Cstatute%2CI%2CB%2C612%2Ca%2C5%2C<\|endoftext\|>	3.890625
378	# Factorize the expression $x^4-(2y-3z)^2$. Given: The given algebraic expression is $x^4-(2y-3z)^2$. To do: We have to factorize the expression $x^4-(2y-3z)^2$. Solution: Factorizing algebraic expressions: Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors. $x^4-(2y-3z)^2$ can be written as, $x^4-(2y-3z)^2=(x^2)^2-(2y-3z)^2$ [Since $x^4=(x^2)^2$] Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $x^4-(2y-3z)^2=(x^2)^2-(2y-3z)^2$ $x^4-(2y-3z)^2=[x^2+(2y-3z)][x^2-(2y-3z)]$ $x^4-(2y-3z)^2=(x^2+2y-3z)(x^2-2y+3z)$ Hence, the given expression can be factorized as $(x^2+2y-3z)(x^2-2y+3z)$. Updated on: 09-Apr-2023 70 Views ##### Kickstart Your Career Get certified by completing the course<\|endoftext\|>	4.46875
464	# Chapter 3 - Derivatives - 3.1 Introducing the Derivative - 3.1 Execises - Page 133: 35 $a.$ The value of the derivative is $f'(5) = -\frac{1}{100}$. $b.$ The equation is $y=-\frac{1}{100}x+\frac{3}{20}$ #### Work Step by Step $a.$ By definition of a derivative at point $a=5$ we have $$f'(5)=\lim_{h\to0}\frac{f(5+h)-f(5)}{h}=\lim_{h\to0}\frac{\frac{1}{(5+h)+5}-\frac{1}{5+5}}{h}=\lim_{h\to0}\frac{\frac{1}{10+h}-\frac{1}{10}}{h}=\lim_{h\to0}\frac{\frac{10-(10+h)}{10(10+h)}}{h}=\lim_{h\to0}\frac{-h}{10h(10+h)}=\lim_{h\to0}\frac{-1}{10(10+h)}=\frac{-1}{10(10+0)}=-\frac{1}{100}.$$ $b.$ We know that the equation of the tangent at point $(a,f(a))$ is given by $y-f(a)=f'(a)(x-a)$. Using $a=5$, $f(a)=1/(5+5)=1/10$ and $f'(a)=-1/100$ we get $$y-\frac{1}{10}=-\frac{1}{100}(x-5)\Rightarrow y-\frac{1}{10}=-\frac{1}{100}x+\frac{1}{20}$$ which gives $$y=-\frac{1}{100}x+\frac{1}{20}+\frac{1}{10}=-\frac{1}{100}x+\frac{3}{20}.$$ After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.<\|endoftext\|>	4.6875
913	{[ promptMessage ]} Bookmark it {[ promptMessage ]} # HW3 - GROUP WORK I SECTION 3.3 Doing a Lot with a Little d... This preview shows pages 1–2. Sign up to view the full content. This preview has intentionally blurred sections. Sign up to view the full version. View Full Document This is the end of the preview. Sign up to access the rest of the document. Unformatted text preview: GROUP WORK I, SECTION 3.3 Doing a Lot with a Little . . d . ‘ . Sect1on 3.3 1ntroduces the Power Rule: E—x" = nx"“‘, where n 15 any real number. The good news IS that x . this rule, combined with the Constant Multiple and Sum Rules, allows us to take the derivative of even the most formidable polynomial with ease! To demonstrate this power, try Problem 1: I. A formidable polynomial: f(x) = x10 + gx9 + %x8 — 5x7 — 0.33x6 + 7rx5 — ﬁx“ — 42 Its derivative: f’(x)= The ability to differentiate polynomials is only one of the things we’ve gained by establishing the Power Rule. Using some basic deﬁnitions, and a touch of algebra, there are all kinds of functions that can be diﬁerentiated using the Power Rule. 2. All kinds of functions: 5 _ 1 1 _ x5 — 3,5 + 2 f<x>=€/§+f2 goo—Fax, 1200— ﬂ Their derivatives: f/ (x) = g’ (x) = h’ (x) = Unfortunately, there are some deceptive functions that look like they should be straightforward applications of the Power and Constant Multiple Rules, but actually require a little thought. 3. Some deceptive functions: f(x) = (2x)4 gov) = (x3)5 Their derivatives: f/(x)= g’(x)= Doing a Lot with a Little The process you used to take the derivative of the functions in Problem 3 can be generalized. In the ﬁrst case, f (x) = (2x)4, we had a function that was of the form (kx)”, where k and n were constants (k = 2 and n = 4). In the second case, g (x) = (x3)5, we had a function of the form (xk)". Now we are going to ﬁnd a pattern, similar to the Power Rule, that will allow us to ﬁnd the derivatives of these functions as well. 4. Show that your answers to Problem 3 can also be written in this form: f’(x) = 4(2x>3 - 2 g/(x) = 5 (x3)4 . 3x2 And now it is time to generalize the Power Rule. Consider the two general functions, and try to ﬁnd expres- sions for the derivatives similar in form to those given in Problem 4. You may assume that n is an integer. 5. Two general ﬁmctions: f(x) = (W g(x) = M)” Their derivatives: f I (x) g” (x) ... View Full Document {[ snackBarMessage ]} ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern<\|endoftext\|>	4.5625
171	Scientists have created a sieve capable of removing salt from seawater using the “wonder material” graphene. Researchers at the University of Manchester developed a graphene membrane to desalinate water and make it drinkable, offering the promise of easy and accessible potable water for millions of people around the world. A study published in the journal Nature Nanotechnology describes how the filtration system works by precisely controling the membrane’s pore size to sieve common salts out of salty water. Phi Beta Iota: Free energy may finally be about to be released from secret patent control. That will be a lot less expensive that refined graphene as a means of distilling unlimited quantities of clean water while also cleansing the Ganges, among other important bodies of water, at scale. Still, this is a positive development.<\|endoftext\|>	3.734375
397	How do you get a submarine to rise from the bottom of the ocean to the surface again? Do you press a button? Flip a switch? Use a lever? How do you initiate the surfacing process? closed as off-topic by BillDOe, Camilo Rada, BHF, Gimelist, bon Apr 14 '18 at 9:12 This question appears to be off-topic. The users who voted to close gave this specific reason: - "This question does not appear to be about earth science, within the scope defined in the help center." – BillDOe, Camilo Rada, BHF, Gimelist, bon Interesting question but more an engineering one. The basic physics is that a submarine modifies its buoyancy by filling various chambers (dedicated for this purpose) with water or pumping the water out and filling the chambers with air. Understanding buoyancy is important: basically, if something weighs less than the water it displaces, it floats; if something weighs more than the water it displaces, it sinks. As more water is allowed to enter the submarine's chambers through the opening of valves, the submarine becomes heavier (or another way to look at it is, the submarine is decreasing its displacement of water by allowing the water to fill its volume) and therefore its buoyancy decreases and it can be caused to sink. When air is pumped into the chambers and water is pumped and then sealed out of the vessel, the submarine becomes lighter and has a larger displacement, so its buoyancy increases and it rises toward the surface. Pressure differentials play a big role in all this. Getting water back into the chambers, getting water out of the chambers and filling them with air, maintaining healthy pressure for the submarine's compartments that house humans - all this involves careful calculation of pressures in different parts of the system and how we can manage and change those pressures to get the submarine to function as desired.<\|endoftext\|>	4.125
2,080	December 11, 2012 – Today children have an increased possibility of getting sick from a varied class of illnesses. Sometimes it is difficult to know how to take care of your child properly and have a sense of security and confidence that you have made the right decision. Here is a list of the most prevalent illnesses with definitions, signs, and symptoms that may lead you in the right direction. The common cold is a viral infection of the upper respiratory tract, nose, and throat. The common cold is usually harmless, although it may not feel that way to a child. There are more than 100 viruses that can cause a common cold therefore signs and symptoms tend to vary. Preschool children are at greatest risk of frequent colds, but even healthy adults can expect to have a few colds each year. Symptoms of a common cold usually appear about one to three days after exposure to a cold-causing virus. Most people recover from a common cold in about a week or two. If symptoms don't improve, see your doctor. Common Cold signs and symptoms may include: runny or stuffy nose, itchy or sore throat, cough, congestion, slight body aches, a mild headache, sneezing, watery eyes, low-grade fever, and mild fatigue. A fever is usually a sign that something out of the ordinary is going on in the body. For an adult, a fever may be uncomfortable, but fever usually isn't dangerous unless it reaches 103 F (39.4 C) or higher. For very young children and infants, a slightly elevated temperature may indicate a serious infection but the degree of fever doesn't necessarily indicate the seriousness of the underlying condition. A minor illness may cause a high fever, and a more serious illness may cause a low fever. Usually a fever goes away within a few days. A number of over-the-counter medications lower a fever, but sometimes it's better left untreated. Fever seems to play a key role in helping your body fight off a number of infections. To be on the safe side consult with your family physician. Fever signs and symptoms may include: sweating, shivering, headache, muscle aches, loss of appetite, dehydration, general weakness. High fevers between 103 F (39.4 C) and 106 F (41.1 C) may cause: hallucinations, confusion, irritability, convulsions, and dehydration. Influenzais a viral infection that attacks the respiratory system — nose, throat and lungs. Influenza, commonly called the flu, is not the same as the stomach "flu" viruses that cause diarrhea and vomiting. Influenza and its complications can be deadly. People at higher risk of developing flu complications include young children, older adults, pregnant women, people with weakened immune systems, and people who have chronic illnesses. The best defence against influenza is to receive an annual vaccination. Dinarello CA, et al. Fever and hyperthermia. In: Fauci AS, et al. Harrison's Online. 17th ed. New York, N.Y.: The McGraw-Hill Companies; 2008. http://www.accessmedicine.com/content.aspx?aid=2871330. Accessed April 28, 2011. 1.Fever. The Merck Manuals: The Merck Manual for Healthcare Professionals. http://www.merckmanuals.com/professional/sec14/ch167/ch167e.html. Accessed April 28, 2011. 2.Torpy J. Fever in infants. Journal of the American Medical Association. 2004;291:1284. 3.Fever, sweats and hot flashes. National Cancer Institute. http://www.cancer.gov/cancertopics/pdq/supportivecare/fever/healthprofessional. Accessed April 28, 2011. 4.Febrile seizures fact sheet. National Institute of Neurological Disorders and Stroke. http://www.ninds.nih.gov/disorders/febrile_seizures/detail_febrile_seizures.htm. Accessed April 28, 2011. 5.Ward MA. Pathophysiology and treatment of fever in infants and children. http://www.uptodate.com/home/index.html. Accessed April 25, 2011. 6.O'Grady N, et al. Guidelines for evaluation of new fever in critically ill adult patients: 2008 update from the American College of Critical Care Medicine and the Infectious Diseases Society of America. Critical Care Medicine, 2008;36:1330. 7.Sullivan JE, et al. Clinical report — Fever and antipyretic use in children. Pediatrics. 2011;127:580. 8.Leggett J. Approach to fever or suspected infection in the normal host. In: Goldman L, et al. Cecil Medicine. 23rd ed. Philadelphia, Pa.: Saunders Elsevier; 2008. http://www.mdconsult.com/books/linkTo?type=bookPage&;isbn=978-1-4160-2805-5&eid=4-u1.0-B978-1-4160-2805-5..50307-4. Accessed May 2, 2011. 9.Bor DH. Approach to the adult with fever of unknown origin. http://www.uptodate.com/home/index.html. Accessed April 25, 2011. Initially, the flu may seem like a common cold with a runny nose, sneezing, and sore throat but colds usually develop slowly, whereas the flu tends to come on suddenly. Although a cold can be a nuisance, you usually feel much worse with the flu. Influenza signs and symptoms may include: fever over 100 F (38 C), aching muscles (especially in your back, arms and legs), chills and sweats, headache, dry cough, fatigue, weakness and nasal congestion. Bronchitis is the inflammation of the lining of the bronchial tubes, which carry air to and from your lungs. Bronchitis may be either acute or chronic. Often developing from a cold or other respiratory infection, acute bronchitis is very common and usually improves within a few days without lasting effects, although a cough may continue for weeks after the bronchitis resolves. Bronchitis signs and symptoms may include: cough, production of mucus (sputum that is clear, white, yellow, or green), fatigue, slight fever and chills, and chest discomfort. Pneumonia is the inflammation of the lungs caused by infection such as bacteria, viruses, fungi, or parasites. Pneumonia can range in seriousness from mild to life threatening and is often a complication of another condition, such as the flu beginning with a cough and a fever. Antibiotics can treat most common forms of bacterial pneumonias, but antibiotic-resistant strains are a growing problem. The World Health Organization (WHO) strongly advocates the routine vaccination to prevent infection. Infants and young children under the age of two are especially at higher risk to get pneumonia due to the following facts: - They attend daycare centres - They have older siblings in daycare centres - They have had a recent ear infection or a recent course of antibiotics - They have an underlying illness or a compromised immune system Pneumonia signs and symptoms may include: coughing, wheezing, fever, rapid and laboured breathing, sweating, shaking chills, chest pain that fluctuates with breathing (pleurisy), headache, muscle pain, and fatigue. Depending on the type and severity of infection, infants may also suffer convulsions, hypothermia, lethargy and feeding problems. Facts about Pneumonia - Over 2 million children die from pneumonia each year globally. - Pneumonia is the leading cause of death among children under the age of 5. - Pneumonia kills more children than any other illness including AIDS, malaria, and measles combined. Protect Your Children Teach your children to cover their nose and mouth with a tissue when they cough or sneeze. The tissue must then be discarded in a bin and hands should be thoroughly washed to help prevent the spread of germs. Teaching your children to wash their hands is the single most effective way to stop the spread of contagious illnesses. Make sure children wash their hands before: - Eating their meals and snacks - Handling food or helping in the kitchen Make sure children wash their hands after: - Playing with toys that were shared with other children - Contact with other people who are sick - Using the washroom facilities - Blowing their nose - Coughing or sneezing - Handling garbage - Visiting a public place - Petting an animal The method to wash hands properly Teach your children to sing a song like "Twinkle Twinkle Little Star" or "Frère Jacque" while they are washing their hands. When the song is over, it's time to rinse their hands. - Wet your hands with warm running water. - Put a small amount of liquid soap in the palm of one hand. (Bar soaps are not as hygienic because they attract germs). - Rub hands together for at least 15 seconds, bringing the soap to a lather. - Rub each palm over the back of the other hand. - Scrub between fingers, under fingernails and on the back of your hands. - Rub around each of your thumbs. - Rinse hands well with running water for at least 10 seconds. - Dry your hands with a clean or disposable towel. Ways You Can Fight the Flu - Getting the annual flu shot between October and December is the most effective way to prevent the flu. - Keep kids at home and away from crowds when they are sick. - Keep common surfaces in your home clean and sanitized. - Do not share personal items, food, and drinks with people that are sick.<\|endoftext\|>	3.796875
4,614	This Double lesson plan has been created specifically for the teachers delivering the 10th grade ‘Keep Talking’ programme in Israeli high schools. I suggest using the lesson in one of the first lessons of the year in order to get the students thinking about their oral proficiency, and to make them aware of the different techniques and strategies available for effective learning. I have included the lesson plan and the PowerPoint for you to use with your classes. The link to the student Self-Assessment Google Form is in Task 3 in the lesson plan itself. I hope you and your students enjoy the lesson as much as I did. I’d love to hear from you in the comments box below. Bullying is brutal. It always has been, and probably always will be an issue that too many children and teenagers have to deal with, both at school and now, due to cyber-bullying, at any time and in any place. I created this lesson after seeing a short JuBafilm – A Piece of Chalk – whilst invigilating a 12th grade Cinema Studies class. The speaking booklet, that I have been using with my 10th grade ‘Keep Talking’ class, has a whole unit devoted to the topic of bullying, so I simply adapted that unit to incorporate ideas that were raised by the film. This 4 lesson unit plan includes a mix of speaking activities: two oral presentations, and another short film – Listen to Me – which deals with the concepts of stereotypes and bullying. The lesson plans are aimed at students from both intermediate and proficiency levels (CEFR levels B1 – C1). I have provided suggested times, however, the lessons may need more or less time depending on the class. As I continue to work with my 10th grade students in the new ‘Keep Talking’ programme, I have created two lesson plans to supplement the approved speaking booklets. The lessons may be used to supplement the travel units, in each of the books, or as a standalone ‘speaking’ unit. The lesson plans are aimed at students from both intermediate and proficiency levels (CEFR levels A2/B1 – C1). I have provided suggested times, however, the lessons may need more or less time depending on the class. My overarching objective in this course is to get the students speaking in pairs and small groups, and to make sure they both increase their lexis and their confidence, in a fun and engaging learning environment. Speaking is the most challenging of the four skills to teach in large heterogeneous classes. As speaking is interactive and demands an almost instant response, the pressure to ‘perform’ is often overwhelming for students. With this in mind I created a fun lesson plan based on the famous song ‘It’s Friday I’m in Love’ by The Cure. Not one of my 10th grade students knew the song but they all quickly ‘fell in love’ with it. My aim was to get the students speaking and using the lexical chunks from the song. However, to my amazement they quickly began singing the song, and asked me to play it again and again. Our students are constantly asking us: “Will it be on the exam?” “Is there a grade for this?” When the answer is no, the next question is often: What’s the point? “So why bother? With this in mind we ran an informal evening event at the ETAI national summer conference in Ashkelon with our panel of experts: Denise Ross Hayne, Penny Ur, Batia Laufer, Amos Paran, Ben Goldstein . Our goal was to pose questions, sourced from the audience via Todaysmeet, for our expert panel, who were asked to give us good reasons for why we should still bother being ‘creative and demanding ELT teachers’ in an age of ‘bottle flipping, finger spinners and Google Translate.’ As the convener of the Q & A session I would like to share with you some of my post-event reflections. Firstly, there was no need for the panel to prepare anything in advance, which enabled them to communicate directly with the audience, and to answer questions spontaneously, on their area of expertise, without investing further time in preparation in contrast to a Pecha Kucha evening (see Pecha Kucha and the Power of (saying) ‘Yes’). Secondly, we decided to use Todaysmeet to source questions from the audience, because it is user friendly and has a good visual layout, and meant we did not need a person running around the huge auditorium with a microphone. Lastly, as the panel members were all experienced conference presenters they understood that the aim of the evening event is to keep things light, fast-paced and informative. Some tips for those of you who might want to use this format: Set up the Todaysmeet room in advance, with a demonstration question, for example: ‘Why bother coming to ETAI when you could go to the beach instead?’JaneCohenEFL Create a slide with instructions and a URL address to source great questions, and enable audience participation, as soon as they enter the auditorium. As the convener, introduce your panel and then go straight to audience questions, otherwise you might expect some feedback like this:“Why bother asking us to write questions if you’re not gonna use them? Anonymous. Or: “Why bother asking us for questions when you’re using yours?” Anon In order to keep the Q & A session fast paced, use a timer, and tell the plenary speakers that they have 2 or 3 minutes maximum to answer a given question. Note, I didn’t do this but would do next time. If you want to remember any of the panel’s answers record the event, as it is really difficult to host and remember what was said. Again, I didn’t do this but definitely will next time as I missed out on so much personal learning. Here is a sample of some of the questions sourced from the audience, and answered by the panel. Why bother giving our students homework when we know they won’t do it anyway? Why bother telling my friends how good ETAI conferences are when they never come? Why bother trying to build up the school English library when kids don’t read books anymore? Why bother teaching a 45 minute lesson when students can’t stay focused for that long? Why bother travelling when we can meet online? Why bother teaching Shakespeare when no one speaks that way anymore? Why bother looking at research on ESL in the US Why bother trying to get pupils to read books, they’ll never read enough books to really improve their English. Why bother teaching English when they plagiarize and use Google translate and don’t understand what’s wrong with it? Why bother correcting them on present perfect errors when there are more people in the world who use it incorrectly than those who do? Why bother teaching vocabulary if the students can use electronic dictionaries? Why bother going to IATEFL conferences abroad? Why bother spending so much time with grammar when the goal is communication? Why bother teaching them literary terms? Why not just deal with the message and the useful vocabulary? Why bother using grammar books with gap-fill activities? Why bother with spelling tests when our pupils will write e mails and use electronic notebooks in their future? Why bother giving written feedback on student drafts when they do not bother correcting their work accordingly? Why bother making kids read books when they don’t even read them in their own language? After seven years in ELT management I decided that I need to live the life that I want to live and not just continue doing what I had been doing for the last seven years, just because of the ‘conditions and status’, or because that is what everybody expected me to do. My resignation from the Open University surprised everybody, but for me it meant going back to teaching, to learning and to working directly with teenagers. On 30 August I rolled up to the pre-teaching in-service day at my new school feeling both excited and nervous. Those feelings were personified two days later when I stood at the door of my first class, waiting for the students to stand. How would they perceive me? Would they behave? Would they understand me? Would they participate in the speaking tasks? How was I going to remember all of their names? I was full of doubts, but as soon as I stepped into the room, greeted the class and started off with a ‘Getting to know you’ icebreaker, those initial doubts evaporated. I left the room feeling energized and excited to be back in my own classroom, after a seven year gap. Since then I have had a few great lessons, many ordinary lessons and some less than good lessons. Following each lesson I reflect on what went well, what could I have done differently, and did I actually meet the learning aims that I had set? I often think to myself, if I were observing this class I might have asked the teacher why she corrected that particular mistake and interrupted the student’s fluency, or, why didn’t she scaffold the task better, or had she noticed that boy in the back row, who was on his phone under the desk, during most of the speaking task. I used to think that if I am the best teacher I can be and plan my lessons really well, the students will respond accordingly. But I now think that this isn’t always enough. I used to think that if I integrate technology effectively to enhance the learning outcomes, the students will be motivated and engaged. But I now think, that sometimes this is true, but quite often, it is not. I used to think that through my teacher training I could have an impact on so many more students than I could in the classroom myself. But I now think that there is nothing more satisfying than engaging with the students themselves, in a large classroom, with all its challenges, and seeing everybody engaged and on task. Hosting the Pecha Kucha evening at the 7th International ETAI conference, 4-6 July 2016, provided me with the opportunity to source the international and local presenters, send them guidelines with a delivery deadline, and then review their presentations, to check they had met the criteria and had the automatic timings set correctly. The presenters in order of their presentations were: When initially approached, some of the presenters immediately gave me an affirmative answer, “Yes, sure” or “Ok”, whilst others were more hesitant. One presenter wrote to me saying, “If truth be told, I’d forgotten that I’d allowed myself to be talked into doing a Pecha Kucha!!! I’d been thinking about chickening out, but …. hey why not! Another presenter stated, “At first I groaned – And then I thought about what I could do …. So, end of moan. I am happy to do something. Am I mad? Yes, I am.” The latter responses were similar to my own when Leo Selivan (Leoxicon) asked me to host the event. “I don’t think I can. I work full time. I’m studying etc.” I responded and then I stopped myself and thought – If Leo is asking me then he must believe I can do it, and in the words of Richard Branson, “If someone offers you an amazing opportunity and you are not sure you can do it, say yes. Then learn how to do it later.” This year’s team of courageous ELT presenters demonstrated that they also subscribe to the Branson philosophy, and as a result each of the presenters put themselves out there, and we, the audience, benefited from their experience and humour, and had a great time. Giving a Pecha Kucha is different from being a plenary or keynote speaker, it seems to fall much more into the ‘edutainment’spere, and the pressure on the presenter to ‘perform’ is not insignificant. For the host, though, once the presenters have agreed to present and their presentations have been received and checked, all that is left to do is to choose the order of the presentations, upload them onto the computer in the auditorium, check the timings, and write some introductory notes about each speaker. There shouldn’t be any surprises. However, on Tuesday 5 July, a few hours before the Pecha Kucha evening was due to start, I bumped into Mel Rosenberg and Andy Curtis, who told me that they had had an idea that they wanted to run by me. “Andy is going to do my Pecha Kucha, sight unseen. What do you think?” I looked at Andy and asked him, “Do you know what Mel’s Pecha Kucha is about?” “No, not a clue.” Andy responded. “Though it would be a great example of creativity, if I presented it without seeing it, don’t you think? Do we have your permission to do this crazy thing?” I thought to myself, Mel’s presentation is not clear to me, and I’ve seen the slides, so how is Andy going to present it? But then I thought, this could be an opportunity to do something different from the ‘traditional’ Pecha Kucha format. So I said, “Yes. I like the idea.” Andy looked a little surprised, as he hadn’t expected me to agree quite so quickly. And thus the first Pecha Kucha ‘Unseen Hack’ was born. Video courtesy of @MelRosenberg As teachers we are always putting our students on the spot in front of their peers, asking them questions, getting them to do presentations, prepare speeches and debates and complete numerous other language tasks, that many of them don’t feel comfortable doing. Our students usually have no choice but to say ‘yes’, as the task often forms part of their summative assessment. As teachers/ELT professionals we must be role models for our students, and also be willing to put ourselves ‘out there’ in front of our peers, even when we may feel uncomfortable about the request, because saying ‘yes’, can be both challenging and rewarding. In fact, Emily Liscom (Education to the Core) would go even further, and say that by using the word ‘Yes’, to our students more than the word ‘No’, we might be surprised to experience improved classroom management and teaching strategies. Thank you to each of the six/seven presenters for saying ‘yes’, when I approached you – each of you were courageous and inspirational, and are great role models to other ELT professionals and students across the globe. “What is a book? According to the Merriam Webster dictionary it is “a set of printed sheets of paper that are held together inside a cover.” At this year’s ETAI winter event more than 50 English teachers were taken on a book discovery tour through the impressive National Library of Israel, in Jerusalem. The 120 year old library has a collection of more than 5,000,000 books, 2000 manuscripts, 700 personal archives and 30,000 hours of recordings which are available to the public, at no cost. As a consequence of my experience I would like to share with you ’10 things I now know about the National Library of Israel’: The map room houses the most significant Holy Land maps’ collection in the world The Ardon Windows (pictured) represent Isaiah’s vision of eternal peace The oldest book in the museum is a Koran, dating back to the ninth century Israel’s ‘Book Law’ requires two copies of all printed matter published in Israel to be deposited in the national library The museum is divided into 4 major collections: Judaica, Israel, Islam & the Middle East and the Humanities Gershom Scholem loved to write notes in the margins of his books, which can be seen in the Gershom Scholem Library (comprising 35,000 items related to the Kabbalah, Jewish Mysticism and Hassidism) ‘Ephemeral’ means transient or short-lived TheTime Travel and EuropeanEphemeralcollections are made up of letters, tickets, posters, postcards etc., and provide a rich resource of life and culture that can be used for engaging our students in the English classroom The library has an educational partnership with the UK, available via an online site, and includes lesson plans and worksheets for use in British classrooms, which could be relevant to our English language classrooms in Israel The National Library has a resource rich Facebook page in English which is regularly updated, and provides authentic materials for English teaching. So why should English teachers teach with Primary Resources? Karen Ettinger, Project Manager for Education at the NLI, explained that primary resources are motivating, relevant, make use of authentic material, enable students to practice 21 century skills, exercise their critical thinking and research skills, whilst connecting them with their past. So if you want to do some, or all of the above I strongly recommend a trip, either physical or virtual, to the National Library of Israel. Thank you to all of the National Library staff who took us on a journey which made me think differently about the role of the library in the English language classroom today. I recently signed up for some college classes to get myself out of the office and back in the classroom as a student. Due to my heavy work load I decided to take it easy and to start with just one class this semester. I also realised that I needed a F2F class, as I was missing the human interaction of a classroom. I have taken numerous online courses over the last few years so knew what both settings have to offer. Introduction to Learning Disabilities, Thursday evenings, 18:00-19:30, Levinsky College: The lecturer, Ahrona Korman Gvaryahu, filled the class with her energy, passion and ability to keep the students’ interest despite the late hour. Ahrona made it clear that the course would be focused on the LD students and on their parents, and not on the teachers. Ahrona didn’t use a PowerPoint, she talked and showed us two YouTube videos. We were given a pre-viewing task and were simply encouraged to actively watch. The inspiring Rita Pierson, bowled me away with her insights and understanding of learning and teaching. “Every kid needs a champion”, is a must watch for all educators. The underlying message of this first class was that the human connection that we make with our students is the critical component for successful teaching. “The teacher is the glue.” I couldn’t agree more and can’t wait for next week’s class. It isn’t often that you get an opportunity to be present at the start of something new, something that has the power to change the way we think about teacher training, about the proficiency of non-native ELT teachers, and about the role and impact of research upon English teachers in the classroom. Today, at the ETAI pre-conference Teacher Training and Development, inaugural Special Interest Group, I was privileged to witness the start of a movement for change. Dr. Lindsey Shapiro Steinberg, opened the day with questions regarding the recruiting of talent, and whether need necessitates compromise. What is a good ELT practitioner? What level of proficiency is required by English teachers? What is learning, and how is learning assessed? Following Dr. Shapiro Steinberg’s opening Dr. Debbie Lifshitz spoke about ‘Shaking Up the Israeli Conventions of Teacher Training.’ With statistics to demonstrate the challenges faced by Non-native English speaking teacher (NESTS) trainees, regarding proficiency at entry and exit of teacher training programmes, and the challenges that lay ahead. Dr. Lifshitz suggested that proficiency levels of NESTS are critical for teacher retention in the schools, in a system where teachers are aging, and more than 40% of newly qualified English teachers never even enter the school system upon graduation. Following the morning presentation participants divided into 3 discussion groups, Proficiency, Methodology and Linguistics, and discussed changes that could be taken by each of these areas, to positively impact upon the proficiency of future NESTS . (Watch the ETAI website for a summary of each group’s suggestions.) The afternoon session was expertly led by Professor Penny Ur who discussed ‘Research and the language teacher.’ Professor Ur asserted that “Research is not the main source of teacher knowledge, but it can enrich it.” She stated that it contributes to teaching in three ways by: Producing evidence, that can be used to create practical principles for teaching Providing new insights / information that would not have occurred to teachers otherwise Contradicting inaccuracies in methodology or firmly held theoretical beliefs Professor Ur provided numerous examples of why research is regarded so highly by the academia and ministries of education, and yet is often seen as trivial, irrelevant or impractical by teachers in the field. The sheer quantity of literature is overwhelming, and therefore needs to be read selectively and critically. Professor Ur suggested that if we want preset and inset teachers to read research there is a need for ‘mediators’, chiefly teacher trainers, who can mediate the research on their behalf. The day closed with an open discussion led by Professor Penny Ur and a thirst for more discussion and dare I say, action. “Professional Development takes place through professional conversation.” Garton and Richards (2011) Today was truly a day of Professional Development at the inaugural Teacher Training & Development SIG.<\|endoftext\|>	3.96875
578	Welcome to Speech Therapy at Home: R Home Articulation Practices. This is your home page for all things R. Please bookmark or add to your favorites so you can refer back to it as your child progress through each level. R...the dreaded R! Many children have difficulty saying R and it happens to be very tricky to teach as well. Even speech therapists struggle with this sound. However, don't let this discourage you! I will walk you through how to say and teach R. R is a relatively later developing sound, usually acquired by 8 years of age. Some children master this sound earlier; but it is normal to make errors up until 8 years of age. Below you will find: In order to teach R to your child effectively, you must know how to say R yourself. It is important to understand the mechanics behind making the sound. Once you have the language to describe how to say R, you will be better able to teach your child. I will teach you tactile, verbal, and visual cues as well. This section is a MUST READ! There are many ways to teach R to a child. First, I will explain tips on how to teach correct tongue and lip positioning. Next, I will explain the elicitation techniques for which I have had the most success during my years of practice. Your child will most likely respond better to some techniques. Feel free to use the ones that help your child the most. Click on Teach R now! Congratulations! Your child can say R. Now, you are ready to work on saying R correctly in words. At this level, it is important to practice saying the sound in the beginning (initial), middle (medial), and end (final) position. Your child's mouth has to coordinate and move muscles differently depending on where the sound falls within a word. This stage may last longer than the previous and that is to be expected. However, don’t get stuck here if your child misses a word here or there. 100% accuracy is rare and not necessary! Once your child is able to say R correctly in words (drill practice) and during flashcards games, you are ready to move on to the functional games listed here. This is the most exciting phase since this is where generalization really starts to happen. Here, your child will learn how to say R correctly in connected speech. In order to encourage carryover of progress, you will be throwing away all speech cards and practicing saying R in sentences while playing natural games! Visit R Sentence Level Games to get started Below you will find a copy of all the materials given throughout the R section. Refer back to this section for quick printing! If you would like to cancel your subscription, you may do so at any time. Click the button below.<\|endoftext\|>	3.984375
1,380	The primary cosmology mission of DESI is to study the nature of dark energy: How does its energy density evolve in time, and how does it affect the clustering of matter? To do this, DESI will use its maps to measure two cosmological effects: baryon acoustic oscillations and redshift-space distortions. The same maps will also provide other opportunities to study cosmology and the physics of galaxies, quasars, and intergalactic gas. Baryon Acoustic Oscillations DESI is optimized to measure a subtle imprint in the maps that is all that remains of an important physical process in the early universe known as the baryon acoustic oscillations (BAO). To explain this effect, we must first account for the cosmic microwave background radiation – the heat left over from the Big Bang – that fills the universe. Today the temperature is only 2.7 degrees above absolute zero, but at earlier times it was much hotter. Prior to 400,000 years after the Big Bang, when the universe was over a billion times denser than it is today, the background radiation was hot enough to cause the hydrogen and helium to be ionized, separating the electrons from atomic nuclei and forming a plasma. The electrons scattered easily with the photons of the background and the photons provided a very strong radiation pressure that resisted compression of the gas. Density perturbations created in the first fraction of a second of the universe thereby traveled as sound waves, or baryon acoustic oscillations, in this plasma. The sound waves traveled through the universe for 400,000 years, coming to a halt when the expanding universe cooled enough that the electrons and nuclei combined into neutral atoms. At this point, the gravitational forces of the density perturbations dominated and started to shape the universe into the large-scale structure that we see today. But the effect of the sound waves can still be measured: Around each overdense region there is a faint spherical imprint of the wave that left that region. This has created a slight tendency for pairs of galaxies to be separated by the distance the waves traveled, which owing to the expanding Universe is about 500 million light-years today. This faint relic imprint has been observed in several data sets, most notably from the Sloan Digital Sky Survey (SDSS) and its Baryon Oscillation Spectroscopic Survey (BOSS).This subtle correlation is now a major player in the study of dark energy. When we observe a sample of galaxies at high redshift and detect their distinctive clustering scale, we know that the scale is 500 million light-years. With that, we can infer the distance to the galaxies. It is as if the redshift survey has given us a map without an obvious conversion of inches to miles – by careful analysis on the spacing of the galaxies we can infer the correct scaling factor. DESI will use the BAO to measure the relationship of distance to redshift over a wide range of redshifts with subpercent precision. This in turn allows us to infer the expansion history and the evolution of dark energy. When DESI measures the redshift of galaxies, this measurement is actually composed of two contributions: the large shift coming from the cosmological expansion of the universe, and a smaller shift that results from the motion of the galaxy due to the gravitational pull from the universe’s surrounding large-scale structure. The latter is known as the peculiar velocity. By measuring the size of the peculiar velocity, scientists can measure the amount of mass in the large-scale structure. Or if that is known from other methods, we can test whether gravitational attractions on scales of hundreds of millions of light-years follows the predictions of Albert Einstein’s theory of general relativity, which relates to gravity. The testing of general relativity on these enormous scales is important as it might reveal alternative explanations, known as modified gravity theories, for the universe’s accelerating expansion rate. On the very largest scales we find that the expansion rate is evolving very differently from what would be predicted from the attractions of known matter and the physics that are so well-tested on solar system scales. Perhaps this breakdown has other signatures in the universe’s large-scale structure. The size of the velocities can be measured in the DESI maps by comparing the characteristic amplitudes of the large-scale clustering of galaxies in directions perpendicular to the line of sight to the clustering signals in directions along the line of sight. This is known as redshift-space distortions. Because the large-scale structure we are measuring is giving rise to the peculiar velocities, there is a systematic change in the appearance of the large-scale structure. Careful analysis of the DESI maps can reveal this signature to high precision. DESI in the Dark Energy Portfolio The enigma of dark energy and the ongoing appeal of cosmology continues to motivate a wide range of cosmological experiments. DESI is one of the most ambitious surveys now under development, and once operational it will quickly become the world’s largest galaxy redshift survey. The exquisite 3D maps from DESI will have great opportunity on their own, but they become even more effective when combined with data from upcoming imaging surveys in optical/infrared, microwave, and X-ray light. One of the hallmarks of modern cosmology is the ability to measure cosmological properties in multiple manners, as the comparison of results can reinforce confidence in the answers and can point the way to new opportunities. We look forward to the prominent and collaborative role that DESI will play in the cosmological portfolio of the next decade, particularly in the study of dark energy. Beyond Dark Energy The DESI maps will be used for many other applications beyond baryon acoustic oscillations and redshift-space distortions. We expect that our measurements of the intermediate-redshift amplitude of large-scale structure will be important to measurements of the mass of the neutrino. Our measurements on the largest scales can test whether the initial perturbations in the universe follow the simplest model, or alternatively show correlations indicative of novel behaviors at the enormous energy scales of the first second of the universe. Detailed measurements of galaxy clustering can use gravitational clustering to test a wide range of extensions to the standard cosmological model. Beyond cosmology, DESI will measure precise distances to millions of galaxies and quasars whose properties and demographics can then be better interpreted. DESI will produce the most detailed map yet of the nearby universe, which will provide the backbone for studies of galaxy groups and clusters as well as of extreme phenomena within those galaxies. And DESI’s stellar spectroscopy will measure the dynamical state of the Milky Way halo and thick disk in great detail.<\|endoftext\|>	3.671875
517	Comet Kopff or 22P/Kopff is a periodic comet in the Solar System. Discovered on August 23, 1906, it was named after August Kopff who discovered the comet. The comet was missed on its November 1912 return, but was recovered on its June 1919 return. The comet has not been missed since its 1919 return and its last perihelion passage was on May 25, 2009. Close approaches to Jupiter in 1938 and 1943 decreased the perihelion distance and orbital period. 22P/Kopff’s next perihelion passage is October 25, 2015. at different epochs 22P/Kopff was discovered at Königstuhl Observatory on Heidelberg, Germany. Kopff analyzed photographic plates which he exposed on August 20, 1903 against pre-discovery images of the same region. On August 23, 1903, Kopff concluded it to be a comet with an estimated apparent magnitude of 11. On mid-September 1906, the short-period nature of the comet was recognized by a team headed by Kiel Ebell of the Berkeley Astronomical Department. The comet was missed when it made a return on November 25, 1912 however on June 25, 1919, astronomers recovered the comet. The comet was located less than three days from the predicted position. Over the next several returns to Earth, none were notable until the 1945 comet’s return when the comet peaked at magnitude 8.5. The increase in brightness was a result of Jupiter altering the comet’s orbit between the years of 1939 to 1945. This change in orbit brought the comet closer to the Sun. The 1951 return was unique due to the comet being 3 magnitudes fainter than what was expected when recovered in April 1951. But the comet still reached magnitude 10.5 in October 1951. A very close pass to Jupiter in 1954 increased the comet’s perihelion distance to 1.52 AU and increased the orbital period to 6.31 years. On November 30, 1994, Carl W. Hergenrother was able to recover the comet at a stellar magnitude of 22.8 using the 1.5-m reflector at the Catalina Sky Survey. The comet reached magnitude 7 during the 1996 perihelion passage. The comet nucleus is estimated to be 3.0 kilometers in diameter with an albedo of 0.05. The nucleus is dark because hydrocarbons on the surface have been converted to a dark, tarry like substance by solar ultraviolet radiation.<\|endoftext\|>	3.9375
1,124	# Division of Literals Division of literals obeys all operation of division of numbers. In arithmetic, we have studied that the division sign ‘÷’ read as ‘by’ between two numbers means that the number on the left of the division sign is to be divided by the number on the right. For example: 15 ÷ 3 means that the number 15 on the left of the division sign is to be divided by the number 3 on the right of division sign. In the case of literal numbers also x ÷ y read as ‘x by y’ means that the literal x is to be divided by the literal y and is written as x/y. Thus 25 divided by a is written as 25/a and y divided by 5 written as y/5. It should be noted that 1/5 of y or y divided by 5 is also written as y/5. Similarly, x divided by 10 is x/10. Quotient of 5 by z is 5/z. Examples on Division of Literals: Write each of the following phrases using numbers, literals and the basic operations of addition, subtraction, multiplication and division: 1. Quotient of x by 4 is added to y. Solution: We have, Quotient of x by 4 = x/4 Therefore, quotient of x by 4 added to y = x/4 + y 2. Quotient of z by 6 is multiplied by y. Solution: We have, Quotient of z by 6 = z/6 Therefore, quotient of z by 6 is multiplied by y = z/6 × y = zy/6 3. Quotient of x by y added to the product of x and y. Solution: We have, Question of x by y = x/y And product of x and y = xy Therefore, quotient of x by y and to product of x and y = x/y + xy. 4. 100 taken away from the quotient of 5m by 2x. Solution: We have, Quotient of 5m by 2x = 5m/2x. Therefore, 100 taken away from the quotient of 5m by 2x = 5m/2x - 100 5. Product of 4 and a divided by the difference of 5 and a. Solution: Product of 4 and a = 4a Difference of 5 and a = (5 - a) Thus, we have 4a ÷ (5 - a) or, 4a/(5 - 4) 6. Express the share of each child algebraically if x apples were equally distributed among four children. Solution: Total number of Apple = x Number of children = 4 Therefore, each children gets x ÷ 4 = x/4 apples. 7. Quotient of x by 2 subtracted from 10 less than x Solution: We have, Quotient of x by 2 = x/2 10 less than x = (x - 10) Thus, we have (x - 10) - x/2 Properties of Division of Literals For any literal number a (i) a ÷ a = 1 (ii) 0 ÷ a = 0 (iii) a ÷ 1 = a Division of literals is neither commutative nor associative. Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need. ## Recent Articles 1. ### Addition and Subtraction of Fractions \| Solved Examples \| Worksheet Jul 18, 24 03:08 PM Addition and subtraction of fractions are discussed here with examples. To add or subtract two or more fractions, proceed as under: (i) Convert the mixed fractions (if any.) or natural numbers 2. ### Worksheet on Simplification \| Simplify Expressions \| BODMAS Questions Jul 18, 24 01:19 AM In worksheet on simplification, the questions are based in order to simplify expressions involving more than one bracket by using the steps of removal of brackets. This exercise sheet 3. ### Fractions in Descending Order \|Arranging Fractions an Descending Order Jul 18, 24 01:15 AM We will discuss here how to arrange the fractions in descending order. Solved examples for arranging in descending order: 1. Arrange the following fractions 5/6, 7/10, 11/20 in descending order. First… 4. ### Fractions in Ascending Order \| Arranging Fractions \| Worksheet \|Answer Jul 18, 24 01:02 AM We will discuss here how to arrange the fractions in ascending order. Solved examples for arranging in ascending order: 1. Arrange the following fractions 5/6, 8/9, 2/3 in ascending order. First we fi… 5. ### Worksheet on Comparison of Like Fractions \| Greater & Smaller Fraction Jul 18, 24 12:45 AM In worksheet on comparison of like fractions, all grade students can practice the questions on comparison of like fractions. This exercise sheet on comparison of like fractions can be practiced<\|endoftext\|>	4.875
2,183	# Search by Topic #### Resources tagged with Working systematically similar to Dicey Addition: Filter by: Content type: Age range: Challenge level: ### There are 321 results Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically ### Twinkle Twinkle ##### Age 7 to 14 Challenge Level: A game for 2 people. Take turns placing a counter on the star. You win when you have completed a line of 3 in your colour. ### Window Frames ##### Age 5 to 14 Challenge Level: This task encourages you to investigate the number of edging pieces and panes in different sized windows. ### Counters ##### Age 7 to 11 Challenge Level: Hover your mouse over the counters to see which ones will be removed. Click to remove them. The winner is the last one to remove a counter. How you can make sure you win? ### Growing Garlic ##### Age 5 to 7 Challenge Level: Ben and his mum are planting garlic. Use the interactivity to help you find out how many cloves of garlic they might have had. ### Triangle Edges ##### Age 5 to 7 Challenge Level: How many triangles can you make using sticks that are 3cm, 4cm and 5cm long? ### Triangle Animals ##### Age 5 to 7 Challenge Level: How many different ways can you find to join three equilateral triangles together? Can you convince us that you have found them all? ### Matching Time ##### Age 5 to 7 Challenge Level: Try this matching game which will help you recognise different ways of saying the same time interval. ### Maths Trails ##### Age 7 to 14 The NRICH team are always looking for new ways to engage teachers and pupils in problem solving. Here we explain the thinking behind maths trails. ### Coloured Squares ##### Age 5 to 7 Challenge Level: Use the clues to colour each square. ### All the Digits ##### Age 7 to 11 Challenge Level: This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures? ### One of Thirty-six ##### Age 5 to 7 Challenge Level: Can you find the chosen number from the grid using the clues? ### How Long Does it Take? ##### Age 7 to 11 Challenge Level: In this matching game, you have to decide how long different events take. ### Teddy Town ##### Age 5 to 14 Challenge Level: There are nine teddies in Teddy Town - three red, three blue and three yellow. There are also nine houses, three of each colour. Can you put them on the map of Teddy Town according to the rules? ### First Connect Three ##### Age 7 to 14 Challenge Level: Add or subtract the two numbers on the spinners and try to complete a row of three. Are there some numbers that are good to aim for? ### Two Spinners ##### Age 5 to 7 Challenge Level: What two-digit numbers can you make with these two dice? What can't you make? ### A Bag of Marbles ##### Age 5 to 7 Challenge Level: Use the information to describe these marbles. What colours must be on marbles that sparkle when rolling but are dark inside? ### Diagonal Sums Sudoku ##### Age 7 to 16 Challenge Level: Solve this Sudoku puzzle whose clues are in the form of sums of the numbers which should appear in diagonal opposite cells. ### Tessellate the Triominoes ##### Age 5 to 7 Challenge Level: What happens when you try and fit the triomino pieces into these two grids? ### Cover the Camel ##### Age 5 to 7 Challenge Level: Can you cover the camel with these pieces? ### Arrangements ##### Age 7 to 11 Challenge Level: Is it possible to place 2 counters on the 3 by 3 grid so that there is an even number of counters in every row and every column? How about if you have 3 counters or 4 counters or....? ### Games Related to Nim ##### Age 5 to 16 This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning. ### Cuisenaire Counting ##### Age 5 to 7 Challenge Level: Here are some rods that are different colours. How could I make a dark green rod using yellow and white rods? ### Getting the Balance ##### Age 5 to 7 Challenge Level: If you hang two weights on one side of this balance, in how many different ways can you hang three weights on the other side for it to be balanced? ### Fault-free Rectangles ##### Age 7 to 11 Challenge Level: Find out what a "fault-free" rectangle is and try to make some of your own. ### Are You Well Balanced? ##### Age 5 to 7 Challenge Level: Can you work out how to balance this equaliser? You can put more than one weight on a hook. ### Latin Lilies ##### Age 7 to 16 Challenge Level: Have a go at this game which has been inspired by the Big Internet Math-Off 2019. Can you gain more columns of lily pads than your opponent? ### Find the Difference ##### Age 5 to 7 Challenge Level: Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it. ### Combining Cuisenaire ##### Age 7 to 11 Challenge Level: Can you find all the different ways of lining up these Cuisenaire rods? ### First Connect Three for Two ##### Age 7 to 11 Challenge Level: First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line. ### Same Length Trains ##### Age 5 to 7 Challenge Level: How many trains can you make which are the same length as Matt's, using rods that are identical? ### Three Ball Line Up ##### Age 5 to 7 Challenge Level: Use the interactivity to help get a feel for this problem and to find out all the possible ways the balls could land. ### ABC ##### Age 7 to 11 Challenge Level: In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication? ### Wonky Watches ##### Age 7 to 11 Challenge Level: Stuart's watch loses two minutes every hour. Adam's watch gains one minute every hour. Use the information to work out what time (the real time) they arrived at the airport. ### Jumping Cricket ##### Age 5 to 7 Challenge Level: El Crico the cricket has to cross a square patio to get home. He can jump the length of one tile, two tiles and three tiles. Can you find a path that would get El Crico home in three jumps? ### Seven Flipped ##### Age 7 to 11 Challenge Level: Investigate the smallest number of moves it takes to turn these mats upside-down if you can only turn exactly three at a time. ### The Pied Piper of Hamelin ##### Age 7 to 11 Challenge Level: This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether! ### 1 to 8 ##### Age 7 to 11 Challenge Level: Place the numbers 1 to 8 in the circles so that no consecutive numbers are joined by a line. ### Octa Space ##### Age 7 to 11 Challenge Level: In the planet system of Octa the planets are arranged in the shape of an octahedron. How many different routes could be taken to get from Planet A to Planet Zargon? ##### Age 5 to 11 Challenge Level: How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done? ### Eight Queens ##### Age 7 to 11 Challenge Level: Place eight queens on an chessboard (an 8 by 8 grid) so that none can capture any of the others. ##### Age 7 to 11 Challenge Level: Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon? ### Room Doubling ##### Age 7 to 11 Challenge Level: Investigate the different ways you could split up these rooms so that you have double the number. ### It Figures ##### Age 7 to 11 Challenge Level: Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all? ### Dienes' Logiblocs ##### Age 5 to 7 Challenge Level: This problem focuses on Dienes' Logiblocs. What is the same and what is different about these pairs of shapes? Can you describe the shapes in the picture? ### Ordered Ways of Working Lower Primary ##### Age 5 to 7 Challenge Level: These activities lend themselves to systematic working in the sense that it helps to have an ordered approach. ### Jumping Squares ##### Age 5 to 7 Challenge Level: In this problem it is not the squares that jump, you do the jumping! The idea is to go round the track in as few jumps as possible. ### Calcunos ##### Age 7 to 11 Challenge Level: If we had 16 light bars which digital numbers could we make? How will you know you've found them all? ### Ice Cream ##### Age 7 to 11 Challenge Level: You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream. ### Mixed-up Socks ##### Age 5 to 7 Challenge Level: Start with three pairs of socks. Now mix them up so that no mismatched pair is the same as another mismatched pair. Is there more than one way to do it? ### 5 on the Clock ##### Age 7 to 11 Challenge Level: On a digital clock showing 24 hour time, over a whole day, how many times does a 5 appear? Is it the same number for a 12 hour clock over a whole day?<\|endoftext\|>	4.4375
600	Climate change is the long-term alteration of temperature and normal weather patterns in a place. This could refer to a particular location or the planet as a whole. Climate change is currently occurring throughout the world as a result of global warming. Global warming is an increase in the planet’s overall temperature due to the burning of fossil fuels, such as natural gas, oil, and coal. Burning these materials releases certain gases into Earth’s atmosphere. These gases trap the heat from the Sun’s rays inside the atmosphere, causing Earth’s average temperature to rise. The warming of the planet impacts local and regional climates. Climate is different from weather because it is measured over a long period of time whereas weather can vary daily, or from year to year. The climate of an area includes averages of seasonal temperatures, rainfall, snowfall, and wind patterns. Different places have different climates. A desert, for example, is referred to as an arid climate because it is dry year round. Other types of climate include tropical, which is hot and humid, and temperate, which has warm summers and cooler winters. The effects of climate change make average temperatures and weather patterns more unpredictable. In an arid region, for example, this may mean higher than average temperatures and more severe or prolonged droughts. In a temperate region, it may mean that there is a lot of rain one year, and then very little rain the next year. These unpredictable weather patterns can therefore make it difficult to grow crops and maintain farmland in temperate zones because the temperatures and rainfall that farmers had come to expect can no longer be relied upon. Climate change has also been connected with other damaging weather events such as more frequent and more intense hurricanes, floods, downpours, and winter storms. In polar regions, the warming global temperatures associated with climate change have meant ice sheets and glaciers are melting at an accelerated rate from season to season. This contributes to sea levels rising in different regions of the planet. Together with expanding ocean waters due to rising temperatures, the resulting rise in sea level has begun to damage coastlines as a result of increased flooding and erosion. Term Part of Speech Definition Encyclopedic Entry atmosphere Noun layers of gases surrounding a planet or other celestial body. Encyclopedic Entry: atmosphere climate change Noun gradual changes in all the interconnected weather elements on our planet. Encyclopedic Entry: Earth's Changing Climate fossil fuel Noun coal, oil, or natural gas. Fossil fuels formed from the remains of ancient plants and animals. mass of ice that moves slowly over land. Encyclopedic Entry: glacier humid Adjective air containing a large amount of water vapor. sea level Noun base level for measuring elevations. Sea level is determined by measurements taken over a 19-year cycle. Encyclopedic Entry: sea level temperate Adjective<\|endoftext\|>	3.75
630	# Finding central angle after slicing cone • Nov 19th 2011, 08:43 AM benny92000 Finding central angle after slicing cone A cone made of cardboard has a vertical height of 8 cm and a radius of 6 cm. If this cone is cut along the slanted height to make a sector, what is the central angle, in degrees, of the sector? I drew a diagram. I thought the central angle was 90 degrees because of the 6,8,10 triangle that included the slant height. The answer says 216 degrees. (Thinking) • Nov 19th 2011, 10:15 AM Soroban Re: Finding central angle after slicing cone Hello, benny92000! The problem is not stated clearly. Quote: A cone made of cardboard has a vertical height of 8 cm and a radius of 6 cm. If this cone is cut along the slanted height to make a sector, what is the central angle, in degrees, of the sector? The side view of the cone looks like this: Code: B * /:\ / : \ / : \ / : \ 10 / :8 \ / : \ / : \ A --------------* C : - 6 - : - 6 - : We see that the slant height is 10 cm. The circumference of the base is: . $2\pi r = 2\pi(6) = 12\pi\text{ cm}$ Now we cut the cone and lay it flat. We have a large sector of a circle of radius 10. Code: B * * * * * * * * * * O * * * * * / \ * 10 / \ 10 * / \ * * / \ * A * * C * * * Length-of-arc Formula: . $s \,=\,r\theta$ . where:. $\begin{Bmatrix}{s &=& \text{length of arc} \\ r &=& \text{radius} \\ \theta &=& \text{central angle}\\[-2mm] && \text{in radians} \end{Bmatrix}$ We have: . $\begin{Bmatrix}s \,=\,\text{arc}(ABC) \,=\, 12\pi \\ r \:=\: 10\end{Bmatrix}$ Therefore: . $12\pi \,=\,10\theta$ . . . . . . . . . . $\theta \:=\:\frac{12\pi}{10}\text{ radians} \:=\:216^o$<\|endoftext\|>	4.46875
218	Penguin calls (vocalizations) are individually identifiable, allowing mates to recognize each other and also their chick. This is important because members of a large colony of penguins are nearly indistinguishable by sight. Research has identified differences in the calls of male and female emperor penguins. These differences probably function in courtship and mate selection. There are three main kinds of penguin calls. The contact call assists in recognition of colony members. The contact call of emperor and king penguins can be heard one kilometer (0.6 mi.) away. - The display call is the most complex of all the calls and is used between partners in a colony. The call must convey information on territorial, sexual, and individual recognition. - The threat call is the simplest and is used to defend a territory and warn other colony members of predators. Penguins communicate by vocalizing and performing physical behaviors called displays. They use many vocal and visual displays to communicate nesting territories, mating information, nest relief rituals, partner and chick recognition, and defense against intruders.<\|endoftext\|>	4.375
918	What Is It? Ichthyosis is the term for severe, persistent problems with dry skin that almost always start in childhood or infancy. Ichthyosis can be genetic (inherited) or can develop later in life. In a large majority of people with the disease, the cause is related to one or more genetic mutations. Under normal circumstances, the body continuously renews its skin surface, building new skin cells and allowing older cells to be shed from the surface. Ichthyosis disrupts this balance either because too many replacement skin cells are produced or because the new skin cells do not separate well from the skin surface when it is their time to drop off. The result is that skin cells accumulate into thick flakes that adhere to the body and can resemble fish scales. Ichthyosis often causes severe cosmetic concerns for the person with the condition. However, it is not a contagious disease. The condition also can interfere with the skin’s critical roles in protecting against infection, preventing dehydration, and regulating body temperature. Most people have ichthyosis vulgaris, the mildest form of the disease. It occurs in one out of every 250 people. Depending on the type of gene abnormality that causes ichthyosis, the skin can show different patterns of flaking. In most cases of ichthyosis vulgaris, for example, skin will flake over the majority of the body, but not on the inside surfaces of joints, in the groin area or on the face. Symptoms of all genetic types of ichthyosis are either noticeable at birth or appear during childhood. Symptoms may include: - Severe dryness of the skin with thickening and flaking, which may be appear only in limited areas or may involve almost the entire skin surface - Mild itching of the skin - Body odor, because the spaces under and between skin flakes can harbor collections of bacteria or fungus - Wax buildup in the ears, causing hearing difficulties Symptoms are usually worse in winter months and in dry climates, because warmth and humidity improve thesymptoms. More than half of people who have ichthyosis vulgaris also have allergic problems, such as allergic nasal congestion, asthma or eczema. A doctor usually can diagnose ichthyosis by looking at the skin. A family history is also very useful. In some cases, a skin biopsy will be done to help to confirm the diagnosis. In a biopsy, a small piece of skin is removed and examined under a microscope. In rare instances, genetic testing may be helpful in making a diagnosis. Because this family of diseases is caused by genetic abnormalities, it is a lifelong condition. However, symptoms can be controlled with treatment and can improve in humid, warm climates. There is no way to prevent ichthyosis. As with other genetic diseases, there is a risk that children of an affected parent will inherit the gene. Treatment of all types of ichthyosis involves restoring the skin’s moisture and facilitating more aggressive shedding of dead skin cells. Regular visits to a dermatologist can help when ichthyosis causes severe symptoms. To get the best results, use moisture-retaining creams or ointments after a bath or shower, so that moisture is sealed into the skin surface. Petrolatum, lanolin or urea-containing preparations are very helpful in maintaining skin moisture. Medicines that promote shedding of skin flakes include lotions or creams that contain lactic acid or other “alpha-hydroxy acids,” or swallowed medicines such as isotretinoin (Accutane). If the ichthyosis leads to scratching that causes skin infections, or if body odor is a major problem, antibiotics may be needed from time to time. When To Call A Professional Because ichthyosis can make the skin a less effective barrier to infection, it is important to contact a physician if you develop a fever or redness of the skin. With continued effective treatment and good advice about skin care, ichthyosis is usually very manageable. Some forms of ichthyosis improve after childhood. Rare forms of ichthyosis can be life threatening, even in infancy, if skin problems are severe. Diseases and Conditions Center All ArmMed Media material is provided for information only and is neither advice nor a substitute for proper medical care. Consult a qualified healthcare professional who understands your particular history for individual concerns.<\|endoftext\|>	3.890625
381	Categories # [Smart Math] Ratio Proportion Problem 28 Here’s and example of a SMART MATH problem for RATIO PROPORTION. ### Problem The ratio of number of ladies to gents in a party was 3 : 2. When however, 20 more gentlemen joined the party, the ratio was reversed. How many ladies were there at the party? 1. 16 2. 24 3. 32 4. 36 5. 40 ### The Usual Method Let ‘x’ be the constant of proportionality. Hence, there will be 3x ladies and 2x gents in the party originally. New total number of gents = 2x + 20 Hence, ratio of ladies to gents = $\frac{3x}{2x+20}=\frac{2}{3}$ $\therefore 9x=4x+40$ $\therefore 5x=40$ $\therefore x=8$ $\therefore$Original number of ladies = 3x = 3 x 8 = 24 (Ans: 2) Estimated Time to arrive at the answer = 45 seconds. ### Using Technique Since the total number of ladies at the party = 3x, the answer should be a multiple of 3. From the options, only option ‘2’ and ‘4’ are multiples of 3. Assuming option ‘2’ to be the right answer, the number of men becomes 2x = 16 (using 3x = 24 $\therefore x=8$) Hence, new total of gents = 20 + 16 = 36 and 24 : 36 = 2 : 3. This satisfies the conditions and so, ‘2’ is the right option.<\|endoftext\|>	4.5625
737	# Subtraction - The Equal Change Algorithm- It's Fun! 3 teachers like this lesson Print Lesson ## Objective SWBAT explain and use the equal change algorithm to solve 2 and 3 digit subtraction problems. #### Big Idea An alternative approach can increase understanding of regrouping! ## Opener 2 minutes I let students know that the algorithm I'm going to teach them today, the Equal Change Algorithm, is an alternative to the standard algorithm that some of them will like and some of them might not. It's important to try out different methods because that's the way that each of them will begin to discern their own particular preferences. Even if this isn't an algorithm that all of them end up using, their is value in teaching it because it forces them to examine their understanding of subtraction from a different perspective. Additionally, as the goal is to move the subtrahend to an even ten or hundred, it helps them with the skill of rounding as well, though it's important to note that in this algorithm the subtrahend doesn't need to be moved to the closest ten or hundred, simply one of the two. (It would work even if you rounded to the nearest 200, but I don't suggest letting students play with that concept). ## Guided Lesson 25 minutes I begin teaching the equal change algorithm for subtraction by showing them examples with 2 digit subtraction. This is a brief video, intended for you (too fast for students) that explains this method. I have observed that some students seem to instantly fall in love with this method - to them it feels like finding the correct interlocking puzzle pieces. Other students see that it produces the correct answer, but don't understand why the add or subtract the same number to "both sides". For that reason, I also demonstrate with place value blocks. I have demonstrated how I would do that in this brief teacher video: Finally, here is a teacher example of how to use the equal change algorithm with a 3-digit subtraction problem. These are some examples for equal change guided practice that I find useful. ## Active Engagement 33 minutes I write up 3 sets of equal change - independent practice problems on the board: 2 digit, 3 digit with some regrouping and 3 digit with regrouping in both the tens and hundreds place. All students need to do at least 2 of the 2 digit problems, but after that if they feel like they are ready, they may move on to column B or column C. I provide a fourth choice -- students who wish to stay at the carpet and work through some additional problems as group practice are welcomed to do so. In my classroom, this is a common choice, and there is no stigma associated with it, as often some of the more successful students choose to stay on for a little bit because they are self-aware enough to know they need a little additional guidance. Other students have observed this and in most cases students make a wise choice. Occasionally there are students who are over-confident or lack sufficient meta-cognitive skills to make this decision for themselves, and I'll ask them to stay on as well. ## Wrap-up 5 minutes I ask students to write the key way in which the equal change algorithm of subtraction differs from the equal and opposite change algorithm for addition. I reteach the word differ if necessary, distinguishing between this the verb form and the noun, difference, having the quality of differing.<\|endoftext\|>	4.40625
587	× Get Full Access to Statistics: Informed Decisions Using Data - 5 Edition - Chapter 7.2 - Problem 6 Get Full Access to Statistics: Informed Decisions Using Data - 5 Edition - Chapter 7.2 - Problem 6 × # ?In Problems 5–12, find the indicated areas. For each problem, be sure to draw a standard normal curve and shade the area that is to be found. Det ISBN: 9780134133539 240 ## Solution for problem 6 Chapter 7.2 Statistics: Informed Decisions Using Data \| 5th Edition • Textbook Solutions • 2901 Step-by-step solutions solved by professors and subject experts • Get 24/7 help from StudySoup virtual teaching assistants Statistics: Informed Decisions Using Data \| 5th Edition 4 5 1 236 Reviews 28 0 Problem 6 In Problems 5–12, find the indicated areas. For each problem, be sure to draw a standard normal curve and shade the area that is to be found. Determine the area under the standard normal curve that lies to the left of (a) z = -3.49 (b) z = -1.99 (c) z = 0.92 (d) z = 2.90 Step-by-Step Solution: Step 1 of 5) For each problem, be sure to draw a standard normal curve and shade the area that is to be found. Determine the area under the standard normal curve that lies to the left of (a) z = -3.49 (b) z = -1.99 (c) z = 0.92 (d) z = 2.90 Coefficient of Determination Problem For the model obtained in Example 2, determine the coefficient of determination and the adjusted R2 . Compare the R2 with the two explanatory variables age and daily saturated fat to the R2 with the single explanatory variable age. Comment on the effect the additional explanatory variable has on the value of the model. Approach Use Minitab to determine the values of the coefficient of determination and the adjusted coefficient of determination. Solution Looking back to the output in Figure 15, we see that R2 = 0.847 = 84.7%. This means that 84.7% of the variation in the response variable is explained by the least-squares regression model. In addition, we find that Radj 2 = 0.820 = 82.0%. Step 2 of 2 ## Discover and learn what students are asking Statistics: Informed Decisions Using Data : Discrete Probability Distributions ?State the criteria that must be met for an experiment to be a binomial experiment. #### Related chapters Unlock Textbook Solution<\|endoftext\|>	4.40625
6,601	## Problem 10 of the 2011 Euclid Math Contest (remix) April 14, 2011 I wrote the Euclid contest this year two days ago, on tuesday. There were 10 problems, and the tenth problem was a nice geometry problem. Three subproblems with a nice neat triangle on the right, with the subproblems getting progressively harder and harder. As I proceeded with the problem, I couldn’t help but imagine it as a Project Euler problem — instead of finding one such integer triangle, it would ask you to find the sum of the perimeters of all such integer triangles with perimeter less than 10^12 or some large number. ### A modified problem In the above diagram, $\angle CAK = 2 \angle KAB$ and $\angle ABC = 2 \angle KAB$. Let $a = CB$, $b = AC$, $c = AB$, $d = AK$, and $x = KB$. Write an algorithm to find triangles satisfying these conditions where a, b, c are all integers. ### Similar triangles It is difficult to try to find integer triangles with such strange requirements as these. It seems that the line $AK$ is completely unnecessary, but if we take it out, there doesn’t seem to be any way to relate the angle ratios to integer side lengths. We can prove that $\triangle CAK$ is similar to $\triangle ABC$. Being an exterior angle, $CKA = 3 \theta$, and also $\angle CAB = 3 \theta$. Both of the triangles have an angle of measure $2 \theta$ and another angle of measure $3 \theta$, thus they are similar. From the similar triangles ratio $\frac{b}{d} = \frac{a}{c}$ We can write d in terms of the three sides of the triangle: $d = \frac{bc}{a}$ Similarly, the side $CK$ can be written as $a-x$. Then we have the ratio $\frac{a}{b} = \frac{b}{a-x}$ Solving for x allows us to express it in terms of the three sides of the triangle, again: $x = \frac{a^2 - b^2}{a}$ ### Constructing another similar triangle Our goal here is to relate the lengths a, b, c with a simple equation, which then the problem turns into a number theory problem. Since we can write the lengths d and x in terms of a, b, and c, we can also relate any of a, b, c, d, x in an equation. Again, there doesn’t seem to be a way to relate all of the variables together, in a way that any solution implies the original angle ratio required — unless we make a construction. Here we extend AB to F, so that $KB = BF$ and $\triangle KBF$ is an isosceles triangle. Again, since the exterior angle here is $2 \theta$, both $\angle BKF = \angle BFK = \theta$. Also with this construction, $\triangle AKF \sim \triangle KBF$, and is also isosceles, hence $d = AK = KF$. With this construction, we can write the ratio $\frac{x}{d} = \frac{d}{c+x}$ Perfect! Cross multiplying and then substituting in the original sides of the triangles gives $(\frac{a^2-b^2}{a})(c+\frac{a^2-b^2}{a}) = (\frac{bc}{a})^2$ Simplifying this gives $(a^2 - b^2) (a^2 - b^2 + ac) = b^2 c^2$ ### Number theory magic Now that we have an equation relating the three side lengths — it’s easy to verify that any three integers satisfying the triangle inequality gives a triangle with the required conditions — we can use number theory to try to generate integers that fit the equation. If we expand the equation, we get $a^4+a^3 c-2 a^2 b^2-a b^2 c+b^4-b^2 c^2 = 0$ It makes sense to solve for c, which can be done using just the quadratic formula: $c = \frac{a^3 - ab^2 \pm \sqrt{(ab^2-a^3)^2 + 4b^2(b^4 + a^4 - 2a^2 b^2)}}{2b^2}$ We need the discriminant D — the expression inside the square root — to be a perfect square, where we are allowed to have integer values for a and b. If we can get D to be a perfect square, then c will turn out to be a rational number. Then multiplying all three variables by a constant gives integer values for all three. So we defined D: $D = (ab^2-a^3)^2 + 4b^2(b^4 + a^4 - 2a^2 b^2)$ Expanding this gives $D = a^6 - 7a^2 b^4 + 2 a^4 b^2 + 4b^6$ Fortunately, this expression has an interesting factorization: $D = (a^2+4b^2) (a+b)^2 (a-b)^2$ Or we can also write $\sqrt{D} = (a+b) (a-b) \sqrt{a^2 + 4b^2}$ We’ve simplified this problem to finding values where $a^2 + 4b^2$ is a perfect square, that is: $a^2 + (2b)^2 = k^2$ This is just finding Pythagorean triples where one of the two sides are even! For instance, in the triple (3,4,5), we have a=3 and b=2. However, substituting a=3, b=2 into the quadratic formula gives c=5. This is almost a solution, only that the sides have to satisfy the triangle inequality (two sides have to add up to more than the third side). The next Pythagorean triple (5,12,13) gives a=5 and b=6. Substituting this in gives c=11/9, which does satisfy the triangle inequality. Multiplying everything by 9 gives a=45, b=54, c=11 as the smallest working combination. With this method, it is possible to quickly find arbitrarily many such triples, using Pythagorean triples as a starting point (which can be generated quickly with known methods). ## 2010 Euclid Contest April 13, 2010 Today quite a few people from my school took the Euclid contest, again from the University of Waterloo. This contest is really meant for grade 12 students, and it’s probably the most important of the contests, as it can be used to apply for universities and scholarships. While most people took the contest last wednesday (April 7), my school took it on April 12. Surprisingly I haven’t found anybody uploading the full solutions, or even all the questions of the euclid. Having took the contest, I’ll do so now. On to the first question! ### Question 1 1. a) $3^x = 27$, find $3^{x+2}$. Solving for x, we get $x=3$. So $x+2 = 5$, and $3^5$ = 243. b) $2^5 \cdot 3^{13} \cdot 5^9 x = 2^7 \cdot 3^{14} \cdot 5^9$, find x Solve for x: $\begin{array}{rcl} x &=& \frac{2^7 \cdot 3^{14} \cdot 5^9}{2^5 \cdot 3^{13} \cdot 5^9} \\ &=& 2^2 \cdot 3 \\ &=& 12 \end{array}$ The value for x is 12. c) The equation of line AB is $y=x+2$, and the equation of BC is $-\frac{1}{2}x+2$. Determine the area of $\triangle ABC$. Because A and C are on the x axis and B is on the y axis, it’s fairly simple to figure out the coordinates of all three points. They are: $\begin{array}{l} A = (-2,0) \\ B=(0,2) \\ C=(4,0) \end{array}$ We have a base and a height, so the area can be calculated to be 6. ### Question 2 2. a) Maria has a red, blue, and green package. The sum of the masses of all three packages, is 60kg, of the red and green packages is 25kg, and of the green and blue packages is 50kg. What is the mass of the green package? Using R, G, and B for the masses of the red, green, and blue packages respectively, we can put together this system of equations: $\begin{array}{rrrrrcl} R&+&G&+&B&=&60 \\ R&+&G&&&=&25 \\ &&G&+&B&=&50 \end{array}$ Solve this by elimination and you should get (R,G,B) = (10,15,35). So the weight of the green package is 15 kg. b) A palindrome is a positive integer reading the same forwards and backwards (151, for example). What is the largest palindrome less than 200 that can be written as the sum of three consecutive integers? Any number that can be written as a sum of three consecutive integers must be divisible by 3, because if n is an integer then the sum is $n + (n+1) + (n+2)$ or $3n + 3$. Palindromes less than 200 are 191, 181, 171, etc. The first one in the sequence that’s divisible by 3 is 171. c) If $(x+1) (x-1) = 8$, determine the value of $(x^2+x) (x^2-x)$. Solving the first quadratic for x: $\begin{array}{l} (x+1)(x-1)=8 \\ x^2-1=8 \\ x^2-9=0 \\ (x+3)(x-3) = 0 \\ x = \pm3 \end{array}$ Substituting either of the two roots into the second expression gives the value as 72. ### Question 3 3. a) Bea (a bee) starts from H (the hive), flies south for 1 hour to F (field), spends 30 minutes at F, flies 45 minutes to G (garden), spends 1 hour at G, then flies back to H. She flies at a constant speed in a straight line. What is the total time (in minutes) that she’s away from her hive? All the times are given, except for the hypotenuse HG, which can easily be calculated using the pythagorean theorem: The sum of the amount of times spent in each segment is 270 minutes. b) $\triangle OPB$ is right angled. Determine all possible values of p. Since $\triangle OPB$ is right angled, P is on a circle with OB as diameter (Thales’ theorem). We can let M be the center of the circle at (5,0): The y coordinate of P is 4, so the equation is $y=4$, intersecting with the circle at two points given by this system of equations: $\begin{array}{l} y=4 \\ (x-5)^2 + y^2 = 25 \end{array}$ $\begin{array}{rcl} (x-5)^2 + 16 = 25 \\ (x-5)^2 = 9 \\ x-5 = \pm 3 \\ x = 5 \pm 3 \end{array}$ So the possible values of x, or p, are 2 and 8. ### Question 4 4. a) Toy goats cost $19 each and toy helicopters$17 each. Thurka spent \$201 on integral amounts of goats and helicopters. How many of each did she buy? Because $201 \equiv 14 \; (\textrm{mod} \; 17)$ and $19 \equiv 2 \; (\textrm{mod} \; 17)$, 201 – 7*19 is divisible by 17. Thus the only solution is 7 goats and 4 helicopters. b) Determine all values of x where $(x+8)^4 = (2x+16)^2$ Rewrite the equation as $(x+8)^4 = 4(x+8)^2$ and let $t=(x+8)^2$: $\begin{array}{l} t^2 = 4t \\ t^2 - 4t = 0 \\ t (t-4) = 0 \\ t=0, t=4 \end{array}$ Substituting the values of t and solving for x: $\begin{array}{l} (x+8)^2 = 0 \rightarrow x = -8 \\ (x+8)^2 = 4 \rightarrow x+8 = \pm 2 \rightarrow x = -8 \pm 2 \end{array}$ So the possible values for x are -10, -8, and -6. ### Question 5 5. a) If $f(x) = 2x+1$ and $g(f(x)) = 4x^2+1$, determine an expression for $g(x)$. $g(x)$ must be a quadratic function in the form $ax^2 + bx + c$. We know $f(x)$, so we can express $g(f(x))$ as such: $\begin{array}{rcl} g(f(x)) &=& g(2x+1) \\ &=& a(2x+1)^2 + b(2x+1) + c \\ &=& (4a)x^2 + (4a+2b)x + (a+b+c) \end{array}$ In $g(f(x))$, a=4, b=0, and c=1. Thus we can solve this system of equations: $\begin{array}{l} 4a=4 \\ 4a+2b=0 \\ a+b+c=1 \end{array}$ The solution is (a,b,c) = (1,-2,2), so g(x) = x^2-2x+2. b) A geometric sequence has 20 terms. The sum of the first two terms is 40, of the first three is 76, and of the first four is 130. Determine how many terms are integers. Because the difference between the sum of the first three and the sum of the first two is known, the forth term is 54 by subtraction. Similarly, the third term is 36. We can calculate the ratio between terms to be $\frac{54}{36} = \frac{3}{2}$. Then the second term can be interpolated by multiplying the third term by $\frac{2}{3}$, and it is 24. The first term is 16. So the sequence goes like this: $16, 24, 36, 54, 81, \cdots$ Notice that in each term the number of 2-factors decreases by 1. For example the first term, 16, is divisible by $2^4$, and the second term is divisible by $2^3$, and so on. So 81 is divisible by $2^0$, and the next term and all following terms are not integers. The sequence contains 5 terms. ### Question 6 6. a) AB is 1cm, find AH. The ratio of AB to AC is $\cos 30$ or $\frac{\sqrt{3}}{2}$. Or $AC = \frac{2}{\sqrt{3}} \cdot AB$. The ratio of AD to AC is the same, and so is the ratio of AE to AD, all because of similar triangles. AH is just the ratio applied six times. It’s $(\frac{2}{\sqrt{3}})^6$ or 64/27. b) AF=4 and DF=2. Determine the area of BCEF. All the triangles: $\triangle AFB$, $\triangle AFD$, and $\triangle DFE$ are similar because they are all right triangles, all having a right angle plus another angle shared. The ratio of the longer side to the shorter side is 4:2 or 2:1, so BF=8 and FE=1. The area of AFB is 16, the area of AFD is 4, and the area of DFE is 1: The area of ABD is 20, which is equal to BCD. Since DFE=1, BCEF must be 19. ### Question 7 7. a) Determine all solutions for this equation: $3^{x-1} \cdot 9^{\frac{3}{2x^2}} = 27$ Using some logarithms, it’s not too difficult to solve: $\begin{array}{rcl} 3^{x-1} \cdot 3^{2 \cdot \frac{3}{2x^2}} &=& 3^3 \\ (x-1) + \frac{3}{x^2} &=& 3 \\ x + 4 + \frac{3}{x^2}&=& 0 \\ x^3 - 4x^2 + 3 &=& 0 \\(x-1) (x^2-3x-3) &=& 0 \end{array}$ Using the quadratic formula we can get the solutions for the second factor. The solutions for x are 1 and 3±√21/2. b) Determine all solutions (x,y) for these equations: $\begin{array}{rcl} y &=& \log_{10} (x^4) \\ y &=& (\log_{10} x)^3 \end{array}$ This first equation can be written like this: $y = 4 \log_{10} x$ Substitute g for $\log{10}x$ and combine the two equations: $\begin{array}{rcl} 4g &=& g^3 \\ g^3 - 4g &=& 0 \\ g (g+2) (g-2) &=& 0 \end{array}$ The solutions for g are 0, -2, and 2. Because $x=10^g$, replace the g values with their respective x values to get 100, 1, and $\frac{1}{100}$. Now calculate the y values by substituting back the x values, and the solutions are (1,0), (100,8), and (1/100, -8). ### Question 8 8. a) Oi-Lam tosses 3 coins, and removes those that come up heads (if any). He then tosses the remaining coins (if any). Determine the probability he tosses exactly one head (on the second toss). Using the pascal triangle, there is a $\frac{1}{8}$ chance of having no coins to toss on the second round, $\frac{3}{8}$ chance of one coin, $\frac{3}{8}$ chance of having two coins, and $\frac{1}{8}$ chance of having no coins. For each case the probability of coming up with exactly one head is given by the first diagonal: Namely the probability of having exactly one head when tossing 3 coins is 3 divided by the sum of the row, giving $\frac{3}{8}$. So adding it up, we have this expression: $\begin{array}{l} \frac{1}{8} \cdot (1 \cdot 0 + 3 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2} + 1 \cdot \frac{3}{8}) \\ = \frac{27}{64} \end{array}$ The total probability is 27/64. This fraction is the same as the answer to 6a, reversed. Coincidence? b) Here P is the center of the larger circle with AC as diameter; Q is the center of the smaller circle with BD as diameter. $\angle PRQ$ is 40. Determine $\angle ARD$. It’s best to solve this problem by first drawing a few additional lines: We also give the variables a, b, c, and d to several of the angles as shown above. Because of isosceles triangles, $\angle RAP = \angle ARP = a+b$. Similarly, $\angle RBQ = \angle BRQ = b+40$. The two must be equal so $a+b = b+40$. From that, a=20. With the exact same logic, d=20. AC is a diameter so ARC is a right angle. We know d so ARD = 110°. ### Question 9 9. a) i) Prove that: $\cot \theta - \cot 2\theta = \frac{1}{\sin 2\theta}$ If you know the trigonometric identities, this question shouldn’t be too difficult: $\begin{array}{l} \cot \theta - \cot 2 \theta \\ = \frac{\cos \theta}{\sin \theta} - \frac{\cos 2 \theta}{\sin 2 \theta} \\ = \frac{\cos \theta}{\sin \theta} - \frac{\cos^2 \theta - \sin^2 \theta}{2 \sin \theta \cos \theta} \\ = \frac{2 \cos^2 \theta - \cos^2 \theta + \sin^2 \theta}{2 \sin \theta \cos \theta} \\ = \frac{1}{2 \sin \theta \cos \theta} \\ = \frac{1}{\sin 2 \theta} \end{array}$ ii) S is given by this formula: $S = \frac{1}{\sin 8^\circ} + \frac{1}{\sin 16^\circ} + \cdots + \frac{1}{\sin 4096^\circ} + \frac{1}{\sin 8192^\circ}$ Determine, without a calculator, the value of $\alpha$: $S = \frac{1}{\sin \alpha}$ It’s important to know to use the formula given in part (i) to solve this problem. Other than that, it’s not too hard: $\begin{array}{l}\frac{1}{\sin 8} + \frac{1}{\sin 16} + \cdots + \frac{1}{\sin 4096} + \frac{1}{\sin 8192} \\ = \cot 4 - \cot 8 + \cot 8 - \cdots + \cot 4096 - \cot 8192 \\ = \cot 4 - \cot 8192 \\ = \cot 4 - \cot 92 \\ = \frac{\cos 4}{ \sin 4} - \frac{\cos 92}{\sin 92} \\ = \frac{\cos 4}{\sin 4} + \frac{\sin 2}{\cos 2} \\ = \frac{\cos 4 \cos 2 + \sin 4 \sin 2}{\sin 4 \cos 2} \\ = \frac{\cos (4-2)}{\sin 4 \cos 2} \\ = \frac{1}{\sin 4} \end{array}$ So the value of $\alpha$ is . b) In $\triangle ABC$, $a < \frac{1}{2} (b+c)$ (where a would be the side opposite of angle A). Prove that $\angle A < \frac{1}{2} (\angle B + \angle C)$. The Sine law states the following: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ From this we can multiply to find the relationship between sides, their angles, and an unknown constant k: $\begin{array}{l} a = k \sin A \\ b = k \sin B \\ c = k \sin C \end{array}$ We now substitute and simplify: $\begin{array}{rcl} k \sin A &<& \frac{1}{2} (k \sin B + k \sin C) \\ \sin A &<& \frac{1}{2} (\sin B + \sin C) \\ \sin A &<& \frac{1}{2} (2 \sin \frac{B+C}{2} \cos \frac{B-C}{2}) \\ \sin A &<& \sin \frac{B+C}{2} \cos \frac{B-C}{2}\end{array}$ It’s obvious that $\cos \frac{B-C}{2} \leq 1$. So if we divide both sides of the equation by $\cos \frac{B-C}{2}$, the right hand side is what we desire, and the left hand side is smaller. Thus, $\begin{array}{rcl} \sin A &<& \sin \frac{B+C}{2} \\ A &<& \frac{1}{2} (B+C) \end{array}$ This is what we set out to prove. ### Question 10 10. Let $T(n)$ be the number of triangles with integer sides and perimeter n (where n is a positive integer). For example $T(6) = 1$ because only (2,2,2) has a perimeter of 6 and nothing else. a) Determine $T(10)$, $T(11)$, and $T(12)$. Triangles with perimeter 10 are (2,4,4) and (3,3,4). So $T(10)$ is 2. Triangles with perimeter 11 are (1,5,5), (2,4,5), (3,3,5), and (3,4,4). So $T(11)$ is 4. Triangles with perimeter 12 are (2,5,5), (3,4,5), and (4,4,4). So $T(12)$ is 3. The answers are 2, 4, and 3 respectively. b) If m is a positive integer and $m \geq 3$, prove that $T(2m) = T(2m-3)$. Let a, b, and c be the sides of the triangle so that $a \leq b \leq c$. Let U be the triangle with perimeter 2m, and V be the triangle with perimeter 2m-3. In order for the triangle to be non-degenerate, $a+b > c$. Because otherwise the ‘triangle’ would be a line, or something (I’m going to refer to it as a degenerate triangle). For every triangle V, there must be at least one triangle U. If we represent the sides of V by (a,b,c), we can add one to each side giving (a+1,b+1,c+1). This triangle can’t be degenerate. So $T(2m) \geq T(2m-3)$. Now we want to prove that the converse is true: for any triangle U with side lengths (a,b,c), a triangle V with side lengths (a-1,b-1,c-1) is also non-degenerate. The only case where this might not be true is when c is exactly one more than a+b, in which case the resulting triangle would be degenerate. In this case, $a+b = c+1$. But this cannot be true. The perimeter of U has to be even, so a+b+c is even. But in the equation, if a+b is even then c is odd, and if a+b is odd then c is even. Therefore it’s impossible for a+b+c to be even and at the same time for the equation $a+b = c+1$ to be true. Thus (a-1,b-1,c-1) is never degenerate. For every triangle U, there must be at least one triangle V, so $T(2m-3) \geq T(2m)$. But we’ve already proved that $T(2m) \geq T(2m-3)$. Combined with the other inequality, $T(2m) = T(2m-3)$ which is what we need. c) Determine the smallest n where $T(n) > 2010$. The formula for $T(n)$ is fairly well known. It’s given by: $\begin{array}{rcll} T(n) &=& round(\frac{n^2}{48}) & \textrm{when n is even} \\ && round(\frac{(n+3)^2}{48}) & \textrm{when n is odd} \end{array}$ Let’s use the even one just because. Solving the inequality $\frac{n^2}{48} > 2010$, we get $n > 310.6$. Since the equation is only relevant for even integers, the first even integer greater than 310.6 is 312. But since we proved that $T(n) = T(n-3)$ for even integers n, $T(309) = T(312)$. There is no smaller n meeting the requirements, so n is 309. This solution uses an external formula and would probably not be considered correct. I don’t know how to do it without this formula. ### Notes Because I took the contest several days later than everyone else, I had a considerable and unfair advantage because some of the questions were available already. So I was able to research the questions and figure out how to solve them before writing it. Nevertheless I wasn’t able to solve all of the questions before the contest. But perhaps this unfair advantage balances my lack of experience (this contest is for grade 12 students). This blog post itself may be useful to anyone taking the euclid after today. Notice that ‘today’ in this blog post is inconsistent because I wrote this blog post over the course of two days. \frac{\cost \theta}{\sin \theta} – \frac{\cos^2 \theta – \sin^2 \theta}{2 \sin \theta \cos \theta}<\|endoftext\|>	4.5
1,188	Monarch butterflies (Danaus plexippus) are among the most widely recognized wild creatures in North America. Their distinctive orange-and-black wings, which warn predators that the butterflies are chock full of toxins from the milkweed they eat, make them easily spotted in backyard flower beds. They’re also known for a massive annual migration, flying thousands of miles between wintering colonies in central Mexico and summer sites across the United States and Canada. More recently, it’s been discovered that female monarchs infected by parasites respond by laying their eggs on food plants that can prevent the parasite from infecting their offspring. Monarchs are also one of the more visible victims of the massive changes humans have made to the world around us. Increased conversion of farmland to corn production has reduced the supply of milkweed, the butterflies’ only food plant, across much of the Midwest. It’s gotten so bad the number of monarchs making the annual migration back to Mexico hit a record low last year, and while things were better in 2014, a nationwide campaign to encourage planting of milkweed in home gardens is only beginning. For all our familiarity with monarchs, we’ve known remarkably little about their evolutionary history. That’s changing rapidly now, as evidenced by a paper published last month in the journal Nature, which uses a big new genetic dataset to trace the origins of some of the monarch’s most distinctive features. The paper’s authors, led by Shuai Zhan, and including collaborators from Shanghai to Córdoba, collected whole-genome DNA sequence data from 80 monarchs collected from across the globe: the famous migratory populations of North America, but also non-migratory monarchs in Australia, New Zealand, and Hawaii, South America, and Europe. By comparing the DNA sequences of all those butterflies to the monarch reference genome sequence, they identified some 32 million single-nucleotide polymorphism markers—single points in the genome where some butterflies carried one DNA letter, and other butterflies carried another. A dataset like that allows some truly high-precision reconstruction of evolutionary history. First, the authors compared the data from the 80 monarchs to DNA sequences from another nine butterflies of different species in the same genus, Danaus, and from the more distantly related silk moth and Heliconius butterflies. In the resulting evolutionary tree, the migratory North American population branched out from the base of the larger cluster of monarchs, which suggests that the common ancestors of all the monarch populations were also migrants. The non-North American populations also showed signs of founder effects, the loss of genetic diversity associated with colonizing new territory. The authors then scanned through the genome-wide dataset to identify regions where monarchs from the migratory population differed from the non-migratory monarchs. A single stretch of about 21,000 DNA letters emerged as the region with the strongest contrast between migrants and non-migrants, based on multiple comparison statistics—and that stretch of sequence included the genetic code for a key muscle-development protein. Muscle endurance is a big part of making the monarchs’ long migration, so it seems quite reasonable to think that these differences in DNA reflect the results of natural selection for migration. And, indeed, the authors compared the metabolic rate of migratory and non-migratory monarchs in flight and found that the migratory monarchs expended less energy doing the same amount of work. In addition to non-migratory monarchs, there are populations of the butterflies that have lost their orange coloration—they’re white with black stripes instead. These nivosus monarchs are found in Hawaii, and they provided another opportunity to unravel monarch’s genetic code by comparison. Similarly to the migration analysis, Zhan et al. scanned through genomic sequence data from several orange Hawaiian monarchs, several nivosus monarchs, and some individuals that were descended from matings between parents of different color forms. Again, a single gene emerged as having different sequences in the different color morphs—this one related to a gene that has been found to affect fur color in mice. All together, Zhan and coauthors report a really impressive volume of population genetic analysis and experimental work in this single paper—I’ve just given a brief rundown. It’s a nice example of what’s possible with really high-precision genetic data, and I hope it’s the first of many population genomic studies in the monarch. This kind of data can reveal the genetic basis of the traits and behaviors that makes monarchs so unique—but they might also help us to make sure monarchs continue their hemisphere-spanning migration for many years to come. Brower L.P. 1988. Avian predation on the monarch butterfly and its implications for mimicry theory, The American Naturalist, 131 (s1) S4. DOI: 10.1086/284763 Zhan S., J.L. Boore and S.M. Reppert. 2011. The monarch butterfly genome yields insights into long-distance migration, Cell, 147 (5) 1171-1185. DOI: 10.1016/j.cell.2011.09.052 Zhan S., K. Niitepõld, J. Hsu, J.F. Haeger, M.P. Zalucki, S. Altizer, J.C. de Roode, S.M. Reppert and M.R. Kronforst. 2014. The genetics of monarch butterfly migration and warning colouration, Nature, 514 (7522) 317-321. DOI: 10.1038/nature13812<\|endoftext\|>	3.8125
370	# GMAT Math : Complex polygons In this chapter we will learn a critical technique relating to GMAT Polygon questions Is the sum of all the interior angles of a polygon less than 14400Β ? 1. The polygon has 9, 10 or 11 sides 2. The polygon has 10, 11 or 12 sides Readers are encouraged to take some time to tackle the above question π β¦ β¦ β¦ β¦ Trick: The sum of the interior angles of ANY polygon with n sides is equal to (n β 2) x 1800 Statement (1) tells us that the polygon has 9, 10 or 11 sides β the implication on the interior angles is 12600, 14400 or 16200 β clearly not sufficient independently Statement (2) tells us that the polygon has 10, 11 or 12 sides β the implication on the interior angles is 14400 or 16200 or 18000β clearly also not sufficient independently Statement (1) and (2) tells us the interior angles can either be 14400 or 16200 β again not sufficient. We go with option (E) How do we know that the formula (n β 2) x 1800 always works β the reason is that any polygon can be decomposed into triangles as shown below. There are n triangles and each triangleβs internal angle sum measure 1800. If we subtract the center angle (around O) measuring 3600 we are left with 1800 x n β 360 = (n – 2) x 1800<\|endoftext\|>	4.65625
603	Good Language Model Remember you are your child’s language model. If your child says a word or sentence incorrectly, repeat it so they hear the correct way of producing the word/sentence. This way they hear the correct way of saying it – you are the model! Try not to make them repeat it or correct it as this tends to make them shy away from speaking. - Simple Language Try use simple language that your child understands rather than using long winded sentences and complex words. Explain any new words, especially when they ask – this is when they are curious and ready to learn. more… Infographic Source: Turner-AC Nielsen Research When kids play on the iPad or the phone, the dopamine levels in their brain increase. This affects the frontal cortex of a child’s brain pretty much the same way cocaine does. This portion of the brain controls memory, reasoning, problem-solving, and impulse control, which is why any addiction leads to a loss of these skills. Children are not born with these skills, but develop them as they grow older which is why they are more susceptible to digital addiction. Rehab clinic expert Mandy Saligari is one of the top addiction experts and warns that The introduction of utensils and the mastery of eating with it is a developmental process just like sitting and crawling. For a lot of babies, this comes naturally, but some might need a little bit of help. I am hoping to shed some light on this topic for you. Independent eating relies on abilities and develops in stages. Abilities which are needed for independent eating: - Chewing or moving food around in the mouth - Swallowing and preparing for the next bite - Full fist grip in order to finger feed and later hold a spoon - Eye-hand coordination to get the spoon in the bowl - Coordination to get the hand or spoon to the mouth - Pincer grip for isolated finger feeding and holding a fork The different stages of eating independently with utensils is as follows: - Ability to maintain an upright sitting posture during meal time - Tolerating textures and eating solid foods while being fed - Ability to grasp utensils and use them appropriately during play - Self-feeding with utensil while parent feeds most of the meal (spoon) - Self-feeding with parent doing hand-over-hand guiding (spoon) - Independent self-feeding of “sticky foods” like yogurts (spoon) - Independent self-feeding of most foods (spoon) - Self-feeding with utensils while parent feeds most of the meal (fork) - Self-feeding with parent doing hand-over-hand guiding (fork) - Independent self-feeding of “sticky foods” like yogurts (fork) - Independent self-feeding of most foods (fork)<\|endoftext\|>	3.921875
419	What is X-ray Fluorescence? X-ray fluorescence (XRF) is an analytical technique that can be used to determine the chemical composition of a wide variety of sample types including solids, liquids, slurries and loose powders. X-ray fluorescence is also used to determine the thickness and composition of layers and coatings. It can analyze elements from beryllium (Be) to uranium (U) in concentration ranges from 100 wt% to sub-ppm levels. What are the benefits of XRF analysis? XRF analysis is a robust technique, combining high precision and accuracy with straightforward, fast sample preparation. It can be readily automated for use in high-throughput industrial environments, plus XRF provides both qualitative and quantitative information on a sample. Easy combination of this ‘what?’ and ‘how much?’ information also makes rapid screening (semi-quantitative) analysis possible. Principles behind XRF XRF is an atomic emission method, similar in this respect to optical emission spectroscopy (OES), ICP and neutron activation analysis (gamma spectroscopy). Such methods measure the wavelength and intensity of ‘light’ (X-rays in this case) emitted by energized atoms in the sample. In XRF, irradiation by a primary X-ray beam from an X-ray tube, causes emission of fluorescent X-rays with discrete energies characteristic of the elements present in the sample. Determination of elemental composition The technology used for the separation (dispersion), identification and intensity measurement of a sample’s X-ray fluorescence spectrum gives rise to two main types of spectrometer: wavelength dispersive (WDXRF) and energy dispersive (EDXRF) systems. We offer a wide range of X-ray Fluorescence solutions for the analysis of elemental composition of a wide range of materials and applications, and comprising both Wavelength and Energy Dispersive solutions. Discover our solution portfolio in the table below:<\|endoftext\|>	3.84375
3,945	Posted on 24 Jun 2021 ## The Basics of GMAT Combinatorics By: Apex GMAT Contributor: Svetozara Saykova Date: 24th June 2021 Combinatorics can seem like one of the most difficult types of questions to come across on the GMAT. Luckily there are not many of them within the exam. Still these questions make up the top level of scoring on the test and therefore it is best if you are well equipped to solve them successfully, especially if you are aiming for a 700+ score. The most important rule to follow when considering this question type is the “Fundamental Counting Principle” also known as the “Counting Rule.” This rule is used to calculate the total number of outcomes given by a probability problem. The most basic rule in Combinatorics is “The Fundamental Counting Principle”. It states that for any given situation the number of overall outcomes is equal to the product of the number of each discrete outcome. Let’s say you have 4 dresses and 3 pairs of shoes, this would mean that you have 3 x 4 = 12 outfits. The Fundamental Counting Principle also applies for more than 2 options. For example, you are at the ice cream shop and you have a variety of 5 flavors, 3 types of cones and 4 choices for toppings. That means you have 5 x 3 x 4 = 60 different combinations of single-scoop ice creams. The Fundamental Counting Principle applies only for choices that are independent of one another. Meaning that any option can be paired with any other option and there are no exceptions. Going back to the example, there is no policy against putting sprinkles on strawberry vanilla ice cream because it is superb on its own. If there were, that would mean that this basic principle of Combinatorics would not apply because the combinations (outcomes) are dependent. You could still resort to a reasoning solution path or even a graphical solution path since the numbers are not so high. Let’s Level Up a Notch The next topic in Combinatorics is essential to a proper GMAT prep is permutations. A permutation is a possible order in which you put a set of objects. Permutations There are two subtypes of permutations and they are determined by whether repetition is allowed or not. • Permutations with repetition allowed When there are n options and r number of slots to fill, we have n x n x …. (r times) = nr permutations. In other words, there are n possibilities for the first slot, n possibilities for the second and so on and so forth up until n possibilities for position number r. The essential mathematical knowledge for these types of questions is that of exponents To exemplify this let’s take your high school locker. You probably had to memorize a 3 digit combination in order to unlock it. So you have 10 options (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) for 3 available slots. The total number of locker passwords you can have is 103 = 1,000. • Permutation without repetition allowed When repetition is restricted in the given GMAT problem, we would have to reduce the number of available choices for each position. Let’s take the previous example and add a restriction to the password options – you cannot have repeating numbers in your locker password. Following the “we reduce the options available each time we move to the next slot” rule, we get 10x9x8 = 720 options for a locker combination (or mathematically speaking permutation). To be more mathematically precise and derive a formula we use the factorial function (n!). In our case we will take all the possible options 10! for if we had 10 positions available and divide them by 7!, which are the slots we do not have. 10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 And when we divide them (7 x 6 x 5 x 4 x 3 x 2 x 1) cancels and we are left with 10 x 9 x 8 = 720. Pro tip: Taking problems and deeply examining them by running different scenarios, and changing some of the conditions or numbers is a great way to train for the GMAT. This technique will allow you to not only deeply understand the problem but also the idea behind it, and make you alert for what language and piece of information stands for which particular concept. So those are the fundamentals, folks. Learning to recognize whether order matters and whether repetition is allowed is essential when it comes to Combinatorics on the GMAT. Another vital point is that if you end up with an endless equation which confuses you more than helps, remember doing math on the GMAT Quant section is not the most efficient tactic. In fact, most of the time visualizing the data by putting it into a graph or running a scenario following your reasoning are far more efficient solution paths. Feeling confident and want to test you GMAT Combinatorics skills? Check out this GMAT problem and try solving it. Let us know how it goes! Posted on 04 Mar 2021 ## Combinations with Restrictions By: Rich Zwelling, Apex GMAT Instructor Date: 4th March, 2021 In our previous post, we discussed how GMAT combinatorics problems can involve subtracting out restrictions. However, we discussed only PERMUTATIONS and not COMBINATIONS. Today, we’ll take a look at how the same technique can be applied to COMBINATION problems. This may be a bit more complicated, as you’ll have to use the formula for combinations, but the approach will be the same. Let’s start with a basic example. Suppose I were to give you the following problem: The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from. How many potential committees are possible? This is a straightforward combination problem. (And we know it’s a COMBINATION situation because we do not care about the order in which the three people appear. Even if we shift the order, the same three people will still comprise the same committee.) We would simply use the combination math discussed in our Intro to Combination Math post: 10! 10C3 = ————- 3! (10-3!) 10! ——— 3! (7!) 1098 ——— 3! 1098 ——— 321 = 120 Combinations However, what if we shifted the problem slightly to look like the following? (As always, give the problem a shot before reading on…): The board of a large oil company is tasked with selecting a committee of three people to head a certain project for the following year. It has a list of ten applicants to choose from, three of whom are women and the remainder of whom are men. How many potential committees are possible if the committee must contain at least one woman? A) 60 B) 75 C) 85 D) 90 E) 95 In this case, there’s a very important SIGNAL. The language “at least one” is a huge giveaway. This means there could be 1 woman, 2 women, or 3 women which means we would have to examine three separate cases. That’s a lot of busy work. But as we discussed in the previous post, why not instead look at what we don’t want and subtract it from the total? In this case, that would be the case of 0 women. Then, we could subtract that from the total number of combinations without restrictions. This would leave behind the cases we do want (i.e. all the cases involving at least one woman). We already discussed what happens without restrictions: There are 10 people to choose from, and we’re selecting a subgroup of 3 people, leading to 10C3 or 120 combinations possible. But how do we consider the combinations we don’t want? Well, we want to eliminate every combination that involves 0 women. In other words, we want to eliminate every possible committee of three people that involves all men. So how do we find that? Well, there are seven men to choose from, and since we are choosing a subgroup of 3, we can simply use 7C3 to find the number of committees involving all men: 7! 7C3 = ————- 3! (7-3!) 7! ——— 3! (4!) 765 ——— 3! = 75 = 35 Combinations involving all men So, out of the 120 committees available, 35 of them involve all men. That means 120-35 = 85 involve at least one woman. The correct answer is C. Next time, we’ll return to probability and talk about how the principle of subtracting out elements that we don’t want can aid us on certain questions. Then we’ll dovetail the two and talk about how probability and combinatorics can show up simultaneously on certain questions. Posted on 18 Feb 2021 ## Review of Example From Last Post Last time, when we started our discussion of GMAT Combinatorics, we gave a brief example of GMAT permutations in which we had five paintings and asked how many arrangements could be made on a wall with those paintings. As it turns out, no complicated combinatorics formula is necessary. You can create an easy graph with dashes and list five options for the first slot, leaving four for the second slot, and so on: _5_ _4_ _3_ _2_ _1_ Then multiply 54321 to get 120 arrangements of the five paintings. Remember you could see this notationally as 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!) ## More Permutation Math But there could be fewer slots then items. Take the following combinatorics practice problem: At a cheese tasting, a chef is to present some of his best creations to the event’s head judge. Due to the event’s very bizarre restrictions, he must present exactly three or four cheeses. He has brought his best cheddar, brie, gouda, roquefort, gruyere, and camembert. How many potential orderings of cheeses can the chef create to present to the judge? A) 120 B) 240 C) 360 D) 480 E) 600 First, as a review, how do we know this is a PERMUTATION and not a COMBINATION? Because order matters. In the previous problem, the word “arrangements” gave away that we care about the order in which items appear. In this problem, we’re told that we’re interested in the “orderings” of cheeses. Cheddar followed by gouda would be considered distinct from gouda followed by cheddar. (Look for signal words like “arrangements” or “orderings” to indicate a PERMUTATION problem.) In this case, we must consider the options of three or four cheeses separately, as they are independent (i.e. they cannot both happen). But for each case, the process is actually no different from what we discussed last time. We can simply consider each case separately and create dashes (slots) for each option. In the first case (three cheeses), there are six options for the first slot, five for the second, and four for the third: _6_ _5_ _4_ We multiply those together to give us 654 = 120 possible ways to present three cheeses. We do likewise for the four-cheese case: _6_ _5_ _4_ _3_ We multiply those together to give us 6543 = 360 possible ways to present four cheeses. Since these two situations (three cheeses and four cheeses) are independent, we simply add them up to get a final answer of 120+360 = 480 possible orderings of cheeses, and the correct answer is D. You might have also noticed that there’s a sneaky arithmetic shortcut. You’ll notice that you have to add 654 + 6543. Instead of multiplying each case separately, you can factor out 654 from the sum, as follows: 654 + 6543 = 654 ( 1 + 3) = 6544 = 3016 OR 2024 = 480 Develop the habit of looking for quick, efficient ways of doing basic arithmetic to bank time. It will pay off when you have to do more difficult questions in the latter part of the test. Now that we have been through GMAT permutations, next time, I’ll give this problem a little twist and show you how to make it a COMBINATION problem. Until then… By: Rich Zwelling, Apex GMAT Instructor Date: 16th February, 2021 Posted on 11 Feb 2021 ## Introduction to Combinatorics & Permutations GMAT Combinatorics. It’s a phrase that’s stricken fear in the hearts of many of my students. And it makes sense because so few of us are taught anything about it growing up. But the good news is that, despite the scary title, what you need to know for GMAT combinatorics problems is actually not terribly complex. To start, let’s look at one of the most commonly asked questions related to GMAT combinatorics, namely the difference between combinatorics and permutations ## Does Order Matter? It’s important to understand conceptually what makes permutations and combinations differ from one another. Quite simply, it’s whether we care about the order of the elements involved. Let’s look at these concrete examples to make things a little clearer: ## Permutations Example Suppose we have five paintings to hang on a wall, and we want to know in how many different ways we can arrange the paintings. It’s the word “arrange” that often gives away that we care about the order in which the paintings appear. Let’s call the paintings A, B, C, D, and E: ABCDE ACDEB BDCEA Each of the above three is considered distinct in this problem, because the order, and thus the arrangement, changes. This is what defines this situation as a PERMUTATION problem. Mathematically, how would we answer this question? Well, quite simply, we would consider the number of options we have for each “slot” on the wall. We have five options at the start for the first slot: _5_ ___ ___ ___ ___ After that painting is in place, there are four remaining that are available for the next slot: _5_ _4_ ___ ___ ___ From there, the pattern continues until all slots are filled: _5_ _4_ _3_ _2_ _1_ The final step is to simply multiply these numbers to get 54321 = 120 arrangements of the five paintings. The quantity 54321 is also often represented by the exclamation point notation 5!, or 5 factorial. (It’s helpful to memorize factorials up to 6!) ## Combinations Example So, what about COMBINATIONS? Obviously if we care about order for permutations, that implies we do NOT care about order for combinations. But what does such a situation look like? Suppose there’s a local food competition, and I’m told that a group of judges will taste 50 dishes at the competition. A first, a second, and a third prize will be given to the top three dishes, which will then have the honor of competing at the state competition in a few months. I want to know how many possible groups of three dishes out of the original 50 could potentially be selected by the judges to move on to the state competition. The math here is a little more complicated without a combinatorics formula, but we’re just going to focus on the conceptual element for the moment. How do we know this is a COMBINATION situation instead of a permutation question? It’s a little tricky, because at first glance, you might consider the first, second, and third prizes and believe that order matters. Suppose that Dish A wins first prize, Dish B wins second prize, and Dish C wins third prize. Call that ABC. Isn’t that a distinct situation from BAC? Or CAB? Well, that’s where you have to pay very close attention to exactly what the question asks. If we were asking about distinct arrangements of prize winnings, then yes, this would be a permutation question, and we would have to consider ABC apart from BAC apart from CAB, etc. However, what does the question ask about specifically? It asks about which dishes advance to the state competition. Also notice that the question specifically uses the word “group,” which is often a huge signal for combinations questions. This implies that the total is more important than the individual parts. If we take ABC and switch it to BAC or BCA or ACB, do we end up with a different group of three dishes that advances to the state competition? No. It’s the same COMBINATION of dishes. ## Quantitative Connection It’s interesting to note that there will always be fewer combinations than permutations, given a common set of elements. Why? Let’s use the above simple scenario of three elements as an illustration and write out all the possible permutations of ABC. It’s straightforward enough to brute-force this by including two each starting with A, two each starting with B, etc: ABC ACB BAC BCA CAB CBA But you could also see that there are 32*1 = 3! = 6 permutations by using the same method we used for the painting example above. Now, how many combinations does this constitute? Notice they all consist of the same group of three letters, and thus this is actually just one combination. We had to divide the original 6 permutations by 3! to get the correct number of permutations. Next time, we’ll continue our discussion of permutation math and begin a discussion of the mechanics of combination math. Contributor: Rich Zwelling, Apex GMAT Instructor<\|endoftext\|>	4.5
7,339	# 2.2.1: Discrete Random Variables $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ ## Probability Mass Functions In this chapter, we develop mathematical tools for describing and analyzing random experiments, experiments whose outcome cannot be determined with certainty. A coin flip and a roll of a die are classical examples of such experiments. The outcome of a random experiment is described by a random variable $$X$$ that takes on a finite number of values, $x_{1}, \ldots, x_{J} \text {, }$ where $$J$$ is the number of values that $$X$$ takes. To fully characterize the behavior of the random variable, we assign a probability to each of these events, i.e. $X=x_{j}, \quad \text { with probability } p_{j}, \quad j=1, \ldots, J .$ The same information can be expressed in terms of the probability mass function (pmf), or discrete density function, $$f_{X}$$, that assigns a probability to each possible outcome $f_{X}\left(x_{j}\right)=p_{j}, \quad j=1, \ldots, J .$ In order for $$f_{X}$$ to be a valid probability density function, $$\left\{p_{j}\right\}$$ must satisfy \begin{aligned} &0 \leq p_{j} \leq 1, \quad j=1, \ldots, J, \\ &\sum_{j=1}^{J} p_{j}=1 \end{aligned} The first condition requires that the probability of each event be non-negative and be less than or equal to unity. The second condition states that $$\left\{x_{1}, \ldots, x_{J}\right\}$$ includes the set of all possible values that $$X$$ can take, and that the sum of the probabilities of the outcome is unity. The second condition follows from the fact that events $$x_{i}, i=1, \ldots, J$$, are mutually exclusive and collectively exhaustive. Mutually exclusive means that $$X$$ cannot take on two different values of the $$x_{i}$$ ’s in any given experiment. Collectively exhaustive means that $$X$$ must take on one of the $$J$$ possible values in any given experiment. Note that for the same random phenomenon, we can choose many different outcomes; i.e. we can characterize the phenomenon in many different ways. For example, $$x_{j}=j$$ could simply be a label for the $$j$$-th outcome; or $$x_{j}$$ could be a numerical value related to some attribute of the phenomenon. For instance, in the case of flipping a coin, we could associate a numerical value of 1 with heads and 0 with tails. Of course, if we wish, we could instead associate a numerical value of 0 with heads and 1 with tails to describe the same experiment. We will see examples of many different random variables in this unit. The key point is that the association of a numerical value to an outcome allows us to introduce meaningful quantitative characterizations of the random phenomenon - which is of course very important in the engineering context. Let us define a few notions useful for characterizing the behavior of the random variable. The expectation of $$X, E[X]$$, is defined as $E[X]=\sum_{j=1}^{J} x_{j} p_{j} .$ The expectation of $$X$$ is also called the mean. We denote the mean by $$\mu$$ or $$\mu_{X}$$, with the second notation emphasizing that it is the mean of $$X$$. Note that the mean is a weighted average of the values taken by $$X$$, where each weight is specified according to the respective probability. This is analogous to the concept of moment in mechanics, where the distances are provided by $$x_{j}$$ and the weights are provided by $$p_{j}$$; for this reason, the mean is also called the first moment. The mean corresponds to the centroid in mechanics. Note that, in frequentist terms, the mean may be expressed as the sum of values taken by $$X$$ over a large number of realizations divided by the number of realizations, i.e. $(\text { Mean })=\lim _{\text {(# Realizations) } \rightarrow \infty} \frac{1}{\text { (# Realizations) }} \sum_{j=1}^{J} x_{j} \cdot\left(\# \text { Occurrences of } x_{j}\right) .$ Recalling that the probability of a given event is defined as $p_{j}=\lim _{(\# \text { Realizations) } \rightarrow \infty} \frac{\left(\# \text { Occurrences of } x_{j}\right)}{\text { (# Realizations) }},$ we observe that $E[X]=\sum_{j=1}^{J} x_{j} p_{j}$ which is consistent with the definition provided in Eq. (9.1). Let us provide a simple gambling scenario to clarity this frequentist interpretation of the mean. Here, we consider a "game of chance" that has $$J$$ outcomes with corresponding probabilities $$p_{j}, j=1, \ldots, J$$; we denote by $$x_{j}$$ the (net) pay-off for outcome $$j$$. Then, in $$n_{\text {plays }}$$ plays of the game, our (net) income would be $\sum_{j=1}^{J} x_{j} \cdot \text { (# Occurrences of } x_{j} \text { ), }$ which in the limit of large $$n_{\text {plays }}\left(=\right.$$ # Realizations) yields $$n_{\text {plays }} \cdot E[X]$$. In other words, the mean $$E[X]$$ is the expected pay-off per play of the game, which agrees with our intuitive sense of the mean. The variance, or the second moment about the mean, measures the spread of the values about the mean and is defined by $\operatorname{Var}[X] \equiv E\left[(X-\mu)^{2}\right]=\sum_{j=1}^{J}\left(x_{j}-\mu\right)^{2} p_{j}$ We denote the variance as $$\sigma^{2}$$. The variance can also be expressed as \begin{aligned} \operatorname{Var}[X] &=E\left[(X-\mu)^{2}\right]=\sum_{j=1}^{J}\left(x_{j}-\mu\right)^{2} p_{j}=\sum_{j=1}^{J}\left(x_{j}^{2}-2 x_{j} \mu+\mu^{2}\right) p_{j} \\ &=\underbrace{\sum_{j=1}^{J} x_{j}^{2} p_{j}}_{E\left[X^{2}\right]}-2 \mu \underbrace{\sum_{j=1}^{J} x_{j} p_{j}}_{\mu}+\mu^{\mu^{2} \sum_{j=1}^{J} p_{j}}=E\left[X^{2}\right]-\mu^{2} . \end{aligned} Note that the variance has the unit of $$X$$ squared. Another useful measure of the expected spread of the random variable is standard deviation, $$\sigma$$, which is defined by $\sigma=\sqrt{\operatorname{Var}[X]} .$ We typically expect departures from the mean of many standard deviations to be rare. This is particularly the case for random variables with large range, i.e. $$J$$ large. (For discrete random variables with small $$J$$, this spread interpretation is sometimes not obvious simply because the range of $$X$$ is small.) In case of the aforementioned "game of chance" gambling scenario, the standard deviation measures the likely (or expected) deviation in the pay-off from the expectation (i.e. the mean). Thus, the standard deviation can be related in various ways to risk; high standard deviation implies a high-risk case with high probability of large payoff (or loss). The mean and variance (or standard deviation) provide a convenient way of characterizing the behavior of a probability mass function. In some cases the mean and variance (or even the mean alone) can serve as parameters which completely determine a particular mass function. In many other cases, these two properties may not suffice to completely determine the distribution but can still serve as useful measures from which to make further deductions. Let us consider a few examples of discrete random variables. ## Example 9.1.1 rolling a die As the first example, let us apply the aforementioned framework to rolling of a die. The random variable $$X$$ describes the outcome of rolling a (fair) six-sided die. It takes on one of six possible values, $$1,2, \ldots, 6$$. These events are mutually exclusive, because a die cannot take on two different values at the same time. Also, the events are exhaustive because the die must take on one of the six values after each roll. Thus, the random variable $$X$$ takes on one of the six possible values, $x_{1}=1, x_{2}=2, \ldots, x_{6}=6 .$ A fair die has the equal probability of producing one of the six outcomes, i.e. $X=x_{j}=j, \quad \text { with probability } \frac{1}{6}, \quad j=1, \ldots, 6,$ or, in terms of the probability mass function, $f_{X}(x)=\frac{1}{6}, \quad x=1, \ldots, 6 .$ (a) realization (b) pmf Figure 9.1: Illustration of the values taken by a fair six-sided die and the probability mass function. An example of outcome of a hundred die rolls is shown in Figure 9.1(a). The die always takes on one of the six values, and there is no obvious inclination toward one value or the other. This is consistent with the fact that any one of the six values is equally likely. (In practice, we would like to think of Figure 9.1(a) as observed outcomes of actual die rolls, i.e. data, though for convenience here we use synthetic data through random number generation (which we shall discuss subsequently).) Figure 9.1(b) shows the probability mass function, $$f_{X}$$, of the six equally likely events. The figure also shows the relative frequency of each event - which is defined as the number of occurrences of the event normalized by the total number of samples (which is 100 for this case) - as a histogram. Even for a relatively small sample size of 100 , the histogram roughly matches the probability mass function. We can imagine that as the number of samples increases, the relative frequency of each event gets closer and closer to its value of probability mass function. Conversely, if we have an experiment with an unknown probability distribution, we could infer its probability distribution through a large number of trials. Namely, we can construct a histogram, like the one shown in Figure 9.1(b), and then construct a probability mass function that fits the histogram. This procedure is consistent with the frequentist interpretation of probability: the probability of an event is the relative frequency of its occurrence in a large number of samples. The inference of the underlying probability distribution from a limited amount of data (i.e. a small sample) is an important problem often encountered in engineering practice. Let us now characterize the probability mass function in terms of the mean and variance. The mean of the distribution is $\mu=E[X]=\sum_{j=1}^{6} x_{j} p_{j}=\sum_{j=1}^{6} j \cdot \frac{1}{6}=\frac{7}{2}$ The variance of the distribution is $\sigma^{2}=\operatorname{Var}[X]=E\left[X^{2}\right]-\mu^{2} \sum_{j=1}^{6} x_{j}^{2} p_{j}-\mu^{2}=\sum_{j=1}^{6} j^{2} \cdot \frac{1}{6}-\left(\frac{7}{2}\right)^{2}=\frac{91}{6}-\frac{49}{4}=\frac{35}{12} \approx 2.9167,$ and the standard deviation is $\sigma=\sqrt{\operatorname{Var}[X]}=\sqrt{\frac{35}{12}} \approx 1.7078$ ## Example 9.1.2 (discrete) uniform distribution The outcome of rolling a (fair) die is a special case of a more general distribution, called the (discrete) uniform distribution. The uniform distribution is characterized by each event having the equal probability. It is described by two integer parameters, $$a$$ and $$b$$, which assign the lower and upper bounds of the sample space, respectively. The distribution takes on $$J=b-a+1$$ values. For the six-sided die, we have $$a=1, b=6$$, and $$J=b-a+1=6$$. In general, we have \begin{aligned} x_{j} &=a+j-1, \quad j=1, \ldots, J, \\ f^{\text {disc.uniform }}(x) &=\frac{1}{J} . \end{aligned} The mean and variance of a (discrete) uniform distribution are given by $\mu=\frac{a+b}{2} \text { and } \quad \sigma^{2}=\frac{J^{2}-1}{12} .$ We can easily verify that the expressions are consistent with the die rolling case with $$a=1, b=6$$, and $$J=6$$, which result in $$\mu=7 / 2$$ and $$\sigma^{2}=35 / 12$$. ##### Proof The mean follows from \begin{aligned} \mu &=E[X]=E[X-(a-1)+(a-1)]=E[X-(a-1)]+a-1 \\ &=\sum_{j=1}^{J}\left(x_{j}-(a-1)\right) p_{j}+a-1=\sum_{j=1}^{J} j \frac{1}{J}+a-1=\frac{1}{J} \frac{J(J+1)}{2}+a-1 \\ &=\frac{b-a+1+1}{2}+a-1=\frac{b+a}{2} . \end{aligned} The variance follows from \begin{aligned} \sigma^{2} &=\operatorname{Var}[X]=E\left[(X-E[X])^{2}\right]=E\left[((X-(a-1))-E[X-(a-1)])^{2}\right] \\ &=E\left[(X-(a-1))^{2}\right]-E[X-(a-1)]^{2} \\ &=\sum_{j=1}^{J}\left(x_{j}-(a-1)\right)^{2} p_{j}-\left[\sum_{j=1}^{J}\left(x_{j}-(a-1)\right) p_{j}\right]^{2} \\ &=\sum_{j=1}^{J} j^{2} \frac{1}{J}-\left[\sum_{j=1}^{J} j p_{j}\right]^{2}=\frac{1}{J} \frac{J(J+1)(2 J+1)}{6}-\left[\frac{1}{J} \frac{J(J+1)}{2}\right]^{2} \\ &=\frac{J^{2}-1}{12}=\frac{(b-a+1)^{2}-1}{12} . \end{aligned} ## Example 9.1.3 Bernoulli distribution (a coin flip) Consider a classical random experiment of flipping a coin. The outcome of a coin flip is either a head or a tail, and each outcome is equally likely assuming the coin is fair (i.e. unbiased). Without loss of generality, we can associate the value of 1 (success) with head and the value of 0 (failure) with tail. In fact, the coin flip is an example of a Bernoulli experiment, whose outcome takes on either 0 or 1 . Specifically, a Bernoulli random variable, $$X$$, takes on two values, 0 and 1 , i.e. $$J=2$$, and $x_{1}=0 \quad \text { and } \quad x_{2}=1 .$ The probability mass function is parametrized by a single parameter, $$\theta \in[0,1]$$, and is given by $f_{X_{\theta}}(x)=f^{\text {Bernoulli }}(x ; \theta) \equiv\left\{\begin{array}{l} 1-\theta, \quad x=0 \\ \theta, \quad x=1 \end{array}\right.$ In other words, $$\theta$$ is the probability that the random variable $$X_{\theta}$$ takes on the value of 1 . Flipping of a fair coin is a particular case of a Bernoulli experiment with $$\theta=1 / 2$$. The $$\theta=1 / 2$$ case is also a special case of the discrete uniform distribution with $$a=0$$ and $$b=1$$, which results in $$J=2$$. Note that, in our notation, $$f_{X_{\theta}}$$ is the probability mass function associated with a particular random variable $$X_{\theta}$$, whereas $$f^{\text {Bernoulli }}(\cdot ; \theta)$$ is a family of distributions that describe Bernoulli random variables. For notational simplicity, we will not explicitly state the parameter dependence of $$X_{\theta}$$ on $$\theta$$ from hereon, unless the explicit clarification is necessary, i.e. we will simply use $$X$$ for the random variable and $$f_{X}$$ for its probability mass function. (Also note that what we call a random variable is of course our choice, and, in the subsequent sections, we often use variable $$B$$, instead of $$X$$, for a Bernoulli random variable.) Examples of the values taken by Bernoulli random variables with $$\theta=1 / 2$$ and $$\theta=1 / 4$$ are shown in Figure 9.2. As expected, with $$\theta=1 / 2$$, the random variable takes on the value of 0 and 1 roughly equal number of times. On the other hand, $$\theta=1 / 4$$ results in the random variable taking on 0 more frequently than 1 . The probability mass functions, shown in Figure $$9.2$$, reflect the fact that $$\theta=1 / 4$$ results in $$X$$ taking on 0 three times more frequently than 1 . Even with just 100 samples, the relative frequency histograms captures the difference in the frequency of the events for $$\theta=1 / 2$$ and $$\theta=1 / 4$$. In fact, even if we did not know the underlying pmf - characterized by $$\theta$$ in this case - we can infer from the sampled data that the second case has a lower probability of success (i.e. $$x=1$$ ) than the first case. In the subsequent chapters, we will formalize this notion of inferring the underlying distribution from samples and present a method for performing the task. The mean and variance of the Bernoulli distribution are given by $E[X]=\theta \quad \text { and } \quad \operatorname{Var}[X]=\theta(1-\theta) .$ Note that lower $$\theta$$ results in a lower mean, because the distribution is more likely to take on the value of 0 than 1 . Note also that the variance is small for either $$\theta \rightarrow 0$$ or $$\theta \rightarrow 1$$ as in these cases we are almost sure to get one or the other outcome. But note that (say) $$\sigma / E(X)$$ scales as $$1 / \sqrt{(\theta)}$$ (recall $$\sigma$$ is the standard deviation) and hence the relative variation in $$X$$ becomes more pronounced for small $$\theta$$ : this will have important consequences in our ability to predict rare events. (a) realization, $$\theta=1 / 2$$ (b) $$\mathrm{pmf}, \theta=1 / 2$$ (c) realization, $$\theta=1 / 4$$ (d) $$\operatorname{pmf}, \theta=1 / 4$$ Figure 9.2: Illustration of the values taken by Bernoulli random variables and the probability mass functions. ##### Proof Proof of the mean and variance follows directly from the definitions. The mean is given by $\mu=E[X]=\sum_{j=1}^{J} x_{j} p_{j}=0 \cdot(1-\theta)+1 \cdot \theta=\theta .$ The variance is given by $\operatorname{Var}[X]=E\left[(X-\mu)^{2}\right]=\sum_{j=1}^{J}\left(x_{j}-\mu\right)^{2} p_{j}=(0-\theta)^{2} \cdot(1-\theta)+(1-\theta)^{2} \cdot \theta=\theta(1-\theta) .$ Before concluding this subsection, let us briefly discuss the concept of "events." We can define an event of $$A$$ or $$B$$ as the random variable $$X$$ taking on one of some set mutually exclusive outcomes $$x_{j}$$ in either the set $$A$$ or the set $$B$$. Then, we have $P(A \text { or } B)=P(A \cup B)=P(A)+P(B)-P(A \cap B) .$ That is, probability of event $$A$$ or $$B$$ taking place is equal to double counting the outcomes $$x_{j}$$ in both $$A$$ and $$B$$ and then subtracting out the outcomes in both $$A$$ and $$B$$ to correct for this double counting. Note that, if $$A$$ and $$B$$ are mutually exclusive, we have $$A \cap B=\emptyset$$ and $$P(A \cap B)=0$$. Thus the probability of $$A$$ or $$B$$ is $P(A \text { or } B)=P(A)+P(B), \quad(A \text { and } B \text { mutually exclusive }) .$ This agrees with our intuition that if $$A$$ and $$B$$ are mutually exclusive, we would not double count outcomes and thus would not need to correct for it. ## Transformation Random variables, just like deterministic variables, can be transformed by a function. For example, if $$X$$ is a random variable and $$g$$ is a function, then a transformation $Y=g(X)$ produces another random variable $$Y$$. Recall that we described the behavior of $$X$$ that takes on one of $$J$$ values by $X=x_{j} \text { with probability } p_{j}, \quad j=1, \ldots, J .$ The associated probability mass function was $$f_{X}\left(x_{j}\right)=p_{j}, j=1, \ldots, J$$. The transformation $$Y=g(X)$$ yields the set of outcomes $$y_{j}, j=1, \ldots, J$$, where each $$y_{j}$$ results from applying $$g$$ to $$x_{j}$$, i.e. $y_{j}=g\left(x_{j}\right), \quad j=1, \ldots, J .$ Thus, $$Y$$ can be described by $Y=y_{j}=g\left(x_{j}\right) \quad \text { with probability } p_{j}, \quad j=1, \ldots, J .$ We can write the probability mass function of $$Y$$ as $f_{Y}\left(y_{j}\right)=f_{Y}\left(g\left(x_{j}\right)\right)=p_{j} \quad j=1, \ldots, J .$ We can express the mean of the transformed variable in a few different ways: $E[Y]=\sum_{j=1}^{J} y_{j} f_{Y}\left(y_{j}\right)=\sum_{j=1}^{J} y_{j} p_{j}=\sum_{j=1}^{J} g\left(x_{j}\right) f_{X}\left(x_{j}\right) .$ The first expression expresses the mean in terms of $$Y$$ only, whereas the final expression expresses $$E[Y]$$ in terms of $$X$$ and $$g$$ without making a direct reference to $$Y$$. Let us consider a specific example. ## Example 9.1.4 from rolling a die to flipping a coin Let us say that we want to create a random experiment with equal probability of success and failure (e.g. deciding who goes first in a football game), but all you have is a die instead of a coin. One way to create a Bernoulli random experiment is to roll the die, and assign "success" if an odd number is rolled and assign "failure" if an even number is rolled. Let us write out the process more formally. We start with a (discrete) uniform random variable $$X$$ that takes on $x_{j}=j, \quad j=1, \ldots, 6,$ with probability $$p_{j}=1 / 6, j=1, \ldots, 6$$. Equivalently, the probability density function for $$X$$ is $f_{X}(x)=\frac{1}{6}, \quad x=1,2, \ldots, 6 .$ Consider a function $g(x)= \begin{cases}0, & x \in\{1,3,5\} \\ 1, & x \in\{2,4,6\}\end{cases}$ Let us consider a random variable $$Y=g(X)$$. Mapping the outcomes of $$X, x_{1}, \ldots, x_{6}$$, to $$y_{1}^{\prime}, \ldots, y_{6}^{\prime}$$, we have \begin{aligned} &y_{1}^{\prime}=g\left(x_{1}\right)=g(1)=0, \\ &y_{2}^{\prime}=g\left(x_{2}\right)=g(2)=1, \\ &y_{3}^{\prime}=g\left(x_{3}\right)=g(3)=0, \\ &y_{4}^{\prime}=g\left(x_{4}\right)=g(4)=1, \\ &y_{5}^{\prime}=g\left(x_{5}\right)=g(5)=0, \\ &y_{6}^{\prime}=g\left(x_{6}\right)=g(6)=1 . \end{aligned} We could thus describe the transformed variable $$Y$$ as $Y=y_{j}^{\prime} \quad \text { with probability } p_{j}=1 / 6, \quad j=1, \ldots, 6 .$ However, because $$y_{1}^{\prime}=y_{3}^{\prime}=y_{5}^{\prime}$$ and $$y_{2}^{\prime}=y_{4}^{\prime}=y_{6}^{\prime}$$, we can simplify the expression. Without loss of generality, let us set $y_{1}=y_{1}^{\prime}=y_{3}^{\prime}=y_{5}^{\prime}=0 \quad \text { and } \quad y_{2}=y_{2}^{\prime}=y_{4}^{\prime}=y_{6}^{\prime}=1 .$ We now combine the frequentist interpretation of probability with the fact that $$x_{1}, \ldots, x_{6}$$ are mutually exclusive. Recall that to a frequentist, $$P\left(Y=y_{1}=0\right)$$ is the probability that $$Y$$ takes on 0 in a large number of trials. In order for $$Y$$ to take on 0 , we must have $$x=1,3$$, or 5 . Because $$X$$ taking on 1, 3, and 5 are mutually exclusive events (e.g. $$X$$ cannot take on 1 and 3 at the same time), the number of occurrences of $$y=0$$ is equal to the sum of the number of occurrences of $$x=1, x=3$$, and $$x=5$$. Thus, the relative frequency of $$Y$$ taking on $$0-$$ or its probability - is equal to the sum of the relative frequencies of $$X$$ taking on 1,3 , or 5 . Mathematically, $P\left(Y=y_{1}=0\right)=P(X=1)+P(X=3)+P(X=5)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2} .$ Similarly, because $$X$$ taking on 2,4 , and 6 are mutually exclusive events, $P\left(Y=y_{2}=1\right)=P(X=2)+P(X=4)+P(X=6)=\frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2} .$ Thus, we have $Y= \begin{cases}0, & \text { with probability } 1 / 2 \\ 1, & \text { with probability } 1 / 2,\end{cases}$ or, in terms of the probability density function, $f_{Y}(y)=\frac{1}{2}, \quad y=0,1 .$ Note that we have transformed the uniform random variable $$X$$ by the function $$g$$ to create a Bernoulli random variable $$Y$$. We emphasize that the mutually exclusive property of $$x_{1}, \ldots, x_{6}$$ is the key that enables the simple summation of probability of the events. In essence, (say), $$y=0$$ obtains if $$x=1$$ OR if $$x=3$$ OR if $$x=5$$ (a union of events) and since the events are mutually exclusive the "number of events" that satisfy this condition - ultimately (when normalized) frequency or probability - is the sum of the individual "number" of each event. The transformation procedure is illustrated in Figure 9.3. Let us now calculate the mean of $$Y$$ in two different ways. Using the probability density of $$Y$$, we can directly compute the mean as $E[Y]=\sum_{j=1}^{2} y_{j} f_{Y}\left(y_{j}\right)=0 \cdot \frac{1}{2}+1 \cdot \frac{1}{2}=\frac{1}{2} .$ Or, we can use the distribution of $$X$$ and the function $$g$$ to compute the mean $E[Y]=\sum_{j=1}^{6} g\left(x_{j}\right) f_{X}\left(x_{j}\right)=0 \cdot \frac{1}{6}+1 \cdot \frac{1}{6}+0 \cdot \frac{1}{6}+1 \cdot \frac{1}{6}+0 \cdot \frac{1}{6}+1 \cdot \frac{1}{6}=\frac{1}{2} .$ Clearly, both methods yield the same mean. This page titled 2.2.1: Discrete Random Variables is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Masayuki Yano, James Douglass Penn, George Konidaris, & Anthony T Patera (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.<\|endoftext\|>	4.53125
487	Bronchodilators are medications that relax the muscles that surround the airways in the lungs. This helps to open the airways, which makes breathing easier. There are several different types of bronchodilators. They can be grouped by the length of their effects: - Short-acting bronchodilators: short-acting bronchodilators are used on an as-needed basis. Their effects last for about four to six hours. Short-acting bronchodilators are usually used for less-severe lung disorders, such as mild asthma or mild chronic obstructive pulmonary disease. - Long-acting bronchodilators: long-acting bronchodilators are taken daily. Their effects last for 12 hours or more. Long-acting bronchodilators are usually used for moderate or more severe lung disorders, such as moderate asthma or severe COPD. Bronchodilators can also be grouped by the way they work. There are three types of bronchodilators. Each type works differently to relax the muscles around the airways: - Beta-agonists: Beta-agonists activate tiny structures called beta-2 receptors on the muscles surrounding the airways in the lungs. Activation of the beta-2 receptors causes the muscles surrounding the lungs to relax. Beta-agonists begin working within a few minutes of administration. For this reason, they are especially helpful for patients who experience short periods of serious shortness of breath. The effects of beta-agonists last for around four hours, so multiple doses may be necessary each day. - Anticholinergenics: Anticholinergenic drugs work by blocking the action of acetylcholine, a substance that transmits signals in the brain and to muscles throughout the body. By blocking acetylcholine, anticholinergenics prevent signals that tell muscles to contract from reaching the muscles surrounding the airways in the lungs. This keeps the airways open. - Theophyllines/theofyllines: Theophyllines cause the muscles surrounding the airways to relax. Many patients take more than one different bronchodilator. In fact, some patients take bronchodilators of every type. The type of bronchodilators a patient is prescribed depends on the severity of their disease and how well their symptoms are controlled.<\|endoftext\|>	3.828125
1,005	The legal term amnesty is related to the word amnesia—loss of memory. Amnesty means forgetting past deeds, consigning them to oblivion so that they may not become an issue in the future. Amnesty has often been used as a means of healing animosities and divisions caused by war. After the American Civil War, President Andrew Johnson granted amnesty to most Southerners who had fought against the Union. His General Amnesty Proclamation, issued in 1865, granted amnesty to many supporters of the Southern Confederacy; and his Universal Amnesty in 1868 did the same for all but 300 Confederates. Amnesty is closely related to another legal term, the pardon; in fact they are often used interchangeably. They are not quite the same, however. The pardon is normally used for a person who has been convicted of a crime. The chief executive officer of a country or state, such as the president or a governor, may pardon a criminal or may prevent an offender from being prosecuted. The most famous pardon in United States history occurred on September 8, 1974, when President Gerald R. Ford pardoned former President Richard M. Nixon “for all offenses which he, Richard Nixon, has committed or may have committed or taken part in” during his terms of office. Both the president and the Congress have the power of amnesty, but only the president has the power to grant a pardon. For hundreds of years amnesty has been used after wars and periods of civil strife. Twelve years after the English Civil War (1642–48), when Charles II was restored to the throne, he proclaimed a general amnesty, excepting only those who had taken part in the execution of his father, Charles I. In more recent history, President Jimmy Carter, in 1977, extended amnesty to draft resisters—men who had chosen to leave the country or be jailed rather than fight in the Vietnam War. President Carter hoped to end the divisions and bad feelings caused by a war that was unpopular among many segments of the population. In 1986 the United States legislature signed a landmark immigration law. The Immigration Reform and Control Act of 1986, which prohibited the hiring of illegal aliens, also offered amnesty (and legal residency) to illegal aliens who were living in the United States. Additionally, it offered a special amnesty to illegal agricultural workers, entitling them to temporary residency and, after a certain number of years, to permanent residency. To bring the problem of political prisoners to the attention of the world, an English lawyer named Peter Benenson founded an organization called Amnesty International in 1961. Its aims were to work for the release of persons imprisoned for political or religious opinions, to seek fair and public trials for such prisoners, to help refugees who had been forced to leave a country by finding them asylum and work, and to work for effective international means of guaranteeing freedom of opinion and conscience. Amnesty International, which was awarded the Nobel Peace Prize in 1977, had 700,000 members in 47 nations by 1990. Members are responsible for maintaining contact with specific prisoners and pleading their cases with the government concerned. With the emergence of a number of totalitarian regimes in the 20th century, amnesty for political prisoners had become a significant issue. In the Soviet Union, China, North Korea, South Korea, Taiwan, Iran, South Africa, Argentina, Chile, the nations of Eastern Europe, and in several other countries, political dissent and the exercise of civil liberties had been severely curtailed. Millions of individuals were put into concentration camps and prisons. In 1989 and 1990 the loosening of some of the restrictions in a few of these totalitarian regimes consequently resulted in a new wave of amnesties. In 1989 the Soviet Union amended the Law on Criminal Liability for Crimes against the State, which was most frequently used to punish dissidents for “anti-Soviet agitation and propaganda.” It reduced the maximum levels of imprisonment and fines on political prisoners. In the same year, authorities in Poland pardoned people imprisoned for specified political offenses. On January 1, 1990, Czechoslovakian president Vaclav Havel granted amnesty to 20,000 political prisoners. Under this declaration, nearly 75 percent of the prison population received either full pardons or reduced sentences. This was the world’s broadest grant of amnesty in 40 years. The reforms of President F.W. de Klerk in South Africa allowed for the release in 1989 of Walter Sisulu, former secretary-general of the African National Congress (ANC), and other political prisoners. In February 1990 Nelson Mandela, black nationalist and most famous member of the ANC, was also released on amnesty. In October 1989 President Carlos Saúl Menem of Argentina pardoned 277 military personnel and civilians. Many of those pardoned had been charged with violating human rights in the “dirty war” of the 1970s, during which more than 9,000 people had died or disappeared in conflicts between Argentinian armed forces and urban leftist guerrillas.<\|endoftext\|>	4.1875
536	South America’s vast Amazon region harbors one of the world’s most diverse collection of tree species, but more than half may be at risk for extinction due to ongoing deforestation to clear land for farming, ranching and other purposes, scientists say. Researchers said on Friday that if recent trends continued, between 36 and 57 percent of the estimated 15,000 Amazonian tree species likely would qualify as threatened with extinction under criteria used by the group that makes such determinations, the International Union for Conservation of Nature. The study covered roughly 2.1 million square miles (5.5 million square km) spanning Brazil, Peru, Colombia, Venezuela, Ecuador, Bolivia, Guyana, Suriname and French Guiana. The researchers analyzed Amazonian forest surveys and data on current and projected deforestation areas. “Many of the species that we suggest may be threatened are used by Amazonian residents on a daily basis, and many others are crucial to Amazonian economies,” conservation ecologist Nigel Pitman of the Field Museum in Chicago. These range from wild populations of economically important food crops like the Brazil nut, açaí fruit and heart of palm, to valuable timber species, to several hundred species that Amazonian residents depend upon for fruits, seeds, thatch, medicines, latex and essential oils, Pitman said. The trees also are important in their ecosystems for erosion control and climate moderation, Pitman said. “Scientists have been raising the alarm about Amazonian deforestation for several decades, and projections indicate that forest loss will continue for the foreseeable future,” said forest ecologist Hans ter Steege of the Naturalis Biodiversity Center in the Netherlands. “The good news is that over the last 10 years the rate of forest loss in the Amazon has dropped dramatically.” Amazonian forests have been shrinking since the 1950s as people cut and burn areas for farming, ranching and development. Until now, there has been no reliable estimate of how many tree species were threatened with extinction. “Yes, the threats are daunting, but it’s important to remember that more than 85 percent of forests in the greater Amazon are still standing,” Pitman said. The researchers said Amazon parks, reserves and indigenous territories, if managed well, should be able to protect most of the threatened species. Previous research found Amazon forests already have dwindled by about 12 percent and will decline up to another 28 percent by 2050. The research was published in the journal Science Advances. (Reporting by Will Dunham; Editing by Sandra Maler)<\|endoftext\|>	3.921875
572	In 1834 Robert Campbell and William Sublette established the first “Fort Laramie” at the confluence of the Laramie and North Platte Rivers in southeast Wyoming. Officially named Fort William, the post was rectangular, and measured only 100 by 80 feet. Hewn cottonwood logs 15 feet high formed the fort’s palisade. With the beaver trade already in decline, the future of the fur trade lay in trading with the Nativ…e population for buffalo robes. Fort William enjoyed a near monopoly on the buffalo trade in this region until a competing trading post, Fort Platte, was built a mile away in 1841. This rivalry spurred Fort William’s owners to replace their own aging fort with a larger, adobe walled structure named Fort John. The Lakota traded tanned buffalo robes here for a variety of manufactured goods. Each spring caravans arrived with trade goods at the fort. In the fall, tons of buffalo hides and other furs were shipped east. Throughout the 1840’s, however, the take of buffalo robes declined and Fort John’s role changed. In 1841, the first of many westward-bound emigrants arrived at Fort John. Tens of thousands of emigrants bound for Oregon, California, and the Salt Lake Valley would eventually stop at the fort. The traders at Fort John did a brisk seasonal business catering to the needs of emigrants. In 1849, the U.S. Army purchased Fort John as part of a plan to establish a military presence along the emigrant trails. The post was officially renamed Fort Laramie, and it began its tenure as a military post. The Army quickly constructed new buildings for stables, officers’ and soldiers’ quarters, a bakery, a guardhouse, and a powder magazine to house and support the fort garrison. As the years went by, the post continued to grow in size and importance. Fort Laramie soon became the principal military outpost on the Northern Plains. Fort Laramie also became the primary hub for transportation and communication through the central Rocky Mountain region as emigrant trails, stage lines, the Pony Express, and the transcontinental telegraph all passed through the post. Fort Laramie played an important role hosting several treaty negotiations with the Northern Plains Indian Nations, the most famous of which were the Horse Creek Treaty of 1851 and the still contested Treaty of 1868. When the Indian Wars came to a close Fort Laramie’s importance diminished. The post was abandoned and sold at public auction in 1890. Over the next 48 years, it nearly succumbed to the ravages of time. Preservation of the site was finally secured, however, when Fort Laramie became part of the National Park System.<\|endoftext\|>	3.71875
273	Courses Courses for Kids Free study material Offline Centres More Store # The lengths of two sides of an isosceles triangle are $15$ and $22$, respectively, what are the possible values of perimeter? Last updated date: 16th Jun 2024 Total views: 413.1k Views today: 10.13k Verified 413.1k+ views Hint: Here we need to apply the concept of isosceles triangle and perimeter of triangle. Isosceles Triangle: The triangle in which the two sides of a triangle are equal, then the angles opposite to equal sides are also equal. Perimeter of triangle = Sum of all sides. As two sides of a triangle are given as $15$ and $22$, respectively. We can fix either the two equal sides as $15$ and the third side is $22$ (or) the two equal sides as $22$ and the third side is $15$. The sides of triangle can be either $15,15,22$ or $15,22,22$ Perimeter of the triangle = $15+22+22=59$ or $15+15+22=52$ Therefore, the possible values of the perimeter are $59\text{ or 52}$ .<\|endoftext\|>	4.40625
1,009	# What is the cube of (−3)? • Last Updated : 03 Sep, 2021 In mathematics, when arithmetic calculations are performed or measuring any objects such as the length of a park, area of a rounded table, etc., numbers are needed. A number is a mathematical value. Different types of numbers are available in mathematics, such as Natural numbers, Whole numbers, Rational numbers, and Irrational numbers, Real numbers, etc. There are four number systems available depending upon different properties 1. Decimal Number System 2. Binary Number System 3. Octal Number System ### Cube A cube is a 3-dimension shape with sides of the same length, height, and width. In geometry, a cube has 6 sides or faces where all the faces are squares, 8 vertices, and 12 edges. It is also called a regular hexahedron. The volume of a cube is calculated as side3, i.e., (side × side × side) cubic units. In arithmetic and mathematics, the third power of a quantity has termed the cube of that amount since the volume of a cube is represented in terms of an edge e, as e3. For example, if the edge of a square box is 5cm, then the volume of a cube is 5 x 5 x 5 = 125 cm3. Some of the examples of cubes in real life are dice, sugar cube, ice cube, Rubik’s cube, etc. ### Cube of a number The product obtained by the number multiplied three times by itself is called the cube of a number. It can be represented as 23 = 2 × 2 × 2 = 8. Cube root of a number The cube root of a number is a number the returns the original number when cubed. It is obtained by using a ‘radical’ symbol with a tiny 3. This is the “cube root” sign, which is the “radical” symbol (used for square roots) plus a tiny three to represent the cube root. ## 3√ For example, the cube root of a number 3√125 = 5 The natural number is said to be a perfect cube if the number is multiplied by itself thrice. Perfect cubes are not defined for negative numbers just like the perfect square. The perfect cube numbers from 1 to 10 are 13 = 1 × 1 × 1 = 1 23 = 2 × 2 × 2 = 8 33 = 3 × 3 × 3 = 27 43 = 4 × 4 × 4 = 64 53 = 5 × 5 × 5 =125 63 = 6 × 6 × 6 = 216 73 = 7 × 7 × 7 = 343 83 = 8 × 8 × 8 = 512 93 = 9 × 9 × 9 = 729 103 = 10 × 10 × 10 = 1000. Some properties of the cube of a number • The cube of a negative number will always result in a negative number. • The cube of a positive number will always result in a positive number. • The cube of an even number is an even number. • The cube of an odd number is an odd number. ### What is the Cube of (−3)? The cube of a number is the number multiplied by itself thrice as mentioned earlier. for the negative numbers, cubes are negative since the negative numbers are multiplied thrice. When two negative numbers are squared, the result is positive. However, when the negative numbers are cubed, the result is negative (-3)3 = (-3) × (-3) = 9 = 9 × (-3) = -27 Hence, the cube of (-3) is -27. ### Similar Problems Question 1: What is the cube of 12? Solution: 123 = 12 × 12 × 12 =1728 Question 2: What is the cube of 10.5? Solution: 10.53 = 10.5 × 10.5 × 10.5 = 1157.625 Question 3: What is the cube of -25? Solution: (-25)3 = -25 × -25 × -25 = -15625 Question 4: What is the cube of 197? Solution: 1973 = 197 × 197 × 197 = 7645373 Question 5: What is the cube of -45.80? Solution: (-45.80)3 = (-45.80) × (-45.80) × (-45.80) = -96071.9119 My Personal Notes arrow_drop_up<\|endoftext\|>	4.71875
3,642	The word energy is derived from the Greek word meaning energos meaning activity. Energy is a characteristic of the system which describes the ability of the system to perform some work. According to the international system of units, in honor of the English physicist James Prescott Joule-in (1818 – 1889), a unit of measure for energy is called the joule (J). Important feature of energy is that energy can neither arise nor perish, and therefore the amount of energy in a closed system is always constant. This energy feature is called the energy preservation law, which was first set in the 19th century. All known natural processes and phenomena can be explained with several forms of energy according to the following definitions: kinetic energy, potential energy, thermal energy, gravity, elasticity, electromagnetic energy, chemical energy, nuclear energy and mass. Energy can be neither created nor be destroyed. Although energy cannot arise or disappear, energy can be transformed from one form to another. Transformation of energy from one form to another is called work or power. In the honor of Scotsman engineer and inventor James Watt (1736 – 1819) unit of measurement for work is called the watt (W). One watt is the work done in one second to transform one joule of energy from one form to another (W = 1 J/s). From this definition it can be seen that watts actually express the rate of transformation of energy from one form to another. Sometimes, as the unit of measurement for energy is also used the unit watt-hour (Wh). One watt-hour is constant work (power) of one watt in the period of one hour, and is therefore 1Wh = 1 J / s * 3600s = 3600J. For the amount of spent or generated electricity there are usually used the multipliers of Wh measuring unit, such as kWh, MWh and GWh (kilowatt-hour, megawatt-hour and gigawatt-hour). The automobile industry typically uses expression horse power for expressing the maximum power used by the engine. There are several definitions of horsepower, but in automobile industry the most important are two of them: mechanical horsepower and metric horsepower. Mechanical horsepower is about 746 W, and metric horsepower is about 735.5 W. Manufacturers usually express the power of automobile in form of mechanical horse power, but sometimes for the purposes of the “beautification” figures are also used in form of metric horsepower, especially for exotic sports cars. THE SUN AS THE ENERGY SOURCE The main energy process in the Sun is nuclear fusion: E = mc² The sun is by far the biggest source of energy in our solar system. Because of its relatively small size in comparison to other stars, and because it is yellow, the sun belongs to the class of stars which are called yellow dwarf. Although it is relatively small if we compare it with other stars, the sun is a giant compared to planets that surround it – the mass of Sun is about 99.8% the mass of our entire solar system. For easy understanding of how big the Sun really is the next exemplary data can be useful: 962,000 copies of entire Earths could fit inside the Sun, and in the gaps between the balls we could still squeeze approximately 340,000 “melted” Earths. Therefore in order to fill volume of the Sun we had to use about 1,300,000 Earths. The main energy process that takes place in the Sun is nuclear fusion, which is melting of the two light atoms into one heavier atom, with the release of energy proportional to difference of masses before and after the reaction (according to Einstein’s formula E = mc²). Nuclear fusion is responsible for conversion of approximately 700,000,000 tons of hydrogen to about 695,000,000 tons of helium every second in the Sun, and the difference of 5,000,000 tons is converted by Einstein’s formula in the energy, in the form of gamma radiation. When these 5,000,000 tons per second we turn into the work, we can calculate how the power of the Sun is around 386 billion billion megawatts. Just for comparison: the largest nuclear power plant in the world is a Japanese nuclear power plant Kashiwazaki, which has seven operating reactors and the total power of 8,212 MW. At any time Earth is receiving 174 petawatts (1015 W) of solar radiation from the Sun. FORMS OF ENERGY As we already mentioned in the introduction all currently known natural processes and phenomena can be explained with a few basic forms of energy. Below are some of these energy forms enumerated and explained in more details. - Potential energy – Potential energy is defined as the work that is done against the given force by changing the position of the observed object in relation to a reference position. The name “potential energy” comes from the assumption that this energy can be easily converted to useful work. This is not quite correct for all systems, but helps the understanding of the potential energy theory. The two most obvious types of potential energy are gravitational potential energy and elastic potential energy. Gravitational potential energy is the energy associated with gravitational force and works between any two objects that have mass. It is proportional to the mass of objects, and inversely proportional to the distance between objects. Elastic potential energy is potential energy of some elastic object, such as springs, catapults, etc. It occurs as a consequence of forces that are trying to move object back to the original position; these are in most cases electromagnetic forces in atoms and molecules that form the object. The best example of exploiting gravitational potential energy are large hydroelectric power plants where the potential energy of water is converted into kinetic energy, which then drives turbines to generate electricity. - Kinetic energy – Kinetic energy or energy of the movement is the energy required to accelerate a certain object to a certain speed or energy of the object at a certain speed in relation to a reference object. According to classical mechanics kinetic energy is proportional to the mass of the object and the square speed of the object. At speeds that are comparable to the speed of the light, kinetic energy can no longer be calculated using equations that apply to regular classical mechanics. Instead of that we must use the theory of relativity. Energy of the object that is moving at speeds comparable to speed of light can be calculated using Lorentz’s transformations under which an object that is moving at the speed of light should have an infinite energy, so it is therefore impossible to accelerate an object to the speed of light. Example of exploiting the kinetic energy is converting wind energy into electricity in windmills. Kinetic energy of an object is the energy that it possesses due to its motion. - Thermal energy – Thermal energy is the energy of random movement of microscopic particles that form the object. Thermal energy of the object increases with temperature. Thermal energy is transferred from one object to another because of differences in temperature. Heat is transferred in three basic ways: conduction, radiation and convection. Heat conduction is the spontaneous transition of thermal energy through matter from warmer to colder parts. Convection is the flow of gases where warmer liquid flows towards the colder liquid transferring the heath to the environment. Warmer body radiates stronger electromagnetic radiation, because the warmer the body is the more energy there is and vibration of electric charge is also increased. This radiation heat can be transferred from one body to another. Thermal energy can be directly used for heating or indirectly to obtain other forms of energy. For instance, the thermal energy stored within the Earth – geothermal energy – can be used to generate electricity. - Electricity – Electricity is a form of potential energy in the Coulomb force field in which the particles of the same charge are repulsed, and particles of the opposite charge are attracted. Electrical energy is undoubtedly the most important form of energy used by humanity since it is relatively easy to transport and most importantly – it can be easily converted into other useful forms of energy such as kinetic and thermal energy. Electricity is currently produced mostly from fossil fuels (mainly from coal). Since fossil fuels have a negative impact on the environment and in limited quantities, there is an increased need to use alternative methods of power generation such as the exploitation of solar energy, water energy, geothermal energy, wind energy and other renewable energy sources. Chemical energy – Chemical energy can be defined as the work that is done by electrical forces during rearrangement of the electrical charges – protons and electrons – in chemical processes. If the chemical energy of the system decreases in the chemical reaction this means that the difference is emitted in the environment in the form of light or heat, and if the chemical energy increases, this means that the system has taken from the environment a certain amount of energy, usually in the form of light or heat. Fire is for instance a form of shifting the chemical energy into heat and light, and can occur only if three basic conditions for a chain reaction are met: the presence of sufficient amounts of oxygen, presence of the burning materials and presence of sufficient amount of heat. Examples of exploiting the chemical energy are fossil fuels. When burning fossil fuels release heat that is then through the pressure converted into kinetic energy, or is used for heating some liquid for the purpose of vaporization of this liquid and to obtain kinetic energy. Coal-fired power plants are example of converting chemical energy into electricity. - Nuclear energy – Nuclear energy is the energy that is produced by the processes of nuclear fusion or nuclear fission. Nuclear fusion is the joining of two or more light atoms into one heavier with the release of certain amounts of energy in the form of various radiations. Nuclear fission also involves releasing specific amounts of energy in the form of various radiations, but this energy is the result from splitting the heavy atoms into two or more lighter atoms. In both these processes the mass before the reaction is always bigger than the mass after the reaction, and the difference in masses is converted into energy according to Einstein’s formula E=mc2. Solar energy is a consequence of constant nuclear fusion that takes place in the center of the star, and then in the form of radiation comes to the surface and is afterwards radiated to space. Researches that could mean better exploit of nuclear fusion on earth are still at an early stage, in the form of the international ITER project, but for now there is no indication that nuclear fusion could be heavily exploited in years to come. But on the other hand nuclear fission is simple enough process that is widely used in nuclear reactors to generate electricity. ENERGY VALUES – COMPARISON Normal (100 g) chocolate has energetic value of about 2.3 MJ. One liter of gasoline fuel has an energy value of about 34 MJ and weighs approximately 730 grams. Thus if the cars could drive on the chocolate one liter of gasoline should be replaced with about a kilogram and a half of chocolate. \|Contained in\|\|Energy in J\|\|Energy in btu\| Liter of gasoline fuel \|34 MJ\|\|32.200 btu\| Liter of diesel fuel \|38.7 MJ\|\|36.720 btu\| One kilogram of chocolate \|23 MJ\|\|21.780 btu\| One barrel of crude oil (around 159 liters) \|6,123 MJ\|\|5.800.000 btu\| Li-ion battery (density) \|540-720 kJ/kg\|\|511 – 682 btu/kg\| kWh of electricity \|3.6 MJ\|\|3,412 btu\| Natural gas (m2) \|38.3 MJ\|\|36,241 btu\| Kilocalories (calories, food) \|4,184 J\|\|3.96 btu\| Metrical ton of coal \|29 GJ\|\|27,467,300 btu\| Ton of uranus-235 \|7,4 x 1016 J\|\|70 x 1012 btu\| A normal AA battery has about 1000J of energy stored in it and this is enough only for very small applications like MP3 players. Large consumers usually use a series of batteries and this principle can be used to drive the car. One of the best produced electric cars was the Nissan Altre that could reach 120 km/h, and with only one filling car range was around 190 km. This car used li-ion battery. Better Li-Ion batteries have the energy density of about 720 kJ / kg and this capacity is the basic flaw of all electric cars. One tank in the average car has about 50l and if you take a car that normally uses gasoline fuel, this means that one tank of fuel has around 1700 MJ of energy. We should have about 2360 kg heavy li-ion battery that could store the same amount of energy as the fuel tank of 50l, but electric motors are much more efficient in using energy than gasoline engines and to obtain the same efficiency they do not require the same amount of energy. This is only approximate comparison in order to obtain an indication of batteries as a power source. All energy sources that have been mentioned so far can be in no way compared with the nuclear energy.Uranium-235 has density of energy per kilogram of material for about six orders of magnitude greater than the density of energy in the coal or any other fossil energy sources. This is because nuclear energy is the result of the mass converting into energy according to the well-known Einstein’s formula E=mc², and for other energy sources the sum of masses remains unchanged after reaction. FOOD ENERGETIC VALUE AND LOSING WEIGHT \|Calories in gram\| In order to describe the energy value of food we mainly use kilocalories or calories (kcal). As with HP, with calories there are also some definitions that differ a little, but the most usual definition of kilocalories is the energy needed to heat one kilogram of water for 1 ° C from 14.5 °C to 15.5 °C on standard atmospheric pressure of one atmosphere – which is about 4.184 kJ. When we want to describe energetic value of the food we usually use term calories instead of proper term kilocalories. A person who is moderate athlete and weighs around 80 kg needs around 28-30 kcal per kilogram of its weight, which is somewhere around 2320 kcal of energy per day. Professional athletes, for example marathon runners, need around 50-60 kcal / kg per day of energy and therefore a professional athlete weighing around 80 kg needs around 4400 kcal of energy per day. In order to calculate energetic value of certain food we’ll take the example of normal (100 grams) chocolate. Energetic value of this chocolate is around 550 kcal (around 2300 kJ), and this is energy needed for 100 W light bulb to give maximum light for slightly more than 6 hours and 23 minutes. This same energy of 2300 kJ is enough to lift a car that weighs one ton from surface of the Earth to a height of 234.45 meters in the air. According to some estimates, a person who weighs 70 kg and is doing more serious jogging consumes about 560 kcal per hour. Let’s assume that a person that weighs 75 kg that is using proper diet to maintain that weight wants to lose five kilos by not changing her/his dietary habits. In this case it is necessary to increase the physical activity of the body in order to create an energy deficit so our person has decided to do one hour of jogging each day. If we assume that person in this case is losing both proteins and fats we can get the figures that in order for this person to lose one gram of his weight an energy deficit of 6.5 kcal is required. By doing one hour of jogging person can lose about 86 grams, and this is the daily amount of weight loss. With this pace person will lose five kilograms in little less than two months. Jogging is defined as the running with average speed of less than 10 kilometers per hour. The modern lifestyle involves much greater use of energy in order to achieve the greater efficiency and comfort, so the energy use is increasing each day. Currently, world energy needs are mostly satisfied by using the fossil fuels that are harmful to environment, and in the future these fuels will have to be replaced with cleaner energy sources, mostly in the form of renewable energy or nuclear energy. As you can see from this article, currently available energy is more than enough to cover all possible future energy needs, all what is needs is to find ways to clean and safe exploitation of various energy sources, of course, with the gradual reduction of the oil lobby influence, which is making life difficult for all energy sources that aren’t under their control. Energy needs are constantly growing affecting everyday life in much of the modern world and this has turn energy into one of the main strategic resources of developed countries. If we take a look at history books we can see how various wars have resulted due to lack of water, lack of food, different religious reasons or because of a desire for territory expansions. Recently wars have also started in order to maintain stable energy supply by occupying areas filled with different energy sources. The best example is the occupation of oil rich Iraq by the US military forces in order to control the oil supply. This occupation, together with ever-increasing energetic needs of developing countries has caused a substantial increase in price of oil products which are later reflected indirectly in the increased prices of nearly all products. Renewable energy sources are likely to become the primary sources of energy in years to come, making wars for energy something that can be found only in history books, which could make entire world much more peace-loving.<\|endoftext\|>	3.734375
3,658	# Dependent Events – Explanation & Examples In probability theory, when we are dealing with multiple events, one event can change the probability of another event. For example, the probability that a randomly chosen person is suffering from some particular disease is usually small. However, if he/she is tested positive for that disease, the probability becomes very high (the probability is still not 1 because no test is perfectly accurate). Such events are termed dependent events. Two events are said to be dependent events if the occurrence of one event changes the probability of occurrence of the other event. After reading this article, you should understand the following: 1. Dependent events 2. Identifying two events are dependent 3. Solving problems related to dependent events 4. Various formulae related to probabilities of dependent events To understand the concept of dependent events, it is advisable to refresh the following topics: ## What Are Dependent Events: Suppose we roll a six-sided fair die. What is the probability of getting a $2?$ Since there are six possible outcomes and each outcome is equally likely, so the probability of getting a $2$ is $\frac16$. Now, let’s suppose we roll a die, and we know that an outcome is an even number. What is the probability that the outcome is $2?$ Now, since we know that the outcome is even, so we are left with three possible outcomes only,i.e., $\{2,4,6\}$. Since $2$ is one of the three possible equally likely outcomes, hence the probability of getting a $2$ is $\frac13$. Note that the event’s occurrence that “the outcome is even” changes the probability of $2$ from $\frac16$ to $\frac13$. Hence, the outcome is $2$, and the event that the outcome is even are dependent events. Note that dependence works both ways. The probability of getting an even number is $\frac{3}{6} = \frac12$. However, if we know that the outcome is $2$, then the probability that the “outcome is even” becomes $1$ (since now we know that the outcome is $2$, which is an even number). Note that again we observe that one event (in this case $2$) has changed the probability of another event (in this case, even numbers). ### How to Tell If Two Events Are Dependent To analyze whether two events are dependent or not, we first need to understand the concept of conditional probability. ### Conditional Probability Conditional probability $P(A\|B)$ is the probability of event A given the information that B has already taken place. For any two events $A$ and $B,$ $P(A\|B)$ is given as $P(A\|B) = \frac{P(A \cap B)}{P(B)}$ Let’s reconsider an example Example 1: We roll a six-sided fair die. Let $E1$ be the event that the outcome is $3$. Let $E2$ be the event that the outcome is odd. Find $P(E1\|E2)$. Solution: Note that the sample space(link) $S = \{1,2,3,4,5,6\}$ has six elements. Also, $E1=\{3\}$ and $E2 = \{1,3,5\}$ and $E1 \cap E2 = \{3\}$, hence $P(E2) = \frac36$ and $P(E1 \cap E2) = \frac16$. Therefore, $P(E1\|E2) = \frac{\frac16}{\frac36} = \frac13$. ### Conditional Probability and dependence Now that we have the tool of conditional probability with us, we can easily define the concept of dependence mathematically. By definition, $A$ and $B$ are dependent if the occurrence of $B$ changes the probability of $A$, and similarly, the occurrence of $A$ affects the probability of $B$. Using the concept of conditional probability, we can write that $A$ AND $B$ are dependent if $P(A\|B) \neq P(A)$ and $P(B\|A) \neq P(B)$. The formula of conditional probability states that $P(A\|B) = \frac{P(A \cap B)}{P(B)}$. Hence, if $P(A\|B) \neq P(A)$, then $P(A \cap B) = P(A\|B)P(B) \neq P(A)P(B)$ Similarly, we can show that if $P(A \cap B) \neq P(A)P(B)$, then $P(B\|A) \neq P(B)$ ### Dependent Events Definition Two events are said to be dependent if the occurrence of one event changes the probability of occurrence of the other event. Mathematically, two events $A$ and $B$ are said to be independent if $P(A \cap B) \neq P(A)P(B)$. ## How to Solve Dependent Events Let us consider an example to see how to solve dependent events using the above definition. Example 2: We toss a coin three times. Let $E1$ is the event that the first toss results in Heads. Let $E2$ is the event, the third toss is Tails, and let $E3$ is when we get an even number of tails. 1. Are $E1$ and $E2$ dependent events? 2. Are $E2$ and $E3$ dependent events? 3. Are $E3$ and $E1$ dependent events? Solution: The sample space of the experiment can be written as follows: $S = \{HHH, HHT, HTH, THH, TTT, TTH, THT, HTT\}$. The event $E1$ can be written as $E1 = \{HHH, HHT, HTH, HTT\}$, $E2$ can be written as $E2 = \{HHT, TTT, THT, HTT\}$ and $E3$ is written as $E3 = \{TTH, THT, HTT\}$ There are $8$ elements in the sample space and $4$ elements in $E1$, $4$ in $E2$ and $3$ in $E3$. Since all elements are equally likely, hence $P(E1) = \frac48 = \frac12$ $P(E2) = \frac48 = \frac12$ $P(E1) = \frac38$ 1) To find if $E1$ and $E2$ are dependent or not, we need to find $P(E1 \cap E2)$. Note that $E1 \cap E2 = \{HTT, HHT\}$, hence $P(E1 \cap E2) = \frac28 = \frac14$. Now $P(E1) \times P(E2) = \frac12 \times \frac12 = \frac14 = P(E1 \cap E2)$, hence $E1$ and $E2$ are NOT dependent events. 2) We note that $E2 \cap E3 = \{THT, HTT\} = \frac28 = \frac14$. Now, $P(E2) \times P(E3) = \frac12 \times \frac38 = \frac{3}{16} \neq P(E1 \cap E2)$, hence $E2$ and $E3$ are dependent events. 3) We find $E1 \cap E3 = \{HTT\}$, hence $P(E1 \cap E3) = \frac18$. Also, $P(E1) \times P(E3) = \frac12 \times \frac38 = \frac{3}{16} \neq P(E1 \cap E3)$, hence $E1$ and $E3$ are dependent events. ### Dependent Events Formula We list the various relations between $A$ and $B$, if $A$ and $B$ are dependent 1. $P(A\|B)$ = $\frac{P(A\cap B)}{P(B)} \neq P(A)$ 2. $P(B\|A) = \frac{P(A \cap B)}{P(A)} \neq P(B)$ 3. $P(A \cap B) = P(A\|B)P(B) = P(B\|A)P(A) \neq P(A)P(B)$ 5. $P(A \cup B) = P(A) + P(B) – P(A \cap B)$ Example 3: The probability that a test for a certain disease comes out positive given that the patient actually has the disease is $95\%$. The probability that a randomly chosen person is infected is $1\%$. The probability that a randomly chosen person is tested positive is $10.85\%$. A person is tested positive, what is the probability that he/she actually has the disease. Solution: Let $P(D)$ is the probability of disease and is equal to $0.01$. $P(T)$ is the probability that the test is positive and is given to be $0.1085$. Let $P(T/D)$ is the probability that the test is positive given that the person actually has the disease and is known to be $0.95$. We are interested in the probability of disease given that the test is positive, i.e., $P(D/T)$. From the formula listed for dependent events, we note that $P(D/T) = \frac{P(D \cap T)}{P(T)}$, we know $P(T)$ but we do not know $P(D \cap T)$. To find $P(D \cap T)$, we use the relation that $P(T/D) = \frac{P(D \cap T)}{P(D)}$, hence $P(D \cap T) = P(T/D)P(D) = 0.95 \times 0.01 = 0.0095$. Now, $P(D/T) = \frac{P(D \cap T)}{P(T)} = \frac{0.0095}{0.1085} = 0.087 = 8.7\%$. Dependent Events and Tree Diagrams: There are many scenarios of interest where we repeat the same experiments many times. For instance, tossing a coin or picking a card from a deck, etc. Tree diagrams offer a useful tool to analyze such events. Let us consider how to make tree diagrams when we know that the multiple attempts or trials of the same experiment are dependent. For simplicity, we consider an experiment with two outcomes only. Let’s call those $A$ and $A’$. When the experiment is performed the first time, we can have two possible outcomes, as shown in the tree diagram below. Therefore, we label the branches of the tree diagram with the probabilities of the events $A1$ and $A’1$, where $1$ denotes that the first attempt. Now we perform the same experiment again. If the outcome of the first attempt was $A$, then we get a tree diagram as shown below. Note that, in this case, we are drawing the branches of the tree diagram for the case when the first outcome is known to be $A$. Hence the branches are labeled with probabilities $P(A2\|A1)$ and $P(A’2\|A1)$. Similarly, we draw the branches when the first attempt was $A’$, as shown below. The overall tree diagram is as follows. Note that the tree diagrams’ leaves (i.e., endpoints or terminal nodes) represent mutually exclusive events. Let’s say we wish to find the probability of the event $A1A2 = A1 \cap A2$, i.e., the first attempt resulted in $A$ and the second attempt also resulted in $A$. We use the fact that $P(A1 \cap A2) = P(A1)\times P(A2\|A1)$. In other words, to find the probability of any event given by the terminal node of the tree diagram, we multiply the probabilities along the branches. Also, if we are interested in the probability of the compound event, say $A1A2 \cup A1A’2$, then we add the probabilities $P(A1A2)$ and $P(A1A’2)$ as the events are mutually exclusive. Example 4: We have two boxes, $B1$ and $B2$. $B1$ contains two red balls and three green balls. $B2$ contains three green and three red balls. We toss a coin, and if the outcome is Heads, we sample a ball from $B1$; otherwise, we sample a ball from $B2$. Draw a tree diagram and find the probability of sampling a red ball. Solution: Hence the probability of drawing a Red ball is $\frac25\times \frac12 + \frac12 \times \frac12 = \frac15 + \frac14 = \frac{9}{20}$. #### Sampling Without Replacement and Dependent Events: In sampling experiments, we collect $n$ objects, and we sample from the collection $k$ times. There are two possible ways of sampling 1. We replace the sampled object in the collection. 2. We do not replace the sample object in the collection (i.e., sampling without replacement). When we are doing sampling without replacement, the outcomes of each sample affect the probabilities of other samples. Hence the trials in sampling without replacement are dependent. Let us consider an example: Example 5: We sample two cards from a deck of 52 cards without replacement. What is the probability that the first card is a King and the second card is a Queen. Solution: There are $4$ kings in a deck of 52 cards. So the probability of drawing a king in the first attempt is given as $P(\textrm{First card is a King}) = \frac{4}{52} = \frac{1}{13}$. There are $4$ Queens in a deck. If we have already drawn a king in the first attempt, so we are left with $51$ cards. Hence, drawing a Queen in the second attempt given that the first attempt was as King is given as $P(\textrm{Second card is a Queen}) \| \textrm{First Card is a King}) = \frac{4}{51}$. Using the formulae described above, we can write $P(\textrm{First King AND Second Queen}) = P(\textrm{First King} \cap \textrm{Second Queen})$ $P(\textrm{First King} \cap \textrm{Second Queen}) = P(\textrm{Second Queen \| First King)}P(\textrm{First King})$ $P(\textrm{First King} \cap \textrm{Second Queen})= \frac{4}{51} \times \frac{1}{13} = 0.006 = 0.6\%$ ### Practice Questions: 1.We roll a fair die twice. Let $E1$ be the event that the sum is even, and let $E2$ be the event that the sum is greater than $5$. Are $E1$ and $E2$ dependent? 2.The probability that Team-A wins a match is $P(W) = 65\%$. However, if Team-A is playing the match in their home ground, then the probability of winning increases to $P(W\|\textrm{Home}) = 70\%$. Team-A plays equal matches in “Home ground” and “Away ground,” so the probability of playing at home is $P(\textrm{Home}) = 50\%$. If it is known that Team-A has won a match, what is the probability that the match was being played at home ground, i.e., $P(\textrm{Home}\|W)$? 3.A box contains $100$ items. There are $4$ items that are defective. We sample two items from the box without replacement. Draw a tree diagram and find the probability of at most one defective item. 4.We sample two cards from a deck of $52$ cards without replacement. What is the probability that a) Both are Kings b) We get at least one King ### Answer Key 1.Yes, $P(E1) = \frac12$, $P(E2) = \frac{13}{18}$, $P(E1 \cap E2) = \frac{7}{18} \neq P(E1) \times P(E2)$. Hence, the events are dependent. 2. $53.8$ 3. $\frac{4}{100}\frac{96}{99} + \frac{96}{100}\frac{4}{99} + \frac{96}{100}\frac{95}{99}$. 4. a) $\frac{1}{13} \times \frac{3}{51}$. b) $1 – (\frac{48}{52}\times\frac{47}{51})$.<\|endoftext\|>	4.90625
327	Ada Lovelace Day, observed on October 13th, celebrates the achievements of women in science, technology, engineering, and math (STEM). The holiday is about sharing stories of women – computer technologists, engineers, scientists, and mathematicians – who have been inspirational in their fields. The aim of the holiday is to create and encourage new role models for girls in these male-dominated fields by acknowledging the many and varied contributions of women in STEM. Why is it named after Ada Lovelace? Ada Lovelace (1815-1852) is widely known for writing an algorithm to calculate Bernoulli numbers for Charles Babbage’s Analytical Engine, an early computer that used punch cards for input and output. This algorithm is widely considered to be the first ever computer program, making Ada Lovelace the world’s first computer programmer. Here are some other interesting facts about Ada Lovelace: - She was the daughter of poet Lord Byron - Her actual married name was Lady Ada King, Countess of Lovelace - Charles Babbage called her “The Enchantress of Numbers” - She was a visionary and predicted that machines like the Analytical Engine could be used to compose music, produce graphics, and be useful for scientific calculations - There is a programming language called Ada, named in her honor - Lovelace’s notes on the Analytical Engine inspired Alan Turing’s work on the first modern computers in the 1940s Feel free to share stories of women in STEM fields who have inspired you in the comments!<\|endoftext\|>	3.8125
339	# 2011 AMC 8 Problems/Problem 8 (diff) ← Older revision \| Latest revision (diff) \| Newer revision → (diff) ## Problem Bag A has three chips labeled 1, 3, and 5. Bag B has three chips labeled 2, 4, and 6. If one chip is drawn from each bag, how many different values are possible for the sum of the two numbers on the chips? $\textbf{(A) }4 \qquad\textbf{(B) }5 \qquad\textbf{(C) }6 \qquad\textbf{(D) }7 \qquad\textbf{(E) }9$ ## Solution 1 By adding a number from Bag A and a number from Bag B together, the values we can get are $3, 5, 7, 5, 7, 9, 7, 9, 11.$ Therefore the number of different values is $\boxed{\textbf{(B)}\ 5}$. ## Solution 2 The sum of an even number added to an odd number is always odd. The smallest possible sum is $3$, and the largest possible sum is $11$. The odd numbers in between can be achieved by replacing chips with ${\displaystyle \pm }2$ within the same bag. Therefore, we can conclude that there are $(11-3)/2(*)+1=\boxed{\textbf{(B)}\ 5}$ possible sums. • Note that we are only accounting for odd numbers, thus the $/2$. ~megaboy6679<\|endoftext\|>	4.46875

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