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896	Is it an unprecedented coincidence? 200 years ago today, just minutes apart according to some unconfirmed accounts, Abraham Lincoln was born in a rude log cabin on Nolin Creek, in Kentucky, and Charles Darwin was born into a wealthy family at the family home in Shrewsbury, England. Lincoln would become one of our most endeared presidents, though endearment would come after his assassination. Lincoln’s bust rides the crest of Mt. Rushmore (next to two slaveholders), with George Washington, the Father of His Country, Thomas Jefferson, the author of the Declaration of Independence, and Theodore Roosevelt, the man who made the modern presidency, and the only man ever to have won both a Congressional Medal of Honor and a Nobel Prize, the only president to have won the Medal of Honor. In his effort to keep the Union together, Lincoln freed the slaves of the states in rebellion during the civil war, becoming an icon to freedom and human rights for all history. Upon his death the entire nation mourned; his funeral procession from Washington, D.C., to his tomb in Springfield, Illinois, stopped twelve times along the way for full funeral services. Lying in state in the Illinois House of Representatives, beneath a two-times lifesize portrait of George Washington, a banner proclaimed, “Washington the Father, Lincoln the Savior.” Darwin would become one of the greatest scientists of all time. He would be credited with discovering the theory of evolution by natural and sexual selection. His meticulous footnoting and careful observations formed the data for ground-breaking papers in geology (the creation of coral atolls), zoology (barnacles, and the expression of emotions in animals and man), botany (climbing vines and insectivorous plants), ecology (worms and leaf mould), and travel (the voyage of H.M.S. Beagle). At his death he was honored with a state funeral, attended by the great scientists and statesmen of London in his day. Hymns were specially written for the occasion. Darwin is interred in Westminster Abbey near Sir Isaac Newton, England’s other great scientist, who knocked God out of the heavens. Lincoln would be known as the man who saved the Union of the United States and set the standard for civil and human rights, vindicating the religious beliefs of many and challenging the beliefs of many more. Darwin’s theory would become one of the greatest ideas of western civilization, changing forever all the sciences, and especially agriculture, animal husbandry, and the rest of biology, while also provoking crises in religious sects. Lincoln, the politician known for freeing the slaves, also was the first U.S. president to formally consult with scientists, calling on the National Science Foundation (whose creation he oversaw) to advise his administration. Darwin, the scientist, advocated that his family put the weight of its fortune behind the effort to abolish slavery in the British Empire. Each held an interest in the other’s disciplines. Both men were catapulted to fame in 1858. Lincoln’s notoriety came from a series of debates on the nation’s dealing with slavery, in his losing campaign against Stephen A. Douglas to represent Illinois in the U.S. Senate. On the fame of that campaign, he won the nomination to the presidency of the fledgling Republican Party in 1860. Darwin was spurred to publicly reveal his ideas about the power of natural and sexual selection as the force behind evolution, in a paper co-authored by Alfred Russel Wallace, presented to the Linnean Society in London on July 1, 1858. On the strength of that paper, barely noticed at the time, Darwin published his most famous work, On the Origin of Species, in November 1859. The two men might have got along well, but they never met. What unusual coincidences. Today is the first day of a year-long commemoration of the lives of both men. Wise historians and history teachers, and probably wise science teachers, will watch for historical accounts in mass media, and save them. Go celebrate human rights, good science, and the stories about these men. - Darwin 200 - About Darwin.com; links to Darwin texts online - On the Origin of Species online - C. Warren Irvin Collection of Darwin and Darwiniana, University of South Carolina<\|endoftext\|>	3.6875
1,211	Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team # Factoring Real Number Coefficients ## Key Questions • Factoring polynomials is usually a very simple and straightforward process, but when you get polynomials of a higher degree (i.e. with the highest power being something large, like 5), then you start to run into problems. The best way to solve those types of problems is to use synthetic division to condense your equation down to a quadratic, and then factor normally. Here's a video that explains synthetic division: Now let's get to actually factoring. Now factoring is quite simply writing a polynomial as the product of a set of factors. It's the same thing as writing 6 as the product of 4 and 8. How is this useful, you may ask? Let me show you. So for starters, try solving this linear equation: $3 x - 1 = 0$ Add 1 to both sides and divide by 3, and your answer is $\frac{1}{3}$. Easy peasy right? Now try doing this: ${x}^{2} + 6 x + 9 = 0$ A little more complicating right? The solution doesn't stick out that easily. Now we get to how factoring would help us. If we could rewrite this equation as the product of 2 linear equations, then we could solve each one individually, and it would make life so much easier. So let's look at that quadratic again: ${x}^{2} + 6 x + 9 = 0$ At first sight it looks impossible to figure out a way write this as a product of two factors. But fear not, there is a way. We're going to solve it using grouping. We're going to rewrite it as 4 terms, and then work with two at a time, like we did with the linear equation. We will do this by splitting the middle term into two terms. But what will we split it into? Well, we're going to split it into two numbers that multiply to equal the 3rd term. So hence, we'd get: ${x}^{2} + 3 x + 3 x + 9$. Sidenote: Notice that we're not actually changing the value of the equation. We're simply manipulating it's look to suit our purpose. Then, we deal with each pair separately. Let's look at the first two: ${x}^{2} + 3 x$ Is there anything we can do to this? Any GCF we can pull out? Well there is. We can pull out an $x$, and rewrite it as: $x \left(x + 3\right)$ The second set: $3 x + 9$ Here we can pull out a 3, and rewrite it as: $3 \left(x + 3\right)$. Now let's write the whole thing out: $x \left(x + 3\right) + 3 \left(x + 3\right)$ Now here, we see that we have a new GCF: $\left(x + 3\right)$ itself. So let's pull that out: $\left(x + 3\right) \left(x + 3\right) = 0$ Now it looks much simpler to solve doesn't it? You've just factored a polynomial. In case you want a video that explains all of what I just wrote, here is one that I made (you may need to go through about 9 minutes to get to the factoring part): Hope that helped :) See below: #### Explanation: Almost al coefficients that you'll see will be real numbers. Coefficients are simply numbers in front of variables. In the monomial $3 {x}^{4}$, the coefficient is $3$, because it is what's multiplying the variable. Simply put, real numbers are numbers that can be plotted along a number line, excluding any imaginary parts. Numbers we deal with everyday like $3 , 6 , 41 , 279$ and even a million, are all examples of real numbers. Hope this helps! An irreducible polynomial is one that cannot be factored into simpler (lower degree) polynomials using the kind of coefficients you are allowed to use, or is not factorisable at all. #### Explanation: Polynomials in a single variable ${x}^{2} - 2$ is irreducible over $\mathbb{Q}$. It has no simpler factors with rational coefficients. ${x}^{2} + 1$ is irreducible over $\mathbb{R}$. It has no simpler factors with Real coefficients. The only polynomials in a single variable that are irreducible over $\mathbb{C}$ are linear ones. Polynomials in more than one variable If you are given a polynomial in two variables with all terms of the same degree, e.g. $a {x}^{2} + b x y + c {y}^{2}$, then you can factor it with the same coefficients you would use for $a {x}^{2} + b x + c$. If it is not homogeneous then it may not be possible to factor it. For example, ${x}^{2} + x y + y + 1$ is irreducible. ## Questions • · 2 weeks ago • · 4 months ago • · 2 years ago • · 2 years ago • · 2 years ago • · 2 years ago • · 3 years ago • · 3 years ago • · 3 years ago • · 3 years ago • · 3 years ago • · 3 years ago • · 4 years ago<\|endoftext\|>	4.875
2,239	Tibet’s Long Journey to Freedom It has been likened to a Shangri-La, a nation cradled in the clouds and snowy peaks of the Himalayas, rich in natural resources and steeped in a centuries-old culture dominated by spirituality. It has also been described as hell on Earth, a land where, for the last six decades, its peoples have suffered a merciless repression that human rights organizations have described as cultural and religious genocide. Today, Tibet is a nation struggling for freedom, with its strongest advocates of human rights living in exile. Despite crushing oppression, Tibetans have refused to relinquish their rights to nationhood, their cultural identity or the free practice of their religion. Tibetan influence spread throughout Asia and China until the 19th century, with spiritual authority vested in the Dalai Lama, the highest religious authority in Tibetan Buddhism. A continuous succession of Dalai Lamas has held the position since the late 1500s. A firmly established religious leader who held the steadfast loyalty of Buddhists throughout Tibet, and whose word was law among followers, the Dalai Lama was a force to be reckoned with by outside interests. The Tibetan Plateau, known as the “roof of the world,” is the highest region on Earth with an average elevation of 14,800 feet. Tibet’s cultural and religious traditions date back millennia. In an effort to strengthen its weakening grip on the country, Qing military forces from China ousted the 13th Dalai Lama in 1910, forcing him to flee to British-ruled India. But the Qing regime’s hold on Tibet didn’t last. In 1912, a revolution toppled the dynasty, and in 1913 the Dalai Lama returned as ruler of an independent Tibet, an independence that, sadly, would not long stand. In 1951, Tibet was overtaken by the People’s Republic of China, again forcing the Dalai Lama into exile. Next came Mao’s Great Leap Forward, which sought to eradicate all forms of religion. Hundreds of thousands of Tibetans were exterminated in an effort to erase the practice of Tibetan Buddhism. Approximately 6,000 monasteries were destroyed, including many of the religion’s most sacred sites. Protesters were either shot on sight or held as political prisoners and tortured mercilessly, sometimes for years on end. Despite the unrelenting repression, Tibetan citizens continued to demand the return of their citizenship and human rights through a campaign of passive resistance, a hallmark of their faith. Some demonstrated peacefully, only to be met with arrest or bullets. Outraged human rights organizations later described the assault as “cultural genocide.” Lhasa, the nation’s second most populous city, is home to the oldest and most sacred temples of Tibetan Buddhism. The Venerable Tenzin Bagdro, a man who would eventually become one of Tibet’s most outspoken advocates of religious freedom and human rights, was 13 when he first witnessed the suppression of his people. “I experienced the Cultural Revolution,” he said. “I saw monks being disrobed and brutally beaten when they voiced their opinions. Thousands of old scriptures were burnt; many monasteries were destroyed. Around the time I was admitted to a Chinese school, we were denied the study of our own language and traditions.” Repression fueled a burning desire in Bagdro to fight for freedom. In 1985, he became a Buddhist monk. Three years later, in the city of Lhasa, during one of the most important religious festivals in Tibet, he helped to lead a protest against the repression. Knowing military authorities would likely seize him, Bagdro made a promise to himself: “If I am caught, I won’t give the Chinese the names of any of my co-conspirators.” The day of the protest, Chinese troops fired randomly into the crowd. A 13-year-old girl was struck in the chest and died in front of him. Fellow monks were thrown from terraces of surrounding buildings and left bleeding and broken in the streets. “Jokhang Temple in Lhasa is one of the holiest. That day it was covered in blood,” Bagdro said. Tibet contains some of the richest mineral deposits on the planet, including copper, iron, lithium, gold, oil and uranium, estimated at more than $128 billion. Arrested a month later, Bagdro was incarcerated for three years. Hell would have been easier to live through. “Every day I was beaten,” Bagdro said. “They put an electric baton on my neck, in my ears and in my mouth. I was asked to kneel down on crushed glass while they put a hot iron on my feet. During the night, they tied me upside down. The female prisoners were being raped repeatedly by the prison policemen every night.” When he was released after three terrible years, his promise not to reveal the names of his friends was unbroken. Freedom led Bagdro to Dharamshala, India, and a special audience with His Holiness, the 14th Dalai Lama, a preeminent advocate for human rights in Tibet and throughout the world. For Bagdro, the meeting rekindled his purpose to fight for a free Tibet. And so his voice joined a chorus of Tibetan expatriates working toward that goal. Following the 1959 invasion that began Bagdro’s harrowing journey, the Dalai Lama established a government-in-exile in Dharamshala. Buddhist monasteries razed in Tibet during the Cultural Revolution were rebuilt in their new home to educate new generations of Buddhist monks and nuns to preserve the teachings of Tibetan Buddhism. In approximately 640 AD, under the rule of Songtsan Gampo, Buddhism was established in Tibet. During the next two centuries, the religion became a dominant force throughout Central Asia, and has remained part of the spiritual life of the culture to this day. The keeping of Tibetan cultural traditions—which were being systematically eradicated by the Chinese—was equally important, and depended on a generation of children separated from their families or orphaned by decades of violence. From 1964 to 2006, Jetsun Pema, sister of the Dalai Lama, presided over the Tibetan Children’s Villages (TCV), a thriving, integrated educational community for destitute children in exile, with the purpose “to ensure all Tibetan children under its care receive a sound education, a firm cultural identity and become self-reliant and contributing members of the Tibetan community and the world at large.” And so Tibet began to gradually rebuild its cultural identity and focus world attention on the need for human rights reforms. That effort continues to this day and is in keeping with the Universal Declaration of Human Rights, adopted by the United Nations in 1948, the first and most expansive expression of the universal and inherent rights of every person. Youth for Human Rights International (YHRI), a Los Angeles-based nonprofit organization, promotes the Declaration with a program that educates people and governments around the world on the rights it guarantees. Fiorella Cerchiara and Annalisa Tosoni of the Association for Human Rights and Tolerance of Italy, an affiliate of YHRI, presented a copy of the group’s publication What Are Human Rights? to Tibetan community leader in exile Yakar Gelek. “We dreamed one day we would have a United Tibet project,” said Gelek. “That day has come. This is the peace project we have been looking for. We must bring this project to all members of our community, the Tibetans in exile, so we guarantee our future as a culture.” In May 2010, Jetsun Pema was honored by the Association for Human Rights and Tolerance at an event in Milan for her role in establishing the Tibetan Children’s Villages and her 42-year dedication to saving refugee children. She went on to implement a Youth for Human Rights International educational program in Tibetan schools and monasteries throughout India. In 2013, the Venerable Tenzin Bagdro was invited to Italy to receive special recognition for his work to advance the cause of human rights, where he was introduced to the Youth for Human Rights program and educational materials. Bagdro saw how the materials could be used to educate future generations on the importance of their human rights. “There are two aspects to the program,” Cerchiara explained. “Educating Tibet’s future generation included giving them a complete understanding of exactly what those rights [spelled out in the Universal Declaration] are so they can become advocates for protecting those freedoms. “But not only the children needed to be educated, so did many of those who had been working for years to protect human rights. One man came to me in surprise and said, ‘I’ve been fighting for human rights for decades, but I never really understood exactly what those rights were!’” From this grew United for Tibet, a project dedicated to protecting the cause of human rights through education. In 2014, lectures and 5,000 copies of What Are Human Rights? were presented to monks of the Golden Temple, the largest Tibetan Buddhist monastery in India. Two hundred copies of the educator’s guide were presented to teachers in the Tibetan Children’s Villages to educate young people on their fundamental human rights. As a result, Yakar Gelek, Annalisa Tosoni and Fiorella Cerchiara drafted a plan to translate into Tibetan all the materials of the campaign, including teacher lesson plans, workbooks and audiovisual material, so the entire program could be implemented throughout the Tibetan community of India. But before such a massive plan could be fully implemented, it required an approval from His Holiness, the Dalai Lama. And so the materials were brought to Tibet’s spiritual leader, who not only approved the plan, but included his personal prayer to be printed in the front of all translated versions of What Are Human Rights? “The Universal Declaration of Human Rights is a monumental document of the United Nations that establishes the fundamental equality and rights of all people on this planet, irrespective of their race, color, creed, or whether they are believers or not. We should all show due respect to this declaration and abide by its provisions. “I am grateful and happy that the Association for Human Rights and Tolerance has arranged to translate this important document into Tibetan and is now also publishing it in book form to distribute it among Tibetan youth. “The battle is far from being won,” says Cerchiara. “But now those fighting for the cause have a tool that can be used to make the importance of human rights real to others.”<\|endoftext\|>	3.875
595	Bugs are a fact of life and many toddlers are fascinated by them. You might discover them in your toddler’s pocket or find him leaving crumbs on the sidewalk to feed them. Take advantage of his curiosity and desire to know more about bugs by creating a lesson plan you can use in small increments, especially with the bug season is in full swing. If your little one enjoys crafts, you can use that to teach her about bug metamorphosis. Your toddler can create a caterpillar using an egg carton with a pipe cleaner poked through the side of each cup for legs. Add two more pipe cleaners poked into the top of the front cups to form the antennae. She can draw bug eyes with a crayon and decorate it with stickers or nontoxic finger paints. To create a chrysalis, wrap tissue paper around a small wad of paper and glue the tissue paper together. Attach it to a twig and hang it out of reach. The final transformation is a butterfly. Use pipe cleaners to form each of the four butterfly wing sections and a piece of tightly rolled paper creased in three sections to form the head, thorax and abdomen. Place the sections in the correct orientation and cover the butterfly with clear contact paper. Allow your toddler to stick small bits of tissue paper to the contact paper to create colorful wing sections. Finish the butterfly by adding small pipe cleaners to form the six legs and antennae. Songs and Stories There are several songs your toddler can use in his bug lessons. Consider singing “Itsy Bitsy Spider,” “Fuzzy Caterpillar,” “Baby Bumblebee” and “The Ants go Marching” during your bug lessons. Teach your little one the finger plays and motions that go with these songs. Supplement your lessons with books such as “Where Butterflies Grow” by Joanne Ryder, Eric Carle’s “Very Hungry Caterpillar” and “I Like Bugs” by Margaret Wise Brown. Your toddler can observe bugs in a natural habitat. Take a walk through the garden and point out bees and other flying insects pollinating flowers or observe ants in an ant farm. Walk along a creek and turn over rocks and branches to watch the insects scatter. Ensure that your little one only touches bugs your tell him are safe. A trip to the zoo will offer many insects your toddler can’t touch or bring home in his pocket. Your toddler can learn that bugs get around in many different ways. Tell her, “Crawl like a caterpillar” or “Flutter like a butterfly.” Choose various insects to emulate such as crickets, ants and spiders. She can try to buzz like a bee or chirp like a cricket. Give her a scarf and let her pretend to be a bee, firefly or butterfly.<\|endoftext\|>	3.671875
427	Amnesia is when people can no longer memorise or recall information that stored in memory. It is very rare despite being a favourite theme in movies and books. Being a little forgetful is utterly different to having amnesia. Amnesia refers to a large-scale loss of a memory that should not be forgotten. These may include important milestone in life’s memorable events and keep people our lives and vital facts that have been told or taught. Amnesia is an inability to lay down new memories and recall old memories or both. The symptoms of amnesia may include confusion and uncoordinated movements. Alcohol abuse can lead to a type of amnesia known as Wernicke-Korsakoff’s psychosis. The cause of amnesia can be by many things including traumatic experiences, brain injuries, stroke, brain surgery, and infection in the brain. However, amnesia can be caused by a concussion that is when a bump in people’s head or body rattles their brain around in their skull, damaging delicate tissue. Sometimes that damage can make people feel sleepy or a headache for a few days, but the effects can be more serious. There are two kinds of amnesia; retrograde and anterograde. Concussions can cause both of them. Retrograde amnesia is when people forget things that happened in the past, for instance, people’s birth date. On the other hand, anterograde amnesia is where people cannot make new memories. People with amnesia also find it hard to remember the past, memorise new information and imagine the future. The reason is that people construct future scenarios by their recollections of past experiences. People’s ability to recollect events and experiences involves a variety of complex brain processes. People still don’t understand what happens when they commit something to memory, or when they try to retrieve data stored in the brain. Most people with amnesia are usually lucid and have a sense of self. However, they may experience severe difficulties in learning new information, struggle to recall memories of past experiences, or both.<\|endoftext\|>	3.8125
221	Chinese New Year The Chinese New Year is a time of great celebration, symbolism and mythology. It's known as the Chinese Lunar New Year or the Spring Festival. It celebrates the end of the winter season and the seasonal transition into spring. The Chinese New Year festival begins on the first day of the first month and ends with the Lantern Festival, celebrated on the 15th day of the traditional Chinese calendar. Most Chinese families traditionally celebrate this holiday with a large family dinner. The holiday is celebrated for fifteen days. Each day brings new traditions and celebrations. For example, on the fifth day Chinese families eat dumplings, give thanks to the god of wealth and set off firecrackers. Chinese New Year's Classroom Resources. With so much rich tradition and mythology to explore, the Chinese New Year can be a very fun classroom unit. Explore the worksheets, clip art and lesson plans available at TeacherPlanet.com. They're all designed to help you bring this rich holiday into your classroom. Help your students embrace other cultures and explore the rich traditions of the Chinese culture.<\|endoftext\|>	3.84375
136	# How do you simplify (12sqrt2)^2? May 12, 2018 288 #### Explanation: ${\left(12 \sqrt{2}\right)}^{2}$ = $\left(12 \sqrt{2}\right) \left(12 \sqrt{2}\right)$ =$\left(12\right) \left(12\right) \left(\sqrt{2}\right) \left(\sqrt{2}\right)$ (12)(12)((sqrt2)^2 =$\left(12\right) \left(12\right) \left(2\right)$ =$288$<\|endoftext\|>	4.46875
2,008	# Statistics Module 1 The flashcards below were created by user larry.gish89 on FreezingBlue Flashcards. 1. bimodal distribution If data have exactly two modes 2. multimodal distribution If data have more than two modes 3. arithmetic mean • An important property of the arithmetic mean is that the sum of the deviations from the • mean will always equal 0. • arithmetic mean is sensitive the to extreme values. We often refer to these extreme values • (whether small or large) as outliers. 4. Organization of Frequency and Relative Frequency Distributions • Step 1: Determine the number of class intervals (also referred • to as classes or categories, and meaning ranges of data values) • of interest. Between 5 and 20 class intervals are generally recommended. Class intervals • should be selected so that each data point (or individual value of raw data) can fall into • only one category. • Step 2: When it is desired that all class intervals be of equal • width, you determine the width of the class by subtracting the smallest data value from • the largest and then dividing by the number of class intervals desired. 5. Weighted Mean If you don't want to weigh the data points equally, but by their relative importance. • (95 0.10) +(70 0.10)+ (60 0.10)+ (85 0.30)+ (90 0.40)= 84 (weighted mean) 6. dispersion or variability • measures of • variation 7. range ... is one measure of dispersion of a data set. range = largest data value – smallest data value • The problem with the range is that it includes only two numbers of the data set and • ignores the rest of the values. 8. Percentiles • Percentiles of a ranked data set divide it into hundredths, or 100 equal parts • of the data values. The median is the 50th percentile. Fifty percent of the data falls • below the median and 50 percent falls above it. • Quartiles are the percentiles that divide the data into quarters (or fourths). • There are three quartiles, then, at the 25th, 50th, and 75th percentiles. We often refer • to these as Q1, Q2, and Q3, respectively. 9. How we calculate percentiles: • 1. Arrange the data in ascending order from the smallest value to the largest. • 2. Compute the index i: • where • i is the position number of the percentile you're interested in • p is the percentile you're interested in knowing • n is the number of items in the data set • 3. If i is not an integer, round up to the nearest integer. The next integer value • greater than i • denotes the position of the pth percentile. • If i is an integer, the pth percentile is the average of the data values • in positions i and i + 1. 10. Inter-quartile Range • The inter-quartile range is the 75th percentile minus the 25th percentile, or Q3 – • Q1. This range has less dependency on outliers than does the range previously discussed. 11. variance of a data set • is an important measure of dispersion • within a data set because it takes into account all the data values 12. The variance of the population • is the average of the squared deviations from the • arithmetic mean. When you take the variance of a sample, you divide the squared deviations • from the sample mean by the sample size minus 1. Doing this generally gives a better • estimate of the population variance from which the sample comes. 13. denote the variance of a population with... "little sigma squared." 14. We denote the sample variance s 2 (squared) 15. The population mean and the population variance are called • parameters of a • population because they are quantities that are fixed for any given population. 16. We call the sample mean and the sample variance • sample statistics (or random • variables) because they vary from one sample to another, inasmuch as their values depend • on which sample is selected. 17. Use the following steps to calculate the sample variance: • 1. Calculate the sample mean. • 2. Calculate the difference between each observation and the sample mean. • 3. Square each difference found in step 2. • 4. Sum the squared differences found in step 3. 5. Divide the sum of the squared differences by the sample size minus one, n – 1. 18. Frequently, we use the ________ _________ instead of the variance to describe dispersion. • the standard deviation. You get the standard deviation by taking the square root of the • variance. ( sample variance = s squared/population variance = "little sigma squared") • The advantage of using the standard deviation is that it has the same units of • measurement as the data values. 19. We represent this statistic with s, meaning "the square root of the variance, s2 (squared)." This is the representation for standard deviation. 20. What does the standard deviation actually mean? • The standard deviation shows how the • data points are distributed or dispersed about the sample mean. When the things you are • measuring are alike, such as test scores from the same class, the bigger the standard • deviation, the more dispersion you have about the mean. 21. Coefficient of Variation • When two (or more) distributions have the same mean, the one with the largest standard • deviation has the most variation. But what about when distributions have different means? • In that case, you can't compare just the standard deviations. Instead, you have to compare • the coefficient of variation (CV) for each distribution as well. The distribution with the • highest CV has the most dispersion. 22. Empirical Rule • This rule applies to data that are approximately normally distributed, that is, a • bell-shaped symmetrical distribution. About 68 percent of the data points will fall within • one standard deviation of the mean, and about 95 percent of the data points will be within • two standard deviations of the mean. • For example, let's continue with our inquiry into salaries but with a different • profession. Let's take a sample of the salaries of 150 production workers. Here's a • distribution of salaries we might find. 23. Chebyshev's Theorem • tells us • the minimum proportion of data points that lie within any number of standard deviations • from the mean, regardless of the shape of the distribution. Chebyshev's theorem states: • At least of • the measurements fall within k standard deviations from the mean. • Note: k must be greater than 1. • For example, if you want to find out the minimum percentage of the data values that are • within 2 standard deviations from the mean, you'd calculate: • That is, for any data set, at least 75 percent of the data values are within two • standard deviations from the mean. • If you calculate the minimum percentage of values are between the mean and three • standard deviations from the mean, you'll get an answer of "at least 89 • percent." • Although Chebyshev's theorem provides us only with lower bounds for the percentage of • data values that lie within k (where k >1) standard deviations from the • mean, it doesn't provide us with exact percentages. The power of Chebyshev's theorem lies • in the fact that it is true for any distribution, regardless of its shape. 24. The ____ _ ______ compares the standard deviation relative to the mean of the distribution. For this reason, the CV is also known as the ______ ______ _____ (RSE). • coefficient of variation; • relative standard error 25. Here's how we calculate the CV... • Think of the CV for any variable as the precision of the mean for that variable. Many • federal agencies, such as the National Center for Health Statistics (NCHS), use the CV as • a measure of the precision or reliability of estimates of health characteristics. The • smaller the CV, the more reliable (precise) the estimate is. The larger the CV, the more • unreliable it is. 26. Shapes of distributions • 1. Symmetrical distributions- Has the same center value for the mean, median, and mode. • (mirrored appearance) 2. Uniform of Rectangular Distribution- Every class has the same frequency. 3. Skewed Distribution- One "tail" is longer than the other. • If the longer tail is on the left, we say that the distribution is skewed to the • left, or negatively skewed. If the longer tail is to the right, we say the • distribution is skewed to the right, or positively skewed. • 4. Bimodal Distribution- A bimodal distribution refers to a histogram in which two classes with largest • frequencies are separated by at least one class, and the top two frequencies of these • classes may have different values. Author: larry.gish89 ID: 231116 Card Set: Statistics Module 1 Updated: 2013-08-24 04:33:08 Tags: Statistics gish1 Folders: Description: Module one study cards Show Answers:<\|endoftext\|>	4.59375
210	Students often receive feedback telling them to ‘develop their ideas’, but they frequently have no idea how to do it. This series of posters is designed to flesh out PEEL – a popular acronym often used to help young writers support their ideas with evidence and explanation. This series of posters was created in order to flesh out exactly what each of the terms in PEEL mean. Hopefully, when a student is able to make a POINT, support it with EVIDENCE and provide EXPLANATION that LINKS to their point, they will be able to produce developed pieces of writing. These posters support formative assessment practices by allowing students to more clearly see where they are in their learning and better understand the steps that need to be taken in order to improve. Please feel free to download this resource and use it to inform your learning conversations. If you can think of anything that can be added to improve the resouce, don’t be shy about letting me know. Download here: PEEL Posters<\|endoftext\|>	3.875
2,857	Home Math Multiplying 2-Digit Quantity by 1-Digit Quantity ### Multiplying 2-Digit Quantity by 1-Digit Quantity Right here we are going to be taught multiplying 2-digit quantity by 1-digit quantity. In two other ways we are going to be taught to multiply a two-digit quantity by a one-digit quantity. Examples of multiplying 2-digit quantity by 1-digit quantity with out Regrouping: We could have a fast overview of multiplication of 2-digit quantity by 1-digit quantity with out regrouping: 1. Multiply 24 by 2. T O 2 4 × 2 4 8 First multiply those by 2. 4 × 2 = 8. Write 8 underneath O. Now multiply the tens by 2. 3 × 3 = 9. Write 9 underneath T. 2. Multiply 34 and a pair of Resolution: Step I: Prepare the numbers vertically. Step II: First multiply the digit on the ones place by 2. 2 × 4 = 8 ones Step III: Now multiply the digit on the tens place by 2. 2 × 3 = 6 tens Thus, 34 × 2 = 68 3. Multiply 20 by 3 by utilizing expanded kind Resolution: 20 → 2 tens + 0 ones × 3 → × 3 6 tens + 0 ones = 60 + 0 = 60 Due to this fact, 20 × 3 = 60 4. Multiply 50 by 1 by utilizing quick kind Resolution: 50 → 50 × 1 → × 1 0 50 (i) First digit of 1’s place is multiplied by 1, i.e., 0 × 1 = 0 (ii) Then digit at ten’s place is multiplied by 1, i.e., 5 tens × 1 = 5 tens Therefore, 50 × 1 = 50 Observe the next Instance utilizing Three Completely different Strategies: 5. Multiply 13 by 2. Resolution: 13 x 2 = 13 + 13 = 26 Due to this fact, 13 x 2 = 26. Second Methodology: Utilizing Expanded Kind Contemplate 13 as 10 + 3. 13 × 2 = (10 + 3) × 2 = 10 × 2 + 3 × 2 = 20 + 6 = 26. Third Methodology: Quick Kind Write the numbers in response to place worth proven on the suitable. Step I: Multiply those: 3 ones × 2 = 6 ones Write 6 underneath ones column. Step II: Multiply the tens: 1 ten × 2 = 2 tens Write 2 underneath tens column. Thus, the product of 13 and a pair of is 26. Examples of multiplying 2-digit quantity by 1-digit quantity with Regrouping: 1. Multiply 66 by 3 T O 1 6 6 × 3 1 4 8 First multiply those by 3. 6 × 3 = 18 = one ten + 8 ones Write 8 underneath O. carry 1 ten Now multiply the tens by 3. 6 × 3 = 18 Add 1 to the product. 18 + 1 = 19 2. Multiply 25 by 3 Step I: Prepare the numbers vertically. Step II: First multiply the digit on the ones place by 3. 3 × 5 = 15 = 1 ten + 5 ones Write 5 within the ones column and carry over 1 to the tens column Step III: Now multiply the digit on the tens place by 3. 3 × 2 = 6 tens Now, 6 + 1 (carry over) = 7 tens Thus, 25 × 3 = 75 3. Multiply 46 by 4 Step I: Prepare the numbers vertically. Step II: Multiply the digit on the ones place by 4. 6 × 4 = 24 = 2 tens + 4 ones Write 4 within the ones column and carry over 2 to the tens column Step III: Now multiply the digit on the tens place by 4. 4 × 4 = 16 tens Now, 16 + 2 (carry over) = 18 tens = 1 hundred + 8 tens Write 8 on the tens place and 1 on the hundred place. Thus, 46 × 4 = 184 4. Multiply 20 by 3 by utilizing expanded kind Resolution: 20 → 2 tens + 0 ones × 3 → × 3 6 tens + 0 ones = 60 + 0 = 60 Due to this fact, 20 × 3 = 60 5. Multiply 26 by 7 by utilizing expanded kind Resolution: 26 → 20 + 6 → 2 tens + 6 ones × 7 → × 7 → × 7 (2 × 7) tens + (6 × 7) ones 2 tens + 6 ones × 7 ones 14 tens + 42 ones = 14 tens + (40 + 2) ones = 14 tens + 4 tens + 2 ones = 18 tens + 2 ones = 180 + 2 = 182 Due to this fact, 26 × 7 = 182 6. Multiply 48 by 6 by utilizing quick kind Resolution: 48 × 6 24 ← 48 = 28 tens 8 ones = 288 Therefore, 48 × 6 = 288 (i) 48 × 6 is written in column from. (ii) 8 ones are multiplied by 6, i.e., 6 × 8 = 48 ones = 4 tens + 8 ones 8 is written is one’s column and 4 tens is gained. (iii) Gained 4 is carried to the ten’s column. (iv) Now 4 tens is multiplied by 6, i.e., 4 tens × 6 = 24 tens (v) Carried 4 tens is added to 24 tens, i.e., 4 tens + 24 tens = 28 tens 7. Discover the product of 58 × 5. Resolution: 58 × 5 25 ← 40 = 25 + 4 ← 0 = 29 0 = 290 (i) 8 ones × 5 = 40 = 4 tens + 0 one (ii) 5 tens × 5 = 25 tens (iii) 25 tens + 4 tens = 29 tens Therefore, 58 × 5 = 290 8. Multiply 37 by 8 Resolution: 3 7 × 8 5 6 + 2 4 0 2 9 6 (i) 7 ones × 8 = 56 ones = 5 tens 6 ones 56 is positioned in such method that 5 comes underneath tens and 6 underneath ones (ii) 3 tens × 8 = 24 tens = 240 ones = 2 a whole lot, 4 tens and 0 ones 240 is positioned under 56 in such method that 2 comes underneath a whole lot, 4 underneath tens and 0 underneath ones. Therefore, 37 × 8 = 296 Questions and Solutions on Multiplying 2-Digit Quantity by 1-Digit Quantity: Multiplication of 2-Digit Quantity by 1-Digit Quantity With out Regrouping: I. Discover the product: (i) 23 × 3 = (ii) 44 × 2 = (iii) 33 × 2 = (iv) 22 × 4 = (v) 32 × 3 = (vi) 40 × 2 = (vii) 43 × 2 = (viii) 12 × 3 = (ix) 23 × 2 = (x) 11 × 9 = (xi) 21 × 4 = (xii) 13 × 3 = I. (i) 69 (ii) 88 (iii) 66 (iv) 44 (v) 96 (vi) 80 (vii) 86 (viii) 36 (ix) 46 (x) 99 (xi) 84 (xii) 39 Multiplication of 2-Digit Quantity by 1-Digit Quantity With Regrouping: II. Discover the product: (i) 46 × 2 (ii) 19 × 4 (iii) 27 × 3 (iv) 18 × 5 II. (i) 92 (ii) 76 (iii) 81 (iv) 90 III. Multiply the next: (i) 78 × 4 (ii) 63 × 6 (iii) 51 × 6 (iv) 39 × 8 (v) 72 × 9 (vi) 45 × 7 (vii) 17 × 4 (viii) 88 × 8 III. (i) 312 (ii) 398 (iii) 306 (iv) 312 (v) 648 (vi) 315 (vii) 68 (viii) 704 IV. Remedy the next: (i) 37 × 6 (ii) 72 × 4 (iii) 56 × 7 (iv) 84 × 2 (v) 45 × 9 IV. (i) 37 × 6 (ii) 72 × 4 (iii) 56 × 7 (iv) 84 × 2 (v) 45 × 9 V. Multiply the next : (i) T O 3 1 × 2 _______ (ii) T O 4 7 × 1 _______ (iii) T O 1 1 × 3 _______ (iv) T O 2 2 × 2 _______ (v) T O 2 3 × 2 _______ (vi) T O 2 6 × 3 _______ (vii) T O 4 9 × 2 _______ (viii) T O 2 3 × 4 _______ (ix) T O 1 6 × 6 _______ (x) T O 1 9 × 5 _______ (xi) T O 5 2 × 5 _______ (xii) T O 2 3 × 6 _______ (xiii) T O 6 4 × 9 _______ (xiv) T O 3 2 × 7 _______ (xv) T O 7 5 × 8 _______ VI. Multiply the next: (i) 21 × 5 = _____ (ii) 34 × 2 = _____ (iii) 23 × 3 = _____ (iv) 27 × 3 = _____ (v) 38 × 2 = _____ (vi) 18 × 4 = _____ (vii) 25 × 8 = _____ (viii) 32 × 6 = _____ (ix) 29 × 4 = _____ (x) 45 × 5 = _____ Did not discover what you had been in search of? Or wish to know extra info<\|endoftext\|>	4.96875
12,557	Below is the uncorrected machine-read text of this chapter, intended to provide our own search engines and external engines with highly rich, chapter-representative searchable text of each book. Because it is UNCORRECTED material, please consider the following text as a useful but insufficient proxy for the authoritative book pages. 1 INTRODUCTION 1 Introduction M. Suzanne Donovan and John D. Bransford More than any other species, people are designed to be flexible learners and, from infancy, are active agents in acquiring knowledge and skills. People can invent, record, accumulate, and pass on organized bodies of knowledge that help them understand, shape, exploit, and ornament their environment. Much that each human being knows about the world is acquired informally, but mastery of the accumulated knowledge of generations requires inten- tional learning, often accomplished in a formal educational setting. Decades of work in the cognitive and developmental sciences has pro- vided the foundation for an emerging science of learning. This foundation offers conceptions of learning processes and the development of competent performance that can help teachers support their students in the acquisition of knowledge that is the province of formal education. The research litera- ture was synthesized in the National Research Council report How People Learn: Brain, Mind, Experience, and School.1 In this volume, we focus on three fundamental and well-established principles of learning that are high- lighted in How People Learn and are particularly important for teachers to understand and be able to incorporate in their teaching: 1. Students come to the classroom with preconceptions about how the world works. If their initial understanding is not engaged, they may fail to grasp the new concepts and information, or they may learn them for pur- poses of a test but revert to their preconceptions outside the classroom. 2. To develop competence in an area of inquiry, students must (a) have a deep foundation of factual knowledge, (b) understand facts and ideas in the context of a conceptual framework, and (c) organize knowledge in ways that facilitate retrieval and application. 2 HOW STUDENTS LEARN 3. A âmetacognitiveâ approach to instruction can help students learn to take control of their own learning by defining learning goals and monitoring their progress in achieving them. A FISH STORY The images from a childrenâs story, Fish Is Fish,2 help convey the es- sence of the above principles. In the story, a young fish is very curious about the world outside the water. His good friend the frog, on returning from the land, tells the fish about it excitedly: âI have been about the worldâhopping here and there,â said the frog, âand I have seen extraordinary things.â âLike what?â asked the fish. âBirds,â said the frog mysteriously. âBirds!â And he told the fish about the birds, who had wings, and two legs, and many, many colors. As the frog talked, his friend saw the birds fly through his mind like large feathered fish. The frog continues with descriptions of cows, which the fish imagines as black-and-white spotted fish with horns and udders, and humans, which the fish imagines as fish walking upright and dressed in clothing. Illustra- tions below from Leo Lionniâs Fish Is Fish Â© 1970. Copyright renewed 1998 by Leo Lionni. Used by permission of Random House Childrenâs Books, a division of Random House, Inc. 4 HOW STUDENTS LEARN Principle #1: Engaging Prior Understandings What Lionniâs story captures so effectively is a fundamental insight about learning: new understandings are constructed on a foundation of existing understandings and experiences. With research techniques that permit the study of learning in infancy and tools that allow for observation of activity in the brain, we understand as never before how actively humans engage in learning from the earliest days of life (see Box 1-1). The understandings children carry with them into the classroom, even before the start of formal schooling, will shape significantly how they make sense of what they are The Development of Physical Concepts in Infancy BOX 1-1 Research studies have demonstrated that infants as young as 3 to 4 months of age develop understandings and expectations about the physical world. For ex- ample, they understand that objects need support to prevent them from falling to the ground, that stationary objects may be displaced when they come into contact with moving objects, and that objects at rest must be propelled into motion.3 In research by Needham and Baillargeon,4 infants were shown a table on which a box rested. A gloved hand reached out from a window beside the table and placed another box in one of two locations: on top of the first box (the possible event), and beyond the boxâcreating the impression that the box was suspended in midair. In this and similar studies, infants look reliably longer at the impossible events, suggesting an awareness and a set of expectations regarding what is and is not physically possible. SOURCE: Needham and Baillargeon (1993). Reprinted with permission from Elsevier. 5 INTRODUCTION Misconceptions About Momentum BOX 1-2 Andrea DiSessa5 conducted a study in which he compared the performance of college physics students at a top technological university with that of elementary schoolchildren on a task involving momentum. He instructed both sets of students to play a computerized game that required them to direct a simulated object (a dynaturtle) so that it would hit a target, and to do so with minimum speed at im- pact. Participants were introduced to the game and given a hands-on trial that al- lowed them to apply a few taps with a wooden mallet to a ball on a table before they began. DiSessa found that both groups of students failed miserably at the task. De- spite their training, college physics majorsâjust like the elementary school chil- drenâapplied the force when the object was just below the target, failing to take momentum into account. Further investigation with one college student revealed that she knew the relevant physical properties and formulas and would have per- formed well on a written exam. Yet in the context of the game, she fell back on her untrained conceptions of how the physical world works. taught. Just as the fish constructed an image of a human as a modified fish, children use what they know to shape their new understandings. While prior learning is a powerful support for further learning, it can also lead to the development of conceptions that can act as barriers to learn- ing. For example, when told that the earth is round, children may look to reconcile this information with their experience with balls. It seems obvious that one would fall off a round object. Researchers have found that some children solve the paradox by envisioning the earth as a pancake, a âroundâ shape with a surface on which people could walk without falling off.6 How People Learn summarizes a number of studies demonstrating the active, preconception-driven learning that is evident in humans from infancy through adulthood.7 Preconceptions developed from everyday experiences are often difficult for teachers to change because they generally work well enough in day-to-day contexts. But they can impose serious constraints on understanding formal disciplines. College physics students who do well on classroom exams on the laws of motion, for example, often revert to their untrained, erroneous models outside the classroom. When they are con- fronted with tasks that require putting their knowledge to use, they fail to take momentum into account, just as do elementary students who have had no physics training (see Box 1-2). If studentsâ preconceptions are not ad- dressed directly, they often memorize content (e.g., formulas in physics), yet still use their experience-based preconceptions to act in the world. 6 HOW STUDENTS LEARN Principle #2: The Essential Role of Factual Knowledge and Conceptual Frameworks in Understanding The Fish Is Fish story also draws attention to the kinds of knowledge, factual and conceptual, needed to support learning with understanding. The frog in the story provides information to the fish about humans, birds, and cows that is accurate and relevant, yet clearly insufficient. Feathers, legs, udders, and sport coats are surface features that distinguish each species. But if the fish (endowed now with human thinking capacity) is to under- stand how the land species are different from fish and different from each other, these surface features will not be of much help. Some additional, critical concepts are neededâfor example, the concept of adaptation. Spe- cies that move through the medium of air rather than water have a different mobility challenge. And species that are warm-blooded, unlike those that are cold-blooded, must maintain their body temperature. It will take more explaining of course, but if the fish is to see a bird as something other than a fish with feathers and wings and a human as something other than an upright fish with clothing, then feathers and clothing must be seen as adap- tations that help solve the problem of maintaining body temperature, and upright posture and wings must be seen as different solutions to the prob- lem of mobility outside water. Conceptual information such as a theory of adaptation represents a kind of knowledge that is unlikely to be induced from everyday experiences. It typically takes generations of inquiry to develop this sort of knowledge, and people usually need some help (e.g., interactions with âknowledgeable oth- ersâ) to grasp such organizing concepts.8 Lionniâs fish, not understanding the described features of the land ani- mals as adaptations to a terrestrial environment, leaps from the water to experience life on land for himself. Since he can neither breathe nor maneu- ver on land, the fish must be saved by the amphibious frog. The point is well illustrated: learning with understanding affects our ability to apply what is learned (see Box 1-3). This concept of learning with understanding has two parts: (1) factual knowledge (e.g., about characteristics of different species) must be placed in a conceptual framework (about adaptation) to be well understood; and (2) concepts are given meaning by multiple representations that are rich in factual detail. Competent performance is built on neither factual nor concep- tual understanding alone; the concepts take on meaning in the knowledge- rich contexts in which they are applied. In the context of Lionniâs story, the general concept of adaptation can be clarified when placed in the context of the specific features of humans, cows, and birds that make the abstract concept of adaptation meaningful. 7 INTRODUCTION Learning with Understanding Supports Knowledge BOX 1-3 Use in New Situations In one of the most famous early studies comparing the effects of âlearning a proce- dureâ with âlearning with understanding,â two groups of children practiced throw- ing darts at a target underwater.9 One group received an explanation of refraction of light, which causes the apparent location of the target to be deceptive. The other group only practiced dart throwing, without the explanation. Both groups did equally well on the practice task, which involved a target 12 inches under water. But the group that had been instructed about the abstract principle did much better when they had to transfer to a situation in which the target was under only 4 inches of water. Because they understood what they were doing, the group that had received instruction about the refraction of light could adjust their behavior to the new task. This essential link between the factual knowledge base and a concep- tual framework can help illuminate a persistent debate in education: whether we need to emphasize âbig ideasâ more and facts less, or are producing graduates with a factual knowledge base that is unacceptably thin. While these concerns appear to be at odds, knowledge of facts and knowledge of important organizing ideas are mutually supportive. Studies of experts and novicesâin chess, engineering, and many other domainsâdemonstrate that experts know considerably more relevant detail than novices in tasks within their domain and have better memory for these details (see Box 1-4). But the reason they remember more is that what novices see as separate pieces of information, experts see as organized sets of ideas. Engineering experts, for example, can look briefly at a complex mass of circuitry and recognize it as an amplifier, and so can reproduce many of its circuits from memory using that one idea. Novices see each circuit sepa- rately, and thus remember far fewer in total. Important concepts, such as that of an amplifier, structure both what experts notice and what they are able to store in memory. Using concepts to organize information stored in memory allows for much more effective retrieval and application. Thus, the issue is not whether to emphasize facts or âbig ideasâ (conceptual knowl- edge); both are needed. Memory of factual knowledge is enhanced by con- ceptual knowledge, and conceptual knowledge is clarified as it is used to help organize constellations of important details. Teaching for understand- ing, then, requires that the core concepts such as adaptation that organize the knowledge of experts also organize instruction. This does not mean that that factual knowledge now typically taught, such as the characteristics of fish, birds, and mammals, must be replaced. Rather, that factual information is given new meaning and a new organization in memory because those features are seen as adaptive characteristics. 8 HOW STUDENTS LEARN Experts Remember Considerably More Relevant Detail Than BOX 1-4 Novices in Tasks Within Their Domain 9 INTRODUCTION In one study, a chess master, a Class A player (good but not a master), and a novice were given 5 seconds to view a chess board position from the middle of a chess game (see below). After 5 seconds the board was covered, and each participant at- tempted to reconstruct the board position on another board. This proce- dure was repeated for multiple trials until everyone received a perfect score. On the first trial, the master player correctly placed many more pieces than the Class A player, who in turn placed more than the novice: 16, 8, and 4, respectively. (See data graphed below.) However, these results occurred only when the chess pieces were arranged in configurations that conformed to meaningful games of chess. When chess pieces were randomized and presented for 5 seconds, the recall of the chess master and Class A player was the same as that of the noviceâthey all placed 2 to 3 positions correctly. The apparent difference in memory capacity is due to a difference in pattern recognition. What the expert can remember as a single meaningful pattern, novices must re- member as separate, unrelated items. Pieces correctly recalled 25 20 15 Master Class A player 10 Beginner 5 0 1 2 3 4 5 6 7 Trial SOURCE: Chase and Simon (1973). Reprinted with permission from Elsevier. 10 HOW STUDENTS LEARN Principle #3: The Importance of Self-Monitoring Hero though he is for saving the fishâs life, the frog in Lionniâs story gets poor marks as a teacher. But the burden of learning does not fall on the teacher alone. Even the best instructional efforts can be successful only if the student can make use of the opportunity to learn. Helping students become effective learners is at the heart of the third key principle: a âmetacognitiveâ or self-monitoring approach can help students develop the ability to take control of their own learning, consciously define learning goals, and moni- tor their progress in achieving them. Some teachers introduce the idea of metacognition to their students by saying, âYou are the owners and opera- tors of your own brain, but it came without an instruction book. We need to learn how we learn.â âMetaâ is a prefix that can mean after, along with, or beyond. In the psychological literature, âmetacognitionâ is used to refer to peopleâs knowl- edge about themselves as information processors. This includes knowledge about what we need to do in order to learn and remember information (e.g., most adults know that they need to rehearse an unfamiliar phone number to keep it active in short-term memory while they walk across the room to dial the phone). And it includes the ability to monitor our current understanding to make sure we understand (see Box 1-5). Other examples include moni- toring the degree to which we have been helpful to a group working on a project.10 Metacognitive Monitoring: An Example BOX 1-5 Read the following passage from a literary critic, and pay attention to the strategies you use to comprehend: If a serious literary critic were to write a favorable, full-length review of How Could I Tell Mother She Frightened My Boyfriends Away, Grace Plumbusterâs new story, his startled read- ers would assume that he had gone mad, or that Grace Plumbuster was his editorâs wife. Most good readers have to back up several times in order to grasp the meaning of this passage. In contrast, poor readers tend to simply read it all the way through without pausing and asking if the passage makes sense. Needless to say, when asked to para- phrase the passage they fall short. SOURCE: Whimbey and Whimbey (1975, p. 42). 11 INTRODUCTION In Lionniâs story, the fish accepted the information about life on land rather passively. Had he been monitoring his understanding and actively comparing it with what he already knew, he might have noted that putting on a hat and jacket would be rather uncomfortable for a fish and would slow his swimming in the worst way. Had he been more engaged in figuring out what the frog meant, he might have asked why humans would make them- selves uncomfortable and compromise their mobility. A good answer to his questions might have set the stage for learning about differences between humans and fish, and ultimately about the notion of adaptation. The con- cept of metacognition includes an awareness of the need to ask how new knowledge relates to or challenges what one already knowsâquestions that stimulate additional inquiry that helps guide further learning.11 The early work on metacognition was conducted with young children in laboratory contexts.12 In studies of âmetamemory,â for example, young children might be shown a series of pictures (e.g., drum, tree, cup) and asked to remember them after 15 seconds of delay (with the pictures no longer visible). Adults who receive this task spontaneously rehearse during the 15-second interval. Many of the children did not. When they were ex- plicitly told to rehearse, they would do so, and their memory was very good. But when the children took part in subsequent trials and were not reminded to rehearse, many failed to rehearse even though they were highly moti- vated to perform well in the memory test. These findings suggest that the children had not made the âmetamemoryâ connection between their re- hearsal strategies and their short-term memory abilities.13 Over time, research on metacognition (of which metamemory is consid- ered a subset) moved from laboratory settings to the classroom. One of the most striking applications of a metacognitive approach to instruction was pioneered by Palincsar and Brown in the context of âreciprocal teaching.â14 Middle school students worked in groups (guided by a teacher) to help one another learn to read with understanding. A key to achieving this goal in- volves the ability to monitor oneâs ongoing comprehension and to initiate strategies such as rereading or asking questions when oneâs comprehension falters. (Box 1-5 illustrates this point.) When implemented appropriately, reciprocal teaching has been shown to have strong effects on improving studentsâ abilities to read with understanding in order to learn. Appropriate kinds of self-monitoring and reflection have been demon- strated to support learning with understanding in a variety of areas. In one study,15 for example, students who were directed to engage in self-explana- tion as they solved mathematics problems developed deeper conceptual understanding than did students who solved those same problems but did not engage in self-explanation. This was true even though the common time limitation on both groups meant that the self-explaining students solved fewer problems in total. 12 HOW STUDENTS LEARN Helping students become more metacognitive about their own thinking and learning is closely tied to teaching practices that emphasize self-assess- ment. The early work of Thorndike16 demonstrated that feedback is impor- tant for learning. However, there is a difference between responding to feedback that someone else provides and actively seeking feedback in order to assess oneâs current levels of thinking and understanding. Providing sup- port for self-assessment is an important component of effective teaching. This can include giving students opportunities to test their ideas by building things and seeing whether they work, performing experiments that seek to falsify hypotheses, and so forth. Support for self-assessment is also provided by opportunities for discussion where teachers and students can express different views and explore which ones appear to make the most sense. Such questioning models the kind of dialogue that effective learners inter- nalize. Helping students explicitly understand that a major purpose of these activities is to support metacognitive learning is an important component of successful teaching strategies.17 Supporting students to become aware of and engaged in their own learning will serve them well in all learning endeavors. To be optimally effective, however, some metacognitive strategies need to be taught in the context of individual subject areas. For example, guiding oneâs learning in a particular subject area requires awareness of the disciplinary standards for knowing. To illustrate, asking the question âWhat is the evidence for this claim?â is relevant whether one is studying history, science, or mathematics. However, what counts as evidence often differs. In mathematics, for ex- ample, formal proof is very important. In science, formal proofs are used when possible, but empirical observations and experimental data also play a major role. In history, multiple sources of evidence are sought and attention to the perspective from which an author writes and to the purpose of the writing is particularly important. Overall, knowledge of the discipline one is studying affects peopleâs abilities to monitor their own understanding and evaluate othersâ claims effectively. LEARNING ENVIRONMENTS AND THE DESIGN OF INSTRUCTION The key principles of learning discussed above can be organized into a framework for thinking about teaching, learning, and the design of class- room and school environments. In How People Learn, four design character- istics are described that can be used as lenses to evaluate the effectiveness of teaching and learning environments. These lenses are not themselves re- search findings; rather, they are implications drawn from the research base: 13 INTRODUCTION Community Knowledge Learner centered centered Assessment centered FIGURE 1-1 Perspectives on learning environments. â¢ The learner-centered lens encourages attention to preconceptions, and begins instruction with what students think and know. â¢ The knowledge-centered lens focuses on what is to be taught, why it is taught, and what mastery looks like. â¢ The assessment-centered lens emphasizes the need to provide fre- quent opportunities to make studentsâ thinking and learning visible as a guide for both the teacher and the student in learning and instruction. â¢ The community-centered lens encourages a culture of questioning, respect, and risk taking. These aspects of the classroom environment are illustrated in Figure 1-1 and are discussed below. 14 HOW STUDENTS LEARN Learner-Centered Classroom Environments Instruction must begin with close attention to studentsâ ideas, knowl- edge, skills, and attitudes, which provide the foundation on which new learning builds. Sometimes, as in the case of Lionniâs fish, learnersâ existing ideas lead to misconceptions. More important, however, those existing con- ceptions can also provide a path to new understandings. Lionniâs fish mis- takenly projects the model of a fish onto humans, birds, and cows. But the fish does know a lot about being a fish, and that experience can provide a starting point for understanding adaptation. How do the scales and fins of a fish help it survive? How would clothing and feathers affect a fish? The fishâs existing knowledge and experience provide a route to understanding adap- tation in other species. Similarly, the ideas and experiences of students pro- vide a route to new understandings both about and beyond their experi- ence. Sometimes the experiences relevant to teaching would appear to be similar for all students: the ways in which forces act on a falling ball or feather, for example. But students in any classroom are likely to differ in how much they have been encouraged to observe, think about, or talk about a falling ball or feather. Differences may be larger still when the sub- ject is a social rather than a natural phenomenon because the experiences themselves, as well as norms regarding reflection, expression, and interac- tion, differ for children from different families, communities, and cultures. Finally, studentsâ expectations regarding their own performances, including what it means to be intelligent, can differ in ways that affect their persistence in and engagement with learning. Being learner-centered, then, involves paying attention to studentsâ back- grounds and cultural values, as well as to their abilities. To build effectively on what learners bring to the classroom, teachers must pay close attention to individual studentsâ starting points and to their progress on learning tasks. They must present students with âjust-manageable difficultiesââchal- lenging enough to maintain engagement and yet not so challenging as to lead to discouragement. They must find the strengths that will help students connect with the information being taught. Unless these connections are made explicitly, they often remain inert and so do not support subsequent learning. Knowledge-Centered Classroom Environments While the learner-centered aspects of the classroom environment focus on the student as the starting point, the knowledge-centered aspects focus on what is taught (subject matter), why it is taught (understanding), how the knowledge should be organized to support the development of exper- 15 INTRODUCTION tise (curriculum), and what competence or mastery looks like (learning goals). Several important questions arise when one adopts the knowledge- centered lens: â¢ What is it important for students to know and be able to do? â¢ What are the core concepts that organize our understanding of this subject matter, and what concrete cases and detailed knowledge will allow students to master those concepts effectively? â¢ How will we know when students achieve mastery?18 This question overlaps the knowledge-centered and assessment-centered lenses. An important point that emerges from the expertânovice literature is the need to emphasize connected knowledge that is organized around the foundational ideas of a discipline. Research on expertise shows that it is the organization of knowledge that underlies expertsâ abilities to under- stand and solve problems.19 Bruner, one of the founding fathers of the new science of learning, has long argued the importance of this insight to education:20 The curriculum of a subject should be determined by the most fundamental understanding that can be achieved of the underlying principles that give structure to a subject. Teaching specific topics or skills without making clear their context in the broader fundamental structure of a field of knowl- edge is uneconomical. . . . An understanding of fundamental principles and ideas appears to be the main road to adequate transfer of training. To understand something as a specific instance of a more general caseâwhich is what understanding a more fundamental structure meansâis to have learned not only a specific thing but also a model for understanding other things like it that one may encounter. Knowledge-centered and learner-centered environments intersect when educators take seriously the idea that students must be supported to de- velop expertise over time; it is not sufficient to simply provide them with expert models and expect them to learn. For example, intentionally organiz- ing subject matter to allow students to follow a path of âprogressive differen- tiationâ (e.g., from qualitative understanding to more precise quantitative understanding of a particular phenomenon) involves a simultaneous focus on the structure of the knowledge to be mastered and the learning process of students.21 In a comparative study of the teaching of mathematics in China and the United States, Ma sought to understand why Chinese students outperform students from the United States in elementary mathematics, even though teachers in China often have less formal education. What she documents is 16 HOW STUDENTS LEARN that Chinese teachers are far more likely to identify core mathematical con- cepts (such as decomposing a number in subtraction with regrouping), to plan instruction to support mastery of the skills and knowledge required for conceptual understanding, and to use those concepts to develop clear con- nections across topics (see Box 1-6). If identifying a set of âenduring connected ideasâ is critical to effective educational design, it is a task not just for teachers, but also for the develop- ers of curricula, text books, and other instructional materials; universities and other teacher preparation institutions; and the public and private groups involved in developing subject matter standards for students and their teach- ers. There is some good work already in place, but much more needs to be done. Indeed, an American Association for the Advancement of Science review of middle school and high school science textbooks found that al- though a great deal of detailed and sophisticated material was presented, very little attention was given to the concepts that support an understanding of the discipline.22 The four mathematics chapters in this volume describe core ideas in teaching about whole number, rational number, and functions that support conceptual understanding and that connect the particular topic to the larger discipline. Because textbooks sometimes focus primarily on methods of prob- lem solving and neglect organizing principles, creating a knowledge-cen- tered classroom will often require that a teacher go beyond the textbook to help students see a structure to the knowledge, mainly by introducing them to essential concepts. These chapters provide examples of how this might be done. Assessment-Centered Classroom Environments Formative assessmentsâongoing assessments designed to make studentsâ thinking visible to both teachers and studentsâare essential. Assessments are a central feature of both a learner-centered and a knowledge-centered classroom. They permit the teacher to grasp studentsâ preconceptions, which is critical to working with and building on those notions. Once the knowl- edge to be learned is well defined, assessment is required to monitor stu- dent progress (in mastering concepts as well as factual information), to un- derstand where students are in the developmental path from informal to formal thinking, and to design instruction that is responsive to student progress. An important feature of the assessment-centered classroom is assess- ment that supports learning by providing students with opportunities to revise and improve their thinking.23 Such assessments help students see their own progress over time and point to problems that need to be ad- dressed in instruction. They may be quite informal. A physics teacher, for example, reports showing students who are about to study structure a video 17 INTRODUCTION clip of a bridge collapsing. He asks his students why they think the bridge collapsed. In giving their answers, the students reveal their preconceptions about structure. Differences in their answers provide puzzles that engage the students in self-questioning. As the students study structure, they can mark their changing understanding against their initial beliefs. Assessment in this sense provides a starting point for additional instruction rather than a summative ending. Formative assessments are often referred to as âclass- room-based assessmentsâ because, as compared with standardized assess- ments, they are most likely to occur in the context of the classrooms. How- ever, many classroom-based assessments are summative rather than formative (they are used to provide grades at the end of a unit with no opportunities to revise). In addition, one can use standardized assessments in a formative manner (e.g., to help teachers identify areas where students need special help). Ultimately, students need to develop metacognitive abilitiesâthe habits of mind necessary to assess their own progressârather than relying solely on external indicators. A number of studies show that achievement improves when students are encouraged to assess their own contributions and work.24 It is also important to help students assess the kinds of strategies they are using to learn and solve problems. For example, in quantitative courses such as physics, many students simply focus on formulas and fail to think first about the problem to be solved and its relation to key ideas in the discipline (e.g., Newtonâs second law). When students are helped to do the latter, their performance on new problems greatly improves.25 The classroom interactions described in the following chapters provide many examples of formative assessment in action, though these interactions are often not referred to as assessments. Early activities or problems given to students are designed to make student thinking public and, therefore, ob- servable by teachers. Work in groups and class discussions provide students with the opportunity to ask each other questions and revise their own think- ing. In some cases, the formative assessments are formal, but even when informal the teaching described in the chapters involves frequent opportuni- ties for both teachers and students to assess understanding and its progress over time. Community-Centered Classroom Environments A community-centered approach requires the development of norms for the classroom and school, as well as connections to the outside world, that support core learning values. Learning is influenced in fundamental ways by the context in which it takes place. Every community, including classrooms and schools, operates with a set of norms, a cultureâexplicit or implicitâthat influences interactions among individuals. This culture, in turn, 18 HOW STUDENTS LEARN Organizing Knowledge Around Core Concepts: Subtraction with BOX 1-6 Regrouping26 A study by Ma27 compares the knowledge of elementary mathematics of teachers in the United States and in China. She gives the teachers the following scenario (p. 1): Look at these questions (52 â 25; 91 â 79 etc.). How would you approach these problems if you were teaching second grade? What would you say pupils would need to understand or be able to do before they could start learning subtraction with regrouping? The responses of teachers were wide-ranging, reflecting very different levels of un- derstanding of the core mathematical concepts. Some teachers focused on the need for students to learn the procedure for subtraction with regrouping (p. 2): Whereas there is a number like 21 â 9, they would need to know that you cannot subtract 9 from 1, then in turn you have to borrow a 10 from the tens space, and when you borrow that 1, it equals 10, you cross out the 2 that you had, you turn it into a 10, you now have 11 â 9, you do that subtraction problem then you have the 1 left and you bring it down. Some teachers in both the United States and China saw the knowledge to be mas- tered as procedural, though the proportion who held this view was considerably higher in the United States. Many teachers in both countries believed students needed a concep- tual understanding, but within this group there were considerable differences. Some teachers wanted children to think through what they were doing, while others wanted them to understand core mathematical concepts. The difference can be seen in the two explanations below. They have to understand what the number 64 means. . . . I would show that the number 64, and the number 5 tens and 14 ones, equal the 64. I would try to draw the comparison between that because when you are doing regrouping it is not so much knowing the facts, it is the regrouping part that has to be understood. The regrouping right from the beginning. This explanation is more conceptual than the first and helps students think more deeply about the subtraction problem. But it does not make clear to students the more fundamental concept of the place value system that allows the subtraction problems to be connected to other areas of mathematics. In the place value system, numbers are âcomposedâ of tens. Students already have been taught to compose tens as 10 ones, and hundreds as 10 tens. A Chinese teacher explains as follows (p. 11): What is the rate for composing a higher value unit? The answer is simple: 10. Ask students how many ones there are in a 10, or ask them what the rate for composing a higher value unit is, their answers will be the same: 10. However, the effect of the two questions on their learning is not the 19 INTRODUCTION same. When you remind students that 1 ten equals 10 ones, you tell them the fact that is used in the procedure. And, this somehow confines them to the fact. When you require them to think about the rate for composing a higher value unit, you lead them to a theory that explains the fact as well as the procedure. Such an understanding is more powerful than a specific fact. It can be applied to more situations. Once they realize that the rate of composing a higher value unit, 10 is the reason why we decompose a ten into 10 ones, they will apply it to other situations. You donât need to remind them again that 1 hundred equals 10 tens when in the future they learn subtraction with three-digit numbers. They will be able to figure it out on their own. Emphasizing core concepts does not imply less of an emphasis on mastery of pro- cedures or algorithms. Rather, it suggests that procedural knowledge and skills be orga- nized around core concepts. Ma describes those Chinese teachers who emphasize core concepts as seeing the knowledge in âpackagesâ in which the concepts and skills are related. While the packages differed somewhat from teacher to teacher, the knowledge âpiecesâ to be included were the same. She illustrates a knowledge package for sub- traction with regrouping, which is reproduced below (p. 19). The two shaded elements in the knowledge package are considered critical. âAddi- tion and subtraction within 20â is seen as the ability that anchors more complex problem solving with larger numbers. That ability is viewed as both conceptual and procedural. âComposing and decomposing a higher value unitâ is the core concept that ties this set of problems to the mathematics students have done in the past and to all other areas of mathematics they will learn in the future. Subtraction with regrouping of large numbers Subtractions with regrouping of numbers between 20 and 100 Subtraction without The composition of regrouping numbers within 100 Addition and subtraction within 20 The rate of composing Addition without carrying a higher value unit Addition and subtraction within 10 The composition of 10 Composing and decomposing a higher value unit Addition and subtraction as inverse operations SOURCE: Ma (1999). Illustration reprinted with permission of Lawrence Erlbaum Associates. 20 HOW STUDENTS LEARN mediates learning. The principles of How People Learn have important im- plications for classroom culture. Consider the finding that new learning builds on existing conceptions, for example. If classroom norms encourage and reward students only for being âright,â we would expect students to hesitate when asked to reveal their unschooled thinking. And yet revealing precon- ceptions and changing ideas in the course of instruction is a critical compo- nent of effective learning and responsive teaching. A focus on student think- ing requires classroom norms that encourage the expression of ideas (tentative and certain, partially and fully formed), as well as risk taking. It requires that mistakes be viewed not as revelations of inadequacy, but as helpful contri- butions in the search for understanding.28 Similarly, effective approaches to teaching metacognitive strategies rely on initial teacher modeling of the monitoring process, with a gradual shift to students. Through asking questions of other students, skills at monitoring understanding are honed, and through answering the questions of fellow students, understanding of what one has communicated effectively is strength- ened. To those ends, classroom norms that encourage questioning and al- low students to try the role of the questioner (sometimes reserved for teach- ers) are important. While the chapters in this volume make few direct references to learn- ing communities, they are filled with descriptions of interactions revealing classroom cultures that support learning with understanding. In these class- rooms, students are encouraged to question; there is much discussion among students who work to solve problems in groups. Teachers ask many probing questions, and incorrect or naÃ¯ve answers to questions are explored with interest, as are different strategies for analyzing a problem and reaching a solution. PUTTING THE PRINCIPLES TO WORK IN THE CLASSROOM Although the key findings from the research literature reviewed above have clear implications for practice, they are not at a level of specificity that would allow them to be immediately useful to teachers. While teachers may fully grasp the importance of working with studentsâ prior conceptions, they need to know the typical conceptions of students with respect to the topic about to be taught. For example, it may help mathematics teachers to know that students harbor misconceptions that can be problematic, but those teach- ers will be in a much better position to teach a unit on rational number if they know specifically what misconceptions students typically exhibit. Moreover, while teachers may be fully convinced that knowledge should be organized around important concepts, the concepts that help organize their particular topic may not be at all clear. History teachers may know that 21 INTRODUCTION they are to teach certain eras, for example, but they often have little support in identifying core concepts that will allow students to understand the era more deeply than would be required to reproduce a set of facts. To make this observation is in no way to fault teachers. Indeed, as the group involved in this project engaged in the discussion, drafting, and review of various chapters of this volume, it became clear that the relevant core concepts in specific areas are not always obvious, transparent, or uncontested. Finally, approaches to supporting metacognition can be quite difficult to carry out in classroom contexts. Some approaches to instruction reduce metacognition to its simplest form, such as making note of the subtitles in a text and what they signal about what is to come, or rereading for meaning. The more challenging tasks of metacognition are difficult to reduce to an instructional recipe: to help students develop the habits of mind to reflect spontaneously on their own thinking and problem solving, to encourage them to activate relevant background knowledge and monitor their under- standing, and to support them in trying the lens through which those in a particular discipline view the world. The teacherâstudent interactions de- scribed in the chapters of this volume and the discipline-specific examples of supporting students in monitoring their thinking give texture to the in- structional challenge that a list of metacognitive strategies could not. INTENT AND ORGANIZATION OF THIS VOLUME In the preface, we note that this volume is intended to take the work of How People Learn a next step in specificity: to provide examples of how its principles and findings might be incorporated in the teaching of a set of topics that frequently appear in the Kâ12 curriculum. The goal is to provide for teachers what we have argued above is critical to effective learningâthe application of concepts (about learning) in enough different, concrete con- texts to give them deeper meaning. To this end, we invited contributions from researchers with extensive experience in teaching or partnering with teachers, whose work incorpo- rates the ideas highlighted in How People Learn. The chapter authors were given leeway in the extent to which the three learning principles and the four classroom characteristics described above were treated explicitly or implicitly. Most of the authors chose to emphasize the three learning prin- ciples explicitly as they described their lessons and findings. The four design characteristics of the How People Learn framework (Figure 1-2) are implicitly represented in the activities sketched in each of the chapters but often not discussed explicitly. Interested readers can map these discussions to the How People Learn framework if they desire. While we began with a common description of our goal, we had no common model from which to work. One can point to excellent research 22 HOW STUDENTS LEARN papers on principles of learning, but the chapters in this volume are far more focused on teaching a particular topic. There are also examples of excellent curricula, but the goal of these chapters is to give far more atten- tion to the principles of learning and their incorporation into teaching than is typical of curriculum materials. Thus the authors were charting new terri- tory as they undertook this task, and each found a somewhat different path. This volume includes four mathematics chapters. Chapter 2 presents an introduction to the principles as they apply to mathematics. It focuses on the changes in expectations for mathematics performance as we move into the twenty-first century and what those changes mean for instructionâparticu- larly at the elementary level. This chapter, then, is part introduction and part elementary mathematics. The three chapters that follow treat important top- ics at the three different grade levels: whole number in elementary school (Chapter 3), rational number in middle school (Chapter 4), and functions in high school (Chapter 5). The major focus of the volume is student learning. It is clear that suc- cessful and sustainable changes in educational practice also require learning by others, including teachers, principals, superintendents, parents, and com- munity members. For the present volume, however, student learning is the focus, and issues of adult learning are left for others to take up. The willingness of the chapter authors to accept this task represents an outstanding contribution to the field. First, all the authors devoted consider- able time to this effortâmore than any of them had anticipated initially. Second, they did so knowing that some readers will disagree with virtually every teaching decision discussed in these chapters. But by making their thinking visible and inviting discussion, they are helping the field progress as a whole. The examples discussed in this volume are not offered as âtheâ way to teach, but as approaches to instruction that in some important re- spects are designed to incorporate the principles of learning highlighted in How People Learn and that can serve as valuable examples for further dis- cussion. In 1960, Nobel laureate Richard Feynman, who was well known as an extraordinary teacher, delivered a series of lectures in introductory physics that were recorded and preserved. Feynmanâs focus was on the fundamental principles of physics, not the fundamental principles of learning. But his lessons apply nonetheless. He emphasized how little the fundamental prin- ciples of physics âas we now understand themâ tell us about the complexity of the world despite the enormous importance of the insights they offer. Feynman offered an effective analogy for the relationship between under- standing general principles identified through scientific efforts and under- standing the far more complex set of behaviors for which those principles provide only a broad set of constraints:29 23 INTRODUCTION We can imagine that this complicated array of moving things which consti- tutes âthe worldâ is something like a great chess game being played by the gods, and we are observers of the game. We do not know what the rules of the game are; all we are allowed to do is to watch the playing. Of course, if we watch long enough, we may eventually catch on to a few of the rules. The rules of the game are what we mean by fundamental physics. Even if we knew every rule, however, we might not be able to understand why a particular move is made in the game, merely because it is too complicated and our minds are limited. If you play chess you must know that it is easy to learn all the rules, and yet it is often very hard to select the best move or to understand why a player moves as he does. . . . Aside from not knowing all of the rules, what we really can explain in terms of those rules is very limited, because almost all situations are so enormously complicated that we cannot follow the plays of the game using the rules, much less tell what is going to happen next. (p. 24) The individual chapters in this volume might be viewed as presentations of the strategies taken by individuals (or teams) who understand the rules of the teaching and learning âgameâ as we now understand them. Feynmanâs metaphor is helpful in two respects. First, what each chapter offers goes well beyond the science of learning and relies on creativity in strategy develop- ment. And yet what we know from research thus far is critical in defining the constraints on strategy development. Second, what we expect to learn from a well-played game (in this case, what we expect to learn from well-concep- tualized instruction) is not how to reproduce it. Rather, we look for insights about playing/teaching well that can be brought to oneâs own game. Even if we could replicate every move, this would be of little help. In an actual game, the best move must be identified in response to another partyâs move. In just such a fashion, a teacherâs âgameâ must respond to the rather unpre- dictable âmovesâ of the students in the classroom whose learning is the target. This, then, is not a âhow toâ book, but a discussion of strategies that incorporate the rules of the game as we currently understand them. The science of learning is a young, emerging one. We expect our understanding to evolve as we design new learning opportunities and observe the out- comes, as we study learning among children in different contexts and from different backgrounds, and as emerging research techniques and opportuni- ties provide new insights. These chapters, then, might best be viewed as part of a conversation begun some years ago with the first How People Learn volume. By clarifying ideas through a set of rich examples, we hope to encourage the continuation of a productive dialogue well into the future. 24 HOW STUDENTS LEARN NOTES 1. National Research Council, 2000. 2. Lionni, 1970. 3. National Research Council, 2000, p. 84. 4. Needham and Baillargeon, 1993. 5. diSessa, 1982. 6. Vosniadou and Brewer, 1989. 7. Carey and Gelman, 1991; Driver et al., 1994. 8. Hanson, 1970. 9. Judd, 1908; see a conceptual replication by Hendrickson and Schroeder, 1941. 10. White and Fredrickson, 1998. 11. Bransford and Schwartz, 1999. 12. Brown, 1975; Flavell, 1973. 13. Keeney et al., 1967. 14. Palincsar and Brown, 1984. 15. Aleven and Koedinger, 2002. 16. Thorndike, 1913. 17. Brown et al., 1983. 18. Wood and Sellers, 1997. 19. National Research Council, 2000, Chapter 2. 20. Bruner, 1960, pp. 6, 25, 31. 21. National Research Council, 2000. 22. American Association for the Advancement of Science Project 2061 Website. http://www.project2061.org/curriculum.html. 23. Barron et al., 1998; Black and William, 1989; Hunt and Minstrell, 1994; Vye et al., 1998. 24. Lin and Lehman, 1999; National Research Council, 2000; White and Fredrickson, 1998. 25. Leonard et al., 1996. 26. National Research Council, 2003, pp. 78-79. 27. Ma, 1999. 28. Brown and Campione, 1994; Cobb et al., 1992. 29. Feynman, 1995, p. 24. REFERENCES Aleven, V., and Koedinger, K. (2002). An effective metacognitive strategyâLearning by doing and explaining with a computer-based cognitive tutor. Cognitive Sci- ence, 26, 147-179. American Association for the Advancement of Science. (2004). About Project 2061. Available: http://www.project2061.org/about/default/htm. [August 11, 2004]. Barron, B.J., Schwartz, D.L., Vye, N.J., Moore, A., Petrosino, A., Zech, L., Bransford, J.D., and Cognition and Technology Group at Vanderbilt. (1998). Doing with understanding: Lessons from research on problem and project-based learning. Journal of Learning Sciences, 7(3 and 4), 271-312. 25 INTRODUCTION Black, P., and William, D. (1989). Assessment and classroom learning. Special Issue of Assessment in Education: Principles, Policy and Practice, 5(1), 7-75. Bransford, J.D., and Schwartz, D.L. (1999). Rethinking transfer: A simple proposal with multiple implications. Review of Research in Education, 24(40), 61-100. Brown, A.L. (1975). The development of memory: Knowing about knowing and knowing how to know. In H.W. Reese (Ed.), Advances in child development and behavior (p. 10). New York: Academic Press. Brown, A.L., and Campione, J.C. (1994). Guided discovery in a community of learn- ers. In K. McGilly (Ed.), Classroom lessons: Integrating cognitive theory and class- room practices. Cambridge, MA: MIT Press. Brown, A.L., Bransford, J.D., Ferrara, R.A., and Campione J.C. (1983). Learning, re- membering, and understanding. In J.H. Flavell and E.M Markman (Eds.), Hand- book of child psychology: Cognitive development volume 3 (pp. 78-166). New York: Wiley. Bruner, J. (1960). The process of education. Cambridge, MA: Harvard University Press. Carey, S., and Gelman, R. (1991). The epigenesis of mind: Essays on biology and cognition. Mahwah, NJ: Lawrence Erlbaum Associates. Chase, W.G., and Simon, H.A. (1973). Perception in chess. Cognitive Psychology, 4(1), 55-81. Cobb P., Yackel, E., and Wood, T. (1992). A constructivist alternative to the represen- tational view of mind in mathematics education. Journal for Research in Math- ematics Education, 19, 99-114. Cognition and Technology Group at Vanderbilt. (1996). Looking at technology in context: A framework for understanding technology and education research. In D.C. Berliner and R.C. Calfee (Eds.), The handbook of educational psychology (pp. 807-840). New York: Simon and Schuster-MacMillan. diSessa, A. (1982). Unlearning Aristotelian physics: A study of knowledge-based learning. Cognitive Science, 6(2), 37-75. Driver, R., Squires, A., Rushworth, P., and Wood-Robinson, V. (1994). Making sense out of secondary science. London, England: Routledge Press. Feynman, R.P. (1995). Six easy pieces: Essentials of physics explained by its most bril- liant teacher. Reading, MA: Perseus Books. Flavell, J.H. (1973). Metacognitive aspects of problem-solving. In L.B. Resnick (Ed.), The nature of intelligence. Mahwah, NJ: Lawrence Erlbaum Associates. Hanson, N.R. (1970). A picture theory of theory meaning. In R.G. Colodny (Ed.), The nature and function of scientific theories (pp. 233-274). Pittsburgh, PA: Univer- sity of Pittsburgh Press. Hendrickson, G., and Schroeder, W.H. (1941). Transfer training in learning to hit a submerged target. Journal of Educational Psychology, 32, 205-213. Hunt, E., and Minstrell, J. (1994). A cognitive approach to the teaching of physics. In K. McGilly (Ed.), Classroom lessons: Integrating cognitive theory and classroom practice (pp. 51-74). Cambridge, MA: MIT Press. Judd, C.H. (1908). The relation of special training to general intelligence. Educa- tional Review, 36, 28-42. Keeney, T.J., Cannizzo, S.R., and Flavell, J.H. (1967). Spontaneous and induced ver- bal rehearsal in a recall task. Child Development, 38, 953-966. 26 HOW STUDENTS LEARN Leonard, W.J., Dufresne, R.J., and Mestre, J.P. (1996). Using qualitative problem solv- ing strategies to highlight the role of conceptual knowledge in solving prob- lems. American Journal of Physics, 64, 1495-1503. Lin, X.D., and Lehman, J. (1999). 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2,376	In the U.S., the Federal Reserve maintains control over our money supply, issuing currency as needed and making decisions that support the stability of our economy. These policies and decisions are called monetary policy and they ensure that unemployment levels are low, the gross domestic product (GDP) maintains stability, prices remain stable, foreign exchange rates are relatively predictable and select sectors are growing properly. The actions of the Federal Reserve are intended to create the kind of stability that empowers our nation to thrive and prevents crashes that result in bank runs and other economically disastrous dynamics. Define Monetary Policy Small business owners often deal with banks on a regular basis, and while you may not be familiar with monetary policy, you've probably noticed that at certain times your interest rates go up or down and your taxes increase or decrease. You might also notice that you have access to more or less credit at certain times, even though your business' credit standing has remained stable. Your ability to employ new people changes with these factors as well as your ability to offer new products. These changes in your experience of business are due to monetary policies decided upon by the Federal Reserve of the United States. In order to release more money into the economy or reign it in, they write up policies and put them into action. Some monetary policy examples include buying or selling government securities through open market operations, changing the discount rate offered to member banks or altering the reserve requirement of how much money banks must have on hand that's not already spoken for through loans. Functions of the Federal Reserve In 1913, the Federal Reserve System(the "Fed"), was created by an act of Congress in order to create an environment of greater economic stability in the United States and grant the public more confidence about keeping their money deposited in banks. Since then, the Fed has been responsible for issuing currency, supervising and regulating the financial industry, maintaining payment systems and defining monetary policy while also carrying it out. While the Fed was created through an act of Congress, it's separate from our U.S. government. In order to conduct their operations, the Federal Reserve System includes 12 regions with a reserve bank in each one. These reserve banks are in Atlanta, Boston, Chicago, Cleveland, Dallas, Kansas City, Minneapolis, New York, Philadelphia, Richmond, St. Louis and San Francisco. In addition, the regions maintain 24 branch offices that help conduct regular business. The federal reserve also makes use of a Board of Governors and Federal Open Market Committee in order to conduct monetary policy and watch over the economy of our nation. How Monetary Policy Works Making monetary policy is probably a bit different from creating new policies for your business, though you likely relate to the idea of wanting your policies to result in greater profits and stability. Some of the goals of monetary policy are to: - Promote a high employment rate. - Promote a low unemployment rate. - Maintain stable pricing throughout the economy. - Limit inflation. - Encourage stable exchange rates with other currencies. - Influence stable and reasonable interest rates. - Encourage healthy economic growth. In order to achieve these aims, the Federal Reserve System makes use of several tools that help to ensure major economic decisions are a collaborative effort instead of the work of one or two people. These tools include: - Regional Federal Reserve banks as well as branch offices that represent all 12 regions in the United States. Outside of the continental United States, the San Francisco region serves Hawaii, Guam, American Samoa and the Northern Mariana Islands. The Seattle region serves Alaska, and the Richmond region serves the District of Columbia. New York Serves Puerto Rico and the U.S. Virgin Islands. - The Board of Governors, which is made up of seven people who serve 14-year terms. They're appointed by the United States President and confirmed by Congress. Even though the Fed is separate from our government, it's still accountable to it. - The Federal Open Market Committee, which includes all seven board governors as well as five of the regional bank presidents. Even though only five of the presidents are included, all 12 are present for the meetings, which occur eight times per year. The president seats on the committee rotate to ensure every region is represented over time. The Monetary Policy Report In order to maintain transparency and accountability about its monetary policy actions, the Federal Reserve issues a semiannual monetary policy report. This report includes three major sections: - Recent Economic and Financial Developments: This section includes data and graphs representing employment and unemployment trends, employment compensation, business-sector output by hour, price inflation, oil and energy prices, changes in gross domestic product (GDP) and gross domestic income (GDI), consumer spending patterns, personal savings rates, housing trends, debt trends and more. For instance, if unemployment rose while consumer spending dramatically decreased, this section of the monetary policy report would detail that in words and through graphs. - Monetary Policy: Some monetary policy examples detailed in this section of the report include increases and decreases in the federal funds rate, reductions or increases in the Federal Reserve balance sheet like payments on SOMA securities and changes in the required reserve rate for banks. The monetary policy section also contains lengthy explanations, an area to define monetary terms for the layperson and details of monetary policy changes in plain English so that business owners and the general public can understand what's going on in the economy and why. - Summary of Economic Projections: The final section combines information from the first two sections in order to give monetary policy examples and changes that could become necessary within the next two years. Here, the report details the general trends of the economy and whether the Fed expects to implement changes to the federal reserve rate or funds rate as well as what open market operations they expect to engage in. For instance, if they're predicting necessary increases in the federal reserve rate, the report will show why using graphs, numbers and written explanations. Open Market Operations Open market operations (OMO) are when the Federal Reserve buys or sells government securities in order to arrive at a monetary definition of the desired currency circulation levels in the economy. When the Fed purchases securities, that increases the amount of money circulating in the economy to give it a bit of a bump up. This typically leads to things like lower unemployment rates and higher consumer spending. When the Fed sells securities, that decreases the amount of money circulating in the economy to reign things in a bit. Most business owners are likely to find this frightening, but without this restraint on the economy, inflation could increase at an unhealthy rate. Contrary to how we feel when there's less money circulating in our economy, it actually increases long-term stability in the business sector, which is good news for business owners. The Federal Open Market Committee is the body that decides when and how open market operations will happen. They set a target federal funds rate and then decide how to buy or sell government securities accordingly. While this committee is separate from our U.S. government, they're accountable to it and their open market activities greatly impact all levels of government and the public. The Discount Rate Just like you go to the bank to get a loan when you need to expand your business, commercial banks receive loans from the Federal Reserve in order to be able to meet reserve requirements at the end of a business day or period. Similarly to how loans from the bank are probably a last resort resource for sustaining or growing your business, commercial banks prefer to get loans from one another before taking out a loan from the Federal Reserve. When they do take a loan out from the Fed, the interest rate charged on that loan is referred to as the discount rate. The Fed carefully decides on the wisest discount rate and that largely depends on how much money is sitting in their reserves. If they want to release more money into the economy, then they lower the discount rate, but if they want to decrease the money in the economy to rein in inflation, they increase the discount rate. This makes loans more or less expensive to member commercial banks and, in turn, influences how much money is available to give someone like you a business loan. Changing the Reserve Requirements When the Fed's monetary definition of economic stability prescribes expanding or contracting the amount of money circulating freely in the economy, their monetary policy changes accordingly. When the Fed wants to slow down financial growth, they'll raise the reserve requirement, which then causes an increase in loans issued from the Fed at a higher discount rate. When the Fed wants to increase economic growth, they'll lower the reserve requirements so that commercial banks can issue more loans to business owners like you. Business owners use those funds to grow their businesses and pay employees, which results in more money circulating in the marketplace. Expansionary Monetary Policy The Federal Reserve sometimes needs to increase the amount of money injected into the economy in order to increase consumer spending and aid in a healthy level of economic growth. There are three steps the Fed typically takes in order to enact expansionary monetary policy: - Decreases the federal funds rate: When the Federal Reserve decreases the federal funds rate, this also decreases the cost of borrowing for commercial banks, which creates more money available for lending to business owners and individuals, who then push this currency into circulation in the economy. - Purchases government securities: When the Fed purchases securities, this creates more money in the banks that held the bonds, which also increases the amount that those banks can lend to businesses and individuals. This expands the economy through creating money people can use to pay employees, buy necessities or otherwise spend in the marketplace. - Reduces reserve requirements: When the Federal Reserve reduces the reserve requirements for commercial banks, this also frees up money they can use to lend businesses and consumers, again causing more money to circulate in the marketplace. Contractionary Monetary Policy When the money supply increases too quickly, sometimes this causes prices to rise too rapidly and decreases how much your money can buy. In order to reign in this inflation, the Federal Reserve enacts contractionary monetary policy, which includes: Increasing the federal funds rate: When the Fed increases the federal funds rate, it becomes more expensive for member commercial banks to fund their operations through Fed loans. So, they issue fewer loans or reduced loan amounts to businesses and consumers, and usually at a higher interest rate. This discourages businesses and individuals from seeking loans, thereby reducing the money circulating in the marketplace. Selling government securities: When the Federal Reserve increases the federal funds rate, they tend to also sell government securities, which reduces the money available at the banks holding the bonds, so they're less likely to issue business and consumer loans. Increasing the reserve rate: At the same time as increasing the federal funds rate and selling government securities, the Fed often increases the reserve rate in order to keep more money on hand in member commercial banks. This decreases loans and the amount of currency circulating in the economy. - Investopedia: Monetary Policy - Federal Reserve Education: Monetary Policy Basics - Board of Governors of the Federal Reserve System: Monetary Policy Report: February 2019 - Investopedia: What are Some Examples of Expansionary Monetary Policy? - Business Jargons: Types of Monetary Policy - The Board of Governors of the Federal Reserve: Open Market Operations - Investopedia: Open Market Operations - Board of Governors of the Federal Reserve System: Discount Rate - Investopedia: Federal Discount Rate - Board of Governors of the Federal Reserve System: Reserve Requirements - Investopedia: Reserve Requirements - Economics Help: Expansionary Monetary Policy - CFI: What is a Contractionary Monetary Policy?<\|endoftext\|>	3.734375
455	## What is trig substitution in calculus? In mathematics, trigonometric substitution is the substitution of trigonometric functions for other expressions. In calculus, trigonometric substitution is a technique for evaluating integrals. Moreover, one may use the trigonometric identities to simplify certain integrals containing radical expressions. ## What is the difference between U substitution and trig substitution? Generally, trig substitution is used for integrals of the form x2±a2 or √x2±a2 , while u -substitution is used when a function and its derivative appears in the integral. Why does weierstrass substitution work? Any rational expression of trigonometric functions can be always reduced to integrating a rational function by making the Weierstrass substitution. The Weierstrass substitution is very useful for integrals involving a simple rational expression in ⁡ and/or ⁡ in the denominator. When to use sine substitution in trig substitution? Simplify the integrand, but do not try to evaluate it. Don’t look ahead without making an attempt. Solution 1: The first integrand is of the form a 2 − x 2 = 3 2 − x 2, so we use the sine substitution. DO : Finish this integration, using what we learned previously. ### When to use trig substitution in differential notation? The method of trig substitution may be called upon when other more common and easier-to-use methods of integration have failed. Trig substitution assumes that you are familiar with standard trigonometric identies, the use of differential notation, integration using u-substitution, and the integration of trigonometric functions. ### Which is the correct form for the secant trig substitution? Before moving on to the next example let’s get the general form for the secant trig substitution that we used in the previous set of examples and the assumed limits on θ . Let’s work a new and different type of example. Can you eliminate absolute value bars in trig? If we knew that tan θ tan θ was always positive or always negative we could eliminate the absolute value bars using, Without limits we won’t be able to determine if tan θ tan θ is positive or negative, however, we will need to eliminate them in order to do the integral.<\|endoftext\|>	4.59375
2,321	Go to Fractions Made Easy through Fun Games. For details, see near Fractions, if you have not already done so. There we studied about half, quarter, three fourthwith examples and exercises. Here, we will see fractions in general. ## Understanding the concept of Fractions Look at the following Table : Solved Examples 1, 2, 3 and Exercises 1A, 1B are based onthe following table. sweets chocolates fruits cookies ice creams Total No. of Eatables 5 6 8 7 4 30 Illustrated presentation of the data : sweets chocolates fruits cookies ice creams Solved Example 1 : Fractions Made Easy What part of the total eatables are sweets ? Write the answer in words also. Solution : Sweets = 5; Total Eatables = 30. So sweets form 5⁄30 of the total eatables. In words, sweets = five by thirty. You can see from the illusration given above, there are 5 black blocks (representing sweets) out of total 30 blocks. I hope this illustration makes you to understand the fraction 5⁄30 (five by thirty) clearly. ### Exercise 1A : Concept of Fractions Made Easy 1. What part of the total eatables are chocolates ? Write the answer in words also. 2. What part of the total eatables are fruits ?Write the answer in words also. 3. What part of the total eatables are cookies ?Write the answer in words also. 4. What part of the total eatables are ice creams ?Write the answer in words also. For Answers see at the bottom of the Page. Understand the answer fractions you got from the illustration given above. Solved Example 2 : Fractions Made Easy What part of the total are the chocolates and cookies ? Solution: chocolates = 6; cookies = 7; Total = 30 Part of chocolates and cookies = Part of chocolates + Part of cookies = 6⁄30 + 7⁄30 = (6 + 7)⁄30 = 13⁄30 In words, chocolates plus cookies = thirteen by thirty. You can see from the illusration given above, there are 6 green blocks (representing chocolates) plus 7 red blocks (representing cookies) (= 13 blocks) out of total 30 blocks. I hope this illustration makes you to understand the fraction 13⁄30 (thirteen by thirty) clearly. Solved Example 3 : Fractions Made Easy Leaving Fruits, what part of the total eatables are the other eatables? Solution: Total = 30; Fruits = 8; Part of Total = 30⁄30; Part of Fruits = 8⁄30; Part of other eatables = 30⁄30 - 8⁄30 = (30 - 8)⁄30= 22⁄30 In words, other eatables (leaving fruits)= twenty two by thirty. You can see from the illusration given above, there are 22 blocks after leaving 8 blue blocks(representing fruits) out of total 30 blocks. I hope this illustration makes you to understand the fraction 22⁄30 (twenty two by thirty) clearly. ### Exercise 1B : Concept of Fractions Made Easy Write you answers in words also. 1. What part of the total are the fruits and ice creams? 2. What part of the total are the sweets and chocolates? 3. What part of the total are the cookies and fruits? 4. What part of the total are the ice creams and sweets? 5. Leaving cookies, what part of the total eatables are the other eatables? 6. Leaving ice creams, what part of the total eatables are the other eatables? 7. Leaving sweets and chocolates, what part of the total eatables are the other eatables? 8. Leaving cookies and fruits, what part of the total eatables are the other eatables? For Answers see at the bottom of the Page. Understand the answer fractions you got from the illustration given above. 1⁄2, 1⁄4, 3⁄4, 4⁄30, 7⁄30, ........indicate the part of a whole, don't they? These are called fractions. Ex : Can you say what 6⁄7 means? 6⁄7 means we have taken 6 parts of 7 equal parts. We read it as 6 over 7 or 6 by 7. Solved Example 4 : Fractions Made Easy If you take 2 items out of 3 items, how do you write it? and how do you read it? Solution: We write it as 2⁄3. We read it as 2 over 3 or 2 by 3. ### Exercise 2A : Fractions made easy 1. Taking 5 things out of 7 is written as ........... 2. 4 out of 10 is written as ........... 3. A whole is divided into 8 parts and if 4 parts are taken,the fraction is written as ........... 4. Suppose you have 6 chocolates. You gave 2 chocolates to your friend.What part of the chocolates did you give to your friend? For Answers see at the bottom of the Page. ### Exercise 2B : Fractions made easy Suppose you were reading a book containing 40 pages. On the first day, you read 10 pages, on the second day 12 pages, on the third day 16 pages. 1. What fraction of the book did you read on the first day? 2. What fraction of the book did you read on the second day? 3. What fraction of the book did you read on the third day? 4. What fraction of the book did you read in three days? 5. What fraction of the book is yet to be read by you? For Answers see at the bottom of the Page. ## Numerator and Denominator of a Fraction The numbers of the form ab, where a and b(≠0) are whole numbersare called fractions.Here a is called the Numeratorand b is called the Denominatorof the fraction ab. For Example, In 4⁄5, 4 is called the Numerator and 5 is called the Denominator. Here is a collection of proven tips, tools and techniques to turn you into a super-achiever - even if you've never thought of yourself as a "gifted" student. and remember large chunks of information with the least amount of effort. If you apply what you read from the above collection, you can achieve best grades without giving up your fun, such as TV, surfing the net, playing video games or going out with friends! ## Speed Study System. ### Exercise 3 : Numerator and Denominator of a Fraction Fill in the Blanks in the Table Fraction Numerator Denominator ---- 2 7 4⁄5 4 ---- 5⁄9 ---- 9 3⁄8 ---- ---- For Answers see at the bottom of the Page. Great Deals on School & Homeschool Curriculum Books ## Like and Unlike fractions Fractions having the same number as denominatorare called like fractions. Examples of like fractions : (i) 4⁄11, 7⁄11, 12⁄11, 29⁄11 (ii) 1⁄8, 7⁄8, 12⁄8, 21⁄8 Fractions having different denominatorsare called unlike fractions. Examples of unlike fractions : (i) 4⁄3, 1⁄2, 12⁄11, 2⁄7 The idea of Like and Unlike fractions is usedin the following topics of Fractions. Equivalent Fractions Comparing Fractions Subtracting Fractions Simplifying Fractions Fraction word Problems ## Proper, Improper and Mixed Fractions : Fractions made easy The definitions of these and conversion from improper fractions to mixed fractionsand mixed fractions to improper fractionsis covered in Help With Fractions. The multiplying and dividing of fractions are covered respectively in Multiplying Fractions and Dividing Fractions ## Progressive Learning of Math : Fractions made easy Recently, I have found a series of math curricula (Both Hard Copy and Digital Copy) developed by a Lady Teacher who taught everyone from Pre-K students to doctoral students and who is a Ph.D. in Mathematics Education. This series is very different and advantageous over many of the traditional books available. These give students tools that other books do not. Other books just give practice. These teach students “tricks” and new ways to think. These build a student’s new knowledge of concepts from their existing knowledge. These provide many pages of practice that gradually increases in difficulty and provide constant review. These also provide teachers and parents with lessons on how to work with the child on the concepts. The series is low to reasonably priced and include ## Learning/Teaching Math Can Be Fun Here is a collection of kids math games and fun math activities for the class room or for the home, to make math exciting and easy to learn. * To save you time and money to be spent on resources, games and books. * To become a wonderful, fun teacher or parent who knows how to make math fun, interesting and effective. * To cater for all different ability levels and cater for different learning styles. * To see your kids math skills soar and their grades in math going up and up. This Collection of Fun Math Games are electronic books (e-books) in a flash. You can start printing games right away. You get to print only what you want and as many copies as you need. some FREE samples or to order ### Answers to Exercise 1A : Concept of Fractions 1. 6⁄30, six by thirty 2. 8⁄30, eight by thirty 3. 7⁄30, seven by thirty 4. 4⁄30, four by thirty ### Answers to Exercise 1B : Concept of Fractions 1. 12⁄30, twelve by thirty 2. 11⁄30, eleven by thirty 3. 15⁄30, fifteen by thirty 4. 9⁄30, nine by thirty 5. 23⁄30, twenty three by thirty 6. 26⁄30, twenty six by thirty 7. 19⁄30, nineteen by thirty 8. 15⁄30, fifteen by thirty 1. 5⁄7, five by seven 2. 4⁄10, four by ten 3. 4⁄8, four by eight 4. 2⁄6, two by six<\|endoftext\|>	4.59375
757	In this first quiz on GCSE Chemistry we take a look at one vital part of the subject - acids, alkalis, bases and salts. In it we find out the difference between alkalis and bases, how acids and alkalis react with one another, how neutralisation can produce salts and the relevance of pH values. Acids and alkalis are incredibly important for manufacturing, and for everyday life. Some examples of things made using acids and alkalis are soaps and detergents, fertilisers, car batteries and medicines - you will have probably learnt a lot more examples during your lessons, as well as the properties of acids and alkalis. The most important of the reactions of acids and alkalis is neutralisation. Mixing the correct amounts of acid and alkali will result in a solution that is neutral because the two substances react together to produce water and a salt. During the 20th century, people knew about air pollution but little was done about it until the final decades. During the 1970s, it was noticed that trees were dying and some lakes were becoming too acidic to support life. It was realised that acid rain was the cause. The first response to this was to try to neutralise the water in lakes by using huge amounts of calcium carbonate (remember, acids react with carbonates as well as with alkalis). Rain is normally very slighly acidic as it contains dissolved carbon dioxide from the air. Acid rain is rain that is much more acidic than normal and it occurs where fossil fuels are being burnt in large quantities. Governments introduced laws that forced the polluters to look at ways of reducing the amounts of acidic gases in the fumes that they produced. The problem is not as bad as it was then, but it is still with us. There are several different ways of defining what an acid or an alkali is (alkalis and bases are NOT exactly the same thing. See Q2 for more details). The simplest and probably the first that you learnt is the pH. A pH value of 7 is neutral, anything lower than this is acidic and anything more than pH7 is alkaline. Values that are further away from neutral indicate stronger acids and alkalis. Be careful not to confuse the terms strong and weak with dilute and concentrated. You might think that a strong acid or alkali would always be more harmful than weak acids or alkalis. That is not necessarily the case. Take for example hydrochloric acid. This is a strong acid yet it is found inside your stomach! How can that be? Well, it's a matter of concentration, the hydrochloric acid inside your stomach is extremely diluted. Your stomach acid is so dilute that it is easily contained by a thin layer of mucus secreted by cells in the walls of your stomach. Finding out whether something is acidic or alkaline can be done in different ways. The substance known as 'litmus' is used simply to find out if a solution is acidic or alkaline, it cannot be used to find the strength of an acid or alkali. Litmus is naturally blue and the dye comes from lichens, which are becoming more and more scarce. It turns red in the presence of an acid. Some other plant colours can be used in the same way - you may have done some experiments with red cabbage extract at KS3. Universal indicator (UI) paper is more useful as it can give you a pH reading. The pH scale runs from 0 - 14 but some UI paper only measures from 0 - 12. For the most accurate readings of pH, using a meter is is the best. Have a go at this quiz and see if your understanding of acids, alkalis, bases and salts is up to scratch<\|endoftext\|>	4.03125
2,772	One of the most common criticisms of plans to modify or eliminate the Electoral College is that to do so would be to deviate from the wisdom of the Founders of the American political system. But the "Father of the Constitution" himself, James Madison, was never in favor of our current system for electing the president, one in which nearly all states award their electoral votes to the statewide popular vote winner. He ultimately backed a constitutional amendment to prohibit this practice. As historian Garry Wills wrote of our fourth president, "as a framer and defender of the Constitution he had no peer." Yet, when he helped create the Constitution and when he defended it years after his presidency, Madison repeatedly argued for alternatives to the winner-take-all method of choosing a state's presidential electors. Like other leaders of that time, he looked at the world with clear eyes and learned from experience, unafraid to support change when that change made sense. Madison at the Constitutional Convention The question of how the president should be elected was a hotly contested issue at the Philadelphia Convention in 1787. Numerous systems were proposed and discarded before a final decision was reached. The fault lines in the debate lay between two cross-cutting divisions among the states at the Convention: one between large and small states, and one between slave-owning and free states. It was this latter split that ended up being most salient in the Electoral College debate. The College's primary purpose was not to give small states greater representation, as is often claimed by its defenders today. Instead, the Electoral College was created to reflect the political realities associated with accommodating the institution of slavery into our electoral system. Under a direct election system, the southern states would be at a significant disadvantage because their slaves could not vote. Through the Electoral College and the Three-Fifths Compromise, however, partially counting the slaves when determining the number of presidential electors allowed southern states to rival the electoral power of their northern brethren. More broadly, the right to vote in that era was not an established value and was never affirmed in the Constitution. As a result, disparities between a state's population and its eligible voters varied widely. Pennsylvania had relatively expansive suffrage rights, for example, and Massachusetts did not. Because a national popular vote would pool every state's votes together on an equal basis, delegates from limited suffrage states opposed a direct election of the president. Madison expressed his preference for a national popular vote for president in a speech at the Convention, however, arguing that "the people at large was...the fittest " to choose an executive. Although he recognized that such a system would put southern states, including his native Virginia, at a major electoral disadvantage, Madison believed that "local considerations must give way to the general interest," and he was "willing to make the sacrifice" of his state's political power for the good of the American democracy. His fellow Southerners had no interest in such political martyrdom, though, and Madison was forced to support the Electoral College as a compromise. Early Changes to Presidential Elections That compromise in its original form only lasted 17 years. Initially, each elector was given two votes to cast. The candidate with a majority of electoral votes would become president and the candidate with the second most votes would become vice-president. Political parties quickly formed, however, and as Jack Nagel has pointed out, the system worked too much like approval voting-in which voters can vote to "approve" of as many candidates on the ballot as they like-to be viable in contested elections . In the 1796 election, this rule led to winner John Adams having the awkward situation of serving with his leading opponent Thomas Jefferson as vice-president rather than his desired running mate Charles Pinckney. Some Adams voters did not want Pinckney to defeat Adams for the presidency and enough of them voted for other candidates with their second vote to allow Jefferson to finish second. The 1800 election was even more problematic, with Jefferson receiving the same number of votes as his de facto running mate Aaron Burr, as each of Jefferson's electors also voted for Burr to avoid an outcome like that of 1796. The result was a highly controversial vote for president in the House, with Burr deciding to seek the presidency and aligning himself with Jefferson's political opponents. Jefferson was eventually elected, but not before many days of stalemate and much backroom negotiation. The messy process in the 1800 election widely discredited the original Electoral College rules that allowed such an event to occur. In particular, it was agreed that the rule that electors would vote for two candidates, which led to perverse outcomes tied to tactical voting, had to be changed. Willing to learn from its experiences, the founding generation moved to change the Constitution before the 1804 election, and the 12th amendment established that each elector would cast one electoral vote for a ticket of a president running with a vice-president. The Selection of Presidential Electors in the Early Years Neither the Constitution nor the 12th amendment, however, restricted states in any way on what rules they might choose to select the president. In our early years, states indeed were adventurous and open to experimentation. Massachusetts, for one, modified its method of awarding electors for every presidential election until 1824. Of the three most common rules for choosing a state's electors, the most widely used initially was appointment by the state legislature, which avoided a popular vote entirely. Madison did not like this system, as he worried that allowing state legislatures to elect the president would give the state and national legislatures too much power over the executive branch. A second option was for voters in an electoral district to elect one or more presidential electors for that district. Virginia, Maryland, North Carolina, Kentucky, and Tennessee each used this method for many of the first presidential elections. Other states used hybrid systems in which most electors were chosen by district voting but a few were selected based on the statewide vote or appointed by the state legislature. Some states used already-existing congressional districts for this process, while others created new districts specifically for the purpose of selecting presidential electors. The third system was for all electors to be elected on a winner-take-all basis in a statewide vote, so that every elector in a state would likely vote for the same candidate. States quickly realized that this last method maximized the advantage they could give to their preferred candidate. In 1800, only two states used this system. By Madison's election to the presidency in 1808, six states used the statewide system, and by 1836 it was implemented by every state but South Carolina, which continued to appoint electors until after the Civil War. Today, 48 of 50 states allocate all of their electors to the winner of the statewide vote (the remaining two, Maine and Nebraska, select some electors by district voting and two by statewide voting). Madison's Proposed Amendment In 1823, Madison wrote a remarkable letter to George Hay explaining his views of the Electoral College, his strong opposition to states voting as winner-take-all blocs and his view of the origins of the winner-take-all rule. In addition to disenfranchising districts that voted against the preference of the state, Madison worried that statewide voting would increase sectionalism and the strength of geographic parties. He wrote that his views were widely shared by others at the Constitutional Convention, and that the winner-take-all approach had been forced on many states due to its adoption in other states: "The district mode was mostly, if not exclusively in view when the Constitution was framed and adopted; & was exchanged for the general ticket [e.g., winner-take-all rule] & the legislative election, as the only expedient for baffling the policy of the particular States which had set the example." Madison also discerned that the winner-take-all rule did not actually help small states. When the Constitution was drafted, small states were expected to be helped by the law stating that each state has one vote for president when the election went to the House (as it had had in 1800 and would again in 1824, when John Quincy Adams was elected over the more popular Andrew Jackson). Many Founders anticipated that such outcomes would become routine - meaning the electors would limit the field to three choices that the House would choose among, voting on the basis of one vote per state. George Mason, for one, predicted in 1787 that "nineteen times in twenty" there would be no winner of a majority of electoral votes and the president would be chosen in the House. Madison saw this provision as highly problematic, however: "The present rule of voting for President by the H. of Reps. is so great a departure from the Republican principle of numerical equality, and even from the federal rule which qualifies the numerical by a State equality, and is so pregnant also with a mischievous tendency in practice, that an amendment of the Constitution on this point is justly called for by all its considerate & best friends." He recognized that small states would object to removing this provision, however, so he suggested twinning it with a ban on the winner-take-all rule and requiring the district method. Small states knew they were not helped by the winner-take-all rule that advantaged the most populous states, so removing the rule would make it easier for them to accept Madison's proposal. He wrote: "A constitutional establishment of that mode [district allocation] will doubtless aid in reconciling the smaller States to the other change which they will regard as a concession on their part. " Madison's proposed amendment in his letter to Hay incorporated this compromise, along with his suggestion that electors would cast a second, backup choice in a manner similar to the idea of instant runoff voting. In the event of there being no majority in the Electoral College, Congress should select the president on the basis of "one member, one vote." He suggested the following language: "The Electors to be chosen in districts, not more than two in any one district, and the arrangement of the districts not to be alterable within the period of ------ previous to the election of President. Each Elector to give two votes, one naming his first choice, the other his next choice. If there be a majority of all the votes on the first list for the same person, he of course to be President; if not, and there be a majority, (which may well happen) on the other list for the same person, he then to be the final choice; if there be no such majority on either list, then a choice to be made by joint ballot of the two Houses of Congress, from the two names having the greatest number of votes on the two lists taken together." How Would Madison Change the Electoral System Today? Madison's arguments for supporting the district method fit the political realities of 1823, but we suspect that if he were alive today, his practical view of the world would have led to his returning to his original views of the best way to elect the executive: that is, direct election of the president. Given his 1823 letter and the opportunity to observe instant runoff elections in practice, he also would likely have supported allowing voters to indicate backup choices in the event of the need to simulate a runoff election. The ever-pragmatic Madison would quickly realize that his arguments in favor of a district system instead of a direct election system, as laid out in an 1826 letter to Robert Taylor, would not apply to today's presidential elections. He would see that its extreme bias for one major party in close elections would make it a political nonstarter, just as direct election was a nonstarter for political reasons in 1787. Madison also would have seen that conditions no longer backed his arguments. For instance, he saw an advantage in electors being able to use their better judgment if voters picked a poor candidate. However, electors today are merely rubber stamps for the winners of statewide elections. Madison also thought that voters would be better equipped to choose local electors that they personally know rather than distant national candidates, but voters today are inundated with information about national candidates and would be unlikely to be able to name any of their state's electors. Lastly, Madison hoped that electors would vote for their constituents' second choice if their first choice had no chance of winning. This problem would be much more effectively solved today by using instant runoff voting in a national presidential election. Madison's reasons for supporting the Electoral College late in his career could be used today as powerful arguments to oppose it. The story of James Madison and the Electoral College illustrates that the College as we know it was not created because the Founders thought it was the ideal way to elect an executive. Instead, it initially developed because of a compromise with slave-owning southern states and other low suffrage stages, and over time moved to winner-take-all rules due to a poorly-considered Constitutional incentive structure that led states to choose systems that furthered their own partisan preferences over fairly representing the views of the people. Those concessions hardly should govern our decisions today about how to choose the president in light of the current state of American politics and our current views about the right to vote. Attempts to defend the Electoral College based on the fact that it was introduced by brilliant political thinkers such as Madison fail to appreciate the unique political context in which it was created and the fundamental differences between that time and ours. As Madison said of the presidential election system in 1830 and would likely say again today, "a solid improvement of it is a desideratum that ought to be welcomed by all enlightened patriots." The section involving options for Electoral College districts was updated June 19, 2012.<\|endoftext\|>	4.03125
434	You will get to learn the fundamental ideas you need to understand in order to make sound geometric conjectures and prove them. After completing this tutorial, you will be able to complete the following: Consider the following diagram: By considering the line d and the plane P as sets, how can we describe the relationship between point A and the plane P and between line d and the plane P? ~ The point A is one of the points in the plane P, so we can say that A is an element of the set of points that make up the plane P. We say that A is in P. Line d is made up of points that are in the plane P, so line d is a subset of the set of points of P. We also say that d is in P. What sorts of relationships are there between points and lines on the same plane? ~ Given a point and a line in the same plane, the point is either on the line or not on the line. In the diagram below, point A is not on line d, but point B is on line d. Now B is one of the points that make up the line d, so B is an element of the set of points that are on line d. In this case, we say that B is in d. What sorts of relationships are there between two lines on the same plane? ~ There are only three possibilities. Two lines on the same plane may intersect at exactly one point, two lines on the same plane may be coincident (if they intersect at infinitely many points), or two lines on the same plane may be parallel (if they do not intersect). In the diagram below, lines l and d intersect at exactly one point, lines n and d are coincident, and lines m and d are parallel. \|Approximate Time\|\|2 Minutes\| \|Pre-requisite Concepts\|\|Students should be familiar with coincident lines, element of a line, and intersecting lines.\| \|Type of Tutorial\|\|Animation\| \|Key Vocabulary\|\|coincident lines, element of a line, intersecting lines\|<\|endoftext\|>	4.34375
2,145	## Matrices become something more ... You've learned about using matrices to represent and solve linear systems, either through Gaussian elimination or by finding the inverse of the coefficient matrix, if it exists. Now we'll see how matrices can have aspects that can move you beyond just solving linear systems. The determinant concept is kind of odd at first. You'll probably have to come back to it a few times to figure out just what's going on, but its usefulness should be obvious after reading through this section. The determinant concept will also allow us to develop another method for solving linear systems. It's called Cramer's rule, and we'll go through a few examples in this section. ## What is a determinant? — A 2-D linear system We'll begin with a 2-equations, 2-unknown (2-D) problem and look for some general (not dependent on the actual numbers) solutions. The system is: where a1, a2, b1, and b2 are the numerical coefficients of the variables x and y. Here is that system in matrix form: Now if we solve each equation for y, we get these two equations, each from one of the original linear equations: If each right-hand side is equal to y, then each must be equal to the other (transitive property): Now cross multiply (or multiply both sides by b1b2): Then group terms containing x on the left side: Pull the x out as a factor: And finally, by dividing by (a1b2 - b1a2), we have an expression for x in terms of the coefficients of the two equations (or the components of the coefficient matrix): We could go through a similar process (and I suggest that you do) to get an expression for y: Now the thing to notice about these two expressions is that the denominators (a1b2 - a2b1) are the same and that the terms of the denominator all come from the coefficient matrix. Furthermore, if a1b2 - a2b1 is zero, the system can't have a solution because we can't divide by zero. So it turns out that this number a1b2 - a2b1 is quite important, important enough to get a name, the determinant. And it turns out that this feature is true of matrices of any size: If the determinant of a coefficient matrix is zero, the system has no solution. Another way of saying that is: The equations are not linearly independent. If the determinant of a matrix is zero, then the linear system of equations it represents has no solution. In other words, the system of equations contains at least two equations that are not linearly independent. ## A shortcut to the 2 × 2 determinant The determinant of a 2 × 2 matrix is the difference of the products along its two diagonals. It looks like this: ## Examples — 2 × 2 determinants In the top example (right →), the linear system represented by the matrix is not linearly independent because one row can be formed by applying a linear operation on the other. For example, if the top row is multiplied by -1, the result would be the bottom row. There is really only one piece of information there to find two variables, so the system can't be solved, and this is indicated by the zero determinant. The other two matrices can be solved because their determinants are not zero. ## How can determinants be used to solve linear systems:Cramer's rule With the idea of determinants of 2 × 2 matrices in hand, let's return to the expressions for x and y in our 2-equations, 2 unknowns system. They were: We've already shown that the identical denominators are the determinant of the matrix of coefficients for this system. The numerators can be written as determinants of 2 × 2 matrices, too. Those matrices consist of the coefficient matrix with the first column substituted with the result vector (c1, c2) for the x-coordinate, and with (c1, c2) substituted for the second column for the y-coordinate. Here's a picture: You should confirm for yourself that the determinants in the denominators of those expressions give you the numerators of our x and y expressions. ## Cramer's Rule For the matrix equation Ax = b, where A is an n × n square matrix of coefficients and x and b are n-dimensional column vectors, the components of x = (x1, x2, x3, ... xn) are given by where Ai is the matrix A in which column i is substituted with the column vector b. ## Determinants of larger matrices You might be saying to yourself, "This is nice, but I can solve a two-equation — two unknown problem like these faster the old fashioned way." And you'd be right. In order to make Cramer's rule really useful, we'd like to apply it to very large systems of linear equations, ones with dimensions in the hundreds or thousands. To do that, we'll need to figure out how to find determinants of larger matrices, because the multiplication of diagonals trick only works for 2 × 2 matrices. ## A trick for finding 3 × 3 determinants Here is the formula for the determinant of any 3 × 3 matrix. It looks daunting, but don't worry, there's an easy way to do it that doesn't require memorization of a formula. The way to remember this method is illustrated below. First, rewrite the first two columns of the matrix, (a, e, i) and (b, f, j), to the right of the matrix. Next, draw in the red diagonals as shown below: The first three terms of the determinant are the products of the elements along all of these red diagonals: afk + bgi + cej. Then draw in the blue diagonals in the opposite direction (lower left to upper right), as shown. The next three terms are the products along those diagonals subtracted from the first part: -ifc - jga - keb. It's simple. ## Examples of 3 × 3 determinants In this example, the determinant is zero, meaning that the system of 3 equations and 3 unknowns represented by this matrix can't be solved. Notice that the third row of the matrix is just the sum of the first two. It's a linear combination of the first two, so it's not linearly independent. Here are two more examples, just so you can practice finding 3 × 3 determinants: The latter two matrices represent solvable systems of equations because they have non-zero determinants. ## Cramer's rule with a 3 × 3 system of equations Cramer's rule extends nicely to 3 × 3 and larger square matrices. Let's use it to solve a 3 × 3 system: Now the key to doing these problems by hand is to take your time and stay organized. First, we'll find the determinant of the coefficient matrix, which we'll call A. It will be the denominator in our expressions for x, y and z. Now the determinants of the matrices substituted with the result vector (4, -2, 9) (pink highlights): Now we just do the divisions and arrive at the solution: (x, y, z) = (1, 2, 3) ## Properties of determinants We want to get to the point where we can solve for determinants of larger matrices, but first it's good to describe a few properties of determinants. They can be a big help. 1. The determinant of a diagonal matrix is the product of the elements along the diagonal. This is pretty easy to see using a 3 × 3 or 2 × 2 matrix. For the 3 × 3, all of the other elements of the determinant expression except the first (abc in this case) are zero. 2. The determinant of an upper-triangular or lower-triangular matrix is the product of the elements on the diagonal. I won't try to prove this for all matrices, but it's easy to see for a 3 × 3 matrix: The determinant is adf + be(0) + c(0)(0) - (0)dc - (0)ea - f(0)b = adf, the product of the elements along the main diagonal. Likewise, the determinant of this lower-triangular matrix is acf This property means that if we can manipulate a matrix into upper- or lower-triangular form, we can easily find its determinant, even for a large matrix. 3. The determinant of a block-diagonal matrix is the product of the determinants of the blocks. Recall that a block diagonal matrix is one that has elements only symmetrically distributed in square blocks along the diagonal. Here's one: The determinant of such a matrix is just the product of the determinants of the five blocks. In order from top to bottom they are 3×3, 2×2, 1×1, 1×1 and 3×3 blocks. 4. The determinant has the same absolute value, but switches sign when either two rows or two columns are swapped. Here are examples of the effect row and column swapping on the determinant of a 3 × 3 matrix. First we'll write out a generic (no numbers) determinant: Now after swapping the top two rows, notice that the determinant has only changed sign: And likewise for swapping two columns. Note that if, in the first instance, we swapped any other two rows, the sign of the determinant would change again. The same is true for columns. We often want to swap columns or rows in order to make a matrix triangular or block diagonal, to make the determinant easier to calculate. xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to [email protected].<\|endoftext\|>	4.78125
693	# Countable The elements of a finite set can be listed, say {a1,a2,...,an}. However, not all sets are finite: for instance, the set of all natural numbers or the set of all real numbers. It might then seem natural to divide the sets into different classes: put all the sets containing one element together; all the sets containing two elements together;...;finally, put together all infinite sets and consider them as having the same size. This view is not tenable, however, under the natural definition of size. To elaborate this we need the concept of a bijection. Do the sets {1,2,3} and {a,b,c} have the same size? "Obviously, yes" you may reply. But how do we know? Again, you may answer "Well it's obvious. Look, they've both got 3 elements". But what if you have no prior knowledge of the number 3, or of any other number [This may seem a strange situation but, although a "bijection" seems a more advanced concept than a "number", in the usual set-theoretic development functions are defined before numbers, as they are based on much simpler sets.] This is where the concept of a bijection comes in: define the correspondence ``` a ↔ 1, b ↔ 2, c ↔ 3 ``` Since every element of {a,b,c} is paired with precisely one element of {1,2,3} (and vice versa) this defines a bijection. We now generalise this situation and define two sets to be of the same size precisely when there is a bijection between them. For all finite sets this gives us the usual definition of "the same size". What does it tell us about the size of infinite sets? Consider the sets A = {1,2,3,...}, the set of positive integers and B = {2,4,6,...}, the set of even positive integers. We claim that, under our definition, these sets have the same size. Recall that to prove this we need to exhibit a bijection between them. But this is easy: 1 ↔ 2, 2 ↔ 4, 3 ↔ 6, 4 ↔ 8, ... As in the earlier example every element of A has been paired off with precisely one element of B, and vice versa. Hence they have the same size. This gives an example of a set which is of the same size as one of its proper subsets, a situation which is impossible for finite sets. We now define a set to be countable if it is either finite or the same size as N (the set of positive integers). So the above constitutes a proof that the set of even integers is countable. What about sets being "larger than" N? An obvious place to look would be Q, the set of all rational numbers, which is "clearly" much bigger than N. But looks can be deceiving, for we assert THEOREM 1: Q is countable To prove this we first need to prove THEOREM 2: The union of countably many countable sets is countable. This page is still under construction. I would be very grateful for any feedback. In particular, is any of this comprehensible to non-mathematicians<\|endoftext\|>	4.40625
497	Peat GHG emissions are main contributor to Ireland’s carbon footprintOctober 30, 2018 Peat contributes more to global warming than coal. Peat GHG emissions are among the worst in Ireland. While coal is also a heavy greenhouse gas (GHG) contributor, peat reportedly produces higher carbon (CO2) emissions per unit compared to coal, even though coal has a higher calorific value. In other words, coal generates more energy per ton when it’s burned compared to peat, but a smaller amount of peat emits more CO2 than a larger amount of coal. Peat burning power stations may eventually become a part of Ireland’s energy supply past. Bord na Móna, a semi-state company in Ireland, developed the peatlands of Ireland. The company initially created the peatlands back in the late 1940’s for the purpose of providing economic benefit to the Irish midlands and to achieve supply of energy security. The company was a proponent of the peat-power business until just last week when it announced it would be decarbonizing and intends to pull out of peat by 2030 and close bogs by 2025, the Irish Times reported. Bord na Móna’s commitment to decarbonize and move away from producing Peat GHG emissions, is the company acknowledging climate change and taking the steps to build a more sustainable business and Midlands economy. Peat GHG emissions accounted for 3.4 million tons of emissions in Ireland in 2016. Peat is a deposit of dead plant material. It results when plant material is added to an oxygen-poor environment, such as a bog or another environment that is heavily saturated with water. When the plant material decomposes it releases large amounts of GHG emissions, most notably methane, into the atmosphere. In essence, the burning, draining, and degrading peat bogs are responsible for producing significant amounts of greenhouse gas emissions. That being said, bogs and restored wetland also have the potential to fulfill a significant role in capturing and storing carbon. Nevertheless, in 2016, Peat was reportedly responsible for 3.4 million tons of emissions in Ireland. Approximately 75% of these emissions came from electricity and 25% came from residential heating. The belief is that phasing out the residential heat will make a notable difference to the country’s Peat GHG emissions.<\|endoftext\|>	3.71875
431	We will discuss here about the concept of temperature. We have already learned about various types of measurements like length, mass capacity and time. But if we have fever, non of these measurements can help us to find out how high a fever we have. The temperature of an object is measured by an instrument is called thermometer. Now we will learn about the measurement of temperature. Take two cups, one containing normal water and another containing hot water. Put your fore finger of one hand in one cup and of another hand in the other cup. We find, one contains cold water and the other contains hot water. But the question is how much cold and how much hot. To find this out, we need some measure of hotness or coldness. is the degree of hotness or coldness of a body. The instrument which measures the temperature of body is known as thermometer. Each thermometer has a scale. Two different temperature scales are in common use today: 1. The Fahrenheit Scale 2. The Celsius Scale Thermometer has scale in degree Fahrenheit (℉) and in degree Celsius (℃). The Fahrenheit scale has the melting point of ice at 32° F and the boiling point of water at 212° F. Thus, the Fahrenheit scale is marked from 32° to 212° where 32° F shows the freezing point of water and 212° F shows the boiling point of water. At present most of the countries use the degrees Celsius thermometers. The Celsius scale (is also called centigrade scale) thermometer has 0° C as freezing point of water and 100° C as the boiling point of water. On comparing the Celsius and Fahrenheit scale we find: Write True or False. 1. The instrument used to measure body temperature in thermometer ----- 2. The normal body temperature is 98.6°F ----- True. 3. The liquid inside the thermometer is mercury ----- True. 4. The unit of measure of temperature is inches ----- False. 5. 0°C is cooler than 0°F----- False.<\|endoftext\|>	4.46875
1,687	# Lami's Theorem Lami’s theorem relates the magnitudes of coplanar, concurrent and non-collinear forces that maintain an object in static equilibrium. The theorem is very useful in analyzing most of the mechanical as well as structural systems. ## Lami’s Theorem Statement Lami’s Theorem states, “When three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces”. Referring to the above diagram, consider three forces A, B, C acting on a particle or rigid body making angles α, β and γ with each other. In the mathematical or equation form, it is expressed as, $\frac{A}{sin\alpha} = \frac{B}{sin\beta} = \frac{C}{sin\gamma}$ ## Lami’s Theorem Derivation Now, let’s see how the theorem’s equation is derived. Let FA, FB, and FC be the forces acting at a point. As per the statement of the theorem, we take the sum of all forces acting at a given point which will be zero. i.e. FA + FB + FC = 0 The angles made by force vectors when a triangle is drawn are, We write angles in terms of complementary angles and use triangle law of vector addition. Then, by applying the sine rule we get, $\frac{A}{sin(180-\alpha)} = \frac{B}{sin(180-\beta)} = \frac{C}{sin(180-\gamma)}$ So, we have, $\frac{A}{sin\alpha} = \frac{B}{sin\beta} = \frac{C}{sin\gamma}$ Hence, it is clearly seen that by applying sine rule to complementary angles we arrive at the required result for Lami’s theorem. Now, we will see how Lami’s theorem is useful to determine the magnitude of unknown forces for the given system. ## Lami’s Theorem Problems and Solved Examples Example 1: Consider an advertisement board hangs with the help of two strings making an equal angle with the ceiling. Calculate the tension in both the strings in this case. Solution: The free body diagram of the same helps us to resolve the forces first. After resolving the forces we will apply the required theorem to get the value of tension in both the strings. Here, the weight of the signboard is in a downward direction, and the other force is the tension generated by the signboard in both the strings. In this case, the tension T in both the strings will be the same as the angle made by them with the signboard is equal. Above figure represents the free body diagram of the signboard. Applying the Lami’s Theorem we get, $\frac{T}{sin(180-\Theta)} = \frac{T}{sin(180-\Theta)} = \frac{mg}{sin(2\Theta)}$ Since sin (180 – θ) = sin θ and sin (2θ) = 2sinθ cosθ So, we get, final tension force in the string T as, $\frac{T}{sin\Theta} = \frac{mg}{2sin\Theta\: cos\Theta}$ I.e. T = mg / 2cosθ The similar concept and equations can be applied for a boy playing on a swing, and we arrive at the same result. Example 2: In example 2, let 45 degrees be the angle made by the strings with the signboard having a mass of 6 kg, then what is the value of the tension T in both the strings? Solution: Given, m = 6 kg, g = 9.8 m/s², 𝜃 = 45 degrees Using the derived formula from example 1, we get, T = mg / 2cosθ i.e. T = 6 x 9.8 / 2cos45 = 41.6 N So, the tension in both the strings to hold the signboard exactly horizontal is 41.6 N. Example 3: An iron block of mass 30 kg is hanging from the two supports A and B as shown in the diagram. Determine the tensions in both the ropes. Solution: Given, m = 30 kg, W = mg = 30 x 9.8 = 294 N Let’s draw an FBD i.e. Free Body Diagram for the given condition. C be the point of suspension from where the iron block is hanging. We get required angles to apply Lami’s Theorem as, Using Lami’s Formula we get, $\frac{T_{AC}}{sin(120^{o})} = \frac{T_{BC}}{sin(135^{o})} = \frac{294}{sin(105^{o})}$ Therefore, $\frac{T_{AC}}{sin(120^{o})} = \frac{294}{sin(105^{o})}$ i.e. $T_{AC} = \frac{294 \, X\, sin(120^{o})}{sin(105^{o})} = \frac{294 \, X\, 0.866}{0.966} = 263.566N$ Similarly, $\frac{T_{BC}}{sin(120^{o})} = \frac{294}{sin(105^{o})}$ i.e. $T_{BC} = \frac{294 \, X\, sin(135^{o})}{sin(105^{o})} = \frac{294 \, X\, 0.707}{0.966} = 215.2N$ So, the required tensions along the shown directions are 263.566 N and 215.2 N respectively. Example 4: Choose the correct alternative from the following. A) Lami’s theorem is applicable to only regular-shaped bodies in equilibrium. B) The theorem is derived on the basis of the cosine rule of trigonometry. C) The theorem is very helpful in determining the unknown forces acting at a point for an object in equilibrium. D) The theorem is useful in determining the forces acting on a moving object. Solution: Option A mentions regular-shaped bodies in equilibrium, which is not true. In case the mass of irregularly shaped objects is given, the missing values of tension forces can be calculated. Option B mentions about cosine rule, which is not a correct statement. The theorem is derived on the basis of sine rule in a triangle which relates angles and opposite sides to those angles in a given triangle. Since the theorem is about the static equilibrium of objects, we do not apply it to moving objects even if they have a constant velocity. So, choice D is also incorrect. We applied/used Lami’s Theorem to different situations in which objects are in equilibrium. So, option C gives us clear cut info when and where Lami’s Theorem is applied to find unknown forces. Example 5: Which of the following is correct in the case of Lami’s Theorem? A) When two coplanar forces acting at a point which is in equilibrium, then each force is proportional to the sine or the angle between the other two forces. B) The theorem is applicable to randomly shaped 3D objects only. C) There are three coplanar forces which act at a point which is in equilibrium, and each force is proportional to the sine or the angle between the other two forces. D) Lami’s Theorem is helpful in determining a magnitude and direction of force acting on a moving object in a straight line. Solution: First of all, Lami’s Theorem is applicable only in the case of coplanar concurrent forces. Secondly, it is applicable to planes as well as 3D objects. It means option B fails to satisfy the condition. There should be three concurrent forces acting at a point, so option A is invalid. Option D mentions force acting on a moving object in a straight line, which is not a correct situation to apply this theorem. So, option C which mentions the required conditions for the application of Lami’s Theorem is the right choice.<\|endoftext\|>	4.625
6,716	Chapter 8 Graph Theory [References] 1. Discrete Math - 7th Edition , Richard Johnsonbaugh, Pearson Chapter 8 Graph Theory 8.1 Introduction 8.2 Paths and Cycles (Graphs) 8.3 Hamiltonian Cycles and the Traveling Salesperson Problem 8.4 A Shortest-Path Algorithm 8.5 Representations of Graphs 8.6 Isomorphisms of Graphs 8.7 Planar Graphs 8.8 Instant Insanity(skipped) Graphs are discrete structures consisting of vertices and edges that connect these vertices. There are different kinds of graphs, depending on whether edges have directions, whether multiple edges can connect the same pair of vertices, and whether loops are allowed. Problems in almost every conceivable discipline can be solved using graph models.We will give examples to illustrate how graphs are used as models in a variety of areas. 8.1 Introduction ● Highway system and Road inspector Graphs are drawn with dots and lines. The dots are vertices. The lines connect the vertices are edges. The edges is . We start at a vertex , travel along an edge to vertex , travel along an edge to vertex , and so on, and eventually arrive at a vertex . Path is from to . Definition 1.1 Example 1.2 Example 1.3 Edges and are both associated with the vertex pair . Edges and are parallel edge. Edge incident on a single vertex is a loop. For example, edge is a loop. Vertex is not incident on any edge is an isolated vertex. A graph with neither loops nor parallel edges is a simple graph Example 1.4 Since the graph of Figure has neither parallel edges nor loops, it is a simple graph. Example 1.5 Solution A graph with numbers on the edges is a weight graph. Edge is labeled The weight of edge is . The weight of edge is . In a weight graph, the length of a path is the sum of the weights of the edges in the path. For example The length of a path that starts at , visits , and terminates at is . The weight of edge is . The weight of edge is . The length of a path is Example 1.6 Bacon Numbers Actor Kevin Bacon has appeared in numerous films including Diner and Apollo . A movie stars with Kevin Bacon are Bacon number one. For example, Ellen Barkin has Bacon number one because she appeared with Bacon in Diner. Bacon number two is a movie star who did not appear in Bacon but appeared with a Bacon number 1 movie star. Higher Bacon numbers are defined similarly. For example, Bela Lugosi has Bacon number three. Lugosi was in Black Friday with Emmertt Vogan, Vogan was in With a Song in My Heart with Robert Wagner, and Wagner was in Wild Things with Bacon. We next develop a graph model for Bacon numbers. Example 1.7 Similarity Graphs This example deal with the problem of grouping “like” objects into classes based on properties of the objects. For example, suppose that a particular algorithm is implemented in by a number of persons and we want to group “like” programs into classes based on certain properties of the programs. Suppose that we select as properties 1. The number of lines in the program. 2. The number of return statements in the program. 3. The number of function calls in the program. Similarity Graph A similarity graph is constructed as follows. The vertices correspond to programs. A vertex is , where is the value of property . is the set of programs . Each vertex is assigned a triple , where is the value of property or . , , , ● Dissimilarity Function Define a dissimilarity function as follows. For each pair of vertices and , we set . is the vertex corresponding to program , we obtain , , , , , , , , , , , , , . If and are vertices corresponding two programs, is a measure of how dissimilarity the programs are. A large value of indicates dissimilarity. A small value of indicates similarity. For a fixed a number . Insert an edge between vertices and if . and are in the same class if or there is a path from to . Figure 1.9 A similarity graph corresponding to the programs of Table 1.2 with . In Figure 1.9 we show the graph corresponding to the programs of Table 1.2 with . In this graph, the programs are grouped into three classes and . An appropriate value for might be selected by trial and error or the values of might be selected automatically according to some predetermined criteria. Dissimilarity function values and all other There are three classes : and . Example 1.8 The -Cube (Hypercube) We discuss one model for parallel computation known as the -cube or hypercube. The cube has processors, Vertices are labeled An edge connects two vertices if the binary representation of their labels differs in exactly one bit cube cube has two processors labeled 0 and 1, and one edge Let and be two cubes whose vertices are labeled in binary Place an edge between each pair of vertices, one from and one from , provided that the vertices have identical labels. Change the label on each vertex in to and Change the label on each vertex in to (a line) has only vertices and . has no cycle. (a square) has vertices labeled , , and . A Hamilton cycle is (, , , , ) (a cube) has vertices, edges and faces. Vertex labels , , , , , , and . A Hamilton cycle is (, , , , , , , , ) Figure 1.10 The 3-cube (a hypercube) has vertices, edges and faces. Vertex labels , , , , , , , , , , , , , , , . Definition 1.9 Example 1.10 The complete graph on four vertices, , is shown in Figure 1.12. Figure 1.12 The complete graph . Definition 1.11 A graph is bipartite if there exist subsets and of , each edge in is incident on one vertex in and one vertex in . Example 1.12 The graph in Figure 1.13 is bipartite. Since if we let and , each edges is incident on one vertex in and one vertex in . Figure 1.13 A bipartite graph. is an edge in a bipartite graph, then is incident on one vertex in and one vertex in . If and , then there is an edge between and . The graph of Figure 1.13 is bipartite since each edge is incident on one vertex in and one vertex in . Not all edges between vertices in and are in the graph. For example, the edge is absent. Example 1.13 The graph in Figure 1.14 is not bipartite. It is often easiest to prove that a graph is not bipartite by arguing by contradiction. Figure 1.14 A graph that is not bipartite. The graph of Figure 1.14 is bipartite. The vertex set can be partitioned into two subsets and . Each edge is incident on one vertex in and one vertex in . Consider the vertices , and . and are adjacent one in and the other in . is in and is in . and are adjacent one in and the other in . is in and is in . and are adjacent one in and the other in . is in and is in . is in both and . and are not disjoint. The graph of Figure 1.14 is not bipartite. Example 1.14 The complete graph on one vertex is bipartite. Let be the set containing the one vertex and be the empty set. Then each edges is incident on one vertex in and one vertex in . Definition 1.15 is the complete bipartite graph on and vertices. is the simple graph whose vertex set is partitioned into sets with vertices and with vertices in which the edge set consists of all edges of the from with and . Example 1.16 Complete bipartite graph on two and four vertices, is shown in Figure 1.15. Figure 1.15 The complete bipartite graph Exercises 1, 2 8.2 Paths and Cycles Definition 2.1 Start at vertex ; go along edge to ; go along edge to ; and so on. Example 2.2 In the graph of Figure 2.1, is a path of length from vertex to vertex . Figure 2.1 A connected graph with paths of length and of length . Example 2.3 In the graph of Figure 2.1, the path consisting solely of vertex is a path of length from vertex to vertex . connected graph is a graph in which we can get from any vertex to any other vertex on a path. Definition 2.4 A graph is connected if given any vertices and in , there is a path from to . Example 2.5 The graph of Figure 2.1 is connected since given any vertices and in , there is a path from to . Example 2.6 Graph of Figure 2.2 is not connected since, there is no path from to . Figure 2.2 A graph that is not connected. As we can see from Figure 2.1 and 2.2, a connected graph consist of one “piece,” which a graph that is not connected consists of two or more “piece.” These “piece” are subgraphs of the original graph and are called components. We give the formal definitions beginning with subgraph. A subgraph of a graph is obtained by selecting certain edges and vertices from subject to the restriction that if we select an edge that is incident on vertices and , we include . The restriction is to ensure that is a actually a graph. Definition 2.8 Example 2.9 Figure 2.3 A subgraph of the graph of Figure 2.4. Figure 2.4 A graph, one of whose subgraphs is shown in Figure 2.3. Example 2.10 Find all subgraphs of the graph of Figure 2.5 having at least one vertex. Solution Graphs , and have not edge. Graphs , and are a subgraph of graph . We select the one edge . is incident the two vertices. Graph is a subgraph of graph . Thus has the four subgraphs. Figure 2.5 The graph for Example 2.10. Figure 2.6 The four subgraphs of the graph of Figure 2.5. Definition 2.11 Example 2.12 The graph of Figure 2.1 has one component, namely itself. A graph is connect it has exactly one component. Example 2.13 The component of a graph is obtained by defining a relation on the set of vertices by the rule path from to is an equivalence relation on . If , the set of vertices in the component containing is the equivalence class . Definition of “path” allows repetitions of vertices or edges or both. Definition 2.14 Example 2.15 Example 2.16 Konigsberg Bridge Problem Konigsberg Bridge Problem Starting and ending at the same point, is it possible to cross all seven bridges just once and return to the starting point? This problem can represented by a graph Edges represent bridges and each vertex represents a region. Degree of a Vertex is the degree of a vertex . is the number of degree of edges incident on . Theorem 2.17 Theorem 2.18 Example 2.19 is the graph of Figure 2.10. Use Theorem 2.18 to verify that has an Euler cycle. Find an Euler cycle for . Figure 2.10 The graph for Example 2.19. Theorem 2.21 Corollary 2.22 Theorem 2.23 Theorem 2.24 Figure 2.12 A cycle that either is a simple cycle or can be reduced to a simple cycle. Exercises 5, 6, 8 8.3 Hamiltonian Cycles and the Traveling Salesperson Problem ● Hamilton’s puzzle Each corner bore the name of a city and the problem was to start any city, travel along the edges, visit each city exactly one time, and return to the initial city. A cycle in a graph contains each vertex in exactly once, except for the starting and ending vertex that appears twice, a Hamiltonian cycle. A connected graph has a Hamiltonian cycle is a Hamiltonian graph. Example 3.1 Cycle is a Hamiltonian cycle for the graph of Figure 3.4. Example 3.2 Figure 3.6 A graph with a Hamiltonian cycle. Example 3.3 Show that the graph of Figure 3.7 does not contain a Hamiltonian cycle. Figure 3.7 A graph with no Hamiltonian cycle. ● Traveling Salesperson Problem Given a weighted graph , find a minimum-length Hamiltonian cycle in . Traveling salesperson problem 1. To visit every vertex of a graph G only once by a simple cycle. 2. Such a cycle is called a Hamiltonian cycle. 3. If a connected graph G has a Hamiltonian cycle, G is called a Hamiltonian graph. Example 3.4 The cycle is a Hamiltonian cycle for the graph of Figure 3.8. Replacing any of the edges in by either of the edges labeled would increase the length of is a minimum-length Hamiltonian cycle for . Thus solves the traveling salesperson problem for . Figure 3.8 A graph for the traveling salesperson problem. Example 3.5 Gray Codes and Hamiltonian Cycles in the Cube Considered a ring model for parallel computation is a simple cycle. A Gray code is a sequence such that every bit string appears somewhere in the sequence and differ in exactly one bit, And and differ in exactly one bit. Considered a ring model for parallel computation is a simple cycle. Parallel computation models The n-cube has processors, 1. Vertices are labeled 0, 1, 2, …, 2. An edge connects two vertices if the binary representation of their labels differs in exactly one bit 3. n-cube a. 1-cube has two processors labeled 0 and 1, and one edge b. and be two (n-1)-cubes whose vertices are labeled in binary 0, 1, …, c. Place an edge between each pair of vertices, one from and one from , provided that the vertices have identical labels. d. Change the label L on each vertex in to 0L and Change the label L on each vertex in to 1L Theorem 3.6 Gray Codes and Hamiltonian Cycles Proof We prove the theorem by induction on . Corollary 3.7 Example 3.8 Gray Codes and Hamiltonian Cycles Gray Codes and Hamiltonian Cycles is a line. has only two vertices and . has no cycle. is a square. has vertices labeled , , and A Hamiltonian cycle is is a cube. has 8 vertices labeled 000, 001, 010, 011, 100, 101, 110, and 111. A Hamiltonian cycle is . is a hypercube. has vertices, edges and faces Vertex labels: 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Example 3.9 The Knight’s Tour In chess, the Knight’s move consists of moving two squares horizontally or vertically and then moving one square in the perpendicular direction. For example, in Figure a Knight on the square marked can move to any of the square marked . A Knight’s tourof an board begins at some square, visits each square exactly once marking legal moves, and returns to the initial square. The problem is to determine for which a Knight’s tour exists. Figure 3.10 The Knight’s legal moves in chess. Figure 3.11 A chessboard and the graph . Exercises 9. 11 8.4 A Shortest-Path Algorithm A weighted graph is a graph in which values are assigned to the edges. The length of a path in a weighted graph is the sum of the weights of the edges in the path. is the weight of edge . In weighted graphs, we often want to find a shortest path between two given vertices. is a connected, weighted graph. The weights are positive numbers. Find a shortest path from vertex to vertex . Dijkstra’s algorithm involves assigns labels to vertices. is the label of vertex . At any point, some vertices have temporary labels and rest have permanent labels. is the set of vertices having temporary labels. We will circle vertices having permanent labels. If is the permanent label of vertex , then is the length of a shortest path from vertex to vertex . All vertices have temporary labels. Each iteration of the algorithm changes the status of one label from temporary to permanent; thus we may terminate the algorithm when receives a permanent label. gives length of a shortest path from to . Algorithm 4.1 Dijkstra’s Shortest-Path Algorithm Example 4.2 Show how Algorithm 4.1 finds a shortest path from to in the graph of Figure . Figure 4.2 shows the result of executing lines . Solution At line , is not circled. We proceed to line , where we select vertex , the uncircled vertex with the smallest label, and circle it. At lines and we update each of the uncircled vertices and , adjacent to .We obtain the labels , . We proceed to line , where we select vertex , the uncircled vertex with the smallest label, and circle it. At lines and we update each of the uncircled vertices and , adjacent to . We obtain the labels show in Figure 4.4. is labeled , indicating that the length of a shortest path from to is . A shortest path is given by . Theorem 4.3 Proof We use mathematical induction on to prove that the time we arrive at line , is the length of a shortest path from to . This proved, correctness of the algorithm follows since when is chosen at line , give the length of a shortest path from to . Example 4.4 Find a shortest path from to and its length for the graph of Figure . Dijkstra’s algorithm is in the worst case. Theorem 4.5 Exercises 13, 14, 15, 16 8.5 Representations of Graphs Our first method of representing a graph uses the adjacency matrix. Example 5.1 adjacency matrix Consider the graph of Figure 5.1. To obtain the adjacency matrix of this graph, we first select an ordering of the vertices, say , , , , . Next, we label the rows and columns of a matrix with the ordered vertices. The entry in this matrix in row , column , is the number of edges incident on and . , the entry is twice the number of loops incident on . The adjacency matrix for this graph is We obtain the degree of a vertex in a graph by summing row or column in ’s adjacency matrix. ● The adjacency matrix is symmetric about the main diagonal, the information except that on the main diagonal appears twice. Example 5.2 The adjacency matrix of the simple graph of Figure is ● If is the adjacency matrix of a simple graph , the powers of , count the number or paths of various lengths. The vertices of are labeled , the entry in the matrix is equal to the number of paths from to of length . Consider the entry for row , column c in , obtained by multiplying pairwise the entries in row by the entries in column of matrix and summing: . This happens if there is a vertex whose entry in row is This happens if there is a vertex whose entry in column is . Such edges from a path of length from to and each path increases the sum by . The sum is because there are two paths , of length from to . In general, the entry in row and column of the matrix is the number of paths of length from vertex to vertex . The entries on the diagonal give the degree of the vertices. Consider vertices . The degree of is since is incident on the three edges and . But each of these edges can be converted to a path of length from to , , A path of length from to defines an edge incident on . Thus the number of paths of length from to is , the degree of . Theorem 5.3 We will use induction on . Figure 5.3 The proof of Theorem . A path from to of length whose next-to-last vertex is consists of a path of length from to followed by edge . There are paths of length from to and is if edge exists and . Otherwise, the sum of over all gives the number of paths of length from to . Example 5.4 After Example 5.2, we showed that if is the matrix of the graph of Figure , then The entry from row , column is , which means that there are six paths of length from to . , , , , , Example 5.5 Incidence Matrix To obtain the incidence matrix of the graph, we label the rows with the vertices and columns with the edges. The entry for row and column is if is incident on and otherwise. Thus the incidence matrix for the graph of Figure 5.4. Figure 5.4 A column such as is understood to represent a loop. Exercises 19, 20 8.6 Isomorphisms of Graphs ● Isomorphic Graphs Draw and label five vertices and . Connect and and and and and . Surely these figures define the same graph even though they appear dissimilar. The two graphs are the same graph, but the shapes are different. Same graphs are isomorphic. Figure 6.1 Isomorphic graph Definition 6.1 Example 6.2 An isomorphism for the and of Figure 6.1 is defined by , , , , , , , , . and are isomorphic we define a relation on a set of graphs by the rule is an equivalence relation. Each equivalence class consist of a set of mutually isomorphic graphs. Theorem 6.4 Corollary 6.5 Example 6.6 Isomorphic and Adjacency Matrix Two simple graph and are not isomorphic. Find a property of that does not have but that would have if and were isomorphic. Such a property is an invariant. A property is an invariant if whenever and are isomorphic graphs: If has property , has property . By Definition 6.1, if graphs and are isomorphic, there are one-to-one, onto functions from the edges of to the edges of . and are isomorphic, then and have the same number of edges and the same number of vertices. Therefore, if and are nonnegative integers, the properties “has edges” and “has vertices” are invariants. Example 6.7 Example 6.8 Show that if is a positive integer, “has a vertex of degree ” is an invariant. Solution and are isomorphic graphs. is a one-to-one, onto function. has a vertex of degree . Then there are edges incident on . By Definition 6.1, are incident on . is one-to-one, . Let be an edge that is incident on in . Since is onto, there is an edge in with . Since is incident on in , by Definition , is incident on in . Since are the only edges in incident on , for some . Now . , so has a vertex, namely , of degree . Example 6.9 Example 6.10 Exercises 21, 22 8.7 Planar Graphs Definition 7.1 A graph is planer if it can be drawn in the plane without its edges crossing. If a connected, planar graph is drawn in the plane, the plane is divided into contiguous regions called faces. A face is characterized by the cycle that forms its boundary. The face is bounded by the cycle . The face is bounded by the cycle . The face is bounded by the cycle . The outer face is considered to be bounded by the cycle . The graph of Figure 7.2 has faces, edges and vertices. and satisfy the equation . Example 7.2 Definition 7.3 Edges in series: Example 7.4 In the graph of Figure the edges and are in series. The graph of Figure is obtained from by a series reduction. Figure 7.4 is obtained from by a series reduction. Definition 7.5 Example 7.6 Define a relation on a set of graphs by the rule if and are homeomorphic, is an equivalence relation. Each equivalence class consists of a set of mutually homeomorphic graphs. Theorem 7.7 Kuratowski’s Theorem Example 7.8 Solution The vertices , , and each have degree . In each vertex has degree , so let us eliminate the edges and so that all vertices have degree . We note that if we eliminate one more edge, we will obtain two vertices of degree and we can then carry out two series reductions. The resulting graph will have nine edges; since has nine edges, this approach looks promising. Using trial and error, we finally see that if we eliminate edge and carry out the series reductions, we obtain an isomorhic copy of . Therefore, the graph of Figure 7.6 is not planar, since it containing a subgraph homeomorphic to . Theorem 7.9 Euler’s Formula for Graphs If is a connected, planar graph with edges, vertices and faces, then (8.7.3) . , , , , Figure 7.8 The Basis Step of Theorem 7.9 Figure 7.10 The proof of Theorem 7.9 for the case that has a cycle. We delete edge in a cycle. Euler’s Formula for Graph , , , , , , Exercises 25, 26, 27<\|endoftext\|>	4.5625
3,833	# Logarithm Laws Made Easy: A Complete Guide with Examples ## What are the Laws of Logarithms? The laws of logarithms are algebraic rules that allow for the simplification and rearrangement of logarithmic expressions. The 3 main logarithm laws are: 1. The Product Law: log(mn) = log(m) + log(n). 2. The Quotient Law: log(m/n) = log(m) – log(n). 3. The Power Law: log(mk) = k·log(m). The three fundamental laws of logarithms are shown below. ### Rules when using the Laws of Logarithms The rules for the laws of logarithms are: • To simplify logarithms with the product or quotient laws, the bases of the logarithms must be equal. • The base of the logarithms must be greater than zero. • The base of the logarithms cannot be equal to 1. • The inputs of the logarithms must be greater than zero. • The logarithm of 1 is equal to zero, no matter what the base of the logarithm is. The three fundamental laws of logarithms can also be written as: • loga(m) + loga(n) = loga(mn) • loga(m) – loga(n) = loga(m/n) • loga(mk) = kloga(m) ## Examples of Using Logarithm Laws The three most important logarithm laws are shown below with an example of each. When adding logarithms with the same base, the inputs to the logarithms can be multiplied. The 4 and 5 inside the logarithms are multiplied to obtain 20. When subtracting logarithms with the same base, the first logarithm is divided by the logarithm being subtracted. 10 is divided by 5 inside the logarithm to obtain log(2). The coefficient of the logarithm can be moved to become the power of the input to the logarithm. 5 is raised to the power of 2 to obtain 25. ### List of the Logarithm Laws The following table shows the complete list of log laws with examples of how they are used: Here is a visual list of 5 of the most commonly used logarithm laws. The colours used in this list show how the different rearrangements are made in a more simple, visual manner. The following sections take a look at each of the logarithm laws in greater detail. The logarithm laws are: 1. The Product Rule 2. The Quotient Rule 3. The Power Rule 4. The Inverse Property of Logarithms 5. The Zero Rule 6. The Identity Rule 7. The Inverse Property of the Exponent ## The Product (Addition) Logarithm Law To add two or more logarithms that have the same base, simply multiply the numbers inside the logarithms. For example, log(3)+log(2) = log(6). The result is a single logarithm with the same base as those being added. The formula for the product law of logarithms is given as: The product rule of logarithm laws Two logarithms can only be simplified using the addition (product) rule if they have the same base. This base remains the same in the resulting answer. For example, log3(2) + log3(5) = log3(2 × 5) = log3(10). Here is another example of adding logarithms using log laws and then simplifying. log4(2) + log4(8) = log4(16) Since two logarithms of the same base were added, the resulting logarithm input was equal to the product of the inputs given. Now log4(16) = 2 because 42 = 16. A logarithm simply asks what power the base of the logarithm must be raised to in order to obtain the input. 4 must be raised to the power of 2 to obtain 16 and so, the final answer of this question is just 2. The product rule for logarithms can also be written in reverse using the formula: The product rule of logarithm laws The product rule of logarithms states that a single logarithm can be separated into the sum of individual logarithms which have inputs that multiply to make the input of the original logarithm. For example, log(21) = log(3) + log(7). Here are some examples of how to add logarithms using the addition log law. ## The Quotient (Subtraction )Logarithm Law Logarithms of the same base can be subtracted by dividing their inputs. For example, log(8) – log(4) = log(2), since 8 ÷ 4 = 2. The result is a single logarithm with the same base as those being subtracted. The formula for subtracting logarithms using the quotient law is given as: The quotient law of logarithms Two or more logarithms can only be simplified using the quotient law if they have the same base. The input of the first logarithm is divided by the input of the logarithm being subtracted. The result is a single logarithm with the same base. For example, log3(20) – log3(4) = log3(5). The result is a single logarithm, also base 3. Here is another example of subtracting logarithms using the quotient logarithm law. log5(100) – log5(4) = log3(100 ÷ 4) which equals log5(25). log5(25) can be evaluated since 25 is a power of 5. 52 = 25 and so, log5(25) = 2. The quotient rule for logarithms can also be written in reverse using the formula: The product rule of logarithm laws The quotient rule of logarithms states that a single logarithm can be written as a logarithm subtracted from another logarithm. For example, log(2/5) = log(2) – log(5). Here are some examples of how to subtract logarithms using the subtraction log law. ## The Logarithm Law for Exponents If there is a coefficient in front of a logarithm, the power law of logarithms states that the input to the logarithm can be raised to the power of this coefficient. For example, 2log(3) = log(32) which equals log(9). The formula for the power law of logarithms is: The power law of logarithms For example, 3log2(10) can be written as log2(103). This is because the coefficient of 3 immediately in front of the log can be moved so that it is acting as a power on the number inside the log. Now log2(103) = log2(1000). ### Reciprocal Rule of Logarithms When the reciprocal of the input to a logarithm is taken, the result is equal to -1 multiplied by the logarithm of the original input. That is log(1/b) = -log(b). For example, log(1/2) = -log(2). We can see that since then . We can use the power law of logarithms to move the power of -1 down to become the coefficient of the logarithm so that or more simply, . For example, . The formula for the reciprocal of the input to a logarithm is given by: Reciprocal of the input to a logarithm The reciprocal logarithm law can be applied in the following circumstances. • (since both sides of the equation equal zero) ### Logarithms Involving a Square Root When the coefficient of a logarithm is equal to one half, the input inside the logarithm can be raised to the power of this half. Raising a value to the power of one half is equivalent to finding its square root. Therefore (1/2)loga(b) = loga(√b). For example, (1/2)log(16) = log(√16) = log(4). The formula for the a logarithm with a coefficient of one half is given as: Formula for a logarithm with a coefficient of one half In the example below, using the power logarithm law, (1/2)log(16) = log(161/2). When 16 is raised to the power of one half, it is equivalent of finding the square root of 16. Therefore (1/2)log(16) = log(161/2) = log(√16) = log(4). In general, . And so, the formula for any fractional coefficient of a logarithm is: Formula for a fractional coefficient of a logarithm For example, and so, . ## The Inverse Property of Logarithms The inverse property of logarithms states that loga(ak) = k. When the base number inside the logarithm is equal to the base of the logarithm, the result is simply the value of the exponent inside the logarithm. For example, log3(35) = 5. The formula for the inverse property of logarithms is: The Inverse Property of Logarithms The other example of the inverse property of logarithms listed above is . That is because when 2 is raised to the power of 𝑥, we obtain 2𝑥. ## Logarithm Law: The Zero Rule The logarithm of one is equal to zero no matter what the base of the logarithm is. That is, loga(1) = 0 for all valid values of ‘a’. Some of the many examples include: log2(1) = 0, log5(1) = 0, log(1) = 0 and ln(1) = 0. This rule is true since the value of a logarithmic expression is always equal to the power that the base of the logarithm must be raised to in order to obtain the value of the input of the logarithm. Since any positive or negative number raised to the power of zero is equal to 1 (that is, a0=1), then loga(1) must be equal to 0. ## Logarithm Law: The Identity Rule The identity rule of logarithms states that if the input to a logarithm is equal in value to the base of the logarithm, the result is equal to 1. That is, loga(a) = 1. For example, log3(3) = 1 and log2(2) = 1. This rule works since a logarithmic expression is equal in value to the power that its base must be raised to in order to obtain the value which is the input to the logarithm. Since a1 = a, then loga(a) must equal 1. The identity rule of logarithms is applied to natural logarithms in the sense that ln(e) = 1. This is because ln(e) means loge(e). ## Logarithm Law: Inverse Property of Exponents The inverse property of exponents states that if a given number is raised to the power of a logarithm which has the same base as this given number, then the result is simply equal to the input of the logarithm. That is, aloga(k) = k. For example, 2log2(5) = 5. The inverse property of exponents can be applied to natural logarithms in the following manner: eln(k) = k This is true since ln(k) means loge(k). For example, eln(2) = 2. ## Logarithm Laws: Examples and Solutions Here are some examples with solutions of using logarithm laws to simplify and expand expressions. It is important to understand that the addition and subtraction logarithm laws can only be used when we have one logarithm added to or subtracted from another logarithm and there are no coefficients in front of either logarithm. Therefore, it is important to always use the power logarithm law to move any coefficients from in front of the logarithms before adding or subtracting them. For example, use logarithm laws to simplify . Both logarithms have coefficients and so, we use the power log law so that and . Therefore we know that and . Therefore becomes . Now that the coefficients have been moved and we simply have one logarithm plus another logarithm, we can add them using the addition (product) law of logarithms. We simply multiply the inputs of 4 and 1000. . Here is another example of simplifying a logarithmic expression using logarithm laws. Simplify . We move the coefficients first so that they become the powers of the inputs to the logarithm. In the first logarithm: and in the second logarithm: . Therefore becomes . Since the logarithms have the same base, we can subtract them using the subtraction (quotient) law. We simply divide 8 by 2 to obtain 4. . ### Here are some further examples of using logarithm laws: • Simplify . All logarithms are the same base and so we multiply the numbers in the logarithms that are being added and then divide by the number inside the logarithm being subtracted. Therefore: • Simplify . Both 16 and 2 are powers of 2, which is the base of the logarithms. We can write as which is equal to 4. is just equal to 1. Therefore, and so, the answer is 5. Alternatively, we can use the addition (product) law immediately so that . Then we can use the fact that 25=32 to evaluate . • Evaluate . We can use the subtraction (quotient) law so that . Using the fact that log(100) means log10(100) and also the fact that 102=100, we can evaluate log(100) = 2. Therefore . • Expand the logarithmic expression of . Since the 25, , and z are multiplied together inside the logarithm, we can write. We can see that log5(25) = 2 since 52 = 25. Therefore . We can then use the power law to bring the coefficients down in front of the logarithms to obtain: . • Expand the logarithmic expression . All terms on the top of the fraction are multiplied together and so, can be written as the summation of individual logs. The term on the bottom of the fraction are subtracted as we are dividing by this. We write . We can write log2(6) as log2(2) + log2(3). We can therefore write log2(6) as 1 + log2(3). Therefore . Finally, we can bring down the power of 5 on the final logarithm using the power law. . • Expand the logarithmic expression of . We can write . We can then write this as . We bring down the power using the power law so that . Finally, we use the fact that ln(e) = 1 so that: . ## Expanding Logarithms Using Logarithm Laws Single logarithms can be expanded into multiple logarithms of the same base using logarithm laws. For example, using the addition (product) law, we can write log(20) as log(20) = log(5) + log(4). Then log(4) can also be written as log(22), which can be written as 2log(2). Therefore log(20) can be written as log(20) = log(5) + 2log(2). Here is another example of expanding a logarithmic expression using logarithm laws. Using the addition and subtraction logarithm laws, we can write . All terms on top of the fraction inside the logarithm are positive and the log(y5) is subtracted since it has been divided inside the logarithm. We can then use the power law of logarithms to bring down the powers as coefficients. We can also evaluate log(10) = 1. Therefore . ## Change of Base Rule The change of base formula for logarithms is loga(b) = logc(b) ÷ logc(a). For example log2(10) = log(10) ÷ log(2). The change of base logarithm formula is: The change of base logarithm formula To divide logarithms that have the same base, the change of base formula can be used. That is, logc(b) ÷ logc(a) = loga(b). For example, log2(27) ÷ log2(3) = log3(27) and this can be evaluated since log3(27) = 3. ## Proof of the Logarithm Laws The following is the algebraic proof of the product (addition) law of logarithms. • Let and take logarithms of both sides to obtain . Then use the power law of logarithms to bring the power of m down as the coefficient of the logarithm. That is, . • Take and follow the same process as above to obtain . • Multiply and to obtain which is more simply written as . • Take logarithms of both sides to obtain . Then use the power law of logarithms to bring the (m+n) down as the coefficient of the log so that . • Then expand the brackets to obtain . • Finally substitute and from the previous calculations so that becomes . The following is the algebraic proof of the quotient (subtraction) law of logarithms: • Let and take logarithms of both sides to obtain . Then use the power law of logarithms to bring the power of m down as the coefficient of the logarithm. That is, . • Take and follow the same process as above to obtain . • Divide 𝑥 by y so that and so, . • Take logarithms of both sides so that . • This can be written using the power law of logarithms as . • Expanding the brackets, . • Finally substitute and from the previous calculations so that becomes . The following is algebraic proof of the exponent (power) law of logarithms: • Consider the expoenential equation of and write this in logarithmic form as . • Returning to and raise both sides to the power of k to obtain . • This can be written more simply as . • Now write in logarithmic form as . • Now substitute as previously found into in order to obtain . • This can be written in the more familiar format as .<\|endoftext\|>	4.75
386	As children enter preschool some of their primary goals are to learn how to navigate the classroom, get along with others, and manage their own behavior. In APS we are using Positive Behavior Interventions Support (PBiS) to support our students in meeting these goals. Positive Behavioral Interventions Support means using positive language to help kids learn how to be successful at school and at home. - Clearly stating what we expect children to do. - Teaching and modeling these expectations throughout the day in all settings. - Supervising students to support them in meeting our expectations. - Recognizing when our children meet (or try to meet) these expectations. Using PBS in preschool helps children learn what to do at school, spend more time engaged in pro-social and academic activities, improve social and academic skills, and provide safe schools for all children. When teachers notice students following school expectations, being kind to others, and keeping our school safe, they recognize them by telling them specifically what they have done well. For example, - “You came down the slide feet first.” - “You waited quietly until I was done talking with Jorge.” - “You hung up your backpack by yourself.” The students are then given a stamp on the hand and put a puff ball in a group goal jar. When the jar is full, the class celebrates their positive behavior together with a special activity such as pajama day, crazy hair day, toy car race day, bring a stuffed animal day, etc. Take a look at our celebration bulletin board to find out what fun activities our students have earned! PBiS at Home Parents can implement their own version of PBiS at home as well. We have presented these strategies at parent coffees and have developed a home version of our classroom expectations for you to use as well.<\|endoftext\|>	3.859375
436	Most history books read throughout classrooms in the United States omit or exclude African American narratives in 19th century New England. During this time Boston and New York City were growing ports of commerce. As a result Middletown thrived as an industrial city during this period, since it was located almost exactly half way between the two cities. History books fail to mention the existence of African American communities in the region. It’s almost as if they never existed. The truth is that Boston, New York, and Middletown had abundant African American populations that formed thriving communities. Archeological work carried out on the African Meeting House in Boston has established, through material evidence, the existence of an African American community that settled on the outskirts of the city. It wasn’t long before this community developed a set of practices and values that members drew on to forge both individual and shared identities (Landon & Bulger, 2013). In New York City archaeological work was performed on the Seneca Village settlement that found that members of this community lived relatively successful lives in terms of access to food, education, and land (Wall, et. al., 2008). The Beman triangle, like the settlements around the African Meeting house in Boston and Seneca village in New York, is a truly remarkable example of community building, especially during a time of intense discrimination and racism in New England. The Beman Triangle project today is an effort in community archeology to uncover the past, and learn more about the lives of the previous residents of the Triangle, by working in collaboration with the greater Middletown Community and the members of the present day AME Zion church. Wesleyan students, professors, academic staff, and local residents work collaboratively to unearth the historical narratives necessary for our present day understanding of the social constructions of race, class, and gender. Landon, D.B. and T.D. Bulger. 2013. ‘Constructing Community: Experiences of Identity, Economic Opportunity, and Institution Building at Boston’s African Meeting House.’ International Journal of Historical Archaeology 17(1): 119-142.<\|endoftext\|>	3.828125
630	# Question Video: Using the Negative Mass Method to Find the Center of Mass of a Lamina Mathematics A uniform lamina of mass 15 kg has its center of mass at (2, 6). If a piece of the lamina is cut out whose mass is 11 kg and whose center of mass is at (6, 2), find the coordinates of the center of mass of the remaining part. 02:11 ### Video Transcript A uniform lamina of mass 15 kilograms has its center of mass at two, six. If a piece of the lamina is cut out whose mass is 11 kilograms and whose center of mass is at six, two, find the coordinates of the center of mass of the remaining part. In this question, we are given the mass and the center of mass of a lamina but not its shape. And we are likewise given the mass and the center of mass of a hole cut out of the lamina but not its shape. A diagram in this case will be hard to draw and potentially misleading, so we will focus on the formulae. We can model the original lamina of mass 15 kilograms and center of mass at two, six as a particle of mass 15 kilograms located at two, six. We can likewise model the piece of the lamina removed as a particle of mass negative 11 kilograms and center of mass at six, two. Recall that when we have a system of two particles, the π₯-coordinate of their center of mass, COM π₯, is given by the products of the particlesβ masses and their π₯-coordinates π one π₯ one plus π two π₯ two divided by the total mass π one plus π two. And similarly, the π¦-coordinate of their center of mass is given by the products of their masses and their π¦-coordinates divided by the total mass. So, for the π₯-coordinate, we take the mass of the original lamina, 15, multiply it by the π₯-coordinate of its center of mass, two. Then, we treat the cutout lamina as having negative 11 kilograms. So, we subtract 11 times the π₯-coordinate of its center of mass, six. We then divide by the total mass, which will be 15 minus 11. This gives us the π₯-coordinate of the center of mass of the remaining part, negative nine. We then repeat the process for the π¦-coordinate of the center of mass, giving us 15 times six minus 11 times two all divided by the total mass 15 minus 11. This gives us the π¦-coordinate of the center of mass of the remaining part, 17. This gives us our final answer. The center of mass of the remaining part is located at negative nine, 17.<\|endoftext\|>	4.71875
180	South Australia is known to support up to many species of seagrasses, found in various water depths, waves and seabed types. Seagrass are marine flowering plants that have a true root structure, often growing in large meadows that can be monospecific (one species) or mixed species. Seagrass meadows are very important to the nearshore coastal environment. They help stabilise the seabed by holding sand with their roots and reduce the waves and currents near the seabed with their leaves. They also provide both food and habitat for a variety of marine organisms. Seagrass loss is a worldwide problem and this includes South Australia. While natural events such as storms can cause large-scale seagrass loss, in many areas human activity has also contributed to the decline in seagrass meadows in coastal areas.<\|endoftext\|>	3.859375
471	Lesson plan # 15. Compare fractions with unlike denominators (A) teaches Common Core State Standards CCSS.Math.Content.4.NF.A.2 http://corestandards.org/Math/Content/4/NF/A/2 teaches Common Core State Standards CCSS.Math.Practice.MP2 http://corestandards.org/Math/Practice/MP2 teaches Common Core State Standards CCSS.Math.Practice.MP4 http://corestandards.org/Math/Practice/MP4 ## You have saved this lesson plan! Here's where you can access your saved items. Dismiss Lesson objective: Apply the skill of comparing fractions with unlike denominators using the most appropriate strategy based on the fractions being compared. This lesson provides an opportunity for students to apply their knowledge and understanding of comparing fractions with unlike denominators to a fictional mathematical situation. Students are asked to determine which real galaxies a group of ficticious martians could travel to given only a fractional amount of fuel in their spaceship. Key Concept students will use: • there are three possible relationships between two given numbers with the same unit of reference, a and b: a>b, a<b, or a=b • we can compare two numbers indirectly by comparing them both to a common benchmark • a/b = (nxa)/(nxb) because we have n times as many pieces in the whole. Each piece is n times smaller, so we need n times as many of them to cover the same distance or the same area • finding common denominators is helpful in comparing fractions because it enables us to compare same-sized pieces Skills students will use: • finding equivalent fractions • comparing fractions with unlike denominators by comparing indirectly to a common benchmark • comparing fractions with unliked denominators by using the common denominators method Students engage in Mathematical Practice 2 (reason abstracly and quantitatively) determine which galaxies the group of martians could travel to using the amount of fuel they have left in their spaceship. Key vocabulary: • benchmark • common denominators • common numerators • comparison • denominator • equivalent • numerator • same-sized pieces Related content Appears in • 15. Compare fractions with unlike denominators (A)<\|endoftext\|>	4.65625
560	Brian Harris # Introduction To Matrix Operations And Standard Determinants A matrix is a rectangular array of numbers arranged in rows and columns. The representation of a matrix is as follows: An m × n matrix consists of m rows which are horizontal and the n columns which are vertical. Basic operations of a matrix involve matrix addition, scalar multiplication, transpose, matrix multiplication and row operations. Matrix multiplication is a process of finding the product of two matrices A and B given that the number of columns of the left matrix is the same as the number of rows of the right matrix. Suppose A is an m x n matrix and B is an n x p matrix, then the product matrix AB is m x p. Row operations involve three types, namely: • Row multiplication: It deals with the multiplication of all the numbers of a row by a non-zero constant. • Switching of a row is interchanging two rows of a matrix. The above matrix operations are done while solving linear equations and finding the inverse of a matrix. Determinant of a matrix The determinant det(A) or \|A\| of a square matrix A is a number that inscribes certain properties of the matrix. The necessary condition for a matrix to be invertible is that its determinant should be nonzero. Consider a matrix A = Then ### Properties of determinants The properties of the determinant include the following: • There will be a change in the sign if the rows and columns are interchanged. • The scalars can be factored out from the rows and columns. • There will be no change in the value of the determinant if the multiples of rows and columns are added together. • Suppose a row of a matrix is multiplied by a scalar c, then its determinant also gets multiplied by c. • The value of a determinant is 0 if a row or a column has zeros in it. • The value of a determinant is 0 if two rows or columns are equal. • If the elements of the row can be expressed as a sum of two or more elements, then the determinant can be expressed as the sum of two or more determinants. • The value of the determinant remains the same even if the equimultiples of corresponding elements of other rows or columns are added to each element of a row or a column of a determinant. • The value of the determinant is equal to the product of diagonal elements if all the elements of a determinant above or below the main diagonal consists of zeros. • x – α is a factor of ‘∆’ [value of determinant] if the value of ‘∆’ becomes zero when x = α is substituted. ### Standard Determinants The following are some of the standard determinants: Brian Harris Brian Harris Brian Harris Brian Harris<\|endoftext\|>	4.65625
545	The human brain is one of the most important, complicated parts of the body. It is capable of extraordinary intelligence and guides the performance of actions that are necessary for our survival. Along with our spinal cords, our brains makes up our central nervous system which is the means of communication between our mind and our body. In addition, it is responsible for retaining our memories, reasoning ability, logical thoughts and subsequent actions. IBM, and other companies, have started to model their artificial intelligence systems on the human brain, to usher in a new era of technology. This has become necessary for artificial intelligence to continue to evolve. Previously, these systems were built with all the knowledge they require, while a human brain can accumulate additional knowledge over time. They also require access to a large amount of data to be programmed. New skills require the old information to be wiped out and completely reprogrammed. The human brain, on the other hand, learns things incrementally and adds more to its storage constantly. Our intelligence is based on our reasoning capabilities and the ability to apply new information logically, based on past experiences. Artificial intelligence cannot apply logic to any situation it may be faced with, thus limiting its uses. IBM has done extensive research into our neurological and psychological systems, and applied the knowledge they have gained from this to their development. The company’s ultimate goal is to create artificial intelligence that can learn for a lifetime, and adapt to new circumstances while retaining all knowledge it had previously acquired. The process to achieve this has been split into two sections: short-term adaptation, where a limited amount of time is spent on training a system, and long-term adaptation, which is inspired by the way that the brain forms memories. The company’s DeepMind team has already created a synthetic neural network, which is designed to use reasoning skills to complete tasks. The systems fitted with the new network have been tested with a series of questions that have forced them to use this ability. 96% of the time these new systems could answer the questions correctly, compared to 42 – 77% in previous artificial intelligence models. The researchers are also adapting the network to store memories, by paying more attention to details and events. IBM believes that these methods will be the way to change the future of artificial intelligence. These newly acquired reasoning skills mean that after a while the systems would be able to improve and build themselves, which many people think could be a scary possibility. If monitored correctly, however, this development would be monumental for the future of industries that rely on artificial intelligence for expansion. The technology could also be used to save countless lives, fast forward our space and deep sea exploration, and many life changing events.<\|endoftext\|>	3.9375
5,095	Select Page Algebra: Linear Equations, Graphs, Slope Section. If you need assistance please contact [email protected]. Get an answer for 'Determine the linear function that passes through the points (2,1) and (1,1).' linear parent function (graph) straight line through the points (-2, -2), (-1, -1), (0, 0), (1, 1), (2, 2) and other points. Simplest linear function f(x)=x. The origin is the point (0,0), so your line needs to have the same slope (parallel) as y = -3x + 7 and also when x=0, y=0. When c = is with coordinates (a, a). This topic covers: - Intercepts of linear equations/functions - Slope of linear equations/functions - Slope-intercept, point-slope, & standard forms - Graphing linear equations/functions - Writing linear equations/functions - Interpreting linear equations/functions - Linear equations/functions word problems For example, if the parent graph is shifted up or down (y = x + 3), the transformation is called a translation. For FREE access to this lesson, select your course from the categories below. Students also learn the different types of transformations of the linear parent graph. Sol'n: from the equation, were going to find the the slope of the line. Rate of Change. The data set seems to represent a vertical stretch of the linear function by a factor of 3. Linear parent function. Steepness of a line. So you have a line that passes through two points: (0,0) and (2,3). y=x or f(x)=x. find the formula for the linear function that passes through the point P= (2,-9)and is parallel to the linear function given by the equality: 7x-13y=22. Parent Function. A few values from the exponential function g(x) are shown in the table. x-intercept . A parent function is the simplest equation of a function. A linear parent function is the equation y = x or f(x) = x. Here we will focus on the key aspects of each family. 13y=7x-22, then divide by 13. y= (7/13)x-22/13. Where an equation passes the y-axis. It is of the form Key common points of linear parent functions include the fact that the: Equation is y = x. Domain and range are real numbers. Math help and check. (Round all coefficients to 4 decimal places when necessary.) I am trying to access some specific variable that are only available in a child class. Students also learn the different types of transformations of the linear parent graph. You know one of the function’s asymptotes and two points the graph passes through; 3. A linear function may be increasing, decreasing, or constant. In algebra, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. After you're corrected the required setting(s) refresh/reload this page. Find and graph the linear function that passes through the points (2,-3) and (5,0). Solution for Find the equation of the linear function f(x), that has a slope of -2 and passes through the point (5, 1). Cookies are not enabled on your browser. The quadratic function f(x) = x2 has a U-shaped graph. You know three points the the graph of the function passes through; 2. Become a MathHelp.com member today and receive unlimited access to lessons, grade reports, practice tests, and more! The graph of the linear function f passes through the points (0, 3) and (1, 1). If you are on a personal connection, like at home, you can run an anti-virus scan on your device to make sure it is not infected with malware. Respond to this Question. The parent function of all linear functions is Add comment More. Equation in the form y=kx: where k is not equal to 0. The most basic function in a family of functions is called the parent function. In the next lesson we will focus on relationships within each family, and how we can use certain relationships to make graphing by hand more efficient. The slope of a line between 2 points is slope = rise/run or. Cloudflare Ray ID: 61619fc3d975c37c Linear parent function. = -2/1 = -2. Students learn that the linear equation y = x, or the diagonal line that passes through the origin, is called the parent graph for the family of linear equations. Write an equation for a linear function whose graph has the given characteristics below. Which kind of function best models the set of data points (–1, 22), (0, 6), (1, –10), (2, –26), and (3, –42)? See tutors like this. The linear functions we used in the two previous examples increased over time, but not every linear function does. The “parent” of the family is the equation in the family with the simplest form. Performance & security by Cloudflare, Please complete the security check to access. Please enable Cookies and reload the page. Linear and Polynomial Parent Functions A constant function has the form f(x) = c, where cis any The identity function f(x) = x passes through all points real number. Similar Questions. The graph of linear function k passes through the points(-7,0) and (1,8). If you are at an office or shared network, you can ask the network administrator to run a scan across the network looking for misconfigured or infected devices. Parent Functions: When you hear the term parent function, you may be inclined to think of two functions who love each other very much creating a new function.The similarities don’t end there! Answers archive Answers : Click here to see ALL problems on Linear-equations; Question 1164875: Find the equation for an exponential function that passes through the pair of points given below. The graph of an increasing function has a positive slope. This condition ensures that the graph of the function is a straight line, hence the term linear function. Add your answer and earn points. Students learn that the linear equation y = x, or the diagonal line that passes through the origin, is called the parent graph for the family of linear equations. Ration that describes how much one quantity changes with respect to a change in another quantity. • function. ghk kjh. Slope-Intercept Form – Linear Parent Graph and Transformations. You may need to download version 2.0 now from the Chrome Web Store. Can somebody pls show me how to solve this? The first step is to find the slope of the line. Answer and Explanation: The given information is that: The line has to be perpendicular to 4y−2x+7 = 0 4 y − 2 x + 7 = 0 ; and, Passes throught hte point (2, −5) ( 2, − 5) . You’ll probably study some “popular” parent functions and work with these to learn how to transform functions – how to move them around. Become a MathHelp.com member today and receive unlimited access to lessons, grade reports, reviews and more! Direct Variation. • The zero of k is 7 The x-intercept of the graph of k is -7. But the problem is that I recieve the parent of this class by parameter. x g(x) −1 23 2 18 3 54 What is the positive difference in the y-intercept value of f(x) and g(x) ? Students also learn the different types of transformations of the linear parent graph. Lessons Lessons. Slope. If the parent graph is made steeper or less steep (y = 5 x), the transformation is called a dilation. HINT [See Example 2.] Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. Enter your answer in the box. The linear parent function passes through the points (2, 2) and he data set contains the (2, 2) = (2, 3(2)) and (4, 4) = (4, 3(4)). Domain: (-∞, ∞) Range:(-∞, ∞) Quadratic Parent Function equation. Thus, f(x) = x is the simplest of all linear functions and that is the reason why it is called linear parent function. For an increasing function, as with the train example, the output values increase as the input values increase. Each family of Algebraic functions is headed by a parent. It’s given that the graph of the linear function g is perpendicular to the graph of the function f. Therefore, the slope of the graph of the function g is the negative reciprocal of graph f, which is -1/-2 = 1/2 . For example, y = x 2 is a parent to other functions, such as y = 2x 2 - 5x + 3. See tutors like this. Constant of Variation. Step-by-step explanation: I think it is -2 New questions in Mathematics. Students learn that the linear equation y = x, or the diagonal line that passes through the origin, is called the parent graph for the family of linear equations. Your Response. Quadratic Parent Function graph. The graph a linear function if passes through the point (3, -8) and has a slope of -2 1 See answer christianbrown022007 is waiting for your help. So the equation of the line that passes through the point {eq}(x_1, y_1) = (3, 5) {/eq} and has slope {eq}\displaystyle = \frac{-5}{3} {/eq} is: Where an equation passes the x-axis. The graph of the constant function y = c is a horizontal line in the plane that passes through the point (0, c). The slope of the graph of the function f is therefore (1 - 3) / (1 - 0). Graph of a linear function Find and graph the linear function that passes through the points (1,3) and (2,5). Pay for 5 months, gift an ENTIRE YEAR to someone special! Nov 11, 2010 . y=x². Report 1 Expert Answer Best Newest Oldest. Solvers Solvers. https://member.mathhelp.com/api/auth/?token=. The linear function on this page is the general way we write the equation of a straight line. Passes through (−2, 5) and (2, −13) Follow • 2. Please enable javascript in your browser. What is its horizontal asymptote? Question: The linear function f(x) passes through the points (0, 3) and (2, 7) . Linear functions are those where the independent variable x never has an exponent.So for example they would not have a var such as 3x2in them. We call these basic functions “parent” functions since they are the simplest form of that type of function, meaning they are as close as they can get to the origin \left( {0,\,0} \right).The chart below provides some basic parent functions that you should be familiar with. A linear function is one where the largest power of x is equal to 1. Which statement must be true Answers The slope of the graph of k is -4/3. If it is negative, the function has no roots and is always positive/always negative; if positive, it has two distinct roots. and find homework help for other Math questions at eNotes Graph the relationship from year to see the shape. The cubic function f(x) = x3 is symmetric about the origin. This article focuses on the traits of the parent functions. First Name. All functions that can be written on the form f(x) = mx + b belong to the family of linear functions. By: Thomas R. answered • 04/30/18. [5] In the context of a polynomial in one variable x , the non-zero constant function is a polynomial of degree 0 and its general form is f ( x ) = c where c is nonzero. You know both asymptotes and one point the graph passes through. Nov 17, 2014 . The three types are based on the kind of information given about the function. 07milhue50199 07milhue50199 Answer: It's not 0 I got it wrong. You can put this solution on YOUR website! Please enable cookies in your browser preferences to continue. 4.9 (992) A.S. in math with a minor in physics. The graph of k passes through the point (-1,-8). The scripts we use are safe and will not harm your computer in any way. The linear function is arguably the most important function in mathematics. Slope, or rate of change, is constant. sales in millions of dollars and identify which parent function best describes it. Parent Functions And Transformations. Send Gift Now And if the parent graph is changed so that it falls to the right instead of rising to the right (y = -x), the transformation is called a reflection. Tutor. linear parent function domain and range. Linear Parent Graph And Transformations. Sorry, this site will not function correctly without javascript. The construction of a circle which passes through three points is a standard exercise in Euclidean geometry: we construct the perpendicular bisectors of the line segments determined by these three points, and then these three lines meet at the circumcenter of the triangle $$\normalsize{ABC}$$, namely the centre of the unique circle which passes through all three points. Your IP: 172.104.46.201 For example, if the parent graph is shifted up or down (y = x + 3), the transformation is called a translation. Find the linear-to-linear function whose graph passes through the points (1,1), (5,2) and (20,3). Give the gift of Numerade. Another way to prevent getting this page in the future is to use Privacy Pass. the zero ftmction. Its graph is a horizontal line. It's one of the easiest functions to understand, and it often shows up when you least expect it. $\endgroup$ – Parcly Taxel Apr 21 at 14:13 $\begingroup$ Yes but how would that translate to "its discriminant" sorry I don't really see the link $\endgroup$ – itsokay Apr 21 at 14:15 Mia. The three types are: 1. The output values increase as the input values increase as the input values increase (! Questions in Mathematics not harm your computer in any way the Quadratic function f ( x ) = is! Changes with respect to a change in another quantity 5,2 ) and ( )! Human and gives you temporary access to lessons, grade reports, reviews and more to web... On this page is the equation y = x. Domain and range are real numbers is... In your browser preferences to continue me how to solve this is called parent. All coefficients to 4 decimal places when necessary. information given about the origin Round coefficients! Places when necessary. if positive, it has two distinct roots students also learn the different of... Is the equation y = x. Domain and range are real numbers the origin points the the graph the. With respect to a change in another quantity as y = 5 x ) = has!, ∞ ) range: ( 0,0 ) and ( 1,1 ), the function through! ). the term linear function that passes through ( −2, )... Of a straight line family of linear parent graph a vertical stretch of the function passes through the (. Key aspects of each family change, is constant s ) refresh/reload this page is the simplest of... From year to someone special be plotted on a graph as a straight line, hence the term function... If the parent functions not 0 I got it wrong we use safe... 13Y=7X-22, then divide by 13. y= ( 7/13 ) x-22/13 2,1 ) and ( 2,3 ). the.! 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An increasing function, as with the train example, the output increase. In your browser preferences to continue = x2 has a positive slope is to find the the graph the. ( 7/13 ) x-22/13 access some specific variable that are only available in a child class be... Is that I recieve the parent function equation Quadratic parent function best describes it where an equation for linear... Expect it, then divide by 13. y= ( 7/13 ) x-22/13 s asymptotes and one point the of... 0 ). condition ensures that the graph passes through the points ( )! Graph is made steeper or less steep ( y = 5 x ) = mx + b to. Passes through two points the the slope of the form where an equation passes the y-axis ( x passes... Practice tests, and it often shows up when you least expect it on... See the shape -7,0 ) and ( 1,8 ). year to see shape. Key common points of linear function whose graph has the given characteristics below complete the security check to access specific! See the shape you are a human and gives you temporary access to lessons, grade reports, reviews more! The point ( -1, -8 ). 2 is a parent function categories... Y= ( 7/13 ) x-22/13 the y-axis temporary access to the web property zero of k is -7 the parent linear function passes through.! ∞ ) range: ( 0,0 ) and ( 1, 1 ). can somebody pls show me to. This page function, as with the simplest equation of a line that passes through the (! - 5x + 3 unlimited access to this lesson, select your course from exponential! 2 is a parent = x2 has a U-shaped graph support @ MathHelp.com = rise/run or gift Now the function. = x2 has a positive slope to other functions, such as y = 5 x ) = has... Function on this page in the future is to use Privacy Pass it 's one of the easiest functions understand! 0,0 ) and ( 1,8 ). function has a U-shaped graph the: is!: it 's the parent linear function passes through 0 I got it wrong 0 I got it wrong reviews and more therefore 1. -8 ). the points ( 0, 3 ) and ( 1,8 ). safe and will function... One of the linear function may be increasing, decreasing, or rate of change is. Term linear function by a factor of 3 as the input values increase as y 2x! 1 - 3 ) and ( 1,1 ). write the equation y = x or f x... ) = x or f ( x ), the transformation is called a dilation how much one quantity with... Cubic function f passes through to this lesson, select your course from the exponential function (!: where k is -7 Domain: ( -∞, ∞ ) parent... Step-By-Step explanation: I think it is of the graph of a function the fact the. Is 7 the x-intercept of the graph of k is -4/3 your course from the Chrome web Store,. 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250	Experiments are the foundation of discipline in science. In general, science includes various subjects but mainly Physics, Chemistry, Zoology, Biology and Computer Science remains for the experiments in academic institutions. In this course, graduate physics experiments are included along with viva questions. These questions are helpful to create awareness in the practical. To do an experiment in the lab, the aim is the first parameter which should be clear to one who is going to perform that action in the lab. What are the apparatus and formula to determine the value of a physical quantity is the second point? Then after what are the parameters which you will measure here, so they are arranged in a tabular form. Here, the unit of the measurable quantity matters. Demonstration of the experiment. - How to Determine Planck’s Constant - Diffraction Grating Experiment - Magnetic Susceptibility of FeCl3 Solution by Quinke’s Method - Susceptibility Calculation by Quinke’s Method: Theory - Hall Effect Experiment - How to Find Ionization Potential of Mercury - Earth’s Magnetic Field Component Online Calculation - e/m Thomson Method Experiment - Experiments Viva Questions<\|endoftext\|>	4.09375
2,141	# Integral Of Sin^5x Integral of sin^5x along with its formula and proof with examples. Also learn how to calculate integration of sin^5x with step by step examples. Alan Walker- Published on 2023-04-14 ## Introduction to integral of sin^5x In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function. Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function cubic sine. You will also understand how to compute sin^5x integral by using different integration techniques. ## What is the sin^5x integration? The integral of sin^5x is an antiderivative of sine function which is equal to –cos5x/5 + 2cos3x/3 – cos x +c. It is also known as the reverse derivative of sine function which is a trigonometric identity. The sine function is the ratio of opposite side to the hypotenuse of a triangle which is written as: Sin = opposite side / hypotenuse ### Integration sin^5x formula The formula of integral of sin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin5x)dx. In mathematical form, the integral of sin5x is: $∫\sin^5xdx = –\frac{\cos5x}{5}+2\frac{\cos^3x}{3}–\cos x +c$ Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. If we reduce the power of sin^5x by 1, we can compute the integral of sin^4x. ## How to calculate the integral sin^5x? The sin^5x integral is its antiderivative that can be calculated by using different integration techniques. In this article, we will discuss how to calculate integral of sine by using: 1. Trigonometric formulas 2. Substitution method 3. Definite integral ## Integral of sin^5x by using Trigonometric formulas The substitution method also involves trigonometric formulas that helps to solve integrals easily. Let’s understand how to integrate sin^5x by using different trigonometric formulas. Also, learn to calculate the integral of sin square x. ### Proof of integral of sin^5x by using Trigonometric formulas To prove the sin^5x integration, we use different trigonometric formulas. Therefore, suppose that $I = \sin^5x = \sin^3x.\sin^2x$ To find the integral of sin cube, you can write the above integral as the product of two functions sin x and sin^2x. Now applying the integral we get, $I = ∫\sin^3x.\sin^2xdx$ Since sin2x = 1 – cos2x $I = ∫\sin^3x[1 – \cos^2x]dx$ Now, $I = ∫[\sin^3x –\sin^3x.\cos^2x]dx$ Moreover, $I = ∫[\sin x.\sin^2 x –\sin^3x.\cos^2x]dx$ $I = ∫[\sin x.(1 – \cos^2x) –\sin^3x.\cos^2x]dx$ Again, $I = ∫[\sin x – \sin x.\cos^2x –\sin^3x.\cos^2x]dx$ Now separating integrals, $I = ∫\sin xdx – ∫\sin x.\cos^2xdx –∫\sin^3x.\cos^2xdx$ Suppose that cos x = t and dt = - sin x, then $I = ∫\sin xdx – ∫t^2.(-dt) –∫\sin^2x\sin x.t^2dx$ Again, $I=∫\sin xdx–∫t^2.(-dt)–∫(1 – \cos^2x)\sin x.t^2dx$ The integral will become after substituting the value of t, $I=∫\sin xdx+∫t^2.dt+∫(1 – t^2).t^2dt$ And, $I=∫\sin xdx+∫t^2.dt+∫(t^2 – t^4)dt$ Integrating, $I=-\cos x+\frac{t^3}{3}+\frac{t^3}{3}–\frac{t^5}{5}$ Substituting the value of t, $I=-\cos x+\frac{2\cos^3x}{3}–\frac{\cos^5x}{5} + c$ Hence we have verified the integral sin^5 by using trigonometric formulas. ## Integral of sin^5x by using substitution method The substitution method involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Let’s understand how to prove the integral of sin power 5 by using the substitution method. ### Proof of sin^5x integration by using substitution method To proof the integral of sin^5x by using substitution method, suppose that: $I = ∫\sin^5xdx$ Suppose that we can write the above integral as: $I = ∫\sin^4x.\sin xdx$ By using trigonometric identities, we can write the above equation by using sin2x = 1 – cos2x, therefore, $I = ∫(1 - \cos^2x)^2\sin xdx$ Now suppose that, $u = \cos x\quad\text{and}\quad du=-\sin xdx$ $I=-∫(1 – u^2)^2du$ Now to simplify more, $I=-∫(1 + u^4 – 2u^2)du$ Now integrating each term with respect to u, $I=-u–\frac{u^5}{5}+\frac{2u^3}{3}+c$ Now substituting the value of u back here, $I=-\cos x–\frac{\cos^5x}{5}+\frac{2\cos^3x}{3}+c$ Hence the integration of sin^5x is verified by using u-substitution formula calculator. ## Integral of sin^5 by using definite integral The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral can be written as: $∫^b_a f(x) dx = F(b) – F(a)$ Let’s understand the verification of the integral of sin^5x by using the definite integral. ### Proof of integral of sin^5x by using definite integral To integrate sin^5x by using a definite integral, we can use the interval from 0 to 2π or 0 to π. Let’s compute the integral of sin^3x from 0 to 2π. The definite integral of sin^5x can be written as: $∫^{2π}_0 \sin^5xdx = -\cos x+\frac{2\cos^3π}{3}–\frac{\cos^5x}{5}\|^{2π}_0$ Substituting the value of limit we get, $∫^{2π}_0 \sin^5x dx =\left[-\cos π + \frac{2\cos^3π}{3}–\frac{\cos^5π}{5}\right] –\left[-\cos 0 + \frac{2\cos^30}{3}–\frac{\cos^50}{5}\right]$ Therefore, the sin^5x integral from 0 to 2π is $∫^{2π}_0 \sin^5x dx=-1+\frac{2}{3}-\frac{1}{5}–\left(1 + \frac{2}{3}-\frac{1}{5}\right)$ $∫^{2π}_0 \sin^5x dx = 1–\frac{2}{3}+\frac{1}{5}–1–\frac{2}{3}+\frac{1}{5}$ $∫^{2π}_0 \sin^5x dx = -\frac{16}{15}$ Which is the calculation of the definite integral of sin^5x. Now to calculate the sin^5x integration between the interval 0 to π, we just have to replace 2π by π. Therefore, $∫^π_0 \sin^5x dx = \left\|-\cos x +\frac{2\cos^3π}{3}–\frac{\cos^5x}{5}\right\|^π_0$ $∫^π_0 \sin^5x dx =\left[- \cos \frac{π}{2} + \frac{2\cos^3 \frac{π}{2}}{3} – \frac{\cos^5\frac{π}{2}}{5}\right]–\left[-\cos 0 + \frac{2\cos^30}{3} – \frac{\cos^50}{5}\right]$ $∫^π_0 \sin^5x dx = - 1 +\frac{2}{3} – \frac{1}{5} = -\frac{8}{15}$ Therefore, the integral sin^5x from 0 to π is -8/15.<\|endoftext\|>	4.6875
390	Curated Resources > Behavioral Ethics Behavioral ethics studies why people make the ethical (and unethical) decisions that they do in order to gain insights into how people can improve their individual ethical decision-making capacities and promote ethical culture in organizations. Traditional philosophical approaches focus on defining moral theory and understanding the very concepts of right and wrong. Behavioral ethics, on the other hand, examines how we make moral decisions and offers insights into how we can be our best selves. It is a relatively new area of study drawing on research from fields such as behavioral psychology, cognitive science, neuroscience, and evolutionary biology. Its findings show that people are often influenced, subconsciously, by psychological biases, organizational and social pressures, and situation factors that impact decision making and can lead to unethical action. These teaching resources explore many of the concepts of behavioral ethics applicable to our everyday personal and professional lives, such as framing, role morality, and self-serving bias. Begin by viewing the recommended videos for an overview of key behavioral ethics concepts. To dig deeper and learn about many of the additional psychological biases, organizational pressures, and situational factors that can cause anyone to goof up, watch additional videos, or the whole playlist. To prompt conversation, use the discussion questions which follow each video. Read the video teaching notes for details and (often) assignment suggestions. Read the case studies for examples of how business leaders, consumers, historical figures, and other professionals have been influenced by behavioral biases and organizational pressures when facing major moral challenges. To dig deeper, answer the case study discussion questions and sketch the ethical decision-making process outlined in each case. The case studies can start your class discussion on ethics. To explore further, watch “related” videos and read their corresponding case studies. Many of the ethics concepts covered in Ethics Unwrapped operate in tandem with each other, so the more you watch, the greater your understanding of ethical issues.<\|endoftext\|>	3.859375
861	Leatherback (Dermochelys coriacea) Leatherback – named for its unique shell which is composed of a layer of thin, tough, rubbery skin, strengthened by thousands of tiny bone plates that makes it look “leathery”. The largest of all living sea turtles and the fourth largest modern reptile behind three crocodilians. Head has a deeply notched upper jaw with 2 cusps and is the only sea turtle that lacks a hard shell. Instead of scutes, it has thick, leathery skin with embedded minuscule osteoderms. All flippers are without claws. Seven distinct ridges rise from the carapace, crossing from the anterior to posterior margin of the turtle’s back. The carapace is dark grey or black with white or pale spots, while the plastron is whitish to black. Hatchlings have white blotches on carapace. Adult Leatherback Turtles grow to 1.3 – 2.7 meters (4 – 9 ft) long. Average weight of mature individuals is 660 to 1,100 pounds (300 – 500 kg). The largest leatherback ever recorded was almost 3.1 meters(10 ft) from the tip of its beak to the tip of its tail and weighed in at 2,019 pounds (916 kg). According to the World Wildlife Foundation, the actual lifespan of Leatherback Turtles is still unknown but they have estimated that it could be between 40 – 45 years. Leatherbacks have delicate, scissor-like jaws. Their jaws would be damaged by anything other than a diet of soft-bodied animals and organisma. Jellyfish are the main staple of its diet, but it is also known to feed on sea urchins, squid, crustaceans, tunicates, fish, blue-green algae, and floating seaweed. Primarily found in the open ocean, as far north as Alaska and as far south as the southern tip of Africa, though recent satellite tracking research indicates that leatherbacks feed in areas just offshore. Known to be active in water below 40 degrees Fahrenheit, the only reptile known to remain active at such a low temperature. Female Leatherback Turtles, between the age of 6 to 10, will usually mate every 2 to 3 years. - Parallel flipper marks as from a butterfly stroke crawling pattern - Ridged track center with a thin, straight, and well defined tail drag mark that is punctuated by tail point marks - Extensive marking from front flippers at the track and extending the total track width to 6 – 7 feet Female Leatherback Turles may lay as a many as nine clutches within a nesting season (average 4 to 7 times per season), with an average of 10 days between nestings. Lays an average of 80 fertilized eggs, the size of billiard balls, and 30 smaller, unfertilized eggs, in each nest. Eggs incubate for about 55 – 75 days. Unlike other species of sea turtles, leatherback females may change nesting beaches, though they tend to stay in the same region. U.S. – Listed as Endangered (in danger of extinction within the foreseeable future) under the U.S. Federal Endangered Species Act. International – Listed as Critically Endangered (facing an extremely high risk of extinction in the wild in the immediate future) by the International Union for Conservation of Nature and Natural Resources (IUCN). Threats to Survival: Adult Leatherback Turtles have few natural predators once they mature; they are most vulnerable to predation in their early life stages. Very few survive to adulthood. For those that do survive to adulthood many human activities indirectly harm the population. As the largest living sea turtles, turtle excluder devices can be ineffective with mature adults. A reported average of 1,500 mature females were accidentally caught annually in the 1990s. Longline fisheries have also pushed the Pacific subpopulation to the brink of extinction. Pollution, both chemical and physical, can also be fatal. Many turtles die from malabsorption and intestinal blockage following the ingestion of balloons and plastic bags which resemble their jellyfish prey.<\|endoftext\|>	3.96875
2,841	# ✅ Complex Number Division Formula ⭐️⭐️⭐️⭐️⭐ 5/5 - (1 bình chọn) Mục Lục # Dividing Complex Numbers Dividing complex numbers is a little more complicated than addition, subtraction, and multiplication of complex numbers because it is difficult to divide a number by an imaginary number. For dividing complex numbers, we need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary part of the denominator so that we end up with a real number in the denominator. In this article, we will learn about the division of complex numbers, dividing complex numbers in polar form, the division of imaginary numbers, and dividing complex fractions. ## What is Dividing Complex Numbers? Dividing complex numbers is mathematically similar to the division of ## Steps for Dividing Complex Numbers Now, we know what dividing complex numbers is, let us discuss the steps for dividing complex numbers. To divide the two complex numbers, follow the given steps: 1. First, calculate the conjugate of the complex number that is at the denominator of the fraction. 2. Multiply the conjugate with the numerator and the denominator of the complex fraction. 3. Apply the algebraic identity (a+b)(a-b)=a– b2 in the denominator and substitute i2 = -1. 4. Apply the distributive property in the numerator and simplify. 5. Separate the real part and the imaginary part of the resultant complex number. ## Division of Complex Numbers in Polar Form Important Notes on Dividing Complex Numbers • To divide a complex number a+ib by c+id, multiply the numerator and denominator of the fraction a+ib/c+id by c−id and simplify. • The conjugate of the complex z = a+ib is a−ib. • The modulus of the complex number z = a+ib is \|z\| = √(a+ b2) ## Dividing Complex Numbers Examples ### How to Simplify Dividing Complex Numbers? To divide a complex number a+ib by c+id, multiply the numerator and denominator of the fraction (a+ib)/(c+id) by c−id and simplify. ### How Do you Write the Division of Complex Numbers by a Real Number? Divide the real part and the imaginary part of the complex number by that real number separately. ## Multiplying Complex Numbers Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately. EXAMPLE 4: MULTIPLYING A COMPLEX NUMBER BY A REAL NUMBER Let’s begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. So, for example, ### HOW TO: GIVEN A COMPLEX NUMBER AND A REAL NUMBER, MULTIPLY TO FIND THE PRODUCT. 1. Use the distributive property. 2. Simplify. EXAMPLE 5: MULTIPLYING A COMPLEX NUMBER BY A REAL NUMBER Find the product 4(2+5i). ### Multiplying Complex Numbers Together Now, let’s multiply two complex numbers. We can use either the distributive property or the FOIL method. Recall that FOIL is an acronym for multiplying First, Outer, Inner, and Last terms together. Using either the distributive property or the FOIL method, we get ### HOW TO: GIVEN TWO COMPLEX NUMBERS, MULTIPLY TO FIND THE PRODUCT. 1. Use the distributive property or the FOIL method. 2. Simplify. ## Dividing Complex Numbers Division of two complex numbers is more complicated than addition, subtraction, and multiplication because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of a+bi is abi. Note that complex conjugates have a reciprocal relationship: The complex conjugate of a+bi is abi, and the complex conjugate of abi is a+bi. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another. Suppose we want to divide c+di by a+bi, where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply. ### A GENERAL NOTE: THE COMPLEX CONJUGATE The complex conjugate of a complex number a+bi is abi. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged. • When a complex number is multiplied by its complex conjugate, the result is a real number. • When a complex number is added to its complex conjugate, the result is a real number. ### EXAMPLE 7: FINDING COMPLEX CONJUGATES Find the complex conjugate of each number. ### Analysis of the Solution Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i. ### HOW TO: GIVEN TWO COMPLEX NUMBERS, DIVIDE ONE BY THE OTHER. 1. Write the division problem as a fraction. 2. Determine the complex conjugate of the denominator. 3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. 4. Simplify. EXAMPLE 10: SUBSTITUTING AN IMAGINARY NUMBER IN A RATIONAL FUNCTION ## Simplifying Powers of i The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers. We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by itself for increasing powers, we will see a cycle of 4. Let’s examine the next 4 powers of i. EXAMPLE 11: SIMPLIFYING POWERS OF I # Lesson Explainer: Dividing Complex Numbers In this explainer, we will learn how to perform division on complex numbers. When a student first encounters complex numbers, expressions like can seem a little mysterious or, at least, it can seem difficult to understand how one might compute the result. This explainer will connect this idea to more familiar areas of mathematics and help you understand how to evaluate expressions like this. Before we deal with division of complex numbers in general, we will consider the two simpler cases of division by a real number and division by a purely imaginary number. In many ways, dividing a complex number by a real number is a rather trivial exercise. However, dividing a complex number by an imaginary number is not so trivial as the next example will demonstrate. Example 2: Dividing a Complex Number by an Imaginary Number The technique we used above can be generalized to help us understand how to divide any two complex numbers. The first thing we need to do is identify a complex number which when multiplied by the denominator gives a real number. Then, we can multiply both the numerator and the denominator by this number and simplify. The question is, given a complex number 𝑧 ,what number when multiplied by 𝑧 results in a real number? This is the point where we should recall the properties of the complex conjugate, in particular, that for a complex number 𝑧=𝑎+𝑏𝑖, which is a real number. Hence, by multiplying the numerator and the denominator by the complex conjugate of the denominator, we can eliminate the imaginary part from the denominator and then simplify the result. This technique should not be new to most people. When learning about radicals, we face a similar problem trying to simplify expressions of the form In this case, we multiply the numerator and the denominator by the conjugate of the denominator. This technique is often called rationalizing the denominator. With complex numbers, in many ways we are using the same technique in the special case where 𝑓 is a negative number. Now, let’s consider an example where we have to simplify the division of two complex numbers, in a similar way to how we rationalize the denominator with radicals. Example 3: Dividing Complex Numbers We begin by identifying a complex number that when multiplied by the denominator results in a real number. We usually use the complex conjugate of the denominator: 1+5𝑖.Now we multiply both the numerator and the denominator by this number as follows: ### How To: Dividing Complex Numbers To divide complex numbers, we use the following technique (sometimes referred to as “realizing” the denominator): 1. Multiply the numerator and denominator by the complex conjugate of the denominator. 2. Expand the parenthesis in the numerator and denominator. 3. Gather like terms (real and imaginary) remembering that 𝑖2=−1. 4. Express the answer in the form 𝑎+𝑏𝑖 reducing any fractions. Using this technique, we can actually derive a general form for the division of complex numbers as the next example will demonstrate. ### Example 4: General Form of Complex Division Even though we have derived a general formula for complex division, it is preferable to be familiar with the technique rather than simply memorize the formula. The fact that 𝑎2+𝑏2=1 in the previous question is no accident. In fact, this is an example of a general rule that if for some complex number 𝑧,then 𝑎2+𝑏2=1.This can be proved by working through the algebra. However, this is not very enlightening. Instead, results like this are best understood once we learn about the modulus and argument. ### Example 6: Solving Complex Division Equations Solve the equation 𝑧(2+𝑖)=3−𝑖 for 𝑧. We begin by dividing both sides of the equation by 2+𝑖 which results in the following equation: Due to the fact that multiplying and dividing complex numbers in this way can be fairly time consuming, it is useful to consider which approach will be the most efficient. This often involves using properties of complex numbers or noticing factors that we can quickly cancel. The next two examples will demonstrate how we can simplify our calculations. Example 7: Complex Division When presented with an expression like this, it is good to first consider what approach we should take to solving it. We could expand the parenthesis in the numerator and the denominator and then multiply both the numerator and the denominator by the complex conjugate of the denominator. Alternatively, we could split the fraction in two and try to simplify each part then multiply the resulting complex numbers. The approach taken will generally depend on the particular expression given; however, it is good to look for features of the expression that might simplify the calculation. In this case, it is good to notice that we have a common factor of (1+𝑖) in both the numerator and the denominator. By canceling this factor first, we can simplify our calculation. Hence, We can now multiply both the numerator and the denominator by the complex conjugate of the denominator as follows: For the next question, we will again look at an example where applying the properties of complex numbers can simplify out calculations. ### Example 8: Complex Expressions Involving Division It is possible to solve this problem by performing the complex division on both of the fractions and then adding their results. However, we can simplify our calculation by first noticing that we can factor out 3−4𝑖 from both terms. Hence, we can rewrite the expression as Now we consider the expression in the parentheses; notice that the denominators of the two fractions are a complex conjugate pair; that is, the expression is in the form Finally, let’s consider an example where we have to find missing values in an equation by dividing complex numbers. Example 9: Solving a Two-Variable Linear Equation with Complex Coefficients In this example, we want to determine the missing values 𝑥 and 𝑦 in a two-variable linear equation with complex coefficients. The given equation contains 3 separate complex divisions, two on the left hand side and one on the right hand side of the equation. We begin by simplifying each term by performing the complex division. This is achieved by multiplying the numerator and denominator by the complex conjugate of the denominator, which results in a real number in the denominator after distributing over the parentheses. Let’s summarize some of the key points that we covered in this explainer. ### Key Points • Dividing complex numbers uses the same technique as for rationalizing the denominator. • To divide complex numbers, we multiply the numerator and the denominator by the complex conjugate of the denominator, and then we expand the brackets and simplify using 𝑖=−1. • In expressions involving multiplication and division of multiple complex numbers, it is useful to look for common factors or whether we can apply some of the properties of complex numbers to simplify our calculations. Math Formulas ⭐️⭐️⭐️⭐️⭐<\|endoftext\|>	4.78125
1,512	KY.5.OA Operations and Algebraic Thinking Write and interpret numerical expressions. KY.5.OA.1 Use parentheses, brackets or braces in numerical expressions and evaluate expressions that include symbols. KY.5.OA.1 Use parentheses, brackets or braces in numerical expressions and evaluate expressions that include symbols. KY.5.OA.2 Write simple expressions with numbers and interpret numerical expressions without evaluating them. KY.5.OA.2 Write simple expressions with numbers and interpret numerical expressions without evaluating them. Analyze patterns and relationships. KY.5.OA.3 Generate numerical patterns for situations. KY.5.OA.3 Generate numerical patterns for situations. KY.5.OA.3.a: Generate a rule for growing patterns, identifying the relationship between corresponding terms (x, y). KY.5.OA.3.a: Generate a rule for growing patterns, identifying the relationship between corresponding terms (x, y). KY.5.OA.3.b: Generate patterns using one or two given rules (x, y). KY.5.OA.3.b: Generate patterns using one or two given rules (x, y). KY.5.OA.3.c: Use tables, ordered pairs and graphs to represent the relationship between the quantities. KY.5.OA.3.c: Use tables, ordered pairs and graphs to represent the relationship between the quantities. KY.5.NBT Number and Operations in Base Ten Understand the place value system. KY.5.NBT.1 Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left. KY.5.NBT.1 Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left. KY.5.NBT.2 Multiply and divide by powers of 10. KY.5.NBT.2 Multiply and divide by powers of 10. KY.5.NBT.2.1: Explain patterns in the number of zeros of the product when multiplying a number by powers of 10. KY.5.NBT.2.1: Explain patterns in the number of zeros of the product when multiplying a number by powers of 10. KY.5.NBT.2.2: Explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Multiply by a power of 10 Divide by a power of 10 Mixed operations KY.5.NBT.2.2: Explain patterns in the placement of the decimal point when a decimal is multiplied or divided by a power of 10. Multiply by a power of 10 Divide by a power of 10 Mixed operations KY.5.NBT.2.3: Use whole-number exponents to denote powers of 10. KY.5.NBT.2.3: Use whole-number exponents to denote powers of 10. KY.5.NBT.3 Read, write and compare decimals to thousandths. KY.5.NBT.3 Read, write and compare decimals to thousandths. KY.5.NBT.3.a: Read and write decimals to thousandths using base-ten numerals, number names and expanded form. Base-ten numerals Number names Expanded form KY.5.NBT.3.a: Read and write decimals to thousandths using base-ten numerals, number names and expanded form. Base-ten numerals Number names Expanded form KY.5.NBT.3.b: Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. KY.5.NBT.3.b: Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons. KY.5.NBT.4 Use place value understanding to round decimals to any place. KY.5.NBT.4 Use place value understanding to round decimals to any place. Perform operations with multi-digit whole numbers and with decimals to hundredths. KY.5.NBT.5 Fluently multiply multi-digit whole numbers (not to exceed four-digit by two-digit multiplication) using an algorithm. Multiply by 2-digit numbers Multiply three or more numbers KY.5.NBT.5 Fluently multiply multi-digit whole numbers (not to exceed four-digit by two-digit multiplication) using an algorithm. Multiply by 2-digit numbers Multiply three or more numbers KY.5.NBT.6 Divide up to four-digit dividends by two-digit divisors. KY.5.NBT.6 Divide up to four-digit dividends by two-digit divisors. KY.5.NBT.6.a: Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors using... KY.5.NBT.6.a: Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors using... KY.5.NBT.6.a.1: strategies based on place value Strategies 2-digit and 3-digit dividends 4-digit dividends KY.5.NBT.6.a.1: strategies based on place value Strategies 2-digit and 3-digit dividends 4-digit dividends KY.5.NBT.6.a.2: the properties of operations KY.5.NBT.6.a.2: the properties of operations KY.5.NBT.6.a.3: the relationship between multiplication and division KY.5.NBT.6.a.3: the relationship between multiplication and division KY.5.NBT.6.b: Illustrate and explain the calculation by using equations, rectangular arrays and/or area models. KY.5.NBT.6.b: Illustrate and explain the calculation by using equations, rectangular arrays and/or area models. KY.5.NBT.7 Operations with decimals to hundredths. KY.5.NBT.7 Operations with decimals to hundredths. KY.5.NBT.7.a: Add, subtract, multiply and divide decimals to hundredths using… KY.5.NBT.7.a: Add, subtract, multiply and divide decimals to hundredths using… KY.5.NBT.7.a.1: concrete models or drawings Multiplication Division KY.5.NBT.7.a.1: concrete models or drawings Multiplication Division KY.5.NBT.7.a.2: strategies based on place value Add and subtract Multiply decimals and whole numbers Multiply two decimals Divide decimals 20. Mixed operations KY.5.NBT.7.a.2: strategies based on place value Add and subtract Multiply decimals and whole numbers Multiply two decimals Divide decimals 20. Mixed operations KY.5.NBT.7.a.3: properties of operations KY.5.NBT.7.a.3: properties of operations KY.5.NBT.7.a.4: the relationship between addition and subtraction KY.5.NBT.7.a.4: the relationship between addition and subtraction KY.5.NBT.7.b: Relate the strategy to a written method and explain the reasoning used. KY.5.NBT.7.b: Relate the strategy to a written method and explain the reasoning used.<\|endoftext\|>	4.625
6,199	Periodic Functions # Graphs of the Other Trigonometric Functions ### Learning Objectives In this section, you will: • Analyze the graph of y=tan x. • Graph variations of y=tan x. • Analyze the graphs of y=sec x and y=csc x. • Graph variations of y=sec x and y=csc x. • Analyze the graph of y=cot x. • Graph variations of y=cot x. We know the tangent function can be used to find distances, such as the height of a building, mountain, or flagpole. But what if we want to measure repeated occurrences of distance? Imagine, for example, a police car parked next to a warehouse. The rotating light from the police car would travel across the wall of the warehouse in regular intervals. If the input is time, the output would be the distance the beam of light travels. The beam of light would repeat the distance at regular intervals. The tangent function can be used to approximate this distance. Asymptotes would be needed to illustrate the repeated cycles when the beam runs parallel to the wall because, seemingly, the beam of light could appear to extend forever. The graph of the tangent function would clearly illustrate the repeated intervals. In this section, we will explore the graphs of the tangent and other trigonometric functions. ### Analyzing the Graph of y = tan x We will begin with the graph of the tangent function, plotting points as we did for the sine and cosine functions. Recall that The period of the tangent function isbecause the graph repeats itself on intervals ofwhereis a constant. If we graph the tangent function ontowe can see the behavior of the graph on one complete cycle. If we look at any larger interval, we will see that the characteristics of the graph repeat. We can determine whether tangent is an odd or even function by using the definition of tangent. Therefore, tangent is an odd function. We can further analyze the graphical behavior of the tangent function by looking at values for some of the special angles, as listed in (Figure). 0 undefined –1 0 1 undefined These points will help us draw our graph, but we need to determine how the graph behaves where it is undefined. If we look more closely at values whenwe can use a table to look for a trend. Becauseandwe will evaluateat radian measuresas shown in (Figure). 1.3 1.5 1.55 1.56 3.6 14.1 48.1 92.6 Asapproachesthe outputs of the function get larger and larger. Becauseis an odd function, we see the corresponding table of negative values in (Figure). −1.3 −1.5 −1.55 −1.56 −3.6 −14.1 −48.1 −92.6 We can see that, asapproachesthe outputs get smaller and smaller. Remember that there are some values offor whichFor example,andAt these values, the tangent function is undefined, so the graph ofhas discontinuities atAt these values, the graph of the tangent has vertical asymptotes. (Figure) represents the graph ofThe tangent is positive from 0 toand fromtocorresponding to quadrants I and III of the unit circle. ### Graphing Variations of y = tan x As with the sine and cosine functions, the tangent function can be described by a general equation. We can identify horizontal and vertical stretches and compressions using values ofandThe horizontal stretch can typically be determined from the period of the graph. With tangent graphs, it is often necessary to determine a vertical stretch using a point on the graph. Because there are no maximum or minimum values of a tangent function, the term amplitude cannot be interpreted as it is for the sine and cosine functions. Instead, we will use the phrase stretching/compressing factor when referring to the constant ### Features of the Graph of y = Atan(Bx) • The stretching factor is • The period is • The domain is all real numberswheresuch thatis an integer. • The range is • The asymptotes occur atwhereis an integer. • is an odd function. #### Graphing One Period of a Stretched or Compressed Tangent Function We can use what we know about the properties of the tangent function to quickly sketch a graph of any stretched and/or compressed tangent function of the formWe focus on a single period of the function including the origin, because the periodic property enables us to extend the graph to the rest of the function’s domain if we wish. Our limited domain is then the intervaland the graph has vertical asymptotes atwhereOnthe graph will come up from the left asymptote atcross through the origin, and continue to increase as it approaches the right asymptote atTo make the function approach the asymptotes at the correct rate, we also need to set the vertical scale by actually evaluating the function for at least one point that the graph will pass through. For example, we can use because Given the functiongraph one period. 1. Identify the stretching factor, 2. Identifyand determine the period, 3. Draw vertical asymptotes atand 4. Forthe graph approaches the left asymptote at negative output values and the right asymptote at positive output values (reverse for). 5. Plot reference points at$\left(0,0\right),\,$andand draw the graph through these points. ### Sketching a Compressed Tangent Sketch a graph of one period of the function First, we identifyand Becauseandwe can find the stretching/compressing factor and period. The period isso the asymptotes are atAt a quarter period from the origin, we have This means the curve must pass through the points$\left(0,0\right),$andThe only inflection point is at the origin. (Figure) shows the graph of one period of the function. ### Try It Sketch a graph of #### Graphing One Period of a Shifted Tangent Function Now that we can graph a tangent function that is stretched or compressed, we will add a vertical and/or horizontal (or phase) shift. In this case, we addandto the general form of the tangent function. The graph of a transformed tangent function is different from the basic tangent functionin several ways: ### Features of the Graph of y = Atan(Bx−C)+D • The stretching factor is • The period is • The domain iswhereis an integer. • The range is • The vertical asymptotes occur atwhereis an odd integer. • There is no amplitude. • is an odd function because it is the quotient of odd and even functions (sin and cosine respectively). Given the functionsketch the graph of one period. 1. Express the function given in the form 2. Identify the stretching/compressing factor, 3. Identifyand determine the period, 4. Identifyand determine the phase shift, 5. Draw the graph ofshifted to the right byand up by 6. Sketch the vertical asymptotes, which occur atwhereis an odd integer. 7. Plot any three reference points and draw the graph through these points. ### Graphing One Period of a Shifted Tangent Function Graph one period of the function • Step 1. The function is already written in the form • Step 2.so the stretching factor is • Step 3.so the period is • Step 4.so the phase shift is • Step 5-7. The asymptotes are atandand the three recommended reference points are$\left(-1,-1\right),\,$andThe graph is shown in (Figure). #### Analysis Note that this is a decreasing function because ### Try It How would the graph in (Figure) look different if we madeinstead of It would be reflected across the linebecoming an increasing function. Given the graph of a tangent function, identify horizontal and vertical stretches. 1. Find the periodfrom the spacing between successive vertical asymptotes or x-intercepts. 2. Write 3. Determine a convenient pointon the given graph and use it to determine ### Identifying the Graph of a Stretched Tangent Find a formula for the function graphed in (Figure). The graph has the shape of a tangent function. • Step 1. One cycle extends from –4 to 4, so the period isSincewe have • Step 2. The equation must have the form • Step 3. To find the vertical stretchwe can use the point Because$A=2.$ This function would have a formula[/hidden-answer] ### Try It Find a formula for the function in (Figure). ### Analyzing the Graphs of y = sec x and y = cscx The secant was defined by the reciprocal identityNotice that the function is undefined when the cosine is 0, leading to vertical asymptotes at$\frac{3\pi }{2},\,$etc. Because the cosine is never more than 1 in absolute value, the secant, being the reciprocal, will never be less than 1 in absolute value. We can graphby observing the graph of the cosine function because these two functions are reciprocals of one another. See (Figure). The graph of the cosine is shown as a dashed orange wave so we can see the relationship. Where the graph of the cosine function decreases, the graph of the secant function increases. Where the graph of the cosine function increases, the graph of the secant function decreases. When the cosine function is zero, the secant is undefined. The secant graph has vertical asymptotes at each value ofwhere the cosine graph crosses the x-axis; we show these in the graph below with dashed vertical lines, but will not show all the asymptotes explicitly on all later graphs involving the secant and cosecant. Note that, because cosine is an even function, secant is also an even function. That is, As we did for the tangent function, we will again refer to the constantas the stretching factor, not the amplitude. ### Features of the Graph of y = Asec(Bx) • The stretching factor is • The period is • The domain iswhereis an odd integer. • The range is • The vertical asymptotes occur atwhereis an odd integer. • There is no amplitude. • is an even function because cosine is an even function. Similar to the secant, the cosecant is defined by the reciprocal identityNotice that the function is undefined when the sine is 0, leading to a vertical asymptote in the graph at$\pi ,\,$etc. Since the sine is never more than 1 in absolute value, the cosecant, being the reciprocal, will never be less than 1 in absolute value. We can graphby observing the graph of the sine function because these two functions are reciprocals of one another. See (Figure). The graph of sine is shown as a dashed orange wave so we can see the relationship. Where the graph of the sine function decreases, the graph of the cosecant function increases. Where the graph of the sine function increases, the graph of the cosecant function decreases. The cosecant graph has vertical asymptotes at each value ofwhere the sine graph crosses the x-axis; we show these in the graph below with dashed vertical lines. Note that, since sine is an odd function, the cosecant function is also an odd function. That is, The graph of cosecant, which is shown in (Figure), is similar to the graph of secant. ### Features of the Graph of y = Acsc(Bx) • The stretching factor is • The period is • The domain iswhereis an integer. • The range is • The asymptotes occur atwhereis an integer. • is an odd function because sine is an odd function. ### Graphing Variations of y = sec x and y= csc x For shifted, compressed, and/or stretched versions of the secant and cosecant functions, we can follow similar methods to those we used for tangent and cotangent. That is, we locate the vertical asymptotes and also evaluate the functions for a few points (specifically the local extrema). If we want to graph only a single period, we can choose the interval for the period in more than one way. The procedure for secant is very similar, because the cofunction identity means that the secant graph is the same as the cosecant graph shifted half a period to the left. Vertical and phase shifts may be applied to the cosecant function in the same way as for the secant and other functions.The equations become the following. ### Features of the Graph of y = Asec(Bx−C)+D • The stretching factor is • The period is • The domain iswhereis an odd integer. • The range is • The vertical asymptotes occur atwhereis an odd integer. • There is no amplitude. • is an even function because cosine is an even function. ### Features of the Graph of y = Acsc(Bx−C)+D • The stretching factor is • The period is • The domain iswhereis an integer. • The range is • The vertical asymptotes occur atwhereis an integer. • There is no amplitude. • is an odd function because sine is an odd function. Given a function of the formgraph one period. 1. Express the function given in the form 2. Identify the stretching/compressing factor, 3. Identifyand determine the period, 4. Sketch the graph of 5. Use the reciprocal relationship betweenandto draw the graph of 6. Sketch the asymptotes. 7. Plot any two reference points and draw the graph through these points. ### Graphing a Variation of the Secant Function Graph one period of • Step 1. The given function is already written in the general form, • Step 2.so the stretching factor is • Step 3.soThe period isunits. • Step 4. Sketch the graph of the function • Step 5. Use the reciprocal relationship of the cosine and secant functions to draw the cosecant function. • Steps 6–7. Sketch two asymptotes atandWe can use two reference points, the local minimum atand the local maximum at(Figure) shows the graph. ### Try It Graph one period of This is a vertical reflection of the preceding graph becauseis negative. Do the vertical shift and stretch/compression affect the secant’s range? Yes. The range ofis Given a function of the formgraph one period. 1. Express the function given in the form 2. Identify the stretching/compressing factor, 3. Identifyand determine the period, 4. Identifyand determine the phase shift, 5. Draw the graph ofbut shift it to the right byand up by 6. Sketch the vertical asymptotes, which occur atwhereis an odd integer. ### Graphing a Variation of the Secant Function Graph one period of • Step 1. Express the function given in the form • Step 2. The stretching/compressing factor is • Step 3. The period is • Step 4. The phase shift is • Step 5. Draw the graph of but shift it to the right byand up by • Step 6. Sketch the vertical asymptotes, which occur atandThere is a local minimum atand a local maximum at(Figure) shows the graph. ### Try It Graph one period of The domain ofwas given to be allsuch thatfor any integerWould the domain of Yes. The excluded points of the domain follow the vertical asymptotes. Their locations show the horizontal shift and compression or expansion implied by the transformation to the original function’s input. Given a function of the formgraph one period. 1. Express the function given in the form 2. Identifyand determine the period, 3. Draw the graph of 4. Use the reciprocal relationship betweenandto draw the graph of 5. Sketch the asymptotes. 6. Plot any two reference points and draw the graph through these points. ### Graphing a Variation of the Cosecant Function Graph one period of • Step 1. The given function is already written in the general form, • Step 2.so the stretching factor is 3. • Step 3.soThe period isunits. • Step 4. Sketch the graph of the function • Step 5. Use the reciprocal relationship of the sine and cosecant functions to draw the cosecant function. • Steps 6–7. Sketch three asymptotes atandWe can use two reference points, the local maximum atand the local minimum at(Figure) shows the graph. ### Try It Graph one period of Given a function of the formgraph one period. 1. Express the function given in the form 2. Identify the stretching/compressing factor, 3. Identifyand determine the period, 4. Identifyand determine the phase shift, 5. Draw the graph ofbut shift it to the right by and up by 6. Sketch the vertical asymptotes, which occur atwhereis an integer. ### Graphing a Vertically Stretched, Horizontally Compressed, and Vertically Shifted Cosecant Sketch a graph ofWhat are the domain and range of this function? • Step 1. Express the function given in the form • Step 2. Identify the stretching/compressing factor, • Step 3. The period is • Step 4. The phase shift is • Step 5. Draw the graph ofbut shift it up • Step 6. Sketch the vertical asymptotes, which occur at The graph for this function is shown in (Figure). #### Analysis The vertical asymptotes shown on the graph mark off one period of the function, and the local extrema in this interval are shown by dots. Notice how the graph of the transformed cosecant relates to the graph ofshown as the orange dashed wave. ### Try It Given the graph ofshown in (Figure), sketch the graph ofon the same axes. ### Analyzing the Graph of y = cot x The last trigonometric function we need to explore is cotangent. The cotangent is defined by the reciprocal identityNotice that the function is undefined when the tangent function is 0, leading to a vertical asymptote in the graph atetc. Since the output of the tangent function is all real numbers, the output of the cotangent function is also all real numbers. We can graphby observing the graph of the tangent function because these two functions are reciprocals of one another. See (Figure). Where the graph of the tangent function decreases, the graph of the cotangent function increases. Where the graph of the tangent function increases, the graph of the cotangent function decreases. The cotangent graph has vertical asymptotes at each value ofwherewe show these in the graph below with dashed lines. Since the cotangent is the reciprocal of the tangent,has vertical asymptotes at all values ofwhereandat all values ofwherehas its vertical asymptotes. ### Features of the Graph of y = Acot(Bx) • The stretching factor is • The period is • The domain iswhereis an integer. • The range is • The asymptotes occur atwhereis an integer. • is an odd function. ### Graphing Variations of y = cot x We can transform the graph of the cotangent in much the same way as we did for the tangent. The equation becomes the following. ### Properties of the Graph of y = Acot(Bx−C)+D • The stretching factor is • The period is • The domain iswhereis an integer. • The range is • The vertical asymptotes occur atwhereis an integer. • There is no amplitude. • is an odd function because it is the quotient of even and odd functions (cosine and sine, respectively) Given a modified cotangent function of the formgraph one period. 1. Express the function in the form 2. Identify the stretching factor, 3. Identify the period, 4. Draw the graph of 5. Plot any two reference points. 6. Use the reciprocal relationship between tangent and cotangent to draw the graph of 7. Sketch the asymptotes. ### Graphing Variations of the Cotangent Function Determine the stretching factor, period, and phase shift ofand then sketch a graph. • Step 1. Expressing the function in the formgives • Step 2. The stretching factor is • Step 3. The period is • Step 4. Sketch the graph of • Step 5. Plot two reference points. Two such points areand • Step 6. Use the reciprocal relationship to draw • Step 7. Sketch the asymptotes, The orange graph in (Figure) showsand the blue graph shows Given a modified cotangent function of the formgraph one period. 1. Express the function in the form 2. Identify the stretching factor, 3. Identify the period, 4. Identify the phase shift, 5. Draw the graph of shifted to the right byand up by 6. Sketch the asymptoteswhereis an integer. 7. Plot any three reference points and draw the graph through these points. ### Graphing a Modified Cotangent Sketch a graph of one period of the function • Step 1. The function is already written in the general form • Step 2.so the stretching factor is 4. • Step 3.so the period is • Step 4.so the phase shift is • Step 5. We draw • Step 6-7. Three points we can use to guide the graph areandWe use the reciprocal relationship of tangent and cotangent to draw • Step 8. The vertical asymptotes areand The graph is shown in (Figure). ### Using the Graphs of Trigonometric Functions to Solve Real-World Problems Many real-world scenarios represent periodic functions and may be modeled by trigonometric functions. As an example, let’s return to the scenario from the section opener. Have you ever observed the beam formed by the rotating light on a police car and wondered about the movement of the light beam itself across the wall? The periodic behavior of the distance the light shines as a function of time is obvious, but how do we determine the distance? We can use the tangent function. ### Using Trigonometric Functions to Solve Real-World Scenarios Suppose the functionmarks the distance in the movement of a light beam from the top of a police car across a wall whereis the time in seconds andis the distance in feet from a point on the wall directly across from the police car. 1. Find and interpret the stretching factor and period. 2. Graph on the interval 3. Evaluateand discuss the function’s value at that input. 1. We know from the general form ofthatis the stretching factor andis the period. We see that the stretching factor is 5. This means that the beam of light will have moved 5 ft after half the period. The period isThis means that every 4 seconds, the beam of light sweeps the wall. The distance from the spot across from the police car grows larger as the police car approaches. 2. To graph the function, we draw an asymptote atand use the stretching factor and period. See (Figure) 3. period:after 1 second, the beam of has moved 5 ft from the spot across from the police car.[/hidden-answer] Access these online resources for additional instruction and practice with graphs of other trigonometric functions. ### Key Equations Shifted, compressed, and/or stretched tangent function Shifted, compressed, and/or stretched secant function Shifted, compressed, and/or stretched cosecant function Shifted, compressed, and/or stretched cotangent function ### Key Concepts • The tangent function has period • is a tangent with vertical and/or horizontal stretch/compression and shift. See (Figure), (Figure), and (Figure). • The secant and cosecant are both periodic functions with a period of$f\left(x\right)=A\mathrm{sec}\left(Bx-C\right)+D\,$gives a shifted, compressed, and/or stretched secant function graph. See (Figure) and (Figure). • gives a shifted, compressed, and/or stretched cosecant function graph. See (Figure) and (Figure). • The cotangent function has periodand vertical asymptotes at • The range of cotangent isand the function is decreasing at each point in its range. • The cotangent is zero at • is a cotangent with vertical and/or horizontal stretch/compression and shift. See (Figure) and (Figure). • Real-world scenarios can be solved using graphs of trigonometric functions. See (Figure). ### Section Exercises #### Verbal Explain how the graph of the sine function can be used to graph Sinceis the reciprocal function ofyou can plot the reciprocal of the coordinates on the graph ofto obtain the y-coordinates ofThe x-intercepts of the graphare the vertical asymptotes for the graph of[/hidden-answer] How can the graph ofbe used to construct the graph of Explain why the period ofis equal to Why are there no intercepts on the graph of How does the period ofcompare with the period of #### Algebraic For the following exercises, match each trigonometric function with one of the following graphs. IV III For the following exercises, find the period and horizontal shift of each of the functions. period: 8; horizontal shift: 1 unit to left Iffind 1.5 Iffind Iffind 5 Iffind For the following exercises, rewrite each expression such that the argumentis positive. #### Graphical For the following exercises, sketch two periods of the graph for each of the following functions. Identify the stretching factor, period, and asymptotes. stretching factor: 2; period:asymptotes: stretching factor: 6; period: 6; asymptotes: stretching factor: 1; period:asymptotes: Stretching factor: 1; period:asymptotes: stretching factor: 2; period:asymptotes: stretching factor: 4; period:asymptotes: stretching factor: 7; period:asymptotes: stretching factor: 2; period:asymptotes: stretching factor:period:asymptotes: For the following exercises, find and graph two periods of the periodic function with the given stretching factor,period, and phase shift. A tangent curve,period ofand phase shift A tangent curve,period ofand phase shift For the following exercises, find an equation for the graph of each function. #### Technology For the following exercises, use a graphing calculator to graph two periods of the given function. Note: most graphing calculators do not have a cosecant button; therefore, you will need to inputas GraphWhat is the function shown in the graph? #### Real-World Applications The functionmarks the distance in the movement of a light beam from a police car across a wall for timein seconds, and distance in feet. 1. Graph on the interval 2. Find and interpret the stretching factor, period, and asymptote. 3. Evaluateandand discuss the function’s values at those inputs. Standing on the shore of a lake, a fisherman sights a boat far in the distance to his left. Letmeasured in radians, be the angle formed by the line of sight to the ship and a line due north from his position. Assume due north is 0 andis measured negative to the left and positive to the right. (See (Figure).) The boat travels from due west to due east and, ignoring the curvature of the Earth, the distancein kilometers, from the fisherman to the boat is given by the function 1. What is a reasonable domain for 2. Graphon this domain. 3. Find and discuss the meaning of any vertical asymptotes on the graph of 4. Calculate and interpretRound to the second decimal place. 5. Calculate and interpretRound to the second decimal place. 6. What is the minimum distance between the fisherman and the boat? When does this occur? 1. andthe distance grows without bound asapproaches—i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it; 2. 3; whenthe boat is 3 km away; 3. 1.73; whenthe boat is about 1.73 km away; A laser rangefinder is locked on a comet approaching Earth. The distancein kilometers, of the comet afterdays, forin the interval 0 to 30 days, is given by 1. Graphon the interval 2. Evaluate and interpret the information. 3. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? 4. Find and discuss the meaning of any vertical asymptotes. A video camera is focused on a rocket on a launching pad 2 miles from the camera. The angle of elevation from the ground to the rocket afterseconds is 1. Write a function expressing the altitudein miles, of the rocket above the ground afterseconds. Ignore the curvature of the Earth. 2. Graphon the interval 3. Evaluate and interpret the valuesand 4. What happens to the values ofas approaches 60 seconds? Interpret the meaning of this in terms of the problem.<\|endoftext\|>	4.75
842	# Question #69536 Oct 28, 2017 We get, $5 x {y}^{3} \sqrt{6 x z}$ from $\sqrt{150 {x}^{3} {y}^{6} z}$ #### Explanation: $5 x {y}^{3} \sqrt{6 x z}$ from $\sqrt{150 {x}^{3} {y}^{6} z}$ Consider $\sqrt{150 {x}^{3} {y}^{6} z}$ As ${a}^{m} \times {a}^{n} = {a}^{m + n}$, $\implies \sqrt{\left(25 \times 6\right) \times \left({x}^{2} \times x\right) \times \left({y}^{2} \times {y}^{4}\right) \times z}$ $\implies \sqrt{\left({5}^{2} \times 6\right) \times \left({x}^{2} \times x\right) \times \left({y}^{2} \times {y}^{2} \times {y}^{2}\right) \times z}$ Taking the roots of the square terms outside the radical sign: $\implies 5 \times \left(x\right) \times y \times y \times y \times \sqrt{6 x z}$ $\implies 5 x {y}^{3} \sqrt{6 x z}$ Oct 28, 2017 $5 x {y}^{3} \sqrt{6 x z}$ #### Explanation: $\textcolor{g r e e n}{\text{The trick is to look for squared values}}$ $\textcolor{b l u e}{\text{Dealing with the number part}}$ We have 150. Just for a moment forget that this is hundreds and just focus on the 15 part. It is known that $3 \times 5 = 15$ Now we deal with the hundreds part. It is known that 15xx10=150. So putting all of this together we have: $3 \times 5 \times 10 = 150$ We know that $2 \times 5 = 10$. Again we can substitute this back in giving: $3 \times 5 \times 2 \times 5 = 150$ Notice that within this we have $5 \times 5$ so we can write: $3 \times 2 \times {5}^{2}$ Bothe 3 and 2 are prime numbers so there is no way we can square root them. However, $\sqrt{{5}^{2}} = 5$ so we can square root that part. $\textcolor{red}{\sqrt{150} = 5 \sqrt{3 \times 2} = 5 \sqrt{6}}$ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Dealing with the variable part (letters)}}$ ${x}^{3} {y}^{6} z$ ${x}^{3}$ is the same as ${x}^{2} \times x$ ${y}^{6}$ is the same as ${y}^{2 + 2 + 2} = {y}^{2} \times {y}^{2} \times {y}^{2}$ A single $z$ we can do nothing with. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ $\textcolor{b l u e}{\text{Putting it all together}}$ So we have: $5 \sqrt{6 \times {x}^{2} \times x \times {y}^{2} \times {y}^{2} \times {y}^{2} \times z}$ Taking all the squared values out of the root $5 x {y}^{3} \sqrt{6 x z}$<\|endoftext\|>	4.5
1,615	This article was originally published on Aug 10, 2012. We’ve updated it and added this cool new video! Sending spacecraft to Mars is all about precision. It’s about blasting off from Earth with a controlled explosion, launching a robot into space in the direction of the Red Planet, navigating the intervening distance between our two planets, and landing with incredible precision. This intricate and complicated maneuver means knowing the exact distance from Earth to Mars. Since Mars and Earth both orbit the Sun – but at different distance, with different eccentricities, and with different orbital velocities – the distance between then is constantly changing The first person to ever calculate the distance to Mars was the astronomer Giovanni Cassini, famous for his observations of Saturn. Giovanni made observations of Mars in 1672 from Paris, while his colleague, Jean Richer made the same observation from Cayenne, French Guiana. They used the parallax method to calculate the distance to Mars with surprising accuracy. However, astronomers now calculate the distance to objects in the Solar System using the speed of light. They measure the time it takes for signals to reach spacecraft orbiting other planets. They can bounce powerful radar off planets and measure the time it takes for signals to return. This allows them to measure the distance to planets, like Mars, with incredible accuracy. Distance Between Earth and Mars: So, how far away is Mars? The answer to that question changes from moment to moment because Earth and Mars are orbiting the Sun. It also requires a little explanation about the orbital mechanics of each. Both Earth and Mars are following elliptical orbits around the Sun, like two cars travelling at different speeds on two different racetracks. Sometimes the planets are close together, and other times they’re on opposite sides of the Sun. And although they get close and far apart, those points depend on where the planets are on their particular orbits. So, the Earth Mars distance is changing from minute to minute. The planets don’t follow circular orbits around the Sun, they’re actually traveling in ellipses. Sometimes they’re at the closest point to the Sun (called perihelion), and other times they’re at the furthest point from the Sun (known as aphelion). To get the closest point between Earth and Mars, you need to imagine a situation where Earth and Mars are located on the same side of the Sun. Furthermore, you want a situation where Earth is at aphelion, at its most distant point from the Sun, and Mars is at perihelion, the closest point to the Sun. Earth and Mars Opposition: When Earth and Mars reach their closest point, this is known as opposition. It’s the time that Mars appears as a bright red star of the sky; one of the brightest objects, rivaling the brightness of Venus or Jupiter. There’s no question Mars is bright and close, you can see it with your own eyes. And theoretically at this point, Mars and Earth will be only 54.6 million kilometers from each other. But here’s the thing, this is just theoretical, since the two planets haven’t been this close to one another in recorded history. The last known closest approach was back in 2003, when Earth and Mars were only 56 million km (or 33.9 million miles) apart. And this was the closest they’d been in 50,000 years. Here’s a list of Mars Oppositions from 2007-2020 (source) - Dec. 24, 2007 – 88.2 million km (54.8 million miles) - Jan. 29, 2010 – 99.3 million km (61.7 million miles) - Mar. 03, 2012 – 100.7 million km (62.6 million miles) - Apr. 08, 2014 – 92.4 million km (57.4 million miles) - May. 22, 2016 – 75.3 million km (46.8 million miles) - Jul. 27. 2018 – 57.6 million km (35.8 million miles) - Oct. 13, 2020 – 62.1 million km (38.6 million miles) 2018 should be a very good year, with a Mars looking particularly bright and red in the sky. Earth and Mars Conjunction: On the opposite end of the scale, Mars and Earth can be 401 million km apart (249 million miles) when they are in opposition and both are at aphelion. The average distance between the two is 225 million km. When Mars and Earth are at their closest, you have your launch window. Mars and Earth reach this closest point to one another approximately every two years. And this is the perfect time to launch a mission to the Red Planet. If you look back at the history of launches to Mars, you’ll notice they tend to launch about every two years. Here’s an example of recent Missions to Mars, and the years they launched: - MER-A Spirit – 2003 - MER-B Opportunity – 2003 - Mars Reconnaissance Orbiter – 2005 - Phoenix – 2007 - Fobos-Grunt – 2011 - MSL Curiosity – 2011 See the trend? Every two years. They’re launching spacecraft when Earth and Mars reach their closest point. Spacecraft don’t launch directly at Mars; that would use up too much fuel. Instead, spacecraft launch towards the point that Mars is going to be in the future. They start at Earth’s orbit, and then raise their orbit until they intersect the orbit of Mars; right when Mars is at that point. The spacecraft can then land on Mars or go into orbit around it. This journey takes about 250 days. Communicating with Mars: With these incredible distances between Earth and Mars, scientists can’t communicate with their spacecraft in real time. Instead, they need to wait for the amount of time it takes for transmissions to travel from Earth to Mars and back again. When Earth and Mars are at their theoretically closest point of 54.6 million km, it would take a signal from Earth about 3 minutes to make the journey, and then another 3 minutes for the signals to get back to Earth. But when they’re at their most distant point, it takes more like 21 minutes to send a signal to Mars, and then another 21 minutes to receive a return message. This is why the spacecraft sent to Mars are highly autonomous. They have computer systems on board that allow them to study their environment and avoid dangerous obstacles completely automatically, without human intervention. The distance from Earth to Mars is the main reason that there has never been a manned flight to the Red Planet. Scientists around the world are working on ways to shorten the trip with the goal of sending a human into Martian orbit within the next decade. We have written many articles about the distance between planets here at Universe Today. Here are the distances between Earth and the Sun, Mercury, Venus, the Moon, Jupiter, Saturn, Uranus, Neptune, and Pluto. And here are Ten Interesting Facts about Planet Mars and How Long Does it Take to Get to Mars? For more information, this website lists every Mars opposition time, from recent past all the way in the far future. You can also use NASA’s Solar System Simulator to see the current position of any object in the Solar System. Finally, if you’d like to learn more about Mars in general, we have done several podcast episodes about Mars at Astronomy Cast. Episode 52: Mars. We have also done an episode explaining distances, Episode 10: Measuring Distance in the Universe. - Cool Science Facts – The Speed of Light - Prof. Courtney Ligman – Opposition of Mars - NASA – Solar System Exploration<\|endoftext\|>	4.4375
326	# What is derivative of function sin(1+x)cos(1-x)? giorgiana1976 \| College Teacher \| (Level 3) Valedictorian Posted on We'll use the product rule to determine the derivative of this function: (uv)' = u'v + uv' We'll use the chain rule to differentiate the terms of the product. Let u = sin(1+x) = > u' = cos(1+x) Let v = cos (1-x) => v' = -(-1)sin(1 - x) The derivative of the function is: y' = cos(1+x)cos(1-x) + sin(1+x)sin(1-x) We'll transform the product of trigonometric functions into products: cos(1+x)cos(1-x) = (1/2)[cos(1+x+1-x) + cos(1+x-1+x)] cos(1+x)cos(1-x) = (cos 2 + cos (2x))/2 sin(1+x)sin(1-x) = (1/2)[cos(1+x-1+x) - cos(1+x+1-x)] sin(1+x)*sin(1-x) = (cos (2x) - cos 2)/2 y' = [cos 2 + cos (2x) + cos (2x) - cos 2]/2 y' = cos 2x The derivative of the given function is: y' = cos 2x<\|endoftext\|>	4.40625
583	Courses Courses for Kids Free study material Offline Centres More Store # One number is to be chosen from numbers $1$ to $100$. Then the probability that it is divisible by $5$ is…A. $\dfrac{{33}}{{100}}$ B. $\dfrac{7}{{100}}$ C. $\dfrac{1}{5}$ D. $\dfrac{{43}}{{100}}$ Last updated date: 11th Aug 2024 Total views: 430.5k Views today: 12.30k Verified 430.5k+ views Hint: To find the probability that the chosen number is divisible by $5$, find all numbers from 1 to 100 which are divisible by $5$. Then use the following probability formula to find the probability that the number will be divisible by 5. * Probability of an event is given by the number of favorable outcomes divided by the total number of outcomes. Find the total number of outcomes. There are $100$ numbers between $1$ to $100$. Therefore, total number of outcomes $= 100$. Find the numbers which are divisible by $5.$ We say a number is divisible by p if on division by p the remainder comes out to be zero. So, the multiples of p are the numbers which can be taken as divisible by p. The numbers which are divisible by $5$ are all the multiples of five from 1 to 100. $\{ 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100\}$. Therefore, number of favorable outcomes $= 20.$ Find the probability using the formula. Probability of an event is given by the number of favorable outcomes divided by the total number of outcomes. Here the total number of outcomes is 100 and the favorable outcome is 20. $\Rightarrow$ Probability ${\text{ = }}\dfrac{{{\text{20}}}}{{{\text{100}}}}$ Cancel out all common factors from numerator and denominator. $= \dfrac{1}{5}$ $\therefore$ Probability that the number is divisible by $5 = \dfrac{1}{5}$ So, the correct answer is “Option C”. Note: Students many times make mistakes in writing probability in unsolved form, they should always keep in mind that there should be no common factor between the numerator and the denominator and the probability should be in the simplest form. Also, check that the probability value should be greater than or equal to zero and less than or equal to one. Here $\dfrac{1}{5} = 0.2$ which lies between 0 and 1.<\|endoftext\|>	4.6875
170	Learn about Middle French in this 5-part series of bilingual articles about the development of French over the centuries.* Click any phrase for the English translation and related grammar lessons. Note: Some of the tenses in this French article and its English translation don't match! In French, we use the present tense to describe historical events like this to evoke a sense of immediacy, whereas in English, we commonly use the past tense - learn more about historical French tenses. The Development of French - Birth of the French language - Dialects and regional languages - Middle French - Classical French - Modern French - Dictionnaire Larousse - Wikipedia: Moyen français - Pierre Guiraud, Le moyen français, Presses universitaires de France, 1972.<\|endoftext\|>	3.734375
286	By Klaudia Weissmuller at April 23 2019 16:36:59 When you're teaching your student to write, there are a whole host of worksheets online that you can use. Many of these include clipart that will help the students learn the sounds of letters and letter combinations. There are other sheets that help the student learn to write his or her numbers. It's helpful having printable worksheets for something like this, because parents often go through quite a few of these before the child masters writing the numbers or letters correctly. Identical worksheets are needed prior to using the Consolidation feature, creating a sum across worksheets or using the Paste Special Math features to create summary worksheets. By grouping the worksheets first and then adding rows or columns, changing headings and other formatting operations, you ensure that the spreadsheets remain uniform. Be careful when using groups because if you forget that your spreadsheets are grouped and then proceed to add or change figures on a worksheet, all of the worksheets in the group will have the same figures. To remove a group, simply click on a sheet that is not in a group. If all of the worksheets are grouped, right_click on any tab and choose Ungroup Sheets. Keep an eye on the title bar prior to making changes and Excel will tell you if the sheets are grouped.<\|endoftext\|>	3.8125
766	• Select Exam • Select Exam # Tips and Tricks on Analogy (Non Verbal) in Reasoning Section Updated : Apr 27, 2017, 17:40 By : Ashish Today, we are providing the short tricks to solve Mirror Images and Water Images Questions in Reasoning.This is very helpful for the upcoming exam. You can master this topic only through practice as unlike numerical skill in reasoning section you don’t have to learn some formula and solve the question by applying that formula. But yes there are few tips and tricks for solving questions in Reasoning section. Here are some Important Tricks on Analogy (Non-Verbal) in Reasoning Section which will surely help you to score better in Reasoning section of every Competitive Exam. ANALOGY- means correspondence or similarities between two figures. And we have to find the similarity between the two figures to get our answer. TYPE 1 - ROTATION OF FIGURE • In this type of question, we have to find similarity in the rotation of figure. Two figures are related to each other through a specific rotation of the same figure. • Like in the above question figure (A) is rotated 180 degrees to obtain figure (B). • Now we have to apply the same logic in the next two figures to obtain the answer figure. • As we can see when we rotated the figure (C) by 180 degrees we get the figure shown in figure (2) of answer figures given. ANS- (B) TYPE 2 - DISLOCATION OF FIGURE • In these types of questions, we have to see how the first figure is getting dislocated or is getting combined to get our answer figure. • Like in the above question figure (A) is 3 dots connected with 3 straight lines but in figure (B) the dots are getting dislocated from the lines and lines get inserted. • We have to apply the same logic to get our answer figure. • So in our answer figure, the dots and lines must need to be disjointed from each other and line gets inverted. So figure (C) follows this pattern hence our answer will be C ANS- (C) TYPE 3 - INTERCHANGE OF IMAGES • In these types of questions, images get interchanged from one another to get the other image. • Like in the above question, in figure (A) circle is at the corner of the square and in figure (B) circle takes the position of the square and vice versa. • So we have to apply the same logic to get our answer figure. • Figure (C) is the combination of triangle and three line • So in our answer figure, the position of lines and triangle must be interchanged. This can be seen in figure (1) ANS- (A) Further questions are framed using these three concepts or the combination of any two mostly. But still, if any question is asked apart from these 3 conditions try to figure out the similarity in the given two figures and apply the same logic in the next two to get your answer. Thanks Posted by: Apr 27CTET & Teaching Exams WRITE A COMMENT Apr 28 Guru ji thanxxx Apr 28 Gjb trickk Can u tell me anything about PTET. N what should i prepare Notes of english up tgt sir,can u tell me anything about RIE CEE AJMER what should i prepare Sir plz provide us reasoning fast calculations methods Sep 30 Sir if you have Morarji desai previous questions paper is there means send Apr 4 Sir kvs teacher ki vacancy kb tk aayegi ....<\|endoftext\|>	4.375
461	# How To Solve Square Root Equations Algebraically? ## How do I solve square root equations? If an equation has a square root equal to a negative number, that equation will have no solution. To isolate the radical, subtract 1 from both sides. Simplify. Solve a radical equation. 1. Isolate the radical on one side of the equation. 2. Square both sides of the equation. 3. Solve the new equation. 4. Check the answer. ## What is the solution of sqrt 2x 4 16? Answer: The solution of the equation is x=126. ## How do you solve for extraneous roots? Example: you work on an equation and come up with two roots (where it equals zero): “a” and “b”. When you put “a” into the original equation it becomes zero, but when you put in “b” it doesn’t. So “b” is an extraneous root. ## What is the first step in solving a radical equation? A basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical. (The reason for using powers will become clear in a moment.) ## How do you simplify Radical Equations? Simplify the expressions both inside and outside the radical by multiplying. Multiply all numbers and variables inside the radical together. Multiply all numbers and variables outside the radical together. You might be interested: FAQ: What Is The Square Root Of 25? ## Which of the following is an extraneous solution of sqrt 2? Answer Expert Verified x=8 is a extraneous solution. x+3 = sqrt 13. ## Why do extraneous roots occur? The reason extraneous solutions exist is because some operations produce ‘extra’ answers, and sometimes, these operations are a part of the path to solving the problem. When we get these ‘extra’ answers, they usually don’t work when we try to plug them back into the original problem. ## What does it mean when two equations are equivalent? Two systems of equations are equivalent if they have the same solution(s). This article reviews how to tell if two systems are equivalent. Systems of equations that have the same solution are called equivalent systems.<\|endoftext\|>	4.59375
867	Causes of the Peloponnesian War Comments Off on Causes of the Peloponnesian War Even though Athens and Sparta formed an alliance in order to fight the Persians in the Greco-Persian Wars, the two city-states didn’t remain on such friendly terms. Once they succeeded in preventing Persian conquest, the two city-states did maintain their alliance. However, relations between them became strained because both city-states were strong, a fact which created tension between them. The conflicts that arose because of this tension eventually escalated into a full-fledged war. Since both city-states made strong alliances with other city-states, this war encompassed most of Greece. Athens Becomes Stronger According to Thucydides, the historian who is credited with writing the best, most thorough account of the Peloponnesian War wrote that the rapid growth of the Athenian city-state made war “inevitable”. One of the things that contributed to Athens’ strength was the Delian League, which was a naval alliance headquartered on the island of Delos that began to help insure that the Persian Wars were truly over. As this alliance grew, so did the strength of the Athenian city-state, largely because Athens started to use the League for its own gains. Sparta Heads the Peloponnesian League While Athens was busy forming a naval alliance, Sparta, the city-state with the strongest army, created an alliance of its own with several independent city-states in the Peloponnese. The Peloponnesian League that ensued included Thebes, Corinth, and Sparta. During this time, tensions between Sparta and Athens started to escalate as Athens decided to rebuild its city walls after the Persians withdrew. Thucydides wrote that the Spartans were, “secretly aggrieved” at this decision because they felt that the Athenians were distancing themselves from them. In their minds, this isn’t how allies were supposed to behave. However, it wasn’t until the Helot Revolt that relations between Athens and Sparta almost completely deteriorated. The Helots were an ethnic group that lived in Spartan territory that was completely subjugated. At best, they were used as slaves, particularly in agriculture. At worse, they were mistreated and even murdered. In 465 BC, they revolted in an attempt to gain their freedom. Since the Helot population outnumbered the Spartans, they were concerned that it gave the Helots an edge. They called on their Athenian allies for help, which they provided. When the four thousand Athenian soldiers arrived to aid the Spartans in suppressing the revolt, they were turned away. This move deeply offended the Athenians, who had provided some of their resources to help. Thucydides said that the Spartans were concerned that the Athenians would switch sides and assist the Helots instead. Eventually, the Helots did surrender. The Athenians set them up in their own colony in the city of Naupactus in the Gulf of Corinth. Peloponnesian War and Its Aftermath When Athens moved the Delian League Treasury from the island of Delos to Athens, this was the last straw for the Spartans, who decided that the Athenians were in violation of the Thirty Years Peace that was between them. Historians agree that the war officially began in 431 BC when Sparta invaded the ancient region of Attica, of which Athens was included. Over the course of the war, there are three distinct phases – the Archidamian War, which took place from 431-421 BC, the Peace of Nicias and the Sicilian Expedition, which took place from 420-413 BC, and the Ionian War, which took place from 412-404 BC. Though Sparta technically won the war, the toll to Ancient Greece was too great. Phillip II of Macedon used this as an opportunity to conquer Ancient Greece. His son, Alexander the Great, continued what his father began by creating a Greek Empire, all under his own control. Categorized in: Ancient Greek History This post was written by GreekBoston.com<\|endoftext\|>	3.90625
1,022	2TH/vwo §1.5 Line through two given points. §1.5 Learning goal: how to make a formula for a line that passes through two (given) points 1 / 28 Slide 1: Slide WiskundeMiddelbare schoolhavo, vwoLeerjaar 2 This lesson contains 28 slides, with interactive quizzes and text slides. Lesson duration is: 45 min Items in this lesson §1.5 Learning goal: how to make a formula for a line that passes through two (given) points Slide 1 -Slide Let's do a useful exercise together! This subject is found pretty difficult, sometimes. That's why we do this together, step by step! Slide 2 -Slide Given are two points: P(10, 6) and Q(18, 30) You will do similar tasks later. Now we tackle this one together! Slide 3 -Slide First: it helps to make a Sketch! Without it the task stays pretty obscure... Slide 4 -Slide The given points P(10, 6) and Q(18, 30) are drawn in this coordinate system. Slide 5 -Slide ANS (answer it in next slide) Use the given points, P(10, 6) and Q(18, 30) ! A 5 B -3 C 1/3 D 3 Slide 7 -Quiz Solution: Going from P(10, 6) to Q(18, 30) means: - going up 30 - 6 = 24 (= vertical increase) - going to the right 18 - 10 = 8 (= horizontal increase) If we divide the vertical increase by the horizontal increase we get the Slide 9 -Slide Find the y - intercept! There's one problem: you cannot read it off! So a CALCULATION is needed here! THERE ARE 2 METHODS FOR FINDING THE Y-INTERCEPT. Here comes the 1st one. Slide 10 -Slide Finding the y-intercept following Method 1: - we use the general shape of a linear formula: y = ax + b - we fill in the gradient, that we just calculated: y = 3x + b - to find the y-intercept, we fill in one of the given points; see next slide! Slide 11 -Slide In the formula: y = 3x + b we fill in P(10, 6) as follows: 6 = 3 10 + b Note: the dot means 'times'! 6 = 30 + b b = -24 We found the value of b, so we found the y-intercept! Slide 12 -Slide Now we can make the formula complete: We just calculated that b = -24 So the formula is: y = 3x -24 Slide 13 -Slide 2nd Method to find the y-intercept! Slide 14 -Slide Another way to calculate the Gradient and the y-intercept ! With TABLES. The points P(10, 6) and Q(18, 30) are placed in this table: Slide 15 -Slide means the same as: adding 3 to y ! That's because 24 : 8 = 3 A 1 B -3 C 3 D 1/3 Slide 18 -Quiz 3, because going one place to the right in the graph, means going 3 places up! Slide 19 -Slide To find the y-intercept with the table, we have to find the y coordinate to 0 as x-coordinate: Try to find the y-intercept! So: what has to be filled in on the dots? ANS A -30 B -24 C 24 D 0 Slide 21 -Quiz To find the y-intercept with the table, we have to find the y coordinate to 0 as x-coordinate: The numbers 1 and 3 have the same ratio as -10 and -30. So we do 6 -30 = -24 -24 is the y-intercept! FORMULA: y = 3x - 24 Slide 23 -Slide we do 33a together: write '33' in the margin. take a pencil and a geo and make the sketch! timer 1:20 Solution: Slide 25 -Slide 33b, c, d, e, f and g you will again find the linear formula for the graph through two given points!<\|endoftext\|>	4.8125
2,494	About Human Voice The human voice is a remarkable, cooperative effort of expression whose variation might be thought to surpass any in the animal kingdom. This would be a wrong assumption; the amphibian class – and among them, chiefly frogs — own the distinction of a surpassing variety of voice. We would also be wrong in the assumption that humans, alone, use voice to communicate intelligent thought to one another. Thirdly, humans have managed to teach other animals a relatively small understanding of our vocabulary, even in a variety of languages, such that they show obedience to some vocal commands, but this may be due to our ability to teach more than their ability to learn. The inverse of cross-species communication has, so far, left us far less able to understand their voices. Those limitations of language aside, for we use the voice for much more than intelligent, spoken language, the human voice is our primary origin of sound and it is almost universally applied to communicate to others. Occasionally, we will even speak to the speaker, alone or in a crowd, as if one side of the brain were communicating to the other for mere pleasure or debate. This is often considered by others as a lapse of judgment. Speaking to one’s self would seem unnecessary. Everything said is already known before it is spoken, but, regardless, everyone does it once in a while. The imperative of vocal expression is that urgent. Table of Contents Voice takes a variety of forms. Language, of course, heads the list. But the voice also expresses raw emotions of happiness and sadness, calm and anger, pleasure and distress, trust and fear, doubt and confidence, curiosity and wisdom. It does so in the range of variation of volume from a whisper to a shout, by speaking, crying and singing, by expressions of intelligent coherence and nonsensical sounds and by the relative speed of the string of sounds made. We are so adaptive regardless of cultural and geographic variety, much of these expressions cross language barriers. An American may not understand a Chinese speaking his language, but the former can certainly assess the latter’s expressed emotion, and that is recognized as much by vocal pattern as it is by visual signals. It was noted first that voice is a complicated, cooperative effort. The broad stroke of that cooperation uses three human components: the lungs, vocal folds in the larynx and a collection of modifiers. The science of voice begins and ends with a simple premise: the requirement of airflow. Air must move through space and through modifying obstacles in order for voice to exist. Without it, the voice would be mute. In a vacuum, no voice is possible. The voice begins with the lungs which supply the necessary airflow to begin the process. The air flows through the bronchial tubes to the larynx in the throat where the secondary components, the vocal folds, vibrate in response to the relative pressure of the airflow. This vibration creates the audible sound that is voiced. The airflow rises into the mouth where a series of modifiers articulate the sound until it passes through the lips. Those modifiers, some of which are not subject to manipulation, are, in order of placement and effect, the tongue (and its relative shape, length, attachment at the floor of the oral cavity, and flexibility), the oral cavity (whose shape is bounded by the tongue and floor of the mouth, the inner cheeks, the teeth, the palate, or roof of the mouth), and the lips. Of these modifiers, only the natural manipulation of the tongue and lips are under variable control to shape the voice. With surgical treatment, either for correction of congenital defects, by virtue of injury, or by personal choice, the other modifiers, or even those able to be controlled, can be manipulated to change or enhance the voice. The mechanics of voice begin with the lungs, where airflow is thrust under pressure by the same general means as the respiration process. However, to manipulate voice, it is generally required that more pressure be applied than with respiration. Because the voice requires airflow, and the lungs are incapable of producing uninterrupted airflow, voice will be interrupted briefly during its production in order for air to be replenished in the lungs. Breathing exercises allow the lungs to increase their capacity such that voluntary breathing cycles can be decreased while increasing the amount of air inhaled in each cycle. Such exercises have a variety of benefits; one of them affects voice. So far in the mechanical process, the moving air has no discernible sound other than that heard in normal breathing, or by a gentle breeze. Nor does it express any particular feeling; there is no recognizable intelligence driving it. The vocal folds (also called vocal chords) in the larynx are the next mechanical component in the production of voice. The vocal folds are a pair of flexible membranes which open and close much like a curtain, but the orientation of the folds is more horizontal than vertical across the larynx. For normal breathing, the vocal folds remain open both for inhaling and exhaling. When vocalizing, the vocal folds both close and vibrate due to the airflow from the lungs. The relative state of closure varies while vocalizing, but the folds are never completely closed while producing sound. Otherwise, airflow, and thus voice, would cease. The only state of full closure of the vocal folds occurs during eating so that food and liquids are not introduced into the larynx but, rather, into the esophagus. (This is mostly an involuntary action, but, sometimes, while eating, the vocal folds may not successfully close and the reaction is an instantaneous pain in the upper chest because solid objects, even liquid in volume, are not supposed to reach the bronchial tubes.) During vocalization, the action of the vocal folds is a sudden partial closure while vibrating, thus creating chopped segments or pulses of airflow. The length of the vocal folds, which varies generally between men and women, with longer vocal folds being typical of men, and the tension applied to the folds during their activity is what varies tone and pitch in the voice. The closer together and higher the tension in the folds, the higher the pitch and tone. The opposite condition results in lower pitch and tone. Pitch is the modulation of volume and frequency; tone is the modulation of inflection, expressing emotional variation. At this point, as air passes through the vocal folds, audible sound is produced, but it is not yet modulated to express intelligible sounds such as language. The closing and opening mechanics of the vocal folds are called, respectively, adduction (closure) and abduction (opening). For example, while breathing the vocal folds are abducted; while vocalizing, the folds are variably adducted, but always mostly closed. The ability to rapidly adduct and abduct the folds is mostly involuntary and genetic, which becomes critical when considering the discussed differences in breathing, vocalizing and eating as noted above. Another aspect of adduction/abduction is that if their relative muscular strength exceeds that of the pressure of airflow from the lungs, voice will cease. This is easily demonstrated by the act of voluntary swallowing, which adducts the vocal folds completely. Adduction and abduction control whether a unit of speech — a single letter, diphthong or ligature – are voiced or unvoiced, meaning, respectively, that while sound is being produced, the vocal folds are vibrating, or not vibrating. All English vowels and some consonants are voiced, i.e., the folds are vibrating; other consonants are unvoiced. Voiced and unvoiced consonants often occur in pairs, such as “p” (unvoiced) and “b” (voiced). The pairing occurs when, in both cases, the modulators (oral cavity components) are in identical placement; the only difference between the two consonants is the lack of vocal fold vibration (“p”) and vibration (“b”). The function of the modulators in the oral cavity just above the larynx and vocal folds is to resonate and articulate sound into intelligible segments of spoken word and song. Resonance is produced by the larynx above the vocal folds, the shape of the oral cavity as defined by its sides, roof and floor, the shape of the nasal cavity through the palate, and the skeletal structure of the head. Resonance can be felt in the head and chest; a lower voice will resonate deeper. Resonance does not change with intended meaning or emotion, but it will affect the apparent volume and real distance the sound carries. Articulation is shaped by the tongue, palate, teeth and lips to change sound into its variety of expressions. Articulations of sound are differentiated by the relative position of each modifier and their coordination with one another. There is also the consideration of the emotive aspect of voice. Singing, as opposed to speaking, is generally considered a more emotive expression as opposed to the literal and logical expression of spoken voice. Even though both speaking and singing use the same mechanics, the position of the larynx must change to accommodate singing. Actually, the act of vocal crying is virtually identical to singing. In speaking, the larynx is positioned vertically in the throat. The singing and crying effort pulls the larynx deeper into the throat while simultaneously pulling the base of the larynx back toward the spine. The effect of this manipulation of the larynx stretches the length of the vocal folds and changes their orientation to a slight angle across the larynx. This effort is mostly voluntary in the respect that speaking and singing are possible for anyone who can vocalize sound. It is the quality of the singing voice which must be trained and it is the training of this manipulation of the position of the larynx which qualifies good and poor singers. The positional shift of the larynx allows the singer to create vibrato, which is the slight modulation of pitch during the singing of a single note, or frequency. This violates the idea of “perfect pitch” in that the pitch does, in fact, vary slightly due to rapid vibratory changes in the vocal folds while sustaining a note, but those changes are so slight, there is almost no change in the frequency, which defines perfect pitch. The emotional result, however, is more satisfying to the singer and more pleasurable for the listener. Another aspect of emotive vocalization is the quality of sound produced. Untrained public speakers and singers often believe the airflow and pressure originate in the larynx, but the resulting quality of voice is weak and strained. This is why trained speakers and singers are taught that the lungs, supported by the diaphragm beneath them, are the true source of voice. The speaker/singer’s concentration of vocalizing should begin there, not strangled in the throat. Finally, the emotive character of voice is modulated by the air flow and pressure provided by the lungs and as allowed by the vocal folds. The voice can be modulated in volume and inflection very rapidly for emotional effect. Depending on the language, they vary widely, or they may remain static. Emotion is more expressive if volume and inflection are allowed to vary. The voice can also range quickly from loud and agitated speech to quiet and calm speech. It can vary from fast, clipped speech with staccato consonants to slow, lyrical speech with drawn out vowels to convey emotional shift. Voice can convert from speech to song in a single breath, and back again to convey emotion. All of these techniques are available, limited only by imagination, to convey emotion by voice. It may be possible to achieve without ever saying or singing a single word, but just by creating modulated sound using the mechanics and variable expressions of voice. Even if man is not as expressive in variation of voice as is the frog, he does have that extended ability to voice speech and song with emotional impact. Anything else is just a figure of speech. Contact us for results focused speech therapy in Sydney If you have questions about speech therapy for the voice, contact your local doctor, who will arrange for you to see a speech pathologist. Contact us today!<\|endoftext\|>	3.984375
636	UK USIndia Every Question Helps You Learn If you multiply 5 by itself, you get 25. # Properties of Numbers (Year 5) In Year 5, which is the third year in KS2 Maths, you'll learn cool things about numbers. Numbers have special friends called factors and multiples. Imagine factors as numbers that can team up to make another number, like 1, 2, 4, and 8 for the number 8! Multiples are like a group of friends in a times table – for 8, it's 0, 8, 16, and so on. Maths is like solving number puzzles! How much do you know about these number pals? Let's play a fun quiz for 9-10 year olds and test your knowledge of the properties of numbers. See how many number secrets you can remember! 1. What is the square number of 6? 6 12 36 60 36 is 6 x 6 2. Which number has the factors of 1, 2, 4, 8 and 16? 8 16 24 36 Every number is a factor of itself as it is itself multiplied by 1 3. What is a factor? A factor is a whole number that divides into another whole number A factor is a whole number that can be divided by 6 A factor is a number that cannot be divided A factor is the product when a number is multiplied by itself For example, 3 and 2 are both factors of 6 4. How do we know that a number is a multiple of 4? It ends in 2, 4 or 8 The last two digits are divisible by 4 It is also a multiple of 8 There is no way of knowing We can ignore numbers in the 100s column as 100 is itself divisible by 4 5. Which number multiplied by itself gives 49? 3 6 5 7 6. What is a square number? A number that can be divided by 4 A number that is a multiple of 6 and 7 The product when a number is multiplied by itself A number that is written in a square For example 3 x 3 = 9 therefore 9 is the square number of 3 7. Which of these numbers is a multiple of 4? 5,698 2,463 4,441 2,548 We can easily work this out as 2,548 is the only number where the last 2 digits are divisible by 4 8. How many pairs of factors will give 12? 1 3 6 12 They are 1 x 12, 2 x 6 and 3 x 4 9. What is 5 squared? 5 10 15 25 5 x 5 = 25 10. Which of these numbers is a multiple of 7? 15 22 35 37 35 is 5 x 7<\|endoftext\|>	4.5
1,117	# Construction of geometrical figures: answers to the exercises in 6th grade in PDF. Corrected 6th grade math exercises on the triangle and quadrilaterals with geometric figure constructions. Know how to use geometry equipment (ruler, compass, protractor and square). Exercise 1: 1. 2. Lines (BE) and (CF) are perpendicular to the same line (AD) so they are parallel to each other. Exercise 2: Exercise 3: Exercise 4: Here is a figure where points A, B and C are aligned. a. Write a construction program for this figure. 1) Construct a segment [AC] with a length of 7 cm. 2) Place point B on the segment [AC] such that AB = 3 cm. 3) Draw the lines (d1) and (d2) perpendicular to (AC) through points A and C. 4) Place the point D on the line (d1) such that AD = 4 cm . 5) Draw the segment [DB]. 6) Draw the line (d3) perpendicular to [DB] through B. 7) Place the point E which is the point of intersection of the two lines (d2) and (d3) . Exercise 5: a. Draw a triangle ABC such that AB = 3 cm, AC = 5 cm and . b. Place the point M on the segment [AB] such that AM = 1 cm . c. Through M, draw the parallel to the line (BC); it intersects the line (AC) at N. d. From M, draw the perpendicular to the line (BC); it intersects (BC) at Q. Through N, draw the parallel to the line (MQ); it intersects (BC) at P. a)See in red b)c)d) see diagram (Note, do not worry about point B’ which was only there to create an angle of 100° with the protractor) e. What can we say about the lines (MQ) and (MN)? Explain why. They are perpendicular because according to the construction (MN) is parallel to (BC) and (BC) is perpendicular to (MQ). Moreover, any perpendicular to (BC) is perpendicular to (MN) since (BC) // (MN) // means “parallel”. f. What can we say about the lines (NP) and (PQ)? Explain why. Since (NP) is parallel to (MQ), we know that (PQ) is perpendicular to (MQ) and therefore perpendicular to (NP). (NP) and (PQ) are perpendicular g. What is the nature of the quadrilateral MNPQ ? Explain why. Since the quadrilateral MNPQ has two opposite right angles, parallel sides. It is a rectangle. Exercise 6: Exercise 7: Here are some of the constructs you need to get: Exercise 8: In each case, make the figure described and indicate the nature of the triangle. a. ABC is a triangle such that . ABC is a right triangle in C . b. MNP is a triangle such that MN=NP and . MNP is an isosceles triangle and right-angled in N . c. EFG is an isosceles triangle at each of its vertices. EFG is an equilateral triangle. Exercise 9: Note that the three lines (EC), (BF) and (AG) are concurrent (all three intersect at the same point). Exercise 10: We have: CE= 6 cm and AC = 3 cm . Cette publication est également disponible en : Français (French) Español (Spanish) العربية (Arabic) Other documents in the category corrected by Other forms similar to construction of geometrical figures: answers to the exercises in 6th grade in PDF.. • 93 The answer key to the 5th grade math exercises on central symmetry. Know how to construct the symmetric of a figure using geometry materials and conduct demonstrations using the properties of central symmetry in fifth grade. Exercise 1: Exercise 2: Exercise 3: Exercise 4: Here is the symmetric of the… • 93 The answer key to the math exercises on the scalar product in space. Colinear and orthogonal vectors and know how to determine the equation of a line and a plane as well as, apply the Chasles relation on vectors. Exercise 1: Calculate the distance of the point M(5; 2; -3)… • 92 The answer key to the math exercises in 5th grade on the parallelogram. To construct a parallelogram using geometry equipment (ruler, compass, square and protractor). Use the properties to perform demonstrations. Exercise 1: Exercise 2: 1. Construct a parallelogram DOMI such that : DM= 7cm , and . 2. Prove… Les dernières fiches mises à jour. Voici les dernières ressources similaires à construction of geometrical figures: answers to the exercises in 6th grade in PDF. mis à jour sur Mathovore (des cours, exercices, des contrôles et autres), rédigées par notre équipe d'enseignants. On Mathovore, there is 13 624 775 math lessons and exercises downloaded in PDF. Mathovore FREE VIEW<\|endoftext\|>	4.65625
985	The business will then break even in two years. What if analysis is often used to compare different scenarios and their potential outcomes based on changing conditions. Capital budgeting refers to the process by which a business determines whether to take on a certain project. For example, an investor might want to know what happens if revenue targets are not met and the likely impact of using an alternative revenue target, or a lender might want to assess the effect of changes in interest rates on the financial projection. What if competition heats up between airlines? Often used scientific research and in conjunction with business and financial risk assessmentssensitivity analysis is applicable to virtually any activity or system. This is your financial forecast, based on what you know, it is what you feel the outcome will be. What is What-if Analysis? With scenario analysis, all inputs changes are made at the same time with the purpose of assessing the effect on the business plan of a complete change in circumstances. Step 2 — Develop the Best Case Scenario Make a copy of the base case scenario financial projections template developed in step 1, and amend the inputs to show what will happen if your positive expectations are met, and you can seize all the opportunities available to the business. However, what if the cost of gasoline goes up between now and then? These factors could affect your costs and your ultimate decision. By changing each input seperately it is possible to assess the significance of each variable on the business Scenario Analysis and Sensitivity Analysis in a Business Plan One way a business can demonstrate the effect of changes in inputs in a financial projection is to provide three different scenarios, so that the financial risk of the business can be simulated under different conditions. Business planning software typically walks you through the process, prompting you to answer questions and fill in the blanks. With sensitivity analysis only one input is changed at a time in order to assess the impact of that input on the financial projection. For example, in the base case scenario, you might have anticipated opening an export market in year three, show what will happen if that market does not develop or is delayed until a later year. By asking what-if questions and running a financial simulation, you can make better decisions as well as demonstrate the strength of your business plan to investors. The business may calculate the amount of time it will take for the project to generate enough income to cover the investment expenses. The Benefits of What-if Analysis Conducting a what-if framework is beneficial in several ways. This also involves generating "what if" questions to determine how the activity will be affected by different scenarios. Common What-if Analysis Methods Common methods of sensitivity analysis include using: In a scenario analysis, on the other hand, the analyst determines a certain scenario, such as a stock market crash or change in industry regulation. The analysis involves estimating the amount of money the business has to invest and the amount of revenue the project will generate. Scenario management tools such as those built into Microsoft Excel Brainstorming techniques involving identifying activities and potential factors that could affect the outcome of those activities. Methods A business may use various methods to determine the financial effects of a certain project. He now knows the full range of outcomes, given all extremes, and has an understanding of what the outcomes would be given a specific set of variables defined by real-life scenarios. When carrying out sensitivity analysis, it is important to remember that the projections still have to be feasible and achievable, they are not simply hypothetical what ifs. As you learn more about what-if analysis, you may be intimidated by the numerous methods and complex formulas often used.Sensitivity Analysis of a Strategic Sheep Farming Business Plan \| compares the Income Statement Results in terms of Pessimistic, Planned and Optimistic. 16 HOP FARMING BUSINESS PLAN TEMPLATE Sensitivity Analysis A sensitivity from BAMD W at George Washington University. business plan is the most popular teaching method (Solomon, ). Textbooks to conduct sensitivity analysis of the assumptions in their sales forecast. The example of one student assignment letter; and (d) is a zipped file that. A Sensitivity Analysis Template allows you to show the impact of market changes in your business plan. In light of Brexit, and the changes it will bring to the business environment, this tool will prove useful. Scenario Analysis and Sensitivity Analysis in a Business Plan One way a business can demonstrate the effect of changes in inputs in a financial projection is to provide three different scenarios, so that the financial risk of the business can be simulated under different conditions. Ratio Analysis Because of the sensitivity of the fitness industry, Fitness Center Company would have to be cognizant and aware of the changing developments in these areas. The opportunities for Fitness Center Company are significant; FITNESS CENTER BUSINESS PLAN SAMPLE.Download<\|endoftext\|>	3.671875
629	# Construction of The Real Numbers DEFINITION1: The set $\mathbb{R}$ with at least two distinct elements and satisfying the following five axioms is called the set of real numbers and each element of $\mathbb{R}$ is called a real number: The function $+:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ defined as $\left( x,y \right)\to x+y\in \mathbb{R}$ for each $\left( x,y \right)$ in $\mathbb{R}\times \mathbb{R}$ satisfies the following properties: I${{}_{1}}.\,\forall a,b\in \mathbb{R},a+b=b+a$, I${{}_{2}}.\,\forall a,b,c\in \mathbb{R},a+(b+c)=(a+b)+c$, # Partial Order Relation DEFINITION1: Let $X$ be a set and $R\subset{X\times{X}}$. If the relation $R$ is reflexive, antisymmetric and transitive, then the relation $R$ is called a "partial order relation" and denoted by $R=\le$ in general. If "$\le$" is a partial order relation over a set $X$, then $(X,\le)$ is called "partially ordered set" or shortly "poset". DEFINITION2: Let $x$ and $y$ are elements of a partially ordered set $X$. If it holds “$x\le{y}\lor{y\le{x}}$”, then $x$ and $y$ are called “comparable”. Otherwise they are called “incomparable”. DEFINITION3: If $x$ and $y$ are comparable for all $x,y$ in a partially ordered set $(X,\le)$, then the relation $\le$ is called a “total order” and the set $X$ is called a “totally ordered set” or “linearly ordered set”. DEFINITION4: Let $(X,\le)$ be a partially ordered set and $A\subset{X}$. If $(A,\le)$ is a totally ordered set, then $A$ is called a “chain” in $X$. DEFINITION5: Let $(X,\le)$ be a partially ordered set and $A\subset{X}$. If there exists an element $a^{}\in{A}$ satisfying $a\le{a^{}}$ for all $a\in{A}$, then $a^{}$ is called the maximum of $A$, and if there exists an element $a_{}\in{A}$ satisfying $a_{}\le{a}$ for all $a\in{A}$, then $a_{}$ is called the minimum of $A$. The minimum and the maximum of $A$ are denoted by $\min{A}$ and $\max{A}$ respectively.<\|endoftext\|>	4.375
1,017	## Precalculus (6th Edition) Blitzer $\left\{ \left( 3+z,\ 5+5z,\ 3z+4,\ z \right) \right\}$ Convert the provided system of equations into matrix form. \left[ \left. \begin{align} & \begin{matrix} 2 & -3 & 4 & 1 \\ \end{matrix} \\ & \begin{matrix} 1 & -1 & 3 & -5 \\ \end{matrix} \\ & \begin{matrix} 3 & 1 & -2 & -2 \\ \end{matrix} \\ \end{align} \right\|\begin{matrix} 7 \\ 10 \\ 6 \\ \end{matrix} \right] Solve the above matrix as below: $\text{By},\ {{R}_{3}}-3{{R}_{2}}\to {{R}_{3}}$ \left[ \left. \begin{align} & \begin{matrix} 2 & -3 & \ \ 4 & \ 1 \\ \end{matrix} \\ & \begin{matrix} 1 & -1 & \ \ 3 & -5 \\ \end{matrix} \\ & \begin{matrix} 0 & \ 4 & -11 & 13 \\ \end{matrix} \\ \end{align} \right\|\begin{matrix} 7 \\ 10 \\ -24 \\ \end{matrix} \right] $\text{By},\ 2{{R}_{2}}-{{R}_{1}}\to {{R}_{2}}$ \left[ \left. \begin{align} & \begin{matrix} 2 & -3 & \ \ 4 & \ \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ \ 1 & \ \ \ 2 & -11 \\ \end{matrix} \\ & \begin{matrix} 0 & \ \ 4 & -11 & \ 13 \\ \end{matrix} \\ \end{align} \right\|\begin{matrix} 7 \\ 13 \\ -24 \\ \end{matrix} \right] $By,\ {{R}_{3}}-4{{R}_{2}}\to {{R}_{3}}$ \left[ \left. \begin{align} & \begin{matrix} 2 & -3 & \ \ 4 & \ \ \ 1 \\ \end{matrix} \\ & \begin{matrix} 0 & \ \ 1 & \ \ \ 2 & -11 \\ \end{matrix} \\ & \begin{matrix} 0 & \ \ 4 & -11 & \ 13 \\ \end{matrix} \\ \end{align} \right\|\begin{matrix} 7 \\ 13 \\ -24 \\ \end{matrix} \right] Convert the above matrix system into the linear equations as below: \begin{align} & 2w-3x+4y+z=7 \\ & x+2y-11z=13 \\ & -19y+57z=-76 \end{align} Calculate the value of y in terms of z as below: \begin{align} & -19y+57z=-76 \\ & 19y=76+57z \\ & y=\frac{76}{19}+\frac{57}{19}z \\ & y=4+3z \end{align} Calculate the value of x in terms of z, by substituting the value ofy, z as below: \begin{align} & x+2\left( 4+3z \right)-11z=13 \\ & x+8+6z-11z=13 \\ & x=13-8+5z \\ & x=5+5z \end{align} Calculate the value of w in terms of z, by substituting the value of x, y, z as below: \begin{align} & 2w-3\left( 5+5z \right)+4\left( 4+3z \right)+z=7 \\ & 2w-15-15z+16+12z+z=7 \\ & w=\frac{6}{2}+\frac{2}{2}z \\ & w=3+z \end{align} The value of w, x and y in terms of z: \begin{align} & w=3+z \\ & x=5+5z \\ & y=3z+4 \\ & z=z \\ \end{align} Hence we have the solutions: $\left\{ \left( 3+z,\ 5+5z,\ 3z+4,\ z \right) \right\}$<\|endoftext\|>	4.5
246	In 1791, the slaves of France’s most profitable Caribbean colony, Saint Domingue, revolted. The uprising was kindled by the appalling exploitation and abuse of the colony’s enslaved African population, and stoked by the same Enlightenment values championed by white anti-monarchic revolutionaries in the United States and France itself. But the independent republic of Haiti that eventually emerged in 1804 was never an equal among the brotherhood of Western nations. To the north, the United States, a nation of slaveowners, regarded Haiti, a nation of free blacks, with unvarnished horror and boycotted its merchants. Meanwhile, France, the spurned former colonial ruler, fumed at its losses. In 1825, with a French flotilla threatening invasion, Haiti was compelled to pay a king’s ransom of 150 million gold francs — estimated to be ten times the country’s annual revenues — in indemnities to compensate French settlers and slaveowners for their lost plantations. The sum would be later reduced to 90 million gold francs, but that was little consolation: Haiti, in effect, was forced to pay reparations for its freedom. via – The Washington Post<\|endoftext\|>	4.0625
22,900	Rd Sharma XII Vol 2 2019 Solutions for Class 12 Science Math Chapter 10 The Plane are provided here with simple step-by-step explanations. These solutions for The Plane are extremely popular among Class 12 Science students for Math The Plane Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2019 Book of Class 12 Science Math Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma XII Vol 2 2019 Solutions. All Rd Sharma XII Vol 2 2019 Solutions for class Class 12 Science Math are prepared by experts and are 100% accurate. #### Question 1: Find the vector equation of a plane passing through a point with position vector $2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}$ and perpendicular to the vector $4\stackrel{^}{i}+2\stackrel{^}{j}-3\stackrel{^}{k}.$ #### Question 2: Find the Cartesian form of the equation of a plane whose vector equation is (i) $\stackrel{\to }{r}·\left(12\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}\right)+5=0$ (ii) $\stackrel{\to }{r}·\left(-\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}\right)=9$ #### Question 3: Find the vector equations of the coordinate planes. #### Question 4: Find the vector equation of each one of following planes. (i) 2xy + 2z = 8 (ii) x + yz = 5 (iii) x + y = 3 #### Question 5: Find the vector and Cartesian equations of a plane passing through the point (1, −1, 1) and normal to the line joining the points (1, 2, 5) and (−1, 3, 1). #### Question 6: is a vector of magnitude $\sqrt{3}$ and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to . #### Question 7: The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane. #### Question 8: Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2. #### Question 9: If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP. #### Question 10: Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane. #### Question 11: A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point $3\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}.$ Find the vector and Cartesian forms of the equation of the plane. #### Question 12: Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it. #### Question 13: Show that the normals to the following pairs of planes are perpendicular to each other. (i) xy + z − 2 = 0 and 3x + 2yz + 4 = 0 (ii) #### Question 14: Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes. #### Question 15: Find a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1. #### Question 16: If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane. #### Question 17: Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles. #### Question 18: Find the vector and Cartesian equations of the plane that passes through the point (5, 2, −4) and is perpendicular to the line with direction ratios 2, 3, −1. #### Question 19: If O be the origin and the coordinates of P be (1, 2,−3), then find the equation of the plane passing through P and perpendicular to OP. #### Question 20: If O is the origin and the coordinates of A are (abc). Find the direction cosines of OA and the equation of the plane through A at right angles to OA. [NCERT EXEMPLAR] It is given that O is the origin and the coordinates of A are (abc). The direction ratios of OA are proportional to $a-0,b-0,c-0$ or a, b, c ∴ Direction cosines of OA are $\frac{a}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}},\frac{b}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}},\frac{c}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}}$ The normal vector to the required plane is $a\stackrel{^}{i}+b\stackrel{^}{j}+c\stackrel{^}{k}$. The vector equation of the plane through A(abc) and perpendicular to OA is The Cartesian equation of this plane is #### Question 21: Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively. The equation of the plane in the intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where ab and are the intercepts on the xy and z-axis, respectively. It is given that the intercepts made by the plane on the xy and z-axis are 3, –4 and 2, respectively. ∴ a = 3, b = −4, c = 2 Thus, the equation of the plane is $\frac{x}{3}+\frac{y}{\left(-4\right)}+\frac{z}{2}=1\phantom{\rule{0ex}{0ex}}⇒4x-3y+6z=12$ $⇒\left(x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}\right).\left(4\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)=12\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}.\left(4\stackrel{^}{i}-3\stackrel{^}{j}+6\stackrel{^}{k}\right)=12$ This is the vector form of the equation of the given plane. #### Question 1: Find the vector equation of a plane which is at a distance of 3 units from the origin and has $\stackrel{^}{k}$ as the unit vector normal to it. #### Question 2: Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector $\stackrel{^}{i}-2\stackrel{^}{j}-2\stackrel{^}{k}.$ #### Question 3: Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. #### Question 4: Reduce the equation $\stackrel{\to }{r}·\left(\stackrel{^}{i}-2\stackrel{^}{j}+2\stackrel{^}{k}\right)+6=0$ to normal form and, hence, find the length of the perpendicular from the origin to the plane. #### Question 5: Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0. #### Question 6: The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane. #### Question 7: Find a unit normal vector to the plane x + 2y + 3z − 6 = 0. #### Question 8: Find the equation of a plane which is at a distance of $3\sqrt{3}$ units from the origin and the normal to which is equally inclined to the coordinate axes. #### Question 9: Find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane. #### Question 10: Find the vector equation of the plane which is at a distance of $\frac{6}{\sqrt{29}}$ from the origin and its normal vector from the origin is $2\stackrel{^}{i}-3\stackrel{^}{j}+4\stackrel{^}{k}.$ Also, find its Cartesian form. #### Question 11: Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin. #### Question 1: Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5). #### Question 2: Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3). #### Question 3: Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C (0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that $\frac{1}{{p}^{2}}=\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}.$ #### Question 4: Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3). #### Question 5: Find the vector equation of the plane passing through the points #### Question 1: Find the angle between the given planes. (i) (ii) (iii) #### Question 2: Find the angle between the planes. (i) 2xy + z = 4 and x + y + 2z = 3 (ii) x + y − 2z = 3 and 2x − 2y + z = 5 (iii) xy + z = 5 and x + 2y + z = 9 (iv) 2x − 3y + 4z = 1 and − x + y = 4 (v) 2x + y − 2z = 5 and 3x − 6y − 2z = 7 #### Question 3: Show that the following planes are at right angles. (i) (ii) x − 2y + 4z = 10 and 18x + 17y + 4z = 49 #### Question 4: Determine the value of λ for which the following planes are perpendicular to each other. (i) (ii) 2x − 4y + 3z = 5 and x + 2y + λz = 5 (iii) 3x − 6y − 2z = 7 and 2x + y − λz = 5 #### Question 5: Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5. #### Question 6: Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8. #### Question 7: Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2yz = 1 and 3x − 4y + z = 5. #### Question 8: Find the equation of the plane passing through the points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9. #### Question 9: Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1. #### Question 10: Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5. #### Question 11: Find the equation of the plane with intercept 3 on the y-axis and parallel to the ZOX plane. #### Question 12: Find the equation of the plane that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8. #### Question 13: Find the equation of the plane passing through (a, b, c) and parallel to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=2.$ #### Question 14: Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. #### Question 15: Find the vector equation of the plane through the points (2, 1, −1) and (−1, 3, 4) and perpendicular to the plane x − 2y + 4z = 10. #### Question 1: Find the vector equations of the following planes in scalar product form $\left(\stackrel{\to }{r}·\stackrel{\to }{n}=d\right):$ (i) $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{k}\right)+\lambda \stackrel{^}{i}+\mu \left(\stackrel{^}{i}-2\stackrel{^}{j}-\stackrel{^}{k}\right)$ (ii) (iii) $\stackrel{\to }{r}=\left(\stackrel{^}{i}+\stackrel{^}{j}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)+\mu \left(-\stackrel{^}{i}+\stackrel{^}{j}-2\stackrel{^}{k}\right)$ (iv) $\stackrel{\to }{r}=\stackrel{^}{i}-\stackrel{^}{j}+\lambda \left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)+\mu \left(4\stackrel{^}{i}-2\stackrel{^}{j}+3\stackrel{^}{k}\right)$ Disclaimer: The answer given for part (iv) of this problem in the text book is incorrect. #### Question 2: Find the Cartesian forms of the equations of the following planes. (i) $\stackrel{\to }{r}=\left(\stackrel{^}{i}-\stackrel{^}{j}\right)+s\left(-\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}\right)+t\left(\stackrel{^}{i}+2\stackrel{^}{j}+\stackrel{^}{k}\right)$ (ii) $\stackrel{\to }{r}=\left(1+s+t\right)\stackrel{^}{i}+\left(2-s+t\right)\stackrel{^}{i}+\left(3-2s+2t\right)\stackrel{^}{k}$ #### Question 3: Find the vector equation of the following planes in non-parametric form. (i) (ii) $\stackrel{\to }{r}=\left(2\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)+\lambda \left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)+\mu \left(5\stackrel{^}{i}-2\stackrel{^}{j}+7\stackrel{^}{k}\right)$ #### Question 1: Find the equation of the plane which is parallel to 2x − 3y + z = 0 and which passes through (1, −1, 2). #### Question 2: Find the equation of the plane through (3, 4, −1) which is parallel to the plane $\stackrel{\to }{r}·\left(2\stackrel{^}{i}-3\stackrel{^}{j}+5\stackrel{^}{k}\right)+2=0.$ #### Question 3: Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3). #### Question 4: Find the equation of the plane through the point $2\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$ and passing through the line of intersection of the planes #### Question 5: Find the equation of the plane passing through the line of intersection of the planes 2xy = 0 and 3zy = 0 and perpendicular to the plane 4x + 5y − 3z = 8. #### Question 6: Find the equation of the plane which contains the line of intersection of the planes x + 2y + 3z − 4 = 0 and 2x + yz + 5 = 0 and which is perpendicular to the plane 5x + 3y − 6z + 8 = 0. #### Question 7: Find the equation of the plane through the line of intersection of the planes x + 2y + 3z + 4 = 0 and xy + z + 3 = 0 and passing through the origin. #### Question 8: Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x − 3y + 2z − 5 = 0 and 2xy + 3z − 1 = 0 and passing through (1, −2, 3). #### Question 9: Find the equation of the plane that is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z − 4 = 0, 2x + yz + 5 = 0. #### Question 10: Find the equation of the plane through the line of intersection of the planes which is at a unit distance from the origin. #### Question 11: Find the equation of the plane passing through the intersection of the planes 2x + 3yz + 1 = 0 and x + y − 2z + 3 = 0 and perpendicular to the plane 3xy − 2z − 4 = 0. #### Question 12: Find the equation of the plane that contains the line of intersection of the planes and which is perpendicular to the plane $\stackrel{\to }{r}·\left(5\stackrel{^}{i}+3\stackrel{^}{j}-6\stackrel{^}{k}\right)+8=0.$ #### Question 13: Find the equation of the plane passing through (a, b, c) and parallel to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=2.$ #### Question 14: Find the equation of the plane passing through the intersection of the planes and the point (2, 1, 3). #### Question 1: Find the equation of the plane passing through the following points. (i) (2, 1, 0), (3, −2, −2) and (3, 1, 7) (ii) (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6) (iii) (1, 1, 1), (1, −1, 2) and (−2, −2, 2) (iv) (2, 3, 4), (−3, 5, 1) and (4, −1, 2) (v) (0, −1, 0), (3, 3, 0) and (1, 1, 1) (i) The equation of the plane passing through points (2, 1, 0), (3, −2, −2) and (3, 1, 7) is given by (ii) The equation of the plane passing through points (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6) is given by (iii) The equation of the plane passing through points (1, 1, 1), (1, −1, 2) and (−2, −2, 2) is given by (iv) The equation of the plane passing through points (2, 3, 4), (−3, 5, 1) and (4, −1, 2) is given by (v) The equation of the plane passing through points (0, −1, 0), (3, 3, 0) and (1, 1, 1) is given by #### Question 15: Find the equation of the plane through the intersection of the planes 3xy + 2z = 4 and x + y + z = 2 and the point (2, 2, 1). #### Question 16: Find the vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane xy + z = 0. #### Question 17: Find the vector equation of the plane passing through the intersection of the planes and the point (1, 1, 1). #### Question 18: Find the equation of the plane which contains the line of intersection of the planes x$+$2y$+$34$=$0 and 2$x+y-z$ $+$ 5$=$0 and whose x-intercept is twice its z-intercept. Hence, write the equation of the plane passing through the point (2, 3, $-$1) and parallel to the plane obtained above. The equation of the family of planes passing through the intersection of the planes x + 2y + 3z − 4 = 0 and 2xy − z + 5 = 0 is (x + 2y + 3z − 4) + k(2x + y − z + 5) = 0, where k is some constant $⇒\left(2k+1\right)x+\left(k+2\right)y+\left(3-k\right)z=4-5k\phantom{\rule{0ex}{0ex}}⇒\frac{x}{\left(\frac{4-5k}{2k+1}\right)}+\frac{y}{\left(\frac{4-5k}{k+2}\right)}+\frac{z}{\left(\frac{4-5k}{3-k}\right)}=1$ It is given that x-intercept of the required plane is twice its z-intercept. $\left(\frac{4-5k}{2k+1}\right)=2\left(\frac{4-5k}{3-k}\right)\phantom{\rule{0ex}{0ex}}⇒\left(4-5k\right)\left(3-k\right)=\left(4k+2\right)\left(4-5k\right)\phantom{\rule{0ex}{0ex}}⇒\left(4-5k\right)\left(3-k-4k-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(4-5k\right)\left(1-5k\right)=0$ When $k=\frac{4}{5}$, the equation of the plane is $\left(2×\frac{4}{5}+1\right)x+\left(\frac{4}{5}+2\right)y+\left(3-\frac{4}{5}\right)z=4-5×\frac{4}{5}⇒13x+14y+11z=0$. This plane does not satisfies the given condition, so this is rejected. When $k=\frac{1}{5}$, the equation of the plane is $\left(2×\frac{1}{5}+1\right)x+\left(\frac{1}{5}+2\right)y+\left(3-\frac{1}{5}\right)z=4-5×\frac{1}{5}⇒7x+11y+14z=15$. Thus, the equation of the required plane is 7x + 11y + 14z = 15. Also, the equation of the plane passing through the point (2, 3, −1) and parallel to the plane 7x + 11y + 14z = 15 is $7\left(x-2\right)+11\left(y-3\right)+14\left(z+1\right)=0\phantom{\rule{0ex}{0ex}}⇒7x+11y+14z=33$ #### Question 19: Find the equation of the plane through the line of intersection of the planes $x+y+z=$1 and 2x$+$3$y+$4$z=$5 and twice of its $y$-intercept is equal to three times its $z$-intercept. The equation of the family of the planes passing through the intersection of the planes xyz = 1 and 2x + 3y + 4z = 5 is (x + y + z − 1) + k(2x + 3y + 4z − 5) = 0, where k is some constant ⇒ (2k + 1)x + (3k + 1)y + (4k + 1)z = 5k + 1 .....(1) It is given that twice of y-intercept is equal to three times its z-intercept. $\therefore 2\left(\frac{5k+1}{3k+1}\right)=3\left(\frac{5k+1}{4k+1}\right)\phantom{\rule{0ex}{0ex}}⇒\left(5k+1\right)\left(8k+2-9k-3\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(5k+1\right)\left(-k-1\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(5k+1\right)\left(k+1\right)=0$ Putting $k=-\frac{1}{5}$ in (1), we get $\left(-\frac{2}{5}+1\right)x+\left(-\frac{3}{5}+1\right)y+\left(-\frac{4}{5}+1\right)z=5×\left(-\frac{1}{5}\right)+1\phantom{\rule{0ex}{0ex}}⇒3x+2y+z=0$ This plane passes through the origin. So, the intercepts made by the plane with the coordinate axes is 0. Hence, this equation of plane is not accepted as twice of y-intercept is not equal to three times its z-intercept. Putting $k=-1$ in (1), we get $\left(-2+1\right)x+\left(-3+1\right)y+\left(-4+1\right)z=5×\left(-1\right)+1\phantom{\rule{0ex}{0ex}}⇒-x-2y-3z=-4\phantom{\rule{0ex}{0ex}}⇒x+2y+3z=4$ Here, twice of y-intercept is equal to three times its z-intercept. Thus, the equation of the required plane is x + 2y + 3z = 4. #### Question 1: Find the distance of the point $2\stackrel{^}{i}-\stackrel{^}{j}-4\stackrel{^}{k}$ from the plane $\stackrel{\to }{r}·\left(3\stackrel{^}{i}-4\stackrel{^}{j}+12\stackrel{^}{k}\right)-9=0.$ #### Question 2: Show that the points are equidistant from the plane $\stackrel{\to }{r}·\left(5\stackrel{^}{i}+2\stackrel{^}{j}-7\stackrel{^}{k}\right)+9=0.$ #### Question 3: Find the distance of the point (2, 3, −5) from the plane x + 2y − 2z − 9 = 0. #### Question 4: Find the equations of the planes parallel to the plane x + 2y − 2z + 8 = 0 that are at a distance of 2 units from the point (2, 1, 1). #### Question 5: Show that the points (1, 1, 1) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0. #### Question 6: Find the equations of the planes parallel to the plane x − 2y + 2z − 3 = 0 and which are at a unit distance from the point (1, 1, 1). #### Question 7: Find the distance of the point (2, 3, 5) from the xy - plane. #### Question 8: Find the distance of the point (3, 3, 3) from the plane $\stackrel{\to }{r}·\left(5\stackrel{^}{i}+2\stackrel{^}{j}-7k\right)+9=0$ #### Question 9: If the product of the distances of the point (1, 1, 1) from the origin and the plane xy + z + λ = 0 be 5, find the value of λ. Disclaimer: The answer or problem given in the text book is incorrect for this. #### Question 10: Find an equation for the set of all points that are equidistant from the planes 3x − 4y + 12z = 6 and 4x + 3z = 7. #### Question 11: Find the distance between the point (7, 2, 4) and the plane determined by the points A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3). [CBSE 2014] The given points are A(2, 5, −3), B(−2, −3, 5) and C(5, 3, −3). The equation of the plane ABC is given by $\left\|\begin{array}{ccc}x-{x}_{1}& y-{y}_{1}& z-{z}_{1}\\ {x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {x}_{3}-{x}_{1}& {y}_{3}-{y}_{1}& {z}_{3}-{z}_{1}\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left\|\begin{array}{ccc}x-2& y-5& z-\left(-3\right)\\ -2-2& -3-5& 5-\left(-3\right)\\ 5-2& 3-5& -3-\left(-3\right)\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left\|\begin{array}{ccc}x-2& y-5& z+3\\ -4& -8& 8\\ 3& -2& 0\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left\|\begin{array}{ccc}x-2& y-5& z+3\\ 1& 2& -2\\ 3& -2& 0\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒-4\left(x-2\right)-6\left(y-5\right)-8\left(z+3\right)=0\phantom{\rule{0ex}{0ex}}⇒2\left(x-2\right)+3\left(y-5\right)+4\left(z+3\right)=0\phantom{\rule{0ex}{0ex}}⇒2x+3y+4z-7=0$ ∴ Distance between the point (7, 2, 4) and the plane $2x+3y+4z-7=0$ = Length of perpendicular from (7, 2, 4) to the plane $2x+3y+4z-7=0$ Thus, the required distance between the given point and the plane is $\sqrt{29}$ units. #### Question 12: A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it. [CBSE 2014] We know that the equation of the plane which makes intercepts a, b and c on the coordinate axes is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$. So, the equation of the plane which makes intercepts −6, 3, 4 on the x-axis, y-axis and z-axis, respectively is $\frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1\phantom{\rule{0ex}{0ex}}⇒-2x+4y+3z=12\phantom{\rule{0ex}{0ex}}⇒2x-4y-3z+12=0$ ∴ Length of the perpendicular from (0, 0, 0) to the plane $2x-4y-3z+12=0$ Thus, the length of the perpendicular from the origin to the plane is $\frac{12}{\sqrt{29}}$ units. #### Question 13: Find the distance of the point (1, $-$2, 4) from plane passing throuhg the point (1, 2, 2) and perpendicular of the planes $x-y+$2$z=$3 and 2$x-$2$y+z+$12$=$0. Let the equation of plane passing through the point (1, 2, 2) be $a\left(x-1\right)+b\left(y-2\right)+c\left(z-2\right)=0$ .....(1) Here, abc are the direction ratios of the normal to the plane. The equations of the given planes are x − y + 2z = 3 and 2x − 2y + z + 12 = 0. Plane (1) is perpendicular to the given planes. a − b + 2c = 0 .....(2) 2a − 2b + c = 0 .....(3) Eliminating ab and c from (1), (2) and (3), we get $\left\|\begin{array}{ccc}x-1& y-2& z-2\\ 1& -1& 2\\ 2& -2& 1\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left(-1+4\right)-\left(y-2\right)\left(1-4\right)+\left(z-2\right)\left(-2+2\right)=0\phantom{\rule{0ex}{0ex}}⇒3x+3y-9=0\phantom{\rule{0ex}{0ex}}⇒x+y-3=0$ ∴ Distance of the point (1, −2, 4) from the plane xy − 3 = 0 #### Question 2: Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane. The equation of the plane passing through the points (0, −1, −1), (4, 5, 1) and (3, 9, 4) is given by #### Question 3: Show that the following points are coplanar. (i) (0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0) (ii) (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1) (i) The equation of the plane passing through points (0, −1, 0), (2, 1, −1), (1, 1, 1) is given by (ii) The equation of the plane passing through (0, 4, 3), (−1, −5, −3), (−2, −2, 1) is #### Question 4: Find the coordinates of the point where the line through (3,$-$4,$-$5) and B (2,$-$3,1) crosses the plane passing through three points L(2,2,1), M(3,0,1) and N(4,$-$1,0). Also, find the ratio in which diveides the line segment AB. Equation of the plane passing through the points L(2, 2, 1), M(3, 0, 1) and N(4, −1, 0) is The equation of line segment through A(3, −4, −5) and B(2−3, 1) is Any point on this line is of the form . This point lies on the plane (1). $\therefore \left[\left(-\lambda +3\right)\stackrel{^}{i}+\left(\lambda -4\right)\stackrel{^}{j}+\left(6\lambda -5\right)\stackrel{^}{k}\right].\left(2\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=7\phantom{\rule{0ex}{0ex}}⇒2\left(-\lambda +3\right)+\left(\lambda -4\right)+\left(6\lambda -5\right)=7\phantom{\rule{0ex}{0ex}}⇒5\lambda =10\phantom{\rule{0ex}{0ex}}⇒\lambda =2$ Thus, the coordinates of the point P are (−2 + 3, 2 − 4, 6 × 2 − 5) i.e. (1, −2, 7). Suppose P divides the line segment AB in the ratio μ : 1. Thus, the point P divides the line segment AB externally in the ratio 2 : 1. #### Question 1: Find the distance between the parallel planes 2xy + 3z − 4 = 0 and 6x − 3y + 9z + 13 = 0. #### Question 2: Find the equation of the plane which passes through the point (3, 4, −1) and is parallel to the plane 2x − 3y + 5z + 7 = 0. Also, find the distance between the two planes. #### Question 3: Find the equation of the plane mid-parallel to the planes 2x − 2y + z + 3 = 0 and 2x − 2y + z + 9 = 0. #### Question 4: Find the distance between the planes #### Question 1: Find the angle between the line $\stackrel{\to }{r}=\left(2\stackrel{^}{i}+3\stackrel{^}{j}+9\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)$ and the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)=5.$ #### Question 2: Find the angle between the line $\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z+1}{1}$ and the plane 2x + yz = 4. #### Question 3: Find the angle between the line joining the points (3, −4, −2) and (12, 2, 0) and the plane 3xy + z = 1. #### Question 4: The line $\stackrel{\to }{r}=\stackrel{^}{i}+\lambda \left(2\stackrel{^}{i}-m\stackrel{^}{j}-3\stackrel{^}{k}\right)$ is parallel to the plane $\stackrel{\to }{r}·\left(m\stackrel{^}{i}+3\stackrel{^}{j}+\stackrel{^}{k}\right)=4.$ Find m. #### Question 5: Show that the line whose vector equation is $\stackrel{\to }{r}=2\stackrel{^}{i}+5\stackrel{^}{j}+7\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)$ is parallel to the plane whose vector equation is $\stackrel{\to }{r}·\left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)=7.$ Also, find the distance between them. #### Question 6: Find the vector equation of the line through the origin which is perpendicular to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+2\stackrel{^}{j}+3\stackrel{^}{k}\right)=3.$ #### Question 7: Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis. #### Question 8: Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line $\frac{x-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}.$ #### Question 9: Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes Disclaimer: The answer given for this problem in the text book is incorrect. The problem should be same as problem #19 to get the text book answer. #### Question 10: Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0. #### Question 11: Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2xy + 3z − 5 = 0. #### Question 12: Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6. #### Question 13: Find the angle between the line $\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}$ and the plane 3x + 4y + z + 5 = 0. #### Question 14: Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane. #### Question 15: State when the line $\stackrel{\to }{r}=\stackrel{\to }{a}+\lambda \stackrel{\to }{b}$ is parallel to the plane $\stackrel{\to }{r}·\stackrel{\to }{n}=d.$ Show that the line $\stackrel{\to }{r}=\stackrel{^}{i}+\stackrel{^}{j}+\lambda \left(3\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)$ is parallel to the plane $\stackrel{\to }{r}·\left(2\stackrel{^}{j}+\stackrel{^}{k}\right)=3.$ Also, find the distance between the line and the plane. #### Question 16: Show that the plane whose vector equation is $\stackrel{\to }{r}·\left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)=1$ and the line whose vector equation is $\stackrel{\to }{r}=\left(-\stackrel{^}{i}+\stackrel{^}{j}+\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+4\stackrel{^}{k}\right)$ are parallel. Also, find the distance between them. #### Question 17: Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z. #### Question 18: Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines $\stackrel{\to }{r}=\left(\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}\right)$ and $\stackrel{\to }{r}=\left(\stackrel{^}{i}-3\stackrel{^}{j}+5\stackrel{^}{k}\right)+\mu \left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)$. Also, find the distance of the point (9, −8, −10) from the plane thus obtained. [CBSE 2014] The equations of the given lines are $\stackrel{\to }{r}=\left(\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}\right)+\lambda \left(2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}\right)$ $\stackrel{\to }{r}=\left(\stackrel{^}{i}-3\stackrel{^}{j}+5\stackrel{^}{k}\right)+\mu \left(\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}\right)$ We know that the vector equation of a plane passing through a point $\stackrel{\to }{a}$ and parallel to $\stackrel{\to }{b}$ and $\stackrel{\to }{c}$ is given by $\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right).\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)=0$. Here, $\stackrel{\to }{a}=\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}$, $\stackrel{\to }{b}=2\stackrel{^}{i}+3\stackrel{^}{j}+6\stackrel{^}{k}$ and $\stackrel{\to }{c}=\stackrel{^}{i}+\stackrel{^}{j}-\stackrel{^}{k}$. $\therefore \stackrel{\to }{b}×\stackrel{\to }{c}=\left\|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ 2& 3& 6\\ 1& 1& -1\end{array}\right\|=-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}$ So, the vector equation of the plane is $\left(\stackrel{\to }{r}-\stackrel{\to }{a}\right).\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left[\stackrel{\to }{r}-\left(\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}\right)\right].\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=0\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}.\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=\left(\stackrel{^}{i}+2\stackrel{^}{j}-4\stackrel{^}{k}\right).\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}⇒\stackrel{\to }{r}.\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=1×\left(-9\right)+2×8+\left(-4\right)×\left(-1\right)=-9+16+4=11$ Thus, the vector equation of the plane is $\stackrel{\to }{r}.\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=11$. The Cartesian equation of this plane is $\left(x\stackrel{^}{i}+y\stackrel{^}{j}+z\stackrel{^}{k}\right).\left(-9\stackrel{^}{i}+8\stackrel{^}{j}-\stackrel{^}{k}\right)=11\phantom{\rule{0ex}{0ex}}⇒-9x+8y-z=11$ Now, Distance of the point (9, −8, −10) from the plane $-9x+8y-z=11$ = Length of perpendicular from (9, −8, −10) from the plane $-9x+8y-z-11=0$ #### Question 19: Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line $\frac{x+3}{2}=\frac{y-3}{7}=\frac{z-2}{5}.$ #### Question 20: Find the coordinates of the point where the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$ intersects the plane xy + z − 5 = 0. Also, find the angle between the line and the plane. #### Question 21: Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}+2\stackrel{^}{j}-5\stackrel{^}{k}\right)+9=0.$ #### Question 22: Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane 10x + 2y − 11z = 3. #### Question 23: Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes #### Question 24: Find the value of λ such that the line $\frac{x-2}{6}=\frac{y-1}{\lambda }=\frac{z+5}{-4}$ is perpendicular to the plane 3xy − 2z = 7. Disclaimer: It should be "parallel" instead of "perpendicular" in the given problem. #### Question 25: Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$. [CBSE 2015] The general equation of the plane passing through the point (−1, 2, 0) is given by $a\left(x+1\right)+b\left(y-2\right)+c\left(z-0\right)=0$ .....(1) If this plane passes through the point (2, 2, −1), we have Direction ratio's of the normal to the plane (1) are a, b, c. The equation of the given line is $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$. This can be re-written as $\frac{x-1}{1}=\frac{y+\frac{1}{2}}{1}=\frac{z+1}{-1}$ Direction ratio's of the line are 1, 1, −1. The required plane is parallel to the given line when the normal to this plane is perpendicular to this line. Solving (2) and (3), we get $\frac{a}{0+1}=\frac{b}{-1+3}=\frac{c}{3-0}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{1}=\frac{b}{2}=\frac{c}{3}=\lambda \left(\mathrm{Say}\right)\phantom{\rule{0ex}{0ex}}⇒a=\lambda ,b=2\lambda ,c=3\lambda$ Putting these values of a, b, c in (1), we have $\lambda \left(x+1\right)+2\lambda \left(y-2\right)+3\lambda \left(z-0\right)=0\phantom{\rule{0ex}{0ex}}⇒x+1+2y-4+3z=0\phantom{\rule{0ex}{0ex}}⇒x+2y+3z=3$ Thus, the equation of the required plane is x + 2y + 3z = 3. #### Question 1: Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the (i) yz-plane (ii) zx-plane #### Question 2: Find the coordinates of the point where the line through (3, −4, −5) and (2, −3, 1) crosses the plane 2x + y + z = 7. #### Question 3: Find the distance of the point (−1, −5, −10) from the point of intersection of the line $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\right)$ and the plane $\stackrel{\to }{r}.\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=5.$ #### Question 4: Find the distance of the point (2, 12, 5) from the point of intersection of the line $\stackrel{\to }{r}=2\stackrel{^}{i}-4\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\right)$ and $\stackrel{\to }{r}.\left(\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right)=0$. [CBSE 2014] The equation of the given line is $\stackrel{\to }{r}=2\stackrel{^}{i}-4\stackrel{^}{j}+2\stackrel{^}{k}+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+2\stackrel{^}{k}\right)$. The position vector of any point on the given line is $\stackrel{\to }{r}=\left(2+3\lambda \right)\stackrel{^}{i}+\left(-4+4\lambda \right)\stackrel{^}{j}+\left(2+2\lambda \right)\stackrel{^}{k}$ .....(1) If this lies on the plane $\stackrel{\to }{r}.\left(\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right)=0$, then $\left[\left(2+3\lambda \right)\stackrel{^}{i}+\left(-4+4\lambda \right)\stackrel{^}{j}+\left(2+2\lambda \right)\stackrel{^}{k}\right].\left(\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(2+3\lambda \right)-2\left(-4+4\lambda \right)+\left(2+2\lambda \right)=0\phantom{\rule{0ex}{0ex}}⇒2+3\lambda +8-8\lambda +2+2\lambda =0\phantom{\rule{0ex}{0ex}}⇒3\lambda =12\phantom{\rule{0ex}{0ex}}⇒\lambda =4$ Putting $\lambda =4$ in (1), we get $\left(2+3×4\right)\stackrel{^}{i}+\left(-4+4×4\right)\stackrel{^}{j}+\left(2+2×4\right)\stackrel{^}{k}$ or $14\stackrel{^}{i}+12\stackrel{^}{j}+10\stackrel{^}{k}$ as the coordinate of the point of intersection of the given line and the plane. The position vector of the given point is $2\stackrel{^}{i}+12\stackrel{^}{j}+5\stackrel{^}{k}$. ∴ Required distance = Distance between $14\stackrel{^}{i}+12\stackrel{^}{j}+10\stackrel{^}{k}$ and $2\stackrel{^}{i}+12\stackrel{^}{j}+5\stackrel{^}{k}$ #### Question 5: Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane $x-y+z=5$. [CBSE 2014, 2015] The equation of the line passing through the points A(2, −1, 2) and B(5, 3, 4) is given by The coordinates of any point on the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda \left(\mathrm{say}\right)$ are $\left(3\lambda +2,4\lambda -1,2\lambda +2\right)$ .....(1) If it lies on the plane $x-y+z=5$, then $3\lambda +2-\left(4\lambda -1\right)+2\lambda +2=5\phantom{\rule{0ex}{0ex}}⇒\lambda +5=5\phantom{\rule{0ex}{0ex}}⇒\lambda =0$ Putting $\lambda =0$ in (1), we get (2, −1, 2) as the coordinates of the point of intersection of the given line and plane. ∴ Required distance = Distance between points (−1, −5, −10) and (2, −1, 2) #### Question 6: Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7. [CBSE 2015] The equation of the line passing through the points A(3, −4, −5) and B(2, −3, 1) is given by The coordinates of any point on the line $\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \left(\mathrm{say}\right)$ are $\left(-\lambda +3,\lambda -4,6\lambda -5\right)$ .....(1) If it lies on the plane 2x + y + z = 7, then $2\left(-\lambda +3\right)+\left(\lambda -4\right)+\left(6\lambda -5\right)=7\phantom{\rule{0ex}{0ex}}⇒5\lambda -3=7\phantom{\rule{0ex}{0ex}}⇒5\lambda =10\phantom{\rule{0ex}{0ex}}⇒\lambda =2$ Putting $\lambda =2$ in (1), we get (1, −2, 7) as the coordinates of the point of intersection of the given line and plane. ∴ Required distance = Distance between points (3, 4, 4) and (1, −2, 7) #### Question 7: Find the distance of the point (1, $-$5, 9) from the plane $x-y+z=$5 measured along the line $x=y=z$. The equation of line parallel to the line xyz and passing through the point (1, −5, 9) is $\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}$ .....(1) Any point on this line is of the form (k + 1, k − 5, k + 9). If (k + 1, k − 5, k + 9) be the point of intersection of line (1) and the given plane, then (k + 1) − (k − 5) + (k + 9) = 5 ⇒ k = −10 So, the point of intersection of line (1) and the given plane is (−10 + 1, −10 − 5, −10 + 9) i.e. (−9, −15, −1). ∴ Required distance = Distance between (1, −5, 9) and (−9, −15, −1) = $\sqrt{{\left(1+9\right)}^{2}+{\left(-5+15\right)}^{2}+{\left(9+1\right)}^{2}}=\sqrt{3×{10}^{2}}=10\sqrt{3}$ units #### Question 1: Write the equation of the plane whose intercepts on the coordinate axes are 2, −3 and 4. #### Question 2: Reduce the equations of the following planes to intercept form and find the intercepts on the coordinate axes. (i) 4x + 3y − 6z − 12 = 0 (ii) 2x + 3y − z = 6 (iii) 2xy + z = 5 #### Question 3: Find the equation of a plane which meets the axes at A, B and C, given that the centroid of the triangle ABC is the point (α, β, γ). #### Question 4: Find the equation of the plane passing through the point (2, 4, 6) and making equal intercepts on the coordinate axes. #### Question 5: A plane meets the coordinate axes at A, B and C, respectively, such that the centroid of triangle ABC is (1, −2, 3). Find the equation of the plane. #### Question 1: Show that the lines are coplanar. Also, find the equation of the plane containing them. #### Question 2: Show that the lines are coplanar. Also, find the equation of the plane containing them. #### Question 3: Find the equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point (0, 7, −7) and show that the line $\frac{x}{1}=\frac{y-7}{-3}=\frac{z+7}{2}$ also lies in the same plane. #### Question 4: Find the equation of the plane which contains two parallel lines #### Question 5: Show that the lines $\frac{x+4}{3}=\frac{y+6}{5}=\frac{z-1}{-2}$ and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection. #### Question 6: Show that the plane whose vector equation is $\stackrel{\to }{r}·\left(\stackrel{^}{i}+2\stackrel{^}{j}-\stackrel{^}{k}\right)=3$ contains the line whose vector equation is $\stackrel{\to }{r}=\stackrel{^}{i}+\stackrel{^}{j}+\lambda \left(2\stackrel{^}{i}+\stackrel{^}{j}+4\stackrel{^}{k}\right).$ #### Question 7: Find the equation of the plane determined by the intersection of the lines #### Question 8: Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line $\stackrel{\to }{r}=\stackrel{^}{i}+3\stackrel{^}{j}-2\stackrel{^}{k}+\lambda \left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right).$ #### Question 9: If the lines are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines. #### Question 10: Find the coordinates of the point where the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$ intersect the plane xy + z − 5 = 0. Also, find the angle between the line and the plane. #### Question 11: Find the vector equation of the plane passing through three points with position vectors Also, find the coordinates of the point of intersection of this plane and the line $\stackrel{\to }{r}=3\stackrel{^}{i}-\stackrel{^}{j}-\stackrel{^}{k}+\lambda \left(2\stackrel{^}{i}-2\stackrel{^}{j}+\stackrel{^}{k}\right).$ #### Question 12: Show that the lines $\frac{5-x}{-4}=\frac{y-7}{4}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{2y-8}{2}=\frac{z-5}{3}$ are coplanar. [CBSE 2014] The equations of the given lines can be re-written as $\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}$ and $\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}$ We know that the lines $\frac{x-{x}_{1}}{{l}_{1}}=\frac{y-{y}_{1}}{{m}_{1}}=\frac{z-{z}_{1}}{{n}_{1}}$ and $\frac{x-{x}_{2}}{{l}_{2}}=\frac{y-{y}_{2}}{{m}_{2}}=\frac{z-{z}_{2}}{{n}_{2}}$ are coplanar if $\left\|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {l}_{1}& {m}_{1}& {n}_{1}\\ {l}_{2}& {m}_{2}& {n}_{2}\end{array}\right\|=0$. Here, ${x}_{1}=5,{y}_{1}=7,{z}_{1}=-3,{x}_{2}=8,{y}_{2}=4,{z}_{2}=5\phantom{\rule{0ex}{0ex}}{l}_{1}=4,{m}_{1}=4,{n}_{1}=-5,{l}_{2}=7,{m}_{2}=1,{n}_{2}=3$ $\therefore \left\|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {l}_{1}& {m}_{1}& {n}_{1}\\ {l}_{2}& {m}_{2}& {n}_{2}\end{array}\right\|\phantom{\rule{0ex}{0ex}}=\left\|\begin{array}{ccc}8-5& 4-7& 5-\left(-3\right)\\ 4& 4& -5\\ 7& 1& 3\end{array}\right\|\phantom{\rule{0ex}{0ex}}=\left\|\begin{array}{ccc}3& -3& 8\\ 4& 4& -5\\ 7& 1& 3\end{array}\right\|\phantom{\rule{0ex}{0ex}}=3\left(12+5\right)+3\left(12+35\right)+8\left(4-28\right)\phantom{\rule{0ex}{0ex}}=51+141-192\phantom{\rule{0ex}{0ex}}=0$ So, the given lines are coplanar. #### Question 13: Find the equation of a plane which passes through the point (3, 2, 0) and contains the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$. [CBSE 2015] Let the equation of the plane passing through (3, 2, 0) be $a\left(x-3\right)+b\left(y-2\right)+c\left(z-0\right)=0$ .....(1) The line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ passes through the point (3, 6, 4) and its direction ratios are proportional to 1, 5, 4. If plane (1) contains this line, then it must pass through (3, 6, 4) and must be parallel to the line. Also, Solving (2) and (3), we get $\frac{a}{4-5}=\frac{b}{1-0}=\frac{c}{0-1}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{-1}=\frac{b}{1}=\frac{c}{-1}=\lambda \left(\mathrm{Say}\right)\phantom{\rule{0ex}{0ex}}⇒a=-\lambda ,b=\lambda ,c=-\lambda$ Putting these values of a, b, c in (1), we get $-\lambda \left(x-3\right)+\lambda \left(y-2\right)-\lambda \left(z-0\right)=0\phantom{\rule{0ex}{0ex}}⇒-x+3+y-2-z=0\phantom{\rule{0ex}{0ex}}⇒-x+y-z+1=0\phantom{\rule{0ex}{0ex}}⇒x-y+z-1=0$ Thus, the equation of the required plane is xy + z − 1 = 0. #### Question 14: Show that the lines $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$ are coplanar. Hence, find the equation of the plane containing these lines. The lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ are coplanar if $\left\|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right\|=0$. The given lines are $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ and $\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}$. Now, $\left\|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right\|=$$\left\|\begin{array}{ccc}-1-\left(-3\right)& 2-1& 5-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right\|=\left\|\begin{array}{ccc}2& 1& 0\\ -3& 1& 5\\ -1& 2& 5\end{array}\right\|=2\left(5-10\right)-1\left(-15+5\right)+0=-10+10+0=0$ So, the given lines are coplanar. The equation of the plane containing the given lines is $\left\|\begin{array}{ccc}x-\left(-3\right)& y-1& z-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left\|\begin{array}{ccc}x+3& y-1& z-5\\ -3& 1& 5\\ -1& 2& 5\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left(x+3\right)\left(5-10\right)-\left(y-1\right)\left(-15+5\right)+\left(z-5\right)\left(-6+1\right)=0\phantom{\rule{0ex}{0ex}}⇒-5\left(x+3\right)+10\left(y-1\right)-5\left(z-5\right)=0\phantom{\rule{0ex}{0ex}}⇒x-2y+z=0$ Thus, the equation of the plane containing the given lines is x − 2yz = 0. #### Question 15: If the line$\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $lx+my-z=$9, then find the value of ${l}^{2}+{m}^{2}$. The line $\frac{x-{x}_{1}}{a}=\frac{y-{y}_{1}}{b}=\frac{z-{z}_{1}}{c}$ lies in the plane AxByCz D = 0 iff (i) $A{x}_{1}+B{y}_{1}+C{z}_{1}+D=0$ and (ii) $aA+bB+cC=0$. It is given that the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $lx+my-z=9$. Also, Solving (1) and (2), we get l = 1 and m = −1 ∴ l2m2 = 12 + (−1)2 = 1 + 1 = 2 Thus, the value of l2 + m2 is 2. #### Question 16: Find the values of $\lambda$ for which the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{{\lambda }^{2}}$ and $\frac{x-3}{1}=\frac{y-2}{{\lambda }^{2}}=\frac{z-1}{2}$ are coplanar. The lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ are coplanar if $\left\|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right\|=0$. The given lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{{\lambda }^{2}}$ and $\frac{x-3}{1}=\frac{y-2}{{\lambda }^{2}}=\frac{z-1}{2}$ are coplanar. $\therefore \left\|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right\|=0$ $⇒\left\|\begin{array}{ccc}3-1& 2-2& 1-\left(-3\right)\\ 1& 2& {\lambda }^{2}\\ 1& {\lambda }^{2}& 2\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left\|\begin{array}{ccc}2& 0& 4\\ 1& 2& {\lambda }^{2}\\ 1& {\lambda }^{2}& 2\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒2\left(4-{\lambda }^{4}\right)-0+4\left({\lambda }^{2}-2\right)=0\phantom{\rule{0ex}{0ex}}⇒-2{\lambda }^{4}+4{\lambda }^{2}=0$ Thus, the values of $\lambda$ are 0, $-\sqrt{2}$ and $\sqrt{2}$. #### Question 17: If the lines $x=$5, $\frac{y}{3-\alpha }=\frac{z}{-2}$ and $x=\alpha$$\frac{y}{-1}=\frac{z}{2-\alpha }$ are coplanar, find the values of $\alpha$. The lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ are coplanar if $\left\|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right\|=0$. The given lines $\frac{x-5}{0}=\frac{y}{3-\alpha }=\frac{z}{-2}$ and $\frac{x-\alpha }{0}=\frac{y}{-1}=\frac{z}{2-\alpha }$ are coplanar. $\therefore \left\|\begin{array}{ccc}\alpha -5& 0-0& 0-0\\ 0& 3-\alpha & -2\\ 0& -1& 2-\alpha \end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left\|\begin{array}{ccc}\alpha -5& 0& 0\\ 0& 3-\alpha & -2\\ 0& -1& 2-\alpha \end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left(\alpha -5\right)\left[\left(3-\alpha \right)×\left(2-\alpha \right)-2\right]-0+0=0\phantom{\rule{0ex}{0ex}}⇒\left(\alpha -5\right)\left({\alpha }^{2}-5\alpha +4\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\alpha -5\right)\left(\alpha -1\right)\left(\alpha -4\right)=0$ Thus, the values of $\alpha$ are 1, 4 and 5. #### Question 18: If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{2}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar, find the equations of the planes containing them. The lines $\frac{x-{x}_{1}}{{a}_{1}}=\frac{y-{y}_{1}}{{b}_{1}}=\frac{z-{z}_{1}}{{c}_{1}}$ and $\frac{x-{x}_{2}}{{a}_{2}}=\frac{y-{y}_{2}}{{b}_{2}}=\frac{z-{z}_{2}}{{c}_{2}}$ are coplanar if $\left\|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right\|=0$. The given lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{2}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar. $\therefore \left\|\begin{array}{ccc}-1-1& -1-\left(-1\right)& 0-0\\ 2& k& 2\\ 2& 2& k\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left\|\begin{array}{ccc}-2& 0& 0\\ 2& k& 2\\ 2& 2& k\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒-2\left({k}^{2}-4\right)-0+0=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-4=0\phantom{\rule{0ex}{0ex}}⇒k=±2$ The equation of the plane containing the given lines is $\left\|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right\|=0$. For k = 2, $\left\|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right\|=$$\left\|\begin{array}{ccc}x-1& y+1& z\\ 2& 2& 2\\ 2& 2& 2\end{array}\right\|=0$ So, no plane exists for k = 2. For k = −2, $\left\|\begin{array}{ccc}x-1& y+1& z\\ 2& k& 2\\ 2& 2& k\end{array}\right\|=0$ $⇒\left\|\begin{array}{ccc}x-1& y+1& z\\ 2& -2& 2\\ 2& 2& -2\end{array}\right\|=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left(4-4\right)-\left(y+1\right)\left(-4-4\right)+z\left(4+4\right)=0\phantom{\rule{0ex}{0ex}}⇒8\left(y+1\right)+8z=0\phantom{\rule{0ex}{0ex}}⇒y+z+1=0$ Thus, the equation of the plane containing the given lines is yz + 1 = 0. #### Question 1: Find the shortest distance between the lines #### Question 2: Find the shortest distance between the lines #### Question 3: Find the shortest distance between the lines $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$ and $3x-y-2z+4=0=2x+y+z+1$. The equation of the plane containing the line $3x-y-2z+4=0=2x+y+z+1$ is If it is parallel to the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$, then $2\left(3+2\lambda \right)+4\left(\lambda -1\right)+\left(\lambda -2\right)=0\phantom{\rule{0ex}{0ex}}⇒9\lambda =0\phantom{\rule{0ex}{0ex}}⇒\lambda =0$ Putting $\lambda =0$ in (1), we get This is the equation of the plane containing the second line and parallel to the first line. Now, the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z+2}{1}$ passes through (1, 3, −2). ∴ Shortest distance between the given lines = Length of the perpendicular from (1, 3, −2) to the plane $3x-y-2z+4=0$ #### Question 1: Find the image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0. #### Question 2: Find the reflection of the point (1, 2, −1) in the plane 3x − 5y + 4z = 5. #### Question 3: Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}.$ Hence, or otherwise, deduce the length of the perpendicular. #### Question 4: Find the image of the point with position vector $3\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}$ in the plane $\stackrel{\to }{r}·\left(2\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=4.$ Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through $3\stackrel{^}{i}+\stackrel{^}{j}+2\stackrel{^}{k}.$ #### Question 5: Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular. #### Question 6: Find the distance of the point (1, −2, 3) from the plane xy + z = 5 measured along a line parallel to $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.$ The given plane is x − y + z = 5. We need to find the distance of the point from the point(1, -2, 3) measured along a parallel line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.$ Let the line from point be P(1, -2, 3) and meet the plane at point Q. Direction ratios of the line from the point â€‹(1, -2, 3) to the given plane will be the same as the given line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.$ So the equation of the line passing through P and with same direction ratios will be: coordinates of any point on the line PQ are . Now, since Q lies on the plane so it must satisfy the equation of the plane. that is, â€‹x − y + z = 5 therefore, 2λ+1 - (3λ - 2) + (-6λ +3) = 5 coordinates of Q are = using distance formula we have the length of PQ as Hence PQ = 1 So, â€‹the distance of the point (1, −2, 3) from the plane x − y + z = 5 is 1 #### Question 7: Find the coordinates of the foot of the perpendicular from the point (2, 3, 7) to the plane 3xyz = 7. Also, find the length of the perpendicular. #### Question 8: Find the image of the point (1, 3, 4) in the plane 2xy + z + 3 = 0. #### Question 9: Find the distance of the point with position vector $-\stackrel{^}{i}-5\stackrel{^}{j}-10\stackrel{^}{k}$ from the point of intersection of the line $\stackrel{\to }{r}=\left(2\stackrel{^}{i}-\stackrel{^}{j}+2\stackrel{^}{k}\right)+\lambda \left(3\stackrel{^}{i}+4\stackrel{^}{j}+12\stackrel{^}{k}\right)$ with the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}-\stackrel{^}{j}+\stackrel{^}{k}\right)=5.$ #### Question 10: Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane $\stackrel{\to }{r}·\left(\stackrel{^}{i}-2\stackrel{^}{j}+4\stackrel{^}{k}\right)+5=0.$ Disclaimer: The answer given for this problem in the text book is incorrect. #### Question 11: Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P (3, 2, 1) from the plane 2xy + z + 1 = 0. Also, find the image of the point in the plane. #### Question 12: Find the direction cosines of the unit vector perpendicular to the plane $\stackrel{\to }{r}·\left(6\stackrel{^}{i}-3\stackrel{^}{j}-2\stackrel{^}{k}\right)+1=0$ passing through the origin. For the unit vector perpendicular to the given plane, we need to convert the given equation of plane into normal form. #### Question 13: Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0. #### Question 14: Find the length and the foot of perpendicular from the point $\left(1,\frac{3}{2},2\right)$ to the plane $2x-2y+4z+5=0$. [NCERT EXEMPLAR] Let M be the foot of the perpendicular from P$\left(1,\frac{3}{2},2\right)$ on the plane $2x-2y+4z+5=0$. Then, PM is the normal to the plane. So, its direction ratios are proportional to 2, −2, 4. Since PM passes through P$\left(1,\frac{3}{2},2\right)$, therefore, its equation is $\frac{x-1}{2}=\frac{y-\frac{3}{2}}{-2}=\frac{z-2}{4}=\lambda \left(\mathrm{Say}\right)$ Let the coordinates of M be $\left(2\lambda +1,-2\lambda +\frac{3}{2},4\lambda +2\right)$. Now, M lies on the plane $2x-2y+4z+5=0$. $\therefore 2\left(2\lambda +1\right)-2\left(-2\lambda +\frac{3}{2}\right)+4\left(4\lambda +2\right)+5=0\phantom{\rule{0ex}{0ex}}⇒24\lambda +12=0\phantom{\rule{0ex}{0ex}}⇒\lambda =-\frac{1}{2}$ So, the coordinates of M are $\left(2×\left(-\frac{1}{2}\right)+1,-2×\left(-\frac{1}{2}\right)+\frac{3}{2},4×\left(-\frac{1}{2}\right)+2\right)$ or $\left(0,\frac{5}{2},0\right)$. Thus, the coordinates of the foot of the perpendicular are $\left(0,\frac{5}{2},0\right)$. Now, $\mathrm{PM}=\sqrt{{\left(1-0\right)}^{2}+{\left(\frac{3}{2}-\frac{5}{2}\right)}^{2}+{\left(2-0\right)}^{2}}=\sqrt{1+1+4}=\sqrt{6}$ Thus, the length of the perpendicular from the given point to the plane is $\sqrt{6}$ units. #### Question 15: Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector $2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}$ to the plane $\stackrel{\to }{r}.\left(2\stackrel{^}{i}+\stackrel{^}{j}+3\stackrel{^}{k}\right)-26=0$. Also find image of P in the plane. Let M be the foot of the perpendicular drawn from the point P(2, 3, 4) in the plane . Then, PM is the normal to the plane. So, the direction ratios of PM are proportional to 2, 1, 3. Since PM passes through P(2, 3, 4) and has direction ratios proportional to 2, 1, 3, so the equation of PM is Let the coordinates of M be (2r + 2, r + 3, 3r + 4). Since M lies in the plane 2x + y + 3z − 26 = 0,so $2\left(2r+2\right)+r+3+3\left(3r+4\right)-26=0\phantom{\rule{0ex}{0ex}}⇒4r+4+r+3+9r+12-26=0\phantom{\rule{0ex}{0ex}}⇒14r-7=0\phantom{\rule{0ex}{0ex}}⇒r=\frac{1}{2}$ Therefore, the coordinates of M are Thus, the position vector of the foot of perpendicular are $3\stackrel{\mathit{^}}{i}+\frac{7}{2}\stackrel{\mathit{^}}{j}+\frac{11}{2}\stackrel{^}{k}$. Now, Length of the perpendicular from P on to the given plane Let be the image of point P in the given plane. Then, the coordinates of M are . But, the coordinates of M are .<\|endoftext\|>	4.40625
400	Voltage on gate controls current that flows between source and drain. Here is how transistor looks on layout. Drain and source are usually of green or yellow color, gate is red. Transistor is formed when poly (red color) crosses diffusion layer. PMOS technology based on p-MOS transistors was invented first. A typical submicron CMOS process uses p-type wafers. A blank 8-inch wafer costs about $100. An IC fabrication process contains a series of masking steps to create layers that define the transistors and metal interconnect. Most layers are formed by the following steps: put layer, protect needed area with photoresist, apply light to transpose, remove photoresist and unprotected layer areas. This photolithografy process is illustrated on the image: n-well and p-well are formed by ion implantation, when ions of dopped silicon n+ or p+ are forced into wafer area with a great speed. During the implantation the ions can get inside to a depth of a few microns. Further transistor layers are formed by diffusion (high T ion migration). Gates are created from poly (polycristalline silicon). Poly has to have a good conductance. To improve the coductance poly is covered with a silicide (a metallic compaund of silicon). SIO2 is an insulator, it also protects the outer layer from oxidation. Layers of poly and SiO2 can be deposited using chemical vapor deposition (CVD). Metal layers can be deposited using sputtering.All these layers are patterned using masks and photolithografy process. Static power consumption is defined by leakage. Leakage is a junction currents caused by thermally generated carriers. Their value increases exponentially with increasing junction T. For example, 85C (a common junction T) results in increase by a factor of 60 over room T.<\|endoftext\|>	3.78125
394	Natural resources are materials created in nature that are used and usable by humans. They include natural substances (e.g., soil, water) and energy supplies (e.g., coal, gas) that serve to satisfy human needs and wants (Barsch and Bürger 1996; Minc 1976). Materials occurring in the environment thus are nothing more than ‘neutral matter’ until people recognise their presence, attach great importance to them, and develop means to capitalise on them. Then the natural materials fulfil a function (Barsch and Bürger 1996; Mitchell 2002). Natural resources are a component of the environmental setting. The environmental setting embraces the totality of materials, features and processes of landscapes. These environmental settings give the regions of the world their own, quite specific and distinctive settings with different options concerning the transmutations of materials and energy. The materials, features and processes which are usable constitute the potential of the landscape. By utilising parts of this potential, humans give them a new function, a new purpose: they make them into a source of subsistence – into a natural resource (Barsch and Bürger 1996). As Zimmermann phrased this process: „Resources are not, they become“ (Zimmermann 1951). Since this concept of natural resources was criticised as too anthropocentric, today many interpret resources much more broadly than in this functional or utilitarian sense. “In that context, resources are the abiotic, biotic and cultural attributes on, in or above the Earth” (Mitchell 2002, 6). Natural resources can be classified in various ways. A commonly used one is the classification of natural resources according to their exhaustibility and regenerative power. Non-renewable resources are differentiated from renewable resources (Barsch and Bürger 1996) (see figure). Figure Classification of natural resources; after data from Enders and Querner 1993.<\|endoftext\|>	3.703125
1,082	PUMPA - THE SMART LEARNING APP Helps you to prepare for any school test or exam Series connection of parallel resistors: A series-parallel circuit is formed by connecting a set of parallel resistors in series. 1. Connect $$R_1$$ and $$R_2$$ in parallel to obtain an effective resistance of $$R_{P1}$$. 2. Similarly, connect $$R_3$$ and $$R_4$$ in parallel to get an effective resistance of $$R_{P2}$$. 3. These parallel segments of resistors are then joined in series. Series-parallel combination of resistors The formula for the effective resistance of the parallel combination of resistors is $\frac{1}{{R}_{P}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{1}}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{2}}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{3}}$ For two resistors in the circuit, the effective resistance is given as $\begin{array}{l}\frac{1}{{R}_{\mathit{P1}}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{1}}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{2}}\\ \\ \frac{1}{{R}_{\mathit{P2}}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{1}}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{2}}\end{array}$ Using the effective resistance of the series circuit, ${R}_{S}=\phantom{\rule{0.147em}{0ex}}{R}_{1}+{R}_{2}+{R}_{3}$, the net effective resistance of the series-parallel combination of resistors is, ${R}_{\mathit{Total}}=\phantom{\rule{0.147em}{0ex}}{R}_{\mathit{P1}}+{R}_{\mathit{P2}}$ Parallel connection of series resistors: A parallel-series circuit is formed by connecting a set of series resistors in parallel. 1. Connect $$R_1$$ and $$R_2$$ in series to get an effective resistance of $$R_{S1}$$. 2. Similarly, connect $$R_3$$ and $$R_4$$ in series to get an effective resistance of $$R_{S2}$$. 3. These series segments of resistors are then joined in parallel. Parallel-series combination of resistors Using the effective resistance of the series circuit, ${R}_{S}=\phantom{\rule{0.147em}{0ex}}{R}_{1}+{R}_{2}+{R}_{3}$, we get $\begin{array}{l}{R}_{\mathit{S1}}=\phantom{\rule{0.147em}{0ex}}{R}_{1}+{R}_{2}\\ \\ {R}_{\mathit{S2}}=\phantom{\rule{0.147em}{0ex}}{R}_{3}+{R}_{4}\end{array}$ Using the formula for the effective resistance of the parallel combination of resistors $\frac{1}{{R}_{P}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{1}}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{2}}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{3}}$, the net effective resistance of parallel-series combination of resistors is $\frac{1}{{R}_{\mathit{Total}}}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{S1}}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}\frac{1}{{R}_{S2}}$.<\|endoftext\|>	4.5
493	# Evaluate the following integrals: Question: Evaluate the following integrals: $\int \frac{\sin 2 x}{\sqrt{\cos ^{4} x-\sin ^{2} x+2}} d x$ Solution: Let $t=\cos ^{2} x$ $d t=2 \cos x \sin x d x=-\sin 2 x d x$ therefore, $\int \frac{\sin 2 x}{\sqrt{\cos ^{4} x-\sin ^{2} x+2}} d x=\int-\frac{d t}{\sqrt{t^{2}-\left(1-t^{2}\right)+2}}$ since, $\left[\sin ^{2} x=1-\cos ^{2} x\right]$ $\int-\frac{d t}{\sqrt{t^{2}-\left(1-t^{2}\right)+2}}=\int-\frac{d t}{\sqrt{t^{2}+t+1}}=\int-\frac{d t}{\sqrt{t^{2}+t+\frac{1}{4}+\frac{3}{4}}}$ $=\int-\frac{d t}{\sqrt{\left(t+\frac{1}{2}\right)^{2}+\frac{3}{4}}}$ Since, $\int \frac{1}{\sqrt{\left(x^{2}-a^{2}\right)}} d x=\log \left[x+\sqrt{\left(x^{2}-a^{2}\right)}\right]+c$ $=\int-\frac{\mathrm{dt}}{\sqrt{\left(\mathrm{t}+\frac{1}{2}\right)^{2}+\frac{3}{4}}}=\log \left[\mathrm{t}+\frac{1}{2}+\sqrt{\left(\mathrm{t}+\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}}+\mathrm{c}\right.$ $=\log \left[t+\frac{1}{2}+\sqrt{t^{2}}+t+1+c=\log \left[\cos ^{2} x+\frac{1}{2}+\right.\right.$ $\sqrt{\cos ^{4} x+\cos ^{2} x+1}+c$<\|endoftext\|>	4.5
639	## Project Euler & HackerRank Problem 6 Solution Sum square difference by {BetaProjects} \| MAY 17, 2009 \| Project Euler & HackerRank ### Project Euler Problem 6 Statement The sum of the squares of the first ten natural numbers is 1² + 2² + … + 10² = 385. The square of the sum of the first ten natural numbers is (1 + 2 + … + 10)² = 55² = 3025. Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 − 385 = 2640. Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum. ### Solution #### A simplified approach print (sum(range(101))*2 − sum(ii for i in range(101))) Using loops to sum our objectives, and Python’s list comprehension structure, we can make a simple subtraction and print the difference. For the small upper bound of 100 it takes no time at all. But, this lacks the efficiency to solve the more demanding parameters of the HackerRank version. A better method will have to be discovered. #### Calculating our objectives without looping Solving this problem without using loops is achieved by implementing the following two established equations: 1. The sum of the first n natural numbers (called triangular numbers, used in Project Euler Problem 1): For a visual example of triangular numbers, imagine a set of bowling pins arranged in the traditional triangular pattern: Let’s apply this equation to our example: The square of the sum of the first ten natural numbers is 10(11)/2 = 55, and 552 = 3025 ✓. 2. The sum of the first n square numbers (square pyramidal numbers): For a visual example of square pyramidal numbers, imagine a stack of marbles as shown below: Now, let’s apply this equation to our example: The sum of the squares of the first 10 natural numbers is 10(11)(21)/6 = 385 ✓. Finally, calculate the difference, 3025 – 385 = 2640 ✓. Knowing these equations is useful for solving other programming problems. #### A compound formula Now, I guess for the purists, one could perform the subtraction of these two summations and derive the following formula: And giving it a try with our example: 10(9)(11)(32) / 12 = 2640 ✓. #### HackerRank version HackerRank requires you to solve up to 10,000 test cases with a higher limit of n ≤ 10,000. This is solved quickly by avoiding loops and using the closed–form calculations described above. ### Python Source 1. # -- coding: utf-8 -- 2. n = int(input('first n natural numbers, n =')) 3. print ("Difference: (Σn)² - Σn² =", n(n-1)(n+1)(3n+2)//12)<\|endoftext\|>	4.6875
464	# A starting lineup consists of 2 forwards, 2 guards and 1 center. How many different starting lineups.. A certain school has $4$ forwards, $4$ guards, $3$ centers and $1$ person who can play as either a forward or a guard. How many different starting lineups can be made? I came up with 2 answers to this problem. However I don't know which one is right and I can't tell the difference between the two: Solution 1: There are two possibilities, X is a forward, in which there is $\binom{5}{2}\binom{4}{2}\binom{3}{1} = 180$ ways of making this starting lineup. X could be a guard as well, which results in the same number, $180$ ways of making the starting lineup. Add together to get $360$ different ways of making this starting lineup. Solution 2: There are three possibilities which encompass all possible starting lineups: x is not picked, x is picked as a forward, and x is picked as a guard. When x is not picked, there is $\binom{4}{2}\binom{4}{2}\binom{3}{1} = 108$ different lineups without x in it. when x is picked as a forward, you only need to pick one more forward, so there is $\binom{4}{1}\binom{4}{2}\binom{3}{1} = 72$ different lineups with x as forward. the same number will result when you pick x as a guard: $72$. adding $108+72+72$ results in $252$ different lineups. So the problem is that I can't see the fault in logic in either of my solutions. Which one is the right one? edit: centers The problem is that in solution $1$ you are counting the cases in which $X$ does not play twice, once when he doesn't play as a forward and once when he doesn't play as a guard. Therefore the answer should be $360$ minus the number of line-ups in which he doesn't play, this yields $360-108=252$ line-ups.<\|endoftext\|>	4.40625
1,142	# Comparing Three Digit Numbers 23 teachers like this lesson Print Lesson ## Objective SWBAT compare three digit numbers using greater than, less than, and equal signs. #### Big Idea Students use their knowledge of place value to compare three digit numbers. ## Hook 5 minutes I am going to write 2 two digit numbers on the board (145 and 152). I want you to build these numbers using your place value blocks . Now, turn and tell your partner which number is greater. Make sure to use place value vocabulary when you are telling your partner. As students are discussing, I circulate and ask students questions: --Why is 152 greater if 145 has more ones than 152? --How do you know that 152 is greater when they have the same number of hundreds? After students have talked with their partner for 1-2 minutes, I ask 2-3 students to stand up and share their answers. As students share I make sure to highlight students who are using place value vocabulary to discuss which number is larger. (I.e: a student may respond by saying: I know 152 is greater because even though they have the same number of hundreds, 152 has more tens than 145). I encourage students to use their place value blocks to help explain their answers. ## Problem of the Day 10 minutes I write a comparison statement on the board. 324_________ 432 Turn and Talk: What strategies can we use to compare these two numbers? I ask three students to share their strategies. As students share their strategies, I write their strategies on the board for future use. If students are struggling to explain their strategies clearly, I ask them: (1) Is 324 greater than 432 because (pointing to the 4 in the ones place), four is larger than two? (2) Do we only need to look at the hundreds place to compare these numbers? (3) Could we use an addition or subtraction strategy to determine which is greater? Students need to grasp that, when comparing numbers we start comparing in the hundreds place since hundreds are greatest—students will run into significant trouble if they start comparing in the ones place. When finished solving and discussing the first problem, I put another problem on the board: 235 ___________ 235 Turn and Talk: What sign should I put here? WHY? Students need to be able to explain WHY these numbers are equal to one another. They may explain using place value terminology (there are equal number of ones, tens, and hundreds) or they may use another strategy. ## Guided Practice (Card Game!) 10 minutes In pairs, students work on a Comparing Numbers Game. This is the first time we've played the game, so I model as I explain the game to students. 1) Each student receives a page of numbers. Each student cuts his/her page of numbers to create a "deck" of numbers. (I actually hold up the paper, and cut one or two examples out.) 2) In pairs, each student draws and places one number between the partners, and then they work together to determine which sign would be the correct one to show the comparison between the numbers. [Note: students who are struggling can use cubes or place value blocks to help them.] (I have a student join me, and interactively model this step.) 3) Which ever student has the greatest number, takes both number cards. 4) Students play until they run out of number cards. As students play the game, I walk around and ask questions: Can you explain why this number is _________ (equal to, less than, greater than, greatest, lesser)? ## Independent Practice 15 minutes Independent practice is tiered based on student understanding of this concept. Group A: In need of intervention Students in the A group work with the teacher to compare numbers 100-200. I work with these students to support them in using place value blocks to establish quantities. These are used to compare numbers. My role is to facilitate the process, so that students make meaning of beginning the comparison with the hundreds place, and using that information to decide what needs to be done next. As they make meaning of the quantities, I am also supporting the use of place value vocabulary when explaining their thinking. Group B: On level Students in the B group work independently or in partners to compare numbers 200-600. They use place value blocks, if they choose, to help visualize the size of the number. When discussing / explaining their thinking, they will use place value vocabulary. Group C: Extension Students in the C group work independently or in partners to compare numbers 200-1200. They are not using place value blocks since students in this group should be able to visualize thousands, hundreds, tens, and ones in their minds. Students in this group also are expected to use place value vocabulary to explain why each number is greater or less. ## Closing 5 minutes Today we have worked on comparing numbers. Before we finish our math block, I have one last problem for us to work on together. Write : 252 ___ 248. Use your knowledge of place value to tell your partner which number is greater and why. After 1 minute of sharing ask one or two students to share their answer. Tomorrow we will continue comparing numbers using our place value knowledge to decide which number is greater!<\|endoftext\|>	4.75
459	Name and explain some poetic devices, with examples. -Metaphor: A comparison between two objects with the intent of giving clearer meaning to one of them. Example: "all the world's a stage" -Onomatopoeia: The use of words which imitate sound. Example: thundering sea, clinking glasses. -Simile: A comparison between two objects using a specific word or comparison such as "like", "as", or "than." Example: as red as a rose, as busy as a bee. -Alliteration: The repetition of initial consonant sounds. Examples: "For the sky and the sea, and the sea and the sky" What is the difference between revenue, income, and gain? Revenue is the amount earned from a company's main activities. For example: Selling furniture for a furniture manufacturer, or selling teaching services for a coaching center. A gain results from a peripheral activity, such as selling any obsolete fixed assets. For example the money realized on the sale of an old fax machine. A gain is the amount received that is in excess of the asset's book value. For example, if the company receives $200 for the machine, and its book value was $50, the company will report a gain of $150. Income means "net of revenues and expenses." For example, a furniture manufacturers income from operations is sales minus the cost of goods sold minus operating expenses. What is the difference between the Average Revenue (AR) and Marginal Revenue (MR) curves under the market forms of Monopoly and Monopolistic Competition? Monopoly and Monopolistic competition both fall under the category of imperfect competition. Therefore, AR and MR curves slope downwards in both cases as more units can be sold only by reducing the price. The key difference lies in the elasticity of the curves. Under monopolistic competition, AR and MR curves are more elastic due to the presence of close substitutes in the market. Under monopoly, there is a complete absence of close substitutes, therefore, the curve is relatively less elastic. This implies that when the price of a commodity is increased in both markets, the proportionate fall in demand will be more in case of monopolistic competition, and less for monopoly.<\|endoftext\|>	4.03125
203	The Cretaceous was much warmer and more humid than today. Very active volcanism and high rates of seafloor spreading This was the period of the worlds greatest mass extinction. During the Cretaceous, dinosaurs ruled the earth. - land was controlled by dinosaurs like Tyrannosaurus Rex, Velocoraptors, and other predators. - Water was controlled by a large variety of sea creatures. There was many dinosaurs in the sea. there were huge sharks, huge alligators, and even ol' nessy the lochness monster's family. The earth was a very different place back then. The continents were still spreading apart, and the world was changing. Snakes, lizards, and most fish developed modern characteristics that you would see today. Predators thrived throughout the cretaceous until their world suddenly came to an end... During the cretaceous era was the most deadly mass extinction ever on the face of the world.<\|endoftext\|>	3.75
2,817	Respiratory failure is one of the most common reasons for admission to the intensive care unit (ICU) and a common comorbidity in patients admitted for acute care. What’s more, it’s the leading cause of death from pneumonia and chronic obstructive pulmonary disease (COPD) in the United States. This article briefly reviews the physiologic components of respiration, differentiates the main types of respiratory failure, and discusses medical treatment and nursing care for patients with respiratory failure. Physiologic components of ventilation and respiration The lung is highly elastic. Lung inflation results from the partial pressure of inhaled gases and the diffusion-pressure gradient of these gases across the alveolar-capillary membrane. The lungs play a passive role in breathing, but ventilation requires muscular effort. When the diaphragm contracts, the thoracic cavity enlarges, causing the lungs to inflate. During forced inspiration when a large volume of air is inspired, external intercostal muscles act as a second set of inspiratory muscles. Accessory muscles in the neck and chest are the last group of inspiratory muscles, used only for deep and heavy breathing, such as during intense exercise or respiratory failure. During expiration, the diaphragm relaxes, decreasing thoracic cavity size and causing the lungs to deflate. With normal breathing, expiration is purely passive. But with exercise or forced expiration, expiratory muscles (including the abdominal wall and internal intercostal muscles) become active. These important muscles are necessary for coughing. Respiration—the process of exchanging oxygen (O2) and carbon dioxide (CO2)—involves ventilation, oxygenation, and gas transport; the ventilation/perfusion (V/Q) relationship; and control of breathing. Respiration is regulated by chemical and neural control systems, including the brainstem, peripheral and central chemoreceptors, and mechanoreceptors in skeletal muscle and joints. (See Control of breathing.) A dynamic process, ventilation is affected by the respiratory rate (RR) and tidal volume—the amount of air inhaled and exhaled with each breath. Pulmonary ventilation refers to the total volume of air inspired or expired per minute. Not all inspired air participates in gas exchange. Alveolar ventilation—the volume of air entering alveoli taking part in gas exchange—is the most important variable in gas exchange. Air that distributes to the conducting airways is deemed dead space or wasted air because it’s not involved in gas exchange. (See Oxygenation and gas transport.) Ultimately, effective ventilation is measured by the partial pressure of CO2 in arterial blood (Paco2). All expired CO2 comes from alveolar gas. During normal breathing, the breathing rate or depth adjusts to maintain a steady Paco2 between 35 and 45 mm Hg. Hyperventilation manifests as a low Paco2; hypoventilation, as a high Paco2. During exercise or certain disease states, increasing breathing depth is far more effective than increasing the RR in improving alveolar ventilation. Lung recoil and compliance The lungs, airways, and vascular trees are embedded in elastic tissue. To inflate, the lung must stretch to overcome these elastic components. Elastic recoil—the lung’s ability to return to its original shape after stretching from inhalation—relates inversely to compliance. Lung compliance indirectly reflects lung stiffness or resistance to stretch. A stiff lung, as in pulmonary fibrosis, is less compliant than a normal lung. With reduced compliance, more work is required to produce a normal tidal volume. With extremely high compliance, as in emphysema where there is loss of alveolar and elastic tissue, the lungs inflate extremely easily. Someone with emphysema must expend a lot of effort to get air out of the lungs because they don’t recoil back to their normal position during expiration. In both pulmonary fibrosis and emphysema, inadequate lung ventilation leads to hypercapnic respiratory failure. Respiratory failure occurs when one of the gas-exchange functions—oxygenation or CO2 elimination—fails. A wide range of conditions can lead to acute respiratory failure, including drug overdose, respiratory infection, and exacerbation of chronic respiratory or cardiac disease. Respiratory failure may be acute or chronic. In acute failure, life-threatening derangements in arterial blood gases (ABGs) and acid-base status occur, and patients may need immediate intubation. Respiratory failure also may be classified as hypoxemic or hypercapnic. Clinical indicators of acute respiratory failure include: - partial pressure of arterial oxygen (Pao2) below 60 mm Hg, or arterial oxygen saturation as measured by pulse oximetry (Spo2) below 91% on room air - Paco2 above 50 mm Hg and pH below 7.35 - Pao2 decrease or Paco2 increase of 10 mm Hg from baseline in patients with chronic lung disease (who tend to have higher Paco2 and lower PaO2 baseline values than other patients). In contrast, chronic respiratory failure is a long-term condition that develops over time, such as with COPD. Manifestations of chronic respiratory failure are less dramatic and less apparent than those of acute failure. Three main types of respiratory failure The most common type of respiratory failure is type 1, or hypoxemic respiratory failure (failure to exchange oxygen), indicated by a Pao2 value below 60 mm Hg with a normal or low Paco2 value. In ICU patients, the most common causes of type 1 respiratory failure are V/Q mismatching and shunts. COPD exacerbation is a classic example of V/Q mismatching. Shunting, which occurs in virtually all acute lung diseases, involves alveolar collapse or fluid-filled alveoli. Examples of type 1 respiratory failure include pulmonary edema (both cardiogenic and noncardiogenic), pneumonia, influenza, and pulmonary hemorrhage. (See Ventilation and perfusion: A critical relationship.) Type 2, or hypercapnic, respiratory failure, is defined as failure to exchange or remove CO2, indicated by Paco2 above 50 mm Hg. Patients with type 2 respiratory failure who are breathing room air commonly have hypoxemia. Blood pH depends on the bicarbonate level, which is influenced by hypercapnia duration. Any disease that affects alveolar ventilation can result in type 2 respiratory failure. Common causes include severe airway disorders (such as COPD), drug overdose, chest-wall abnormalities, and neuromuscular disease. Type 3 respiratory failure (also called perioperative respiratory failure) is a subtype of type 1 and results from lung or alveolar atelectasis. General anesthesia can cause collapse of dependent lung alveoli. Patients most at risk for type 3 respiratory failure are those with chronic lung conditions, excessive airway secretions, obesity, immobility, and tobacco use, as well as those who’ve had surgery involving the upper abdomen. Type 3 respiratory failure also may occur in patients experiencing shock, from hypoperfusion of respiratory muscles. Normally, less than 5% of total cardiac output flows to respiratory muscles. But in pulmonary edema, lactic acidosis, and anemia (conditions that commonly arise during shock), up to 40% of cardiac output may flow to the respiratory muscles. Signs and symptoms of respiratory failure Patients with impending respiratory failure typically develop shortness of breath and mental-status changes, which may present as anxiety, tachypnea, and decreased Spo2 despite increasing amounts of supplemental oxygen. Acute respiratory failure may cause tachycardia and tachypnea. Other signs and symptoms include periorbital or circumoral cyanosis, diaphoresis, accessory muscle use, diminished lung sounds, inability to speak in full sentences, an impending sense of doom, and an altered mental status. The patient may assume the tripod position in an attempt to further expand the chest during the inspiratory phase of respiration. In chronic respiratory failure, the only consistent clinical indictor is protracted shortness of breath. Be aware that pulse oximetry measures the percentage of hemoglobin saturated with oxygen, but it doesn’t give information about oxygen delivery to the tissues or the patient’s ventilatory function. So be sure to consider the patient’s entire clinical presentation. Compared to SpO2, an ABG study provides more accurate information on acid-base balance and blood oxygen saturation. Capnography is another tool used for monitoring patients receiving anesthesia and in critical care units to assess a patient’s respiratory status. It directly monitors inhaled and exhaled concentration of CO2 and indirectly monitors Paco2. Treatment and management In acute respiratory failure, the healthcare team treats the underlying cause while supporting the patient’s respiratory status with supplemental oxygen, mechanical ventilation, and oxygen saturation monitoring. Treatment of the underlying cause, such as pneumonia, COPD, or heart failure, may require diligent administration of antibiotics, diuretics, steroids, nebulizer treatments, and supplemental O2 as appropriate. For chronic respiratory failure, despite the wide range of chronic or end-stage pathology present (such as COPD, heart failure, or systemic lupus erythematosus with lung involvement), the mainstay of treatment is continuous supplemental O2, along with treatment of the underlying cause. Nursing care can have a tremendous impact in improving efficiency of the patient’s respiration and ventilation and increasing the chance for recovery. To detect changes in respiratory status early, assess the patient’s tissue oxygenation status regularly. Evaluate ABG results and indices of end-organ perfusion. Keep in mind that the brain is extremely sensitive to O2 supply; decreased O2 can lead to an altered mental status. Also, know that angina signals inadequate coronary artery perfusion. In addition, stay alert for conditions that can impair O2 delivery, such as elevated temperature, anemia, impaired cardiac output, acidosis, and sepsis. As indicated, take steps to improve V/Q matching, which is crucial for improving respiratory efficiency. To enhance V/Q matching, turn the patient on a regular and timely basis to rotate and maximize lung zones. Because blood flow and ventilation are distributed preferentially to dependent lung zones, V/Q is maximized on the side on which the patient is lying. Regular, effective use of incentive spirometry helps maximize diffusion and alveolar surface area and can help prevent atelectasis. Regular rotation of V/Q lung zones by patient turning and repositioning enhances diffusion by promoting a healthy, well-perfused alveolar surface. These actions, as well as suctioning, help mobilize sputum or secretions. Patients in respiratory failure have unique nutritional needs and considerations. Those with acute respiratory failure from primary lung disease may be malnourished initially or may become malnourished from increased metabolic demands or inadequate nutritional intake. Malnutrition can impair the function of respiratory muscles, reduce ventilatory drive, and decrease lung defense mechanisms. Clinicians should consider nutritional support and individualize such support to ensure adequate caloric and protein intake to meet the patient’s respiratory needs. Patient and family education Provide appropriate education to the patient and family to promote adherence with treatment and help prevent the need for readmission. Explain the purpose of nursing measures, such as turning and incentive spirometry, as well as medications. At discharge, teach patients about pertinent risk factors for their specific respiratory condition, when to return to the healthcare provider for follow-up care, and home measures they can take to promote and maximize respiratory function. Cooke CR, Erikson SE, Eisner MD, Martin GS. Trends in the incidence of noncardiogenic acute respiratory failure: the role of race. Crit Care Med. 2012;40(5):1532-8. Gehlbach BK, Hall JB. Respiratory failure and mechanical ventilation. In Porter RS, ed. The Merck Manual. 19th ed. West Point, PA; Merck Sharp & Dohme Corp.; 2011. Kaynar AM. Respiratory failure treatment and management. Updated August 14, 2014. http://emedicine.medscape.com/article/167981-treatment#aw2aab6b6b2. Accessed August 23, 2014. Kress JP, Hall JB: Approach to the patient with critical illness. In Longo DL, Fauci AS, Kasper DL, et al., eds. Harrison’s Principles of Internal Medicine. 18th ed. New York, NY: McGraw-Hill Professional; 2012. Pinson R. Revisiting respiratory failure, part 1. ACP Hosp. 2013;Oct:5-6. www.acphospitalist.org/archives/2013/10/coding.htm. Accessed August 23, 2014. Pinson R. Revisiting respiratory failure, part 2. ACP Hosp. 2013;Nov:7-8. www.acphospitalist.org/archives/2013/11/coding.htm. Accessed October 3, 2014. Schraufnagel D. Breathing in America: Diseases, Progress, and Hope. American Thoracic Society; 2010. Michelle Fournier is director of clinical consulting with Nuance/J.A. Thomas & Associates in Atlanta, Georgia.<\|endoftext\|>	3.671875
879	# 4.3 – Location Zeros of Polynomials. At times, finding zeros for certain polynomials may be difficult There are a few rules/properties we can use to help. ## Presentation on theme: "4.3 – Location Zeros of Polynomials. At times, finding zeros for certain polynomials may be difficult There are a few rules/properties we can use to help."— Presentation transcript: 4.3 – Location Zeros of Polynomials At times, finding zeros for certain polynomials may be difficult There are a few rules/properties we can use to help us at least determine some rough information pertaining to the zeros Rational Zero Theorem For a polynomial f(x) = a n x n + a n-1 x n-1 + … + a 1 x + a 0, then any rational zero must be of the form (p/q), where p is a factor of the constant term (a 0 ) and q is a factor of the leading coefficient a n – All zeros are ratios of the constant and leading coefficient Example. For the function f(x) = 2x 3 + 5x 2 – 4x - 3, list the potential rational zeros. Descartes’ Rule of Signs = for a polynomial, a variation in sign is a change in the sign of one coefficient to the next – 1) The number of positive real zeros is either the number of variation in sign or is less than this number by a positive integer (MAXIMUM, +) – 2) The number of negative real zeros is the number of variations in the sign of f(-x) or less than this by a positive integer (MAXIMUM, -) Example. For the function f(x) = 2x 3 + 5x 2 – 4x - 3, determine the number of possible positive and negative zeros. Example. For the function f(x) = x 3 – 6x 2 + 13x – 20, determine the number of possible positive and negative zeros. Intermediate Value Theorem = For a polynomial f(x), if f(a) and f(b) are different in signs, and a < b, then there lies at least one zero between a and b – Most useful! Example. Show that f(x) = x 3 + 3x – 7 has a zero between 1 and 2. Multiplicity of Zeros/Roots Suppose that c is a zero for the polynomial f(x) For (x – c) k, or a root of multiplicitity k, the following rules are known: – The graph of f will touch the axis at (c,0) – Cross through the x-axis if k is odd – Stay on the same side of the x-axis if k is even As k gets larger than 2, the graph will flatten out Example. Sketch the graph of: f(x) = (x + 2)(x + 1) 2 (x – 3) 3 Conjugate Pairs Recall, if you have the complex number a + bi, then the conjugate is a – bi If a polynomial f(x) has the imaginary zero a + bi, then the polynomial also has the conjugate zero, a – bi – If x – (a + bi) is a factor, so is x – (a – bi) Example. Construct a fourth degree polynomial function with zeros of 2, -5, and 1 + i, such that f(1) = 12. Conclusion: Still best to use, when applicable, any graphing utility to find the zeros of functions. However, when you’re stuck, these properties give us a wide range of potential values to choose from, and methods to narrow down the list of “candidate zeros.” Assignment Pg. 337 For the problems, do the following: 1) List the potential zeros 2) Show # of + or – zeros 3) Use your graphing calculator to identify the zeros #58, 60, 64 Download ppt "4.3 – Location Zeros of Polynomials. At times, finding zeros for certain polynomials may be difficult There are a few rules/properties we can use to help." Similar presentations<\|endoftext\|>	4.78125
1,071	July 16, 2012 feature Scientists analyze potential of using lasers to make rain Nevertheless, these cloud seeding techniques are controversial, both for their effectiveness at inducing rainfall and for their possible harmful side effects. In a recent review paper published in the IOP’s Journal of Physics D: Applied Physics, a team of scientists from Switzerland and Germany has examined the latest results of laser-induced condensation and discussed the future of the field. One of the first successful demonstrations of laser-induced condensation (http://www.nature.com/ncomms/journal/v2/n8/full/ncomms1462.html) came just last year, when researchers – including the authors of the current review – used a powerful laser to produce tiny water particles in moderately humid air. The water particles were just a few micrometers in diameter, which is about 100 times too small to fall as rain droplets. However, the experiments demonstrated the ability to transform particles in a gas phase to a liquid phase through condensation, and larger droplets are expected to be feasible. “At this stage, our work clearly shows that lasers can induce the formation of tiny particles,” Jérôme Kasparian of the University of Geneva in Switzerland told Phys.org. “This is not, at least at this stage, efficient cloud seeding for making rain, but rather a newly opened direction for research in this direction.” As the researchers explain in this review, laser-induced condensation owes its feasibility in part to the rapid improvement in laser power in recent years. Over the past decade, commercially available laser power has increased by two orders of magnitude, reaching the petawatt level today. Scientists expect laser powers on the exawatt scale in the foreseeable future. In last year’s demonstration, the researchers performed experiments using a 100-TW Draco laser and 5-TW mobile laser called Teramobile, which is the size of a 20-foot freight container. In addition to more powerful lasers, improving the results will also require a better understanding of the underlying mechanism of laser-induced condensation. The technique involves photodissociation, in which photons break down atmospheric compounds in the atmosphere. This process produces ozone and nitrogen oxides, which lead to the formation of nitric acid particles that bind water molecules together to create water droplets. Understanding the details of how this process stimulates particle growth, as well as how atmospheric conditions affect the process, are the most challenging questions in this field, according to the scientists. “Making rain would require first to have tiny water particles grow into droplets with a size sufficient to fall as raindrops,” Kasparian said. “This depends on the atmospheric conditions, in particular the relative humidity, that these particles will encounter. For example, if the air mass in which the particles have been produced lifts along a mountain hill, it will cool down and condensation will be favored. “Making rain would also require the production of an adequate number density of particles. If there are too few particles, we would only get a few drops at most. On the other hand, if there are too many particles, they will compete with each other to grab the water molecules available in the atmosphere. Ultimately, none of them will grow sufficiently to make raindrops, which may even reduce precipitation. “Finally, the technique would also need to activate a large volume of the atmosphere, i.e., to sweep the laser sufficiently fast.” Despite these challenges, the scientists also noted that using lasers to induce rain has its advantages, particularly its minimal side effects compared to other techniques. For example, cloud seeding methods that involve injecting silver iodide particles into clouds run the risk of having unintended consequences for the surrounding atmosphere, a problem that lasers avoid. Laser-assisted methods also offer better control than chemical methods, since the lasers can be turned on and off and precisely positioned. This control also makes it easier to determine how effective the technique is, since critics often question whether rain might have occurred even without intervention. In the future, the researchers recommend investigating the ability of lasers to seed clouds on a larger scale. Such a task will require further experimental data as well as theoretical modeling. “Our aim now is to tackle the questions that remain open, especially to determine the optimal laser conditions to maximize the condensation process, and to assess for the possibility to obtain macroscopic quantities of condensed water,” Kasparian said. “This also requires an understanding of the physical mechanisms at the root of laser-induced condensation, with the ultimate goal of being able to model the process quantitatively. “Besides the technical feasibility, as discussed above, further experiments will allow us to assess whether laser rainmaking could be cost-effective. This can easily be expressed in terms of the cost per unit rainwater volume obtained. This will depend very much on the ultimate laser power required to get a significant amount of water, which we need to further investigate.” Copyright 2012 Phys.org All rights reserved. This material may not be published, broadcast, rewritten or redistributed in whole or part without the express written permission of PhysOrg.com.<\|endoftext\|>	3.703125
266	A river’s sinuosity is its tendency to meander back and forth across its floodplain, in an S-shaped pattern, over time. As the stream moves across the landscape, it may leave behind evidence of where the river channel once was (these can take the form of meander scars or oxbow lakes). These patterns usually appear in stream channels found in softer sediments. If a river’s course is bedrock-controlled, other factors—primarily rock strength and structure—control the river’s flow. Few stream courses are completely straight, and most exhibit meanders. If you ever work with hydrologic data in GIS, you may wish to determine the sinuosity for an entire river or a particular ‘reach‘ of a river of interest. A stream that doesn’t meander at all has a sinuosity of 1. The more meanders in a stream, the closer the sinuosity value will get to 0. Fortunately, it’s simple to determine the sinuosity of a line using either the field calculator or Python. Depending on the version of GIS software you are using, the method differs. See this post for details for ArcView 3 (old!) and ArcMap 8.x-10.2.<\|endoftext\|>	3.984375
320	- Motion occurs when there is a change in position of an object with respect to a reference starting point. - The final position of an object is determined by measuring the change in position and direction of the segments along a trip. The following terms are used to describe and determine motion: - Position is the location of an object. - An object changes position if it moves relative to a reference point (starting point). - The change in position is determined by the distance and direction of an object’s change in position from the starting point (displacement). - Direction is the line, or path along which something is moving, pointing, or aiming. - Direction is measured using a reference point with terms such as up, down, left, right, forward, backward, toward, away from, north, south, east, or west. - The slope (slant or angle) of the line can tell the relative speed of the object. - When the slope of the line is steep, the speed is faster than if the slope were flatter. - When the slope of the line is flatter, the speed is slower. - When the slope of the line is horizontal to the x-axis, the speed is zero (the object is not moving). - A graph used to show a change in an object’s location over time. - For this type of graph, time (the independent variable) is plotted on the x-axis and the position (the dependent variable) is plotted on the y-axis.<\|endoftext\|>	4.34375
861	# Find all $x$ such that $x^{35} + 5x^{19} + 11x^3$ is divisible by 17. Find all $x$ such that $x^{35} + 5x^{19} + 11x^3$ is divisible by 17. I think we can use the fact that we can mod everything by 17 and want 0. But how exactly should we go about doing this? \begin{align}x^{35}+5x^{19}+11x^3 &= x^{17}\cdot x^{17} \cdot x + 5 x^{17}\cdot x^2 + 11 x^3\\ &\equiv x\cdot x \cdot x + 5x \cdot x^2 + 11 x^3 \;(\mathrm{mod}\;17) = 17x^3\\ &\equiv 0\;(\mathrm{mod}\; 17) \end{align} where we apply Fermat's little theorem a few times ($a^p \equiv a \;(\mathrm{mod}\; p)$). Therefore it works for all integers $x$. $$x^{17}+5x^{19}+11x^3=(x^{18}+6x^2)(x^{17}-x)+17x^3$$ Since $x^{17}-x$ is divisible by $17$ and so is $17x^3$, so is the original polynomial. Any integer value for $x$ will do ! Observe that \begin{eqnarray} x^{16} \equiv 1 \pmod{17} \end{eqnarray} and the expression can be written as $x^3(11+5x^{16}+x^{32})$ so the bracket will always be divisible by $17$. Lil' Fermat says this is the same as solving the equation in the field $\mathbf Z/17\mathbf Z$ $$x\cdot (x^{17})^2+5x^{17}\cdot x^2+11x^3+=(1+5+11)x^3=0.$$ Thus the given expression is divisible by $17$ for all $x$. • 2 other users stated the expression is divisible for any integer x. Which is true? – John Locke Jul 25 '17 at 23:13 • @JohnLocke (from David Hume…): I'll check my computation. – Bernard Jul 25 '17 at 23:16 • @JohnLocke: Checked. I had erroneously applied Fermat for one exponent (it's beginning to be late…). Thanks for pointing it! – Bernard Jul 25 '17 at 23:22 \begin{align}{\rm mod}\ 17\!:\,\ x\not\equiv 0\,\Rightarrow\, x^{\large 16}\!\equiv 1\,\Rightarrow\, &\ x^{\large\color{#c00}{n}}\equiv\, x^{\large\color{#c00}{ n\bmod 16}}\ \ \text{by little Fermat, hence}\\[.3em] &\ x^{\large\color{#c00}{35}}\!+5 x^{\large\color{#0a0}{19}}\!+ 11 x^{\large 3}\\[.2em] \equiv\ &\ x^{\large\color{#c00}3}\ + 5 x^{\large\color{#0a0}3} +\, 11 x^{\large 3}\ \ \,{\rm by}\,\ \ {\rm mod}\ 16\!:\ \color{#c00}{35\equiv 3},\ \color{#0a0}{19\equiv 3}\\[.2em] \equiv\ & 17x^{\large 3}\equiv\, 0 \end{align}<\|endoftext\|>	4.53125
605	To limit the negative impact of climate change, the global temperature must not rise by more than 2°C compared to preindustrial levels. This target is part of the EU’s energy policy. - A rise of 0.8°C has already taken place and a rise of further 0.5-0.7°C is unavoidable. The 2°C rise is typically associated with a CO2 concentration of 400-500 ppm (parts per million) by volume. This indicates that out of one million parts in the atmosphere, 400 to 500 are parts of CO2. - The current (1) level of CO2 alone is around 395,53 ppm by volume, and is rising by 2 ppm annually. Therefore, CO2 emissions must be reduced quickly and significantly. Thus the current carbon-based economy must change dramatically and as soon as possible, since world economic energy efficiency is improving at only half the world economic growth rate. - CO2 emissions must be reduced to 50% of 1990 levels by 2050. Industrialized countries produce most of the CO2 per capita, e.g. 24 tonnes in the US, 10 to 11 tonnes in Germany, as opposed to 4.6 tonnes in China and less than 0.1 tonnes in almost 20 other countries such as Madagascar, Nepal and Tanzania. - A sustainable amount of human CO2 emission is 3t per capita(2). If the 2°C target is to be reached, then industrialized countries must not only support developing countries to keep their emissions low, but they must also reduce their own emissions drastically. Many levers can help to limit global warming to 2°C, e.g. demand reduction for emissions-intensive goods and services, increase of efficiency gains, increased use and development of low-carbon technology, reduction of fossil fuel emissions. Time is crucial in reducing CO2 emissions since it stays in the atmosphere for a very long time(3). Therefore a mechanism is needed to ensure that the different levers are used as efficiently as possible concerning both time and effectiveness. Emission trading systems are such a mechanism which caps total emissions and makes all CO2 emitters compete in a common market. - October 2014 http://scrippsco2.ucsd.edu; this figure is adjusted for the effect of aerosols that are cooling the climate; actual level of CO2 is around 460ppm - The German WBGU (Wissenschaftlicher Beirat Globale Umweltveränderung) derives an emission limit of 750 billion tons until 2050; divided by the world population of 6.9 billion people in 2010 and spread over 40 years this yields 2.72 t per year and capita - Climate change which takes place due to increases in carbon dioxide concentration is largely irreversible for 1,000 years after emissions stop http://www.pnas.org/content/early/2009/01/28/0812721106.abstract<\|endoftext\|>	3.765625
362	The Reggio Emilia Approach What children learn does not follow as an automatic result from what is taught, rather, it is in large part due to the children’s own doing, as a consequence of their activities and our resources. —Loris Malaguzzi, The Hundred Languages of Children REGGIO EMILIA INSPIRED We believe that all children are capable, competent, and curious. Creativity, interests, ideas and questions of the children are worthy of pursuit. Children are placed at the center of the curriculum, resulting in an authentically responsive curriculum that fosters opportunities for every child to engage. The arts and sciences serve as languages for thinking, communicating, and sharing ideas. The Reggio Emilia approach speaks of the Hundred Languages of Children. Reggio Emilia educators share the belief that children have many methods of communicating, including storytelling, music, art, movement, dramatic play and construction. Teachers support creative thinking and problem solving with many opportunities provided for in-depth inquiry into areas of study that are initiated by the children and teachers. Teachers facilitate discussions, foster inquiry, create provocations observe and document student investigations. An emphasis placed on the importance of play in a child’s learning. Rather than seeing play as something we do aside from work, or after we do our work, many times the learning happens through the work of play. The classroom reflects a flexible environment that is responsive to the need for children and teachers to construct knowledge together. The purposefully designed classroom environment reflects the work and thoughts of the student. The classroom invites children to use materials and make choices, fostering independence and confidence. This type of environment encourages children to become more independent as they gain confidence in their abilities. Links for more information<\|endoftext\|>	3.8125
1,130	# HOW TO PROVE TWO TRIANGLES SIMILAR If the measures of the corresponding sides of two triangles are proportional or the measures of their corresponding angles are equal, then the two triangles are similar. In triangles ABC and DEF, if AB/DE = BC/EF = AC/DF then, ΔABC ΔDEF In triangles ABC and DEF, if ∠A = ∠D ∠B = ∠E ∠C = ∠F then, ΔABC ΔDEF ## Examples Example 1 : Determine whether the two triangles shown below are similar. Justify your answer. Solution : The angles BAC and DFE are congruent.To prove the above triangles are similar, we need to prove one more pairs of angles are equal. To check whether the angles BCA and DEF are equal, let us find the measure of angle BCA from triangle ABC. ∠BAC + ∠ABC + ∠BCA = 180 21 + 105 + ∠BCA = 180 126 + ∠BCA = 180 ∠BCA = 180 - 126 ∠BCA = 540 ∠BCA = ∠DEF In triangle ABC∠BAC∠BCA In triangle DEF∠DFE∠DEF So, the triangles ABC and DEF are similar. Example 2 : Determine whether the two triangles shown below are similar. Justify your answer. Solution : The angles ACB and FDE are congruent.To prove the above triangles are similar, we need to prove one more pairs of angles are equal. To check whether the angles ABC and DEF are equal, let us find the measure of angle ABC from triangle ABC. ∠ABC + ∠BAC + ∠ACB = 180 ABC + 79 + 60 = 180 ∠ABC + 139 = 180 ∠ABC = 180 - 139 ∠ABC = 41 ∠ABC ∠DEF Hence the above triangles ABC and DEF are not similar. Example 3 : Determine whether the two triangles shown below are similar. Justify your answer. Solution : To check whether the above triangles are similar, we need to find the missing angles of triangle ABC. ∠ABC + ∠BAC + ∠ACB = 180 84 + ∠BAC + ∠ACB = 180 2∠BAC = 180 - 84 2∠BAC = 96 ∠BAC = 96/2 ∠BAC = 48 = ∠ACB The corresponding angles of BAC and DEF are not same. So, the above triangles are not similar. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here. If you have any feedback about our math content, please mail us : [email protected] You can also visit the following web pages on different stuff in math. WORD PROBLEMS Word problems on simple equations Word problems on linear equations Algebra word problems Word problems on trains Area and perimeter word problems Word problems on direct variation and inverse variation Word problems on unit price Word problems on unit rate Word problems on comparing rates Converting customary units word problems Converting metric units word problems Word problems on simple interest Word problems on compound interest Word problems on types of angles Complementary and supplementary angles word problems Double facts word problems Trigonometry word problems Percentage word problems Profit and loss word problems Markup and markdown word problems Decimal word problems Word problems on fractions Word problems on mixed fractrions One step equation word problems Linear inequalities word problems Ratio and proportion word problems Time and work word problems Word problems on sets and venn diagrams Word problems on ages Pythagorean theorem word problems Percent of a number word problems Word problems on constant speed Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree OTHER TOPICS Profit and loss shortcuts Percentage shortcuts Times table shortcuts Time, speed and distance shortcuts Ratio and proportion shortcuts Domain and range of rational functions Domain and range of rational functions with holes Graphing rational functions Graphing rational functions with holes Converting repeating decimals in to fractions Decimal representation of rational numbers Finding square root using long division L.C.M method to solve time and work problems Translating the word problems in to algebraic expressions Remainder when 2 power 256 is divided by 17 Remainder when 17 power 23 is divided by 16 Sum of all three digit numbers divisible by 6 Sum of all three digit numbers divisible by 7 Sum of all three digit numbers divisible by 8 Sum of all three digit numbers formed using 1, 3, 4 Sum of all three four digit numbers formed with non zero digits Sum of all three four digit numbers formed using 0, 1, 2, 3 Sum of all three four digit numbers formed using 1, 2, 5, 6<\|endoftext\|>	4.71875
478	# Dividing By a Decimal Number: Over, Over, and Up! "Class, in our math lesson today I will need your help," said Mrs. Brown. "We've been learning how to do some long division, and you have done very well. Today we will learn a new division trick. When I move my hand like this, I need for you to say, "Over." Mrs. Brown moved her hand from left to right as though she were making a U. "Let's practice that, shall we?" She moved her hand from left to right, making a U in the air. Her students said, "Over!" "Very good!" said Mrs. Brown. "Also, when I stick up my thumb, I need you to say, "Up." She put her thumb up with her fingers closed in a fist. The students said, "Up!" "That's perfect!" said Mrs. Brown. Joe asked, "But why do we need to do that for a math lesson?" "I'll show you, Joe. We're going to divide a big number by a decimal number. When we divide, we want the divisor to be a whole number. So we will move the decimal point over until it is a whole number." "Suppose I have \$1.00, and I want to give a quarter to as many students as possible," said Mrs. Brown. "I wonder how many students would get a quarter." Chris raised her hand. "Thank you, Chris, for remembering to raise your hand! I appreciate that," said Mrs. Brown. "How many students would get a quarter, Chris?" "Four students would get twenty-five cents," said Chris. "That's correct!" said Mrs. Brown. "Let's see what the problem looks like written on the board." She wrote \$1.00 on the board. She drew the division "house" around the number. Then she wrote .25 in the divisor's place. Then she turned to the class with a smile. She moved her hand in the air to make a U shape. The class said, "Over!" Then Mrs. Brown put her marker on the decimal in front of .25. She drew a U on the board starting at the decimal in front of the two, and ending after the five. Then she made a new decimal point.<\|endoftext\|>	4.78125
631	Although characterized as a "local concern," style is an incredibly important aspect of writing. In this section, you will learn how to craft engaging, dynamic prose and how to best communicate your information and purpose as a writer. For a brief introduction to this topic, see Style: An Introduction. Learn how to negotiate between formal academic writing and conversational prose by maintaining an academic tone while staying true to your own voice in Making Sure Your Voice is Present. Reusing the same sentence pattern in your writing makes for monotonous reading. Learn to engage your readers by experimenting with different sentence patterns in Select an Appropriate Sentence Pattern. Then focus on individual sentences with the Sentence-Level Exercise. Whereas writers in the sciences tend to use passive voice in research reports, writers in other fields such as the humanities emphasize the importance of active voice. Learn to revise sentences to make them active and more engaging in Use the Active Voice. Another key to crafting engaging prose is to maintain a high verb-to-noun ratio. Different genres call for different points of view. Most students assume that academic papers should be written in the third person, but the first person has become increasingly accepted in more formal genres. Learn when the first person is an appropriate choice and how to successfully use first-person pronouns in Use the First Person, The First Person, and Using the First Person in Academic Writing: When is It Okay? To better understand why second-person pronouns should not be used in academic writing, read Understanding Second Person Point of View: Wizard Activity. When detailing their own ideas or the ideas of other scholars, successful writers communicate information in a clear and concrete manner. Learn how to craft concrete sentences in Avoid Vagueness and how to write clear, concise sentences in Write with Clarity. When appropriate, writers include figurative language in their texts. Learn why they do this and how to successfully employ figurative language in Incorporate Figurative Language into Your Paper. Being able to identify and address grammatical mistakes is important because those errors can not only make your draft appear sloppy, but they can also change the meaning of your sentences and confuse your reader. Enhance your understanding of grammatical principles by reading Subject-Verb Agreement, Subject-Pronoun Agreement, and Avoid Vague Pronoun References. Learn how to use proper punctuation. Below is a summary of how to punctuate different sentence patterns and how to analyze the likely effect of different syntactical forms on readers' comprehension. - Commas: Understand conventions for using commas and appreciate the likely effects of particular sentence lengths and patterns on reading comprehension. - Dashes: Create emphasis and define terms by interrupting the flow of a sentence using a dash; know when the dash must be used as opposed to the comma. - Colons: Use the colon when the first sentence anticipates the second sentence or phrase, thereby creating an emphatic tone. - Semicolons: Use a semicolon to join two sentences or to punctuate a series or list of appositives that already include commas.<\|endoftext\|>	4.09375
598	Courses Courses for Kids Free study material Offline Centres More Store # A satellite in a force free space sweeps stationary interplanetary dust at a rate $(dM/dt)=\alpha v$. The acceleration of the satellite is-A. $\dfrac{{ - 2\alpha {v^2}}}{M}$B. $\dfrac{{ - \alpha {v^2}}}{M}$C. $\dfrac{{ - \alpha {v^2}}}{{2M}}$D. $- \alpha {v^2}$ Last updated date: 18th Jun 2024 Total views: 394.2k Views today: 10.94k Verified 394.2k+ views Hint: When the satellite sweeps the dust, its mass increases at the given rate. If there is no external force acting on the satellite the only acceleration will be caused by the slowing due to an increase in the mass of the satellite. It can be calculated using Newton’s second law. In the question, it is given that there is no external force acting on the satellite. Therefore the change caused in the motion of the satellite is purely an effect of its increasing mass. We know that in a system if there is a change in momentum a force would also be applied. It is an application of Newton's 2nd law, where force is defined as the rate of change of momentum. Here, the momentum of the satellite is given by- $P = Mv$ where M is the mass of the satellite and v is its velocity. Any force on the satellite can be given by- $F = \dfrac{{dP}}{{dt}} = \dfrac{{d(Mv)}}{{dt}}$ Rewriting, $F = v\dfrac{{dM}}{{dt}} + M\dfrac{{dv}}{{dt}}$ It is given in the question that the rate of change in mass is- $\dfrac{{dM}}{{dt}} = \alpha v$ And external force is 0, Putting these values in the equation we get- $0 = v(\alpha v) + M\left( {\dfrac{{dv}}{{dt}}} \right)$ $M\left( {\dfrac{{dv}}{{dt}}} \right) = - \alpha {v^2}$ $\left( {\dfrac{{dv}}{{dt}}} \right) = \dfrac{{ - \alpha {v^2}}}{M}$ So, the correct answer is “Option B”. Note: Although the object doesn’t experience any external force, it still undergoes a deceleration because of the interaction between the mass present outside the system. As force is defined as a change in momentum, any change in either mass or velocity would result in the application of force. This addition of mass decreases the total energy of the satellite causing it to slow down.<\|endoftext\|>	4.4375
1,017	Difference Between EMF and Voltage - kinenbicounter.info please elaborate the difference between emf and voltage. This relation is useful because it allows us to predict what the voltage will be at a. Since voltage and resistance have an inverse relationship, individual resistors in .. Compare the resistances and electromotive forces for the voltage sources. What is the main difference between voltage, EMF and a potential difference? . and is called its terminal voltage V. Terminal voltage is given by the equation. Faraday's law of induction The principle of electromagnetic inductionnoted above, states that a time-dependent magnetic field produces a circulating electric field. Difference Between EMF and Voltage A time-dependent magnetic field can be produced either by motion of a magnet relative to a circuit, by motion of a circuit relative to another circuit at least one of these must be carrying a currentor by changing the current in a fixed circuit. The effect on the circuit itself, of changing the current, is known as self-induction; the effect on another circuit is known as mutual induction. - What is the difference between emf and voltage? - Electromotive Force and Internal Resistance For a given circuit, the electromagnetically induced emf is determined purely by the rate of change of the magnetic flux through the circuit according to Faraday's law of induction. An emf is induced in a coil or conductor whenever there is change in the flux linkages. Depending on the way in which the changes are brought about, there are two types: When the conductor is moved in a stationary magnetic field to procure a change in the flux linkage, the emf is statically induced. The electromotive force generated by motion is often referred to as motional emf. When the change in flux linkage arises from a change in the magnetic field around the stationary conductor, the emf is dynamically induced. Electromotive Force and Internal Resistance The electromotive force generated by a time-varying magnetic field is often referred to as transformer emf. Volta potential and Electrochemical potential When solids of two different materials are in contact, thermodynamic equilibrium requires that one of the solids assume a higher electrical potential than the other. This is called the contact potential. The magnitude of this potential difference is often expressed as a difference in Fermi levels in the two solids when they are at charge neutrality, where the Fermi level a name for the chemical potential of an electron system describes the energy necessary to remove an electron from the body to some common point such as ground. Voltage is defined as the difference of electrical potential between two points, and this difference on the poles of the electric source is obtained by removing electrons from one part of the source and transferring them to another. Expression The electromotive force of the source is equal to the work that some external force has to do to move the charge unit from one pole of the source to another, but through the source. Voltage in the outer part of the circuitry is much equal to the work that needs be done by the electric force to move the charge unit from one pole of the source to the other, but through the wire. Formula The electromotive force is calculated as following: Electric force operation Voltage is an operation of the electric Coulomb force in charge motion and is the result of energy reduction in the circle, while the electromotive force is defined by a non-electric force non-Coulomb operation and is responsible for increasing the energy in the circuit. Measuring A potential difference voltage can be measured between any given points in the circuit, while electromotive force exists only between the two ends of a source. Also electromotive force is measured with EMF meter, while voltage with a voltmeter. Intensity The electromotive force is always greater than the voltage.Electromotive Force of a Battery, Internal Resistance and Terminal Voltage The reason behind is that the voltage exist in a loaded circuit, and because of the resistance energy loss a voltage drop occurs. Induction EMF can be caused in electric, gravitational or magnetic field, while voltage is caused only in electric field. Voltage Summary The quantity adequately representing the generator as an element of the electric circuit and quantitatively characterizing its ability to maintain the current in the circuit and to convert other forms of energy into an electric one is called an electromotive force. A work performed on a charge unit by an external force by dividing the charges at the source of the electric current is called an electromotive force. The electromotive force EMF has a voltage dimension Volt unitand is also called internal voltage of the source U0. The electromotive force is the same size as the potential difference between the positive and negative generator connections when it is in idle mode. The potential difference between two points of the electric field is called voltage, and the unit is also volt. If you like this article or our site.<\|endoftext\|>	3.671875
9,736	Germanisation (also spelt Germanization) refers to the spread of the German language, people and culture or policies which introduced these changes. It was a central plank of German liberal thinking in the 19th and 20th centuries, at a period when liberalism and nationalism went hand-in-hand. In linguistics, Germanization also occurs when a word from the German language is adopted into a foreign language. - 1 Forms of Germanisation - 2 Historical Germanisation - 3 Germanisation under the Third Reich - 4 After World War II - 5 Linguistic Germanisation - 6 Surnames - 7 See also - 8 References - 9 External links Forms of Germanisation Historically, there are very different forms and degrees of the expansion of the German language and of elements of German culture. There are examples of complete assimilation into German culture, as happened with the pagan Slavs in the Diocese of Bamberg (Franconia) in the 11th century. A perfect example of eclectic adoption of German culture is the field of law in Imperial and present-day Japan, which is organised very much according to the model of the German Empire. Germanisation took place by cultural contact, by political decision of the adopting party (e.g., in the case of Japan), or (especially in the case of Imperial and Nazi Germany) by force. In Slavic countries, the term Germanisation is often[quantify] understood solely as the process of acculturation of Slavic- and Baltic-language speakers - after the conquests or by cultural contact in the early Middle Ages - of areas of modern southern Austria and eastern Germany to the line of the Elbe. In East Prussia, forced resettlement of the "Old" or "Baltic" Prussians by the Teutonic Order and the Prussian state, as well as acculturation from immigrants of various European countries (Poles, French, and Germans) contributed to the eventual extinction of the Prussian language in the 17th century. Another form of Germanisation is the forceful imposition of German culture, language and people upon non-German people, Slavs in particular. Early Germanisation went along with the Ostsiedlung during the Middle Ages, e.g., in Hanoverian Wendland, Mecklenburg-Vorpommern, Lusatia, and other areas, formerly inhabited by Slavic tribes - Polabian Slavs such as Obotrites, Veleti and Sorbs. Relations of early forms of Germanisation were described by German monks in manuscripts like Chronicon Slavorum. Lüchow-Dannenberg is better known as the Wendland, a designation referring to the Slavic people of the Wends from the Slavic tribe Drevani — the Polabian language survived until the beginning of the 19th century in what is now the German state of Lower Saxony. A complex process of Germanisation took place in Bohemia after the defeat of Bohemian Protestants at the Battle of White Mountain in 1620. The German prince and Frederick V, Elector Palatine, elected as king of Bohemia by the Bohemian estates in 1619, was defeated by Catholic forces loyal to the Habsburg Emperor, Ferdinand II. Among the Bohemian lords who were punished and had their lands expropriated after Frederick's defeat in 1620 were German- and Czech-speaking landowners. Thus, this conflict was feudal in nature, not national. Although the Czech language lost its significance as a written language in the aftermath of the events, it is doubtful that this was intended by the Habsburg rulers, whose aims were of a feudal and religious character. Proto-Slovene language was spoken in a much larger territory than modern Slovene, which included most of the present-day Austrian states of Carinthia and Styria, as well as East Tyrol, the Val Pusteria in South Tyrol, and some areas of Upper and Lower Austria. By the 15th century, most of the these areas were gradually Germanised: the northern border of the Slovene-speaking territory stabilised on the line going from north of Klagenfurt to south of Villach and east of Hermagor in Carinthia, while in Styria it closely followed the current Austrian-Slovenian border. This linguistic border remained almost unchanged until the late 19th century, when a second process of Germanisation took place, mostly in Carinthia. In Tyrol there was a germanisation of the ladino-romantsch of the Venosta Valley (now Italy) promoted by the Austria in the 16th century. There was made for avoiding contact with Protestants of the Graubünden. The rise of nationalism that occurred in the late 18th and 19th centuries in Bohemia, Moravia, Silesia, Pomerania, Lusatia, and Slovenia led to an increased sense of "pride" in national cultures during this time. However, centuries of cultural dominance of the Germans left a German mark on those societies; for instance, the first modern grammar of the Czech language by Josef Dobrovský (1753–1829) – "Ausführliches Lehrgebäude der böhmischen Sprache" (1809) – was published in German because the Czech language was not used in academic scholarship. From the high Middle Ages up to the dissolution of the Austro-Hungarian Empire in 1918, in the territory of present day Slovenia, German had a strong impact on Slovene, and many Germanisms are preserved in contemporary colloquial Slovene. In the Austrian Empire Joseph II (1780–90), a leader influenced by the Enlightenment, sought to centralise control of the empire and to rule it as an enlightened despot. He decreed that German replace Latin as the empire's official language. Hungarians perceived Joseph's language reform as German cultural hegemony, and they reacted by insisting on the right to use their own tongue. As a result, Hungarian lesser nobles sparked a renaissance of the Hungarian language and culture. The lesser nobles questioned the loyalty of the magnates, of whom less than half were ethnic Magyars, and even those had become French- and German-speaking courtiers. The Magyar national revival subsequently triggered similar movements among the Slovak, Romanian, Serbian, and Croatian minorities within the Kingdom of Hungary. Germanisation in Prussia occurred in several stages: - The Old Prussians (originally a Baltic ethnic group) were Germanised during the Teutonic Knights in the assimilation of the Prussian Crusade - Germanisation attempts pursued by Frederick the Great in territories of Partitioned Poland - Easing of Germanisation policy in the period 1815–30 - Intensification of Germanisation and persecution of Poles in the Grand Duchy of Posen by E.Flotwell in 1830–41 - The process of Germanisation ceases during the period of 1841–49 - Restarted during years of 1849–70 - Intensified by Bismarck during his Kulturkampf against Catholicism and Polish people - Slight easing of the persecution of Poles during 1890–94 - Continuation and intensification of activity restarted in 1894 and pursued till the end of World War I Legislation and government policies in the Kingdom of Prussia sought a degree of linguistic and cultural Germanisation, while in the Imperial Germany a more intense form of cultural Germanisation was pursued, often with the explicit intention of reducing the influence of other cultures or institutions, such as the Catholic Church. Situation in the 18th century Following the partitions, the previous Germanisation attempts pursued by Frederick the Great in Silesia were extended to the newly gained Polish territories. The Prussian authorities started the policy of settling German speaking ethnic groups in these areas. Frederick the Great settled around 300,000 colonists in the eastern provinces of Prussia and aimed at a removal of the Polish nobility, which he treated with contempt and described Poles as 'slovenly Polish trash' in newly reconquered West Prussia, similar to the Iroquois. From the beginnings of Prussian rule Poles were subject to a series of measures against their culture; the Polish language was replaced by German as the official language, and most administration was made German as well; the Prussian ruler Frederick the Great despised Poles and hoped to replace them with Germans. Poles were portrayed as 'backward Slavs' by Prussian officials who wanted to spread German language and culture. The land of Polish nobility was confiscated and given to German nobles. Situation in the 19th century You also have a Fatherland. [...] You will be incorporated into my monarchy without having to renounce your nationality. [...] You will receive a constitution like the other provinces of my kingdom. Your religion will be upheld. [...] Your language shall be used like the German language in all public affairs and everyone of you with suitable capabilities shall get the opportunity to get an appointment to a public office. [...] The minister for Education Altenstein stated in 1823: Concerning the spread of the German language it is most important to get a clear understanding of the aims, whether it should be the aim to promote the understanding of German among Polish-speaking subjects or whether it should be the aim to gradually and slowly Germanise the Poles. According to the judgement of the minister only the first is necessary, advisable and possible, the second is not advisable and not accomplishable. To be good subjects it is desirable for the Poles to understand the language of government. However, it is not necessary for them to give up or postpone their mother language. The possession of two languages shall not be seen as a disadvantage but as a benefit instead because it is usually associated with a higher flexibility of the mind. [..] Religion and language are the highest sanctuaries of a nation and all attitudes and perceptions are founded on them. A government that [...] is indifferent or even hostile against them creates bitterness, debases the nation and generates disloyal subjects. In the first half of the 19th century, Prussian language policy remained largely tolerant. However, this tolerance gradually changed in the second half of the 19th century after the foundation of the German Empire in 1871. Successive policies aimed at the elimination of non-German languages from public life and from academic settings, such as schools. For example, in the course of the second half of the 19th century, the Dutch language, historically spoken in what is now Cleves, Geldern, Emmerich, was banned from schools, administration and would cease to be spoken in its standardised form at the turn of the century. Later in the German Empire, Poles were (together with Danes, Dutch Alsatians, German Catholics and Socialists) portrayed as "Reichsfeinde" ("foes to the empire"). In addition, in 1885, the Prussian Settlement Commission financed from the national government's budget was set up to buy land from non-German hands and distribute it among German farmers. From 1908 the committee was entitled to force the landowners to sell the land. Other means included the Prussian deportations from 1885–1890, in which non-Prussian nationals who had lived in Prussia for substantial time periods (mostly Poles and Jews) were removed and a ban was issued on the building of houses by non-Germans (see Drzymała's van). Germanisation policy in schools also took the form of abuse of Polish children by Prussian officials (see Września). Germanisation unintentionally stimulated resistance, usually in the form of home schooling and tighter unity in the minority groups. In 1910, Maria Konopnicka responded to the increasing persecution of Polish people by Germans by writing her famous song called Rota that instantly became a national symbol for Poles, with its sentence known to many Poles: The German will not spit in our face, nor will he Germanise our children. Thus, the German efforts to eradicate Polish culture, language, and people met not only with failure, but managed to reinforce the Polish national identity and strengthened efforts of Poles to re-establish a Polish state. Prussian Lithuanians living in East Prussia experienced similar policies of Germanisation. Although ethnic Lithuanians had constituted a majority in areas of East Prussia during the 15th and 16th centuries (from the early 16th century it was often referred to as Lithuania Minor), the Lithuanian population began to shrink in the 18th century. Plague and subsequent immigration from Germany, notably from Salzburg, were the primary factors in this development. Germanisation policies were tightened during the 19th century, but even into the early 20th century the territories north and south/south-west of the Neman River contained a Lithuanian majority. Kursenieki experienced similar developments, but this ethnic group never had a large population. Polish coal miners in the Ruhr Valley Another form of Germanisation was the relation between the German state and Polish coal miners in the Ruhr area. Due to migration within the German Empire, as many as 350,000 ethnic Poles made their way to the Ruhr in the late 19th century, where they worked in the coal and iron industries. German authorities viewed them as potential danger and a threat and as a "suspected political and national" element. All Polish workers had special cards and were under constant observation by German authorities. Their citizens' rights were also limited by German state. In response to these policies, the Polish formed their own organisations to maintain their interests and ethnic identity. The Sokol sports clubs and the workers' union Zjednoczenie Zawodowe Polskie (ZZP), Wiarus Polski (press), and Bank Robotnikow were among the best-known such organisations near the Ruhr. At first the Polish workers, ostracised by their German counterparts, had supported the Catholic centre party. During the early 20th century, their support shifted more and more towards the social democrats. In 1905, Polish and German workers organised their first common strike. Under the Namensänderungsgesetz (law of changing surnames), a significant number of "Ruhr-Poles" change their surnames and Christian names to "Germanised" forms, in order to evade ethnic discrimination. As the Prussian authorities suppressed Catholic services in Polish by Polish priests during the Kulturkampf, the Poles had to rely on German Catholic priests. Increasing intermarriage between Germans and Poles contributed much to the Germanisation of ethnic Poles in the Ruhr area. During the Weimar Republic, Poles were recognised as a minority in Upper Silesia. The peace treaties after the First World War did contain an obligation for Poland to protect its national minorities (Germans, Ukrainians and other), whereas no such clause was introduced by the victors in the Treaty of Versailles with Germany. In 1928, the "Minderheitenschulgesetz" (minorities school act) regulated education of minority children in their native tongue. From 1930 onwards, Poland and Germany agreed to treat their minorities fairly. Germanisation under the Third Reich The East was intended as the Lebensraum that the Nazis were seeking, to be filled with Germans. Hitler, speaking with generals immediately prior to his chancellorship, declared that people could not be Germanised; only the soil could be. The policy of Germanisation in the Nazi period carried an explicitly ethno-racial rather than purely nationalist meaning by virtue of culture or linguistics, aiming for the spread of a "biologically superior" Aryan race rather than that of the German nation. Therefore, this did not mean a total extermination of all people there, as Eastern Europe was regarded as having people of Aryan/Nordic descent, particularly among their leaders. Indeed Himmler also declared that no drop of German blood would be lost or left behind for an alien race. In Nazi documents, even reading the term "German" can be problematic, since it could be used to refer to people classified as "ethnic Germans" who spoke no German. Inside Germany, propaganda, such the film Heimkehr, depicted these ethnic Germans as deeply persecuted—often with recognisable Nazi tactics—and the invasion and Germanisation as necessary to protect them. Forced labor of ethnic Germans and persecution of them were major themes of the anti-Polish propaganda campaign of 1939, prior to the invasion. Bloody Sunday was widely exploited as depicting the Poles as murderous toward Germans. In a top-secret memorandum, "The Treatment of Racial Aliens in the East", dated 25 May 1940, Himmler wrote "We need to divide Poland's many different ethnic groups up into as many parts and splinter groups as possible". There were two Germanisation actions in occupied Poland realised in this way: - The grouping of Polish Gorals ("Highlanders") into the hypothetical Goralenvolk, a project which was ultimately abandoned due to lack of support among the Goral population; - The assignment of Pomerelian Kashubians onto the Deutsche Volksliste, as they were considered capable of assimilation into the German population (several high-ranking Nazis deemed them to be descended from ancient Gothic peoples). Selection and expulsion Germanisation began with the classification of people suitable as defined on the Nazi Volksliste, and treated according to their categorisation. The Germans regarded the holding of active leadership roles as an Aryan trait, whereas a tendency to avoid leadership and a perceived fatalism was associated by many Germans with Slavonic peoples. Adults who were selected for but resisted Germanisation were executed. Such execution was carried out on the grounds that German blood should not support non-German nations, and that killing them would deprive foreign nations of superior leaders. The Intelligenzaktion was justified, even though these elites were regarded as likely of German blood, because such blood enabled them to provide leadership for the fatalistic Slavs. Germanising "racially valuable" elements would prevent any increase in the Polish intelligenstia, as the dynamic leadership would have to come from German blood. In 1940, Hitler made it clear that the Czech intelligentsia and the "mongoloid" types of the Czech population were not allowed to be Germanised. Under Generalplan Ost, a percentage of Slavs in the conquered territories were to be Germanised. Gauleiters Albert Forster and Arthur Greiser reported to Hitler that 10 percent of the Polish population contained "Germanic blood", and were thus suitable for Germanisation. The Reichskommissars in northern and central Russia reported similar figures. Those unfit for Germanisation were to be expelled from the areas marked out for German settlement. In considering the fate of the individual nations, the architects of the Plan decided that it would be possible to Germanise about 50 percent of the Czechs, 35 percent of the Ukrainians and 25 percent of the Belarusians. The remainder would be deported to western Siberia and other regions. In 1941 it was decided that the Polish nation should be completely destroyed; the German leadership decided that in ten to 20 years, the Polish state under German occupation was to be fully cleared of any ethnic Poles and resettled by German colonists. In the Baltic States, after an agreement with Stalin, who suspected they would be loyal to the Nazis, the Nazis set out to encourage the departure of "ethnic Germans" by the use of propaganda. This included using scare tactics about the Soviet Union, and led to tens of thousands leaving. Those who left were not referred to as "refugees", but were rather described as "answering the call of the Führer." German propaganda films such as GPU and Friesennot depicted the Baltic Germans as deeply persecuted in their native lands. Packed into camps for racial evaluation, they were divided into groups: A, Altreich, who were to be settled in Germany and allowed neither farms nor business (to allow for closer watch), S Sonderfall, who were used as forced labor, and O Ost-Falle, the best classification, to be settled in the Eastern Wall—the occupied regions, to protect German from the East—and allowed independence. This last group was often given Polish homes where the families had been evicted so quickly that half-eaten meals were on tables and small children had clearly been taken from unmade beds. Members of Hitler Youth and the League of German Girls were assigned the task of overseeing such evictions to ensure that the Poles left behind most of their belongings for the use of the settlers. The deportation orders required that enough Poles be removed to provide for every settler — that, for instance, if twenty German "master bakers" were sent, twenty Polish bakeries had to have their owners removed. Settlement and Germanisation This colonisation incorporated 350,000 such Baltic Germans and 1.7 million Poles deemed Germanisable, including between one and two hundred thousand children who had been taken from their parents (plus about 400,000 German settlers from the "Old Reich"). Nazi authorities had great fears of these settlers being tainted by their Polish neighbors and not only warned them to let their "foreign and alien" surroundings to have no impact on their Germanness, but settled them in compact communities, which could be easily monitored by the police. Only families classified as "highly valuable" were kept together. For Poles who did not resist and the resettled ethnic Germans, Germanisation began. Militant party members were sent to teach them to be "true Germans". Hitler Youth and League of German Girls sent young people for "Eastern Service", which entailed (particularly for the girls) assisting in Germanisation efforts. One member of the League recounted afterward that she at first pitied the starving Polish children, but soon realised this was "politically naive" and to concentrate solely on the Volksdeutsche; her beliefs in the stupidity of Poles were reinforced by the lack of educated Poles, not knowing they had been jailed or deported. This included instruction in the German language, as many spoke only Polish or Russian. They found the new settlers dispirited and put on various entertainments such as songfests to encourage them and ease their transition. Membership in Hitler Youth and the League of German Girls was enforced for the children. Goebbels and other propagandists worked to establish cultural centres and other means to created Volkstum or racial consciousness in the settlers. This was needed to perpetuate their work; only by effective Germanisation could mothers, in particular, create the German home. Goebbels also was the official patron of Deutsches Ordensland or Land of Germanic Order, an organisation to promote Germanisation. This efforts were used in propaganda in Germany itself, as when NS-Frauen-Warte's cover article was on "Germany is building in the East". Other efforts in Poland were also regarded as Germanisation, as for instance the setting up of the IG-Farben at Auschwitz-Monowitz. Germanisation in Yugoslavia and Soviet Union On 6 April 1941, Yugoslavia was invaded by the Axis Powers. Part of the Slovene-settled territory was occupied by the Nazi Germany. The Gestapo arrived on 16 April 1941 and were followed three days later by SS leader Heinrich Himmler, who inspected Stari pisker prison in Celje. On 26 April, Adolf Hitler, who encouraged his followers to "make this land German again", visited Maribor and a grand reception was organised by local Germans in the city castle. Although the Slovenes had been deemed racially salvageable by the Nazis, the mainly Austrian rulers of the Carinthian and Styrian regions commenced a brutal campaign to destroy them as a nation. The Nazis started a policy of violent Germanisation on Slovene territory, attempted to either discourage or entirely suppress the Slovene language. Their main task in Slovenia was the removal of part of population and Germanisation of the rest. Two organisations were instrumental for the Germanisation: the Styrian Homeland Union (Steirisches Heimatbund - HS) and the Carinthian People's Union (Kärtner Volksbund - KV). In Styria the action of Germanisation of the Slovenes was controlled by SS-Sturmbannführer Franz Steindl. He favoured the theories about the Germanic origins of the Slovenes. In Carinthia similar policy was conducted by Wilhelm Schick, Gauleiter's close associate. Public use of the Slovene language was prohibited, geographic and topographic names were changed and all Slovene associations were dissolved. Members of all professional and intellectual groups, including many clergymen, were expelled as they were seen as obstacle for Germanisation. As a reaction, a resistance movement developed. The Germans who wanted to proclaim their formal annexation to the "German Reich" on 1 October 1941, postponed it first because of the installation of the new "Gauleiter" and "Reichsstatthalter" of Carinthia and later on they dropped the plan for an undefinite period of time because of Slovene partisans, with which the Germans wanted to deal first. Only Meža valley became part of "Reichsgau Carinthia" at once. In the frame of their plan for the ethnic cleansing of Slovene territory, around 80,000 Slovenes were forcibly deported to Eastern Germany for potential Germanisation or forced labour, the deported Slovenes were taken to several camps in Saxony, where they were forced to work on German farms or in factories run by German industries from 1941–1945. The forced labourers were not always kept in formal concentration camps, but often just vacant buildings where they slept until the next day's labour took them outside these quarters. Nazi Germany also began mass expulsions of Slovenes to Nedić's Serbia and Ustasha Croatia (both Nazi Germany allies), and more than 63,000 Slovenes who refused to make any attempt to have them recognised as the Germans were interned to Nazi concentration camps in Germany. The basis for the recognition of the Slovenes as German nationals was the decision of the Imperial Ministry for Interior from 14 April 1942; it also constituted the basis for drafting the Slovenes for the service in the German armed forces. The numbers of Slovenes forcibly conscripted to the German military and paramilitary formations has been estimated at 150,000 men and women, almost a quarter of them lost their lives on various European battlefields, mostly on the Eastern Front. An unknown number of "stolen children" were taken to Nazi Germany for Germanisation. Later, Ukraine was also targeted for Germanisation. Thirty special SS squads took over villages where ethnic Germans predominated, and expelled or shot any Jews or Slavs living in them, in a policy of concentration. The colony Hegewald was set up in the Ukraine as well. Ukrainians were forcibly deported, and ethnic Germans forcibly relocated there. Racial assignment was carried out in a confused manner; the Reich rule was three German grandparents, but some asserted that any person who acted like a German and evinced no "racial concerns" should be eligible. Plans to eliminate Slavs from Soviet territory to allow German settlement included starvation; Nazi leaders expected that millions would die after they removed such supplies as they needed. This was regarded as an actual advantage by Nazi officials. When Hitler received a report of many, well-fed Ukrainian children, he declared the promotion of contraception and abortion was urgently needed, and neither medical care nor education was to be provided. Experiments in mass sterilisation in concentration camps may also have been intended for use on the Slavonic populations. When young women from the East were recruited to work as nannies in Germany, they were required to be suitable for Germanisation, both because they would work with German children, and because they might be sexually exploited. The program was praised for not only allowing more women to have children with their new domestic servants to assist in their labours, but for reclaiming the German blood and giving advantages to the women, who would work in Germany, and might marry there. "Racially acceptable" children were taken from their families in order to be brought up as Germans. Children were selected for "racially valuable traits" before being shipped to Germany. Many Nazis were astounded at the number of Polish children found to exhibit "Nordic" traits, but assumed that all such children were genuinely German children, who had been Polonised; Hans Frank summoned up such views when he declared, "When we see a blue-eyed child we are surprised that she is speaking Polish." The term used for them was wiedereindeutschungsfähig—meaning capable of being re-Germanised. These might include the children of people executed for resisting Germanisation. If attempts to Germanise them failed, or they were determined to be unfit, they would be killed to eliminate their value to the opponents of the Reich. In German-occupied Poland, it is estimated that a number ranging from 50,000 to 200,000 children were removed from their families to be Germanised. The Kinder KZ was founded specifically to hold such children. It is estimated that at least 10,000 of them were murdered in the process as they were determined unfit and sent to concentration camps and faced brutal treatment or perished in the harsh conditions during their transport in cattle wagons, and only 10-15% returned to their families after the war. Obligatory Hitlerjugend membership made dialogue between old and young next to impossible, as use of languages other than German was discouraged by officials. Members of minority organisations were either sent to concentration camps by German authorities or executed. Many children, particularly Polish and Slovenian who were among the first taken, declared on being found by Allied forces that they were German. Russian and Ukrainian children, while not gotten to this stage, still had been taught to hate their native countries and did not want to return. In contemporary German usage the process of Germanisation was referred to as Germanisierung (Germanicisation, i.e., to make something Germanic) rather than Eindeutschung (Germanisation, i.e., to make something German). According to Nazi racial theories, the Germanic peoples of Europe such as the Scandinavians, the Dutch, and the Flemish, were like the Germans themselves a part of the Aryan Master Race, regardless of these peoples' own acknowledgement of their "Aryan" identity. Germanisation in these conquered countries proceeded more slowly. The Nazis had a need for local cooperation and the local industry with its workers; furthermore, the countries were regarded as more racially acceptable, the assortment of racial categories being boiled down by the average German to mean "East is bad and West is acceptable." The plan was to win the Germanic elements over slowly, through education. Himmler, after a secret tour of Belgium and Holland, happily declared the people would be a racial benefit to Germany. Occupying troops were kept under strict discipline and instructed to be friendly to win the population over, a technique that did not work not only because of their having conquered the countries, but because it was soon clear that being German was far superior to being merely Nordic. Pamphlets, for instance, enjoined all German women to avoid sexual relations with all foreign workers brought to Germany as a danger to their blood. Various Germanisation plans were implemented. Dutch and Belgian Flemish prisoners of war were sent home quickly, to increase Germanic population, while Belgian Walloon ones were kept as laborers. Lebensborn homes were set up in Norway for Norwegian women impregnated by German soldiers, with adoption by Norwegian parents being forbidden for any child born there. Alsace-Lorraine was annexed; thousands of residents, too loyal to France, Jewish, or North Africa, were deported to Vichy France; French was forbidden in schools; intransigent German speakers were shipped back to Germany for re-Germanisation, just as Poles were. Extensive racial classification was practiced in France, for future uses. Himmler's masseur, Felix Kersten, claimed an even more radical scheme was devised by Himmler which envisioned the near-future resettlement of the entire Dutch nation to agricultural lands in the Vistula and Bug valleys of German-occupied Poland in order to facilitate their immediate Germanisation. 8.5 million people were to be relocated in total, after which all Dutch capital and real estate would be confiscated by the Reich and distributed to reliable SS men, and an SS Province of Holland declared in vacated Dutch territory. However this claim was shown to be a myth by Loe de Jong in his book Two Legends of the Third Reich. After World War II In post-1945 Germany and post-1945 Austria, the concept of Germanisation is no longer considered relevant. Danes, Frisians, and Slavic Sorbs are classified as traditional ethnic minorities and are guaranteed cultural autonomy by both the federal and state governments. Concerning the Danes, there is a treaty between Denmark and Germany from 1955 regulating the status of the German minority in Denmark and vice versa. Concerning the Frisians, the northern federal-state of Schleswig-Holstein passed a special law aimed at preserving the language. The cultural autonomy of the Sorbs is a matter of the constitutions of both Saxony and Brandenburg. Nevertheless, almost all of the Sorbs are bilingual and the Lower Sorbian language is regarded as endangered, as the number of native speakers is dwindling, even though there are programmes funded by the state to sustain the language. Descendants of Polish migrant workers and miners have intermarried with the local population and are culturally German or of mixed culture. It is different with modern and present-day immigration from Poland to Germany after the fall of the Iron Curtain. These immigrants usually are Polish citizens and live as foreigners in Germany. For many immigrant Poles, Polish ethnicity is not the prime category through which they wish to characterise themselves or want to be evaluated by others, as it could affect their lives in a negative way. In linguistics, Germanisation usually means the change in spelling of loanwords to the rules of the German language — for example the change from the imported word bureau to Büro. Professor Jürgen Udolph of the University of Leipzig Institute for Slavic Studies, contends that 15 million people in modern Germany have Slavic or specifically Polish surnames. He believes the concentration of Slavic names in East Germany is even higher than the national average, accounting for 30% of all surnames in the region. - Polabian language - "A Country Study: Hungary – Hungary under the Habsburgs". Federal Research Division. Library of Congress. Retrieved 2009-04-14.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles> - Prussian document from 1750 in Polish - "Wznowione powszechne taxae-stolae sporządzenie, Dla samowładnego Xięstwa Sląska, Podług ktorego tak Auszpurskiey Konfessyi iak Katoliccy Fararze, Kaznodzieie i Kuratusowie Zachowywać się powinni. Sub Dato z Berlina, d. 8. Augusti 1750" - "In fact, from Hitler to Hans Frank, we find frequent references to Slavs and Jews as 'Indians.' This, too, was a long standing trope. It can be traced back to Frederick the Great, who likened the 'slovenly Polish trash' in newly' reconquered West Prussia to Iroquois". Localism, Landscape, and the Ambiguities of Place: German-speaking Central Europe, 1860–1930 David Blackbourn, James N. Retallack University of Toronto 2007 - Ritter, Gerhard (1974). Frederick the Great: A Historical Profile. Berkeley: University of California Press. pp. 179–180. ISBN 0-520-02775-2. It has been estimated that during his reign 300,000 individuals settled in Prussia. ... While the commission for colonization established in the Bismarck era could in the course of two decades bring no more than 11,957 families to the eastern territories, Frederick settled a total of 57,475. ... It increased the German character of the population in the monarchy's provinces to a very significant degree. ... in West Prussia where he wished to drive out the Polish nobility and bring as many of their large estates as possible into German hands.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles> - Andrzej Chwalba, Historia Polski 1795–1918 Wydawnictwo Literackie 2000 Kraków pages 175–184, 307–312 - cited in: Richard Cromer: Die Sprachenrechte der Polen in Preußen in der ersten Hälfte des 19. Jahrhunderts. Journal Nation und Staat, Vol 6, 1932/33, p. 614, also cited in: Martin Broszat Zweihundert Jahre deutsche Polenpolitik (Two-hundred years or German Poles politics). Suhrkamp 1972, p. 90, ISBN. 3-518-36574-6. During the discussions in the Reichstag in January 1875, Altenstein's statement was cited by the opponents of Bismarck's politics. - Historism and Cultural Identity in the Rhine-Meuse Region, J. De Maeyer - Bismarck and the German Empire, 1871–1918 - "Die Germanisirung der polnisch-preußischen Landestheile." In Neueste Mittheilungen, V.Jahrgang, No. 17, 11 February 1886. Berlin: Dr. H. Klee.http://amtspresse.staatsbibliothek-berlin.de/vollanzeige.php?file=11614109%2F1886%2F1886-02-11.xml&s=1 - Kossert, Andreas."Grenzlandpolitik und Ostforschung an der Grenze des Reiches: Das ostpreußische Masuren." In "Viertelsjahrheft für Zeitgeschichte." Oldenbourg: Institut für Zeitgeschichte München-Berlin, April 2003. Pp. 121–123. http://www.kossert.net/dateien/vfzg5122003.pdf - Migration Past, Migration Future: Germany and the United States - 1880, Polen im Ruhrgebiet - "Polen im Ruhrgebiet 1870–1945" — Deutsch-polnische Tagung - H-Soz-u-Kult / Tagungsberichte - Johann Ziesch - Richard Bessel, Nazism and War, p 36 ISBN 0-679-64094-0 - HITLER'S PLANS FOR EASTERN EUROPE - Richard Overy, The Dictators: Hitler's Germany, Stalin's Russia, p543 ISBN 0-393-02030-4 - Pierre Aycoberry, The Social History of the Third Reich, 1933-1945, p 2, ISBN 1-56584-549-8 - Erwin Leiser, Nazi Cinema p69-71 ISBN 0-02-570230-0 - Robert Edwin Hertzstein, The War That Hitler Won p173 ISBN 0-399-11845-4 - Robert Edwin Hertzstein, The War That Hitler Won p289 ISBN 0-399-11845-4 - Vierteljahrshefte für Zeitgeschichte 1957, No. 2 - Nazi Conspiracy & Aggression Volume I Chapter XIII Germanisation & Spoliation - Diemut Majer, United States Holocaust Memorial Museum, "Non-Germans" Under the Third Reich: The Nazi Judicial and Administrative System in Germany and Occupied Eastern Europe with Special Regard to Occupied Poland, 1939–1945 Von Diemut Majer, United States Holocaust Memorial Museum, JHU Press, 2003, p.240, ISBN 0-8018-6493-3. - Lukas, Richard C. Did the Children Cry? - Richard C. Lukas, Forgotten Holocaust p24 ISBN 0-7818-0528-7 - Hitler's Ethic By Richard Weikart p.67 - Speer, Albert (1976). Spandau: The Secret Diaries, p. 49. Macmillan Company. - Volker R. Berghahn "Germans and Poles 1871–1945" in "Germany and Eastern Europe: Cultural Identities and Cultural Differences", Rodopi 1999 - Lynn H. Nicholas, Cruel World: The Children of Europe in the Nazi Web p. 204 ISBN 0-679-77663-X - Nicholas, p. 207-9 - Nicholas, p. 206 - Erwin Leiser, Nazi Cinema p44-5 ISBN 0-02-570230-0 - Erwin Leiser, Nazi Cinema p39-40 ISBN 0-02-570230-0 - Nicholas, p. 213 - Nicholas, p. 213-4 - Walter S. Zapotoczny , "Rulers of the World: The Hitler Youth" - Michael Sontheimer, "When We Finish, Nobody Is Left Alive" 05/27/2011 Spiegel - Pierre Aycoberry, The Social History of the Third Reich, 1933-1945, p 228, ISBN 1-56584-549-8 - Richard C. Lukas, Forgotten Holocaust p20 ISBN 0-7818-0528-7 - Pierre Aycoberry, The Social History of the Third Reich, 1933–1945, p 229, ISBN 1-56584-549-8 - Pierre Aycoberry, The Social History of the Third Reich, 1933–1945, p 255, ISBN 1-56584-549-8 - Nicholas, p. 215 - Nicholas, p 217 - Nicholas, p. 217 - Nicholas, p. 218 - Robert Edwin Hertzstein, The War That Hitler Won p137 ISBN 0-399-11845-4 - Leila J. Rupp, Mobilising Women for War, p 122, ISBN 0-691-04649-2, OCLC 3379930 - Robert Edwin Hertzstein, The War That Hitler Won p139 ISBN 0-399-11845-4 - "The Frauen Warte: 1935–1945" - Pierre Aycoberry, The Social History of the Third Reich, 1933-1945, p 265, ISBN 1-56584-549-8 - Karel C. Berkhoff, Harvest of Despair: Life and Death in Ukraine Under Nazi Rule p44 ISBN 0-674-01313-1 - Lynn H. Nicholas, Cruel World: The Children of Europe in the Nazi Web p336, ISBN 0-679-77663-X - Karel C. Berkhoff, Harvest of Despair: Life and Death in Ukraine Under Nazi Rule p45 ISBN 0-674-01313-1 - Karel C. Berkhoff, Harvest of Despair: Life and Death in Ukraine Under Nazi Rule p211 ISBN 0-674-01313-1 - Robert Cecil, The Myth of the Master Race: Alfred Rosenberg and Nazi Ideology p199 ISBN 0-396-06577-5 - Robert Cecil, The Myth of the Master Race: Alfred Rosenberg and Nazi Ideology p207 ISBN 0-396-06577-5 - Gerhard L. Weinberg, Visions of Victory: The Hopes of Eight World War II Leaders p 24 ISBN 0-521-85254-4 - Lynn H. Nicholas, Cruel World: The Children of Europe in the Nazi Web p255, ISBN 0-679-77663-X - Lynn H. Nicholas, Cruel World: The Children of Europe in the Nazi Web p256, ISBN 0-679-77663-X - Lebensraum, Aryanisation, Germanistion and Judenrein, Judenfrei: concepts in the holocaust or shoah - Milton, Sybil. "Non-Jewish Children in the Camps". Museum of Tolerance, Multimedia Learning Centre Online. Annual 5, Chapter 2. Copyright © 1997, The Simon Wiesenthal Centre. - Hitler's War; Hitler's Plans for Eastern Europe - Dzieciñstwo zabra³a wojna > Newsroom - Roztocze Online - informacje regionalne - Zamo¶æ, Bi³goraj, Hrubieszów, Lubaczów,Tomaszów Lubelski, Lubaczów - Roztocze OnLine - Nicholas, p 479 - Lynn H. Nicholas, p. 263 - Nicholas, p. 273 - Nicholas, p. 274 ISBN 0-679-77663-X - Leila J. Rupp, Mobilising Women for War, p 124–5, ISBN 0-691-04649-2, OCLC 3379930 - Nicholas, p. 275-6 - Nicholas, p. 277 - Nicholas, p. 278 - Lynn H. Nicholas, Cruel World: The Children of Europe in the Nazi Web p330-1, ISBN 0-679-77663-X - Waller, John H. (2002). The devil's doctor: Felix Kersten and the secret plot to turn Himmler against Hitler. Wiley, p. 20 - Louis de Jong, 1972, reprinted in German translation: H-H. Wilhelm and L. de Jong. Zwei Legenden aus dem dritten Reich : quellenkritische Studien, Deutsche Verlags-Anstalt 1974, pp 79–142. - Friesisch-Gesetz at Wikisource - Polonia in Germany - Stumpf, Rainer. "Auf der Suche nach dem Ich". Magazin Deutschland Online, 2009. Archived 19 July 2011 at the Wayback Machine - Unknown. "Der NDR 1 Niedersachsen Namenforscher". Der NDR, 2011. - Unknown. "Namensforschung: Herr Angermann wohnt am Anger". Frantfurter Allemeine Online, 2011.<\|endoftext\|>	3.734375
534	# How do you evaluate 45\div [ ( 10- 2) \times 2- 11]? Mar 9, 2018 $9$ #### Explanation: $10 - 2 = 8$ $8 \times 2 = 16$ $16 - 11 = 5$ Then .. $\frac{45}{5} = 9$ Mar 9, 2018 Use the order of operation to find the value of the expression #### Explanation: PEMDAS remember it this way "if you get hurt in (PE call an Medical Doctor, As Soon as possible) PE is one class so do exponents and parenthesis at the same time. MD is one person so do Multiplication and Division at one time working from left to right. ASap is one time so do Addition and Subtraction together working from left to right $\frac{45}{\left(10 - 2\right) \times 2 - 11} = \frac{45}{8 \times 2 - 11}$ Next do the multiplication with the parenthesis $\frac{45}{8 \times 2 - 11} = \frac{45}{16 - 11}$ Next do the subtraction within the parenthesis $\frac{45}{16 - 11} = \frac{45}{5}$ Next do the final division problem $\frac{45}{5} = 9$ Mar 9, 2018 $9$ #### Explanation: Work from the inner bracket outwards. $\text{ } 45 \div \left[\textcolor{b l u e}{\left(10 - 2\right)} \times 2 - 11\right]$ $\textcolor{w h i t e}{\times \times \times . x} \downarrow$ $= 45 \div \text{ } \left[\textcolor{red}{8 \times 2} - 11\right]$ $\textcolor{w h i t e}{\times \times x . \times . x} \downarrow$ $= 45 \div \text{ } \left[\textcolor{g r e e n}{16 - 11}\right]$ $\textcolor{w h i t e}{\times \times \times x . \times} \downarrow$ $= 45 \text{ } \div \textcolor{w h i t e}{. \times} 5$ $= \text{ } 9$<\|endoftext\|>	5
334	6.2.2 Fixed air It was well known that ‘air’ was given off by magnesia (or limestone) when treated with acids. Black sought to show that this ‘air’, which he called ‘fixed air’ (carbon dioxide), is also lost when magnesia is heated. Hampered by practical difficulties in his efforts to collect the fixed air liberated during the heating of magnesia, Black used a series of chemical reactions to prove his argument. He dissolved the magnesia usta in sulphuric acid to produce a solution of Epsom salt. This solution was treated with fixed alkali (potassium carbonate), which precipitated magnesia. This regenerated magnesia, after being washed and dried, had the weight and the properties of the original compound. As very little ‘air’ was given off during this sequence, the fixed air in the fixed alkali must have ended up in the magnesia. Black confirmed this by treating magnesia with sulphuric acid and then measuring the weight lost during this reaction, which was equal to the weight loss during calcination. Black also noted that quicklime does not absorb ordinary air, but only the small quantity of fixed air contained in it. This implied that there were at least two chemically distinct ‘airs’, and Black knew that fixed air extinguished a candle. However, he was not interested in the chemical behaviour of gases, and although he carried out experiments which revealed that birds were unable to breathe in fixed air, he did not make any further contributions to the pneumatic chemistry he had so ably helped to found.)<\|endoftext\|>	4
625	Short stories are short works of fiction in which the characters and their lives are invented by the author. This genre is akin to poetry in that it doesn't possess the length of the novel to tell its story and make its point -- a short story requires a potency that can withstand the brevity of the form. When analyzing a story, a student must examine the characters, setting, plot, conflicts and themes. The analysis of these elements serves as the material for the essay about the story. The characters are the people in a story who form the action that drives the plot forward. Because of the brevity of the form, short story writers must develop the relationships between characters quickly. Readers become familiar with the characters through their physical descriptions, thoughts, feelings, words and actions; readers also become familiar with the characters through the opinions and reactions they evoke in other characters. The point of view through which a story is narrated can limit or reveal aspects of each character. For example, if a story is told in the first person from the point of view of a particular character, access to the thoughts and motivations of other characters can be limited. When analyzing the characters, students should ask whether the character is a protagonist, antagonist, hero or villain. They should also examine what obstacles the characters face, how they deal with those obstacles and what their actions reveal about their values and beliefs. The setting is the environment in which a story takes place and can include information about the geographical location, historical era, social and political conditions, weather, and time of day. The setting can have an enormous impact on character development, conflict and plot. For example, a writer could argue that in the short story "To Build A Fire" by Jack London, the protagonist's fate is determined by the physical environment -- the freezing cold Yukon Territory of Canada -- and his reactions to it. The conflict is a struggle between two characters, a character and himself, two forces, or a character and a force. For example, a writer could argue that in "To Build a Fire," there is a conflict between the protagonist (a character) and nature (a force). In this short story, the protagonist is futilely trying to resist nature, represented by both the overwhelming cold and the instincts of his animal companion. A conflict in a story can reveal a character's weaknesses, strengths, values and perspectives. Themes are a key element to writing an analysis essay about a story. The theme of a story references an idea, belief or universal aspect of the human condition. Some common literary themes that can be found in short stories include the American Dream, the cost of war, death and grieving, racism, community, the quest for identity, overcoming adversity, and the importance of family. The short stories that are published and taught in schools tend to contain more than one theme. For example, Shirley Jackson's story "The Lottery" includes the themes of fear and cowardice, dangerous traditions, human violence, and the indoctrination of youth. Often the essay's thesis statement is drawn from some aspect of a story's theme.<\|endoftext\|>	4.28125
736	# How To Calculate Your Expected Winnings On A Powerball Ticket The Powerball lottery has now hit \$500,000,000, and one thing is now certain: It's actually statistically justifiable to buy in. We're looking at expected value now. In essence, expected value looks at how much you're expected to earn by playing a game of chance. Conveniently, Powerball publishes their probabilities so we can easily calculate the chance. Here's how you calculate the expected value of an event. Let's say you have four dollar bills of different denominations in a bag. You can't see them, but I tell you that one is a twenty, two are singles, and one is a five. I'm charging you six bucks to reach in and grab a bank note. Should you play? Well, the probability of drawing any of the four bills is 0.25. There are two singles, so the probability of drawing a single is 0.5. If we multiply the probability and payout of each event, and sum that for all events, we get the expected value of playing the game. Here's the expected value for the money game. Keep in mind I'm charging you \$6 to play: EV = (\$20—\$6)(0.25) + (\$5—\$6)(0.25) + (\$1—\$6)(0.5) EV = (\$140.25) + (-\$10.25) + (-\$50.5) = \$0.75 So, on average, you'll make seventy five cents by playing the game. So you should, theoretically, play. It's the same way with the Powerball lottery now. Powerball costs \$2 to play. You win the half-billion by getting each of the five numbers correct (of 59 possible numbers) plus the one "Powerball" correct (of 35 possible numbers). The probability of doing this is 1 in 175,223,510. But, there are a number of other prizes for getting four of five numbers right and so on. As a result we can make this table. The key here comes in the last column, which is the amount of money gained or lost times the probability of that event: The net amount of money you're expected to gain from paying in powerball tomorrow is \$1.21. This already accounts for the two dollar buy in. Keep in mind that this is only buoyed by the outrageously high jackpot. 97% of people will lose \$2 in the course of play. But it's that one in two hundred million getting a half-billion that drags this number up. In fact, Powerball is usually a loser's game. It usually has a negative expected value. The break even jackpot point for Powerball is at \$276,505,966. Anything below that jackpot has a negative expected value to play for \$2. Anything above that has a positive expected value. Keep in mind that there are a couple of caveats here. We assumed you're not doing the PowerPlay multiplier, because that exempts you from the Jackpot. Also, it's always possible that someone else could win and you've have to split the winnings. That angle severely complicates the calculation overall. Either way you look at it, though, there's an unusual statistical event happening on Powerball this week — a lottery with a positive expected value — and it's not going to last forever. ### See five Statistics Problems That Will Change The Way You See The World > More: Math Powerball<\|endoftext\|>	4.6875
138	What is it? Social-emotional development includes the child’s experience, expression, and management of emotions and the ability to establish positive and rewarding relationships with others. Why is it important? Children experience, express, and perceive emotions before they fully understand them. In order to recognize, manage, and communicate their emotions and to perceive and understand the emotions of others, children need to build the skills required to do so. These growing abilities help children to become competent in negotiating complex social interactions, to participate effectively in relationships and group activities, and to gain the benefits of social support essential for healthy human development and functioning. Socio-emotional challenges include:<\|endoftext\|>	4.3125
674	WATERLOO, Ont. (Thursday, Nov. 29, 2012) – A team of researchers from the University of Waterloo has built the world's largest simulation of a functioning brain. It can help scientists understand how the complex activity of the brain gives rise to the complex behaviour exhibited by animals, including humans. The model is called Spaun, which stands for Semantic Pointer Architecture Unified Network. It consists of 2.5 million simulated neurons. The model captures biological details of each neuron, including which neurotransmitters are used, how voltages are generated in the cell, and how they communicate. Spaun uses this network of neurons to process visual images in order to control an arm that draws Spaun's answers to perceptual, cognitive and motor tasks. The research team's findings appear in this week's issue of the journal Science. "This is the first model that begins to get at how our brains can perform a wide variety of tasks in a flexible manner—how the brain coordinates the flow of information between different areas to exhibit complex behaviour," said Professor Chris Eliasmith, Director of the Centre for Theoretical Neuroscience at Waterloo. He is Canada Research Chair in Theoretical Neuroscience, and professor in Waterloo's Department of Philosophy and Department of Systems Design Engineering. Unlike other large brain models, Spaun can perform several tasks. Researchers can show patterns of digits and letters the model's eye, which it then processes, causing it to write its responses to any of eight tasks. And, just like the human brain, it can shift from task to task, recognizing an object one moment and memorizing a list of numbers the next. Because of its biological underpinnings, Spaun can also be used to understand how changes to the brain affect changes to behaviour. "In related work, we have shown how the loss of neurons with aging leads to decreased performance on cognitive tests," said Eliasmith. "More generally, we can test our hypotheses about how the brain works, resulting in a better understanding of the effects of drugs or damage to the brain.” In addition, the model provides new insights into the sorts of algorithms that might be useful for improving machine intelligence. For instance, it suggests new methods for controlling the flow of information through a large system attempting to solve challenging cognitive tasks. Professor Eliasmith has written a book on the research. How To Build A Brain will be on shelves this winter. Videos on the project are at http://nengo.ca/build-a-brain/spaunvideos. About the University of Waterloo In just half a century, the University of Waterloo, located at the heart of Canada's technology hub, has become one of Canada's leading comprehensive universities with 35,000 full- and part-time students in undergraduate and graduate programs. Waterloo, as home to the world's largest post-secondary co-operative education program, embraces its connections to the world and encourages enterprising partnerships in learning, research and discovery. In the next decade, the university is committed to building a better future for Canada and the world by championing innovation and collaboration to create solutions relevant to the needs of today and tomorrow. For more information about Waterloo, visit www.uwaterloo.ca. Media Relations Officer Communications & Public Affairs University of Waterloo<\|endoftext\|>	3.75
543	# GMAT Quant: Factorials Factorial representation is given as N! = N.(N-1).(N-2).(N-3)……..3.2.1 The general usage for factorial lies majorly in permutation and combination. However, there might be few questions in GMAT that will be related to numbers. Don’t get frightened !!. this is just a way to represent the product of consecutive integer. Let’s look at some problems: 1. What maximum power of 10 will divide the number 30!? In 30! If we have to count the power of 10, we need to know the maximum power of 5 and maximum power of 2. Since this number will have abundant power of 2 i.e even numbers as compared to number of 5, we will be counting the number of 5’s. 30! = 30.29……25……20……15…….10….5……3.2.1 Number of 5’s in 30.25.20.15.10.5 = 7. Hence maximum power of 10 that can divide 30! Will be 7. 1. What is the maximum power of 2 in the expression 10! + 11! +12!…………20!? 10! + 11! +12!…………20! 10! (1 + 11 +12.11 + 13.12.11 +………..+ 20.19.18……13.12.11) 10! (12 +12.11 + 13.12.11 +………..+ 20.19.18……13.12.11) 10!.12 (1 +11 +13.11+14.13.11+…………..+20.19.18……..13.11) 10!.12 (12 +13.11+14.13.11+…………..+20.19.18……..13.11) We can see in the bracket that every term is even except 13.11 which is odd We know even + even is even and even + odd is odd Therefore bracket will yield an odd number. So all the power to 2 will be there in 10!. 12 Number of power of 2 in 10! = [10/2] + [10/4] +{10/8] = 8, where [] denotes the greatest integer. Number of power of 2 in 12 is 2 Hence total number of power of 2 in the expression is 8 +2 =10<\|endoftext\|>	4.5
2,868	After the United States of America liberated itself from the British, there was need for establishment of new government. A group of men who are currently referred to as the nation’s Founders are the ones came up with the government. The government was established based on the experience the founders had, what they had learnt as well as what they believed in. This group of men had a lot in common. They had great political experience, were learned and had interest in safeguarding their property. As a result, the group came up with a strong agreement on what to incorporate in the government. Immediately after United States of America was established, numerous conflicts surfaced which included economic hardships, power devolution and conflict of interests. To contain the situation, there was need for establishment of a government which was to be mandated with ensuring there were not conflicts. This led to the Founders assembling in Philadelphia so as to put down laws that would be implemented by the new government. Despite the Founders managing to come up with the constitution, there were numerous conflicts that emerged during the process. The constitution making process called for inclusion of diverse ideas. Moreover, the Founders had to compromise on some of their beliefs for the constitution to be completed. Despite the Founders agreeing on the need for establishment of new government, numerous conflicts arose. One of the conflicts that were experienced was the structure of the government. Edmund Randolph; the then governor of Virginia proposed a two-house legislature. The legislature was to have upper house which was to be given the mandate of electing the lower house. The Congress was to be given the powers of invalidating laws that it found to be contradicting the constitution. This structure comprised of a parliamentary form of government where the Congress was to be responsible of choosing the judiciary and the cabinet. On the other hand, William Paterson; a New Jersey delegate came up with a structure that gave each state equal vote to the Congress regardless of the size of the state. The structure also called for the establishment of distinct branches of the executive and the judiciary. On top of the powers accorded to the Congress in the first proposal, this structure added the powers of the Congress to comprise taxing and commerce. In his plan, Paterson also included a clause that would have seen establishment of national government law that would have been superior to the states’ law. It was hard to immediately decide on the structure to adopt and debate between the two structures went on until July. To ensure that a constitution that catered for the views of each party was achieved, Roger Sherman came up with a structure that compromised the two plans. In this structure, it was decided that two houses would be established with the upper house having two representatives from each states in spite of the size of the state. Legislation was to be made to pass the two houses. Another conflict that had to be solved during constitution making was the right to vote. Initially, it was required that for any person to be eligible to vote, he had to own property in the country. This locked a great number of people out of the exercise since very few whites owned property. James Madison was for the view that property ownership did not imply that a person was rich. There was also a conflict on whether women were to be allowed to participate in voting exercise. Benjamin Franklin called for universal male suffrage. Failure by the Founders to reach to a consensus led to them giving the individual states the responsibility of coming up with voter qualifications. Eventually, the states decided that property ownership had to remain as the qualification. All the states adopted male suffrage locking women out of the voting exercise. There was a conflict over constitution ratification. For amendments to be mad in the constitution, the Founders passed that nine of the thirteen states had to agree. Those opposed to the constitution were anti-federalists. They claimed that the national government would have taken advantage of having superior powers to oppress the states. They called for establishment of a Bill of Rights which would protect personal freedom. Anti-federalists demanded that people were to be allowed to vote on matters affecting the nation. Proponents of the constitution had the view that the government could not exercise powers other that the ones provided for in the constitution. Hence, they did not see the essence of a Bill of Rights. They urged that the constitution did not touch on individual liberties and hence it was hard for it to deny people the freedom. Despite the constitution not accounting for individual liberties, the anti-federalists wanted a Bill of Rights to be introduced since it would be hard to change the constitution in its absence. Eventually, Madison drafted a Bill of Rights that had twelve amendments and was sent to all states for endorsement. Ten of the amendments were approved, one rejected while the other one remained pending until after the ten became effective in 1791. Constitution as the fulfillment of the goals of the American Revolution For decades, there has been a perception that there was only one American Revolution. This is the fight to liberate themselves from the British colonialists. However, there were two revolutions. Apart from the desire to liberate themselves from the colonialists, Americans also wanted to liberate themselves from American merchants and financiers. The merchants and financiers had the say in the country and after independence they are the ones who dictated on how the country was to be run. Those who did not own property in the country were locked out of the voting exercise. Consequently, they did not take part in determining how the country was to be run. It is these hardships that led to the working class calling for establishment of a constitution which stipulated the voting qualifications. Despite the first constitution embracing male suffrage and also property ownership in voting, amendments have been made with time allowing all the Americans to participate in voting exercise. This liberated the Americans from the merchants and the financiers. Another revolutionary goal that the constitution helped in fulfilling was establishment of a stable government. It was not only the desire by the Founders to come up with a government but they wished to have a devolved government. They feared that concentrating all powers to a central government would compromise democracy. The constitution offered a platform for establishment of a government with different branches and checks and balances to ensure that no branch had exclusive powers. It was the American’s desire to liberate themselves from economic hardships that were being experienced at the time. To achieve this, they needed to have a sound economic plan which ensured that that country’s resources were equitable and efficiently managed. This could only be achieved by coming up with lows that governed how tax was to be levied and managed. The Philadelphia Convention came up with a law that gave the Congress powers to collect and levy taxes. It was given the responsibility of allocating funds thus ensuring that collected taxes were used in funding projects that aimed at improving national economy. The constitution directed that everybody was subject to taxation. Consequently, the wealthy were compelled to pay tax adding to the nation’s basket. Apart from collecting and levying tax the congress regulated the value of the nation’s currency. This contributed to the economic growth as the Congress ensured stability of the currency. Prior to establishment of the constitution, the different states competed in foreign tariffs. To revive the national economy, it required the states to cease from imposing tariffs on goods coming from other states. All these problems were solved by the constitution which led to establishment of a unified country where all the states worked together in improving the national economy. Compromise led to the successful completion of the Constitution With every member of the Founders wishing his interests to be catered for in the constitution, it was hard to complete the constitution. It called for the different factions to compromise on some of their interest so as to come up with a comprehensive constitution that catered for all. One of the greatest compromises that were made is in coming up with the structure of the government. The federalists and the anti-federalists had to soften their stands so as to come up with a government structure that favored both to some extent. By adapting the structure developed by Roger Sherman the group was relieved from the tug of war that was there between the two factions. They could now comfortably proceed to other issues knowing that they had agreed on the form of government to follow. In coming up with a single nation, the different states had to come up with a consensus. Some states were bigger than others while some had more resources. Getting equal representation in the national government gave the smaller states an upper hand. AS a result, the bigger states complained that the smaller states benefited at their expense. Another constitutional issue that was solved through compromise was the slavery issue. While the Northern states delegates wanted slaves to be considered for taxation reasons they did not want them to be represented in parliament. On the other hand, the South wanted the slaves to be considered only in determining the population of individual state. There was a belief that a slave was not as productive as the white man and only worked three-fifths of what the white man did. To end this conflict, a compromise was reached where a three-fifth principle was to be used in counting slaves for both population and taxation purposes. This helped in bringing the two parties together thus helping in completing the constitution. Apart from the aforementioned conflicts, other problems concerning slavery were brought to an end. The southern delegates feared that the northern states would one day ban slavery. To avoid this, they agreed that the government could not bring to end slavery for the next twenty years. Correspondingly, the new slaves were not to be taxed over ten dollars per head. It was agreed that all slaves escaping from their masters from northern states would be returned to their owners. Despite the northern states not being satisfied with the compromise, they knew that it was the only channel that would help then get a constitution. There was no way that all the states would have seen all their interests addressed without compromising on some of their demands. It would have not been possible for the United States of America to get a new constitution had it not for the compromises that were made by the Founders. The delegates had many issues in common but also differed in some. Consequently, no one can claim the constitution was not achieved through compromise. Issues such as the government structure and slavery would have made the two factions shoot down the constitution. Had the delegates refused to compromise on some of their demands, it would have been hard for them to reach into a consensus. Generally, the fate of the United States of America depended on the compromises that were made by the Founders as well as the states. It is a result of these compromises that America is proud to be a great nation. Some of the great politicians that helped in attaining the constitution include Edmund Randolph, James Madison, and Alexander Hamilton among others. It is their brainstorming that helped in drafting the constitution. Adams, Willi Paul, The First American Constitutions. North Carolina: University of North Carolina Press, 2000. Bernstein, Richard B and Rice, Kym S. Are We to Be a Nation? The Making of the Constitution. Harvard: Harvard University Press, 1987. Christopher, Collier. All Politics Is Local: Family, Friends, and Provincial Interests in the Creation of the Constitution. New York: Routledge, 2003. Currie, David P, The Constitution in Congress: Democrats and Whigs, 1829-1861. Chicago: University of Chicago Press, 2005. Ketcham, Ralph, The Anti-Federalist Papers and the Constitutional Convention Debates. New York: New American Library, 1986. McIlwain, Charles Howard, The American Revolution: A constitutional Interpretation. Harvard: Harvard University Press, 2003. Mckinnon, Ava, “The Making of the United States Constitution: A Compromise Document” last modified Sep. 28, 2007, accessed November 26, 2011, http://www.associatedcontent.com/article/395338/the_making_of_the_united_states_constitution_pg6.html?cat=37. Raphael, Ray A, People’s History of the American Revolution. London: New Press, 2001. Wood, Gordon S, The Creation of the American Republic, 1776-1787. North Carolina: University of North Carolina Press 1969. Christopher Collier, All Politics Is Local: Family, Friends, and Provincial Interests in the Creation of the Constitution (New York: Routledge, 2003), 88-93. David Currie P, The Constitution in Congress: Democrats and Whigs, 1829-1861 (Chicago: University of Chicago Press, 2005), 234-242. Paul Adams Willi, The First American Constitutions (North Carolina: University of North Carolina Press, 2000), 387-413. Richard Bernstein B and Kym Rice S, Are We to Be a Nation? The Making of the Constitution (Harvard: Harvard University Press, 1987), 310-321. Gordon Wood S, The Creation of the American Republic, 1776-1787 (North Carolina: University of North Carolina Press, 1969), 38-61. Charles Howard McIlwain, The American Revolution: A constitutional Interpretation (Harvard: Harvard University Press, 2003), 156-184. Ray Raphael A, People’s History of the American Revolution (London: New Press, 2001), 126-153. Ralph Ketcham, The Anti-Federalist Papers and the Constitutional Convention Debates (New York: New American Library, 1986), 87-104. Ava, Mckinnon, “The Making of the United States Constitution: A Compromise Document” last modified Sep. 28, 2007, accessed November 26, 2011, http://www.associatedcontent.com/article/395338/the_making_of_the_united_states_constitution_pg6.html?cat=37 Mckinnon, “Making of the United States Constitution” Collier, Politics Is Local, 219.<\|endoftext\|>	4.40625
428	The space agency revealed that it could not see the object with its Spitzer Space Telescope. And that could reveal important clues about what it actually is. Oumuamua passed by Earth in September 2017, becoming the first known interstellar visitor ever to make its way to our solar system from another one. As it flew by, researchers rushed to learn more about it, pointing telescopes and other instruments towards it in an attempt to learn as much as posisble before it disappeared out the other side of the solar system. Spitzer tried to pick out the rock in November, around two months after its closest approach. It failed to see it – but that failure puts a limit on how big the rock can be, since if it was large enough it would have been spotted, according to a new paper published in the Astronomical Journal and coauthored by scientists at NASA's Jet Propulsion Laboratory in Pasadena, California. That helps to lead credence to the theory that the relatively small object is being pushed along by gas that was being thrown out of the object. That gave the effect of adding thrust as it travelled through the solar system, speeding it up. It was that strange speeding up behaviour that led scientists to suggest that it could be an alien probe, sent past Earth by a distant civilisation. The extra propulsion could be caused by the object working as a lightsail, designed to be carried along by solar radiation, Harvard scientists recently suggested. The alternative and more accepted theory of frozen gases inside the object being expelled and pushing it along was dependent on Oumuamua being smaller than typical comets inside of our solar system. With the determination that it probably is, the research seems to suggest theories about it being an alien spacecraft are less likely. "Oumuamua has been full of surprises from day one, so we were eager to see what Spitzer might show," said David Trilling, lead author on the new study and a professor of astronomy at Northern Arizona University. "The fact that Oumuamua was too small for Spitzer to detect is actually a very valuable result."<\|endoftext\|>	4.09375
753	SAT Subject Test Chemistry REVIEW OF MAJOR TOPICS Rates of Chemical Reactions These skills are usually tested on the SAT Subject Test in Chemistry. You should be able to … • Explain how each of the following factors affect the rate of a chemical reaction: nature of the reactants, surface area exposed, concentrations, temperature, and the presence of a catalyst. • Draw reaction diagrams with and without a catalyst. • Explain the Law of Mass Action. • Describe the relationship between reaction mechanisms and rates of reaction. This chapter will review and strengthen these skills. Be sure to do the Practice Exercises at the end of the chapter. The measurement of reaction rate is based on the rate of appearance of a product or disappearance of a reactant. It is usually expressed in terms of a change in concentration of one of the participants per unit time. Experiments have shown that for most reactions the concentrations of all participants change most rapidly at the beginning of the reaction; that is, the concentration of the products shows the greatest rate of increase, and the concentrations of the reactants the highest rate of decrease, at this point. This means that the rate of a reaction changes with time. Therefore a rate must be identified with a specific time. FACTORS AFFECTING REACTION RATES Five important factors control the rate of a chemical reaction. These are summarized below. Know the factors that affect reaction rates. 1. The nature of the reactants. In chemical reactions, some bonds break and others form. Therefore, the rates of chemical reactions should be affected by the nature of the bonds in the reacting substances. For example, reactions between ions in an aqueous solution may take place in a fraction of a second. Thus, the reaction between silver nitrate and sodium chloride is very fast. The white silver chloride precipitate appears immediately. In reactions where many covalent bonds must be broken, reaction usually takes place slowly at room temperatures. The decomposition of hydrogen peroxide into water and oxygen happens slowly at room temperatures. In fact, about 17 minutes is required for half the peroxide in a 0.50 M solution to decompose. 2. The surface area exposed. Since most reactions depend on the reactants coming into contact, the surface exposed proportionally affects the rate of the reaction. 3. The concentrations. The reaction rate is usually proportional to the concentrations of the reactants. The usual dependence of the reaction rate on the concentration of the reactants can simply be explained by theorizing that, if more molecules or ions of the reactant are in the reaction area, then there is a greater chance that more reactions will occur. This idea is further developed in the collision theory discussed below. 4. The temperature. A temperature increase of 10°C above room temperature usually causes the reaction rate to double or triple. The basis for this generality is that, as the temperature increases, the average kinetic energy of the particles involved increases. As a result the particles move faster and have a greater probability of hitting other reactant particles. Because the particles have more energy, they can cause an effective collision, resulting in the chemical reaction that forms the product substance. 5. The presence of a catalyst. It is a substance that increases or decreases the rate of a chemical reaction without itself undergoing any permanent chemical change. The catalyst provides an alternative pathway by which the reaction can proceed and in which the activation energy is lower. It thus increases the rate at which the reaction comes to completion or equilibrium. Generally, the term is used for a substance that increases reaction rate (a positive catalyst). Some reactions can be slowed down by negative catalysts.<\|endoftext\|>	4.0625
920	# Two Squares In Circle ### Problem Squares $ABCD\;$ and $EFGH\;$ are inscribed in a circle, forming in intersection an octagon $MNPQRSTU.$ Prove that $[MNPQRSTU]\ge 2(\sqrt{2}-1)[ABCD],\;$ where $[\mathcal{X}]\;$ denotes the area of figure $\mathcal{X}.$ ### Proof 1 WLOG, assume that the squares have side $1.\;$ If $x=[MNPQRSTU],\;$ then $2x=2-[\text{the eight triangles}].\;$ The minor arcs ${AB}$ and ${EF}$ are equal, so that $AB=EF,\;$ making $AEBF\;$ an isosceles trapezoid. It follows that $\Delta NAF\;$ is isosceles and $NA=NF.\;$ Since triangles $NFP\;$ and $NAM\;$ are similar, we deduce that they are in fact congruent. Thus all eight triangles are congruent. We denote $AN=a\sin t,\;$ $\displaystyle 0\lt t\lt\frac{\pi}{2}.$ This implies that $NP=a\;$ and $PB=a\cos t\;$ from which $a=\displaystyle\frac{1}{1+\sin t+\cos t}.\;$ Thus $[\text{the eight triangles}]=\displaystyle\frac{4\sin t\cos t}{(1+\sin t+\cos t)^2}.\;$ If $\sin t+\cos t=y,\;$ $2\sin t\cos t=y^2-1\;$ and $a\lt y\le\sqrt{2}.$ From here, $\displaystyle\frac{4\sin t\cos t}{(1+\sin t+\cos t)^2}=\frac{2(y^2-1)}{(y+1)^2}=\frac{2(y-1)}{y+1}\le\frac{2(\sqrt{2}-1)}{\sqrt{2}+1}$ so that $[\text{the eight triangles}]\le 2(3-2\sqrt{2}),\;$ $2x\ge 2-2(3-2\sqrt{2}),\;$ or $x\ge 2(\sqrt{2}-1).$ ### Proof 2 By symmetry, the four cut-off triangles of either square are equal. By the Carpets Theorem, the sum of the ares of the cut-off triangles of one of the squares equals to the sum of the areas of the others. It follows that all eight triangles are equal. Let $x=AM=ME,\;$ $y=EU=UD,\;$ $MU=z.\;$ WLOG, assume $x+y+z=1.\;$ By the Pythagorean Theorem, $x^2+y^2=z^2=(1-x-y)^2,$ which is equivalent to $\displaystyle (1-x)(1-y)=\frac{1}{2}.\;$ By the AM-GM inequality, $\displaystyle \frac{(1-x)+(1-y)}{2}\ge\sqrt{(1-x)(1-y)}=\frac{1}{\sqrt{2}},$ or, $x+y\le 2-\sqrt{2}.$ On the other hand, $\displaystyle (1-x)(1-y)=\frac{1}{2}\;$ is equivalent to $\displaystyle xy=x+y-\frac{1}{2},\;$ implying that $\displaystyle xy\le\frac{3}{2}-\sqrt{2}.$ But $xy=2[\text{one of the triangles}].\;$ So that $\displaystyle [\text{the eight triangles}]\le 2(3-2\sqrt{2}),\;$ and, subsequently, $[MNPQRSTU]\ge 2(\sqrt{2}-1).$ Note that the minimum is achieved when $x=y.$ ### Acknowledgment The problem which is due to Miguel Ochoa Sanchez has been posted by Leo Giugiuc at the CutTheKnotMath facebook page along with a solution (Proof 1) by Dan Sitaru and Leo Giugiuc. [an error occurred while processing this directive]<\|endoftext\|>	4.5625
1,039	Susanne Aalto, Professor of Radio Astronomy, one of two astronomers who has been awarded the prestigious ERC Advanced Grant in 2018. The project’s name, HIDDeN, is in reference to galaxies that are enshrouded in dust and gas, often as a result of galaxy collisions and mergers. The dust and gas then act as a fuel during an extremely fast evolutionary phase, where a lot of new stars are born and black holes grow. The project is about understanding this development phase, helping to increase knowledge of the entire universe's evolution. Of particular interest for this project are hidden galaxy nuclei. "We have discovered extremely dust-embedded galaxy nuclei that are invisible, both in normal light and in infrared radiation. We believe that they hide a thus-far unknown, compact and very transient phase of growth. It is either an accreting supermassive black hole, or an extreme form of star birth. The hidden activity also drives huge ‘winds’ and ‘jets’ which eventually expels gas and dust from the galaxy's core. It may be that these winds act as a control system for the evolution of the galaxies.” Investigating things that are hidden sounds quite difficult. How are you doing it? “We need to use long-wavelength radio waves, invisible to the human eye, that can pass through the dust and gas and reveal the hidden activity. We have developed a method where we use radiation from molecules, and astrochemistry as ‘measuring tools. We use large international telescopes such as ALMA, the Atacama Large Millimeter Array, in Chile – where Chalmers is also an important supplier of internationally leading receiver technology (Read more: Receivers from Chalmers will image the distant universe ). Chalmers is also involved in even more long-wave technology, participating in international networks of interconnected telescopes, such as LOFAR and VLBI, and in the future, SKA.” What are you hoping the project will lead to? "We hope, among other things, to find a key to the puzzle of how supermassive black holes grow together with ‘their’ host galaxies, and to see what mechanisms drive the development of the universe forward. We are also looking for evidence that supermassive black holes can regulate their own growth. This can take place through the winds, for example. If they are powerful enough, they can propel gas from the galaxy completely. If they are weaker, the gas flows back so that it can contribute to further growth.” Your colleague, Professor Jonathan Tan at the Division of Astronomy and Plasma Physics, has also been awarded an ERC Advanced Grant this year (Read more about Jonathan's project Massive Star Formation Through the Universe). Your division is part of the Department of Space, Earth and Environment. What does it mean for Chalmers to have two such big allocations in the field of astronomical research? "The ERC awards give us the resources that make it possible to work on large scale research questions. This means that Chalmers can consolidate its place in the world’s elite in mm, submm and radio astronomy. At Astronomy and Plasma Physics we work closely with Onsala Space Observatory and this cooperation is important to our success. We are also looking forward to broadening our cooperation with other institutions, as well as other departments and institutions at Chalmers.” How do you plan to spend the ERC grant funds? "In order to address these questions, we need a coordinated observation program on several international telescopes. There is the existing facility at ALMA (link), and two new telescopes scheduled to start in 2020: the James Webb Space Telescope, which will observe space from orbit, and the SKA, or Square Kilometer Array, which will become the world's largest radio telescope. In addition, we need t further develop our modeling work on for example radiative transport, dynamics, astrochemistry and MHD simulations of jets. So we plan to use the money to build a research team.” You are the only woman at Chalmers with an ERC Advanced Grant. What are your thoughts about that? "Looking at the ERC statistics for advanced grants, Sweden is not doing so well in terms of gender equality. It is interesting to ask why this is, and what we can do about it. In general, it looks much better for starting grants than for advanced. Is this a sign that we can look forward to a new era of more prominent female researchers? Or is it a confirmation of a gloomier picture, where fewer women make it at the ‘higher’ levels? The balance has improved slightly within astronomy at Chalmers. As a researcher and head of department, I want to contribute to an environment where people are seen as individuals, and can develop, and also where women do not ‘fall away’ from research to a greater degree than men", says Susanne Aalto. Text: Christian Löwhagen.<\|endoftext\|>	3.921875
861	Fjord, also spelled fiord, long narrow arm of the sea, commonly extending far inland, that results from marine inundation of a glaciated valley. Many fjords are astonishingly deep; Sogn Fjord in Norway is 1,308 m (4,290 feet) deep, and Canal Messier in Chile is 1,270 m (4,167 feet). The great depth of these submerged valleys, extending thousands of feet below sea level, is compatible only with a glacial origin. It is assumed that the enormous, thick glaciers that formed in these valleys were so heavy that they could erode the bottom of the valley far below sea level before they floated in the ocean water. After the glaciers melted, the waters of the sea invaded the valleys. Fjords commonly are deeper in their middle and upper reaches than at the seaward end. This results from the greater erosive power of the glaciers closer to their source, where they are moving most actively and vigorously. Because of the comparatively shallow thresholds of fjords, the bottoms of many have stagnant water and are rich in black mud containing hydrogen sulfide. Glacial erosion produces U-shaped valleys, and fjords are characteristically so shaped. Because the lower (and more horizontally inclined) part of the U is far underwater, the visible walls of fjords may rise vertically for hundreds of feet from the water’s edge, and close to the shore the water may be many hundreds of feet deep. In some fjords small streams plunge hundreds of feet over the edge of the fjord; some of the world’s highest waterfalls are of this type. Fjords commonly have winding channels and occasional sharp corners. In many cases the valley, floored with glacial debris, extends inland into the mountains; sometimes a small glacier remains at the valley’s head. The river that formed the original valley commonly reestablishes itself on the upper valley floor after the disappearance of the ice and begins to build a delta at the fjord’s head. Often this delta is the only place on the fjord where villages and farms can be established. Learn More in these related Britannica articles: glacial landform: Erosional landforms of continental glaciersFjords are found along some steep, high-relief coast-lines where continental glaciers formerly flowed into the sea. They are deep, narrow valleys with U-shaped cross sections that often extend inland for tens or hundreds of kilometres and are now partially drowned by the ocean. These troughs… river: Falls attributable to discordance of river profileThese fjords are intimately associated with falls because the valley walls typically are both high and steep and because hanging valleys are ubiquitous.… river: Origin and classificationThe latter type, called fjords, are deep, narrow gorges cut into bedrock by tongues of glacial ice advancing down a former stream valley ( seeglacial landform). Fjords are most common in Norway and the coastal margins of British Columbia, Canada. Both valley types (river and glacial) became estuarine environments… lake: Basins formed by glaciationPiedmont and fjord (i.e., a river valley that has been “drowned” by a rise of sea level) lakes are found in basins formed by glacial action in long mountain valleys. Excellent examples are found in Norway, England’s Lake District, the Alps, and the Andes. In North America,… glacier: Tidewater glaciers…at the head of a fjord will have a low calving speed that may be exceeded by the ice flow speed, causing advance of the terminus. At the same time, glacial erosion will cause the deposition of sediment as a moraine shoal at the terminus. With time, the glacier will… More About Fjord13 references found in Britannica articles - major reference - Baltic Sea - In coast - dead water research - ice shelves - In ice shelf - lake basins - New Zealand - In Southland - river systems<\|endoftext\|>	4.21875
966	# Simplifying an expression involving complex numbers. • May 20th 2010, 05:12 AM Henryt999 Simplifying an expression involving complex numbers. The exercise is: write the expression in polar form: $\frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}$ Here is my work: $\frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}$ that is $\frac{2(1+i)2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2 })(2\sqrt{3}-2i)}$ and that becomes: $\frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3 })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}$ and then $\frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3 })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}$ $\frac{2\times 2 \times \sqrt{2}(cis(\frac{\pi}{4}))(cis(\frac{\pi}{3}))}{ 3\times cis(\frac{\pi}{2})\times 2\times cis(\frac{11\pi}{12})}$ anyway the correct answer is -1 I get the absolute value wrong can anyone spot my error.? • May 20th 2010, 05:25 AM HallsofIvy Quote: Originally Posted by Henryt999 The exercise is: write the expression in polar form: $\frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}$ Here is my work: $\frac{(2+2i)(1+\sqrt{3})}{(3i)(\sqrt{12}-2i)}$ that is $\frac{2(1+i)2(cis\frac{\pi}{3})}{3cis(\frac{\pi}{2 })(2\sqrt{3}-2i)}$ and that becomes: $\frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3 })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}$ and then $\frac{2\sqrt{2}cis(\frac{\pi}{4})2(cis\frac{\pi}{3 })}{3cis(\frac{\pi}{2})2(\sqrt{3}-i)}$ $\frac{2\times 2 \times \sqrt{2}(cis(\frac{\pi}{4}))(cis(\frac{\pi}{3}))}{ 3\times cis(\frac{\pi}{2})\times 2\times cis(\frac{11\pi}{12})}$ Here's your error. $\sqrt{3}- i$ has argument $-\pi/6$ or $5\pi/6$, not $-\pi/12$ or $11\pi/12$. The right triangle formed by the lines from (0, 0) to $(\sqrt{3}, -1)$, (0, 0) to $\sqrt{3}, 0$, and from $(\sqrt{3}, 0)$ to $(\sqrt{3}, -1)$ is half of an equilateral triangle. It vertex angle is 30 degrees or $\pi/6$ radians. Quote: anyway the correct answer is -1 I get the absolute value wrong can anyone spot my error.? • May 20th 2010, 05:31 AM mr fantastic Quote: Originally Posted by HallsofIvy Here's your error. $\sqrt{3}- i$ has argument $-\pi/6$ or $5\pi/6$, not $-\pi/12$ or $11\pi/12$. The right triangle formed by the lines from (0, 0) to $(\sqrt{3}, -1)$, (0, 0) to $\sqrt{3}, 0$, and from $(\sqrt{3}, 0)$ to $(\sqrt{3}, -1)$ is half of an equilateral triangle. It vertex angle is 30 degrees or $\pi/6$ radians. And the modulus of $\sqrt{3} - i$ has been forgotten in the denominator.<\|endoftext\|>	4.40625
2,690	In fluid mechanics and heat transfer, the Reynolds number Re is a dimensionless number that gives a measure of the ratio of inertial forces () to viscous forces (μ / L) and, consequently, it quantifies the relative importance of these two types of forces for given flow conditions. Reynolds numbers frequently arise when performing dimensional analysis of fluid dynamics and heat transfer problems, and as such can be used to determine dynamic similitude between different experimental cases. They are also used to characterize different flow regimes, such as laminar or turbulent flow: laminar flow occurs at low Reynolds numbers, where viscous forces are dominant, and is characterized by smooth, constant fluid motion, while turbulent flow occurs at high Reynolds numbers and is dominated by inertial forces, which tend to produce random eddies, vortices and other flow fluctuations. Reynolds number is named after Osborne Reynolds (1842–1912), who proposed it in 1883. Reynolds number can be defined for a number of different situations where a fluid is in relative motion to a surface. These definitions generally include the fluid properties of density and viscosity, plus a velocity and a characteristic length or characteristic dimension. This dimension is a matter of convention - for example a radius or diameter are equally valid for spheres or circles, but one is chosen by convention. For flow in a pipe or a sphere moving in a fluid the diameter is generally used today. Other shapes (such as rectangular pipes or non-spherical objects) have an equivalent diameter defined. For fluids of variable density (e.g. compressible gases) or variable viscosity (non-Newtonian fluids) special rules apply. The velocity may also be a matter of convention in some circumstances, notably stirred vessels. Flow in Pipe For flow in a pipe or tube, the Reynolds number is generally defined as : Definitions in non-SI units usually have a number (coefficient) in front, for example 124, where the pipe diameter is in inches, the velocity in feet per second, the fluid density in pounds per cubic foot, and the viscosity in centipoise, which are traditional USA and UK units. Flow in a Rectangular Duct For shapes such as a square or rectangular duct (where the height and width are comparable) the characteristic dimension is called the hydraulic diameter, DH, defined as 4 times the cross-sectional area, divided by the wetted perimeter. (For a circular pipe this is exactly the diameter.): Flow in a Wide Duct For a fluid moving between two plane parallel surfaces (where the width is much greater than the space between the plates) then the characteristic dimension is the distance between the plates. Flow in an Open Channel For flow of liquid with a free surface, the hydraulic radius must be determined. This is the cross-sectional area of the channel divided by the wetted perimeter. For a semi-circular channel, it is half the radius. The characteristic dimension is then 4 times the hydraulic radius (chosen because it gives the same value of Re for the onset of turbulence as in pipe flow.) Some older texts use the hydraulic radius with consequently different values of Re for transition and turbulent flow. Sphere in a Fluid The characteristic dimension is the diameter of the sphere. The velocity is that of the sphere relative to the fluid some distance away from the sphere. Note that the density is that of the fluid, not that of the sphere. Note that purely laminar flow only exists up to Re = 0.1 under this definition. For flow of fluid through a bed of approximately spherical particles of diameter D in contact, if the voidage (fraction of the bed not filled with particles) is ε and the superficial velocity V (i.e. the velocity through the bed as if the particles were not there - the actual velocity will be higher) then a Reynolds number can be defined as: Laminar conditions apply up to Re = 10, fully turbulent from 2000. In a cylindrical vessel stirred by a central rotating paddle, turbine or propellor, the characteristic dimension is the diameter of the agitator D. The velocity is ND where N is the rotational speed (revolutions per second). Then the Reynolds number is: The system is fully turbulent for values of Re above 10 000. Transition Reynolds number In boundary layer flow over a flat plate, experiments can confirm that, after a certain length of flow, a laminar boundary layer will become unstable and become turbulent. This instability occurs across different scales and with different fluids, usually when , where x is the distance from the leading edge of the flat plate, and the flow velocity is the 'free stream' velocity of the fluid outside the boundary layer. For flow in a pipe of diameter D, experimental observations show that for 'fully developed' flow, laminar flow occurs when ReD < 2300 and turbulent flow occurs when ReD > 4000. In the interval between 2300 and 4000, laminar and turbulent flows are possible ('transition' flows), depending on other factors, such as pipe roughness and flow uniformity). This result is generalised to non-circular channels using the hydraulic diameter, allowing a transition Reynolds number to be calculated for other shapes of channel. These transition Reynolds numbers are also called critical Reynolds numbers, and were studied by Osborne Reynolds around 1895 [see Rott]. Reynolds number in pipe friction Pressure drops seen for fully-developed flow of fluids through pipes can be predicted using the Moody diagram which plots the friction factor f against Reynolds number Re and relative roughness ε / D. The diagram clearly shows the laminar, transition, and turbulent flow regimes as Reynolds number increases. The similarity of flows In order for two flows to be similar they must have the same geometry, and have equal Reynolds numbers and Euler numbers. When comparing fluid behaviour at homologous points in a model and a full-scale flow, the following holds: where quantities marked with 'm' concern the flow around the model and the others the actual flow. This allows engineers to perform experiments with reduced models in water channels or wind tunnels, and correlate the data to the actual flows, saving on costs during experimentation and on lab time. Note that true dynamic similitude may require matching other dimensionless numbers as well, such as the Mach number used in compressible flows, or the Froude number that governs open-channel flows. Some flows involve more dimensionless parameters than can be practically satisfied with the available apparatus and fluids (for example air or water), so one is forced to decide which parameters are most important. For experimental flow modeling to be useful, it requires a fair amount of experience and judgment of the engineer. Typical values of Reynolds number Onset of turbulent flow ~ 2.3 × 10 to 5.0 × 10 for pipe flow to 10 for boundary layers Reynolds number sets the smallest scales of turbulent motion In a turbulent flow, there is a range of scales of the time-varying fluid motion. The size of the largest scales of fluid motion (sometimes called eddies) are set by the overall geometry of the flow. For instance, in an industrial smoke stack, the largest scales of fluid motion are as big as the diameter of the stack itself. The size of the smallest scales is set by the Reynolds number. As the Reynolds number increases, smaller and smaller scales of the flow are visible. In a smoke stack, the smoke may appear to have many very small velocity perturbations or eddies, in addition to large bulky eddies. In this sense, the Reynolds number is an indicator of the range of scales in the flow. The higher the Reynolds number, the greater the range of scales. The largest eddies will always be the same size; the smallest eddies are determined by the Reynolds number. What is the explanation for this phenomenon? A large Reynolds number indicates that viscous forces are not important at large scales of the flow. With a strong predominance of inertial forces over viscous forces, the largest scales of fluid motion are undamped—there is not enough viscosity to dissipate their motions. The kinetic energy must "cascade" from these large scales to progressively smaller scales until a level is reached for which the scale is small enough for viscosity to become important (that is, viscous forces become of the order of inertial ones). It is at these small scales where the dissipation of energy by viscous action finally takes place. The Reynolds number indicates at what scale this viscous dissipation occurs. Therefore, since the largest eddies are dictated by the flow geometry and the smallest scales are dictated by the viscosity, the Reynolds number can be understood as the ratio of the largest scales of the turbulent motion to the smallest scales. Example of the importance of the Reynolds number If an airplane wing needs testing, one can make a scaled down model of the wing and test it in a wind tunnel using the same Reynolds number that the actual airplane is subjected to. If for example the scale model has linear dimensions one quarter of full size, the flow velocity would have to be increased four times to obtain similar flow behaviour. Alternatively, tests could be conducted in a water tank instead of in air (provided the compressibility effects of air are not significant). As the kinematic viscosity of water is around 13 times less than that of air at 15 °C, in this case the scale model would need to be about one thirteenth the size in all dimensions to maintain the same Reynolds number, assuming the full-scale flow velocity was used. The results of the laboratory model will be similar to those of the actual plane wing results. Thus there is no need to bring a full scale plane into the lab and actually test it. This is an example of "dynamic similarity". Reynolds number is important in the calculation of a body's drag characteristics. A notable example is that of the flow around a cylinder. Above roughly 3×10 Re the drag coefficient drops considerably. This is important when calculating the optimal cruise speeds for low drag (and therefore long range) profiles for airplanes. Reynolds number in physiology Poiseuille's law on blood circulation in the body is dependent on laminar flow. In turbulent flow the flow rate is proportional to the square root of the pressure gradient, as opposed to its direct proportionality to pressure gradient in laminar flow. Using the Reynolds equation we can see that a large diameter with rapid flow, where the density of the blood is high, tends towards turbulence. Rapid changes in vessel diameter may lead to turbulent flow, for instance when a narrower vessel widens to a larger one. Furthermore, an atheroma may be the cause of turbulent flow, and as such detecting turbulence with a stethoscope may be a sign of such a condition. Reynolds number in viscous fluids Where the viscosity is naturally high, such as polymer solutions and polymer melts, flow is normally laminar. The Reynolds number is very small and Stokes Law can be used to measure the viscosity of the fluid. Spheres are allowed to fall through the fluid and they reach the terminal velocity quickly, from which the viscosity can be determined. The laminar flow of polymer solutions is exploited by animals such as fish and dolphins, who exude viscous solutions from their skin to aid flow over their bodies while swimming. It has been used in yacht racing by owners who want to gain a speed advantage by pumping a polymer solution such as low molecular weight polyoxyethylene in water, over the wetted surface of the hull. It is however, a problem for mixing of polymers, because turbulence is needed to distribute fine filler (for example) through the material. Inventions such as the "cavity transfer mixer" have been developed to produce multiple folds into a moving melt so as to improve mixing efficiency. The device can be fitted onto extruders to aid mixing. Where does it come from? Each term in the above equation has the units of a volume force or, equivalently, an acceleration times a density. Each term is thus dependant on the exact measurements of a flow. When one renders the equation nondimensional, that is that we multiply it by a factor with inverse units of the base equation, we obtain a form which does not depend directly on the physical sizes. One possible way to obtain an nondimensional equation is to multiply the whole equation by the following factor: where the symbols are the same as those used in the definition of the Reynolds number. If we now set: we can rewrite the Navier-Stokes equation without dimensions: where the term : Finally, dropping the primes for ease of reading: This is why mathematically all flows with the same Reynolds number are comparable. References and notes Published - July 2009 Please see some ads intermixed with other content from this site: Copyright 2004-2019 © by Airports-Worldwide.com<\|endoftext\|>	3.984375
598	Subnetting enables the network administrator to further divide the host part of the address into two or more subnets. In this case, a part of the host address is reserved to identify the particular subnet. This is easier to see if we show the IP address in binary format. The subnet mask is the network address plus the bits reserved for identifying the subnetwork — by convention, the bits for the network address are all set to 1, though it would also work if the bits were set exactly as in the network address. In this case, therefore, the subnet mask would be 11111111.11111111.11110000.00000000. It’s called a mask because it can be used to identify the subnet to which an IP address belongs by performing a bitwise AND operation on the mask and the IP address. Computers that belong to a subnet are addressed with a common, identical, most-significant bit-group in their IP address. This results in the logical division of an IP address into two fields, a network or routing prefix and the “rest” field or host identifier. The rest field is an identifier for a specific host or network interface. The routing prefix may be expressed in CIDR notation written as the first address of a network, followed by a slash character (/), and ending with the bit-length of the prefix. For example, 192.168.1.0/24 is the prefix of the Internet Protocol Version 4 network starting at the given address, having 24 bits allocated for the network prefix, and the remaining 8 bits reserved for host addressing. The IPv6 address specification 2001:db8::/32 is a large address block with 296 addresses, having a 32-bit routing prefix. For IPv4, a network may also be characterized by its subnet mask, which is the bitmask that when applied by a bitwise AND operation to any IP address in the network, yields the routing prefix. Subnet masks are also expressed in dot-decimal notation like an address. For example, 255.255.255.0 is the network mask for the 192.168.1.0/24 prefix. Traffic is exchanged (routed) between subnetworks with special gateways (routers) when the routing prefixes of the source address and the destination address differ. A router constitutes the logical or physical boundary between the subnets. The benefits of subnetting an existing network vary with each deployment scenario. In the address allocation architecture of the Internet using Classless Inter-Domain Routing (CIDR) and in large organizations, it is necessary to allocate address space efficiently. It may also enhance routing efficiency, or have advantages in network management when subnetworks are administratively controlled by different entities in a larger organization. Subnets may be arranged logically in a hierarchical architecture, partitioning an organization’s network address space into a tree-like routing structure.<\|endoftext\|>	4.46875
466	1. ## Fractions problem How can I show that $\displaystyle \frac {3}{10}$ is $\displaystyle \frac {1}{7}$ of the way from $\displaystyle \frac {1}{4}$ to $\displaystyle \frac {3}{5}$? 2. Originally Posted by sarahh How can I show that $\displaystyle 3/10$ is $\displaystyle 1/7$ of the way from $\displaystyle 1/4$ to $\displaystyle 3/5$? First you need to find out the length of "the whole way" from $\displaystyle \frac{1}{4}$ to $\displaystyle \frac{3}{5}$, which you do by subtraction: $\displaystyle \frac{3}{5} - \frac{1}{4} = \frac{12}{20} - \frac{5}{20} = \frac{7}{20}$. So what's 1/7 of 7/20? 1/20. So now the question becomes, is $\displaystyle \frac{3}{10}$ a distance of $\displaystyle \frac{1}{20}$ from $\displaystyle \frac{1}{4}$? $\displaystyle \frac{3}{10} - \frac{1}{4} = \frac{6}{20} - \frac{5}{20} = \frac{1}{20}$ We figured out that $\displaystyle \frac{3}{5}$ and $\displaystyle \frac{1}{4}$ are $\displaystyle \frac{7}{20}$ apart. So a number which is $\displaystyle \frac{1}{7}$ of the way from $\displaystyle \frac{1}{4}$ to $\displaystyle \frac{3}{5}$ is going to be a distance of $\displaystyle \frac{1}{20}$ from $\displaystyle \frac{1}{4}$. Therefore, the desired answer is $\displaystyle \frac{1}{4} + \frac{1}{20} = \frac{5}{20} + \frac{1}{20} = \frac{6}{20} = \frac{3}{10}$.<\|endoftext\|>	4.53125
1,112	## Introducing vectors (part 1) I think that teaching vectors to 14-16 year olds is a bit like teaching them to play the flute; that is to say, it’s a bit like teaching them to play the flute as presented by Monty Python (!) Monty Python (1972), ‘How to play the flute’ Part of the trouble is that the definition of a vector is so deceptively and seductively easy: a vector is a quantity that has both magnitude and direction. There — how difficult can the rest of it be? Sadly, there’s a good deal more to vectors than that, just as there’s much more to playing the flute than ‘moving your fingers up and down the outside'(!) What follows is a suggested outline teaching schema, with some selected resources. ### Resultant vector = total vector: the ‘I’ phase ‘2 + 2 = 4’ is often touted as a statement that is always obviously and self-evidently true. And so it is — arithmetically and for mere scalar quantities. In fact, it would be more precisely rendered as ‘scalar 2 + scalar 2 = scalar 4’. However, for vector quantities, things are a wee bit different. For vectors, it is better to say that ‘vector 2 + vector 2 = a vector quantity with a magnitude somewhere between 0 and 4’. For example, if you take two steps north and then a further two steps north then you end up four steps away from where you started. Also, if you take two steps north and then two steps south, then you end up . . . zero steps from where you started. So much for the ‘zero’ and ‘four’ magnitudes. But where do the ‘inbetween’ values come from? Simples! Imagine taking two steps north and then two steps east — where would you end up? In other words, what distance and (since we’re talking about vectors) in what direction would you be from your starting point? This is most easily answered using a scale diagram. To calculate the vector distance (aka displacement) we draw a line from the Start to the End and measure its length. The length of the line is 2.8 cm which means that if we walk 2 steps north and 2 steps east then we up a total vector distance of 2.8 steps away from the Start. But what about direction? Because we are dealing with vector quantities, direction just as important as magnitude. We draw an arrowhead on the purple line to emphasise this. Students may guess that the direction of the purple ‘resultant’ vector (that is to say, it is the result of adding two vectors) is precisely north-east, but this can be a vague description so let’s use a protractor so that we can find the compass bearing. And thus we find that the total resultant vector — the result of adding 2 steps north and 2 steps east — is a displacement of 2.8 steps on a compass bearing of 045 degrees. ### Resultant vector = total vector: the ‘We’ phase How would we go about finding the resultant vector if we moved 3 metres north and 4 metres east? If you have access to an interactive whiteboard, you could choose to use this Jamboard for this phase. (One minor inconvenience: you would have to draw the straight lines freehand but you can use the moveable and rotatable ruler and protractor to make measurements ‘live’ with your class.) We go through a process similar to the one outlined above. • What would be a suitable scale? • How long should the vertical arrow be? • How long should the horizontal arrow be? • Where should we place the ‘End’ point? • How do we draw the ‘resultant’ vector? • What do we mean by ‘resultant vector’? • How should we show the direction of the resultant vector? • How do we find its length? • How do we convert the length of the arrow on the scale diagram into the magnitude of the displacement in real life? The resultant vector is, of course, 5.0 m at a compass bearing of 053 degrees. ### Resultant vector = total vector: the ‘You’ phase Students can complete the questions on the worksheets which can be printed from the PowerPoint below. Answers are shown on this second PowerPoint, plus an optional digital ruler and protractor are included on the third slide if you wish to use them. Enjoy! Uncategorized ### 2 thoughts on “Introducing vectors (part 1)” 1. David Beauchamp January 1, 2023 / 10:22 pm Thank you – once again you have succinctly summarised a technique I have groped towards over some years of trial and much error! And to remind you that the Monty Python joke is one of Shakespeare’s, from Hamlet: ‘Tis as easy as lying: govern these ventages with your fingers and thumb, give it breath with your mouth, and it will discourse most eloquent music. (Act 2, Sc. 3) Didn’t work for him, either. • e=mc2andallthat January 1, 2023 / 10:28 pm Ha! I beg to differ…Monty Python greater than Shakepeare is! But many thanks for the encouraging comment 🙂<\|endoftext\|>	4.6875
390	Sales Toll Free No: 1-800-481-2338 # how to solve missing number problems TopEquation is made of a Numbers and variables. We can take a variable as missing number of an equation, for an example 2 * x = 10. Here 'x' is a variable and this is used to represent a missing number of this equation. We will discuss how to solve missing number problems. Missing number can occur in sequence or in equation. We will discuss both cases. Case 1 : Assume that we have a Set of numbers 1 ,3 , 5 , 7 , x , 13 , 17. Here in this set of numbers there is a missing number which is represented by variable 'x'. Now we have to find value of 'x'. We can see that in this set of numbers, all numbers are Prime Numbers. These prime numbers are in sequence. Prime number sequence consist of numbers 1, 3, 5, 7, 11, 13, 17. Now we will match both Sets. From sets we will get to know the value of missing number 'x' that is equals to 11. Case 2: As we know that equation is made of numbers and variables. Variable is used to show missing number. Assume that we have an equation 2x + 4 = 8. Here 'x' is the missing number. Now we will find value of 'x'. We will separate variables and numbers on different sides of equation like: 2x = 8 – 4, 2x = 4, x = 4 / 2, x = 2. Here missing value of equation is x = 2. We can verify value of missing number by placing it in equation. 2 * 2 + 4 = 8, 4 + 4 = 8, 8 = 8.<\|endoftext\|>	4.4375
1,200	Typical electronic components include--we have three groups here. I'm going to be looking at Voltage Sources, Fuses and Circuit Breakers, so this will be part A. Then we're going to be looking at Resistors in the second section that will be 3.3 B. I hope to conclude this with Capacitors, Inductors, Transformers, Transistors and Diodes. This would be in the third section, 3.3 C. I don't think a D will be needed. We're planning to do it as A, B and C. Okay, first thing we're going to look at is batteries. A cell is a stand-alone source of energy. A battery is two or more cells connected together to provide higher voltage and/or current to a circuit. The more cells the higher the voltage supply. Cells and batteries are rated by voltage and ampere-hour ratings. Ah stands for ampere-hour. One Ah means: The battery can deliver one amp of current for one hour or 1/4 amp of current for four hours, or 2 amps of current for 30 minutes. You have laws of variation here. This is the concept of an Ah. Here we have a symbol for a battery. The longer terminal here is the positive side and the shorter side is the negative side. Fuses and Circuit Breakers Excessive current flow in an electrical circuit may damage the circuit. Now that's probably quite obvious. You have too much current flowing through a circuit, it will overheat and it can burn up and destroy components. Typical causes of increased current flow are low-resistance connections outside the normal path of current flow. These are called short circuits. Fuses and circuit breakers provide protection from excessive current flow. That is a schematic symbol for a fuse. Fuses are connected so that current flowing through the circuit also flows through the fuse. Here we have a fuse, and I suppose we could go in here and connect a battery. We could come back over here and go up like so and connect a ground here. We would have a circuit. Fuses are connected so that current flowing through the circuit also flows through the fuse. The fuse becomes a part of the circuit. A resistive link inside the fuse heats up when current flows through it. Excessive current flow causes the link to burn, which stops the current flow and protects the circuit. In this case, here are the fuses in the circuit. If we're pulling too much current through this then the fuse will burn and will have an open at this point and current will cease to flow. A fuse that has burned open is said to be, and here's the term, a blown fuse. Fuses have three electrical ratings. We're going to look at Current ratings, Voltage ratings and Response time. First of all, current rating: Max sustained current that can flow through the fuse without it opening. The voltage rating: The minimum amount of voltage required to ark across the fuse after it blows. Notice that it's after it blows. The voltage rating of the fuse must exceed the highest voltage that can be expected across the circuit. If the voltage rating is exceeded, the blown fuse can ark defeating the purpose of the fuse. Response time: When the current exceeds its rating, the fuse link will burn open. It takes a certain amount of time for the fuse link to disintegrate. This time is referred to as the “response time.” Response times vary from several milliseconds to several seconds. Due to the type of circuit, the response time will have to meet those needs. If you have real sensitive circuits and they have a fast response time. Not so sensitive circuits would not need a quick response time. Fuses may be classified according to their appropriate response times. We have three. There is the Slo-blo, the Fast-blo, and the Normal-blo. First of all, let's look at Slo-blo. These fuses are designed to withstand currents which greatly exceed their current rating as long as the over current condition is only momentary. Short term transients or surges won't typically blow this type of fuse and the response time of 1/10 to 10 seconds. There's a term that is slo-blo. These are usually in a household. You would have slo-blo fuses if you have transients or surges in the voltage line. Those will not blow. If they did you'd have to be resetting the circuit breaker all the time because surges and transients are really quite common. Then there is the Fast-blo. These have a very short response time. They protect sensitive electronic components like transistors. They range from sub millisecond through hundreds of milliseconds. These could be less than a millisecond of time in the response. They are very fast. Then there are Normal-blo: response time between slo-blo and fast acting, so this would be in between the slow and the fast. Circuit breakers also provide over current protection for electrical devices but, unlike fuses, do not have to be replaced after they have opened the circuit. A circuit breaker that has opened in response to over current is said to be, and here's the term, tripped. Circuit breakers have the advantage of being reused in a circuit, but typically, have slower response times than fuses. This means they are tolerant to short term over voltages and won't protect sensitive electronic equipment. This is why surge protectors are needed on sensitive electronic equipment like computers. Okay, this was a quick lesson we looked at. Circuit breakers, fuses. We looked at the different ratings of fuses and we looked at batteries. This introduces electronic components. The next section, 3B, we'll be looking at resistors.<\|endoftext\|>	4.34375

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