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Talk about chops! Comprehend the key structure like never before. Instantly transpose any progression into each and every key. Easily Take complexity out of learning your invaluable chord theory. Most study of music theory begins with scales. With a little research, you'll discover that scales will often be explained in terms of notation (the lines and dots), 'whole-step/half-step' patterns (which refer to the distance between notes in the scale) or a combination of the two. Unfortunately both methods are rather complex and usually result in confusion. Let's avoid this frustration by taking a step back and looking at scales simply in terms of how chords are constructed.As you may know, a scale is simply a set of notes (usually seven) that take you from a given note through a series of other notes until it reaches the same note again an octave higher. Though the initial note is much lower in tone than the finishing note, the way our ears hear music they sound almost identical. (Think of the two "Do's" in "Do, Ra. Mi, Fa, So, La, Ti, Do.") This pattern repeats over and over, up and down. This circular behavior is common in music theory and is a principal reason the Chord Wheel is constructed the way it is. The most fundamental scale is the Major Scale. (The "Do, Ra, Mi, etc." example we referred to earlier is the Major Scale.) Instead on concentrating on how the scale is devised, let's focus on how chords are constructed from a scale. A 'C Major Scale' starts from a C note and climbs through seven of the eleven notes in-between until it reaches the C an octave higher. By assigning numbers starting with the first note and moving upwards counting, we reach the octave C where it would be counted '8' (or another '1' if we consider it to be starting over).The most basic chord is called a triad as it contains three notes. Triads are put together by first taking any note in a scale as a starting point (called the root of the chord). From the root we skip a note in the scale and add the next note instead. So if we were starting from the first note in the C Major Scale (which, of course, is a C and we earlier numbered as '1') we would add the note numbered '3.' By once again skipping a note (the one we numbered '4') and adding the '5' note we have three notes and our first chord!Our three notes, the '1, 3 & 5' notes of the scale provide us with the first chord of the Family of Chords for the C Major Scale. By constructing chords from each note in the scale (using the 'every other note' pattern) we come up with seven basic triads (i.e. three note chords). Whenever we use the C Major Scale and the chords we have derived from it, we are said to be working in the 'Key of C.'Confused yet? If you are, fear not! Just read on, follow along and read it again later on. Like any good theory, it just takes getting used to. The beauty of actually using the Chord Wheel is that much of this knowledge will be just handed to you. However, it's important to have access to the actual theory behind what the Chord Wheel will make so easy for you to practice. © Copyright 1995 – 2010, by The Chord Wheel. All rights reserved.

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- Historic Sites The Sparck Of Rebellion Badly disguised as Indians, a rowdy group of patriotic vandals kicked a revolution into motion Winter 2010 | Volume 59, Issue 4 On September 5 representatives from every colony except Georgia met in Philadelphia at what came to be known as the first Continental Congress. Radicals called for Samuel Adams’s trade embargo, while moderates led by John Jay of New York and Joseph Galloway of Pennsylvania supported a strongly worded protest. All agreed that some form of action had to be taken. On behalf of the radicals, Joseph Warren of Massachusetts introduced the Suffolk Resolves, declaring the Intolerable Acts to be in violation of the colonists’ rights as English citizens and urging the creation of a revolutionary colonial government. Much to his surprise, the resolves passed, if just barely. George III was infuriated at the whole business. To him, the very calling of the Continental Congress was proof of perfidy. “The New England governments are in a state of rebellion,” he told North. Gen. Thomas Gage, the British commander and now Massachusetts governor, received orders to strike a blow at the New England rebels. Gage learned of their whereabouts and sent troops to seize them and then destroy the supply facility at Concord. But that night Boston silversmith Paul Revere rode the 20 miles to Lexington to warn the radical leaders and everyone else along the way that the British were coming. When British troops reached the town on April 19, 1775, they encountered an armed force of 70, some of them “minutemen,” a local militia formed by an act of the provincial congress the previous year. Tensions were high and tempers short on both sides, but as the Lexington militia’s leader, Capt. John Parker, would state after the battle, he had not intended to “make or meddle” with the British troops. In fact it was the British who were advancing to form a battle line when a shot rang out—whether it was from a British or colonial musket, no one knows to this day. At the time neither side realized it was the first blast of the American Revolution. The lone shot was followed by volleys of bullets that killed eight and wounded 10 minutemen before the British troops marched on to Concord and burned the few supplies the Americans had left there. But on their march back to Boston, the British faced the ire of local farmers organized into a well-trained embryo army, which outnumbered the British five to one and shot at them from every house, barn, and tree. By nightfall total casualties numbered 93 colonists and 273 British soldiers, putting a grim twist on Samuel Adams’s earlier exclamation to John Hancock, “What a glorious morning for America is this!” A declaration of independence was no longer a pipe dream but a revolutionary plan in the making. From the Tea Party to the bloody fields of Concord, the thirteen colonies had proved that direct action was the surest way to free themselves from British tyranny.

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Researchers from the School of Biological Sciences discover first new chlorophyll in 60 years 20 August 2010 Found by accident in stromatolites from Western Australia's Shark Bay, the new pigment named chlorophyll f can utilise lower light energy than any other known chlorophyll. The historic study published online in Science, challenges our understanding of the physical limits of photosynthesis - revealing that small-scale molecular changes to the structure of chlorophyll allows photosynthetic organisms to survive in almost any environment on Earth. The new chlorophyll was discovered deep within stromatolites - rock-like structures built by photosynthetic bacteria, called cyanobacteria - by lead author Dr Min Chen from the University of Sydney. A team of interdisciplinary scientists, including Dr Martin Schliep and Dr Zhengli Cai (University of Sydney); Associate Professor Robert Willows (Macquarie University); Professor Brett Neilan (University of New South Wales) and Professor Hugo Scheer (University of Munich), characterised the absorption properties and chemical structure of chlorophyll f, making it the fifth known type of chlorophyll molecule on Earth. Chlorophyll is the essential molecule in oxygenic photosynthesis - the process that enables plants, algae and some bacteria to convert carbon dioxide into sugar and oxygen by using free energy from sunlight. Until recently, oxygenic photosynthesis was thought only to occur in light that is visible to human eyes, between 400nm to 700nm, as chlorophyll was strictly limited to absorbing light in this range. This was overturned in 1996 when scientists found a cyanobacterium that could photosynthesise using light just outside the visible spectrum - at 710nm, in the infrared region - using a modified chlorophyll molecule, named chlorophyll d. Since this discovery, scientists around the world have been puzzled by how chlorophyll d is able to get enough energy from infrared light for photosynthesis. Now the rules of photosynthesis need to be rewritten again, with the discovery of a new chlorophyll that can absorb light of even lower photon energy - 720nm - making it the most red-shifted chlorophyll to date. In ecological terms, chlorophyll f allows cyanobacteria living deep within stromatolites to photosynthesise using low-energy infrared light, the only light able to penetrate into the structure, which challenges further our understanding of the physical limits of photosynthesis. Dr Chen explains: "Finding the new chlorophyll was totally unexpected - it was one of those serendipitous moments of scientific discovery. "I was actually looking for chlorophyll d, which we knew could be found in cyanobacteria living in low light conditions. I thought that stromatolites would be a good place to look, since the bacteria in the middle of the structures don't get as much light as those on the edge." After obtaining a sample of stromatolite from Hamelin Pool, Dr Chen looked for chlorophyll d by culturing the cyanobacterial sample in infrared light of 720nm. This ensured only the survival of cyanobacteria that had chlorophylls able to absorb and use infrared light. High performance liquid chromatography of the cultured sample performed six months later revealed not only trace amounts of chlorophyll d, but also a new chlorophyll not seen before. Testing the optical absorption spectrum of the new chlorophyll revealed that it could absorb much longer wavelengths of light than any other known chlorophyll - 10nm longer than chlorophyll d and more than 40nm longer than chlorophyll a. Sequential mass spectral analysis revealed the molecular weight of the new pigment to be 906 mass units. Then nuclear magnetic resonance (NMR) spectroscopy was performed to determine the chemical structure of the chlorophyll. Results indicated that chlorophylls a, b, d and f have very similar chemical structures, differing only in the position of a substitution. Yet these tiny differences in structure give the chlorophylls very different spectral properties, and hence can function in very different light environments. "Discovering this new chlorophyll has completely overturned the traditional notion that photosynthesis needs high energy light," Dr Chen said. "It is amazing that this new molecule, with a simple change to its chemical structure, can absorb extremely low energy light. This means that photosynthetic organisms can utilise a much larger portion of the solar spectrum than we previously thought and that the efficiency of photosynthesis is much greater than we ever imagined. "Chlorophyll f, and its ability to absorb infrared light, can have numerous applications to industries like plant biotechnology and bioenergy. "For us, the next challenge is to work out the function of this new chlorophyll in photosynthesis. "Is its job to capture additional red light and pass it on to another chlorophyll, like chlorophyll a, in the reaction centre for photosynthesis? "Or is it the only chlorophyll responsible for photosynthesis in the cyanobacterium? And if it is, then we will be speechless wondering how this molecule can get enough energy from infrared light to make oxygen from water. "Whatever happens next, the fact that we have discovered a cyanobacterium that exploits a tiny modification in its chlorophyll molecule to photosynthesise in light that we cannot see, opens our mind to the seemingly limitless ways that organisms adapt to survive in their environment." Media enquiries: Rachel Gleeson, 0403 067 342, 9351 4312, [email protected] Contact: Carla Avolio Phone: 02 9351 4543

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Ever heard that “you’re born with all the brain cells you’ll ever have”? It turns out that could be a good thing – if it were true. A new study shows that at least in some circumstances, neurogenesis actually impairs memory performance. To understand why this might be the case, consider that adults are constantly generating new neurons in a long-term memory structure – the hippocampus. This region requires a large number of neurons to store episodic memories accumulated over a lifetime (and understandably so!). Similarly, to be able to store experiences that may have occurred very quickly, neurons in the hippocampus learn very fast. Finally, to reduce interference from other similar memories, neural activity there is very sparse and highly non-overlapping (so that the hippocampus doesn’t accidentally confuse two similar experiences). These characteristics make this region particularly useful for quickly learning relationships between stimuli. Accordingly, damage to the hippocampus, and disruption of neurogenesis in particular, will usually result in impaired learning. However, Saxe et al. showed that disrupting neurogenesis actually improves performance on some of the more complex learning tasks we know of! Saxe et al. placed mice in the middle of an asterisk-shaped “radial arm maze” with 8 arms; at the end of each arm was a morsel of food. Then, one arm was opened. After mice had retrieved the food from the end of that arm, the same thing was repeated with another arm. Finally, after mice had retrieved food from the second arm, the mice were given a 30- second delay, and then given the option to return to the first arm or to an adjacent arm. In this case, returning to the first arm would reflect a memory failure – that arm had already been cleared of food. Mice without neurogenesis (as dirsupted by X-Rays and a genetic manipulation) actually did better on this task than normal mice – they were more likely to go to the adjacent arm in search of food, rather than return to the original arm from which they had already cleared the food! Mice with and without neurogenesis performed equally on a task where the delay was reduced to 5-seconds. A similar pattern emerged on another version of the task in which mice were only tested on the first arm vs. an adjacent arm, demonstrating that the memory advantage of disrupting neurogenesis is not due simply to less interference from the memory of the second arm. To explain this thoroughly confusing result, the authors suggested two explanations: 1) Mice are less likely to suffer from the “interfering memory” of visiting the first arm because having less neurogenesis results in worse short-term memory – there is no memory to cause interference! However, this explanation seems unlikely both because of previous results (showing no short-term memory defiicts resulting from hippocampal irradiation) and because short-term memory could just as easily help performance as hurt it in the way they imply. The second explanation is far more interesting, though somewhat counterintuitive: 2) Less neurogenesis has no effect on short-term memory, but reduces interference by making overlap between cells less likely. “But wait, didn’t you say the purpose of neurogenesis was to create less overlap, by increasing the total number of available neurons?” Ultimately, new neurons do result in less overlap between representations, but young neurons are quite different from older neurons in that they are far more excitable. In other words, very young neurons will fire to almost anything, and will cause increased interference from previous memories. So less really can be more: an abundance of new neurons can result in overexcitability in the hippocampus, a region that requires very sparse, low-level patterns of activity in order to store memories without the threat of interference. Over the long term, the young neurons become less excitable and do ultimately increase the capacity of memory (indeed, they may do so even earlier for highly distinct memories). An interesting question for future research will be the kind of cues that are required to recruit non-overlapping regions of the hippocampus to make a memory “highly distinct” and thus resistant to the initially deleterious effects of hippocampal neurogenesis.

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× Get Full Access to Elementary Statistics - 12 Edition - Chapter 10.6 - Problem 9ccq Get Full Access to Elementary Statistics - 12 Edition - Chapter 10.6 - Problem 9ccq × # 9 CQQ The exercises are based on the following sample data ISBN: 9780321836960 18 ## Solution for problem 9CCQ Chapter 10.6 Elementary Statistics | 12th Edition • Textbook Solutions • 2901 Step-by-step solutions solved by professors and subject experts • Get 24/7 help from StudySoup virtual teaching assistants Elementary Statistics | 12th Edition 4 5 1 345 Reviews 11 1 Problem 9CCQ ?9 CQQ The exercises are based on the following sample data obtained from different second-year medical students who took blood pressure measurements of the same person (based on data from Marc Triola, MD). S y s t o l i c D i a s t o l i c If you had computed the value of the linear correlation coefficient to be 3.335, what should you conclude? Step-by-Step Solution: Solution 9 CQQ Answer : Step 1 : The value of correlation coefficient must lies in between -1 to 1. If the value correlation is -1, then it is negatively correlated, if it is 1 then we can say that positively correlated, if the value is o then there is no correlation. The correlation coefficient value be 3.335, we can say that there must be error in computation, the correlation value does not exceed more than one on the positive side. Step 2 of 1 ##### ISBN: 9780321836960 This textbook survival guide was created for the textbook: Elementary Statistics, edition: 12. Since the solution to 9CCQ from 10.6 chapter was answered, more than 400 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 9CCQ from chapter: 10.6 was answered by , our top Statistics solution expert on 03/15/17, 10:30PM. This full solution covers the following key subjects: based, data, marc, computed, conclude. This expansive textbook survival guide covers 121 chapters, and 3629 solutions. Elementary Statistics was written by and is associated to the ISBN: 9780321836960. The answer to “?9 CQQ The exercises are based on the following sample data obtained from different second-year medical students who took blood pressure measurements of the same person (based on data from Marc Triola, MD). S y s t o l i c D i a s t o l i c If you had computed the value of the linear correlation coefficient to be 3.335, what should you conclude?” is broken down into a number of easy to follow steps, and 68 words. #### Related chapters Unlock Textbook Solution

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188.8.131.52 Floods and droughts Documented trends in floods show no evidence for a globally widespread change. Although Milly et al. (2002) identified an apparent increase in the frequency of ‘large’ floods (return period >100 years) across much of the globe from the analysis of data from large river basins, subsequent studies have provided less widespread evidence. Kundzewicz et al. (2005) found increases (in 27 cases) and decreases (in 31 cases) and no trend in the remaining 137 cases of the 195 catchments examined worldwide. Table 1.3 shows results of selected changes in runoff/streamflow, lake levels and floods/droughts. Other examples of changes in floods and droughts may be found in Table SM1.2. Globally, very dry areas (Palmer Drought Severity Index, PDSI ² -3.0) have more than doubled since the 1970s due to a combination of ENSO events and surface warming, while very wet areas (PDSI ³ +3.0) declined by about 5%, with precipitation as the major contributing factor during the early 1980s and temperature more important thereafter (Dai et al., 2004). The areas of increasing wetness include the Northern Hemisphere high latitudes and equatorial regions. However, the use of PDSI is limited by its lack of effectiveness in tropical regions. Table 1.3 shows the trend in droughts in some regions. Documented trends in severe droughts and heavy rains (Trenberth et al., 2007, Section 3.8.2) show that hydrological conditions are becoming more intense in some regions, consistent with other findings (Huntington, 2006).

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# How Many Candy Corns Are in a Bag? Exploring the Sweet Science Behind the Perfect Proportion ## Introduction Candy corn is one of the most beloved treats during the Halloween season. But have you ever wondered just how many candy corns are in a bag? If so, you’ve come to the right place! In this article, we’ll explore the sweet science behind calculating the exact amount of candy corns that fit into a bag. We’ll cover topics such as estimating the volume of a bag, calculating candy corn density, measuring candy corn size, understanding the mechanics and physics of candy corns, and tips on counting up the perfect number. So grab a bag of candy corn and let’s get started! ## Calculating How Many Candy Corns Fit in a Bag The first step in determining how many candy corns fit in a bag is to estimate the volume of the bag. This can be done by measuring the length, width, and height of the bag and multiplying those three numbers together. For example, if the bag measures 10 inches long, 5 inches wide, and 3 inches tall, then the volume of the bag would be 150 cubic inches. The next step is to calculate the density of candy corn. This can be done by dividing the weight of the candy corn by its volume. For example, if a bag of candy corn weighs 1 pound and has a volume of 150 cubic inches, then the density of the candy corn would be 6.67 pounds per cubic inch. The third step is to measure the size of each candy corn. This can be done by measuring the length, width, and height of a single candy corn. For example, if the candy corn measures 0.5 inches long, 0.25 inches wide, and 0.125 inches tall, then the volume of the candy corn would be 0.015625 cubic inches. ## The Sweet Science Behind How Many Candy Corns are in a Bag Now that we know the volume of the bag, the density of the candy corn, and the size of each candy corn, it’s time to dive into the sweet science behind how many candy corns fit into a bag. To do this, we need to understand the mechanics, physics, and chemistry of candy corn. The mechanics of candy corn involve understanding the forces that act upon it when it is placed in a bag. These forces include gravity, friction, and air pressure. Gravity causes the candy corn to settle at the bottom of the bag while friction causes the candy corn to stick together. Air pressure also plays a role in how the candy corn is distributed in the bag. The physics of candy corn involve understanding the properties of the material itself. Candy corn is made up of sugar, water, corn syrup, and various flavorings. When these ingredients are mixed together, they create a substance with certain physical properties such as viscosity, surface tension, and elasticity. All of these factors play a role in how the candy corn behaves when placed in a bag. Finally, the chemistry of candy corn involves understanding the chemical reactions that occur when the ingredients are mixed together. Different ingredients will react differently, creating different flavors and textures. The type of ingredients used in the candy corn will affect how it looks, tastes, and behaves in a bag. ## A Guide to Knowing the Ideal Amount of Candy Corns for Your Bag Now that we understand the sweet science behind how many candy corns fit in a bag, it’s time to figure out the ideal amount of candy corn for your bag. The first step is to identify your needs. Do you want a bag of candy corn for a party or for yourself? Do you want a bag that’s full or one that’s half-full? Once you’ve identified your needs, you can move on to the next step. The next step is to find the right size bag. The size of the bag will determine how much candy corn you can fit in it. If you’re planning on buying a large bag, then you’ll need to account for the extra space. On the other hand, if you’re buying a small bag, then you’ll need to make sure there’s enough room for the candy corn. The final step is to determine the optimal proportion of candy corn. This can be done by dividing the volume of the bag by the volume of a single candy corn. The result will give you the ideal number of candy corns for your bag. For example, if you have a bag that’s 150 cubic inches and each candy corn is 0.015625 cubic inches, then the optimal proportion would be 9,600 candy corns. ## Estimating the Perfect Proportion of Candy Corns for a Bag Once you’ve determined the optimal proportion of candy corn for your bag, it’s time to estimate the perfect proportion. There are several factors that can affect the perfect proportion including the shape of the bag, the size of the candy corn, and the type of candy corn. For example, if the bag is square, then the optimal proportion may be higher than if the bag is round. Similarly, if the candy corn is larger, then the optimal proportion may be lower than if the candy corn is smaller. There are also several techniques that can be used to estimate the perfect proportion. One technique is to fill the bag with a test batch of candy corn and measure the amount that fits. This method is best for estimating the perfect proportion for a specific type of candy corn. Another technique is to calculate the exact volume of the bag and divide it by the volume of a single candy corn. This method is best for estimating the perfect proportion for any type of candy corn. Finally, there are some tips that can be used to ensure an accurate measurement when estimating the perfect proportion. For example, it’s important to make sure the bag is completely filled with candy corn and that all of the pieces are the same size. Additionally, it’s important to make sure the measurements are precise and that the measurements are taken from the same point in the bag. ## Counting Up the Perfect Number of Candy Corns for a Bag Once you’ve estimated the perfect proportion of candy corn for your bag, it’s time to count up the exact number. There are several strategies that can be used to count the exact number of candy corn. One strategy is to count each piece individually. This method is best for small bags where the pieces can be easily seen. Another strategy is to use a scale to weigh the bag and then divide the weight by the weight of a single candy corn. It’s also important to keep track of the number of candy corn. One method is to mark off each piece as it’s counted. This method works best for small bags where the pieces can be easily seen. Another method is to use a tally system to keep track of the number of pieces. This method works best for larger bags where the pieces can’t be easily seen. Finally, there are some tricks that can be used to ensure an accurate count. For example, it’s important to make sure all of the pieces are the same size and to count each piece twice to make sure the count is correct. Additionally, it’s important to count the pieces in batches to reduce the chance of missing pieces. ## Conclusion In conclusion, determining the exact number of candy corns that fit in a bag is not as straightforward as it may seem. It requires an understanding of the mechanics, physics, and chemistry of candy corn as well as an understanding of the size of the bag and the size of the candy corn. By following the steps outlined in this article, you should now be able to calculate the exact number of candy corns that fit in a bag. So grab a bag of candy corn and start counting! To summarize, this article explored the science behind how many candy corns fit in a bag. We discussed topics such as estimating the volume of a bag, calculating candy corn density, measuring candy corn size, understanding the mechanics and physics of candy corns, and tips on counting up the perfect number. Armed with this knowledge, you should now be able to calculate the exact number of candy corns that fit in a bag.

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# Testing Students of high school have 10 points for each good solved task. The wrong answer is deducted by 5 points. After solving 20 tasks, student Michael had 80 points. How many tasks did he solve correctly and how many wrong? Correct result: a =  12 b =  8 #### Solution: a+b = 20 10a - 5b = 80 a+b = 20 10•a - 5•b = 80 a+b = 20 10a-5b = 80 a = 12 b = 8 Our linear equations calculator calculates it. We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you! Tips to related online calculators Do you have a linear equation or system of equations and looking for its solution? Or do you have quadratic equation? ## Next similar math problems: • Trip cost In September, the trip cost CZK 12,000. How many crowns did the trip cost in June of the same year, when they have since reduced the price by a quarter and by another CZK 1,200? Susan thought, "If I read 15 pages a day, I will read the whole book in 8 days. “How many pages would she have to read a day if she wanted to finish the book on the 6th day from the start of reading? And how many pages does the book have? • Worker salary The worker had a salary of CZK 18,000. During the year, his salary was increased by a quarter. He earned a total of CZK 247,500 for the whole year. From which month was his salary increased? If I read 15 pages a day, I will read the whole book in 18 days. How long will it take to read a book if I only read 9 pages every day? • Sum of seven The sum of seven consecutive odd natural numbers is 119. Determine the smallest of them. • The sum The sum of five consecutive odd numbers is 75. Find out the sum of the second and fourth of them. Suzan reads a book. If she read half an hour a day, she would read it in nine days. How many minutes does he have to read a day if he wants to read it three days earlier? • The tourist The tourist walked a quarter of the way on the first day, a third of the rest on the second day, and 20 km on the last day. How many km did he walk in three days? • What time is it? What time is it, when there is 4 times less time left until midnight than the time that has elapsed since noon? What time is it, when one-fifth of the hours that have passed since midnight is equal to one-third of the hours that are missing by 12 o'clock? • Second side Calculate the length of the other side of the rectangle if its circumference is 60 cm and one side is 10 cm long. • Work together Two bricklayers plastering a wall. The first would plaster it in 8 hours, the second in 12 hours. How many hours will they be done with the work if they work together? • Diggers The excavation was carried out in 4 days and 5 workers worked on it for 7 hours a day. Determine how long it would take to excavate if 7 workers worked on it 8 hours a day. • Tourist route How long is the tourist route when tourists crossed four-sevenths of the way on foot, crossed the bus twice less than on foot and passed the last 14 kilometers by boat. • The farmer The farmer calculated that the supply of fodder for his 20 cows was enough for 60 days. He decided to sell 2 cows and a third of the feed. How long will the feed for the rest of the peasant's herd last? • Job applicants Job applicants: three-fourths of applicants had experience for a position. The number that did not have prior experience was 36. How many people applied for the job? • Library New books were purchased for the library. Five-eighths were professional books, one-fifth were encyclopedias, and 231 books were dictionaries. How many professional books were there? • Large family The average age of all family members (children, mother, father, grandmother, grandfather) is 29 years. The average age of parents is 40 years, grandparents 66 years and all children are 5 years. How many children are there in this family?

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"It's a really amazing-quality genome," says David Reich of Harvard Medical School in Boston. "It's as good as modern human genome sequences, from a lot of ways of measuring it." The pinky belonged to a girl who lived tens of thousands of years ago. Scientists aren't sure about the exact age. She is a member of an extinct group of humans called Denisovans. The name comes from Denisova cave in Siberia, where the pinky was found. Two years ago, scientists at the Max Planck Institute for Evolutionary Anthropology in Leipzig reported that they had been able to get just enough DNA from the fossil to make a rough sequence of her DNA. But Matthias Meyer developed a far more efficient way of recovering ancient DNA, so he went back to the tiny amount of DNA left over from the first effort, and reanalyzed it. "And from this little leftover, we were able to determine the sequence of the Denisova genome 30-fold over," says Meyer. What that means is they were able to look at every single location along all of this girl's chromosomes 30 times to be absolutely certain that they had the right DNA letter in the right spot. The new results appear in the online edition of the journal Science. The high-quality sequence gives scientists valuable new data for studying ancient humans. Researchers have begun, for example, to explore which modern human populations may have inherited genes from Denisovans.

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I'm a huge believer in hands-on education. But you have to have the right tools. If I'm going to teach my daughter about electronics, I'm not going to give her a soldering iron. And similarly, she finds prototyping boards really frustrating for her little hands. So my wonderful student Sam and I decided to look at the most tangible thing we could think of: Play-Doh. And so we spent a summer looking at different Play-Doh recipes. And these recipes probably look really familiar to any of you who have made homemade play-dough — pretty standard ingredients you probably have in your kitchen. We have two favorite recipes — one that has these ingredients and a second that had sugar instead of salt. And they're great. We can make great little sculptures with these. But the really cool thing about them is when we put them together. You see that really salty Play-Doh? Well, it conducts electricity. And this is nothing new. It turns out that regular Play-Doh that you buy at the store conducts electricity, and high school physics teachers have used that for years. But our homemade play-dough actually has half the resistance of commercial Play-Doh. And that sugar dough? Well it's 150 times more resistant to electric current than that salt dough. So what does that mean? Well it means if you them together you suddenly have circuits — circuits that the most creative, tiny, little hands can build on their own. (Applause) And so I want to do a little demo for you. So if I take this salt dough, again, it's like the play-dough you probably made as kids, and I plug it in — it's a two-lead battery pack, simple battery pack, you can buy them at Radio Shack and pretty much anywhere else — we can actually then light things up. But if any of you have studied electrical engineering, we can also create a short circuit. If I push these together, the light turns off. Right, the current wants to run through the play-dough, not through that LED. If I separate them again, I have some light. Well now if I take that sugar dough, the sugar dough doesn't want to conduct electricity. It's like a wall to the electricity. If I place that between, now all the dough is touching, but if I stick that light back in, I have light. In fact, I could even add some movement to my sculptures. If I want a spinning tail, let's grab a motor, put some play-dough on it, stick it on and we have spinning. (Applause) And once you have the basics, we can make a slightly more complicated circuit. We call this our sushi circuit. It's very popular with kids. I plug in again the power to it. And now I can start talking about parallel and series circuits. I can start plugging in lots of lights. And we can start talking about things like electrical load. What happens if I put in lots of lights and then add a motor? It'll dim. We can even add microprocessors and have this as an input and create squishy sound music that we've done. You could do parallel and series circuits for kids using this. So this is all in your home kitchen. We've actually tried to turn it into an electrical engineering lab. We have a website, it's all there. These are the home recipes. We've got some videos. You can make them yourselves. And it's been really fun since we put them up to see where these have gone. We've had a mom in Utah who used them with her kids, to a science researcher in the U.K., and curriculum developers in Hawaii. So I would encourage you all to grab some Play-Doh, grab some salt, grab some sugar and start playing. We don't usually think of our kitchen as an electrical engineering lab or little kids as circuit designers, but maybe we should. Have fun. Thank you. (Applause)

Hands-on science with squishy circuits

# 2/3 times 5 in fraction form 2/3 x 5 = 10/3. 2/3 times 5 is equal to 10/3 in fraction and 3.3333 in decimal form. 2/3 times 5 in fraction form: 2/3 x 5 = ? Arrange the fractions in the product expression form as like the below: = 2/3 x 5/1 = (2 x 5)/(3 x 1) Check the numerator and denominator and cancel if anything cancelled each other: = 10/3 2/3 x 5 = 10/3 Hence, 2/3 times 5 as a fraction is equal to 10/3. 2/3 times 5 as a decimal: 2/3 x 5 = 10/3 2/3 x 5 = 3.3333 2/3 times 5 as a decimal is 3.3333 where, 2/3 is the multiplicand, 5 is the multiplier, 10/3 is the simplest form of 2/3 times 5, 3.3333 is the decimal form of 2/3 times 5. Important Notes: 2/3 * 5 All the following questions represent 2/3 times 5 in fraction form, so it's very much important to observe the different variations of this question. 1. what is 2/3 times 5 in fraction form? 2. what is 5 times 2/3 as a fraction? 3. 2/3 x 5 = ? 4. 5 x 2/3 = ? 5. 2/3 * 5 = ? 6. 5 * 2/3 = ? For values other than 2/3 times 5, use this below tool:

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This is a 109-page pack that includes digitally designed anchor charts for 3rd grade topics/concepts. You can print these and compile them into a notebook or you can hang them in your room when you're covering a certain topic/concept. There are four subject areas covered: - Language Arts/Writing - Science/Social Studies Here are the concepts covered in each area: * Reading * what readers do, where readers read, just right books, characters, setting, problem and solution, main idea and details, text connections, summary, theme vs. plot, fiction vs. nonfiction, reference materials, fluency, accuracy, comprehension, expression, point of view, narrator, sequence of events, author and illustrator, title page, table of contents, index, glossary, transition words, cause and effect * Language Arts/Writing * where writers write, the writing process, types of writing, synonyms and antonyms, rhyming words, blends, conjunctions, figurative language, capitalization, punctuation, commas, ABC order, subjects and predicates, types of sentences, simple vs. compound sentences, nouns, adjectives, verbs, possessive nouns, common vs. proper nouns, comparative and superlative adjectives, adverbs, prefixes, base words, suffixes, Latin and Greek roots, homophones, multiple meaning words, syllables, quotation marks * Mathematics * addition/subtraction with regrouping, rounding to the nearest ten/hundred, place value, odd and even numbers, standard form, expanded form, word problems, multiplication strategies, division strategies, missing/unknown factor, probability/reasonableness, linear measurement, mass/volume, area and perimeter, fractions, money, graphs, shapes, time, AM/PM, elapsed time, number patterns, lines, angles, triangles * Science/Social Studies * Paul Revere, Susan B. Anthony, Frederick Douglass, Mary McLeod Bethune, Franklin D. Roosevelt, Eleanor Roosevelt, Lyndon B. Johnson, Cesar Chavez, map grids, compass/directions, latitude and longitude, Equator and Prime Meridian, branches of national government, habitats, pollution, conservation, rocks, minerals, soil, fossils, magnets, heat, economics, landforms, timelines The final page is a huge THANK YOU to all the fabulous artists who create clip art and fonts for all of us to use! Thank you to KPM Doodles, Ginger Snaps, Fancy Dog Studio, and First Grade a la Carte for their talents!!! ***Printing - if you're worried about using a bunch of color ink, try printing the pages in black and white. Print them on colored paper! That will save you TONS on colored ink! :) I hope you and your students enjoy these anchor charts. They are a fabulous resource for review and mastery.

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## MSTAR Interventions ### Common Misconceptions and How to Prevent Them #### Examples for Preventing or Correcting Some students believe ratios always compare a part to a whole, like fractions. Provide an example in which students compare a part to a part (e.g., 1 green apple to 3 red apples). Explain that ratios can compare part to part, not only part to whole. The ratio 1 to 3 compares part of the apples to another part of the apples. Some students struggle to simplify ratios. Demonstrate an example of simplification, explicitly describing the steps: 1. Check for common factors. (The easiest thing to remember is to use the prime numbers 2, 3, 5, 7, and 11.) 2. Divide each part of the ratio by the common factor. Example: Show students 8 red and 4 yellow counters. Ask, “What is the ratio of red to yellow counters?” () Ask, “How do I know whether this ratio will simplify?” (Both numbers must divide by a common factor.) Ask, “What divides into 8 and 4 evenly?” (2 and 4) Say, “When 2 factors divide evenly, to simplify, choose the largest factor.” (4) Say, “When I divide by 1 whole, the ratio stays the same. Because 4 is a common factor, I need to divide by  , which is 1 whole." Ask, “What is 8 divided by 4? What is 4 divided by 4?” (2, 1) Say, “So  becomes  when it is simplified.” Ask, “What is the common factor for the ratio 10 to 15?” (5) Some students believe that ratios and rates can always be simplified to a mixed number or a whole number. Remind students that ratios and rates must compare 2 quantities; therefore, you often cannot simplify a ratio or a rate to a mixed number. Examples: 15 boys to 5 bats becomes 3 boys to 1 bat, not 3. 14 girls to 4 boys becomes 7 girls to 2 boys, not 3 .

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### GRE : Quantitative Comparison Problem Directions This is the Quantitative Comparison Problem-1. Each of the following questions consists of two quantities, one in Column A and one in Column B. You are to compare these two quantities and choose A if the quantity in Column A is greater B if the quantity in Column B is greater C if the two quantities are equal D if relationship cannot be determined from the information given Since there are only four choices, never mark E for these type of questions. This is the Quantitative Comparison Problem-1.Understand the basic mathematical principles behind this problem and other problems. Example-1 J – K = 2 K – 6 = 4 Column A : J Column B : 10 Form the Second equation, we get K = 6 + 4 = 10. Substituting this value in the first equation, we get J = 2 + K = 2 + 10 = 12. Since J (12) is greater than 10, we can choose A as the answer. This is the Quantitative Comparison Problem-1.Understand the basic mathematical principles behind this problem and other problems. From Quantitative Comparison Problem-1 to HOME PAG

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Diabetes is a common condition that can affect adults & children. It occurs when the body cannot produce enough insulin or stops producing insulin completely. Insulin helps control the level of glucose (sugar) in the blood. There are 2 types of diabetes – Type 1 and Type 2. Symptoms of diabetes include increased thirst, passing urine more often – especially at night, tiredness, weight loss, thrush and blurred vision. Some people may not experience any symptoms. What is the difference between Type 1 & Type 2 Diabetes? Type 1 Diabetes: In Type 1 Diabetes, the body stops producing insulin. Type 1 Diabetes usually develops in children and young adults, although it can occur at any age. The management of Type 1 Diabetes requires treatment with insulin, combined with healthy eating and exercise. Type 2 Diabetes: In Type 2 Diabetes, the body still produces insulin, but not enough for the body’s needs and the insulin it does produce is not used effectively (this is known as insulin resistance). Over three quarters of people with diabetes have Type 2 Diabetes. There is a strong link between this type of diabetes and being overweight. A simple blood test will indicate if you are at risk of developing diabetes and will also help confirm a diagnosis. Should you have any concerns please make an appointment to see our practice nurse who will assess your individual case and follow up with an appointment in our diabetic clinic if Taking control of your Diabetes If you discover that you have diabetes, you may be worried about what the future will hold and what changes you have to make to your life. Our team are here to give you the advice and support that you need. You play a very important role in the management and control of your diabetes. Weight loss, changes to your diet, regular physical activity, and medication if required, will all help control the glucose level in your blood and keep it as near normal as possible. Poorly controlled diabetes can, over time, cause damage to your eyes, nerves, kidneys, blood vessels and heart. To reduce the risk of damage here are some of the steps you can take: Diet: Eat a healthy balanced diet Weight: It is important to keep your weight at the right level for your height. Smoking: You need to stop smoking Foot care: Take care of your feet and check them daily. Physical activity: Regular activity promotes food health. it helps to lower your blood glucose, lose weight and reduce blood pressure. Monitoring: You will play an important role in monitoring your blood glucose levels. Education: Learn as much as possible about your condition as this will help you to manage it. It is important that you take these measures to manage your diabetes.

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During this crucial period, the United States pursued a policy of expansion based on “manifest destiny,” the ideology that Americans were in fact destined to extend their nation across the continent. Illustration of “Manifest Destiny” ( John Gast, 1872) The United States even proved to be willing to go to war to secure new territories. While it managed to negotiate an agreement with Great Britain to secure the Oregon territory, acquiring the valuable territory south of it—including California and its important Pacific harbors—required the use of force, and, in 1845, the United States embarked on its first offensive war by invading Mexico. In addition to advancing westward, the United States also continued to expand economically through investment in foreign markets and international trade. With these growing commercial interests, came a larger navy and increased international presence. The United States began to turn to the Pacific for new economic opportunities, establishing a presence in China, and opening Japan and Korea to western commercial interests.

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From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid. [Take π = 3.14.] Height of solid cylinder = h = 8 cm Radius of solid cylinder = r = 6 cm Volume of solid cylinder = πr2h = 3.14 × 6 × 6 × 8 cm3 = 904.32 cm3 Curved Surface area of solid cylinder = 2πrh Height of conical cavity = h = 8 cm Radius conical cavity = r = 6 cm Volume of conical cavity = 1/3 πr2h = 1/3 × 3.14 × 6 × 6 × 8 cm3 = 301.44 cm3 Let l be the slant height of conical cavity l2 = r2 + h2 l2 = (62 + 82) cm2 l2 = (36 + 64) cm2 l2 = 100 cm2 l = 10 cm Curved Surface area of conical cavity = πrl Since, conical cavity is hollowed out from solid cylinder, so, volume and total surface area of remaining solid will be found out by subtracting volume and total surface area of conical cavity from volume and total surface area of solid cylinder. Volume of remaining solid = Volume of solid cylinder – Volume of conical cavity Volume of remaining solid = 904.32 cm3 – 301.44 cm3 = 602.88 cm3 Total surface area of remaining solid = Curved Surface area of solid cylinder + Curved Surface area of conical cavity + Area of circular base Total surface area of remaining solid = 2πrh + πrl + πr2 = πr × (2h + l + r) = 3.14 × 6 × (2 × 8 + 10 + 6) cm2 = 3.14 × 6 × 32 cm2 = 602.88 cm2 Rate this question : How useful is this solution? We strive to provide quality solutions. Please rate us to serve you better.

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# How do you find the derivative of [x+(x+sin^2x)^3]^4? $\frac{d}{\mathrm{dx}} {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4}$ $= 4 {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left[1 + 3 {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + \sin 2 x\right)\right]$ Take note: $\sin 2 x = 2 \cdot \sin x \cdot \cos x$ #### Explanation: From the given ${\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4}$ $\frac{d}{\mathrm{dx}} {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4} =$ $4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{4 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x + {\left(x + {\sin}^{2} x\right)}^{3}\right)$ $= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{3 - 1} \cdot \frac{d}{\mathrm{dx}} \left(x + {\sin}^{2} x\right)\right)$ $= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + 2 \cdot \sin x \cdot \cos x\right)\right)$ $= 4 \cdot {\left[x + {\left(x + {\sin}^{2} x\right)}^{3}\right]}^{3} \cdot \left(1 + 3 \cdot {\left(x + {\sin}^{2} x\right)}^{2} \cdot \left(1 + \sin 2 x\right)\right)$ God bless....I hope the explanation is useful.

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Quercus robur (English oak) Unrivalled king of the forest in Britain, English oak (pedunculate oak) is synonymous with strength, size and longevity. Acorns and leaves of Quercus robur (Photo: Wolfgang Stuppy) Quercus robur L. English oak, pedunculate oak, common oak Least Concern (LC) according to IUCN Red List criteria. Timber, tanning leather, food for livestock. Tannic acid in the leaves is poisonous to horses if consumed in excess, damaging the kidneys. Acorns are poisonous to horses and cattle, though swine can consume them safely in moderation. About this species English oak is probably the most well-known and best-loved of the tree species native to Britain. This king of the forest can live for more than a millennium according to some sources, and grow up to 40 m high. Mature specimens are usually home to many species of wildlife. Since the Druids, and probably long before, the oak has played an important role in British culture. Couples were still wed under ancient oaks as late as Oliver Cromwell's time and the Yule log, kept from one year to another to warm the Christmas celebrations, was traditionally cut from oak. Geography and distribution Quercus robur is native to Asia Minor (an area corresponding to the western two-thirds of Turkey), North Africa, the Caucasus (a geopolitical region at the border of Europe and Asia) and Europe. One of English oak's most recognisable characteristics is the shape of its leaves. Pale green in colour, they have four or five lobes on each side and are attached to the branches with almost no stalk. In contrast, the acorns are borne on long stalks known as peduncles, hence the name ‘pedunculate oak’. Large, old oaks often have dead branches at the top. This 'stag head' is not necessarily a sign that the tree is dying. When water is short the oak simply stops supplying its upper extremities, and as a result extends its lifespan. For such a huge, long-lived and widespread tree, the oak is surprisingly bad at reproducing naturally. First, it can be a full 50 years before the first crop of acorns (seeds) is produced. Second, most of the tens of thousands of acorns dropped are eaten by animals, or simply rot. It is therefore left to the buried acorns of forgetful or dying squirrels or jays to ensure the continued lifecycle of this giant of the countryside. Threats and conservation Oakwoods have diminished greatly in the UK over the last few hundred years, but the Forestry Commission is now actively promoting woodland planting with native tree species, such as the English oak. A fungal disease known as ‘sudden oak death’ has recently arrived in the UK, apparently from Europe and although the disease affects many other species including Rhododendron, Viburnum, beech, sweet chestnut and holm oak, at present native oaks appear to be relatively resistant to it. However, it may pose a threat to oaks in the future – particularly if the trees become stressed by climate change. Another, more worrying ‘disease’ known as ‘acute oak decline’ has recently been identified in the UK. This does kill our native oaks and its causes have yet to be understood. English oak's role in biodiversity Quercus robur supports a wealth of organisms which benefit from the food, support and shelter it supplies. Its acorns are rich in starch and provide food for birds such as the Eurasian jay (Garrulus glandarius) and small mammals such as squirrels. An abundance of insects live on the leaves, buds and bark, and even inside the acorns. In autumn the soft leaves break down easily to form a rich leaf mould beneath the tree, which supports a wealth of invertebrates and fungi (such as the edible beefsteak fungus Fistulina hepatica). The trunk and branches offer physical support to mosses, ferns and lichens, and the open leaf canopy lets light through to the woodland floor, meaning flowers such as violets, bluebells and primroses can thrive. The green oak moth (Tortrix viridana) feeds on oak leaves in its larval (caterpillar) form and then rolls itself up in a leaf to pupate into the adult. This in turn is parasitised by at least four species of ichneumon wasps, which lay their eggs inside the defenceless pupae. Gall wasps lay their eggs inside the oak leaves, where their larvae then secrete a chemical causing the leaf to mutate and form a gall (oak apple), providing them with protection and shelter – though not enough to prevent them from being parasitised by other species of ichneumon wasp! In years when populations of caterpillar or moth larvae are high almost all the leaves on a tree can be eaten, yet the English oak is able to survive this trauma. Birds such as great-spotted woodpeckers (Dendrocopos major) come to feed on insects in the bark. Acorns that have fallen to the ground provide food for rooks, wood pigeons and mice, which in turn attract birds of prey such as owls, buzzards and sparrowhawks. Fungi, such as the orange oak bolete (Leccinum quercinum), form symbiotic associations with English oak roots, the fungi helping the tree to extract nutrients from the soil whilst in turn benefiting from sugars provided by the oak. Even after its death the English oak continues to support biodiversity. The decaying wood can host a vast array of fungi and the hollow trunks provide roosting sites for owls and hibernation sites for bats. Quercus robur was named for its robust or sturdy nature (robur means strength in Latin), and since iron tools were first made, people have been felling this mighty tree for its strong and durable timber. It can take as long as 150 years before an oak is ready for use in construction, but it is well worth the wait. Until the middle of the 19th century, when iron became the material of choice for building ships, thousands upon thousands of oaks were felled every year. Oak has many other uses; oak bark has been used in leather tanning and in dyeing, insect galls have been used to make black ink, and the acorns are valued as food for livestock. Pig grazing rights in oak woods, known as 'pannage', have been jealously guarded through the centuries. In the New Forest in southern England these rights are still exploited by Commoners (local people with rights to the forest’s resources), though on a smaller scale than in the past. Acorns have also been roasted to produce a coffee substitute, apparently quite inferior to the real thing. The majority of the oaks propagated at Kew are grown from seed. This method is preferred for the scientific collections as it is the easiest and safest means of collecting oaks in the wild without damaging their populations. Acorns are collected green and sown directly into the Arboretum Nursery field during the autumn. The seeds germinate the following spring. Oak seed does not store well and quickly becomes non-viable as it dries out. Mice and squirrels present a threat to the seeds, and netting or wire mesh is used to keep them at bay during autumn and winter. Rare oak specimens are sown in pots in a frost-free glasshouse for added protection. Occasionally, young trees are also brought in from nurseries, for example to give instant impact in heritage plantings on the vistas. These are grown in a reputable nursery for 12 months where they can be inspected for pests and diseases before they are transplanted to Kew. The fresh foliage of young oaks often suffers from mildew. Trees affected in this way are not treated because once they are planted out the mildew no longer presents a problem. It is often necessary to stake young trees. Tony Hall, who manages the oak collection at Kew, says that the Arboretum is treated as an extension of the nursery, in the sense that young trees there are carefully monitored during their early years. The ‘Coubertin oak’ at Kew The ‘Coubertin oak’ at Kew is one of a ‘ribbon’ of 40 English oaks (Quercus robur) planted between the Olympic Park in London and Much Wenlock in Shropshire, to celebrate the London 2012 Olympic Games. It was grown from one of 40 acorns collected by students in Much Wenlock. The acorns were from a tree planted in 1890 in honour of Baron Pierre de Coubertin who was invited to visit the Wenlock Olympian Games, a local version of the Ancient Greek Olympic Games, by William Brookes. Brookes hoped to establish a modern version of the Games and following this meeting, the Baron went on to set up the International Olympic Committee and the first modern Olympics in Greece in 1896. Students at William Brookes School grew the ‘Coubertin oak’ acorns into young trees which were then moved to the Arboretum Nursery at Kew where they stayed for two years to develop their root systems. They were then transferred to a nursery in Cambridgeshire for a further two years before the planting in April 2012. English oak and Turner's oak at Kew Quercus robur is by far the most numerous species of tree at Kew and well over 1,000 specimens grow around the Gardens. The hybrid Quercus × turneri (Turner's oak) can also be seen at Kew – a cross between the English oak (Quercus robur) and the holm oak (Quercus ilex). The semi-deciduous Turner's oak has been growing at Kew since the 18th century and was raised by Mr Turner (a nurseryman from Essex) in 1783. It was then planted in the original five-acre arboretum at Kew in 1798, where it stands today by the Princess of Wales Conservatory. Compared to some oaks, Quercus × turneri is relatively small, growing to around 17 m. It generally develops a low, domed shape, with its twisting, horizontal branches sometimes barely above the ground. This gives an open feel to the tree, and creates a perfect opportunity for the enthusiastic tree climber! Although the leaves of this hybrid are similar in colour to those of other oaks, they have a different shape. Instead of the familiar rounded leaf lobes of the English oak, those of Turner's oak form almost sharp, forward-pointing teeth. The hurricane-like storm which ravaged the south of England, as well as parts of northern France, on the night of 15 October 1987 brought wind speeds of up to 185 km/h. It caused extensive damage to buildings and blew down around 15 million trees. At Kew the entire root plate of the historic Turner's oak lifted, before settling back into the ground. Far from damaging the tree, this upheaval seemed to rejuvenate it. Previously it had been showing signs of stress and decline due to compaction of the root plate where so many people had taken shelter beneath the tree in the past. Indeed, Kew have learned from this and now routinely de-compact the root plates of the mature trees in the arboretum. Kennedy, C. E. J. & Southwood, T. R. E. (1984). The number of species of insects associated with British trees: a re-analysis. Journal of Animal Ecology 53: 455–478. Logan, W. B. (2005). Oak: the Frame of Civilization. W.W. Norton, New York/London. Tansley, A. G. (1952). Oaks and Oak Woods. Methuen, London Kew Science Editor: William Milliken Kew contributors: Tony Hall Copyediting: Emma Tredwell Although every effort has been taken to ensure that the information contained in these pages is reliable and complete, notes on hazards, edibility and suchlike included here are recorded information and do not constitute recommendations. No responsibility will be taken for readers’ own actions.

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Butterflies are flying colored wing insects that vary in color and pattern from individual to another individual. It has wings covered with overlapping rows of scales. Most of butterflies have developed mechanisms to avoid predators making disguise coloration blending like leaf or bark of the tree. Some releases chemicals as a defense mechanism wherein butterfly evolved to have toxic chemicals. But recent finding due to extreme weather events and trend linked to ongoing anthropogenic climate change species shifts its dynamics. Droughts occur more often in larger spatial scale which has an effect on insects. Generally, drier and warmer climatic conditions have an impact either positive or negative to insect populations. The aim of this research is to address the knowledge gap using multi-decadal dataset of 163 butterfly species. All of this butterflies experienced millennium-scale drought. Impacts of droughts on Butterflies To know the faunal dynamics, investigation of phenology, species richness and diversity with its elevation gradient has been conducted. In which linear model used to understand differential sensitivity of butterflies to climate change at low and high elevation. A decade of dataset of 163 butterfly species across elevational gradient in Northern California has been considered. Results showed that a prolonged shift towards spring flight during drought years and change in phenology is evident across elevations. It also happened that the total flight window expanded at lower elevations while at higher elevation shifted and compressed. This leads the notion that fewer overall flight days at higher sites. The millennium drought in California created across site with elevation-specific changes in flight windows and species richness. This resiliency reveals that lowest elevations are less detrimental than biotic-abiotic association at higher elevations. Most of the researchers hypothesized a mismatch between trophic levels as a result of climate change. But, results of butterflies from low elevation would suggest that at consumer trophic level need not always have negative impacts. Additionally, species at lowest elevations have access to agricultural lands though irrigation does not correlate the population dynamics during drought. Thus, there is a possibility that low elevation population buffered by irrigated crops or agricultural margin during drought. Indeed, that at high elevation butterflies declined in number and become sensitive to dry years with warmer temperatures. Contrary to the theory that mountains offer microclimatic refugia and adapt species for climatic changes. It has been known that high latitude environments are warming faster with negative consequences to several species. But positive or have a neutral effect for other species. Consequently, this research suggests more thorough investigation about organismal responses to extreme weather. As well as on the extent wherein different habitat type may or may not buffer species populations against climate change. Source: Prepared by Joan Tura from Springer BMC Climate Changes Responses Volume 5:3 26 January 2018

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April 27th, 2009 Members of the University at Albany Community: As many of you are aware the White House has declared a public health emergency due to concerns of a new influenza strain originating from pigs – referred to as the “Swine Flu.” This virus is a new strain of the Influenza A virus which causes typical flu-like symptoms. In Mexico, at least 1,400 people have been clearly diagnosed with this virus, though it is believed many more may have been infected but were not diagnosed. A number of people have died. There have been 20 confirmed cases of the Swine Flu in the U.S. In none of these cases did the infected individual have anything more than a mild illness. None of these patients required any treatment other than the routine treatment for seasonal (regular) influenza. None were hospitalized. Confirmed cases of Swine Flu have been reported in Mexico, New York City, Ohio, Texas, Kansas and throughout Canada. Please be advised that anyone with a history of travel to these areas in the past seven to ten days with symptoms of a respiratory illness may be infected with Swine Flu and should be tested. The University’s Health Center is able to collect specimens for testing which are then sent to an outside lab for analysis and will provide further advice and treatment as warranted. The symptoms of the Swine Flu are exactly the same as a typical flu and include sore throat, nasal congestion, cough, fever and muscle aches. Even the mildest form of the flu, including Swine Flu, will result in symptoms similar to a simple cold with the possibility of fever and muscle aches in severe forms. At this point, the University is monitoring the situation with the assistance of the Albany County Department of Health, the New York State Department of Health, the Centers for Disease Control and the World Health Organization. As with all respiratory infections the best way to minimize risk of illness is by avoiding the virus. The flu is spread when an infected person coughs or sneezes and other people breathe in the infected air droplets. To avoid catching or spreading the flu, you are encouraged to take the following steps: - Avoid close contact with people who are sick. When you are sick, keep your distance from others (minimum three to six feet) to protect them from getting sick too. - Seek medical attention when you are sick. This will help prevent others from catching your illness. - Cover your mouth and nose with a tissue when coughing or sneezing. It may prevent those around you from getting sick. Coughing or sneezing into the bend of your elbow, when tissues are not available, directs the cough/sneeze downward and is helpful. - Clean your hands often. Washing your hands often will help protect you from germs. Twenty seconds of hand washing with hot water and soap or use of a disinfectant hand lotion is recommended. - Avoid touching your eyes, nose or mouth. Germs are often spread when a person touches something that is contaminated with germs and then touches his or her eyes, nose, or mouth. Information from the Center for Disease Control on Swine Flu is available at: http://www.cdc.gov/swineflu/investigation.htm I will update you should additional information become available. If you have any questions not answered by the above resources, please send them to [email protected]. We will do our best to answer your questions in a timely manner. Peter Vellis, DO University Health Center

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# Which counting numbers are also whole numbers? A number is a mathematical value used for measuring, counting, and performing different arithmetic calculations such as addition, subtraction, division, multiplication. A number system is characterized as an arrangement of writing to express numbers. It is the mathematical notation for addressing numbers. It provides a unique representation of every number. Numbers have different categories such as natural numbers, Whole numbers, Integers, Rational Numbers, Irrational numbers, Real numbers, Complex Numbers. Examples are 1, 50, -50, 0.25, 8.9999 etc. ### Natural Numbers Counting Numbers are called Natural Numbers. These are a part of the number system, including all the positive integers from 1 to infinity. They are called counting numbers because they do not include zero or negative numbers. Examples of natural numbers are 1, 2 , 9 ,1023 etc. A whole number is a set of numbers that includes all positive integers and 0. These are a part of real numbers Which do not include fractions, decimals, or negative numbers. Natural Numbers along with 0 are called whole numbers. Examples of whole numbers are 0, 1, 2 , 9 ,1023 etc. ### Which counting numbers are also whole numbers? All the counting Numbers such as 1, 2, 3, 4 . . . . . are whole numbers. But all whole numbers are not counting numbers. Whole Numbers excluding 0 can be called counting numbers. In order to put it in the simplest way, all natural numbers are counting numbers and all whole numbers are natural numbers except 0, therefore, apart from 0, all whole numbers can be considered as counting numbers. N = W – {0} ### Sample Problems Question 1: Choose Natural Numbers from the following set of numbers, 11, 0, 25, 0.387, 1.20, 2506. All counting numbers are natural numbers, hence 11, 25, 2506 are natural numbers here. Question 2: Choose whole Numbers from the following set of numbers? 1.1, 0, 125, 0.387, 1.20, 256, 69 All counting numbers along with 0 are whole numbers, hence 0, 125, 256, 69 are whole numbers here. Question 3: Can all decimal numbers be counting numbers? Numbers including decimals are not counting numbers. Question 4: Which one of the given numbers is a natural number? 0, 2.3, 22/7, 9.87, 12, 11.2

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KS3: On entering the school students are able to study French or Spanish. They will study this in two periods per week. Classes are set based on mixed ability and the teachers ensure that work is set at an appropriately challenging level for the ability of each student. All students should by a bilingual dictionary for the language they will be studying as this will enable each students to produce at home a better independent learning homework. The objectives of the Year 7 languages course are to enable all students to have a basic vocabulary and knowledge of grammar in order to be able to use familiar language in familiar contexts up to National Curriculum Level 4a on a series of topics. Year 8 and Year 9 follow similar lesson patterns and the aim is for students to consolidate their knowledge of grammar to be able to independently manipulate language to create paragraphs in the future and past tenses. The most able will also include complex features and tenses. The teaching of Languages at GAC is very good and the department is staffed by very enthusiastic, experienced linguists. We devise our own Schemes of Work designed to make language learning enjoyable and engaging, whilst developing a strong linguistic foundation in knowledge of grammar. We carry out projects including researching areas and traditions of the French and Spanish speaking worlds, film analysis and story writing alongside the traditional topic areas of Home and Town, Family, Hobbies, Healthy Living. GAC believes in the importance of studying languages to enhance cultural awareness, to extend literacy in our native language and to build communication and problem solving skills. A very small number of students who would benefit from baseline Literacy and Numeracy support are withdrawn from MFL classes but this is at the discretion of the school. Students who make a real effort and work well in class and at home achieve very strong results. GCSE: We currently use the Edexcel examination board and build on the sure vocabulary and grammar foundation built at Key Stage 3. The GCSE course is begun in the final term of Year 9 and after students have made their options. We aim that all students can produce short passages from memory, accurately applying grammar rules and knowledge about a range of key subjects laid down by the examination boards. Students studying a language at GCSE level will have to produce two pieces of written coursework and two spoken pieces. A rough outline is given below and letters will be sent detailing the dates and titles for full GCSE assessment pieces in advance of each test taking place. Information regarding the new GCSE structure will be added at a later stage. As at Key Stage 3, the department devises its own Schemes of Work for Key Stage 4 but supplements this heavily with the examination board endorsed Textbooks ‘Edexcel GCSE for French’ and ‘Edexcel GCSE for Spanish’.

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Yesterday I mentioned that the IceCube Neutrino observatory doesn’t detect neutrinos directly, but instead detects the flash of light that occurs after a neutrino collides with a water molecule. As you can imagine, the light emitted with a single neutrino collides with a single molecule is pretty faint. So how do we observe such faint light? We use a photomultiplier tube (or PMT), some of which are so sensitive they can detect the light of a single photon. A PMT is based on something known as the photoelectric effect. Early observations of the photoelectric effect date back to the late 1800s, but it was made famous by Einstein’s 1905 paper, for which he won the Nobel prize. The basic idea of the photoelectric effect is that a negatively charged metallic plate can emit electrons when ultraviolet light is shined on it. Einstein demonstrated that this was due to photons striking the electrons and knocking them free of the metal. Of course a single photon can only knock out one electron, which would be very difficult to observe. So we need to use another effect known as secondary emission. In this case, fast moving electron striking a metal plate can knock free several slower moving electrons, kind of how a bowling ball can knock over several pins. In a PMT, a photon strikes a negatively charged metal plate called the photocathode, which releases the first electron. This is then accelerated toward another metal plate (known as a dynode) which it strikes, releasing a few electrons. These are then accelerated to another dynode, releasing more electrons. The result is a kind of cascade effect (seen in the image above), where a single photon causes an avalanche of electrons to reach the final metal plate (the anode) where an electric current is created that we can measure easily. So a photomultiplier tube allows faint light to trigger an electric current strong enough for us to measure. Different types of PMTs have different light sensitivities, depending on the type of photocathode they use. Some are sensitive to visible wavelengths, while others are sensitive to infrared or ultraviolet. Under the right conditions, we really can detect the light of a single photon.

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There are three types of veins in the human body: superficial veins, perforating veins and deep veins. Deep veins are connected to the vena cava, which is the largest vein in the body and is directly connected to the lungs and heart. Deep vein thrombosis (DVT) occurs when a blood clot forms in a deep vein. DVT is most likely to occur in the calf, pelvis or thigh, though it can occur in other areas such as the chest or arm. While half of patients with DVT will not experience symptoms, the other half is likely to experience: - Sudden swelling or pain - Skin warmth - Leg pain that worsens during walking or standing - Red or blue skin discoloration DVT is a particularly dangerous condition because it can result in a pulmonary embolism. This occurs when a blood clot in one of the deep vein breaks away and travels to the lungs. Once there, the clot can block blood flow, straining the lungs and heart. A pulmonary embolism is considered an emergency and can quickly be fatal if the clot is large enough. The most common symptom of a pulmonary embolism is shortness of breath. Other symptoms may include: - Chest pain that might reach into the shoulder, arm, jaw and neck - Rapid breathing or heartbeat - Restlessness and anxiety - Coughing up blood - Lightheadedness and fainting DVT can result from a problem with the body's clotting system. In this case, a small clot causes inflammation that then allows larger clots to form. Additionally, poor blood flow in the leg veins increases the risk for DVT, especially during periods of prolonged motionlessness. Because of this, DVT has been called Economy Class Syndrome, since clots may form during long airplane trips where there is limited leg room. The truth, however, is that long flights rarely cause DVT. Most often, the condition is found in hospitalized, bedbound patients. Other factors increasing the risk for DVT include: - History of stroke, heart attack or congestive heart failure - Inflammatory bowel syndrome

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# Is f(x)=(1-e^(2x))/(2x-4) increasing or decreasing at x=-1? Jan 5, 2016 Increasing. #### Explanation: Find the sign of the derivative at $x = - 1$. To find $f ' \left(x\right)$, use the quotient rule. $f ' \left(x\right) = \frac{\left(2 x - 4\right) \frac{d}{\mathrm{dx}} \left(1 - {e}^{2 x}\right) - \left(1 - {e}^{2 x}\right) \frac{d}{\mathrm{dx}} \left(2 x - 4\right)}{2 x - 4} ^ 2$ The derivative of each part: The first requires the chain rule:$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$ $\frac{d}{\mathrm{dx}} \left(1 - {e}^{2 x}\right) = - 2 {e}^{2 x}$ $\frac{d}{\mathrm{dx}} \left(2 x - 4\right) = 2$ Thus, $f ' \left(x\right) = \frac{\left(2 x - 4\right) \left(- 2 {e}^{2 x}\right) - 2 \left(1 - {e}^{2 x}\right)}{2 x - 4} ^ 2$ $f ' \left(x\right) = \frac{- 4 x {e}^{2 x} + 10 {e}^{2 x} + 2}{2 x - 4} ^ 2$ Find $f ' \left(- 1\right)$ If it's greater than $0$, the function is increasing at that point. If it's less than $0$, the function is decreasing at that point. $f ' \left(- 1\right) = \frac{- 4 \left(- 2\right) {e}^{-} 2 + 10 {e}^{-} 2 + 2}{- 2 - 4} ^ 2 = \frac{\frac{18}{e} ^ 2 + 2}{36}$ Since this is $> 0$, the function is increasing when $x = - 1$. graph{(1-e^(2x))/(2x-4) [-5.93, 6.56, -2.596, 3.647]}

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# For what values of x, if any, does f(x) = 1/((12x+3)sin(pi+(6pi)/x) have vertical asymptotes? Aug 7, 2017 $x = - \frac{1}{4}$ and $x = \frac{6}{k}$ for integer $k$ #### Explanation: $12 x + 3 = 0$ at $x = - \frac{1}{4}$ sin(pi+(6pi)/x) = 0# where $\pi + \frac{6 \pi}{x} = k \pi$ for integer $k$ And that happens where $\frac{6 \pi}{x} = k \pi$ for integer $k$ Which is at $x = \frac{6}{k}$ for integer $k$

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GMAT Club Forumhttps://gmatclub.com:443/forum/ In the figure above, QRS is a straight line and QR = PR. Ishttps://gmatclub.com/forum/in-the-figure-above-qrs-is-a-straight-line-and-qr-pr-is-143843.html Page 1 of 1 Author: JJ2014 [ 09 Dec 2012, 11:43 ] Post subject: In the figure above, QRS is a straight line and QR = PR. Is Attachment: Screen shot 2012-12-09 at 1.40.06 PM.png [ 19.42 KiB | Viewed 24232 times ] In the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ parallel?(1) Length PQ = Length PR(2) Line TR bisects angle PRS Author: mikemcgarry [ 09 Dec 2012, 15:58 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is In the figure, QRS is a straight line. QR=PR. Are TR and PQ parallel?1) Length PQ = Length PR2) Line TR bisects angle PRS From the prompt, we know that triangle QPR is isosceles, with QR = PR. By the Isosceles Triangle theorem, we know that angle Q = angle P. Statement #1 PQ = PR.This is enough to guarantee that triangle QPR is equilateral, but we don't know anything about ray RT, so we have no idea whether that is parallel to anything else. This statement, alone and by itself, is not sufficient. Statement #2 Line TR bisects angle PRSA fascinating statement. Let's think about this. We already know angle Q = angle P. Call the measure of that angle M. Look angle QRP --- call the measure of that angle K. Clearly, within triangle QPR, M + M + K = 180, but Euclid's famous theorem. Now, look at angle PRS. This is what is known as the "exterior angle" of a triangle, and there's a special theorem about this. The Remote Interior Angle Theorem:If the exterior angle of a triangle is adjacent to the angle of the triangle, then the measure of the exterior angle is equal to the sum of the two "remote" interior angles of the triangle --- that is, the two angles of the triangle which the exterior angle is not touching. If you think about this, it has to be true, because(angle Q) + (angle P) + (angle PRQ) = 180, because they're the three angles in a triangle(angle PRQ) + (exterior angle PRS) = 180, because they make a straight lineSubtract (angle PRQ) from both sides of both equations:(angle Q) + (angle P) = 180 - (angle PRQ) (exterior angle PRS) = 180 - (angle PRQ) Since the two things on the left are equal to the same thing, they are equal to each other. (angle Q) + (angle P) = (exterior angle PRS) Now, going back to the letters we were using ---- if (angle Q) = (angle P) = M, this means (exterior angle PRS) = 2M. If we bisect exterior angle PRS, each piece will have a measure of M. Thus, (angle PRT) = (angle TRS) = MWell, now we know that (angle Q) = (angle TRS) = M. If corresponding angles are congruent, then the lines must be parallel. PQ must be parallel to TR. This statement, alone and by itself, is sufficient to answer the prompt question. Answer = BDoes all this make sense?Mike Author: shanmugamgsn [ 10 Dec 2012, 17:53 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is Hi Mike well explained,I do have a doubt ! now we know that (angle Q) = (angle TRS) = M. If corresponding angles are congruent, then the lines must be parallel. PQ must be parallel to TR. This statement, alone and by itself, is sufficient to answer the prompt question. Correspoding angle means the angle made on its side? angle Q ==> side PQangle TRS ==> side TRvery basic question but i understood complex things Author: mikemcgarry [ 10 Dec 2012, 18:09 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is shanmugamgsn wrote:I do have a doubt ! Corresponding angle means the angle made on its side? No. "Corresponding angles" is a technical term from Euclidean Geometry. It's the name of a particular pair of angles formed when a transversal crosses a pair of parallel lines.Attachment: parallel line diagram.JPG [ 22.96 KiB | Viewed 23480 times ] The following pairs are corresponding angles1 & 52 & 63 & 74 & 8Corresponding angles are congruent if and only if the lines are parallel. The following pairs are alternate interior angles3 & 64 & 5Alternate interior angles are congruent if and only if the lines are parallelThe following pairs are alternate exterior angles1 & 82 & 7Alternate exterior angles are congruent if and only if the lines are parallelThe following pairs are same side interior angles3 & 54 & 6Same side interior angles are supplementary if and only if the lines are parallelThe following pairs are same side exterior angles1 & 72 & 8Same side exterior angles are supplementary if and only if the lines are parallelThose are all the names relating pairing an angle at one vertex with an angle at the other vertex, when a transversal intersects a pair of parallel lines. Does all this make sense?Mike Author: priyamne [ 11 Dec 2012, 04:40 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is JJ2014 wrote:Attachment:Screen shot 2012-12-09 at 1.40.06 PM.pngIn the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ parallel?(1) Length PQ = Length PR(2) Line TR bisects angle PRSAns: For lines TR and PQ to be parallel angle PQR= angle TRS. From statement 1 we get angle PQR=x=60 but nothing about angle TRS.From statement 2 we get PRQ=180-2X , therefore PRS=180-(180-2x)=2x and TR bisects it so angle TRS=x which is equal to PQR. Therefore the answer is (B). Author: jlgdr [ 22 Apr 2014, 08:36 ] Post subject: Re: In the figure above, QRS is a straight line and QR = PR. Is Let's solve. So we need to know if TR is parallel to PQ. Now then, let's hit the first statement. We are told that PQ=QR. Now we know that PQR is an equilateral triangle but still no info on TR. Therefore, insufficient. Statement 2, we have that TR bisects PRS. Now let's see. So we know that QR=PR from the question stem. Hence angle QRP is 180-2x, 'x' being the angles P and Q respecively. Therefore angle PRS will be 2x since QRS is a straight line with total measure of 180 degrees. Now if TRS bisects then angle TRS is x only. Which means that the angles Q and R are equal and thus PQ // TR. B standsCheers!J Author:  BrainLab [ 22 Dec 2015, 04:29 ] Post subject:  Re: In the figure above, QRS is a straight line and QR = PR. Is JJ2014 wrote: Attachment: The attachment Screen shot 2012-12-09 at 1.40.06 PM.png is no longer available In the figure above, QRS is a straight line and QR = PR. Is it true that lines TR and PQ parallel? (1) Length PQ = Length PR (2) Line TR bisects angle PRS (1) this tells us that all 3 sides are equal = Equilateral Triangle. Not sufficient, as there is no info about TR (2) Now look at the attachment, I'll use the property of the "Exterior Angles of a Triangle" Angle Y is the sum of the angles on the both sides of TR Exterior angle of a triangle is equal to the sum of the opposite interior angles Angle Y=2X and if TR bisects angle Y it means that Angle TRS = $$\frac{Y}{2}=\frac{2X}{2}=X$$ and when this two angles of to different triangles are equal then QP and TR are parallel.

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# How to Convert 10 MG to ML? Milligrams (mg) and milliliters (ml) are two common units of measurement used in medicine and science. Milligrams are used to measure the mass of a substance, while milliliters are used to measure the volume of a liquid. When working with medications, it’s important to understand the conversion between milligrams and milliliters, as many medicines are prescribed in milligrams but administered in milliliters. The conversion between milligrams and milliliters can be done using the conversion factor of 1 mg = 0.001 ml. This means that one milligram is equivalent to 0.001 milliliters. To convert milligrams to milliliters, multiply the number by 0.001. ### Let’s understand 10 MG to ML conversion: For example, to convert 10 milligrams to milliliters, you would perform the following calculation: 1 mg = 0.001 ml 10 mg = 10 * 0.001 ml = 0.01 ml So, 10 mg is equivalent to 0.01 ml ## Know more about how to Convert 10 MG to ML? It’s important to note that the conversion factor between MG to ML will vary depending on the measured substance, as different meanings have different densities. However, the conversion factor of 1 mg = 0.001 ml can be used as a general guideline in most cases. In conclusion, converting 10 mg to ml is a simple mathematical calculation that can be done using the conversion factor of 1 mg = 0.001 ml. Understanding the conversion between 10 mg to ml is vital in many fields, including medicine and science, as it allows for the accurate measurement and administration of substances. ### FAQs: • #### What is the conversion factor from milligrams to milliliters? The conversion factor from milligrams (mg) to milliliters (ml) is 1 mg = 0.001 ml. • #### How do I convert 10 MG to ML? To convert 10 milligrams to milliliters, multiply 10 by 0.001: 10 mg * 0.001 = 0.01 ml. So, 10 mg is equal to 0.01 ml. • #### Can 1 mg = 0.001 ml conversion factor be used for all substances? The conversion factor of 1 mg = 0.001 ml is a general guideline and can be used for many substances, but it may not be accurate for all senses due to differences in densities. • #### Why is it important to understand the conversion between milligrams and milliliters? It is important to understand the conversion between milligrams and milliliters in fields such as medicine and science, as many medications are prescribed in milligrams but administered in milliliters. Accurate measurement and administration of substances are crucial in these fields. • #### What happens if I use the wrong conversion factor when converting milligrams to milliliters? Using the wrong conversion factor when converting milligrams to milliliters can result in an incorrect measurement and administration of a substance. This can lead to adverse effects and should be avoided. It is important to use the correct conversion factor, or to consult a professional, to ensure accurate measurements.

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The word shock can be used in a range of ways, but when used in a first aid context it describes a physical condition that results from a loss of circulating body fluid. It should not be confused with emotional shock that might occur, say, when a person has received bad news (although the external signs are very similar). What happens in cases of shock A severe loss of body fluid will lead to a drop in blood pressure. Eventually the blood's circulation around the body will deteriorate and the remaining blood flow will be directed to the vital organs such as the brain. Blood will therefore be directed away from the outer areas of the body, so the casualty will appear paler than previously and the skin will feel cold and clammy. As blood flow slows, so does the amount of oxygen reaching the brain. The casualty may appear to be confused, weak and dizzy, and may eventually deteriorate into unconsciousness. To try to compensate for this lack of oxygen, the heart and breathing rates both speed up, gradually becoming weaker, and may eventually cease. Potential causes of shock include: severe internal or external bleeding; burns; severe vomiting and diarrhoea, especially in children and the elderly; problems with the heart. First Aid Treatment Keep the casualty warm but do not allow her to get overheated. If you are outside, try to get something underneath the casualty if you can do so easily. Wrap blankets and coats around her, paying particular attention to the head, through which much body heat is lost. Maintain a careful eye on the casualty's airway and be prepared to turn her into the recovery position if necessary, or even to resuscitate if breathing stops. Try to clear back bystanders and loosen tight clothing to allow maximum air to the casualty. Keep the casualty still and preferably sitting or lying down. If the casualty is very giddy, lay her down with her legs raised to ensure that maximum blood and therefore maximum oxygen is sent to the brain. Reassure the casualty but keep your comments realistic. Do not say that everything is going to be fine when it is obvious that there is something seriously wrong. Let the casualty know that everything that can be done is being done and that help has been called for. If she has other worries then try to resolve these. Treat the cause of the shock and aim to prevent further fluid loss. Ensure that appropriate medical help is on the way. Signs and symptoms Shock kills, so it is vital that you can recognise these signs and symptoms. With internal bleeding in particular, shock can occur some time after an accident, so if a person with a history of injury starts to display these symptoms coupled with any of the symptoms of internal bleeding, advise her to seek urgent medical attention, or take or send her to hospital. (c) Health-care-clinic.org All rights reserved Disclaimer: Health-care-clinic.org website is designed for educational purposes only. It is not intended to treat, diagnose, cure, or prevent any disease. Always take the advice of professional health care for specific medical advice, diagnoses, and treatment. We will not be liable for any complications, or other medical accidents arising from the use of any information on this web site. Please note that medical information is constantly changing. Therefore some information may be out of date.

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The use of solar power became very popular in the 1970s, but has fallen in and out of favour since depending on the potential savings when compared with fossil-fuel energy costs. The photovoltaic effect is when photo cells convert sunlight directly into electricity - this has been used for sometime to power certain calculators, for example. Photovoltaic cells (PV's) can be used as roof tiles. They cover the roof of a house and take advantage of the light coming from the Sun. This is trapped by the cell and turned into electricity. Another way to take advantage of the energy from the sun is to design buildings so they can collect the heat. They do this by designing the building sensibly and facing it in a way where it can use the sun to the maximum benefit. Large glass windows help with this, especially during the winter when the Sun is very low. In the summer, balconies and trees protect the building from getting too much heat. More common method in the UK is using the benefits of the sun to heat our water pipes. Painting the thin pipes black and putting them in a 'greenhouse' type insulator can heat our water supply and therefore reduce the cost of using electricity to heat it. As well as the fact that energy from the Sun is readily available, there are many other benefits. By locating photovoltaic cells on top of houses, no extra land space is needed and they can also be situated in urban areas. The technology now needed is about 90% cheaper than it was in the 1970s. Houses with solar roof tiles can in fact generate more electricity than is needed at certain times in the day, and can sell this back to local electricity companies. However the UK is behind many other countries in Europe and the rest of the World when it comes to using solar power technologies. In Japan and the USA, billions has been spent on developing PV over a number of years, and more recently, Germany has started to push lots of money into the development of it for projects there. China has pledged to throw its economic might behind a national solar plan and the government has committed to raising its share of renewable energy to 6%, from the current 1.5%. Alternative energy sources:

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Trigonometric Functions on Any Angle 1 / 28 # Trigonometric Functions on Any Angle - PowerPoint PPT Presentation Trigonometric Functions on Any Angle. Section 4.4. Objectives. Determine the quadrant in which the terminal side of an angle occurs. Find the reference angle of a given angle. I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described. ## Trigonometric Functions on Any Angle Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - Presentation Transcript ### Trigonometric Functions on Any Angle Section 4.4 Objectives • Determine the quadrant in which the terminal side of an angle occurs. • Find the reference angle of a given angle. • Determine the sine, cosine, tangent, cotangent, secant, and cosecant values of an angle given one of the sine, cosine, tangent, cotangent, secant, or cosecant value of the angle. Vocabulary • reference angle • sine of an angle • cosine of an angle • terminal side of an angle • initial side of an angle • tangent of an angle • cotangent of an angle • secant of an angle • cosecant of an angle Reference Angle A reference angle is the smallest distance between the terminal side of an angle and the x-axis. All reference angles will be between 0 and π/2. continued on next slide Reference Angle There is a straight-forward process for finding reference angles. Step 1 – Find the angle coterminal to the given angle that is between 0 and 2π. continued on next slide Reference Angle There is a straight-forward process for finding reference angles. Step 2 – Determine the quadrant in which the terminal side of the angle falls. continued on next slide Reference Angle There is a straight-forward process for finding reference angles. Step 3 – Calculate the reference angle using the quadrant-specific directions. continued on next slide θ Reference Angle For quadrant I, the shortest distance from the terminal side of the angle to the x-axis is the same as the angle θ. Thus where the reference angle is continued on next slide θ Reference Angle For quadrant II, the shortest distance from the terminal side of the angle to the x-axis is shown in blue. This is the rest of the distance from the terminal side of the angle to π. Thus This distance is the reference angle. Note: Here put subtracted the angle from π since the angle was smaller than π. This gave us the positive reference angle. If we had subtracted π from the angle, we would have needed to take the absolute value of the answer. continued on next slide θ Reference Angle This distance is the reference angle. For quadrant III, the shortest distance from the terminal side of the angle to the x-axis is shown in blue. This is the distance from the π to the terminal side of the angle. Thus continued on next slide θ Reference Angle Note: Here put subtracted π from the angle since the angle was larger than π. This gave us the positive reference angle. If we had subtracted the angle from π, we would have needed to take the absolute value of the answer. continued on next slide θ Reference Angle This distance is the reference angle. For quadrant IV, the shortest distance from the terminal side of the angle to the x-axis is shown in blue. This is the rest of the distance from the terminal side of the angle to 2π. Thus continued on next slide θ Reference Angle Note: Here put subtracted the angle from 2π since the angle was smaller than 2π. This gave us the positive reference angle. If we had subtracted 2π from the angle, we would have needed to take the absolute value of the answer. continued on next slide Reference Angle Summary Step 1 – Find the angle coterminal to the given angle that is between 0 and 2π. Step 2 – Determine the quadrant in which the terminal side of the angle falls. Step 3 – Calculate the reference angle using the quadrant-specific directions indicated to the right. In which quadrant is the angle? To find out what quadrant θ is in, we need to determine which direction to go and how far. Since the angle is negative, we need to go in the clockwise direction. The distance we need to go is one whole π and 1/6 of a π further. This red part is approximately 1/6 of a π further. This blue part is one whole π in the clockwise direction Now that we have drawn the angle, we can see that the angle θ is in quadrant II. continued on next slide What is the reference angle, , for the angle ? Using our summary for finding a reference angle, we start by finding an angle coterminal to θ that is between 0 and 2π. Thus we need to start by adding 2π to our angle. continued on next slide What is the reference angle, , for the angle ? The next step is to determine what quadrant our coterminal angle is in. We really already did this in the first question of the problem. Coterminal angles always terminate in the same quadrant. Thus our coterminal angle is in quadrant II. continued on next slide Finally we need to use the quadrant II directions for finding the reference angle. Thus the reference angle is Evaluate each of the following for . To solve a problem like this, we want to start by finding the reference angle for θ. Since our angle is bigger than 2π, we need to subtract 2π to find the coterminal angle that is between 0 and 2π. continued on next slide Our next step is to figure out what quadrant is in. You can see from the picture that we are in quadrant II. Evaluate each of the following for . To find the reference angle for an angle in quadrant II, we subtract the coterminal angle from π. This will give us a reference angle of continued on next slide We will now use the basic trigonometric function values for The only thing that we will need to change might be the signs of the basic values. Remember that the sign of the cosine and tangent functions will be negative in quadrant II. The sign of the sine will still be positive in quadrant II. Evaluate each of the following for . continued on next slide Evaluate each of the following for . Once again, we will use our reference angle to determine the basic trigonometric function value. The only difference between the basic value and the value for our angle may be the sign. continued on next slide Evaluate each of the following for . Once again, we will use our reference angle to determine the basic trigonometric function value. The only difference between the basic value and the value for our angle may be the sign. continued on next slide Evaluate each of the following for . Once again, we will use our reference angle to determine the basic trigonometric function value. The only difference between the basic value and the value for our angle may be the sign.

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Analysis of literature: “Black Art” By Amiri Baraka There are much articles and literature in the American history that speaks of racism, racial identity, and race as a concept. “The Black Arts” by Amiri Baraka is a unique piece of literature that interconnects art with racial identity. The poem is well connected with the sensitivity of racism among Black Africans and the association with different forms of art. The Black Arts Movement of the 1960's which is also commonly recognized as BAM is the artistic aspect of the Black Art Movement of the mid-nineteenth century. The Black Art Movement gained popularity as a movement that promoted arts and craft for the Black community and allowed them participation to the full extent. The term gained more momentum and significance after the launch of free jazz in the year 1960s and then in 1980s for the introduction of wireless connections with musicians and the development of new musical instruments at large. It was the period from 1990s onwards that the world witnessed significant changes on the musical front. The initial style of this music resembled the Anglo-African style of music by the name Soul. It was a rugged, tough and rockfish style of music that gained massive appeal from the music lovers. The initial developments on the musical front also included the launch of CDs by the New Yorkers live music. The emphasis of these steps was aimed at encouraging the promotion of music and other forms of art without discriminating on the basis of color and creed. These steps were also included in the Pan-Africanism movement that overlapped with the Black Arts Movement. The similarities of time and space between the followers of each movement allowed for sharing their cultural beliefs and other perspectives with each other. They also shared common intellectual spaces that contributed towards promoting intellectual values in their society in addition to harmonization of their cultural traditions. The women community also participated in this cultural exchange and demonstrated their unique skills and talents at various platforms under this movement. The women involved in each of these movements competed against each other in various art exhibitions, theatres, stage shows, and on television and radio mediums. These female leaders read literature debating the existence of a definable and tangible black existence and raised pertinent questions concerning the impact of gender on this black identity. The Black Art Movement is the most significant revolution ...

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Courtesy of NASA Viking was the culmination of a series of missions to explore the planet Mars; they began in 1964 with Mariner 4, and continued with the Mariner 6 and 7 flybys in 1969, and the Mariner 9 orbital mission in 1971 and 1972. Viking was designed to orbit Mars and to land and operate on the planet's surface. Two identical spacecraft, each consisting of a lander and an orbiter, were built. NASA's Langley Research Center in Hampton, Virginia, had management responsibility for the Viking project from its inception in 1968 until April 1, 1978, when the Jet Propulsion Laboratory assumed the task. Martin Marietta Aerospace in Denver, Colorado, developed the landers. NASA's Lewis Research Center in Cleveland, Ohio, had responsibility for the Titan- Centaur launch vehicles. JPL's initial assignment was development of the orbiters, tracking and data acquisition, and the Mission Control and Computing Center. NASA launched both spacecraft from Cape Canaveral, Florida -- Viking 1 on August 20, 1975, and Viking 2 on September 9, 1975. The landers were sterilized before launch to prevent contamination of Mars with organisms from Earth. The spacecraft spent nearly a year cruising to Mars. Viking 1 reached Mars orbit on June 19, 1976; Viking 2 began orbiting Mars on August 7, 1976. After studying orbiter photos, the Viking site certification team considered the original landing site for Viking 1 unsafe. The team examined nearby sites, and Viking 1 landed on July 20, 1976, on the western slope of Chryse Planitia (the Plains of Gold) at 22.3°N latitude, 48.0° longitude. The site certification team also decided the planned landing site for Viking 2 was unsafe after it examined high-resolution photos. Certification of a new landing site took place in time for a Mars landing on September 3, 1976, at Utopia Planitia, at 47.7°N latitude, and 225.8° longitude. The Viking mission was planned to continue for 90 days after landing. Each orbiter and lander operated far beyond its design lifetime. Viking Orbiter 1 exceeded four years of active flight operations in Mars orbit. The Viking project's primary mission ended November 15, 1976, 11 days before Mars's superior conjunction (its passage behind the Sun). After conjunction, in mid-December 1976, controllers re-established telemetry and command operations, and began extended-mission operations. The first spacecraft to cease functioning was Viking Orbiter 2 on July 25, 1978; the spacecraft had used all the gas in its attitude-control system, which kept the craft's solar panels pointed at the Sun to power the orbiter. When the spacecraft drifted off the Sun line, the controllers at JPL sent commands to shut off power to Viking Orbiter 2's transmitter. Viking Orbiter 1 began to run short of attitude-control gas in 1978, but through careful planning to conserve the remaining supply, engineers found it possible to continue acquiring science data at a reduced level for another two years. The gas supply was finally exhausted and Viking Orbiter 1's electrical power was commanded off on August 7, 1980, after 1,489 orbits of Mars. The last data from Viking Lander 2 arrived at Earth on April 11, 1980. Lander 1 made its final transmission to Earth Nov. 11, 1982. Controllers at JPL tried unsuccessfully for another six and one-half months to regain contact with Viking Lander 1. The overall mission came to an end May 21, 1983. With a single exception -- the seismic instruments -- the science instruments acquired more data than expected. The seismometer on Viking Lander 1 would not work after landing, and the seismometer on Viking Lander 2 detected only one event that may have been seismic. Nevertheless, it provided data on wind velocity at the landing site to supplement information from the meteorology experiment, and showed that Mars has very low seismic background. The three biology experiments discovered unexpected and enigmatic chemical activity in the Martian soil, but provided no clear evidence for the presence of living microorganisms in soil near the landing sites. According to mission biologists, Mars is self-sterilizing. They believe the combination of solar ultraviolet radiation that saturates the surface, the extreme dryness of the soil and the oxidizing nature of the soil chemistry prevent the formation of living organisms in the Martian soil. The question of life on Mars at some time in the distant past remains open. The landers' gas chromatograph/mass spectrometer instruments found no sign of organic chemistry at either landing site, but they did provide a precise and definitive analysis of the composition of the Martian atmosphere and found previously undetected trace elements. The X-ray fluorescence spectrometers measured elemental composition of the Martian soil. Viking measured physical and magnetic properties of the soil. As the landers descended toward the surface they also measured composition and physical properties of the Martian upper atmosphere. The two landers continuously monitored weather at the landing sites. Weather in the Martian midsummer was repetitious, but in other seasons it became variable and more interesting. Cyclic variations appeared in weather patterns (probably the passage of alternating cyclones and anticyclones). Atmospheric temperatures at the southern landing site (Viking Lander 1) were as high as -14°C (7°F) at midday, and the predawn summer temperature was -77°C (-107°F). In contrast, the diurnal temperatures at the northern landing site (Viking Lander 2) during midwinter dust storms varied as little as 4°C (7°F) on some days. The lowest predawn temperature was -120°C (-184°F), about the frost point ofcarbon dioxide. A thin layer of water frost covered the ground around Viking Lander 2 each winter. Barometric pressure varies at each landing site on a semiannual basis, because carbon dioxide, the major constituent of the atmosphere, freezes out to form an immense polar cap, alternately at each pole. The carbon dioxide forms a great cover of snow and then evaporates again with the coming of spring in each hemisphere. When the southern cap was largest, the mean daily pressure observed by Viking Lander 1 was as low as 6.8 millibars; at other times of the year it was as high as 9.0 millibars. The pressures at the Viking Lander 2 site were 7.3 and 10.8 millibars. (For comparison, the surface pressure on Earth at sea level is about 1,000 millibars.) Martian winds generally blow more slowly than expected. Scientists had expected them to reach speeds of several hundred miles an hour from observing global dust storms, but neither lander recorded gusts over 120 kilometers (74 miles) an hour, and average velocities were considerably lower. Nevertheless, the orbiters observed more than a dozen small dust storms. During the first southern summer, two global dust storms occurred, about four Earth months apart. Both storms obscured the Sun at the landing sites for a time and hid most of the planet's surface from the orbiters' cameras. The strong winds that caused the storms blew in the southern hemisphere. Photographs from the landers and orbiters surpassed expectations in quality and quality. The total exceeded 4,500 from the landers and 52,000 from the orbiters. The landers provided the first close-up look at the surface, monitored variations in atmospheric opacity over several Martian years, and determined the mean size of the atmospheric aerosols. The orbiter cameras observed new and often puzzling terrain and provided clearer detail on known features, including some color and stereo observations. Viking's orbiters mapped 97 percent of the Martian surface. The infrared thermal mappers and the atmospheric water detectors on the orbiters acquired data almost daily, observing the planet at low and high resolution. The massive quantity of data from the two instruments will require considerable time for analysis and understanding of the global meteorology of Mars. Viking also definitively determined that the residual north polar ice cap (that survives the northern summer) is water ice, rather than frozen carbon dioxide (dry ice) as once believed. Analysis of radio signals from the landers and the orbiters -- including Doppler, ranging and occultation data, and the signal strength of the lander-to-orbiter relay link -- provided a variety of valuable information. Other significant discoveries of the Viking mission include: Space History Viking Spacecraft Introduction Copyright © 1997-2000 by Calvin J. Hamilton. All rights reserved. Privacy Statement.

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# Calculator search results Formula Factorize the expression $x ^{ 4 } +3x ^{ 3 } -7x ^{ 2 } -27x-18$ $\left ( x - 3 \right ) \left ( x + 1 \right ) \left ( x + 2 \right ) \left ( x + 3 \right )$ Arrange the expression in the form of factorization.. $\color{#FF6800}{ x } ^ { \color{#FF6800}{ 4 } } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ - } \color{#FF6800}{ 7 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 27 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 }$ Do factorization $\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 } \right )$ $\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 3 } } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 18 } \right )$ Do factorization $\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$ $\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 9 } \right )$ Factorize to use the polynomial formula of sum and difference $\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right )$ $\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right )$ Sort the factors $\left ( x + 1 \right ) \left ( x + 2 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$ $\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$ Sort the factors $\left ( x + 1 \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$ $\left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$ Sort the factors $\left ( \color{#FF6800}{ x } \color{#FF6800}{ - } \color{#FF6800}{ 3 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 1 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 2 } \right ) \left ( \color{#FF6800}{ x } \color{#FF6800}{ + } \color{#FF6800}{ 3 } \right )$ Solution search results

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GeeksforGeeks App Open App Browser Continue ## Related Articles • NCERT Solutions for Class 9 Maths # Class 9 NCERT Solutions- Chapter 5 Introduction to Euclid’s Geometry – Exercise 5.1 ### Question 1: Which of the following statements are true and which are false? Give reasons for your answers. (i) Only one line can pass through a single point. (ii) There is an infinite number of lines which pass through two distinct points. (iii) A terminated line can be produced indefinitely on both the sides. (iv) If two circles are equal, then their radii are equal. (v) In Given fig, if AB = PQ and PQ = XY, then AB = XY. Solution: (i) False Reason: If we mark a point O on the surface of a paper. Using pencil and scale, we can draw infinite number of straight lines passing through O. (ii) False Reason: Through two distinct points there can be only one line that can be drawn. Hence, the statement mentioned above is False. (iii) True Reason: A line that is terminated can be indefinitely produced on both sides as a line can be extended on both its sides infinitely. Hence, the statement mentioned above is True. (iv) True Reason: The radii of two circles are equal when the two circles are equal. The circumference and the centre of both the circles coincide; and thus, the radius of the two circles should be equal. Hence, the statement mentioned above is True. (v) True Reason: According to Euclid’s 1st axiom- “Things which are equal to the same thing are also equal to one another”. Hence, the statement mentioned above is True. ### Question 2: Give a definition for each of the following terms. Are there other terms that need to be defined first? What are they and how might you define them? (i) Parallel lines (ii) Perpendicular lines (iii) Line segment (v) Square Solution: Yes, there are other terms which need to be defined first, so that we understand better: • Plane: Flat surfaces in which geometric figures can be drawn are known are plane. A plane surface is a surface which lies evenly with the straight lines on itself. • Point: A dimensionless dot which is drawn on a plane surface is known as point. A point is that which has no part. • Line: A collection of points that has only length and no breadth is known as a line. And it can be extended on both directions. A line is breadth-less length. (i) Parallel lines: Two lines l and m in a plan are said to be parallel if they have a no common point and we can write them as l || m. (ii) Perpendicular lines: Two lines A and B are said to be perpendicular if the form a right angle and we can write them as A ⊥ B. (iii) Line Segment: A line segment is a part of line and having a definite length. It has two end-points. In the figure, a line segment is shown having end points P and Q. Write it as arrow over PQ. (iv) Radius of Circle: The distance from the centre to a point on the circle is called the radius of the circle. In the figure, l is centre and m is a point on the circle, then lm is the radius of the circle. (v) Square: A quadrilateral in which all the four angles are right angles and all the four sides are equal is called a Square In the given figure ABCD is a Square. ### Question 3: Consider two ‘postulates’ given below: (i) Given any two distinct points A and B, there exists a third point C which is in between A and B. (ii) There exist at least three points that are not on the same line. Do these postulates contain any undefined terms? Are these postulates consistent? Do they follow from Euclid’s postulates? Explain. Solution: Yes, these postulates contain undefined terms such as ‘Point and Line’. Undefined terms in the postulates are: • There are many points that lie in a plane. But, in the postulates as given here, the position of the point C is not given, as of whether it lies on the line segment joining AB or it is not joining line segment. • On top of that, there is no information about whether the points are in same plane or not. And Yes, these postulates are consistent when we deal with these two situations: • Point C is lying on the line segment AB in between A and B. • Point C does not lie on the line segment AB. No, they don’t follow from Euclid’s postulates. They follow the axioms i.e “Given two distinct points, there is a unique line that passes through them.” ### Question 4: If a point C lies between two points A and B such that AC = BC, then prove that AC = 12 AB, explain by drawing the figure. Solution: AC = BC (Given) As we have studied in this chapter “If equals are added to equals then there wholes are also equal”. Therefore, AC + BC = BC + AC ⇒ 2AC = BC+AC As we have studied in this chapter, we know that, BC+AC = AB (as it coincides with line segment AB) ∴ 2 AC = AB (If equals are added to equals, the wholes are equal). ⇒ AC = (½)AB. ### Question 5: In Question 4, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point. Solution: Let, AB be the line segment as given in the Question. Assume that points C and D are the two different mid points of line segment AB. Therefore, C and D are the midpoints of AB. Now, as C and D are mid points of Ab we have, AC = CB and AD = DB CB+AC = AB (as it coincides with line segment AB) Now, Adding AC to the L.H.S and R.H.S of the equation AC = CB We get, AC+AC = CB+AC (If equals are added to equals, the wholes are equal.) ⇒ 2AC = AB — (i) Similarly, 2 AD = AB — (ii) From equation (i) and (ii), Since R.H.S are same, we equate the L.H.S we get, 2 AC = 2 AD (Things which are equal to the same thing are equal to one another.) ⇒ AC = AD (Things which are double of the same things are equal to one another.) Thus, we conclude that C and D are the same points. This contradicts our assumption that C and D are two different mid points of AB. Thus, it is proved that every line segment has one and only one mid-point. Hence, Proved. ### Question 6: In Figure, if AC = BD, then prove that AB = CD. Solution: According to the question, AC = BD From the given figure we can conclude that,, AC = AB+BC BD = BC+CD ⇒ AB+BC = BC+CD  (AC = BD, given) As we have studied, according to Euclid’s axiom, when equals are subtracted from equals, remainders are also equal. Subtracting BC from the L.H.S and R.H.S of the equation AB+BC = BC+CD, we get, AB+BC-BC = BC+CD-BC AB = CD Hence Proved. ### Question 7: Why is Axiom 5, in the list of Euclid’s axioms, considered a ‘universal truth’? (Note that the question is not about the fifth postulate). Solution: Euclid’s fifth axiom states that “the whole is greater than the part.” For Example: A cake. When it is whole or complete, assume that it measures 2 pounds but when a part from it is taken out and measured, its weight will be smaller than the previous measurement. So, the fifth axiom of Euclid is true for all the materials in the universe. Hence, Axiom 5, in the list of Euclid’s axioms, is considered a ‘universal truth’. 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# What is rounded to the nearest 100? ## What is rounded to the nearest 100? The rule for rounding to the nearest hundred is to look at the tens digit. If it is 5 or more, then round up. If it is 4 or less, then round down. Basically,in each hundred, all numbers up to 49 round down and numbers from 50 to 99 round up to the next hundred. What is the greatest 3 digit number that round off to 100? 999 100 is the smallest 3-digit number and 999 is the greatest 3-digit number. What is 246 to the nearest 100? 200 To round a number to the nearest hundred, you make it into the hundred that is closest. Take the number 246, for example, which is between 200 and 300. It is closer to 200. So 246 rounded to the nearest hundred is 200. ### What number rounded to the nearest 100 is 500? 4. 509 is between 500 and 600 and rounded to the nearest 100 is 500 . What is the smallest number that rounds to 100? The smallest whole number that if we rounded to the nearest hundred will give us a result of 95,200 is the number 95,150. What is 739 rounded to the nearest hundred? 700 739 rounded to the nearest hundred is 700. When you round to the nearest hundred, your answer is always going to be an even hundred. #### How do you round 1000? Rounding to the nearest 1000 To round a number to the nearest 1000, look at the hundreds digit. If the hundreds digit is 5 or more, round up. If the hundreds digit is 4 or less, round down. The hundreds digit in 4559 is 5. What does 750 round to? 750 is exactly half-way between 700 and 800, and 791 is between 750 and 800. So 791 is closer to 800 than 700 on the number line and it rounds up to 800. How do you round off numbers to the nearest 100? This makes the question easier to understand. The easiest way to learn about rounding off numbers is to use place value and a number line. Click here to learn about Place Value. Click here for Rounding Off to the Nearest 100. The first step is to make sure your child knows how to count by 5s. ## What is the smallest possible number that will round off to 700? From the number line we can see that the smallest possible number that will round off to 700 when rounded to the nearest ten is 695. To check our answer, we round off 694 to the nearest ten. We will get 690 not 700. So 695 is the correct answer. What is 7089 rounded to the nearest hundred? When your child has understood the process of rounding a number, we can use a short cut. I will use the number 7089 to illustrate. Underline the last 3 digits – 7 089. The digit 0 is in the hundreds place, so 7089 will become either 7 000 or 7 100 after rounding off to the nearest hundred. What does it mean to round up to the nearest integer? Rounding up, sometimes referred to as “taking the ceiling” of a number means rounding up towards the nearest integer. For example, when rounding to the ones place, any non-integer value will be rounded up to the next highest integer, as shown below: 5.01. ⇒.

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A new study has found that racial disparities in incidence of diabetes have more to do with living conditions than just genetics. Researchers from the Hopkins Centre for Health Disparities Solutions and Case Western Reserve University School of Medicine showed that when African Americans and whites live in similar environments and have similar incomes, their diabetes rates are similar. In recent decades the United States has seen a sharp increase in diabetes prevalence, with African Americans having a considerably higher occurrence of type 2 diabetes and other related complications compared to whites. "While we often hear media reports of genes that account for race differences in health outcomes, genes are but one of many factors that lead to the major health conditions that account for most deaths in the United States," said Dr Thomas LaVeist, director of the Hopkins Centre for Health Disparities Solutions and lead author of the study. "I don't mean to suggest that genetics play no role in race differences in health, but before we can conclude that health disparities are mainly a matter of genetics we need to first identify a gene, polymorphism or gene mutation that exists in one race group and not others. "And when that gene is found we need to then demonstrate that that gene is also associated with diabetes," LaVeist added. The study's authors said their findings support the need for future health disparities research and creative approaches to examining health disparities within samples that account for socioeconomic and social environmental factors. The study appears in the Journal of General Internal Medicine.

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One such complex issue deals with harvesting natural gas from West Virginia’s abundant Marcellus Shale reserves. The Marcellus Shale is a formation of sedimentary rocks that lies under a 95,000-square-mile area that includes part of southwestern New York, most of western Pennsylvania, eastern Ohio and nearly all of West Virginia. Marcellus gas reserves lie between 3,000 feet and 9,000 feet under the land’s surface. Formed in the Appalachian Basin over 300 million years ago, the Marcellus Shale formation has recently become an economically viable source of natural gas due to technological advances in horizontal drilling and the hydraulic fracturing process. Hydraulic fracturing, or “fracking,” involves injecting a mixture of water and chemicals under high pressure to release the natural gas reserves contained in the shale. While this industry could potentially provide significant economic benefits to the state, lawmakers are currently grappling with a myriad of issues to ensure that this is done in a balanced manner beneficial to all citizens and stakeholders. According to a recently published study produced by West Virginia University and the Oil and Natural Gas Association, between 2002 and 2008, West Virginia led the nation in the number of gas drilling permits issued. More than 2,800 permits were issued for new drilling in 45 of the state’s 55 counties. The industry-funded study focuses on the economic impacts of our growing efforts to extract natural gas from the Marcellus. In 2009, the study shows, West Virginia’s natural gas industry generated more than $12 billion in business, created more than 24,000 jobs in the state and paid more than $550 million in wages. The report also notes that it is possible that Marcellus development created between 7,600 and 8,500 additional jobs in West Virginia in 2010. According to the report, by the year 2015, West Virginia could see 19,000 more jobs because of Marcellus development and related activities. With any new industry the positive economic outlook must be balanced with proper care for the environment and a respect for all citizens’ rights. This industrial process has created many questions for lawmakers to consider with regard to both the environment and landowner rights. Other advocates contend that tapping into the Marcellus shale field comes at a high cost, ranging from contamination of ground water to damage to local roadways from moving heavy equipment to the drilling sites to infringement on landowners’ rights. The Legislature is considering how to regulate it responsibly as the U.S. Environmental Protection Agency is examining the fracking process to determine whether it endangers supplies of drinking water. In order to address some of these concerns through regulation the “Hydraulic Fracturing and Horizontal Drilling Act” was introduced in both the Senate (SB 258) and the House (HB 2878) in late January. This bill came out of extensive work by interim committees during the months leading up to session. The bill is now being considered in the Judiciary Committee of each body. It is a comprehensive bill that will regulate Marcellus Shale development in areas including land use and surface owner rights, water quality and quantity, disclosure of chemicals used in the hydraulic fracturing process and waste management and disposal. Clearly this is a very complex, detailed and highly important issue. Lawmakers will be hard at work during the second half of this legislative session and beyond to come up with real solutions that strike the appropriate age-old balance between industry, environment and citizens’ rights. Senate Bill 200 will correct the names of state institutions of higher education. The bill will correctly name West Liberty State University as West Liberty University. Senate Bill 342 will appropriate $8 million for the purpose of financing the special elections to fill a vacancy in the office of Governor. House Bill 2853 will provide for a primary and special election to fill the vacancy in the office of the Governor. The bill would call for the primary election to be held on May 14 and the general election to be held on Oct. 4. Senate Bill 78 would require parental consent and accompaniment for a minor using a tanning device. The bill would require minors between the ages of 14 to 17 to present a parental consent form, and minors under the age of 14 must present the consent form, and also be accompanied by a parent or legal guardian. Senate Bill 186 would establish the West Virginia State Police as the entity that is authorized to issue administrative subpoenas to Internet service providers in cases of suspected child pornography. This bill gives the West Virginia State Police the authority to define offenses and set penalties and fees. Senate Bill 195 would adjust the requirements for an individual to become a magistrate. The bill would require magistrates to possess a bachelor’s degree, an associate’s degree in criminal justice or at least four years of prior experience as a magistrate. The bill would go in effect in 2014. Senate Bill 254 would make a supplementary appropriation of federal funds to the Development Office and the Division of Human Services - Energy Assistance. Senate Bill 255 would make supplementary appropriations of remaining moneys to various accounts, such as the Governor’s Office, DHHR, Division of Rehabilitation Services, amoung other agencies. Senate Bill 256 would require sex offenders to verify their e-mail and other online identities in the same way as they register their physical address. Senate Bill 281 would make it a crime for a person to put certain types of invasive software such as, spyware or a virus, on mobile devices of another person without his/her consent or knowledge. Senate Bill 349 would make it a requirement for a bittering agent to be placed in certain engine coolants and antifreezes to prevent personal injury or death of human beings and animals. If implemented, violation of this bill would be considered a misdemeanor. Senate Bill 438 would change the election process for magistrates to be elected by division. Senate Bill 265 would allow certain offenders to have contact with children, but only when the court finds it in the best interest of the child. Senate Bill 419 would create the “Health Care Choice Act” by seeking to increase the availability of health insurance coverage by allowing insurers authorized to sell insurance in Kentucky, Ohio, Maryland, Pennsylvania and Virginia to issue accident and sickness policies in West Virginia. Senate Bill 429 would provide educational scholarships for the children of war veterans. The scholarship would include tuition, institutional fees and standard room and board allowance. The bill would provide that the West Virginia Division of Veterans’ Affairs administer the scholarship program. House Bill 2013 would increase the training that a dispatcher must have to work at a 911 center, adding a new 40 hour course in “emergency medical dispatch.” The bill further would require each dispatch center to develop protocols for dispatching their emergency medical calls. House Bill 2368 would require the Board of Barbers and Cosmetologists to establish an apprenticeship program. This bill would update the existing system for training and procedures relating to the practice of beauty care. House Bill 2503 would authorize the Board of Barbers and Cosmetologists to require government identification to be presented prior to issuance of licenses. The bill also authorizes the board to retain all information regarding licensees. House Bill 2562 would move the State Athletic Commission under the Lottery Commission to assist in administrative functions. The bill would also make mixed martial arts a licensed sport with regulations set forth by the State Athletic Commission. House Bill 2663 would require the Public Service Commission to be present in hearings in which it has retained the right to serve as the initial fact finder in the case. This would include any associated public protest hearings. House Bill 2708 would remove the 12-month limitation on the length of agreements between law-enforcement agencies. Rather than expiring after a year, agreements under this proposed bill would remain in effect unless and until the agreement is changed or withdrawn by the head of one of the law-enforcement agencies. House Bill 2750 would make the commission of sexual assault a consideration when issuing the permanent or temporary end to a parent-child relationship. This bill would allow a judge to take into consideration sexual assault or sexual abuse when deciding whether or not to remove a child from the home. House Bill 2752 would increase the age of persons applying for appointment to a position on a police force within certain cities. The age would be increased from 35 to 40 years old. This increase will pertain to any person(s) applying for a position in a Class I or Class II city. House Bill 2757 would provide for the evaluation of professional personnel within the public school systems. This bill would require the State Board of Education to establish a task force to address rule changes for personnel evaluations. The bill would also require the state board to report the evaluations to the Legislative Oversight Commission on Education Accountability. House Bill 2787 would transfer the licensing of private investigators and security guards from the Secretary of the State to the Division of Justice and Community Services. All procedures, rules and regulations for these professions would be transferred to the Division of Justice. House Bill 2864 would create a misdemeanor crime of unlawful restraint in the first or second degree. First-degree unlawful restraint would be considered intentionally restraining someone by use without proper authority. This would carry a maximum sentence of one year in jail and a $500 fee. Second-degree restraint is defined in this bill as restraint based on intimidation. This crime would carry a maximum of six months in jail and a $100 fine. House Bill 2871 would provide that the Brownfield Economic Development districts comply with local planning laws. This compliance must take place before an application for such districts would be approved. House Bill 2936 would change the date of the canvassing votes in a primary election. This bill would change the date from the Friday following a primary election to the Monday following a primary election.

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## Algebra 1 $\frac{1}{8}$ If a=-4, and b=1, in order to evaluate the expression $\frac{-b}{2a}$ all we must do is simply plug in the givens into the expression. $\frac{-b}{2a}$=$\frac{-(1)}{2(-4)}$=$\frac{-1}{-8}$=$\frac{1}{8}$

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GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 18 Oct 2019, 01:43 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # When positive integer x is divided by 44, the remainder is 28 Author Message TAGS: ### Hide Tags GMAT Club Legend Joined: 12 Sep 2015 Posts: 4007 When positive integer x is divided by 44, the remainder is 28  [#permalink] ### Show Tags 30 Mar 2017, 08:35 2 1 Top Contributor 2 00:00 Difficulty: 5% (low) Question Stats: 87% (01:34) correct 13% (02:01) wrong based on 117 sessions ### HideShow timer Statistics When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11? A) 1 B) 3 C) 5 D) 7 E) 9 *kudos for all correct solutions _________________ Test confidently with gmatprepnow.com Math Expert Joined: 02 Aug 2009 Posts: 7977 Re: When positive integer x is divided by 44, the remainder is 28  [#permalink] ### Show Tags 30 Mar 2017, 08:51 4 1 GMATPrepNow wrote: When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11? A) 1 B) 3 C) 5 D) 7 E) 9 *kudos for all correct solutions Good Question Brent.. When x is divided by 44, remainder is 28... So the number x is 44t+28, where t is integer.. When y is divided by 22, remainder is 14.. So the number y is 22s+14, where s is integer.. N=x+y = 44t+28+22s+14 So when N is divided by 11, $$\frac{44t+22s+42}{11}$$ 44t and 22s are div by 11.. 42=11*3+9.. So 9 is the remainder E _________________ ##### General Discussion VP Joined: 05 Mar 2015 Posts: 1003 Re: When positive integer x is divided by 44, the remainder is 28  [#permalink] ### Show Tags 30 Mar 2017, 10:46 2 GMATPrepNow wrote: When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11? A) 1 B) 3 C) 5 D) 7 E) 9 *kudos for all correct solutions x = 44a+28( a = any integer) Let a =1 then x = 44+28= 72 similarly y=22b+14 (b=any integer) let b=1 then y= 36 thus N=x+y =72+36=108 108/11 = 9 9/11 ans E Manager Status: GMAT...one last time for good!! Joined: 10 Jul 2012 Posts: 54 Location: India Concentration: General Management GMAT 1: 660 Q47 V34 GPA: 3.5 Re: When positive integer x is divided by 44, the remainder is 28  [#permalink] ### Show Tags 30 Mar 2017, 11:46 1 GMATPrepNow wrote: When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11? A) 1 B) 3 C) 5 D) 7 E) 9 *kudos for all correct solutions x=44a+28;y=14+22b For a=1 and b=1 x=72,y=36 n=x+5=108 remainder-9 _________________ Kudos for a correct solution VP Joined: 07 Dec 2014 Posts: 1222 Re: When positive integer x is divided by 44, the remainder is 28  [#permalink] ### Show Tags 30 Mar 2017, 11:54 2 GMATPrepNow wrote: When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11? A) 1 B) 3 C) 5 D) 7 E) 9 *kudos for all correct solutions least values of x and y are 28 and 14 respectively x+y=42=N N/11 gives a remainder of 9 E Current Student Joined: 07 Jan 2016 Posts: 1088 Location: India GMAT 1: 710 Q49 V36 Re: When positive integer x is divided by 44, the remainder is 28  [#permalink] ### Show Tags 30 Mar 2017, 13:09 1 remainder 28 when divided by 44 so 44a + 28 remainder 14 when divided by 22 so 22b + 14 now N = x+y 44a + 22b + 42 N/11 as we know 44 and 22 are divisible by 11 only 42 is remaining 42/11 remainder = -2 or 9 44- 2 = 42 33+9 = 42 E Manager Joined: 12 Jun 2016 Posts: 212 Location: India WE: Sales (Telecommunications) Re: When positive integer x is divided by 44, the remainder is 28  [#permalink] ### Show Tags 04 Aug 2017, 09:45 HI! Did it by number picking Possible value of x = 28, 72, 100... Possible values of y = 14, 36, 58.... Since there can be only one unique solution, add to find the smallest possible value of n = 28 + 14 = 42. Remainder of 42/11 = 9 Is the Kudos for correct solution still valid? _________________ My Best is yet to come! Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4775 Location: India GPA: 3.5 Re: When positive integer x is divided by 44, the remainder is 28  [#permalink] ### Show Tags 20 Jan 2019, 09:14 GMATPrepNow wrote: When positive integer x is divided by 44, the remainder is 28. When positive integer y is divided by 22, the remainder is 14. If N = x+y, what is the remainder when N is divided by 11? A) 1 B) 3 C) 5 D) 7 E) 9 *kudos for all correct solutions Let $$x = 28$$ and $$y = 14$$ So, $$N = 28 + 14$$ => $$42$$ $$\frac{N}{11} = 11*3 + 9$$, Thus Answer must be (E) 9 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Re: When positive integer x is divided by 44, the remainder is 28   [#permalink] 20 Jan 2019, 09:14 Display posts from previous: Sort by

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1. ## librarian A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all. a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135 b) if 10 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.096 c) If 20 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.0042 any idea? 2. Originally Posted by swoopesjr01 A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all. a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135 Binomial distribution problem. Out of every $100$ books signed out $5$ are returned damaged and $1$ is not returned at all. So of every $99$ books returned $5$ are damaged so the probability that a returned book is damaged is $5/99\approx 0.0505$ Of $4$ books returned the probability that exactly $2$ are damaged is: $P={4 \choose 2} 0.0505^2 (1-0.0505)^2$ $=\frac{4!}{2!\ 2!} 0.0505^2\ 0.9495^2\approx 0.01379$ Ooops - the above is the probability that of four returns two are damaged irrespective of how many were signed out. - No in retrospect this is right I just managed to confuse myself when rereading the explanation as it is not clear enough. There is a clearer (at least I think so) explanation in one of my later posts. Simulation shows that the proper probability has about $95\%$ chance of being in the interval $[0.01376,0.01396]$ Now if the question had been: a) if 4 people each sign out one book, what is the probability that exactly 2 of the books will be returned damaged? $P_1={4 \choose 2} 0.05^2 (1-0.05)^2=\frac{4!}{2!\ 2!} 0.05^2\ 0.95^2\approx 0.0135$ RonL 3. Originally Posted by swoopesjr01 c) If 20 people each sign out one book, what is the probability that between 2 and 4 (inclusively) are returned damaged and exactly 2 are never returned at all? 0.0042 Again, this is binomial probability thus, $\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$ The other one is, ${20 \choose 2}.05^2.95^{18}=.0159$ Since this is a conjunction probability (this and that) You need to multiply them to get, $\approx 0.00415944$ 4. Originally Posted by ThePerfectHacker Again, this is binomial probability thus, $\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$ The other one is, ${20 \choose 2}.05^2.95^{18}=.0159$ Since this is a conjunction probability (this and that) You need to multiply them to get, $\approx 0.00415944$ The trouble is that the two clauses in the conjunction are not independent - but multiplying may well be OK-ish here but I would like to be more confident than I am. RonL 5. Originally Posted by CaptainBlack The trouble is that the two clauses in the conjunction are not independent - but multiplying may well be OK-ish here but I would like to be more confident than I am. RonL I was troubled by that too, I just saw that my hypothesis matched his answers. So I ignored that. 6. Originally Posted by ThePerfectHacker I was troubled by that too, I just saw that my hypothesis matched his answers. So I ignored that. I was distrubed by the given answers to all three questions in the original post, that is why I stopped after doing (a) - I had to go away and think If I get the chance (and I probably will as my research budget has been frozen due to contract delays) I plan to see if simulation will throw any light on the correctness of the given answers. (Rule 1 of Applied Probability: When in doubt about the correctness of a probability calculation - simulate) RonL 7. Originally Posted by ThePerfectHacker Again, this is binomial probability thus, $\sum^4_{k=2} {20 \choose k}.05^k.95^{20-k}=.2616$ The other one is, ${20 \choose 2}.05^2.95^{18}=.0159$ Typo here, ${20 \choose 2}0.01^2\ 0.99^{18}=.0159$ RonL 8. Originally Posted by CaptainBlack I was distrubed by the given answers to all three questions in the original post, that is why I stopped after doing (a) - I had to go away and think If I get the chance (and I probably will as my research budget has been frozen due to contract delays) I plan to see if simulation will throw any light on the correctness of the given answers. (Rule 1 of Applied Probability: When in doubt about the correctness of a probability calculation - simulate) RonL Simulations done. For b) they give p~0.00023 for c) they give p~0.00357 RonL 9. Originally Posted by swoopesjr01 A librarian has estimated that 5% of the books that people sign out are resturned damaged in some way, while 1% of the books are never returned at all. a) if 4 people each sign out one book and all are returned, what is the probability that exactly 2 of the books will be returned damaged? 0.0135 Given that a book is returned the probability that it is damaged is $0.05/0.99 \approx 0.50505$, and the probability that it is not damages is $1-0.050505 \approx 0.949495$. So the probability that if four books are returned from four checked out, that exactly two are returned damaged is: $ P_1={4 \choose 2} 0.050505^2 (1-0.050505)^2=$ $\frac{4!}{2!\ 2!} 0.050505^2\ 0.949495^2\approx 0.013798 $ RonL

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There is a species of bees called “commercial” bees. These bees are kept by beekeepers to pollinate crops such as tomatoes, sweet peppers, and oilseed. This population of managed bees is coming down with “fast evolving viruses”, according to the University of Exeter in Science Daily News. Then there are “wild” bees, free to fly around, not employed by beekeepers. The viruses that the commercial bees have are starting to spread to the wild bee population. Currently, researchers are “calling for new measures” to protect the wild pollinators, and confine the commercial, diseased population. In the article, Dr. Lena Wilfert said this can be controlled by beekeepers keeping a vigil eye and monitoring the commercial bees they own. It is their “responsibility” to do so. Also, interesingly enough, the international transport of these commercial bees must have more checks and security. They must be screened better, in order to get a better sense of how many have a disease, so they know not to set any of the commercial bees free into the wild. The major cause of the spread is the Varroa mite. This spreads viruses, such as the Deformed Wing Virus, and may increase the power of the viral spread. It significantly weakens bees, causing their RNA to deteriorate. The article says that it has been “identified as an emerging disease in pollinators,” and there is a connection between wild bumblebees who have it, and commercial honeybees. The poor management of the commercial bee community is the cause of this horrible break out of diseases among innocent wild bees. In the future, researchers plan to investigate which species of commercial bees are the major cause of the breakout and spread. The wild bee population is extremely important for our environment, and beekeepers need to realize that, and make sure their bee farm does not spread disastrous diseases. *Additional information is found through the last two hyperlinks.* *Original article is the first hyperlink.*

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Orodromeus was a small, herbivorous, or plant-eating, dinosaur that inhabited North America during the late Cretaceous period, about 65 to 98 million years ago. Orodromeus is classified as a member of the family Hypsilophodontidae, which was one of the most successful dinosaur families, flourishing for about 100 million years from the late Jurassic through the early Cretaceous periods. The Hypsilophodontidae belong to the order Ornithischia (the bird-hipped dinosaurs) and the suborder Ornithopoda, which consists of dinosaurs that exhibited primarily bipedal, or two-legged, locomotion. The first fossil evidence of Orodromeus was discovered in the 1980s while a crew of paleontologists was examining nests of hadrosaur dinosaurs in Montana. At nearby sites that were later dubbed Egg Mountain and Egg Island, they discovered a large number of eggs of a new ornithopod dinosaur among the hadrosaur remains. The new dinosaur was named Orodromeus, which means “mountain runner,” because of both the location of the find and the animal’s presumed swiftness, based on its long hind limbs and slender build. The eggs and nests were remarkably well preserved; one complete nest contained 19 eggs laid in a precise spiral. After radiographic examination revealed embryonic bones inside the eggs, paleontologists carefully opened one egg and removed the preserved embryo for further study. The condition of the nesting site, along with structural aspects of the embryo, shed light on the presumed parenting habits of Orodromeus. The eggs were less trampled than those found in nests of other dinosaur genera, suggesting that the young left the site soon after emerging from their shells—therefore not trampling the eggs. The presumption that Orodromeus and other hypsilophodonts engaged in only minimal parental care was further corroborated by the bone structure of the embryos. The joints of the hatchlings were well developed, suggesting that the young were capable of immediately leaving the site to fend for themselves. In contrast, the offspring of the hadrosaurid Maiasaura had poorly developed leg joints, making them incapable of leaving the nest soon after hatching and therefore in need of more parental care. (See also Maiasaura.) Orodromeus grew to about 8 feet (2.4 meters) in length. Its head was small and light, with a birdlike beak. The hind legs were long and slender, and the arms ended in hands with five short, blunt-clawed fingers. Orodromeus was bipedal, meaning that it stood and walked on two legs. Along with its confamilials—that is, animals belonging to the same family—Orodromeus had effective grinding teeth and cheeks that held food during chewing. Some evidence suggests that they may have traveled in herds. Horner, John, and Dobb, Edwin. Dinosaur Lives: Unearthing an Evolutionary Saga (HarperCollins, 1997). Lambert, David, and the Diagram Group. Dinosaur Data Book: The Definitive Illustrated Encyclopedia of Dinosaurs and Other Prehistoric Reptiles (Gramercy, 1998). Lessem, Don, and Glut, D.F. The Dinosaur Society’s Dinosaur Encyclopedia (Random, 1993). Lockley, Martin. Tracking Dinosaurs: A New Look at an Ancient World (Cambridge Univ. Press, 1991). Norell, M.A., and others. Discovering Dinosaurs in the American Museum of Natural History (Knopf, 1995). Norman, David. The Illustrated Encyclopedia of Dinosaurs (Crescent, 1985). Sattler, H.R. The New Illustrated Dinosaur Dictionary (Lothrop, 1990). Weishampel, D.B., and others, eds. The Dinosauria (Univ. of Calif. Press, 1990). Dixon, Dougal. Questions and Answers About Dinosaurs (Kingfisher, 1995). Farlow, J.O. On the Tracks of Dinosaurs (Watts, 1991). Gohier, François. 165 Million Years of Dinosaurs (Silver Burdett, 1995). Green, Tamara. Looking at: The Dinosaur Atlas (Gareth Stevens, 1997). Sokoloff, Myka-Lynne. Discovering Dinosaurs (Sadlier-Oxford, 1997). Theodorou, Rod. When Dinosaurs Ruled the Earth (Thomson Learning, 1996). Unwin, David. The New Book of Dinosaurs (Copper Beech, 1997).

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The Sahara Desert and in particular its northern regions have attracted its share of attention from Atlantis investigators. However unlikely it may appear as a possible location for Atlantis it must be kept in mind that the Sahara of prehistory was very different from what we see today. Not only was it wetter at various periods in the past, but also there is clear evidence for the existence of a large inland sea extending across the borders of modern Algeria and Tunisia. This evidence is in the form of the chotts or salt flats in both countries. This proposed sea is considered by some to have been the Lake Tritonis referred to by classical writers. It is suggested that some form of tectonic/seismic activity, common in the region, was responsible for isolating this body of seawater from the Mediterranean and eventually turning it into the salt flats we see today. An even more extensive inland sea, further south, was proposed by Ali Bey el Abbassi and based on his theory a map was published in 1802 which can be viewed online(c). More recently, Riaan Booysen has published an illustrated paper on the ancient inland Saharan seas as indicated on the 16th century maps of Mercator and Ortelius(i). King’s College London runs The Sahara Megalakes Project which studies the Megalakes and the Saharan Palaeoclimate record(m). A 2013 report in New Scientist magazine(d) revealed that 100,000 years ago the Sahara had been home to three large rivers that flowed northward, which probably provided migration routes for our ancestors. Other studies(h) have shown the previous existence of a huge river system in the Western Sahara, which flowed into the Atlantic on the Mauritanian coast. An article in the Sept. 2008 edition of National Geographic pointed out that the Saharan climate has been similar for the past 70,000 years except for a period beginning 12,000 years ago when a number of factors combined to alter this fact. A northerly shift by seasonal monsoons brought additional rain to an area the size of contiguous USA. This period of a greener Sahara lasted until around 4,500 years ago. More recent studies claim that “there’s geologic evidence from ocean sediments that these orbitally-paced Green Sahara events occur as far back as the Miocene epoch (23 million to 5 million years ago), including during periods when atmospheric carbon dioxide was similar to, and possibly higher, than today’s levels. So, a future Green Sahara event is still highly likely in the distant future.” More recent studies claim that “there’s geologic evidence from ocean sediments that these orbitally-paced Green Sahara events occur as far back as the Miocene epoch (23 million to 5 million years ago), including during periods when atmospheric carbon dioxide was similar to, and possibly higher, than today’s levels. So, a future Green Sahara event is still highly likely in the distant future.” (p) Henri Lhote contributed an article to the Reader’s Digest’s, The World’s Last Mysteries , regarding the ‘green’ Sahara that existed prior to 2500 BC. An interesting question might be; what happened circa 2500 BC to cause this reversal? Some have suggested a connection between the aridification of the Sahara and the destruction of Atlantis! More recently, human activity has been blamed as a major contributory factor for the desertification of the Sahara region less than 10,000 years ago.(n) Related to the above is a recent study of sediments off the west coast of Africa, which resulted in the discovery of what was “primarily a new “beat,” in which the Sahara vacillated between wet and dry climates every 20,000 years, in sync with the region’s monsoon activity and the periodic tilting of the Earth.” (o) In 1868, it was proposed by D.A. Godron, the French botanist, that the Sahara was the location of Atlantis. In 2003, the non-existent archaeologist Dr.Carla Sage announced that she was hoping to lead an international expedition to the Sahara in search of Atlantis. Her contention was that “Atlantis was the capital of a vast North African empire with ports on the Gulf of Sidra”. This report is now confirmed to have been a hoax! I am indebted to Stel Pavlou for uncovering the origin of this story(e). Gary Gilligan, the well-known catastrophist, wrote a thought-provoking article(k) on the origin of the Saharan sands, which he claims are extraterrestrial in origin and expands on the idea in his 2016 book Extraterrestrial Sands . David Mattingly, an archaeologist at Leicester University has found that an ancient people known as the Garamantes had an extensive civilisation in the Sahara(l). He has evidence of at least three cities and twenty other settlements. The Garamantes reached their peak around 100 BC and then gradually diminished in influence as fossil water supplies reduced until in the 7th century AD they were subjected to Islamic domination. Some researchers such as Frank Joseph have identified the Garamantes as being linked with the Sea Peoples. Bob Idjennaden has published short but informative Kindle books about both the Garamantes and the Sea Peoples , without a suggestion of any connection between the two. The discovery of the megalithic structures discovered at Nabta Playa (Nabta Lake) in the Egyptian Sahara has provided evidence for the existence of a sophisticated society in that area around 5000 BC. In the same region, near the Dakhleh Oasis, archaeologists have produced data that supports the idea that pre-Pharaonic Egypt had Desert Origins rather than being an importation from Mesopotamia or elsewhere(a). Nabta Playa is not unique, in fact the largest megalithic ellipse in the world is to be found at Mzorah, 27 km from Lixus in Morocco(b). It appears that the construction methods employed at both Mezorah and Nabta Playa are both similar to that used in the British Isles. An even more impressive site is Adam’s Calendar in South Africa which has been claimed as 75,000-250,000 years old. West of Cairo near the border with Libya is the Siwa Oasis, where it has now been demonstrated that “it is in fact home to one of Ancient Egypt’s astounding solar-calendar technologies– the solar equinox alignment between the Timasirayn Temple and the Temple of Amun Oracle in Aghurmi.”(j). I think we can expect further exciting discoveries in the Sahara leading to a clearer picture of the prehistoric cultures of the region and what connections there are, if any, with Plato’s Atlantis. In the meanwhile in the Eastern Egyptian Desert, Douglas Brewer, a professor of archaeology at the University of Illinois, has discovered over 1,000 examples of rock art, including numerous depictions of boats although the sites, so far undisclosed, are remote from water. Even more remarkable is the report(e) of March 2015 that a survey of the Messak Settafet escarpment in the central Sahara revealed that there were enough discarded stone tools in the region “to build more than one Great Pyramid for every square kilometre of land on the continent”! Coincidentally, around the same time it was reported that over a thousand stone tools had been found in the Northern Utah Desert(g). What the Utah discovery lacked in quantity was made up for in quality with the finding of the largest known Haskett point spear head, measuring around nine inches in length. (a) Saudi Aramco World (2006, Vol. 57, No.5 p.2-11) (d) NewScientist.com, 16 September 2013, https://tinyurl.com/mg9vcoz (l) See: Archive 3268 The Garamantes, referred to by Herodotus, are generally considered to have been the first Libyan empire. However, attempts to link them to the Atlantis story would appear to be undermined by the fact that they flourished around the middle of the first millennium BC, which would appear to be far too recent to fit any interpretation of Plato’s 9,000 ‘years’, be they solar, lunar or seasonal. Frank Joseph identifies the Garamantes with the Sea Peoples whom he considers to have been Atlantean refugees. However, the Garamantes are generally accepted as having developed urban centres in what is now southern Libya(a)(b), which is not what you might expect from a maritime culture. Recent satellite images(c) offer new information on the extent of the Garamantes domain. Further information is available on the Temehu.com website(d), but perhaps the most intriguing theory is that the Garamantes originally came from the Carpatho–Danubian region of eastern Europe sometimes referred to as Dacia. It is suggested that the fair-skinned Berbers of North Africa are the descendants of those European invaders! Bob Idjennaden, a Belgian living in Ireland, has published a short Kindle book(f) on the Garamantes. He has also authored a series of Kindle books on various aspects of ancient African history, including one about the Sea Peoples, co-authored with Taklit Mebarek. (f) https://www.amazon.co.uk/Buried-Kingdoms-Garamantes-Forgotten-Civilisations-ebook/sim/B007Q239OE/2 (link broken Sept. 2020)<

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XML elements can have attributes, just like HTML. Attributes are designed to contain data related to a specific element. XML Attributes Must be Quoted Attribute values must always be quoted. Either single or double quotes can be used. For a person's gender, the <person> element can be written like this: or like this: If the attribute value itself contains double quotes you can use single quotes, like in this example: or you can use character entities: XML Elements vs. Attributes Take a look at these examples: In the first example gender is an attribute. In the last, gender is an element. Both examples provide the same information. There are no rules about when to use attributes or when to use elements in XML. My Favorite Way The following three XML documents contain exactly the same information: A date attribute is used in the first example: A <date> element is used in the second example: An expanded <date> element is used in the third example: (THIS IS MY FAVORITE): Avoid XML Attributes? Some things to consider when using attributes are: - attributes cannot contain multiple values (elements can) - attributes cannot contain tree structures (elements can) - attributes are not easily expandable (for future changes) Don't end up like this: to="Tove" from="Jani" heading="Reminder" body="Don't forget me this weekend!"> XML Attributes for Metadata Sometimes ID references are assigned to elements. These IDs can be used to identify XML elements in much the same way as the id attribute in HTML. This example demonstrates this: <body>Don't forget me this weekend!</body> <body>I will not</body> The id attributes above are for identifying the different notes. It is not a part of the note itself. What I'm trying to say here is that metadata (data about data) should be stored as attributes, and the data itself should be stored as elements.

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Start a 10-Day Free Trial to Unlock the Full Review Why Lesson Planet? Find quality lesson planning resources, fast! Share & remix collections to collaborate. Organize your curriculum with collections. Easy! Have time to be more creative & energetic with your students! The color sight word "green" is the focus of this language arts worksheet. Learners cut and past pictures of green objects to match sentences about those green objects. A color as well as a black and white copy are provided. 3 Views 13 Downloads

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# Probabilities about drwing lucky cards I wanna to calculate a probabilities of buying a box of lucky cards. However I am messed up with the equations. The question: Let there is a box with 30 cards. and 6 of them are lucky cards. Calculate the probabilities of buying cards one by one until 6 cards are all brought. At which number would the probability that the chance is higher than 80%? - Presumably each purchase is independent and can buy each card with equal probability $\frac{1}{30}$. Suppose $P(n,k)$ is the probability that after $n$ purchases you are missing $k$ lucky cards. Then $$P(n+1,k)=\frac{30-k}{30}P(n,k)+\frac{k+1}{30}P(n,k+1)$$ starting at $P(0,6)=1$ and $P(0,k)=0$ for $k \not = 6$. This can easily be calculated, for example using a spreadsheet. You are interested in the smallest $n$ such that $P(n,0) \gt 0.8$. It turns out that $P(97,0) \approx 0.79402$ while $P(98,0) \approx 0.80031$. As a check on the calculations, the expected number of cards needed to be bought to get all the lucky cards is both $\displaystyle\sum_{n=0}^\infty (1-P(n,0)) = 73.5$ and $30\left(\frac16 + \frac15 + \frac14 + \frac13 + \frac12 + \frac11\right)=73.5$. If purchases were without replacement (meaning that at $30$ purchases you have all the cards, lucky and unlucky), the recurrence becomes $$P(n+1,k)=\frac{30-n-k}{30-n}P(n,k)+\frac{k+1}{30-n}P(n,k+1).$$ You now find $P(29,0) = 0.8$. This is not a surprise as the probability the final card purchased is lucky is $\frac{6}{30}=0.2$. - Thanks for answering the question, How about if the box is fixed to 30 cards and fixed lucky cards are 6. Once the lucky card is picked and no replacement. Will the answer be different? – Kitw Dec 18 '12 at 7:17 Shall the equation become, P(n+1,k)=(30−k)/(24+k) * P(n,k)+ (k+1)/(24+k) * P(n,k+1) – Kitw Dec 18 '12 at 7:39 Followed above, the result turns a bit wield, am I miss something? <a href="picturepush.com/public/11714784"><img src="www1.picturepush.com/photo/a/11714784/img/11714784.png"; border="0" alt="Image Hosted by PicturePush - Photo Sharing" /></a> – Kitw Dec 18 '12 at 7:53 @Kitw: Not quite, instead see my addition. The fact that your rows of probabilities do not add up to 1 should suggest an issue. – Henry Dec 18 '12 at 7:55 Many thanks for solving my mess :) – Kitw Dec 18 '12 at 8:03 If you buy cards one by one, the box of 30 doesn't matter. How many total cards are there? You have the Coupon collector's problem but need to specify the conditions better before there can be a specific solution. - Thanks for the answer, Had just read the Coupon collector's problem, does it replace everytime? If the box size and number of cards are fixed, will the math be the same – Kitw Dec 18 '12 at 7:19 The fact is to find the expected value of how many times is the best to stop buying more. On the same condition, What is the probabilities that first 12 pick will get half cards and so on. – Kitw Dec 18 '12 at 7:22 @Kitw: In the classic Coupon collector's problem, each coupon is randomly selected from the universe, so yes, it replaces every time. You can use the classic coupon collector's problem for 6 cards and multiply by 5, because $\frac 45$ of the time you don't get a lucky card and can ignore it. You want the number of draws to complete your set of 6. – Ross Millikan Dec 18 '12 at 14:27

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Short-sightedness – or myopia – is a vision problem in which close objects can be seen clearly but objects far away are blurred. Short-sightedness occurs if the eyeball is too long or the cornea has too much curvature. As a result, the light entering the eye can’t focus correctly and distant objects appear blurred and distorted. Short-sightedness is a common condition and affects around 30% of the U.S. population. It usually first occurs in school-age children and, because the eye continues to grow during childhood, it will progress until about age 20. What Causes Short-sightedness? The exact causes of the condition are unknown, but research indicates that two factors may be primarily responsible for its development: - Visual stress There is significant evidence to suggest that many people inherit short-sightedness, or at least a tendency to develop it. If one or both parents are short-sighted, there is a greater likelihood their children will be short-sighted. Even though the tendency to develop the condition may be inherited, its development may be affected by how the eyes are used. Those who spend a lot of time doing close visual work, such as reading or working in front of a computer, may be more likely to develop short-sightedness. Symptoms of short-sightedness may also be a sign of a blood sugar condition, such as diabetes or an early indication of a developing cataract. Diagnosis and Treatment A comprehensive eye test, using several procedures, by your optometrist will determine if you have short-sightedness. There are several options available for persons with short-sightedness to regain good distance vision. These are: - Contact lenses - LASIK surgery - Vision therapy for persons with stress-related short-sightedness Eyeglasses are the primary choice for correcting short-sightedness. A single vision lens is usually prescribed to provide good vision at all distances. But patients over 40, and children or adults whose short-sightedness is due to intense close vision work may have bifocal or progressive addition lenses prescribed for them. These lenses provide different strengths or powers throughout the lens to provide for distant and close vision. Some patients will find that contact lenses offer better vision than eyeglasses and may suit them better cosmetically. Although they provide a wider field of vision, they require regular cleaning and care to safeguard eye health. Orthokeratology involves the fitting of a series of rigid gas permeable contact lenses to reshape the cornea. The contact lenses are worn overnight during sleep, and removed for the entire day allowing patients to go about their daily activities eyewear-free. Your optometrist can advise you further about the process and whether or not it would suit your particular requirements. Using a laser beam of light, short-sightedness can be corrected by reshaping the cornea in a process called LASIK surgery. Although this surgery has generally proved successful, it is still a developing therapy in which some problems have occurred. Speak to your optometrist to acquire full information about this process and how it is likely to affect you. Vision therapy is an option for patients whose blurred distance vision is caused by a spasm of the muscles which control eye focusing. Various exercises are used to improve the ability to focus and regain good vision. People with short-sightedness have a number of options to correct their vision problem. Your optometrist will help you select the treatment that best meets your visual and lifestyle needs. Related – Medical Reasons for Wearing Sunglasses, Stevens-Johnson Syndrome, Glaucoma Information, Onchocerciasis: River Blindness, Presbyopia, Posterior Vitreous Detachment, Eye Anatomy: Important Definitions, Duane Syndrome, Helpful Glaucoma Information. 0

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# Conditional probability regarding multiple choice Q: A multiple choice exam has 4 choices for each question. A student has studied enough so that the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible. If they know the answer they will get the question right. If not they have to guess from the 3 or 4 choices. As the teacher you want the test to measure what the student knows. If the student answers a question correctly what’s the probability they knew the answer? A: Stuck in the process of trying to answer this. The way I thought about it was using Baye's rule and treating it as a conditional probability. In that, $$P(answer correct | knew answer) = \frac{P(know answer | answer correct) \cdot P(answer correct)}{P(knew answer)}$$ The probability that the student knows the answer is 0.5 so that gives us the denominator. Figuring out probability for the answer being correct and knowing the answer given a correct answer is a bit more confusing for me. Would appreciate guidance. Let A represent the event that the question is answered correctly Let X represent the event that the student knows the correct answer Let Y represent the event that the student is able to eliminate one choice Let Z represent the event that the student is not able to eliminate any choice I interpret the statement .... the probability they will know the answer to a question is 0.5, the probability that they will be able to eliminate one choice is 0.25, otherwise all 4 choices seem equally plausible. ... to mean ... $$P(X)=0.5; P(Y)=.25;p(Z)=.25$$ So ... $$\begin{eqnarray*} P(X|A) &=& \frac{P(A \cap X)}{P(A)} \\&=& \frac{P(A \cap X)}{P(A\cap X)+P(A\cap Y)+P(A\cap Z)} \\&=& \frac{P(A |X)P(X)}{P(A |X)P(X)+P(A |Y)P(Y)+P(A |Z)P(Z)} \\&=& \frac{1 \cdot \frac 12}{1 \cdot \frac 12+\frac 14 \cdot \frac 13+\frac 14 \cdot \frac 14} \\&=& \frac {\frac 12} {\frac{24}{48}+\frac4{48}+\frac3{48}} =\frac{24}{31} \end{eqnarray*}$$ • Ah, yes, I interpreted the problem statement differently in my Answer, but I bet you have the correct interpretation! +1 – Bram28 Jul 12 '17 at 19:36 With $A$: students answers correctly $B$: student knew answer what you want is $P(B|A)$, rather than $P(A|B)$ Now, we are given that: $$P(B)= \frac{1}{2}$$ and hence $$P(B^C)=1-\frac{1}{2}=\frac{1}{2}$$ Also: $P(A|B)=1$ (if they knew the answer they give the correct answer) and $$P(A|B^C)=\frac{1}{4}\cdot \frac{1}{3}+\frac{3}{4}\cdot \frac{1}{4}=\frac{1}{12}+\frac{3}{16}=\frac{4}{48}+\frac{9}{48}=\frac{13}{48}$$ (When they don't know the answer there is a $\frac{1}{4}$ probability they can eliminate one of the answers and thus have a $\frac{1}{3}$ probability of guessing correctly between the remaining 3, and there is a $\frac{3}{4}$ probability they can't eliminate any one answer in which case they have a $\frac{1}{4}$ probability of guessing correctly) Next, we have that: $$P(A \cap B)=P(A|B)\cdot P(B)=1\cdot \frac{1}{2}=\frac{1}{2}$$ and $$P(A \cap B^C)=P(A|B^C)\cdot P(B^C)=\frac{13}{48}\cdot \frac{1}{2}=\frac{13}{96}$$ and thus: $$P(A)=P(A \cap B)+P(A \cap B^C)=\frac{1}{2}+\frac{13}{96}=\frac{61}{96}$$ So finally: $$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{1}{2}}{\frac{61}{96}}=\frac{48}{61}$$

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# Problem of the Week Problem A and Solution Balancing Act ## Problem Bailey is in charge of sending out boxes from a distribution centre. The contents of the boxes are identified by shapes stamped on them: a heart, a moon, or a sun. All boxes with the same stamp have the same mass, and the cost of sending a box depends on its mass. Bailey has a balance scale and a few standard weights to help with the job. The following diagrams show what Bailey observed when arranging some of the boxes and standard weights on the scales. Find the mass of each box. ## Solution From the diagrams we notice the following. • One heart box has a mass of $$2$$ kg. • One moon box and one sun box have a total mass of $$24$$ kg. • One moon box and one sun box have the same total mass as one moon box and two heart boxes. From this, we can conclude that one moon box and two heart boxes have a total mass of $$24$$ kg. Also, two heart boxes have the same mass as one sun box. Since one heart box has a mass of $$2$$ kg, then two heart boxes have a mass of $$4$$ kg. Therefore, one sun box has a mass of $$4$$ kg. This means $$4 \text{kg} + \text{(mass of a moon box)} = 24$$ kg. Since $$4 + 20 = 24$$, we can determine that one moon box must have a mass of $$20$$ kg. Teacher’s Notes The idea of a balance scale is a nice analogy for an algebraic equation. We can represent the information in the problem using equations with variables to represent the masses of the different types of boxes. Here is one way to solve the problem algebraically. Let $$x$$ represent the mass of a heart box. Let $$y$$ represent the mass of a sun box. Let $$z$$ represent the mass of a moon box. From the information in the diagrams, we can write the following equations: \begin{align} x &= 2 \tag{1}\\ y + z &= 24 \tag{2}\\ y + z &= 2x + z \tag{3} \end{align} From equation $$(1)$$, we already know that a heart box has a mass of $$2$$ kg. From equations $$(2)$$ and $$(3)$$, we notice that the left sides are the same, so the right sides of the equations must be equal to each other. This means we know: $2x + z = 24 \tag{4}$ Now, substituting $$x = 2$$ into equation $$(4)$$, we get $2(2) + z = 24$ Subtracting $$4$$ from each side of this equation, we get $z = 20$ Finally, substituting $$z = 20$$ into equation $$(2)$$, we get $y + 20 = 24$ Subtracting $$20$$ from each side of this equation, we get $y = 4$ So, a heart box has a mass of $$2$$ kg, a sun box has a mass of $$4$$ kg, and a moon box has a mass of $$20$$ kg.

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# The value of $a\:\:and\:\:b$ for which $f:R\rightarrow R$ is defined as $f(x)=ax+b$ where $a,b\in R$ is bijection with $fof$ as its identity is ? $\begin{array}{1 1} a=1\;and\;b \in R \\ a=1,b=0\;or\;a=-1 ,b \in R \\ a= \pm 1,b \in R \\ a= \pm 1, b=0 \end{array}$ Since $f(x)=ax+b$ is linear, it is either increasing or decreasing. $\Rightarrow\:f'(x)>0\:\:or\:\:f'(x)<0$ $\Rightarrow\:a>0\:\:or\:\:a<0$ and $a\neq 0$ because $f$ is bijection. Also given that $fof=I_(x)=x$ $\Rightarrow f(x)=f^{-1}(x)$ Let $y=f(x)=ax+b\:\:\Rightarrow\:x=f^{-1}(y)=\large\frac{y-b}{a}$ $\Rightarrow\:f^{-1}(x)=\large\frac{x-b}{a}$ Since $f=f^{-1}$, $ax+b=\large\frac{x-b}{a}$ Equating the coefficients on both the sides we get $a=\large\frac{1}{a}$ and $b=-\large\frac{b}{a}$ $\Rightarrow\:a^2=1\:\:and\:\:a=-1\:or\:b=0$ $\Rightarrow a=1\:and\:b=0\:\:or\:a=-1\:and\:b\in R$

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Return to Biology Index Gupta, Raj Schurz High School 1. To develop the concept of diffusion. 2. Demonstrate diffusion of molecules of matter in various states. Apparatus and materials needed: Peppermint oil, potassium permanganate, evaporating dish, 250 ml. beaker, glasses, water, powdered drink mix, straws, raisins, pipette, spatula, sandwich bag, marble and teaspoon. 1. Orally review the structure of the cell by using the sandwich bag filled with water and a marble to represent an analogy of the cell. The students know about the presence of openings in cell membranes and about the possibility of molecules moving through these openings. 2. To demonstrate diffusion, pour sufficient peppermint oil to cover the bottom of an evaporating dish. As each student detects the odor, he or she will raise his hand. Continue until all members of the class have detected the odor of diffusing molecules. At the conclusion of the demonstration, the air and oil molecules have spread out and mixed evenly, they continue to bump and move. This spreading out is called 3. To observe diffusion of a solid, fill a 250 mL beaker with distilled water and add several crystals of potassium permanganate. Watch the results as the crystals settle to the bottom. Discuss this phenomenon with the students. 4. To observe diffusion of a liquid, place a glass of water at each student's desk. Do not touch the water; keep the water still for this test. Carefully drop a teaspoonful of powdered drink mix into the glass of water. Watch it for a few minutes without touching the glass. It will go to the bottom and start diffusing. 5. To observe diffusion in raisins, place several raisins in a glass of water. Let them sit overnight. Keep some raisins dry for a control. Distribute to each student some soaked raisins and dry raisins. Let them compare the two kinds of raisins and answer the questions on the Answer the following questions to show the understanding of diffusion. 1. What is diffusion? 2. Refer to the raisin experiment and explain the difference between the soaked raisins and the dry raisins. 3. What is the evidence that something passed out of the raisins but 4. What is the difference between the bag's cell membrane and the raisin's

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The First Personality Reading Based On The Sacred Geometry Of Your Name Symbols. # Anatomy of the number 9 ### Pythagorean numerology of number 9 "Ordo ab chao" "Order out of chaos ## Properties of the number 9 • Divisors : 3 • Number of divisors: 1, 3 and 9 • 1st Prime number: no • ɸ Fibonacci number : no • Triangular numbers : no • Square numbers : yes 1, 4, 9, 16, etc • 2n Power of 2 : no • 3n Power of 3: yes 1, 3, 9, 27, etc. ## Numerology of the number 9 • Letters corresponding to the number 9: I and R ~ RI • Complementary number to the number 9 : number 1 • Theosophical addition of the number 9 : 1 + 2 + ... + 9 = 45 • Secret Value of the number 9: 45 = 4 + 5 = 9 • Numbers with the number 1 as a reduction: 18, 27, 36, 45, 54, 63, 72, 81, 90, etc • Source numbers at the origin of the number 9 : 18, 27, 36, 45 and 90. ## How to get the number 9 in numerology ### Numbers that produce 9 In numerology, the number 9 is obtained by adding the digits of a number or a word until obtaining a single digit equal to 9. Here is how to get a number 9 in numerology: If you have a number between 10 and 99, you can get a number 9 by adding its digits together. For example, for the number 18, you add 1 + 8 = 18. And 18 = 1 + 8 = 9. If you have a number greater than 99, you can get a number 9 by repeating the operation until you get a single digit. For example, for the number 576, you add 5 + 7 + 6 = 18, then 1 + 8 = 9. For the number 1242, you add 1 + 2 + 4 + 2 = 9. If you have a word, you can get a number 9 by adding the numbers corresponding to each letter of this word, according to the correspondence of the letters of the alphabet with the numbers of numerology. For example, for the word "sun" you add 1 (s) + 6 (o) + 3 (l) + 5 (e) + 9 (i) + 3 (l) = 27, then 2 + 7 = 9. The numerical reduction of the word "sun" is therefore 9. In summary, to obtain a number 9 in numerology (see table opposite), there are 5 possibilities, or "source numbers": 18, 27, 36, 45 and 90. ## Geometric shapes associated with the number 9 ### Symbols of unity, principle and beginning In sacred geometry, the number 9 can be associated with many different geometric shapes and symbolic meanings. Here are some examples of the geometric shapes most commonly associated with the number 9 in sacred geometry: • The enneagon: the enneagon is a geometric figure formed by the union of nine sides of equal length. It is a polygon with nine sides. • The enneagram: the enneagram is a geometric figure formed by the union of nine straight segments of the same length. It is a nine-pointed star. • The centered octagram: The centered octagram is a geometric figure formed by the union of two squares whose vertices are all connected at the center. The first way to obtain the number 9 in geonumerology comes from the source number 18 obtained by the addition of a unit to an octagram, that is to say the decomposition 1 + 8. The decomposition 1 + 8 corresponds geometrically to a centered octagram. ### The Centered Octagram The second way to obtain the number 9 in geonumerology comes from the source number 27 obtained by the addition of a line to a centered hexagram. The decomposition 2 + 7 corresponds to the family of the mixed polygons in Geonumerology. ### The Star of Destiny The third way to obtain the number 9 in geonumerology comes from the source number 36 obtained by adding a triangle to a hexagon, i.e. 3 + 6. The decomposition 3 + 6 corresponds to the family of mixed polygons in Geonumerology. ### The emotional Triangle The fourth way to obtain the number 9 in geonumerology comes from the source number 45 obtained by adding a square and a pyramid. The decomposition 4 + 5 corresponds to the family of mixed polygons in Geonumerology. The fifth way to obtain the number 9 in geonumerology comes from the source number 90 obtained by adding an enneagon to a circle. The decomposition 9 + 0 corresponds geometrically to the Enneagon, or 9-sided polygons. ## What is the meaning of the number 9 in numerology ### Personality of number 9 In numerology, the number 9 is considered an end of cycle number, as it is the last number in classical numerology. It is associated with the end of something and the beginning of something new, as well as completion, conclusion and resolution. It is associated with wisdom, universality and altruism. The number 9 also symbolizes compassion and generosity, as well as the ability to give without expecting anything in return. People with the number 9 in their numerological profile tend to be empathetic and caring people. ### The Benefactor Here are some characteristics that might be associated with the number 9: • Loving and generous • Likes to help others and do good • Can be idealistic and humanitarian • May have difficulty saying no and putting their own needs first • Can be very emotional and sensitive • Likes justice and peace ## Number 9 in excess or in lack ### What does an excess of 9 mean in numerology In numerology, an excess of any number can indicate a tendency to overestimate or underestimate the qualities associated with that number. In the case of the number 9, an excess may indicate a person who has a tendency to be selfless, overly generous and put the needs of others before their own, or who has a tendency to exhaust themselves by giving too much. An excess of 9 can also indicate a misanthropic person who prefers to champion a cause. ### What does a lack of 9 mean in numerology The number 9 is considered the number of completion, the end of the cycle and wisdom. If you have a lack of 9 in your numerology, it may mean that you have difficulty accepting the end of things and that you have difficulty letting go of the past. It may also mean that you have a hard time recognizing your own worth and accepting compliments from others. You may have a hard time accepting your own limitations and acknowledging that you can't do everything on your own. To work on this lack of 9, it can be helpful to take time to refocus on yourself and your own needs and values. You can also try to work on accepting the end of things and how to let go of the past. A person with a lacking 9 may have difficulty finding meaning in their life and relationships with others. ### With which numbers is the 9 compatible in numerology In numerology, each number has its own personality traits and vibrations. According to numerology, some numbers are more compatible with each other than others. However, it is important to note that the compatibility between numbers does not only depend on their numerological meaning, but also on the way they are positioned in the matrix or geometrically composed. For example, 1 and 9 are often considered compatible because of their strong energy and ability to focus on accomplishing their goals. The numbers 1 and 9 are at the same ends of the matrix and their union gives unity: 1 + 9 = 10 = 1 ## Location of the number 9 in the matrix of 9 numbers ### The Spirit of the Place #### Duty in the service of an Idea Second attribute of the Natural or Physical World • The number 9 is located in the third row: Natural or Physical World. • The number 9 is located in the third column: Natural or Physical Attribute. • The number 9 is the representative number of the Natural World through energy, etc. • Duty in the service of an Idea: my commitments in the community, etc. • Associated terms: power, duty, transmission, altruism, etc. • Law in play: law of transmutation. • Topology: the Bottom of the Bottom ## Expression path of the number 9 ### Path of involution (Spiritualize the Matter) where you must express Energy to the community Check to see if these numbers or symbols are present in your profile. Examples : hello world! In numerology, if your symbol of expression corresponds to the number 9, you will need to rely on numbers (to be identified among your personal numbers) coming from the two higher worlds, the Spiritual and the Human, in order to realize the unity within yourself, to link the Upper, the Self (your aspirations) and the Lower, the I (your concrete achievements). The path of expression of the number 9 invites you to express your generosity in the community, to be a person of heart. It can also relate to being a teacher, guide or mentor to others (with 1), being a spokesperson for social justice (with 2) and being an advocate for humanitarian causes (with 3). The paths indicated with a "compass rose" are particularly beneficial and offer the possibility of full fulfillment for a person with the number 9 in expression: • Arrow 159: The Determined. • Arrow 369: The Therapists - corresponds to the symbol 9_36, the Healer. To know your expression symbol, as well as your other personal symbols, you just have to fill in the form in the top banner. ## Symbolism of the number 9 ### The Astrologer The 9 heavenly bodies. ### The Initiate The Powers of 3. The progression of the powers of 3: 1, 3, 9, 27, etc. is found in the ramifications of the Three Worlds Tree as well as in the Sierpinski triangle. ### The Magician The polygons of Magic, all from the enneagram. The enneagram generates by duplication and rotation the 18, 36 and 72-sided polygons related to the 72 genies, the 36 talismans and the 18 genies for each of the 4 Elements. ### The Egyptian In ancient Egyptian philosophy, the Ennead was a group of nine gods and goddesses who represented the principles of the universe. The number 9 thus came to symbolize the unity of the divine and the natural world. The ennead or group of 9 gods : Atum, Tefnut, Shu, Geb, Nut, Osiris, Isis, Set and Nephthys. ### The Religious The 9 choirs of angels are divided into three groups: Seraphim, Cherubim, Thrones / Dominations, Virtues, Powers / Principality, Archangels, Angels. ### The Alchemist Fractal transmutation, etc. ### The Scientist The 9 months of gestation, etc ### The Psychologist The personality types of the enneagram, etc ### The Philosopher The Nine Muses: In Greek mythology, the Nine Muses were goddesses of the arts and sciences, and the number 9 symbolized inspiration and creativity. THE SACRED GEOMETRY OF YOUR NAME

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# How do you calculate the height of a column of carbon tetrachloride, C Cl_4(l), with a density of 1.59 g/mL that exerts the same pressure as a 15.2 cm column of Mercury, Hg(l), that has a density of 13.6 g/ml? Nov 2, 2016 You can do it like this: #### Explanation: The pressure caused by the fluid is given by: $\textsf{P = \rho g h}$ $\textsf{\rho}$ is the density $\textsf{g}$ is the acceleration due to gravity $\textsf{h}$ is the height of the column Since the pressures are equal we can write: $\textsf{{\rho}_{1} \cancel{g} {h}_{1} = {\rho}_{2} \cancel{g} {h}_{2}}$ $\therefore$sf(h_(1)=(rho_(2)h_(2))/(rho_(1)) $\textsf{{h}_{1} = \frac{13.6 \times 15.2}{1.59} = 130 \textcolor{w h i t e}{x} \text{cm}}$

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What is bullying? A form of aggressive behavior that is intentional, abusive, intimidating, malicious and/or threatening. It generally involves abuse of power and can be physical and/or psychological in nature. It may include verbal aggression, demeaning behaviors, or threats to do harm (especially if one does not forfeit something, such as homework or money). - Bullying is noted to occur mostly in the school years (middle school especially), but it also carries over into adulthood and in the workplace. - Bullying is the result of a power differential (size and strength are non-discriminate factors). - Programs are only as effective as the environment in which they are implemented. Schools, and especially adults in schools, must create a culture of respect. - The existing research and programs are slow to acknowledge that bystanders are at the root of maintaining the culture, because they fuel the power differential. Programs need to target bystanders and empower them to act. - Research indicates that it is likely that most kids both bully and are bullied; especially when considering relational aggression. - There are laws and/or policies in all 50 states against bullying, which can result in incarceration of the perpetrator. - Awareness, prevention and intervention are steps needed for an anti-bullying program to be effective. - Bullying YouTube video–A survivor’s inspirational story This information was provided by Maria Cuddy-Casey, Ph.D., an associate professor and chair of the Psychology Department at Immaculata University, as well as a clinical psychologist specializing in children and adolescents.

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# Astrophysics: Magnitude of sun when Jupiter crosses over it. 1. Mar 19, 2012 ### Xyius 1. The problem statement, all variables and given/known data An observer 5pc away observes the sun in the plane of Jupiter orbit. He cannot resolve either object but he notices a slight dimming of the star when Jupiter passes across the sun in his vision. Find the magnitude of the sun with and without Jupiter in front of it. 2. Relevant equations Magnitude formula $$m_1-m_2=2.5log\left( \frac{\Phi_1}{\Phi_2} \right)$$ (Where Phi is the flux) I do not know if this is the only equation or not. Absolute Magnitude of Sun = 4.77 3. The attempt at a solution So the case when Jupiter is NOT crossing over seems simple enough. $$m_1-m_2=2.5log\left( \frac{\Phi_1}{\Phi_2} \right)$$ $$=m-M=2.5log\left( \frac{d^2}{10^2} \right)$$ $$=m-4.77=2.5log\left( \frac{5^2}{10^2} \right), m=3.26$$ So now I just need to find the magnitude when Jupiter is crossing the sun. My professor said to find the area of the solar disk that is visible when Jupiter is crossing by simply doing.. $$\pi (R_{sun}^2-R_{Jupiter}^2)$$ But I do not know where this fits in to the above equation. How can I relate this to the flux ratio? Any help would be appreciated :] 2. Mar 19, 2012 ### Andrew Mason Start with the definition of flux: $\Phi$. How is flux related to area of the source? How does Jupiter passing in front of the sun affect the effective area of the light source (ie the sun). AM 3. Mar 19, 2012 ### Xyius Okay so Flux is.. $$\Phi=\frac{L}{4\pi d^2}$$ Where d is the distance from the star to the observer. I guess the only thing that changes here would be the Luminosity. Luminosity is defined to be.. $$L=4\pi R^2 \sigma T^4$$ Where R is the radius of the star. So would this be on the right track? My concern is this isn't a "disk" as my professor hinted towards. :\ 4. Mar 21, 2012 ### Andrew Mason It may not be a disk. But a sphere's cross-sectional area is what one sees from a distance. How is the viewable cross sectional area of the sun affected when Jupiter passes in front of it? When does the effect reach a maximum? AM

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All categories Categories All categories Not categorized (CC) Counting and Cardinality (G) Geometry (MD) Measurement and Data (NBT) Number and Operation in Base Ten (OA) Operations and Algebraic Thinking Prioritized Standards Page:  1  2  3  (Next) ALL # Counting and Cardinality ## Narrative for the (CC) Counting and Cardinality Counting and Cardinality and Operations and Algebraic Thinking are about understanding and using numbers. Counting and Cardinality underlies Operations and Algebraic Thinking as well as Number and Operations in Base Ten. It begins with early counting and telling how many in one group of objects. Addition, subtraction, multiplication, and division grow from these early roots. Students usually know or can learn to say the counting words up to a given number before they can use these numbers to count objects or to tell the number of objects. In Kindergarten, students develop understanding of the relationship between numbers and quantities and connect counting to cardinality - to count a group of objects, they pair each word said with one object.  Students also develop fluency with counting to 100 by ones and tens, count to answer "how many" questions, use counting strategies to compare groups of objects, and apply counting strategies when solving addition and subtraction problems. ## Calculation Method for Domains Domains are larger groups of related standards. The Domain Grade is a calculation of all the related standards. Click on the standard name below each Domain to access the learning targets and rubrics/ proficiency scales for individual standards within the domain. #### MAT-00.CC.01 Kindergarten (MAT) Targeted Standard (CC) Counting Cardinality Cluster: Know number names and the count sequence. #### MAT-00.CC.02 Under Development MAT-00 Targeted Standards(CC) Domain: Counting and Cardinality Cluster: Know number names and the count sequence. MAT-00.CC.02 Count forward beginning from a given number within the known sequence (instead of having to begin at 1). Count backward from a given number within 10. • I can • I can • I can • I can ## Proficiency (Rubric) Scale Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. ## Resources ### Vocabulary • List #### MAT-00.CC.03 Kindergarten (MAT) Targeted Standard (CC) Counting Cardinality Cluster: Know number names and the count sequence. #### MAT-00.CC.04 Under Development MAT-00 Targeted Standards(CC) Domain: Counting and Cardinality Cluster: Count to tell the number of objects.. MAT-00.CC.04 Understand the relationship between numbers and quantities; connect counting to cardinality. • I can • I can • I can • I can ## Proficiency (Rubric) Scale Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. ## Resources ### Vocabulary • List #### MAT-00.CC.04.a Under Development MAT-00 Targeted Standards(CC) Domain: Counting and Cardinality Cluster: Count to tell the number of objects. MAT-00.CC.04.a When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object. • I can • I can • I can • I can ## Proficiency (Rubric) Scale Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. ## Resources ### Vocabulary • List #### MAT-00.CC.04.b Under Development MAT-00 Targeted Standards(CC) Domain: Counting and Cardinality Cluster: Count to tell the number of objects. MAT-00.CC.04.b Understand that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted. • I can • I can • I can • I can ## Proficiency (Rubric) Scale Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. ## Resources ### Vocabulary • List #### MAT-00.CC.04.c Under Development MAT-00 Targeted Standards(CC) Domain: Counting and Cardinality Cluster: Count to tell the number of objects. MAT-00.CC.04.c Understand that each successive number name refers to a quantity that is one larger. • I can • I can • I can • I can ## Proficiency (Rubric) Scale Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. ## Resources ### Vocabulary • List #### MAT-00.CC.05 Kindergarten (MAT) Targeted Standard (CC) Counting Cardinality Cluster: Know number names and the count sequence. ##### MAT-00.CC.05 Count to answer how many questions. a. Tell how many objects up to 20 are in an arranged pattern (e.g., a line or an array) or up to 10 objects in a scattered configuration. b. Represent a number of objects up to 20 with a written numeral. c. Given a number from 1-20, count out that many objects. #### MAT-00.CC.06 Kindergarten (MAT) Targeted Standard (CC) Counting Cardinality Cluster: Compare numbers. #### MAT-00.CC.07 Under Development MAT-00 Targeted Standards(CC) Domain: Counting and Cardinality Cluster: Compare numbers. MAT-00.CC.07 Compare two numbers between 1 and 10 presented as written numerals. • I can • I can • I can • I can ## Proficiency (Rubric) Scale Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. • List # Geometry ## Narrative for the (G) Geometry Understanding and describing shapes and space is one of the two critical areas of Kindergarten mathematics. Students develop geometric concepts and spatial reasoning from experience with two perspectives on space: the shapes of objects and the relative positions of objects. In the domain of shape, students learn to match two-dimensional shapes even when the shapes have different orientations. They learn to name shapes such as circles, triangles, and squares, whose names occur in everyday language, and distinguish them from non-examples of these categories, often based initially on visual models. Students also begin to name and describe three-dimensional shapes with mathematical vocabulary, such as “sphere,” “cube,” “cylinder,” and “cone.” They identify faces of three-dimensional shapes as two-dimensional geometric figures and explicitly identify shapes as two-dimensional (“flat” or lying in a plane) or three-dimensional. A second important area for kindergartners is the composition of geometric figures. Students not only build shapes from components, but also compose shapes to build pictures and designs. Finally, in the domain of spatial reasoning, students discuss not only shape and orientation, but also the relative positions of objects, using terms such as “above,” “below,” “next to,” “behind,” “in front of,” and “beside.”  They use these spatial reasoning competencies, along with their growing knowledge of three-dimensional shapes and their ability to compose them, to model objects in their environment. ## Calculation Method for Domains Domains are larger groups of related standards. The Domain Grade is a calculation of all the related standards. Click on the standard name below each Domain to access the learning targets and rubrics/ proficiency scales for individual standards within the domain. #### MAT-00.G.01 Under Development MAT-00 Targeted Standards(G) Domain: Geometry Cluster: Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). MAT-00.G.01 Describe objects in the environment using names of shapes and solids (squares, circles, triangles, rectangles, cubes, and spheres). • I can • I can • I can • I can ## Proficiency (Rubric) Scale Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated. ## Resources ### Vocabulary • List #### MAT-00.G.02 Kindergarten (MAT) Targeted Standard (G) Geometry Cluster: Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). #### MAT-00.G.03 Under Development MAT-00 Targeted Standards(G) Domain: Geometry Cluster: Identify and describe shapes (squares, circles, triangles, rectangles, hexagons, cubes, cones, cylinders, and spheres). MAT-00.G.03 Identify shapes and solids (squares, circles, triangles, rectangles, cubes, and spheres) as two-dimensional or three-dimensional. • I can • I can • I can • I can ## Proficiency (Rubric) Scale Score Description Sample Activity 4.0 Student is able to - 3.5 In addition to Score 3.0 performance, the student demonstrates in-depth inferences and applications regarding the more complex content with partial success. 3.0 “The Standard.” Student is able to - 2.5 No major errors or emissions regarding 2.0 content and partial knowledge of the 3.0 content. 2.0 Student is able to - 1.5 In addition to 1.0 content, student has partial knowledge of the 2.0 and/or 3.0 content. 1.0 Student is able to - 0.5 Limited or no understanding of the skill id demonstrated.

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Arenavirus is a genus of virus. The type species is Lymphocytic choriomeningitis virus (LCMV); it also includes the species responsible for Lassa fever. Whitewater Arroyo virus causes Hemorrhagic fever in the Southwest USA Some arenaviruses are zoonotic pathogens and are generally associated with rodent-transmitted disease in humans. Each virus usually is associated with a particular rodent host species in which it is maintained. The virus particles are spherical and have an average diameter of 110-130 nanometers. All are enveloped in a lipid membrane. Viewed in cross-section, they show grainy particles that are ribosomes acquired from their host cells. It is this characteristic that gave them their name, derived from the Latin "arena," which means "sandy." Their genome, or genetic material, is composed of RNA only, and while their replication strategy is not completely understood, we know that new viral particles, called virions, are created by budding from the surface of their hosts’ cells.

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# Google Interview: Arrangement of Blocks You are given N blocks of height 1…N. In how many ways can you arrange these blocks in a row such that when viewed from left you see only L blocks (rest are hidden by taller blocks) and when seen from right you see only R blocks? Example given `N=3, L=2, R=1` there is only one arrangement `{2, 1, 3}` while for `N=3, L=2, R=2` there are two ways `{1, 3, 2}` and `{2, 3, 1}`. How should we solve this problem by programming? Any efficient ways? - The first example should be `{1,2,3}` and not `{2,1,3}` –  Cratylus Oct 7 '11 at 20:54 @user384706 Not at all. `{1,2,3}` has `3` visible blocks from the left and `1` visible block from the right. –  PengOne Oct 7 '11 at 20:56 @user384706 no, if {1,2,3} L would be 3. –  daniloquio Oct 7 '11 at 20:56 Not really. {1,2,3} implies L=3. Thank you all the same. –  Terry Li Oct 7 '11 at 20:56 @user384706 Looking at {2,1,3} from the left, you will see the block of height `2` and the block of height `3` but not the block of height `1` as it is obstructed by the block of height `2`. –  PengOne Oct 7 '11 at 21:04 This is a counting problem, not a construction problem, so we can approach it using recursion. Since the problem has two natural parts, looking from the left and looking from the right, break it up and solve for just one part first. Let `b(N, L, R)` be the number of solutions, and let `f(N, L)` be the number of arrangements of `N` blocks so that `L` are visible from the left. First think about `f` because it's easier. APPROACH 1 Let's get the initial conditions and then go for recursion. If all are to be visible, then they must be ordered increasingly, so ``````f(N, N) = 1 `````` If there are suppose to be more visible blocks than available blocks, then nothing we can do, so ``````f(N, M) = 0 if N < M `````` If only one block should be visible, then put the largest first and then the others can follow in any order, so ``````f(N,1) = (N-1)! `````` Finally, for the recursion, think about the position of the tallest block, say `N` is in the `k`th spot from the left. Then choose the blocks to come before it in `(N-1 choose k-1)` ways, arrange those blocks so that exactly `L-1` are visible from the left, and order the `N-k` blocks behind `N` it in any you like, giving: ``````f(N, L) = sum_{1<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * (N-k)! `````` In fact, since `f(x-1,L-1) = 0` for `x<L`, we may as well start `k` at `L` instead of `1`: ``````f(N, L) = sum_{L<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * (N-k)! `````` Right, so now that the easier bit is understood, let's use `f` to solve for the harder bit `b`. Again, use recursion based on the position of the tallest block, again say `N` is in position `k` from the left. As before, choose the blocks before it in `N-1 choose k-1` ways, but now think about each side of that block separately. For the `k-1` blocks left of `N`, make sure that exactly `L-1` of them are visible. For the `N-k` blocks right of `N`, make sure that `R-1` are visible and then reverse the order you would get from `f`. Therefore the answer is: ``````b(N,L,R) = sum_{1<=k<=N} (N-1 choose k-1) * f(k-1, L-1) * f(N-k, R-1) `````` where `f` is completely worked out above. Again, many terms will be zero, so we only want to take `k` such that `k-1 >= L-1` and `N-k >= R-1` to get ``````b(N,L,R) = sum_{L <= k <= N-R+1} (N-1 choose k-1) * f(k-1, L-1) * f(N-k, R-1) `````` APPROACH 2 If you work the problem the opposite way, that is think of adding the smallest block instead of the largest block, then the recurrence for `f` becomes much simpler. In this case, with the same initial conditions, the recurrence is ``````f(N,L) = f(N-1,L-1) + (N-1) * f(N-1,L) `````` where the first term, `f(N-1,L-1)`, comes from placing the smallest block in the leftmost position, thereby adding one more visible block (hence `L` decreases to `L-1`), and the second term, `(N-1) * f(N-1,L)`, accounts for putting the smallest block in any of the `N-1` non-front positions, in which case it is not visible (hence `L` stays fixed). This recursion has the advantage of always decreasing `N`, though it makes it more difficult to see some formulas, for example `f(N,N-1) = (N choose 2)`. This formula is fairly easy to show from the previous formula, though I'm not certain how to derive it nicely from this simpler recurrence. Now, to get back to the original problem and solve for `b`, we can also take a different approach. Instead of the summation before, think of the visible blocks as coming in packets, so that if a block is visible from the left, then its packet consists of all blocks right of it and in front of the next block visible from the left, and similarly if a block is visible from the right then its packet contains all blocks left of it until the next block visible from the right. Do this for all but the tallest block. This makes for `L+R` packets. Given the packets, you can move one from the left side to the right side simply by reversing the order of the blocks. Therefore the general case `b(N,L,R)` actually reduces to solving the case `b(N,L,1) = f(N,L)` and then choosing which of the packets to put on the left and which on the right. Therefore we have ``````b(N,L,R) = (L+R choose L) * f(N,L+R) `````` Again, this reformulation has some advantages over the previous version. Putting these latter two formulas together, it's much easier to see the complexity of the overall problem. However, I still prefer the first approach for constructing solutions, though perhaps others will disagree. All in all it just goes to show there's more than one good way to approach the problem. What's with the Stirling numbers? As Jason points out, the `f(N,L)` numbers are precisely the (unsigned) Stirling numbers of the first kind. One can see this immediately from the recursive formulas for each. However, it's always nice to be able to see it directly, so here goes. The (unsigned) Stirling numbers of the First Kind, denoted `S(N,L)` count the number of permutations of `N` into `L` cycles. Given a permutation written in cycle notation, we write the permutation in canonical form by beginning the cycle with the largest number in that cycle and then ordering the cycles increasingly by the first number of the cycle. For example, the permutation ``````(2 6) (5 1 4) (3 7) `````` would be written in canonical form as ``````(5 1 4) (6 2) (7 3) `````` Now drop the parentheses and notice that if these are the heights of the blocks, then the number of visible blocks from the left is exactly the number of cycles! This is because the first number of each cycle blocks all other numbers in the cycle, and the first number of each successive cycle is visible behind the previous cycle. Hence this problem is really just a sneaky way to ask you to find a formula for Stirling numbers. - What if it were a construction problem? It smells like an NP-Hard one. –  Terry Li Oct 7 '11 at 21:23 You can construct the solutions for `f` from the descriptions above, and then use that to construct solutions for `b`. The problem is simpler than that of constructing all permutations. –  PengOne Oct 7 '11 at 21:27 As to complexity, each iteration decrements `L` and `R` by `1` each, so `O(N^2)` is an upper bound. Perhaps things are even better, but I'd have to think a bit on that. –  PengOne Oct 7 '11 at 21:29 @Ivan `(N choose k)` is the binomial coefficient giving the number of `k` element subsets of an `N` element set. You compute it by `N!/(k! * (N-k)!)`. –  PengOne Oct 14 '11 at 17:05 In second approach, it should be `b(N,L,1) = f(N-1,L-1)`. As, the tallest block should be fixed at the last position(which is always visible from left and right), so there is only one visible block when seen from right. –  Priyank Bhatnagar Mar 12 '12 at 10:20 well, just as an empirical solution for small N: blocks.py: ``````import itertools from collections import defaultdict def countPermutation(p): n = 0 max = 0 for block in p: if block > max: n += 1 max = block return n def countBlocks(n): count = defaultdict(int) for p in itertools.permutations(range(1,n+1)): fwd = countPermutation(p) rev = countPermutation(reversed(p)) count[(fwd,rev)] += 1 return count def printCount(count, n, places): for i in range(1,n+1): for j in range(1,n+1): c = count[(i,j)] if c > 0: print "%*d" % (places, count[(i,j)]), else: print " " * places , print def countAndPrint(nmax, places): for n in range(1,nmax+1): printCount(countBlocks(n), n, places) print `````` and sample output: ``````blocks.countAndPrint(10) 1 1 1 1 1 1 2 1 2 3 1 2 6 3 3 3 1 6 11 6 1 6 22 18 4 11 18 6 6 4 1 24 50 35 10 1 24 100 105 40 5 50 105 60 10 35 40 10 10 5 1 120 274 225 85 15 1 120 548 675 340 75 6 274 675 510 150 15 225 340 150 20 85 75 15 15 6 1 720 1764 1624 735 175 21 1 720 3528 4872 2940 875 126 7 1764 4872 4410 1750 315 21 1624 2940 1750 420 35 735 875 315 35 175 126 21 21 7 1 5040 13068 13132 6769 1960 322 28 1 5040 26136 39396 27076 9800 1932 196 8 13068 39396 40614 19600 4830 588 28 13132 27076 19600 6440 980 56 6769 9800 4830 980 70 1960 1932 588 56 322 196 28 28 8 1 40320 109584 118124 67284 22449 4536 546 36 1 40320 219168 354372 269136 112245 27216 3822 288 9 109584 354372 403704 224490 68040 11466 1008 36 118124 269136 224490 90720 19110 2016 84 67284 112245 68040 19110 2520 126 22449 27216 11466 2016 126 4536 3822 1008 84 546 288 36 36 9 1 `````` You'll note a few obvious (well, mostly obvious) things from the problem statement: • the total # of permutations is always N! • with the exception of N=1, there is no solution for L,R = (1,1): if a count in one direction is 1, then it implies the tallest block is on that end of the stack, so the count in the other direction has to be >= 2 • the situation is symmetric (reverse each permutation and you reverse the L,R count) • if `p` is a permutation of N-1 blocks and has count (Lp,Rp), then the N permutations of block N inserted in each possible spot can have a count ranging from L = 1 to Lp+1, and R = 1 to Rp + 1. From the empirical output: • the leftmost column or topmost row (where L = 1 or R = 1) with N blocks is the sum of the rows/columns with N-1 blocks: i.e. in @PengOne's notation, `b(N,1,R) = sum(b(N-1,k,R-1) for k = 1 to N-R+1` • Each diagonal is a row of Pascal's triangle, times a constant factor K for that diagonal -- I can't prove this, but I'm sure someone can -- i.e.: `b(N,L,R) = K * (L+R-2 choose L-1)` where `K = b(N,1,L+R-1)` So the computational complexity of computing b(N,L,R) is the same as the computational complexity of computing b(N,1,L+R-1) which is the first column (or row) in each triangle. This observation is probably 95% of the way towards an explicit solution (the other 5% I'm sure involves standard combinatoric identities, I'm not too familiar with those). A quick check with the Online Encyclopedia of Integer Sequences shows that b(N,1,R) appears to be OEIS sequence A094638: A094638 Triangle read by rows: T(n,k) =|s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind (1<=k<=n; in other words, the unsigned Stirling numbers of the first kind in reverse order). 1, 1, 1, 1, 3, 2, 1, 6, 11, 6, 1, 10, 35, 50, 24, 1, 15, 85, 225, 274, 120, 1, 21, 175, 735, 1624, 1764, 720, 1, 28, 322, 1960, 6769, 13132, 13068, 5040, 1, 36, 546, 4536, 22449, 67284, 118124, 109584, 40320, 1, 45, 870, 9450, 63273, 269325, 723680, 1172700 As far as how to efficiently compute the Stirling numbers of the first kind, I'm not sure; Wikipedia gives an explicit formula but it looks like a nasty sum. This question (computing Stirling #s of the first kind) shows up on MathOverflow and it looks like O(n^2), as PengOne hypothesizes. - Your observation on the diagonals follows from the formulas I gave above. –  PengOne Oct 7 '11 at 21:45 I'm horrid with combinatorial algebra so I'll have to trust you on that one. :-) –  Jason S Oct 7 '11 at 22:59 @JasonS +1 for your honesty. I didn't have the courage to admit it. Now I do. Thanks. –  Terry Li Oct 7 '11 at 23:18 +1 For the added research. I'm much more a mathematician than a programmer, but I appreciate that both types are useful :-) –  PengOne Oct 7 '11 at 23:23 @JasonS@PengOne Theory and practice combine to make a good answer! –  Terry Li Oct 7 '11 at 23:26 Based on @PengOne answer, here is my Javascript implementation: ``````function g(N, L, R) { var acc = 0; for (var k=1; k<=N; k++) { acc += comb(N-1, k-1) * f(k-1, L-1) * f(N-k, R-1); } return acc; } function f(N, L) { if (N==L) return 1; else if (N<L) return 0; else { var acc = 0; for (var k=1; k<=N; k++) { acc += comb(N-1, k-1) * f(k-1, L-1) * fact(N-k); } return acc; } } function comb(n, k) { return fact(n) / (fact(k) * fact(n-k)); } function fact(n) { var acc = 1; for (var i=2; i<=n; i++) { acc *= i; } return acc; } \$("#go").click(function () { }); `````` - +1 for enriching our code base. Thanks @Ivan! –  Terry Li Oct 17 '11 at 14:45 Here is my construction solution inspired by @PengOne's ideas. ``````import itertools def f(blocks, m): n = len(blocks) if m > n: return [] if m < 0: return [] if n == m: return [sorted(blocks)] maximum = max(blocks) blocks = list(set(blocks) - set([maximum])) results = [] for k in range(0, n): for left_set in itertools.combinations(blocks, k): for left in f(left_set, m - 1): rights = itertools.permutations(list(set(blocks) - set(left))) for right in rights: results.append(list(left) + [maximum] + list(right)) return results def b(n, l, r): blocks = range(1, n + 1) results = [] maximum = max(blocks) blocks = list(set(blocks) - set([maximum])) for k in range(0, n): for left_set in itertools.combinations(blocks, k): for left in f(left_set, l - 1): other = list(set(blocks) - set(left)) rights = f(other, r - 1) for right in rights: results.append(list(left) + [maximum] + list(right)) return results # Sample print b(4, 3, 2) # -> [[1, 2, 4, 3], [1, 3, 4, 2], [2, 3, 4, 1]] `````` - One way to solve this might be as follows: given pillars of height {1,2,3,...,N}, and L,R > 0 s.t. L+R-1 <= N, do the following: take the L largest numbers from {1,2,3,...,N}, i.e. {N-L+1,N-L+2,...,N} and keep them in this order. Now take the next R-1 largest numbers from the remaining numbers and sort them in the decreasing order, i.e. : {N-L,N-L-1,...,N-L-R+2}. Now append this sequence to the end of the L length sequence constructed earlier, and put all the remaining numbers between any two elements of this bigger sequence(say the 1st and the 2nd), to get: ``````N-L+1, 1, 2, 3,..., N-L-R+3, N-L+2, N-L+3,..., N, N-L, N-L-1,..., N-L-R+1, N-L-R+2. `````` This sequence satisfies the required properties. - this makes a single example but does not show how many examples of this there are (which is the actual questin) –  PengOne Oct 14 '11 at 5:21 We derive a solution F(N, L, R) examining F(10, 4, 3). We first consider 10 in the leftmost possible position, the 4th ( _ _ _ 10 _ _ _ _ _ _ ). Then we find the product of the number of valid sequences in the left and in the right of 10. Next, we'll consider 10 in the 5th slot, calculate another product and add it to the previous one. This process will go on until 10 is in the last possible slot, the 8th. We'll use a variable "pos" to keep track of N's position. Now suppose pos = 6 ( _ _ _ _ _ 10 _ _ _ _ ). In the left of 10, there are 9C5 = (N-1)C(pos-1) sets of numbers to be arranged. Since only the order of these numbers matters, we could look at 1, 2, 3, 4, 5. To construct a sequence with these numbers so that 3 = L-1 of them are visible from the left, we can begin by placing 5 in the leftmost possible slot ( _ _ 5 _ _ ) and follow similar steps to what we did before. So if F were defined recursively, it could be used here. The only difference now is that the order of numbers in the right of 5 is immaterial. To resolve this issue, we’ll use a signal, INF (infinity), for R to indicate its unimportance. Turning to the right of 10, there will be 4 = N-pos numbers left. We first consider 4 in the last possible slot, position 2 = R-1 from the right ( _ _ 4 _ ). Here what appears in the left of 4 is immaterial. But counting arrangements of 4 blocks with the mere condition that 2 of them should be visible from the right is no different than counting arrangements of the same blocks with the mere condition that 2 of them should be visible from the left. (In other words instead of counting sequences like 3 1 4 2, one can count sequences like 2 4 1 3.) So the number of valid arrangements in the right of 10 is F(4, 2, INF). Thus the number of arrangements when pos=6 is 9C5 * F(5, 3, INF) * F(4, 2, INF) = (N-1)C(pos-1) * F(pos-1, L-1, INF)* F(N-pos, R-1, INF). Similarly, in F(5, 3, INF), 5 will be considered in a succession of slots with L = 2 and so on. Since the function calls itself with L or R reduced, it must return a value when L = 1, that is F(N, 1, INF) must be a base case. Now consider the arrangement _ _ _ _ _ 6 7 10 _ _. The only slot 5 can take is the first, and the following 4 slots may be filled in any manner; thus F(5, 1, INF) = 4!. Then clearly F(N, 1, INF) = (N-1)!. Other (trivial) base cases and details could be seen in the C implementation below. Here is a link for testing the code. ``````#define INF UINT_MAX long long unsigned fact(unsigned n) { return n ? n * fact(n-1) : 1; } unsigned C(unsigned n, unsigned k) { return fact(n) / (fact(k) * fact(n-k)); } unsigned F(unsigned N, unsigned L, unsigned R) { unsigned pos, sum = 0; if(R != INF) { if(L == 0 || R == 0 || N < L || N < R) return 0; if(L == 1) return F(N-1, R-1, INF); if(R == 1) return F(N-1, L-1, INF); for(pos = L; pos <= N-R+1; ++pos) sum += C(N-1, pos-1) * F(pos-1, L-1, INF) * F(N-pos, R-1, INF); } else { if(L == 1) return fact(N-1); for(pos = L; pos <= N; ++pos) sum += C(N-1, pos-1) * F(pos-1, L-1, INF) * fact(N-pos); } return sum; } `````` -

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# USING CONTEXT TO UNDERSTAND THE RATIONAL NUMBER SYSTEM STRATEGY The introduction of the rational numbers in Grade 6 requires that we broaden our outlook on the number line when making comparisons of numbers and quantities. Simply explaining to students that the numbers to the left of zero are negative values is not enough to develop true understanding of why these numbers are mathematically necessary. In our world we experience applications of the number line on both sides of zero, and these applications will aid our students in understanding what those negative values represent and how we can use them mathematically. Ordering magnitudes, for a sixth grade student, is really nothing new once the magnitudes have been meaningfully determined. They have previously compared numbers and quantities by their magnitudes as those numbers and quantities represented very specific contexts, such as the number of different animals in the zoo, or the heights of their classmates. However, in many real-world and mathematical situations, comparison of magnitudes is insufficient. For example, let’s compare deposits into an account; if we compare the magnitudes of these deposits we will find the greatest deposit made. Similarly, we can compare withdrawals made from an account; if we compare the magnitudes of these withdrawals we will find the greatest withdrawal from the account. If we look at the overall history of account transactions and compare their magnitudes, we can find the greatest transaction; however, it will not provide us with the most favorable transaction. This is because the actions of depositing and withdrawing money have opposite effects on the account balance. Deposits live on the positive side of the number line and withdrawals live on the negative side. It is for this reason that in such situations, we must instead compare numbers via their positions along the entire number line, not just to one side of zero in the case of comparing magnitudes. There are countless applications involving numbers on both halves of the number line. There are also countless applications of comparing magnitudes where the numbers coexist on one-half of a number line. It is the context of the situation that helps us to determine which of these two outlooks is more appropriate. For instance, we can compare depths of objects below sea level using magnitude since each object is described as being a given distance, which must be positive, below sea level. How would we describe the depth of a position on a dock that lies above sea level? If I describe such a position as being x feet above sea level, I’ve created a new context by changing the description to above sea level which requires a distinct new number line. It is often more useful in this situation to consider the elevations of the given objects with a set reference point at sea level. In this case, the objects are considered as positions along the number line with a fixed reference point of zero. The sign of the number representing the elevation of an object determines whether the object is positioned above or below that reference point. Absolute value tells us exactly how far the number is from the reference point; however it is the sign of the number that sets our description of the number’s position in reference to zero. What is important for us to realize is that the context of a situation assigns direction to the quantities involved, and that direction may not always be the same. Consider a football coach who is comparing his plays from a game and which of them are more effective than others. If the coach is focused on his offense, he considers the plays which yield the greatest gains in yardage as favorable, but not every play will result in a gain. Sometimes the offensive team is pushed in the opposite direction. A play that yields a 7-yard gain is far more favorable than a play resulting in a 7-yard loss. However, there is great meaning in both of these plays to the coach however. The number representing it on the number line is what provides that meaning. Being able to compare forward motion of the plays along the number line not only helps him to see which plays are effective, but also shows him which plays are not unfolding as they were designed. If the coach is focused on his defense, then he should take the opposite outlook on those numbers since a 7-yard gain from the opposing team’s offense is instead a 7-yard loss by his defense. Going one step further, sports commentators often provide viewers with an average yards per play statistic for a team’s offense in a given time period. Is this statistic calculated by magnitudes only? Do the statisticians insert the descriptions of gain and loss into the calculation? Of course not. Using both sides of the number line is what provides the most accurate statistic about a team’s offensive capabilities Before any discussion about using operations with signed numbers, it is important that students are fluent in understanding how to represent quantities in given situations using signed numbers. They should be able to accurately locate those numbers along the number line, use the locations of those numbers on the number line for comparison, and be able to contextualize such a comparison to make meaning from the mathematical model. This is a post by Eureka Math team members Beau Bailey, writer for Grades 6, 7, & 10, and Erika Silva, writer for Grades 6–8. By: Debby Grawn ### ¡EUREKA! By: GUEST AUTHOR — CARLOS SANCHEZ FROM PORT CHESTER PUBLIC SCHOOLS (NY)

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# Prove that $(x+1)^x-x^x(x-1)$ only has one (real) root For $$x>0$$, I want to prove that $$(x+1)^x-x^x(x-1)$$ only has one root • Perhaps not super relevant, but note that $x^x>0$, assuming $x>0$. Commented Jun 6, 2019 at 0:58 • Also, I'd be fascinated to know what context this emerged from. Commented Jun 6, 2019 at 1:05 • Why the vote too close? I am willing to edit to prevent closure. How did it emerge? I was just playing around with numbers and came across this and thought it was a fascinating graph.... nothing more. Commented Jun 6, 2019 at 1:13 • The reason I asked for context was purely out of my own curiosity. I could not imagine a situation where this came up. Playing around is a good thing, though, and I support it. As for the close vote... it may be that no effort appears on your part to solve the problem or get started. Just a thought. Commented Jun 6, 2019 at 1:15 • $1$ is not a root. The root is roughly $x=3.40298$ (using a plotting function). This, however, is not a rigorous proof. Commented Jun 6, 2019 at 2:26 My answer is definitely going to pull from some ideas of Robert Israel. So first we re-write the equation as $$(1+1/x)^x = x-1$$ as previously suggested. Then we consider these as two separate functions and ask when they intersect. Let $$f(x) = (1+1/x)^x$$ and let $$g(x)=x-1$$. Motivated from the following answer How to prove $(1+1/x)^x$ is increasing when $x>0$? we see that $$\log(f(x))' = \log(1+\frac{1}{x})-\frac{1}{x+1}$$ is increasing and hence $$f(x)$$ is too. Likewise one can easily show that $$\log(f(x))'' = -\frac{1}{x(1+x)^2}<0$$ this shows that $$f$$ is log-convex and hence convex (this follows from the following The composition of two convex functions is convex since $$Exp$$ is convex and $$\log(f)$$ is convex hence $$f=Exp[\log(f)]$$ is too). $$g(1)=0$$ and $$f(1)=2$$ so $$g at $$x=1$$. Now that we have established $$f$$ is convex it follows that once $$g$$ (a straight line) is greater than $$f$$ it is always greater. It follows there is one and only one solution for the equation equivalently $$g$$ intersects $$f$$ exactly once. $$(1+1/x)^x = x-1$$ Note that $$x > 1$$ is required for both sides to have the same sign, and then $$x \log(1+1/x) - \log(x-1) = 0$$ Call the left side $$g(x)$$. Now show that 1. $$g(x)$$ is convex. 2. $$\lim_{x \to 1+} g(x) = +\infty$$ 3. $$\lim_{x \to \infty} g(x) = -\infty$$. EDIT: As Clclstdnt comments, $$g'' = \frac{x^3+x^2+3x-1}{x(x^2-1)^2}$$. For $$x > 1$$, both numerator and denominator are positive, so $$g$$ is convex on $$(1,\infty)$$. Since $$\lim_{x \to \infty} g(x) = -\infty$$, this implies $$g' < 0$$ (i.e. if $$g'(b) \ge 0$$ for some $$b$$, we'd have $$g'(x) \ge 0$$ for $$x \ge b$$, and then $$\lim_{x\to\infty} g(x)$$ couldn't be $$-\infty$$). That tells you $$g$$ has at most one zero. On the other hand, (2) and (3) and the Intermediate Value Theorem say there is at least one. • Can you please explain why showing these three conditions is sufficient to prove the result? Also I don't think that g(x) is convex (everywhere) because the second derivative of it is $\frac{x^3+x^2+3x-1}{x(x^2-1)^2}$ Commented Jun 6, 2019 at 1:21 • I really feel this answer needs more explanation. All this shows is that $g(x)$ is convex and then that $h(x)=(1+1/x)^x - x+1$ is log-convex and hence convex. But $h(x)$ is not convex... in fact, if anything, it is concave for $x>2.5$ (that's not a precise bound). So not only do I not see why you did this... I also don't see that your argument follows. Commented Jun 6, 2019 at 2:23 • You don't need $h$ to be convex. The zeros of $h$ are the zeros of $g$. Commented Jun 6, 2019 at 12:31

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Our Level 2 Volcanoes textbook contains information on volcanoes found around the world. It includes sixteen lessons on volcanoes. Lessons include Types of Volcanoes, Types of Eruptions, Pyroclastic Flows, Ring of Fire, Mount Saint Helens and Notorious Volcanoes. Each lesson contains 3-4 pages of written material about volcanoes, a quiz and a kids science activity. Myrna Martin introduces each lesson in the textbook on a video that can be purchased with our packages and courses. Myrna covers not only the main points in the lesson she also includes extra information on the topic that is not contained in the written material. The kids science activities are fun and easy to do. Science activities include Mount Saint Helens Flip Book, Notorious Eruptions, Fantasy Island, Seafloor Magnetism, and Frothy Eruptions. Student Edition eBook link Teacher's Edition eBook link Teacher’s Edition Level 2 Volcanoes Textbook The teacher’s textbook is an exact copy of the student textbook. It includes the answers to the quizzes on each quiz page. A Teacher’s Notes page is located before each lesson with the following information 1. Lesson Content 2. Lesson Objectives 3. Activity information and materials needed 4. Vocabulary and definitions 5. Correlation with the National Science Standards More information on volcanoes that you may not know. All volcanoes are mountains of fire because they erupt hot molten lava from a vent. Composite volcanoes (stratovolcanoes) form in subduction zones on the continental side of the subduction zone. The great shield volcanoes of the Hawaiian Islands formed over a persistent hot spot in the middle of the Pacific Plate. Iceland volcanoes formed as the North American Plate and the Eurasian Plate are moving apart. Iceland not only sits about two separating crustal plates but it also sits atop a persistent hot spot in the middle of the Atlantic Ocean. The Iceland volcanoes are fed by both the separating plates and the persistent hot spot. 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Great stratovolcanoes have formed on the continental side of these subduction zones. Mt. Pinatubo, Mt. St. Helens and Mt. Wrangell are examples of volcanoes made of andesite that surround the Pacific Ocean. The vent of a volcano forms when magma (molten rock) first reaches the Earth's surface and erupts as a lava flow or tephra. The magma chamber is the place where molten rock collects before an eruption. When sufficient gases in the magma chamber collect in the molten rock it begins moving upward until it reaches the volcano's vent. Curtains of Fire The vent of a volcano continues to be at the summit of most volcanoes as they increase in size. Shield volcanoes often have vents on the flanks of the volcanoes when fissures form on their sides. These vents often are long and create "curtains of fire" when the volcanoes erupts along a fissure shooting molten rock into the air. 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Earth Science materials Why Educators Choose Our Earth Science Curriculum Homeschool Science Materials Our Earth Science curriculum has been recommended by a number of authors including: All of our science materials have been used by homeschooling families throughout the United States since 1998. Recommended by AFMS Junior Chair Ring of Fire Science materials were recommended by Jim Brace-Thompson of the American Federation of Mineralogical Societies. He found our materials contain a wealth of information for the junior members of the society. He wrote that our books are beautifully designed and illustrated with easy-to-follow instructions for kids. Cambridge Who's Who named Myrna Martin their Science Textbook Publishing Professional of the Year. She is the author of all of our textbooks. Read more about our family business.

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The Fair Employment Practices Committee created *On this date in 1941, the Fair Employment Practices Committee (FEPC) was created. President Franklin D. Roosevelt created the FEPC by signing Executive Order 8802. This was a great advancement for African-America initiated mainly by three people/organizations. Brotherhood of Sleeping Car Porters President A. Philip Randolph, NAACP Executive Secretary Walter White, and National Youth Administration Minority Affairs Director Mary McLeod Bethune were instrumental in forcing FDR to address the issue. The order banned racial discrimination in any defense industry receiving federal contracts by declaring "there shall be no discrimination in the employment of workers in defense industries or government because of race, creed, color, or national origin." The order also empowered the FEPC to investigate complaints and take action against alleged employment discrimination. Randolph, working with other civil rights activists, organized a 1941 March on Washington Movement to protest racial discrimination in the defense industry and the military. This threatened to bring 250,000 African Americans to Washington to demonstrate against congressional resistance to fair employment. FDR sent his wife Eleanor and New York City Mayor Fiorello LaGuardia to negotiate with March on Washington leaders. His wife returned, telling her husband that their plans were firm, that only an anti discrimination ordinance would prevent what promised to be the largest demonstration in the capital's history. Mrs. Roosevelt urged her husband to act for both moral and political reasons. He agreed, but would only go so far. He agreed to have the FEPC prohibit discrimination in defense plants, but he refused to address the issue of segregation in the military, which had been Randolph's original concern. Library of Congress 101 Independence Avenue S.E. Washington D.C. 20540 Today in American History

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The year is 2033, and intelligent life has been found on Mars! Little green Martians have found their way to Earth and have appointed you as the Ultimate Math Ambassador to their planet. You must help them understand how Earthlings use math. 1.Two Martians, Splott and Fizzle, have solved the equation 2x + 4 = –3x + 14. Examine the work of Splott and Fizzle. Identify any errors in the Martians’ calculations and explain, using complete sentences, what corrections they should make. 2x + 4 = –3x + 14 2x + 4 = –3x + 14 2x + 4 + 3x = –3x + 14 + 3x 2x + 4 – 2x = –3x + 14 – 2x 5x + 4 = 14 4 = 14 – 5x 5x + 4 + 4 = 14 + 4 4 – 14 = 14 – 5x – 14 5x = 18 –12 = –5x 5 x divided by 5 is equal to 18 divided by 5 –12 + 5 = –5x + 5 x = 18 over 5 –7 = x 2.Create your own function to teach the Martians about functions. Your function must contain at least two different operations. 3.Using complete sentences, prove to Splott and Fizzle that your function is a legitimate function. 4.Using your function, explain to the Martians how to solve for f(3). Show your work and explain each step using complete sentences. 5.Using complete sentences, describe to the Martians how to find the inverse of your function. 6.The Martians ask you to explain one last thing, Ultimate Math Ambassador. They ask you to create a new function, h(x). Then assign any number to x. Using complete sentences, explain whether f(h(x)) and h(f(x)) will always result in the same number. You will use the function f(x) that you created in problem number 2.

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Leviathan melvillei, a giant toothed whale, is described in a new paper, published online June 29 in Nature (Scientific American is part of Nature Publishing Group). The researchers have studied one specimen's fossil remains—several teeth and portions of the jaw. From this, the team estimates that the skull would have been at least three meters long—and the body longer than 13 meters. The newly described beast lived some 12 million to 13 million years ago and was first discovered two years ago in modern-day Peru. The teeth of this raptorial whale (meaning it fed on large prey) were enormous, measuring more than 36 centimeters long. The authors of the study, led by Olivier Lambert of Department de Paleontologie at the Institut Royal des Sciences Naturelles de Belgique in Brussels, hypothesize that the L. melvillei would have been a fearsome predator on the high Miocene seas, eating prey including baleen whales. This shapes up to be a very different feeding strategy and diet than today's sperm whales (Physeter macrocephalus), which have many smaller teeth, though a larger literary reputation. Image of artists reconstruction of an L. melvillei attacking a baleen whale courtesy of C. Letenneur/MNHN

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The lives of girls and women have changed dramatically over the past quarter century. The pace of change has been astonishing in some areas, but in others, progress toward gender equality has been limited—even in developed countries. This year's World Development Report: Gender Equality and Development argues that gender equality is a core development objective in its own right. It is also smart economics. Greater gender equality can enhance productivity, improve development outcomes for the next generation, and make institutions more representative. The Report also focuses on four priority areas for policy going forward: (i) reducing excess female mortality and closing education gaps where they remain, (ii) improving access to economic opportunities for women (iii) increasing women's voice and agency in the household and in society and (iv) limiting the reproduction of gender inequality across generations. Download the full report and related materials below. Download by Chapter: The Report has nine chapters in three parts. Part I Taking stock of gender equality Part 1:Taking stock of gender equality—presents the facts that will then provide the foundation for the rest of the Report. It combines existing and new data to document changes in key dimensions of gender equality over the past quarter century and across regions and countries. Its main message is that very rapid and, in some cases, unprecedented progress has been made in some dimensions of gender equality (chapter 1), but that it has not reached all women or been uniform across all dimensions of gender equality (chapter 2). Part II What has driven progress? What impedes it? The contrast between the patterns and trends described in the first two chapters of the Report prompts one to ask what explains the progress or lack of it. Part 2—What has driven progress? What impedes it?—constitutes the analytical core of the Report. It presents the conceptual framework and uses it to examine the factors that have fostered change and the constraints that have slowed progress. The analysis focuses on gender differences in education and health (chapter 3), agency (chapter 4), and access to economic opportunities (chapter 5)—discussing the roles of economic growth, households, markets, and institutions in determining outcomes in these three spheres. Part 2 concludes with a discussion of the impact of globalization on gender inequality, paying attention to the opportunities and challenges created by new economic and social trends (chapter 6). The analysis in these four chapters leads to the identification of four priority areas for action: reducing gender gaps in human capital endowments, promoting higher access to economic opportunities among women, closing gender gaps in household and societal voice, and limiting the intergenerational reproduction of gender inequality. Part III The role of and potential for public action Part 3—The role and potential for public action—presents policy recommendations, examines the political economy of reforms for gender equality, and proposes a global agenda for action. The discussion starts with a detailed description of policy options addressing the four priority areas, complemented with concrete illustrations of successful interventions in different contexts (chapter 7). An examination of the political economy of gender reforms follows, with an emphasis on the issues that distinguish reform in this area from other types of redistributive or equality-enhancing reforms (chapter 8). Global action on gender equality should focus on complementing country efforts on the four priority areas identified in the Report (chapter 9).

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Your skin’s dermis supports and enables the epidermis to thrive. It also houses connective tissue, blood vessels, oil and sweat glands, nerves and hair follicles. Your dermis has two layers — the papillary layer and the reticular layer — which merge without clear demarcation. The reticular layer contains dense connective tissue and bundles of elastic fibers while the papillary layer is thinner and composed of loose connective tissue. The epidermis is the outermost of two major layers of skin and is a tough, waterproof barrier. It has no blood vessels within it and is the thinnest of the two skin layers, averaging just 0.10 mm in thickness. It covers almost all of the body surface and is continuous with — but structurally distinct from — the mucous membranes that line the mouth, anus, urethra, and vagina. It is made of keratinized stratified squamous epithelium and has four or five layers of cells, depending on the location of the skin. The innermost layer, called the stratum basale, contains stem cells that divide to form all the keratinocytes of the epidermis. The next layer, the stratum spinosum, is thicker and contains immunologically active Langerhans cells and spiny keratinocytes. The final layer, the stratum corneum, is dry and dead and gives the skin its characteristic scaly appearance. This layer protects the underlying tissues from microbes, chemicals, and UV radiation. It also serves as a mechanical barrier against abrasion. The dead cells in this layer can be so hardy that they form keratinous materials such as hooves, claws, horns, and feathers that resist impact (as in hooves), external attack, and aerodynamic forces (as in horns). This keratinization helps to explain the protective properties of these structures. The layer beneath the dermis is called subcutaneous tissue (synonymous with hypodermis). It contains fat and connective tissue, and it has a thicker texture than the skin on other parts of your body. It also includes hair follicles and blood vessels. It is responsible for protecting the skin and insulates the body from cold temperatures. In addition, it stores energy and helps regulate your internal temperature. Fat cells, called adipocytes, make up the majority of this tissue, and its thickness varies throughout the body. It is thickest in the buttocks, palms of your hands and soles of your feet. The subcutaneous tissue houses larger blood vessels and nerves than those in the dermis. It also functions as a buffer between deeper muscles and bones, protecting them from shock and helping to control body temperature. This tissue is the primary site for the absorption of many medications, which is why injectable treatments can be used to deliver certain drugs. The loss of this fat tissue as you age can reduce the ability of your body to retain heat, which could lead to hypothermia. The subcutis is the third layer of skin, located beneath the dermis. Also known as superficial fascia, cutaneous fascia or panniculus adiposus, the subcutis is a dense network of fat cells and connective tissue. It functions as a heat-insulator and shock absorber, helping to protect the organs below. In addition, it stores adipose tissue for energy. The subcutis also contains blood vessels, hair follicles and lymphatic tissue. It is thickest on the abdomen, arms, lower back and shoulders. This layer is yellowish in color due to the presence of a pigment called carotene. The fibroblasts of the subcutis produce collagen that gives structure to the skin. They also create the proteoglycan and elastin that give the skin its toughness and elasticity. The upper dermal layer, the papillary layer, contains a thin arrangement of collagen fibers. It supplies nutrients to select layers of your epidermis and regulates temperature. Its vascular system — which functions like other vascular systems in your body — can constrict and expand to control how much blood moves through your skin. This layer also contains touch and pain receptors that transmit sensations of pain, itch and pressure to your brain for interpretation and to trigger shivering to generate additional body heat. The deeper reticular layer is dense, irregular connective tissue with thick bundles of collagen and elastic fibers that extend in all directions but are mainly parallel to your skin’s surface. Its extracellular matrix includes a gel-like substance made of proteins and a network of blood vessels that provides nourishment to cells in the papillary and reticular layers and removes wastes from those cells. This layer supports hair follicles, sweat glands and sebaceous (oil) glands. It also carries lymphatic vessels, nerves and a network of blood vessels. Light microscopy of H&E-stained skin samples shows a pattern of alternating dermal papillae and rete ridges. These fingerlike projections increase the strength of the attachment between the epidermis and the dermis. The reticular layer of the dermis is thicker than the papillary layer and consists of dense irregular connective tissue with mats of dense collagen and elastin fibers. This layer is responsible for much of the strength and elasticity of the skin. It also contains hair follicles, sweat glands, oil glands, nerves and blood vessels. It is also known for the ridges that appear as our fingerprints and give the layer its name – dermal papillae. The papillae are full of capillaries that supply the epidermis with blood. Meissner’s corpuscles – which sense light touch – are found in this layer, while Pacinian corpuscles are located deeper in the reticular layer and sense vibration and pressure. The reticular layer also contains fat cells which provide the skin with a cushion and a layer of insulation against heat loss through conduction, radiation and evaporation. In addition, this layer contains lipids (fatty acids) which act as an energy reserve and are the source of vitamin A. The hypodermis, a subcutaneous layer of fat that is highly vascularized and innervated, lies below the reticular layer.

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Introduction to LAN Protocols This article introduces the various media-access methods, transmission methods, topologies, and devices used in a local-area network (LAN). Topics addressed focus on the methods and devices used in Ethernet/IEEE 802.3, Token Ring/IEEE 802.5, and Fiber Distributed Data Interface (FDDI). Subsequent articles in LAN Protocols address specific protocols in more detail. Figure: Three LAN Implementations Are Used Most Commonly illustrates the basic layout of these three implementations. Figure: Three LAN Implementations Are Used Most Commonly What Is a LAN? A LAN is a high-speed data network that covers a relatively small geographic area. It typically connects workstations, personal computers, printers, servers, and other devices. LANs offer computer users many advantages, including shared access to devices and applications, file exchange between connected users, and communication between users via electronic mail and other applications. LAN Protocols and the OSI Reference Model LAN protocols function at the lowest two layers of the OSI reference model, as discussed in article Internetworking Basics between the physical layer and the data link layer. Figure: Popular LAN Protocols Mapped to the OSI Reference Model illustrates how several popular LAN protocols map to the OSI reference model. Figure: Popular LAN Protocols Mapped to the OSI Reference Model LAN Media-Access Methods Media contention occurs when two or more network devices have data to send at the same time. Because multiple devices cannot talk on the network simultaneously, some type of method must be used to allow one device access to the network media at a time. This is done in two main ways: carrier sense multiple access collision detect (CSMA/CD) and token passing. In networks using CSMA/CD technology such as Ethernet, network devices contend for the network media. When a device has data to send, it first listens to see if any other device is currently using the network. If not, it starts sending its data. After finishing its transmission, it listens again to see if a collision occurred. A collision occurs when two devices send data simultaneously. When a collision happens, each device waits a random length of time before resending its data. In most cases, a collision will not occur again between the two devices. Because of this type of network contention, the busier a network becomes, the more collisions occur. This is why performance of Ethernet degrades rapidly as the number of devices on a single network increases. In token-passing networks such as Token Ring and FDDI, a special network frame called a token is passed around the network from device to device. When a device has data to send, it must wait until it has the token and then sends its data. When the data transmission is complete, the token is released so that other devices may use the network media. The main advantage of token-passing networks is that they are deterministic. In other words, it is easy to calculate the maximum time that will pass before a device has the opportunity to send data. This explains the popularity of token-passing networks in some real-time environments such as factories, where machinery must be capable of communicating at a determinable interval. For CSMA/CD networks, switches segment the network into multiple collision domains. This reduces the number of devices per network segment that must contend for the media. By creating smaller collision domains, the performance of a network can be increased significantly without requiring addressing changes. Normally CSMA/CD networks are half-duplex, meaning that while a device sends information, it cannot receive at the time. While that device is talking, it is incapable of also listening for other traffic. This is much like a walkie-talkie. When one person wants to talk, he presses the transmit button and begins speaking. While he is talking, no one else on the same frequency can talk. When the sending person is finished, he releases the transmit button and the frequency is available to others. When switches are introduced, full-duplex operation is possible. Full-duplex works much like a telephone-you can listen as well as talk at the same time. When a network device is attached directly to the port of a network switch, the two devices may be capable of operating in full-duplex mode. In full-duplex mode, performance can be increased, but not quite as much as some like to claim. A 100-Mbps Ethernet segment is capable of transmitting 200 Mbps of data, but only 100 Mbps can travel in one direction at a time. Because most data connections are asymmetric (with more data traveling in one direction than the other), the gain is not as great as many claim. However, full-duplex operation does increase the throughput of most applications because the network media is no longer shared. Two devices on a full-duplex connection can send data as soon as it is ready. Token-passing networks such as Token Ring can also benefit from network switches. In large networks, the delay between turns to transmit may be significant because the token is passed around the network. LAN Transmission Methods LAN data transmissions fall into three classifications: unicast, multicast, and broadcast. In each type of transmission, a single packet is sent to one or more nodes. In a unicast transmission, a single packet is sent from the source to a destination on a network. First, the source node addresses the packet by using the address of the destination node. The package is then sent onto the network, and finally, the network passes the packet to its destination. A multicast transmission consists of a single data packet that is copied and sent to a specific subset of nodes on the network. First, the source node addresses the packet by using a multicast address. The packet is then sent into the network, which makes copies of the packet and sends a copy to each node that is part of the multicast address. A broadcast transmission consists of a single data packet that is copied and sent to all nodes on the network. In these types of transmissions, the source node addresses the packet by using the broadcast address. The packet is then sent on to the network, which makes copies of the packet and sends a copy to every node on the network. LAN topologies define the manner in which network devices are organized. Four common LAN topologies exist: bus, ring, star, and tree. These topologies are logical architectures, but the actual devices need not be physically organized in these configurations. Logical bus and ring topologies, for example, are commonly organized physically as a star. A bus topology is a linear LAN architecture in which transmissions from network stations propagate the length of the medium and are received by all other stations. Of the three most widely used LAN implementations, Ethernet/IEEE 802.3 networks-including 100BaseT-implement a bus topology, which is illustrated in Figure: Some Networks Implement a Local Bus Topology. Figure: Some Networks Implement a Local Bus Topology A ring topology is a LAN architecture that consists of a series of devices connected to one another by unidirectional transmission links to form a single closed loop. Both Token Ring/IEEE 802.5 and FDDI networks implement a ring topology. Figure: Some Networks Implement a Logical Ring Topology depicts a logical ring topology. Figure: Some Networks Implement a Logical Ring Topology A star topology is a LAN architecture in which the endpoints on a network are connected to a common central hub, or switch, by dedicated links. Logical bus and ring topologies are often implemented physically in a star topology, which is illustrated in the following figure. A tree topology is a LAN architecture that is identical to the bus topology, except that branches with multiple nodes are possible in this case. Figure: A Logical Tree Topology Can Contain Multiple Nodes illustrates a logical tree topology. Figure: A Logical Tree Topology Can Contain Multiple Nodes Devices commonly used in LANs include repeaters, hubs, LAN extenders, bridges, LAN switches, and routers. |Note:||Repeaters, hubs, and LAN extenders are discussed briefly in this section. The function and operation of bridges, switches, and routers are discussed generally in "Bridging and Switching Basics" and "Routing Basics."| A repeater is a physical layer device used to interconnect the media segments of an extended network. A repeater essentially enables a series of cable segments to be treated as a single cable. Repeaters receive signals from one network segment and amplify, retime, and retransmit those signals to another network segment. These actions prevent signal deterioration caused by long cable lengths and large numbers of connected devices. Repeaters are incapable of performing complex filtering and other traffic processing. In addition, all electrical signals, including electrical disturbances and other errors, are repeated and amplified. The total number of repeaters and network segments that can be connected is limited due to timing and other issues. Figure: A Repeater Connects Two Network Segments illustrates a repeater connecting two network segments. Figure: A Repeater Connects Two Network Segments A hub is a physical layer device that connects multiple user stations, each via a dedicated cable. Electrical interconnections are established inside the hub. Hubs are used to create a physical star network while maintaining the logical bus or ring configuration of the LAN. In some respects, a hub functions as a multiport repeater. A LAN extender is a remote-access multilayer switch that connects to a host router. LAN extenders forward traffic from all the standard network layer protocols (such as IP, IPX, and AppleTalk) and filter traffic based on the MAC address or network layer protocol type. LAN extenders scale well because the host router filters out unwanted broadcasts and multicasts. However, LAN extenders are not capable of segmenting traffic or creating security firewalls. Figure: Multiple LAN Extenders Can Connect to the Host Router Through a WAN illustrates multiple LAN extenders connected to the host router through a WAN. Figure: Multiple LAN Extenders Can Connect to the Host Router Through a WAN Q - Describe the type of media access used by Ethernet. A - Ethernet uses carrier sense multiple access collision detect (CSMA/CD). Each network station listens before and after transmitting data. If a collision is detected, both stations wait a random time before trying to resend. Q - Describe the type of media access used by Token Ring. A - Token Ring passes a special type of packet called a token around the network. If a network device has data to send, it must wait until it has the token to send it. After the data has been sent, the token is released back on the network. Q - Describe unicast, multicast, and broadcast transmissions. A - A unicast is a transmission from one source to one destination. A multicast is a transmission from one source to many stations that register to receive the traffic. A broadcast is a transmission from one source to every station on the local network segment. For More Information Cisco's web site (www.cisco.com) is a wonderful source for more information about these topics. The Documentation section includes in-depth discussions on many of the topics covered in this article. Teare, Diane. Designing Cisco Networks. Indianapolis: Cisco Press, July 1999.

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# Thread: determine slope of y=2x^2 - 4x + 2 1. ## determine slope of y=2x^2 - 4x + 2 i have to determine the slope of y=2x^2 - 4x +2 isn't that a parabola? how can it have a slope or does it have two different ones? but how can a curve have a slope? there is also another problem similar to this y=x^2 from (-1,1) to (x,y) and it wants me to determine change in y/change in x in terms of x Thanks! 2. thank you 3. Originally Posted by apm i have to determine the slope of y=2x^2 - 4x +2 isn't that a parabola? how can it have a slope or does it have two different ones? but how can a curve have a slope? If we were to look for a tangent line, we end up differentiating the function. This gives us the slope of the function. However, the function has an infinite number of slopes, unless a particular x value is specified. In our case, $\displaystyle y=2x^2-4x+2$. When we differentiate it, we get $\displaystyle y'=4x-4$ This is our slope. Since there is no x value specified, there are an infinite number of slopes values that depend on the value of x, where $\displaystyle x\in\mathbb{R}$. there is also another problem similar to this y=x^2 from (-1,1) to (x,y) and it wants me to determine change in y/change in x in terms of x Thanks! I would assume that they are requesting the average slope... The line is defined over the interval $\displaystyle (-1,x_0)$ Thus, $\displaystyle \frac{\Delta y}{\Delta x}=\frac{f(b)-f(a)}{b-a}\implies \frac{\Delta y}{\Delta x}=\frac{x_0^2-1}{x_0+1}\implies\color{red}\boxed{\frac{\Delta y}{\Delta x}=x_0-1}$ I hope this makes sense. --Chris 4. Sorry for posting in a kind-of-old thread, but I have the same problem and I can't figure it out. The posted problem does not give a particular point to find the slope of. Are there any examples of solving such problems, step by step and preferably with explanations? p.s! I hope its not too difficult to understand what I just typed, my first language has loose word order and I just spent half a day translating to that. :P e: Thanks to mr. fantastic for rewording my post. I got . What will happen to 2h? , , ### y=2x 2 slope Click on a term to search for related topics.

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The fallowing of a field can be intentional as a measure to conserve moisture or a result of prevented planting. In arid areas of the western United States, fallowing for a year or more is a common practice to help build soil water reserves. In the Midwest, flooding and continuous rainfall can, on occasion, prevent the planting of a crop. Regardless of the cause, corn planted into previously fallowed fields may exhibit nutrient deficiencies, particularly phosphorus (P) (Figure 1) and zinc (Zn) (Figure 2), and is generally referred to as fallow syndrome.1 Should a field not be planted, farmers should be aware that these nutrient issues may occur and be proactive in their management to help reduce the potential for these deficiencies. Cause of Fallow Syndrome Nutrient Deficiencies Corn is dependent on a symbiotic relationship with vesicular arbuscular mycorrhizae (VAM) fungi because of their ability to assist in the absorption of water and non-mobile P and Zn. The VAM fungi grow inside and outside of the roots. External hyphae (long filamentous branches) of VAM fungi can be dramatically longer than plant roots and serve as root extensions to aid in the uptake of water and nutrients. When fields are absent of crops or weed growth, the population of VAM fungi can decrease because they cannot survive in the absence of acceptable (most grass species) root tissue. The fungi are also unable to survive in the presence of brassica roots (canola, cabbage, broccoli, and others) or sugar beets; therefore, corn planted after these crops may exhibit fallow syndrome symptoms. Additionally, the fungi cannot survive when fields are saturated or flooded. Fallow Syndrome Management Options Options to help reduce the potential for fallow syndrome include allowing weeds to grow freely, the planting of cover crops in the fall, use of starter fertilizer, and the planting of an alternative crop. Weed growth in fallowed fields may maintain a healthy VAM population; however, in the arid regions, weed growth can use valuable water reserves. In all areas, weeds can add to the weed seed bank if allowed to reproduce. With the increasing potential for herbicide-resistant weed species, maintaining a weed-free environment is a best management recommendation. In non-arid areas, the planting of VAM-supporting cover crops may be a consideration to help maintain and increase the VAM population. Additionally, cover crops can: scavenge nitrogen, provide erosion protection, improve soil tilth, provide livestock feed, and help reduce weed growth.1 If herbicides are used to destroy the cover crop, plant-back or rotational restrictions MUST be followed. If corn is to be planted on fallowed acres, a banded application of 5 gl/acre of 10-34-0 and 1 qt/acre of chelated Zn on silt or silty-clay loams can help reduce the effects of P and Zn deficiencies, respectively.2,3 If applications are to be made to sandy soils, the 10-34-0 rate should be reduced to 3 gl/acre.3 The starter fertilizer should be applied 2 inches below and 2 inches to the side of the seed row to avoid seed and seedling injury.3 Soybean and sorghum are more tolerant to low VAM populations than corn. Where possible, these crops may be alternatives. 1Stahl, L. 2014. Reduce risk of fallow or flooded soil syndrome with cover crops. University of Minnesota. http://blog-crop-news.extension.umn.edu. 2Stahl, L., Fernandez, F., and Kaiser, D. 2018. Reduce risk of fallow or flooded soil syndrome with cover crops. University of Minnesota. https://extension.umn.edu/. 3Sutradhar, A.K., Kaiser, D.E., and Rosen, C.J. 2016. Zinc for crop production. University of Minnesota Extension https://extension.umn.edu/. Additional sources: Ellis, J.R. 1998. Post flood syndrome and vesicular-arbuscular mycorrhizal fungi. Journal of Production Agriculture. Vol 11, no. 2: 200-204. Sawyer, J., Mallarino, A.P., and Al-Kaisi, M. 2011. Flooded soil syndrome. Flood Recovery for Cropland. University of Nebraska-Lincoln and Iowa State University Extension. Web sources verified 02/04/2020. 4013_S3

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NCERT Exemplar - MCQ Chapter 8 Class 10 Introduction to Trignometry Serial order wise (A) 2/3   (B) 1/3    (C) 1/2  (C) 3/4 Transcript Question 15 If 4 tan θ = 3, then ((4 sin⁡〖𝜃 − cos⁡𝜃 〗)/(4 sin⁡〖𝜃 + cos⁡𝜃 〗 )) is equal to (A) 2/3 (B) 1/3 (C) 1/2 (C) 3/4 Given 4 tan θ = 3 tan θ = 𝟑/𝟒 Now, (4 sin⁡〖𝜃 − cos⁡𝜃 〗)/(4 sin⁡〖𝜃 + cos⁡𝜃 〗 ) Dividing numerator and denominator by cos θ = (4 sin⁡𝜃/cos⁡𝜃 − co𝑠⁡𝜃/cos⁡𝜃 )/(4 sin⁡𝜃/cos⁡𝜃 + co𝑠⁡𝜃/cos⁡𝜃 ) = (𝟒 𝐭𝐚𝐧⁡〖𝜽 − 𝟏〗)/(𝟒 𝐭𝐚𝐧⁡〖𝜽 + 𝟏〗 ) Putting tan θ = 𝟑/𝟒 = (4 × 3/4 − 1)/(4 × 3/4 + 1) = (3 − 1)/(3 + 1) = 2/4 = 𝟏/𝟐 So, the correct answer is (C)

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Read this riddle and see if you can work out how the trees must be planted. First or two articles about Fibonacci, written for students. This article for the young and old talks about the origins of our number system and the important role zero has to play in it. Models, images, representations, diagrams. The mathematics education literature is replete with terms suggesting the importance of making mathematical ideas manifest. Common across all these terms is an emphasis on some sort of physicality but I’m not clear about the subtle differences between them. Isn’t a diagram an image? Is ‘representations’ a more encompassing term than ‘models’? Does it help to have a range of terms at all? For example, ‘models and images’ are often ‘bundled’ together. Presumably this arises from the everyday distinction that models are three-dimensional and images two-dimensional. Models, largely, can thus be ‘acted upon’. Images, in this everyday sense, have a more static quality to them - while my eye may roam around the image of the Mona Lisa, I’m not encouraged to pick it up and physically act upon it. But in the mathematics classroom, number lines, arrays and similar images are to be ‘acted upon’, so in an attempt to simplify things, I want to lump everything together under the one heading of models, in the sense of a model being something that ‘stands-in-for’ something else, irrespective of whether it is two- or three-dimensional. Professionals who make models include all sorts of designers - theatrical set designers, architects, furniture makers - where models act as ‘proto-types’ for the real thing. In the mathematics classroom models can be proto-concepts. I choose to use model to emphasis the interactive nature of working over the subtly more passive nature of images. And I can engage learners in the actions of modelling (as in creating a model, not walking down the catwalk!) whereas the activity of ‘imaging’ is more difficult to envisage (although I do value the importance of ‘imagining’). With this definition the images and diagrams that learners create and work with become models: quickly sketching an empty number line and using movement along the line to represent addition or subtraction is an act of modelling. The work from the Dutch Realist Mathematics Education at the Freudenthal Institute provides a framework for thinking about how we can help, and monitor, learners working with models. The researchers there distinguish between: Tools for Classroom artifacts such as number lines, 100 squares or base 10 blocks, start life in the classroom as models of, in the sense that the teacher works with these in ways that imbue the models with meaning before learners may have come to ‘read’ the models in the same way. To take a simple example, young children have informal strategies for solving addition problems. Adding four and three they may hold out four fingers and then another three and count the total. The teacher could model this by drawing a number line, counting along to four on the line and then counting on three more ‘steps’ to land on seven. This model of the addition is not a direct mirroring of what the children did: they were counting objects (fingers), the teacher is counting jumps and it takes time for the children to appreciate the links between their activity and that of the teacher. Through this type of joint activity learners solve problems in their own way and the teacher provides a mathematical model of their solutions, then, over time, the learners begin to appropriate the teacher’s model and to use it for themselves; there is a shift from the teacher providing a model of the activity to learners taking this on as model to use themselves. The Dutch research shows that eventually, if the model is a good one, then learners come to be able to work with the model, but without needing to make actual marks on paper: the model is imagined and has become a tool for thinking with. This move from model for to tool for is, I think, the key thing to consider when choosing models to introduce into the classroom. For example when I began teaching it was the norm that base-ten blocks were used to teach multi-digit addition and subtraction. Although still around in classrooms they are less prominent than they used to be. That may be because of fashion, but I think it is also to do with this move from model for to tool for: base ten blocks are great to actually manipulate but recent research suggests that they may not be the sort of image that we work with mentally. For example the work of Stanislas Dehaene suggests that the mind deals with numbers more akin to ordinal models - the number line - than cardinal models - piles of bricks . A mark of a good model for/tool for thinking with is that it can help learners gain insight into mathematical structure, not simply get correct answers. Here is where base ten blocks can provide a helpful model. Rather than the teacher modelling how to exchange, say, ten unit cubes for a single ten stick by counting out and swapping, the cubes could be used to model the powers of ten structure of the place value system. Handling and talking about the cubes in ways that invite learners to imagine that a single cube is scaled up by a factor of ten to make a ten stick which similarly is scaled up by a factor of ten to make a hundred ‘flat’ and so forth encourages learners to ‘read’ the cubes in this multiplicative fashion rather then additively (as the count out and exchange modelling encourages). It is then easier to use the cubes to model the decimal system by extending the idea of scaling up to scaling down to become ten times smaller. As this example shows, we mustn’t assume that the mathematics is ‘in’ a model simply waiting to be recognised or discovered. The mathematics that emerges depends upon what features of the model are stressed or ignored, to use John Mason’s terms . Stressing the ‘scale’ properties of base ten blocks means playing down the additive properties. This is where the role of the ’significant’ other that Vygotsky raised becomes important. A sensitive other (and that may be a peer as much as the teacher) will gradually lead the learner to this noticing, in ways that hopefully encourage an ‘aha’ moment of insight. Part of provoking ‘stressing’ comes through careful choice of examples. Suppose a learner is able to work out mentally that 200 − 165 = 35. Given 212 − 177 they may have to work that out differently, perhaps using paper and pencil, as it does not look immediately as amenable to a mental calculation as 200 − 165. Getting, again, the answer 35, the learners who throughout their mathematics lessons have been encouraged to stress looking for patterns (even though equal results may simply be coincidental) are encouraged to pause and set up a quick model to check out if there is a relationship between the two answers. The empty number line provides a model that supports thinking about what is going on. Setting up the initial condition of 200 − 165 as finding the difference between the two numbers (as opposed to ‘taking away’ the 165 from 200) gives an initial model to play with: Putting the difference between 212 and 177, perhaps below the line to keep things clear, has the potential for the learner to notice that as the 200 and 165 were both increased by 12, then the difference between the two new numbers must be the same. This ‘noticing’ is an active attending to the model - the learners’ attention has to be primed to look for connections - the model does not in and of itself make the connection obvious (in contrast to the way that, as I write, my attention has just been drawn by the sound on the roof that it has started raining). Mind and model thus interact in bringing the mathematics into being - constant difference, in this example, is not ‘in’ the model waiting to be ‘discovered’ any more than it is ‘in the mind’ waiting to be revealed. The skill in teaching here is the subtle focusing of the learners attention so that they experience that moment of insight: ‘Aha, of course the answer must be the same, as each number has been made equally larger’. On the other hand, too much ’scaffolding’ and there is a danger of provoking a ‘duh’ moment instead: ‘Duh, why didn’t I notice that?’ There’s encouragement in the literature to expose learners to a range of models. While there is no doubt that any one model will have limitations, my work with teachers suggests that this encouragement to introduce a variety of models is sometimes interpreted differently; that if learners don’t ‘get it’ with a particular model, try a different one. As indicated above the Dutch research shows that ‘getting it’ takes time and that we shouldn’t abandon models too soon. The question to ask when choosing models is less one of how many models to introduce and more one of how ‘rich’ a model is in its potential to extend into different aspects of mathematics. The array is one such particularly rich model. Arrays are beginning to be used quite a lot in introducing multiplication and in the ‘grid’ method as a bridge to introducing the algorithm for multi-digit multiplication. But the potential of arrays does not end there. The array can also be used to help make clear the link between multiplication and division. Suppose you want to figure out 176 ÷ 8. We can set this up as an array with the value of one side missing. Using known multiplication facts the value of the missing side can be built up. And what about fractions? Consider for example the calculation 2/3 x 2/5. Learners brought up on ‘multiplication as repeated addition’ can come unstuck when faced with such a calculation - how can you add 2/3 two-fifth number of times. A quick sketch of an array can move things forward. Start with an array that is divided into thirds one way and fifths the other. Learners who have come to think of multiplication in terms of arrays will be comfortable in thinking of the required piece as being the part of the array marked out by the intersection of the 2/3 and 2/5. Finally, here’s an insight from working with this model that I only recently came across. Take the calculation 7/8 x 4/7. The array model looks like: From this we can see that the overall array has been divided into 8 x 7 smaller parts, hence the denominator is 56. And the shaded part is 7 x 4 = 28. So the answer is 28/56 or 1/2. So far, this may seem like a sledgehammer to a nut, but here’s the insight. We can take, in our imagination, the shaded part of the array and, without damage to the calculation, rotate it through 90 degrees. Now we can see the answer 1/2; quite clearly. And more. What this manipulation of the model establishes, is that when multiplying two fractions it’s ok to swap over the numerators - the answer will be the same. So 7/8 x 4/7 = 4/8 x 7/7 = 1/2 x 1 = 1/2. Neat eh? 1. Gravemeijer, K., How emergent models may foster the constitution of formal mathematics. Mathematical Thinking and Learning, 1999. 1(2): p. 155-177. 2. Dehaene, S., The number sense: How the mind creates mathematics. 1999, Oxford: Oxford University Press. 3. Mason, J., Burton, L. & Stacey, K. Thinking Mathematically. 2010. 2nd edition. Harlow: Pearson Education Ltd.

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Rabies is prohibited matter. Under Queensland legislation, if you suspect the presence of this disease in any species of animal, you must report it to Biosecurity Queensland on 13 25 23 or contact the Emergency Disease Watch Hotline on 1800 675 888. Rabies is a viral disease that affects humans and other mammals. It affects the central nervous system and is of great public health and veterinary concern. There are several variants of rabies virus, each adapted to a specific reservoir host. Rabies causes progressive inflammation of the brain and spinal cord (encephalitis) resulting in death. Over 55,000 people die of rabies worldwide each year. More than 95% of human deaths occur in Asia and Africa. Although rabies virus is exotic to Australia, there have been 2 confirmed human rabies deaths here (1987 and 1990). Both people were infected while overseas. Distribution in Queensland Rabies is present in most parts of the world including Europe, Africa, the Americas, the Middle East and most of Asia. Australia, New Zealand, the United Kingdom, Japan, Singapore, Papua New Guinea and the Pacific Islands are free of rabies. A similar lyssavirus (Australian bat lyssavirus) is present in Australia. Since December 2008, rabies in dogs has spread to previously uninfected islands in the Indonesian archipelago, including spreading to the popular tourist destination of Bali. Signs of illness in animals can develop anywhere between 10 days and several months to years after infection. Once signs of illness arise, death typically occurs within 10 days. In people, the incubation period is typically 1-3 months but may vary from less than 1 week to several years. Dogs are the most significant hosts for rabies. Dog-variant rabies is responsible for most human infections. Other variants of rabies virus are also maintained in wildlife species such as foxes, raccoons, skunks, wolves and bats. Other mammals such as cats, horses, cattle and other domestic animals may be affected by rabies, and may transmit the infection to humans, but do not have their own variants of the virus or sustain ongoing cycles of infection within their populations. The clinical signs of rabies are variable, depending on the effect on the brain. Rabies often causes sudden behavioural changes, followed by progressive paralysis, coma and death. Behavioural changes include the following. Dogs and cats - Symptoms range from a depressed quiet form, where the animal remains quiet and only bites when provoked - a furious form with unusual restlessness, snapping at imaginary objects and eating strange objects such as sticks and stones. - Become depressed - Stop producing milk - May grind their teeth - Have increased sexual activity - May attack other animals - Become increasingly paralysed, lose balance, finally cannot rise, become comatose, die - Often have multiple cases existing at the same time in a flock, suggesting a rabid animal attack - Appear restless then depressed, dying within about 3 days - Show abnormal behaviour, such as hiding and then biting if provoked - Develop a crazed appetite - Kill piglets - Are increasingly dull - Become paralysed - Changes range from the quiet form, where there is depression and difficulty swallowing (owners may think the animal has something caught in its throat) - the furious form, where there is marked excitation and it is dangerous to get close to the animal. Once symptoms develop, there is no cure and death is almost certain. However, with timely medical intervention, the disease is easily preventable. For this reason, all potential rabies exposures should be treated as a medical urgency. How it is spread Rabies is transmitted through the saliva of an infected animal. Infection usually occurs when infectious saliva comes into contact with fresh wounds (e.g. bites and scratches) and unprotected mucous membranes (e.g. eyes and mouth) of non-vaccinated animals and people. Most (95%) human cases of rabies are due to bites by infected dogs. Human-to-human transmission of rabies can occur as a result of organ transplant from an infected person. Prevention in animals In countries where rabies is endemic, control relies on vaccination programs and the management of stray animal populations. Rabies vaccines that prevent the development of rabies disease are available for dogs, cats, horses, cattle, sheep and ferrets. Licensed oral vaccines are used for mass immunisation of wildlife. Prevention in humans Rabies infection can be prevented through several simple courses of action: - Seek medical advice about pre-exposure vaccination before travelling to a region with endemic rabies, particularly if contact with wildlife or dogs is likely. - If bitten or scratched by an animal in a country that is not free of rabies, immediately clean the wound, apply a disinfectant, seek urgent medical advice. Treatment for humans - Clinical disease is almost invariably fatal. - If you are ill with signs consistent with rabies, seek urgent medical advice. - Learn more about the World Health Organisation response to rabies. - Learn more about rabies on the mOIE Rabies Portal (World Organisation for Animal Health). - See the AUSVETPLAN for rabies (PDF, 538KB). - Learn about Australian bat lyssavirus. - Last reviewed: 01 Jul 2016 - Last updated: 01 Jul 2016

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A Concise History of Macedonia Although Macedonia is a young state, since it became independent in 1991, its roots run deep into history. The name “Macedonia” is in fact the oldest surviving name of a country in the continent of Europe. Archaeological evidence shows that old European civilization flourished in Macedonia between 7000 and 3500 BC. The region Macedonia is located in the heart of the Balkans, north of ancient Greece, east of Illyria, and west of Thrace. The ancient Macedonians were an ethnically, linguistically, and culturally distinct nation. The origins of the Macedonians are in the ancient Brygian substratum which occupied the whole of Macedonian territory and in Indo-European superstratum, which settled here at the end of the 2nd millennium. Table of Contents - Ancient Macedonia - Roman and Medieval Macedonia - Ottoman Macedonia - The Independence Movement - Macedonia and European Cartography until the end of the 19th century - The Partition of Macedonia and World War I - World War II and the Liberation - The Greek Civil War and the Macedonians in Greece (Aegean Macedonia) - The Macedonians in Bulgaria (Pirin Macedonia) - Republic of Macedonia - Latest Developments The history of the ancient Macedonian kingdom begins with Caranus, who was the first known king (808-778 BC). The Macedonian dynasty Argeadae originated from Argos Orestikon, a city in located in south western Macedonia region of Orestis (App.,Syr., 63;Diod. ,VII, 15; G. Sync., I, 373). Alexander I Philhellene (498-454 BC) expanded the kingdom and by the 5th century BC the Macedonians had forged a unified kingdom. Alexander was a Persian ally in the Greek-Persian wars. As Macedonia appears on the international scene, the first coins with the king’s name on them are made. Around the year 460 BC, Herodotus sojourns in Macedonia and gives an interpretatio macedonica of the Greek-Persian wars (Her.5.17-22, 9.44-45). Alexander’s son Perdiccas II (453-413 BC) worked on starting a war between the Athens maritime power and Sparta which lead the Peloponnesian League (Thucyd.Pel.I.57), and initiated the creation of an Olynthian league from the Greek colonies neighboring Macedonia on Chalcidice, for a war against Athens (Thucyd.I.58). During the Peloponnesian War, Perdiccas II is one moment on the side of Athens and the next on the side of Sparta, depending of Macedonia’s best interests, not wanting either of them to become too powerful, while keeping its country’s sovereignty at the expense of the Greek quarrel. It was Archelaus (413-399 BC) who made Macedonia a significant economic power. Archelaus made roads, built fortresses, and reorganized the Macedonian army (Thucyd.II.100). He moved the Macedonian capital Aigae to Pella and founded Macedonian Olympian Games in Dion (the holy city of the Macedonians), among other reasons also because of the fact that the Greek Olympic Games were forbidden to the barbarians (non-Greeks), including the Macedonians (Her.V.22). In the year 406 BC the Macedonian poet Adaius wrote an epitaph for the grave stone of Euripides (Anth.Pal.7,5,1; A. Gellius, Noct. Att, XV, 20, 10) who was staying in the Macedonian palace of Archelaus. Euripides besides the apologetic work Archelaus also wrote the well known play Bachae inspired by the Macedonian cult for the God Dionysus. The Macedonian council refused to give Euripides’ body to his birthplace Athens (Gell.Noct.Att.XV.20). During the years 407-6 BC Archelaus from Athens received the titles proxenos and euergetes. Amyntas III reigned 393-370/369 BC and led a policy of exhausting and weakening of the Greek city states. His two of his sons, Alexander II and Perdiccas III, reigned later only briefly. Alexander II however, had an expansionist policy and invaded northern Greece. In Thessaly he left Macedonian garrisons in the cities and refused to evacuate them. The Thebans, having at the time the most powerful military, intervened and forced the removal of the garrisons. Alexander II’s youngest brother Philip was taken as hostage to Thebes. After the death of Alexander II, his other brother Perdiccas III took the throne. But Perdiccas III was killed with his Macedonian soldiers in a battle with the Illyrians, and Amyntas’ third son, Philip II (later the Great) now became the next Macedonian king. Philip II (359-336 BC), the greatest man that Europe had given by then (Theop.F.GR.H. f, 27), liberated and unified Macedonia, turning it into the most powerful European state – an armed nation with a common national ideal. In 338 BC, the Greeks unified to prevent Philip II from penetrating the northern Greek states, but to no avail; the Macedonians defeated the the united Greek states at the battle at Chaeronea that summer. Philip became a hegemon to the Greeks who had no choice but to ratify his peace agreement koine eirene (mutual peace). The four Macedonian garrisons – strategically positioned at Corinth, the Theban Cadmeia, Chalcis on Euboea and Ambracia – were to guarantee the Macedonian hold of Greece. This mutual peace, was not a league at all (it did not have the word symachia). But the conqueror of Greece was assassinated before he could lead the Macedonians in the conquest of the Persian Empire. His son Alexander III the Great (356-323 BC), succeeded Phillip II at the age of 20. Encouraged by Alexander’s young age, the Thracians, the Illyrians, and the Greeks, revolted upon hearing of Philip’s death. His first victories in Greece, however, were a warning sign to those that were to defy the deal with his father. Next, at the head of Macedonian and allied Greek, Illyrian, and Thracian troops, he invaded Persia. Alexander believed that his Macedonian soldiers were the backbone of his army, the driving force behind his expeditions, for they were motivated in their battles by permanent values (Q. C. Rufus, Alexander III, 10, 4-10). Alexander’s victories at Granicus, Issus, and Gaugamela put an end to the Persian Empire, which was then replaced by the Macedonian Empire stretching from Europe, to Egypt and India. In order to strengthen his kingdom, Alexander wed the daughter of the last Persian king Darius III’s in a wedding ceremony where 10,000 marriages were performed between Macedonian soldiers and Persian women. Similarly, as he later viewed himself not only as a Macedonian king, but as a king of all nations living in his empire, he often wore non-Macedonian outfits too. Until the rise of Rome, the Macedonians shaped the events in this vast space for the following three centuries. Following Alexander’s death, a series of rebellions took place, in order to free the Greek states from the Macedonian yoke. Such upheavals ended unsuccessfully (Diodorus, 18.7.3-9, 18.10.1-3, 11, 12, 15, 17.5). Unfortunately, the Macedonian leading generals engaged in a dreadful conflict over the rule of the Empire. By 300 BC, the Macedonian Empire was carved up between the dynasties of Antigonus I One-Eye (Macedonia and Greece), Ptolemy I (Egypt), and Seleucus I (Asia). Under Antigonus II Gonatas (276-239), the grandson of Antigonus I, Macedonia achieved a stable monarchy and strengthened its occupation of Greece. His grandson Philip V (222-179 BC), clashed with Rome which was now expanding eastwards, and fought the Macedonian Wars against the Romans. After the Roman army defeated Philip in Thessaly, Macedonia was reduced to its original borders. In the third Macedonian War, Rome finally defeated the Macedonian army under the last king the Philip V’s son Perseus (179-168 BC) and at the Battle of Pydna, 20,000 Macedonian soldiers died defending their land. Perseus’ death in Italy marked the ending of the Macedonian kingdom, and by 146 BC Macedonia became a Roman province. By 65 BC Rome conquered the Seleucid Macedonian kingdom in Asia under its last king Antiochus VII. Finally, the defeat of Cleopatra VII in 30 BC, brought an end to the last of the Macedonian descendants (the Ptolemy dynasty) in Egypt, and with it, the last remains of the Macedonian Empire that was once the mightiest in the world disappeared from the face of the earth. Roman and Medieval Macedonia In 51 AD for the first time on European soil, in the Macedonian towns Philippi, Salonika and Beroea, the Apostle Paul preached Christianity (Acta apos., XVI, id. XVII). In 52 and 53 he sent epistles to the people of Salonika (Epist. Thess); in 57 he came to Macedonia again, and in 63 he sent epistles to the people of Philippi (Epist. Philipp). During the 3rd and 4th centuries, Macedonia was divided into two provinces, Macedonia Prima and Macedonia Salutaris. Since the east-west split of the Roman Empire in 395 AD, Macedonia was ruled by the Eastern Roman Empire (Byzantine Empire). It is interesting to note that the Emperor Iustinian was born near Scupi (nowadays Skopje). In the 5th century Macedonia was divided again into Macedonia Prima and Macedonia Secunda. In the 6th century, the Southern Slavs entered Macedonia, mixing with the locals, laying the foundations of the modern Macedonian nation. In the 9th century, the brothers Sts. Cyrilus and Methodius of Salonika created the first Slavic alphabet, the Glagolitic alphabet, thus paving the way for Slavic literacy. Soon after, the two brothers from Salonika translated Christian scriptures in the language used by the local Slavs. In the years to come, Sts. Cyrilus and Methodius spread literacy and Christianity among the Slavic peoples, starting from the southern-most Slavs, reaching as far as the Slavs of Moravia (in the Czech and Slovak Republics). Their disciples St. Clement and St. Naum established the first Slavic University, the Ohrid Literary School, located in Plaoshnik (Imaret), a part of Ohrid’s Old Town. Here St. Clement reformed St. Cyril’s alphabet, naming it the Cyrilic alphabet in honor of his teacher; St. Clement’s alphabet closely resembles the modern alphabet used by the Macedonians, as well as the Russians, Serbs, Montenegrins, and Bulgarians. In addition to that, 3,500 teachers, clergy, writers, and other literary figures emerged from this Ohrid Literary School. Their activity was crowned with the laying of foundations of a Slavonic cultural, educational and ecclesiastical organization, where the Slavonic alphabet was used and the Old Slavonic language was introduced in religious services. The establishment of the first Slavic bishopric, later to become an Ohrid Archbishopric during the reign of Tsar Samuil, marked the beginning of the Macedonian Orthodox Church. In the first half of the 10th century, the Bogomil teaching appeared in Macedonia. Bogomilism had grown into a large-scale popular movement engulfing the Balkans and Europe. The 10th century also marked the beginning of the first Macedonian medieval state, the Empire of Tsar Samuil (976-1014). Towards the end of the 10th century, with the weakening of the Byzantine Empire, and with the first Bulgarian Empire apart, Tsar Samuil created a strong Macedonian medieval kingdom with its center at Ohrid. Soon he conquered parts of Greece, a large part of Bulgaria, Albania, Serbia, Bosnia, Montenegro and Dalmatia. Tsar Samuil was defeated in 1014 by Basil II at the battle on Mount Belasica, near Strumica, capturing 15,000 of his soldiers. Basil’s punishment was punishment an exceptionally cruel one: Samuil’s soldiers were all blinded, except for every hundreth one, who had one eye left, so as to lead his fellow men to Samuil in Prilep, who escaped death at Belasica. At the site of his blinded soldiers, Samuil suffered a heart attack, dying two days later on October 6, 1014. The tradition of the Tsar Samuil’s state has remained deeply rooted in the minds of the Macedonian people, as he is still greatly praised in numerous folk tales and folk songs. For four centuries after the fall of the kingdom, rebellions and frequent changes of rule disrupted Macedonia’s development. In the 11th century, there were two major uprisings against Byzantine rule, one led by Petar Deljan in 1040, Samuil’s grandson, and the other by Gjorgji Vojteh in 1072. The 12th century saw the rise of the Macedonian feudal lords Dobromir Hrs in 1201, and Strez in 1211. Despite the numerous rebellions, and the short-lived Serbian and Bulgarian states in the 13th and 14th centuries, Macedonia remained a Byzantine territory until the Ottoman Turks conquered it in 1389. The Turks firmly established themselves not only in Macedonia, but in the whole region. Ottoman rule will last for the following five centuries. The first significant resistance movements against the Turkish occupation were the Mariovo-Prilep Rebellion (1564-1565), and the Karposh Uprising in 1689. In the 18th century, under the pressure of the Greek Patriarch in Istanbul, the Turks abolished the Ohrid Archbishopric, which had been keeping the Macedonians spiritually alive since Tsar Samuil. In the 19th century, Greece, Serbia, and Bulgaria freed themselves from the Turkish rule, after which they actively displayed territorial aspirations on the Macedonian land. Thus, the so-called Macedonian Question appeared, which is nothing but a competition for a new conquest of Macedonia by its neighbors. But the Macedonians strove to develop their own national consciousness and begun organizing themselves for fight against the Ottomans at the same time. Thus, the 19th century is a period of growing national awareness among the Macedonian people and their quest for free and independent Macedonia. The Independence Movement The following years marked the flourishment of the Macedonian literary movement, laying the foundations of modern Macedonian literary language. The leading activists were Kiril Pejchinovich, Joakim Krchovski, Partenija Zografski, Georgija Puleski, Jordan Hadzhi Konstantinov – Dzhinot, Dimitar and Konstantin Miladinov, Grigor Prlichev, Marko Cepenkov, and Kuzman Shapkarev. The second half of the 19th century was marked by the beginning of the national revolutionary struggle for the liberation of Macedonia. The Razlovci and Kresna Uprisings, in 1876 and 1878 respectively, had a strong influence on the growth of Macedonian national awareness. An important role in the movement was played by Bishop Theodosius of Skopje, who started a campaign for an independent Macedonian Orthodox Church by restoring the Ohrid Archbishopric, which had been abolished in 1767. Unfortunately, the strong Bulgarian lobby effectively destroyed the idea. In 1893, the Macedonian revolutionary organization known as VMRO (Internal Macedonian Revolutionary Organization) was founded in Salonika, with Goce Delchev as its leader. Its objectives were national freedom and the establishment of an autonomous Macedonian state, as outlined in the organization’s slogan Macedonia for the Macedonians. He believed that such a Macedonian state should be won by the people of Macedonia, not its neighbors: those who believe that answer of our national liberation lies in Bulgaria, Serbia or Greece might consider themselves a good Bulgarian, good Serb or a good Greek, but not a good Macedonian. The Macedonia that Delchev had in mind was to be the homeland of all its religious and ethnic groups, an idea that was embedded later on in the Manifesto of the Krushevo Republic. After all, Delchev was a known cosmopolitan, who said I understand the world only as a ground for cultural competition among all nations, and Macedonia was to be such a ground. In response to the rapidly worsening situation of the Macedonians under Ottoman rule, on August 2, 1903, VMRO launched the Ilinden Uprising against the Ottoman rulers and declared Macedonian independence. The revolutionaries liberated the town of Krushevo, and established the Republic of Krushevo with its own government. The founding principles of the first republic on the Balkans is outlined in the The Manifesto of the Krushevo Republic, outlining the founding principles of a modern democracy. Unfortunately, the uprising was brutally crushed by the Ottoman military; the Macedonian Question, however, aroused an ever increasing international concern. The Great Powers made several attempts to impose reform on the Ottoman Porte, including the sending of their own officers to supervise the gendarmerie – in effect, the first international peacekeeping force. And although the revolt was suppressed, Macedonians remember the brief victory as a key date in the country’s history and the event is enshrined in Macedonia’s constitution. That same year Krste Misirkov from Pella (Postol), one of the most outstanding names in the history of Macedonian culture, and the founder of the modern Macedonian literary language and orthography, published his On the Macedonian Matters, in which he outlined criteria for the establishment of the Macedonian literary language. Macedonia and European Cartography until the end of the 19th century Macedonia has been an interesting destination for European travelers for the simple reason that it was the link to the Orient. For that reason, detailed maps of Macedonia show up as early as the 15th century, the dawn of modern cartography. For instance, the first known map of Macedonia made by modern cartographers is that of S. Ptolomaios (Tabula decima et Ultima Europae Alexandria) from 1477; however, the map only depicts the ancient cities. Following Ptolomaios’ example, later maps continued putting the ancient toponyms (ex: I.Laurenbergio: Macedonia Alexandri M. patria illustris. from 1647). Later maps, like G. Cantelli da Vignola: La Macedonia from 1689 and N. Sanson: Estats de l’Empire des Turqs en Europe from 1696, included the first modern toponyms. European cartographers kept their lively interest in Macedonia in the 18th, as well as the 19th centuries. The Partition of Macedonia and World War I The Young Turk movement, lead by the Young Turk Committee, had the aim of reforming the Turkish country, and consequently making social and political reforms in Macedonia. The Macedonian revolutionary organization, through Jane Sandanski and the newly formed national federal party, actively took part in the Young Turk movement for achieving autonomy for Macedonia within the frontiers of a truly democratic Turkish Republic. Such ideas were also promulgated by the father of the modern Turkish State, Mustafa Kemal Ataturk, who was well acquainted with the situation in Macedonia, as he was raised and educated in Macedonia. Four years after the Young Turk Revolution, in 1912, Greece, Serbia, and Bulgaria joined forces and defeated the Turkish forces in Macedonia. Since the reforms of the Turkish state did not keep pace with reality, over 100,000 Macedonians also participated in the actions against the Ottomans; the victors, however, forgot the Macedonians’ support. The Treaty of London (May 1913), which concluded the First Balkan War, left Bulgaria dissatisfied with the partition of Macedonia. Bulgaria’s attempt to enforce a new partition in a Second Balkan War failed, and the Treaty of Bucharest (August 1913) essentially confirmed the existing boundaries. Having failed to achieve independence in 1903, the Macedonians, now divided, were left to their new masters. Greece took the biggest, southern half of Macedonia (Aegean Macedonia), calling it Northern Greece; Bulgaria annexed the Pirin region, similarly abolishing the Macedonian name. And finally, Serbia took over the Vardar region and renamed it to Southern Serbia. An intensive campaigning took place in all three parts of Macedonia to impose foreign identities upon the population that suited the interests of the controlling states. In Vardar Macedonia, the Serbs labeled the Macedonians with the name Southern Serbs; in Aegean Macedonia, the Greeks labeled them as Slavophone Greeks, Makedo-Slavs, and other names; while in Pirin Macedonia, the Macedonians were simply called Bulgarians. In 1914, World War I erupted. Bulgaria sided with the Central powers and by 1915 it occupied the Serbian-held part of Macedonia. But the defeat of the Central powers and the end of World War I in 1918 saw the partition of 1913 reconfirmed and Macedonia was left divided. At the Paris Peace conference the demands of the Macedonians for independent and united Macedonia were ignored. Vardar Macedonia was re-incorporated with the rest of Serbia and into the new Kingdom of the Serbs, Croats, and Slovenes, later Yugoslavia. Since 1913, Greece has been trying to banish Macedonian language and the Macedonian toponyms in Aegean Macedonia. The Macedonian language was forbidden, despite the fact under the supervision of the League of Nations Greece had recognized its existence as distinct language when it published the primer Abecedar for the needs of the Macedonian children in 1924. This practice followed in the following decades. Yet despite the triple persecution the Macedonians didn’t change their identity. The period between the two world wars was also filled with constant endeavors to change the situation of Macedonia and annul the division of the country and its people. In 1925 VMRO (United) was founded in Vienna. Their main objective was to free Macedonia within its geographical and economical borders and create an independent political entity based on the principles of the Krushevo Republic, that will become an equal member of the future Balkan Federation. In 1935, MANAPO (Macedonian National Movement) was founded in the Vardar part of Macedonia. In 1938 The first collection of poems Fire (Ogin) from Venko Markovski was published in Macedonian. In 1939 publication of White Dawns (Beli Mugri), a collection of poems in Macedonian from the first modern Macedonian poet Kocho Racin. In 1940, the democratic groups in Macedonia defined the political program for the national and social liberation of the country. World War II and the Liberation With the World War II burning throughout Europe, Yugoslavia was invaded by the German army in April of 1941. Bulgaria, now an ally of Nazi Germany and Fascist Italy, again occupied almost all of Macedonia (both the Vardar and Aegean portions). On October 11, 1941, the Macedonians launched a war for the liberation of Macedonia from the Bulgarian occupation. By 1943, the anti-fascist sentiment lent support for the growing communist movement and soon thereafter, the Communist Party of Macedonia was established. In the same year, the first unit of the Army of Macedonia was founded. Bodies of government, such as national liberation councils, were formed over the whole territory of Macedonia. The Headquarters of the National Liberation Army (NOV) published the manifesto of the goals of the war of liberation. The first session of the Anti-Fascist Assembly of the National Liberation of Macedonia (ASNOM) was held in the monastery of St. Prohor Pchinski on August 2nd 1944 on the 41st anniversary of the Ilinden uprising. Representatives from all parts of Macedonia, including the Pirin and the Aegean parts of the country, gathered for the occasion and decided on the constitution of a modern Macedonian state as a member of the new Yugoslav federation under the name of People’s Republic of Macedonia. The ASNOM presidium was formed with Metodija Andonov Chento as its first President. In April 1945 the first Macedonian government was founded with Lazar Kolishevski as its first President. Although first calls for the restoration of the Macedonian Orthodox Church emerged at the end of 1944, and beginning of 1945, the Ohrid Archbishopric was restored in 1958, and its autocephaly was declared in 1967. The Macedonians were finally free in one of the three parts of Macedonia. Still, the Macedonian Question remained open for the years to come. The Greek Civil War and the Macedonians in Greece (Aegean Macedonia) Soon after the Varkisa agreement (December 1945), the use of the Macedonian name and the Macedonian language were once again prohibited in the Aegean part of Macedonia. In the period between 1945-46 alone, according to partial statistics: 400 murders were registered; 440 women and girls were raped; 13,529 interned on the Greek islands; 8,145 imprisoned in the Greek prisons; 4,209 indicted; 3,215 sentenced to prison; 13 driven mad by the torture in the prisons; 45 villages abandoned; 80 villages pillaged; 1,605 families plundered; and 1,943 families evicted. Therefore, during the Greek Civil War that followed World War II (1946-1949), the Macedonians of Aegean Macedonia fought on republican side, for they were promised human rights after the war. Of the 35,000 soldiers of DAG, about half were Macedonians. The liberated territory, covered a large portion of Aegean Macedonia. 87 Macedonian schools were opened for 100,000 pupils, the newspapers in Macedonian were published (Nepokoren, Zora, Edinstvo, Borec), and cultural and artistic associations were created. As the DGA lost the war, and the Macedonians once again were stripped of their human rights. The defeat of DAG resulted in terrible consequences for the Macedonians. 28,000 Aegean Macedonian children, known as child refugees, were separated from their families and settled in eastern Europe and Soviet Union in an attempt to save them from the terror that followed. Thousands of Macedonians lost their lives for the liberty of their people and a great number of the Macedonian villages were burned to the ground. The Macedonians in Bulgaria (Pirin Macedonia) The political changes after the capitulation of fascist Bulgaria and the coup d’etat of September 9, 1944 positively influenced the historical status of the Macedonians from the Pirin part of Macedonia. The Communist Party of Bulgaria, under the leadership of Geogi Dimitrov, on August 10, 1946 officially recognized the Macedonian nation and the right of the Pirin part of Macedonia to be attached to the People’s Republic of Macedonia. The demography data from 1946 revealed that the majority of the population in the Pirin part of Macedonia declared itself as Macedonian in a free census. The Macedonian literary language and the national history were introduced into the educational process. In 1947 in Gorna Djumaja (Blagoevgrad nowadays) the first Macedonian bookstore and reading room were opened, as well as the Regional Macedonian National Theater. The newspapers in Macedonian such as Pirinsko delo, Nova Makedonija, Mlad borec, and others, were also published. In the Bulgarian census of 1956, 63,8% of the population in Pirin declared itself as Macedonian. However, since 1956 Bulgaria has altered its attitude, negating again the existence of the Macedonian nation and forbidding the expression of Macedonian nationality and language. As result of this, in the census of 1965, the number of Macedonians dropped to only 8,750 and in the district of Blagoevgrad, less than 1%. Yet the Macedonians in Bulgaria begun organizing themselves and in 1989, United Macedonian Organization – Ilinden (OMO Ilinden) was formed, demanding cultural and national autonomy for the Macedonians in Pirin. Republic of Macedonia As federal Yugoslavia was disintegrating at the beginning of 1990’s, on September 8, 1991 in a referendum, 95% of eligible voters approved the independence and sovereignty of the Republic of Macedonia. Macedonia gained its independence peacefully, unlike the other former Yugoslav republics. Kiro Gligorov was elected the first president of independent Macedonia. The new Constitution determined the Republic of Macedonia a sovereign, independent, civil, and democratic state, and it recognized the complete equality of the Macedonians and the ethnic minorities:Macedonia is constituted as a national country of the Macedonian people which guarantees complete civil equality and permanent mutual living of the Macedonian people with the Albanians, Turks, Vlachs, Roma and the other nationalities living in the Republic of Macedonia. Although the European Community acknowledged that Macedonia had fulfilled the requirements for official recognition, due to the opposition of Greece, already a member of the community, the EU decided to postpone the recognition. Greece insisted that the new nation has no right to use of the name Macedonia and use the emblem of ancient Macedonia on its flag. In July of 1992 there were demonstrations in the capital Skopje over the failure to receive recognition. Macedonia was admitted to the United Nations under the temporary reference (not an official name) the former Yugoslav Republic of Macedonia in 1993. Full diplomatic relations with a number of EU nations followed, while Russia, China, Turkey, Bulgaria and most nations, recognized Macedonia under its constitutional name Republic of Macedonia. In response to the refusal of the Macedonian President Gligorov to rename the country, nation, and language, and change the Constitution because Article 47 specifies that the Republic of Macedonia cares for the statue and rights of those persons belonging to the Macedonian people in neighboring countries, as well as Macedonian expatriates, assists their cultural development and promotes links with them, Greece imposed a unilateral trade embargo on Macedonia on February 1994. Unfortunately, the embargo had devastating impact on Macedonia’s economy; the country was cut-off from the port of Salonika and became landlocked because of the UN embargo on Yugoslavia to the north, and the Greek embargo to the south. Later, the signing of the Interim accord between Greece and Macedonia marked the increased cooperation between the two neighboring states. However, there are still open issues. Due to the positive role that Macedonia played in the latest turbulent developments in the region, as well as the good human rights record, at the beginning of 2000 the EU opened negotiations with Macedonia on a Stabilization and Association Agreement. At the end of the year, the EU and Macedonia initialized the agreement at the Zagreb Summit, which puts Macedonia at the forefront for closer relationship from the other Balkan non-member states. In addition to that, as a sign of good will, the EU also adopted a set of exceptional trade measures with the Republic of Macedonia. - Ancient Greek and Roman historians: Arrian, Plutarch, Diodorus, Justin, Herodotus, Polybius, Curtius, Thracymachus, Livius, Demosthenes, Isocrates, Thucydides, Pseudo-Herod, Medeios of Larisa, Pseudo-Calisthanes, Pausanius, Ephoros, Pseudo-Skylax, Dionysius son of Kaliphon, Dionisyus Periegetes, Ptolemy of Alexandria (Geography) and Strabo.</li? - In the Shadow of Olympus (1990) and Makedonika (1995) – Eugene N. Borza - Macedonia and Greece in Late Classical and Early Hellenistic Times – Ernst Badian, 1980 - Alexander of Macedon 356-323 B.C.: A Historical Biography by Peter Green, 1991 - Philip and Alexander of Macedon – David G. Hogarth, 1897 - Krste Misirkov – On the Macedonian Matters 1903 - St. Petersburg periodical Nakedonski Glas (Macedonian Voice) 1913-1914 - Fields of Wheat, Hills of Blood: Passages to Nationhood in Greek Macedonia, 1870-1990 – Anastasia N. Karakasidou, 1997 - The Macedonian Conflict – Loring M. Danforth, 1995 - Denying Ethnic Identity: The Macedonians of Greece by Human Rights Watch Helsinki, 1995 - Encyclopaedia Britannica Countrywatch - The Official censa of the Republic of Macedonia - Macedonia and Greece: the Struggle to Define a new Balkan Nation – John Shea, 1997 Republic of Macedonia Home Page Here at Virtual Macedonia, we love everything about our country, Republic of Macedonia. We focus on topics relating to travel to Macedonia, Macedonian history, Macedonian Language, Macedonian Culture. Our goal is to help people learn more about the "Jewel of the Balkans- Macedonia" - See more at our About Us page.

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Home   »   Reasoning Questions For DSSSB Exam 2017 # Reasoning Questions For DSSSB Exam 2017 Directions (1-2): In each of the following questions, find the odd word/number-pair form the given alternatives. Q1. (a) Influenza (b) Scurvy (c) Rickets (d) Night-blindness Q2. (a) Raisin (b) Rain (c) Shower (d) Sleet Directions (3-4): Read the following information carefully and answer the questions which follows : ‘A × B’ means ‘A is brother of B’. ‘A – B’ means ‘A is daughter of B’. ‘A ÷ B’ means ‘A is wife of B’. ‘A + B’ means ‘A is son of B’. Q3. How is P related to S in the expression ‘S × R + Q ÷ P’? (a) Father (b) Grandson (c) Son (d) Grandfather Q4. Which of the following means ‘P is sister of S’? (a) P + Q ÷ R – S (b) P + Q ÷ R × S (c) P × Q – R ÷ S (d) None of these Q5. Select the correct combination of mathematical signs to replace * sings and to balance the given equation. 15 * 24 * 3 * 6 * 17 (a) – ÷ + = (b) – × = + (c) + × = ÷ (d) + ÷ – = Q6. In a certain code language, ‘481’ means ‘sky is blue’, ‘246’ means ‘sea is deep’ and ‘698’ means ‘sea looks blue’. What number is the code for ‘deep’? (a) 2 (b) 4 (c) 6 (d) 9 Q7. If MADRAS can be written as ARSMDA, how can ARKONAM be written in that code? (a) ROAAKNM (b) ROAKANM (c) ROAKNNM (d) ROAKNAM Q8. Choose the odd numeral pair/group in this question? (a) 45, 27 (b) 30, 18 (c) 20, 10 (d) 15, 12 Q9. In this question, certain pairs of words are given, out of which the words in all pairs except one, bear a certain common relationship. Choose the pair in which the words are differently related. (a) Mercury : Sun (b) Moon : Earth (c) Star : Galaxy (d) Wheel : Axle Q10. Choose the word which is least like the other words in the group. (a) Bake (b) Peel (c) Boil (d) Roast Solutions: S1. Ans.(a) Sol. Expert Influenza, all others are diseases caused by a deficiency of vitamins. S2. Ans.(a) Sol. Expect Raisin, all others are different forms of precipitation. S3. Ans.(a) Sol. S4. Ans.(d) Sol. All the options do not satisfy the required condition. S5. Ans.(d) Sol. 15 * 24 * 3 * 6 * 17 ⇒ 15 + 24 ÷ 3 – 6 = 17 ⇒ 15 + 8 – 6 = 17 S6. Ans.(a) Sol. ‘481’ – ‘sky is blue’, ‘246’ – ‘sea is deep’ ‘698’ – ‘sea looks blue’ From 1st and 2nd statement, ‘is’-4 And from 2nd and 3rd statement, ‘sea’ – 6 So, ‘deep’ – 2 S7. Ans.(a) Sol. The word formed by first even number place letters then odd number place letters. So, ARKONAM ->ROAAKNM S8. Ans.(c) Sol. In all other pairs, the two numbers have 3 as the common factor. S9. Ans.(c) Sol. In all other pairs, first moves about the second. S10. Ans.(b) Sol. Here, all except Peel are different forms of cooking. Hence, the answer is (b).

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Join Peggy Fisher for an in-depth discussion in this video Power sets, part of Programming Foundations: Discrete Mathematics. - [Voiceover] Next let's talk about power sets.…A power set is the set of all subsets…for a particular set including the empty set.…The cardinality of a power set…is two to the n, where n is the cardinality…of the original set.…For example, if the cardinality of set A…is equal to three, which means that A has three elements,…then the cardinality of the power set of A,…which is denoted with a script P…in front of the A, is equal to eight,…because two-cubed is equal to eight.… All right, if we're given the set…A equals the set containing one, two, and three,…let's write the power set of A.…So we know, since there's three items,…there's going to be eight sets.…I'll start with the empty set so I don't forget that one.…And this is going to be a set of sets.…Next, I'll do each individual number.…A set containing one,…two, three.… Now I'll start to do combinations of two.…One and two.…I definitely need to include a set that has…all the original elements, and I think I'm done.…If we count them we have one, two, three,…four, five, six, seven, and eight.… This course relies on an open-source SML (standard machine language) library to demo the concepts behind discrete math. Peggy Fisher shows you how to manipulate sets of data, write proofs and truth tables, analyze data sequences, and visualize data using graph theory. Challenges at the end of every chapter allow you to test your knowledge. By the end of the course, you should be able to make the leap from theory to using discrete math in practice: saving time and resulting in code that's cleaner and easier to maintain in the long run. - Real-world discrete math - Objects as sets - Set notation and operations - Standard machine language (SML) setup - Working with data types, strings, and functions in SML - Analyzing data sequences - Writing truth tables - Identifying and evaluating predicates - Validating arguments - Writing proofs: subset, conditional, and biconditional proofs - Visualizing data with graphs - Advanced discrete math techniques Skill Level Intermediate Programming Foundations: Design Patternswith Elisabeth Robson2h 19m Intermediate 1. Discrete Math Uses 3. Setting Up SML 4. Analyzing Data Sequences 5. Effective Arguments and Defensible Decisions 6. Proofs Made Easy 7. Advanced Discrete Math Topics - Mark as unwatched - Mark all as unwatched Are you sure you want to mark all the videos in this course as unwatched? This will not affect your course history, your reports, or your certificates of completion for this course.Cancel Take notes with your new membership! Type in the entry box, then click Enter to save your note. 1:30Press on any video thumbnail to jump immediately to the timecode shown. Notes are saved with you account but can also be exported as plain text, MS Word, PDF, Google Doc, or Evernote.

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# How do you find all the factors of 28? Dec 15, 2016 2,14,4,7,1,28 #### Explanation: Find biggest and smallest factor Your biggest factor is the number itself, $28$. With the biggest number, you also have the smallest factor, $1$. For the rest of the factors; Divide the number by $2$. $\frac{28}{2} = 14$. Hence you should be looking for factors no bigger than $14$. You already have $14$ and $2$. Find factors of smaller factors What is a factor of $14$? $7$! If $7$ is a factor of $14$, it should be a factor of $28$. If so, what times $7$ becomes $28$? 4! Every factor has its pair If you've found a factor, ie: 2, It should have its pair. Just divide the number by 2. If a factor is 7, divide 28 by 7 and you get 4.

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1. Really Really Urgent I REALLY NEED HELP WITH THESE TWO PROBLEMS, please someone help solveeeeee i'm begginggg you, I keep getting them wrong everytime I do them How would I multiply this problem? the small numbers are twos http://img397.imageshack.us/img397/7599/math13ep.jpg How would I perform the indicated operation on this problem? http://img397.imageshack.us/img397/9062/math20me.jpg 2. For number 2: It's just a division/cancellation problem. $\frac{a^{2}-a-30}{a+5}$ Factor the numerator and you will quickly see the cancellations. If you're having a rough time factoring, ask yourself, what 2 numbers when multiplied equal -1 and when added equal -30 (-6)(5)=-30 and -6+5=-1 $a^{2}-6a+5a-30$ $(a^{2}-6a)+(5a-30)$ $a(a-6)+5(a-6)$ $(a+5)(a-6)$ Now, you finish?. 3. Originally Posted by galactus For number 2: It's just a division/cancellation problem. $\frac{a^{2}-a-30}{a+5}$ Factor the numerator and you will quickly see the cancellations. If you're having a rough time factoring, ask yourself, what 2 numbers when multiplied equal -1 and when added equal -30 (-6)(5)=-30 and -6+5=-1 $a^{2}-6a+5a-30$ $(a^{2}-6a)+(5a-30)$ $a(a-6)+5(a-6)$ $(a+5)(a-6)$ Now, you finish?. how would I finish? I'm not trying to act stupid would it be A - 6 for the answer? 4. You ought to go see your teacher if you're still lost from this point on. Here's all is left: $\frac{(a-6)(a+5)}{a+5}$ Now cancel what needs cancelled and that's it. 5. Originally Posted by galactus You ought to go see your teacher if you're still lost from this point on. Here's all is left: $\frac{(a-6)(a+5)}{a+5}$ Now cancel what needs cancelled and that's it. Its homeschool and its self-taught, I'm going into a normal school next year, I Just need these two problems and im finished with this grade 6. Originally Posted by happyboy how would I finish? I'm not trying to act stupid would it be A - 6 for the answer? Yes, you got it! . You updated your post while I was writing mine. Anytime you see a division problem set up like the one you have, rewrite it the way I did and see whether or not anything factors. If it does, they'll more than likely be some cancellations and you'll be on your way. 7. Originally Posted by galactus Yes, you got it! . You updated your post while I was writing mine. Anytime you see a division problem set up like the one you have, rewrite it the way I did and see whether or not anything factors. If it does, they'll more than likely be some cancellations and you'll be on your way. I just have one more problem and i'm done, I don't get how to do the other one at all tho 8. Originally Posted by happyboy I just have one more problem and i'm done, I don't get how to do the other one at all tho I don't understand question #1 do you want us to factor $3x^2-2xy+2y^2$? 9. Originally Posted by Quick I don't understand question #1 do you want us to factor $3x^2-2xy+2y^2$? I don't understand it either, the x - y under it is apart of the problem too. and all it says is to multiply 10. Originally Posted by happyboy I don't understand it either, the x - y under it is apart of the problem too. and all it says is to multiply So you want us to do $\left(3x^2-2xy+2y^2\right)\times\left(x-y\right)$? Alright, write it out... $\left(x-y\right)\left(3x^2-2xy+2y^2\right)$ use the distributive property: $\left(3x^2\left(x-y\right)-2xy\left(x-y\right)+2y^2\left(x-y\right)\right)$ use the distibutive property again: $3x^2\left(x\right)-3x^2\left(y\right)-2xy\left(x\right)+2xy\left(y\right)+2y^2\left(x \right)-2y^2\left(y\right)$ multiply: $3x^3-3x^2y-2x^2y+2y^2x+2y^2x-2y^3$ group like terms: $3x^3-(3x^2y+2x^2y)+(2y^2x+2y^2x)-2y^3$ add/subtract: $\boxed{3x^3-5x^2y+4y^2x-2y^3}$ I think that's the answer you want (if I understand the question correctly) 11. Originally Posted by Quick So you want us to do $\left(3x^2-2xy+2y^2\right)\times\left(x-y\right)$? I'm guessing that is it, all the problem said was multiply and x-y was under it. this is a really confusing problem

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After reading Dr. Seuss’s ABC book, create a silly, creature-filled alphabet class book of your own. Print and trim the mini letter tiles. (I used a Seuss-font!) Toss them into a Seuss hat. Students choose one, glue it to their page and think of a creature that starts with that letter. should be made up creatures like the ones that Dr. Seuss thinks of, so remind students to be really creative. Younger students can simply name their animal and draw a picture. Encourage and challenge older students to make a rhyming sentence, with plenty of tongue-twisting alliteration that Dr. Seuss is famous for. The packet includes a page for students to write on, as well as one for the teacher; plus a sample page, and 26 Seuss-font letter tiles.

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1. Recycling conserves resources Using recycled content in manufacturing products reduces reliance on virgin resources. The European Recycling Council3 says, “Paper recycling is perceived by the public as being the most effective way to reduce environmental impacts of using paper. For the industry, recycled fibres are an indispensable source of raw materials, supporting industry’s resource efficiency.” Recycling one tonne of paper can save around 17 trees. 2. Recycling saves energy Recycling paper can save around 60% of the energy required to manufacture from virgin raw materials4. Not only does the process of procuring virgin resources require significant energy and labour, it can also result in other negative impacts like water or air pollution, and localised impacts on communities and ecosystems5. Recycling one ton of paper can result in 4,100KWh of energy savings6.

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1. ## Seating arrangements. quite confused with this particular question... How many ways can 9 students(5 girls and 4 boys) be seated in a row if there are no restrictions and if the first 3 must be boys? i think its the difference in number of boys and girls that is confusing me 2. When there is no restriction, the possible number of ways is simply 9!. When the first 3 must be boys, these 3 boys can sit in 3! ways and the rest 6 can sit in 6! ways. Hence the total number of possible ways in this case is 3!*6!. 3. ok i used what you showed and got farther into the question, but am not stuck at alternating seating girl then boy AND boy then girl? would it be 5!*4!*9!? does it change if going from girl > boy instead of boy >girl? 4. Hello, slowlearner! How many ways can 5 girls and 4 boys be seated in a row if the first 3 must be boys? Select three of the four boys: . ${4\choose3} \,=\,4$ choices. Seat them in the first three chairs: . $3! \,=\,6$ ways. Seat the other six children: . $6! = 720$ ways. Therefore, there are: . $4\cdot 6\cdot 720 \:=\:17,\!280$ ways. 5. How would i find out how many ways the 5 girls and 4 boys could be seated if the seating must alternate girl then boy? im quite confused here, is it 5!4!(9*8*7*6*5)? that is just my shot in the dark, im not quite sure how to solve this scenario 6. Originally Posted by slowlearner How would i find out how many ways the 5 girls and 4 boys could be seated if the seating must alternate girl then boy? im quite confused here, is it 5!4!(9*8*7*6*5)? that is just my shot in the dark, im not quite sure how to solve this scenario Note that the way you have written the question the row must begin with a girl. So the answer is $(5!)(4!)$. Sit a girl in ever other seat, then put a boy between them. 7. thank you! so if you were to seat it boy then girl then it would just be (4!)(5!)? or can this not be done since it would end up being girl next to girl at the end due to only 4 boys? 8. Originally Posted by slowlearner thank you! so if you were to seat it boy then girl then it would just be (4!)(5!)? or can this not be done since it would end up being girl next to girl at the end due to only 4 boys? But then you have two girls sitting together at the last two seats. Do you allow that? 9. thats what i was asking myself :P its got me quite confused. probability and statistics have been killing me all semester! 10. Originally Posted by slowlearner thats what i was asking myself :P its got me quite confused. probability and statistics have been killing me all semester! It is not "probability and statistics have been killing" you but the meaning of the language. It says that the girls and boys must alternate. So it got be $gbgbgbgbg$. 11. i agree! i think it could possibly be 6! * 4! * 3 * 2 * 1? 12. Originally Posted by slowlearner i agree! i think it could possibly be 6! * 4! * 3 * 2 * 1? That is wrong. 13. Originally Posted by Plato That is wrong. can you elaborate? 14. Originally Posted by slowlearner can you elaborate?

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# Solving a trig equation? How would I solve the following equation? $$1+ \sin (x)=2 \cos(x)$$ I am having difficult with it. - We have $1 + \sin(x) = 2 \cos(x)$. Recall that $1 - \sin^2(x) = \cos^2(x)$. Hence, we get that $$(1 + \sin(x))(1- \sin(x)) = \cos^2(x)$$ i.e. $$2 \cos(x) (1 - \sin(x)) = \cos^2(x)$$ If $\cos(x) \neq 0$, then we get that $$1 - \sin(x) = \dfrac{\cos(x)}2$$ Hence, have \begin{align} 1 + \sin(x) & = 2\cos(x)\\ 1 - \sin(x) & = \dfrac{\cos(x)}2 \end{align} This gives us that $2 = \dfrac52 \cos(x) \implies \cos(x) = \dfrac45$. This gives us $\sin(x) = \dfrac35$. Hence, we get one possible solution as $$x = 2n \pi + \theta$$ where $\sin(\theta) = \dfrac35$ with $0 < \theta < \dfrac{\pi}2$. If $\cos(x) = 0$, then we need $\sin(x) = -1$. Hence, $x = 2n\pi + \dfrac{3 \pi}2$. - ,according to my method, $x=\arccos \frac45=\arcsin\frac45$ is another solution. – lab bhattacharjee Dec 15 '12 at 5:19 @labbhattacharjee Try plugging your answer in... $1 + \frac{4}{5} \neq 2 \cdot \frac{4}{5}$, so it looks like something went wrong. – andybenji Dec 15 '12 at 5:51 @labbhattacharjee Thanks. Yes. the other solution is $x = \arcsin(3/5) = \arccos(4/5)$. – user17762 Dec 15 '12 at 6:03 @andybenji, sorry there was terrible typo--it should be $\arcsin\frac35=\arccos\frac45$ as Marvis has noted. – lab bhattacharjee Dec 15 '12 at 6:58 Hint: Square both sides, and express $\cos^2 x$ in terms of $\sin^2 x$. You will get a quadratic equation in $\sin^2 x$. After finding the solutions of this equation, make sure to substitute into the original equation to check whether they are indeed roots of the original equation. The squaring process could produce "spurious" roots. (It happens not to, but checking is cheap.) - I have a quick question I squared both sides and got. 1+2sin(x)+sin^2(x)=4cos^2(x) so I made 4cos^2(x) into 4-sin^2(x) Is this correct. – Fernando Martinez Dec 15 '12 at 4:32 It would be $4-4\sin^2 x$. The equation then becomes $5\sin^2 x+2\sin x-3=0$. You can then use the Quadratic Formula or factor as $(5\sin x-3)(\sin x+1)$. – André Nicolas Dec 15 '12 at 4:48 Oh thanks very much. – Fernando Martinez Dec 15 '12 at 4:53 This may be the hard way, but if you let $q=\tan(x/2)$, then $\sin x=2q/(1+q^2)$, and $\cos x=(1-q^2)/(1+q^2)$. - Let me take a stab at this. I had to double check a few identities. First, we use the identities $\cos(\frac\pi 2-x)=\sin x$ and $\sin(\frac\pi 2-x)=\cos x$. $$1+\cos(\frac\pi 2-x)=2\sin(\frac\pi 2-x)$$ $$\frac{1+\cos(\frac\pi 2-x)}{\sin(\frac\pi 2-x)}=2$$ Next, we use the half-angle identity $\tan\frac\theta 2=\frac{\sin\theta}{1+\cos\theta}$. $$\cot(\frac\pi 4-\frac x2)=2$$ $$\tan(\frac\pi 4-\frac x2)=\frac12$$ From here, take the inverse tangent and the rest is algebra. - $\cos\left(\frac\pi4-\frac x2\right)$ can also be $0$ – lab bhattacharjee Dec 15 '12 at 9:38 @labbhattacharjee I see how I might have lost some roots by not making sure $\cos x$, or $\sin(\frac\pi 2-x)$ if you prefer, was 0 before dividing by it. Did I lose roots in any other step somehow? – Mike Dec 15 '12 at 18:22 We can try to avoid squaring to avoid tests for extraneous root. $$2\cos x-\sin x=1$$ Putting $r\cos\theta=2,r\sin\theta=1$ where $r>0$ Squaring and adding we get $r^2=2^2+1^2\implies r=\sqrt5$ On division, $\tan \theta=\frac12$ where $0<\theta<\frac\pi2$ as $\sin\theta,\cos\theta>0$ we get $$\cos\theta\cos x-\sin\theta\sin x=\sin\theta$$(cancelling $r$ in either sides ) or, $\cos(x+\theta)=\cos(\frac\pi2-\theta)$ So, $$x+\theta=2n\pi\pm\left(\frac\pi2-\theta\right)$$ where $n$ is any integer. Taking $'-'$ sign, $$x+\theta=2n\pi-\left(\frac\pi2-\theta\right)\implies x=2n\pi-\frac\pi2$$ Taking $'+'$ sign, $$x+\theta=2n\pi+\left(\frac\pi2-\theta\right)\implies x=2n\pi+\frac\pi2-2\theta=2n\pi+\frac\pi2-2\arctan \frac12$$ Now, we know $\cos2y=\frac{1-\tan^2y}{1+\tan^2y},\sin2y=\frac{2\tan y}{1+\tan^2y},\tan2y=\frac{2\tan y}{1-\tan^2y}$ If $\arctan \frac12=y,\tan y=\frac 12$ $$\cos2y=\frac35,\sin2y=\frac45,\tan2y=\frac43$$ So, $$2y=2\arctan \frac12=\arccos\frac35=\arcsin\frac45=\arctan \frac43$$ So, $$x=2n\pi+\frac\pi2-\arccos\frac35=2n\pi+\arcsin\frac35$$ as $\arcsin z+\arccos z=\frac12$ for $-1\le z\le1$ Similarly, $$x=2n\pi+\arccos\frac45=2n\pi+\arccot\frac43$$ - It is not clear we should avoid squaring, since checking is cheap. A good reason to avoid squaring is if degree is increased beyond the comfortable. But degree $2$ is comfortable. – André Nicolas Dec 15 '12 at 5:48 @AndréNicolas, squaring is easing the calculation a lot for $A\cos x+B\sin x=C$ w/o introducing much intricacies. There is no extraneous root as such,right?. We just need to identify the quadrant of $x$ by getting the sign of $\cos x$ from the given equation, if we have the quadratic equation in $\sin x$ or vice versa. Is $\arccot$ not a valid latex? – lab bhattacharjee Dec 15 '12 at 9:44 @labbhattacharjee Apparently not. Try "\mathrm{arccot}" – Thomas Dec 15 '12 at 10:32

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About 60% of the adult human body is water. However, we constantly lose this, usually through sweat and urine. Although opinions differ about how much water we should consume daily, the recommended intake is eight 8-ounce glasses a day. Drinking water has many health benefits. Apart from preventing dehydration, water helps improve brain function, relieve constipation, prevent hangovers, and help with weight loss. As important as water is, however, it’s surprising that a number of people still don’t have ready access to clean drinking water. According to the World Health Organization, 423 million people worldwide get their water from unprotected wells and springs. In the United States, a majority of Americans get their tap water from public water systems. Protecting Your Water The problem with some plumbing systems, especially older ones, is that contaminants can mix with the water. Some old pipes still contain lead. Consumption of lead can affect health negatively and lead to such conditions as behavioral problems and learning disabilities in children. Children six years old and below are most at risk because their brains are still developing. Although not all pipes contain lead, it can still get through the pipes if these are not treated with protective barrier materials. To get safe tap water, water treatment has to meet certain standards such as the use of protective barrier materials and have non-metallic potable water materials. According to industry professional Zebron, barrier materials used for public water systems must meet the NSF/ANSI 61. Did You Know That…? Apart from those mentioned above, here are some other facts about drinking water: - Only 1% of the world’s water is drinkable. Nearly 97% of the world’s water is salty or otherwise undrinkable while the other 2% is locked in ice caps and glaciers. - In 2010, the United Nations declared that access to safe and clean drinking water is a human right - Humans can live for a month without food but only a week without water - Over 89 billion liters of bottled water are sold each year - The United Nations declared March 22 as World Water Day The need for safe and clean drinking water is undeniable; yet far too many people still don’t have access to it. We all must do our part, so everyone can access this basic right.

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The Weather Channel and Climate Central show what sea level rise would mean for Katrina's storm surge in 2065. U.S. power plants emitted less carbon dioxide in April than at any point in the prior 27 years. Katrina helped galvanize hurricane-climate change research, and 10 years later, strides have been made. Alyson Kenward talks danger days in Charleston, W.Va. on West Virginia Morning radio. The role climate change may have played in Hurricane Katrina is a difficult question to answer, even 10 years later. Ten years after Katrina, New Orleans is dealing with sinking land as sea levels rise. Wildfires are making it hard to enjoy the scenic vistas National Parks are famous for. The White House on Monday announced policies designed to help homeowners act on climate change. After a record warm July, 2015 looks to beat 2014 as the warmest year on record. A new study pegs climate change as a bully that's come for California's water. Scientists may have overestimated China's carbon dioxide emissions since 2000. Daily record highs are outpacing record lows across the U.S. due to climate change. The EPA has proposed new rules limiting methane emissions from new fracked oil wells. New West Antarctica research shows that IPCC sea level rise estimates will have to be revised upward. Rising seas, stronger storms and coastal development are taking their toll on sandy shorelines. Warming temperatures are about to push U.S. cities into a new regime where danger days are more common. Researchers have recorded worldwide rapid rises in meltwater and alarming rates of glacial retreat. Record heat continues across much of the West, helping to deepen drought and fuel wildfires. Hawaii could see more hurricanes in its vicinity thanks to the influence of global warming. As renewable energy growth projections rise, the EPA is raising its climate expectations.

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Calculate Length of Side using Trigonometry More Lessons for Grade 9 Math Videos, worksheets, solutions, and activities to help students learn how to find the missing side of a right triangle given an angle and side using trigonometry. Right Triangle Trigonometry This video explores basic right triangle trigonometry used to locate a missing side in a right triangle. Using Cosine ratio to find side length Basic introduction to sin, cos and tan including finding the missing side when given one side and one angle and a right angle. Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations. You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

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If only prejudices, stereotypes, and discrimination were totally diminished from this world, needless to say that peace is something that we could always breathe in and out easily, effortlessly. If only everyone did not overemphasizing differences, everybody will live in harmony by putting aside all differences and paid more attention the commonality and similarity that tightens the values of humanity. We often hear and imagine this will come true. Yet, the world is filled with discrimination in almost every aspect in human life. One can be discriminated because of having different religion, sexual orientation, race, ideology, physical ability, and so forth. The discrimination practices are still a problem for the reason as follows: “It is problematic because it categorizes and classifies, serves to no positive purposes and mostly used negatively. Though the law protects the people and we are to be treated equally, it is the people’s minds that won’t change and what is causing this problem to still go on” (Rothernberg, 2005, p. 1)”. Regarding this problematic issue, it is important to understand how discrimination can occur. Discrimination occurs when prejudice and stereotypes take place. At first, prejudice is initiated by having interpersonal judgment, then, it creates certain stereotypes related to the judgment. Putting prejudice and stereotypes into a practice; that is when discrimination takes over as what Blumenfeld states in Prejudice and Discrimination that discrimination is the result of prejudiced feelings or beliefs that move into the realm of behavior (Adams, et al, 2000). Nonetheless, it is inconceivable to not having prejudice or stereotypes because we live in a society where it “influence behavior and have enormous power to reward or penalize its member to whom it reinforces existing prejudice and discrimination” (Adams, et al, 2000, p. 22). Thus, avoiding prejudice is as hard as putting aside our subjectivity. It is this subjectivity that often causes misinterpretation and misjudgment to which it sometimes distracts our objectivity. Thus, both of our objectivity and subjectivity are challenged in a more complex way as we live in global world nowadays. Globalization and its technology have made everything borderless. New information that carries within it diverse ideas, perspectives, ideologies, concepts, knowledge, and insight torrentially come from every direction. Globalization nowadays enables us to see what is different outside our own “bubble”. Furthermore, globalization has brought competition, to which every country in the world is trying to be the best among others. Competition, it is conducted fairly, will create healthy environment. Yet, it often conducted unfairly, by practicing and implementing dirty tricks and politics that bring casualties, immolation and victims on the table. To be able to implement this, often the practices of discrimination is taken as an attempt to create the sense exclusiveness. Indubitably, within this environment, having prejudice and stereotyping is foreseeable because it is derived from the suspicion towards one and another – fearing that one might threaten and destroy the exclusiveness that has been maintained so far. Thus Similar to discrimination practices in the nation – level, its practices in societal and personal level has also tremendously called for direct causality, victims, and immolation. At this point, the simplest way to describe this pattern is by introducing two major “actors” – the oppressor and the oppressed. In addition, there are two major acts introduced along the way – superior versus inferior. Beforehand, it is important to initiate the discussion with the core of discrimination – identity. Why? It is because this is where everything is started. By understanding what it means to have identity and how identity is constructed, it will provide the framework of understanding discrimination as a whole rather than just partially. When we learn to see things as a whole, we learn as well to balance our subjectivity and objectivity that will help us to have a better understanding. Considering such challenging issues in discrimination, this paper suggests the way of understanding the issues in diversity related to discrimination. This paper will discuss about intertwined aspects in discrimination: identity, oppression, superiority – inferiority, prejudice, stereotyping, and deculturalization. In addition, this paper is focused more on how perceiving and understanding discrimination from bottom-up viewpoint rather than top-down way of thinking. Intertwined aspects in identity Understanding discrimination from bottom up perspective, it is important to take an account of how it is preliminary constituted – identity. Our identity is made up of many traits, values, and characters to which each possesses within us many different selves. When we were born, we are just like a frying pan waiting for “a recipe” to be cooked. A recipe refers to a set of culture that carries values, norms, traits, behavior, attitudes, customs, beliefs, and knowledge that shape our personal identity as an individual. Along the way in our personal growth, our identity has been embedded into layers that further define who and what we are. According to Tatum in The Complexity of Identity: Who Am I, identity is described as a complex and multidimensional because “it is shaped by individual characteristics, family dynamics, historical factors, and social and political contexts…other people are the mirror in which we see ourselves” (Adams, et al, 2000, p. 9). Accordingly, our prior multidimensional knowledge has given us the sense to how we perceive ourselves and others based on designated perspectives given by, say for example, our family and the society where we live. In that case, we have been told to some extent how to see others based on our designated standard. Tatum (2000) asserts that our designated standard is multidimensional to which is “mediated by other dimensions of one-self: male or female; young or old; wealthy, middle class, or poor; gay, lesbian, bisexual, transgender, or heterosexual; able-bodied or with disabilities; Christian, Muslim, Jewish, Buddhist, Hindu, or Atheist” (p. 9). Such designated standard generates other standards considered as otherness to other standards that do not fit our standard. Regarding this, if a person starts to consciously aware about their standards that give the sense of who they are, they will start to have the sense of identity. To some extent, when we encounter “otherness”, we tend to say that they are abnormal rather than just different. Moreover, if other identities arouse the feeling of discomfort, the simplest and fastest anticipation is to oppose it with justification through prejudice and stereotyping. Since it is the simplest and fastest anticipation, there is a lack of experience and direct interaction with the otherness to which a mere subjectivity takes over. According to Jaime Wurzel (1986) in Blumenfeld (2000), there are four functions of prejudice: the utilitarian, the protective, the value-expressive, and the cognitive function. It is very interesting to examine these functions of prejudice to “understand what role prejudice plays in human interactions” (Adams et al, 2000, p. 25). Adding to that of the functions, “the utilitarian refers to its function as gaining rewards, and avoiding punishments; the protective refers to its functions as a psychological shield from their own inadequacies and fears; the value-expressive refers to its functions as retaining their own supposed values of a pure social stock; and the cognitive refers to its functions as relating to others and digesting new information” (Adams et al, 2000, p. 26). To challenge our own subjectivity and countering prejudice, it is highly likely to step out from the given prior knowledge to really experience the differences and otherness in a different way where we need to actually engage in actual experience, communication, and interaction with the otherness. When we are able to experience otherness in a different way that we build our own perception based on the actual experience and knowledge, we will be able to have better understanding and compromise with differences in a more moderate way. Yet, if we have not yet been able to do so, it is rather hard to reach the level of understanding where willingness and sincerity take part. Inasmuch as discrimination manifests itself in that people are “pre-judged” based on superficial characteristics, we must honestly conclude that all people “suffer” from this on various levels. When we don’t know an individual well, we consciously or unconsciously begin to characterize him or her based on what we see. Again, this is due to our ignorance of the person’s real character and personality. We will form opinions, often based along stereotypical lines: “all people of such and such race are. . .” We can fill in the blanks with such expectations that certain races are intellectually superior, others are full of greediness, another is more artistically or athletically inclined, still another has members who are apt to be dishonest and etc. These ideas have been formed from society, media, and our own upbringing. The issue of discrimination has always become a problematic discussion. The problematic dimension is rooted by complexity that is often overlapping and confusing. What has made discrimination perpetuated is the societal and cultural standard of the dominated group. Furthermore, the role of society in establishing pattern and setting is strongly seen as the “moderator” for discrimination practices – it enables, preserves, and sustains the standards and parameter for values and norms that should be followed, obliged, and implemented. As a result, for those who do not “fit” in the current setting will be considered as outsider whom will experience discrimination in the most direct ways. Starting from the early concept of classifying human based on race to which it is implied that there is hierarchical division within racial system itself. The archaic racial system has evolved into further forms of discrimination that nowadays has become daily life practices – consciously or unconsciously. The impact of this classification based on hierarchical racial division is the inevitable practices of discrimination towards an individual or a group. Discrimination does not only reach the area of race but also almost every part of humanity. In Readings for Diversity and Social Justice, there are terms for “isms” mostly associated with discrimination practices nowadays: racism, sexism, misogyny, ethnocentrism, ageism, ableism, xenophobia, anti-Semitism, religious prejudice, chauvinism, classism, heterosexism, and homophobia (Adams, et al, 2000). The practices of discrimination have become a humiliation in the history of human civilization – a vicious cycle of a never-ending debate about how different we are, as opposed to put aside our differences and finding what is similar between one and another. Furthermore, discrimination is conducted based on negative prejudgment, opinion, or hate towards an individual and or group from certain racial category. Subsequently, the negative prejudgment, opinion, or hate creates what is termed as prejudice and stereotype. What is more, discrimination practices have never been a simple matter. It is constructed by so many complicated variables that intertwined one and another. The description of discrimination is clear in theory; but, it is so hard to unpack this matter in the real life because to some extent, it is kept hidden subliminally. To illustrate, people can deny having biases, yet, their behavior displays their biases. A good example of this is the concept of white privilege. Being born as white, there is a propensity to take white privilege for granted. Being white is subconsciously unaware of the ascribed status to which for so long has given comfort and safety in the society as what has described in White Privilege that: “We are all much more likely to disregard attributes that seldom produce a ripple than are those that subject us to discomfort. For most whites, race – or more precisely, their own race – is simply part of the unseen, unproblematic background” (Rothernberg, 2005, p. 17). To put it another way, white privilege is a birthright privilege, “a set of advantages one receives simply by being born with features that society values especially highly” (Rothernberg, 2005, p. 18). Additionally, the “invisible” existence of white privilege is “reinforced as whites cultivated the practice of denying the subjectivity of blacks (the better to dehumanize and oppress), of relegating them to the realm of the invisible” (Rothernberg, 2005, p. 21). What can be learned so far from white privilege is how domination foster and perpetuate discrimination and further create oppression that deculturalize humanity values. Oppression – Superiority versus Inferiority Admittedly, when discrimination has served as a means for dominant group to maintain its control or power, forms of oppression is emerged. There is always victims cause by this oppression – the inferiority and the oppressed. As Pincus (2000) described this situation in United States as an illustration where superiority refers to dominant group named as white, especially white males, whereas the inferiority refers to people of color and women as well as non-Christian religious groups like Jews and Muslims. The oppression has become structurally mechanized in such a way that foster what is termed as collective norms. In this case, the mechanism of oppression is further described as follows: “Oppression refers to systemic constraints on groups that are not necessarily the result of the intentions of a tyrant. Oppression in this sense is structural, rather than the result of a few people’s choices or policies. Its causes are embedded in unquestioned norms, habits, and symbols, in the assumptions underlying institutional rules and the collective consequences of following those rules” (Adams et al, 2000, p. 36) As oppression mechanized both in personal and societal level, there are five faces of oppression as determined by Young in Reading for Diversity and Social Justice (2000): exploitation, marginalization, powerlessness, cultural imperialism, and violence. These forms of oppression carry its own central issue in which it characterizes the degree of discrimination in its mechanism. The central issue in exploitation is the steady process of taking advantage by one group over another. Compared to exploitation, marginalization is likely to be the most dangerous form of oppression because it prevents a person or a group from useful participation in social life and often leads to severe material deprivation and even extermination. Another case is powerlessness. In powerlessness, the injustices are often associated with inhibition of capacity building or one’s capability development, lack of decision-making power, and disrespectful treatment. Alike exploitation, marginalization, and powerlessness in which the root is a concrete power of structural and institutional in relation to others – a direct impact of master and labor, cultural imperialism is rather different. It is different because of the invisibility of impact where the oppressor’s culture has a subliminally indirect impact to the extent where it is unnoticeable, yet the effect is indeed undermining. Apart from the four-forms of oppressions, violence is rather recognizable because it appears to emphasize on direct victimization and violation, such as rape, beating, killing, and harassment (Adams et al, 2000). Referring to five forms of oppression above, the next question will be how to disrupt the cycle of oppression and discrimination as an attempt to interlock the practices of discrimination and to lessen the pressure of oppression. To really arouse willingness of interrupting the cycle of discrimination, we should be able to confront ourselves, our status quo in which is rather hard to make judgment towards ourselves rather than to anybody else. The longer we ignore issues around discrimination, the larger they will become. Ignoring them will not make them go away. The longer inequity exists, the longer we are losing voices that–had they been given equal opportunity–may have helped to solve the other problems we’re facing. In fact, discrimination has deculturalized humanity values and rights to be treated as equal. The concept of deculturalization is clearly portrayed as “cultural genocide where it demonstrates how cultural prejudice, racism, and religious bigotry can be intertwined with democratic beliefs – it combines education for democracy and political equality with cultural genocide” (Spring, 2004, p. 3). It is important to understand ourselves as having multiple identities where we can be part of both the dominated group and the minority. By having understanding towards our identity, we are encouraged to have positive perspective towards our own identity and the concept of otherness in our life. To do so, we are encouraged to engage in an open-minded dimension of communication and interaction to be part of our own-constructed experience in order to rebuild our understanding. Knowing that discrimination appears in almost every aspects in our life as a human, it is important to always be aware and conscious about its form, such as exploitation, marginalization, powerlessness, cultural imperialism, and violence. Interlocking the pressure of oppression as such is like peeling layer by layer to which are interconnected and precipitated by history, time, and life experiences. In doing so, the objectivity and subjectivity to see this matter should go along the way where it should not be overlapping. Objectivity is needed to encounter our subjectivity that might lead us to resentment, skepticism, or even cynicism caused by our own experience regarding this matter. Wini, Athens, Ohio. 2008 Adams, Maurianne., et al. (2000). Readings for diversity and social justice. New York: Routledge. Rothernberg, Paula S. (2005). White Privilege: essential readings on the other side of racism. New York: Worth Publishers Spring, Joel. (2004). Deculturalization and the struggle for equality: A brief history on the education of dominated cultures in the United States. NewYork: The McGraw-Hill Companies, Inc.

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Notes on Unitary Method | Grade 7 > Compulsory Maths > Unitary Method | KULLABS.COM Notes, Exercises, Videos, Tests and Things to Remember on Unitary Method Please scroll down to get to the study materials. • Note • Things to remember • Videos • Exercise • Quiz Unitary Method It is a technique in mathematics for solving a problem finding the value of the single unit and then finding the necessary value by multiplying the single unit value. If the quantities are in direct proportion, unit value is obtained by division. Similarly, if the quantities are in inverse proportion, unit value is obtained by multiplication. For examples, Suppose the cost of 5 pens is Rs. 50. Then, the cost of 1 pen is Rs. $$\frac{50}{5}$$ = Rs. 10 Here, 1 pen is the unit of quantity and Rs. 10 is the unit cost. The number of pens and their cost is in direct proportion. Again, suppose 4 men can finish a work in 6 days. Then, 1 man can finish the work in 4 × 6 days = Rs. 24 days. Here, the number of men and their working days are in inverse proportion. Time and Work In less time we do less amount of and in more time we do the greater amount of work. So, Time and amount of work done are in direct proportion. In this case, the amount of work done in unit time is obtained by division. For example, In 8 days, Hari can do 1 work. In 1 day, Hari can do $$\frac{1}{8}$$ part of work. $$\frac{1}{8}$$ parts of work are done in 1 day. Similarly, in 2 days, Hari can do$$\frac{2}{8}$$ =$$\frac{1}{4}$$ parts of the work In 3 days, Hari can do$$\frac{3}{8}$$ parts of the work. In 4 days, Hari can do$$\frac{4}{8}$$ =$$\frac{1}{2}$$ parts of work and so on. •  If the quantities are in direct proportion, unit value is obtained by division. • If the quantities are in inverse proportion, unit value is obtained by multiplication. •  The amount of work done in unit time is obtained by division. . #### Click on the questions below to reveal the answers Solution: The cost of 12 m of clothes = Rs 600 The cost od 1 m of clothes = Rs $$\frac{600}{12}$$ = Rs 50 The cost of 8 m of clothes = 8 × Rs 50 = Rs 400 So, the required cost of 8 m of clothes is Rs 400. Solution: Here, the length of the floor (l) = 14 m the breadth of the floor (b) = 8.5m ∴ Area of the floor = l × b = 14m × 8.5m = 119 m2 Now, the cost of carpeting  119 m2 is 119 × Rs 75 = Rs  8925 So, the required cost of carpeting the floor is Rs 8925. Solution: Here, Rs 2 is the tax for the income of Rs 100 Re 1 is the tax for the income of Rs $$\frac{100}{2}$$ = Rs 50 Rs 525 is the tax for the income of Rs 525 × 50 = Rs 26250 So, the required income of the man is Rs 26250 Solution: Here, In 20 days, 7 workers build a wall In 1 days, 20 × 7 workers build the wall In 14 days, $$\frac{20 × 7}{14}$$ workers build the wall = 10 workers build the wall ∴ The required number of workers to be added = 10 - 7 = 3 workers Solution: Here, In 8 days, Hari can do 1 work In 1 day, Hari can do $$\frac{1}{8}$$ parts of the work. In 6 days, Hari can do $$\frac{6}{8}$$ parts of the work = $$\frac{3}{4}$$ parts of the work. Now, Remaining parts of work = (1 - $$\frac{3}{4}$$) = $$\frac{1}{4}$$ parts  of the work So, Gopal finished $$\frac{1}{4}$$ part of the work 0% Rs 843 Rs 873 Rs 863 Rs 835 • ### If a bus covers 277.8 km in 6 hours, find the speed of the bus in km per hour. 56.4 km per hour 44.3 km per hour 36.54 km per hour 46.3 km per hour 9 men 11 men 10 men 12 men 11 days 12 days 10 days 9 days Rs 24450 Rs 27250 Rs 25350 Rs 26250 Rs 1000 Rs 875 Rs 925 Rs 750 (frac{2}{5}) (frac{3}{5}) (frac{1}{5}) (frac{4}{5}) • ### Sarmila can finish a piece of work in 12 days. How much work can she finish in 4 days? (frac{2}{6}) (frac{4}{12}) (frac{3}{9}) (frac{1}{4}) (frac{3}{5}) (frac{4}{5}) (frac{2}{5}) (frac{1}{5}) • ### Kumar can finish a piece of work in 12 days. How much work can he finish in 6 days. (frac{3}{4}) (frac{2}{6}) (frac{1}{12}) (frac{1}{2}) ## ASK ANY QUESTION ON Unitary Method No discussion on this note yet. Be first to comment on this note

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The curie (symbol Ci) is a non-SI unit of radioactivity, named after Pierre Curie, but probably also after Marie Curie. It was originally defined as 'the quantity or mass of radium emanation in equilibrium with one gram of radium (element)' but is currently defined as: 1 Ci = 3.7 × 1010 decays per second after more accurate measurements of the activity of 226Ra (which has a specific activity of 3.66 x 1010 Bq/g.) - 1 Ci = 3.7 × 1010 Bq = 37 GBq - 1 Bq ≅ 2.703 × 10−11 Ci ≅ 27 pCi While its continued use is discouraged by NIST and other bodies, the curie is still widely used throughout the government, industry and medicine in the United States and in other countries. The curie is a large amount of activity, and was intentionally so. According to Bertram Boltwood, Marie Curie thought that 'the use of the name "curie" for so infinitesimally small (a) quantity of anything was altogether inappropriate.' The typical human body contains roughly 0.1 μCi (14 mg) of naturally occurring potassium-40. A human body containing 16 kg of carbon (see Composition of the human body) would also have about 24 nanograms or 0.1 μCi of carbon-14. Together, these would have an activity of approximately 0.2 μCi or 7400 Bq inside the person's body. Curie as a measure of quantity Units of activity (the curie and the becquerel) also refer to a quantity of radioactive atoms. Because the probability of decay is a fixed physical quantity, for a known number of atoms of a particular radionuclide, a predicable number will decay in a given time. The number of decays that will occur in one second in one gram of atoms of a particular radionuclide is known as the specific activity of that radionuclide. The activity of a sample decreases with time because of decay. The rules of radioactive decay may be used to convert activity to an actual number of atoms. They state that 1 Ci of radioactive atoms would follow the expression: - N (atoms) × λ (s−1) = 1 Ci = 3.7 × 1010 (Bq) - N = 3.7 × 1010 / λ, where λ is the decay constant in (s−1). We can also express activity in moles: Here are some examples: |Isotope||Half life||Mass of 1 curie||Specific activity (Ci/g)| |232Th||×1010 years 1.405||9.1 tonnes||×10−7 (110,000 pCi/g, 0.11 µCi/g) 1.1| |238U||×109 years 4.471||2.977 tonnes||×10−7 (340,000 pCi/g, 0.34 µCi/g) 3.4| |40K||×109 years 1.25||140 kg||×10−6 (7,100,000 pCi/g, 7.1 µCi/g) 7.1| |235U||×108 years 7.038||463 kg||×10−6 (2,160,000 pCi/g, 2.2 µCi/g) 2.2| |129I||×106 years 15.7||5.66 kg||0.00018| |99Tc||×103 years 211||58 g||0.017| |239Pu||×103 years 24.11||16 g||0.063| |240Pu||6563 years||4.4 g||0.23| |226Ra||1601 years||1.01 g||0.99| |241Am||432.6 years||0.29 g||3.43| |14C||5730 years||0.22 g||4.5| |238Pu||88 years||59 mg||17| |137Cs||30.17 years||12 mg||83| |90Sr||28.8 years||7.2 mg||139| |241Pu||14 years||9.4 mg||106| |60Co||1925 days||883 μg||1132| |210Po||138 days||223 μg||4484| |3H||12.32 years||104 μg||9621| |131I||8.02 days||8 μg||125000| |123I||13 hours||0.5 μg||2000000| Radiation Related Quantities The following table shows radiation quantities in SI and non-SI units. |Exposure (X)||roentgen||R||esu / 0.001293 g of air||1928| |Absorbed dose (D)||erg•g−1||1950| |Activity (A)||curie||Ci||3.7 × 1010 s−1||1953| |Dose equivalent (H)||roentgen equivalent man||rem||100 erg•g−1||1971| |Fluence (Φ)||(reciprocal area)||cm−2 or m−2||1962| - Geiger counter - Ionizing radiation - Radiation exposure - Radiation poisoning - Radiation burn - United Nations Scientific Committee on the Effects of Atomic Radiation - Rutherford, Earnest (6 October 1910). "Radium Standards and Nomenclature". Nature 84 (2136): 430–431. - Delacroix, D (2002). Radionuclide and Radiation Protection Data Handbook 2002. RADIATION PROTECTION DOSIMETRY Vol. 98 No 1: Nuclear Technology Publishing. p. 147. - "SI units for ionizing radiation: becquerel". Resolutions of the 15th CGPM (Resolution 8). 1975. Retrieved 3 July 2015. - Nist Special Publication 811, paragraph 5.2. - Frame, Paul (1996). "How the Curie Came to Be". Health Physics Society newsletter. Retrieved 3 July 2015.

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A/an and one both refer to one thing and are used with singular countable nouns. We usually use one if we want to emphasise the number. Compare: It took just an hour to complete. It took just one hour to complete. We also use one with day when we are thinking of one particular day but we don’t say exactly which. eg. I’ll go to the doctor one day next week. We use one in phrases with one… other/another/the next eg. He moves from one job to the next with no plan for the furure. NB: We use ‘a’ with singular countable nouns in exclamations. eg. What a lovely day! Few, A Few, Quite, Quite a Few Quite a few = a considerable number eg. Quite a few people turned up at the free concert last night. A few = a small number eg. Peter and Jane only invited a few close friends to their wedding. Few = not many or not enough She has few friends. (formal) = She doesn’t have many friends. (informal) NB: little/a little is used with uncountable nouns in the same way as few/a few is used with countable nouns. We can make comparatives with fewer with countable nouns. We use less with uncountable nouns.

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# Sunday Function Well, it’s Gabriel, Gabriel playin’! Gabriel, Gabriel sayin’ “Will you be ready to go When I blow my horn?” – Cole Porter, Anything Goes The commenters in last week’s Sunday Function proposed an excellent idea for this week. As we did then, we’ll start simple and work up to it. Graph the curve f(x) = 1/x. Now take the horizontal axis and think of it as an axle, one that can rotate smoothly. Hook a motor onto it and spin it up to a nice fast clip. The graph of the function will trace out a surface in three-dimensional space. In the business, we call it a surface of revolution. It looks like this: Now I’ve truncated the plot at x = 4 for reasons of space. But in reality you can imagine that in fact the surface keeps going on forever in that direction in the shape of an ever-narrowing trumpet. That shape is called Gabriel’s Horn, and though I’m leaving off the actual explicit specification for the expression describing it, it’s still a function (of two parametric parameters or of y and z) nonetheless. There’s two things we might want to calculate. First, the volume contained in the trumpet. Imagine you have a penny, and you drop it into the trumpet such that it acts as a plug or a manhole cover. It’s perfectly perpendicular to the x-axis. The penny will have a radius of f(c) = 1/c, because at the particular point c that it sticks it clearly must have the same radius as our original function f(x) evaluated at c. What’s the volume of that penny? It’s the area, pi*r^2, times the thickness of the penny. Knowing that r = 1/x for whatever generic x the penny sticks at and calling the thickness dx, we have a volume (call it dV) for the penny of dV = pi/x^2 dx. So if we fill the entire trumpet with custom pennies of appropriately varying radii and add up their volumes, we’ll have the volume of the trumpet. Looks like a job for calculus! So despite being infinitely long, the trumpet has a finite volume. It’s a little counterintuitive, but it’s perfectly true nonetheless. What about the surface area? It’s a little harder to find the expression for surface area of a solid of revolution, but you can take my word that it’s this, with the third expression having our particular function plugged in: That integral looks fairly tricky to evaluate directly. I think it could probably be done, but we won’t have to. The term under the square root is bigger than 1, and so the 1/x term is a lower limit for the behavior of the integrand. But 1/x diverges, so the integral is infinite. Our infinitely long horn has infinite surface area. Infinite surface area and finite volume. That’s weird. As people like to say about this shape, you can fill it with paint, but you can’t paint it. It’s true, but it’s not so implausible as it sounds because physical paint isn’t just a pure 2d surface. Real paint has thickness. If you poured real paint into a real Gabriel’s Trumpet, eventually it would fail to keep coating the interior once it narrowed down to a thinner diameter than a paint molecule. Even Gabriel might think that’s pretty cool. March 15, 2009 That’s coolio that the surface area diverges while the volume converges. 2. #2 Miko March 15, 2009 It’s actually not too hard to evaluate the SA integral directly. Clear the denominators in the square root to get (1/x^3)sqrt(x^4+1), use integration by parts with u=sqrt(x^4+1), dv=1/x^3, and the resulting integral is a simple inverse trig substitution. You end up with indefinite integral = -sqrt(x^4+1)/(2x^2)+arcsin(x^2)/2. 3. #3 Matt Springer March 15, 2009 I believe that should be an arcsinh in your last term, but other than that you’re right. Thanks! 4. #4 jrpb March 15, 2009 Very interesting, thanks! 5. #5 Peter Morgan March 15, 2009 Your example of painting the surface could be expanded. For any finite thickness of paint on the outside of the trumpet, the volume of paint would be infinite. The inside of the trumpet is at distance 1/x from the axis. The outside of the paint lying on the trumpet would be described by a different equation, (1/x)+c. Hence, paint volume infinite. If we painted with more sensitivity to context, so that there would always be paint, but not overpoweringly so, the outside of the paint might be described by c/x, c slightly greater than 1, then the volume of paint would be finite (the difference between the volumes of two Gabriel’s horns). We could also model the outside of the paint by 1/(x-c), c slightly greater than zero, which would use a different volume of paint. As you say, paint molecules are of finite size, but, given that Gabriel’s Horn is a Godly or mathematical object, we might use equally mathematical paint. 6. #6 Uncle Al March 15, 2009 7. #7 Paul Murray March 15, 2009 I thought gabriel’s trumpet was that hypersphere thingo – the tractrix. I suppose anything that’s kind of trumpetlike manages to acquire the name. Speaking of which – how do you calculate the curvature of a surface? 8. #8 Kobra March 16, 2009 Your first radical sign has a 90 degree angle, the second has is greater than 90 degrees. Oh, the OCD! 9. #9 karl March 23, 2009 WELL….ALL I CAN SAY IS IT WAS JUST A RESULT OF CALCULATIONS IT MAYBE SOMEHOW HAVE DIFERRENCES WITH NATURAL QUALITATIVE ANALYSIS”” 10. #10 Jérôme ^ April 5, 2009 Another fun fact: the trumpet is the same (IIRC) as the pseudosphere, which means that it has constant negative Gaussian curvature along its surface. 11. #11 bsz July 27, 2010 It is a convoluted way to look at it, but if you want to paint the outside of the trumpet you need an infinite number of cans of paint; if you want to paint the inside of the trumpet, the paint will be part of the inside volume, so you need only a finite volume of paint. 12. #12 Harry July 28, 2010 The trumpet is quite extraordinary really. You will see an infinite number of cans of paint on the outside. The volume of paint is quite finite in this regard.

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Influenza is a respiratory disease of high importance due to its fast transmission and zoonotic potential. Alternative names: Porcine Influenza, influenza typeA Porcine influenza is caused by several Influenza Type A viruses closely related, characterized to have the ability to change the antigenic structure and create new strains. Each serotype is identified through surface proteins called "H" and "N". The three most common serotypes affecting pigs are H1 N1, H1 N2 and H3 N2. There are also different strains of these serotypes that have different pathogenicity and no crossed protection. The incubation period is very short, 12 to 48 hours. It affects the respiratory system. The virus can infect a farm in few days. This high transmission is almost the main feature helping in the diagnosis. There are three important periods in which the disease can cause infertility, and all of them are caused by fever and its consequences depending on the pregnancy stage because the virus is not systemic and does not cross the placenta. Thus, fetuses are not infected. - First, if sows get ill during the 21 first days of the pregnancy, the virus can prevent implantation of developing embryos, which produces and increase in returns at day 21. If gestation is already established 14 to 16 days after insemination there will be late returns. - Second, if the infection occurs during the first 5 weeks of gestation, embryo death and reabsorption can occur, which lead to pseudo-pregnant and empty sows. Litter size can be affected at this time due to the embryo reabsorption. By the end of gestation, there will be abortions or late mummified piglets at farrowing. - The third main effect is on the boars, in which increased body temperature can affect semen and reduce fertility during 4 to 5 weeks. In big farms, influenza can end up being endemic with intermittent outbreaks of clinical symptoms and infertility. Different strains can consecutively affect one farm. - The elevated temperature may cause abortions. - General coughing. When the virus enters a farm for the first time, 2 or 3 animals show symptoms of disease during the first 2 days, followed by: - A quick outbreak in which almost all animals stop eating, and appear clinically very ill at 2 to 3 days of the virus introduction. - The effects on the reproductive system are produced after the onset of the respiratory disease with coughing, pneumonia, fever and loss of appetite. - Acute respiratory distress without subjacent infections persists during 7 to 10 days (depending on how long contact between groups lasts). In the rest of the farm: - Quick onset of the acute disease in sows. - Coughing and pneumonia which quickly disseminates. - Animals go back to normality in 7 to 10 days. - Late returns after weaning. - Increase of returns at 21 days. - Increase of returns outside the normal cycle. - Increase in the number of sows that arrive empty to the farrowing area. - Increase in the number of abortions, especially during the last 3 months of pregnancy. - Increase in the number of stillborns and slow farrowings. - Occasionally an increase in the number of mummies. - Symptoms in piglets are rare, unless the disease enters the farm for the first time. - Colostrum can prevent the infection during the lactation period. Weaners and growers - Usually pigs suddenly lie down. - Difficulty to breath. - Severe coughing. - Most of the animals look like dying, but survive without a treatment, unless the farm already has a respiratory problem. - Influenza causes severe pneumonia on its own, but when combined with other infections, such as Actinobacillus pleuropneumoniae and PRRS, a chronic severe respiratory syndrome difficult to treat can occur. Severely affected pigs must be treated with antibiotics to prevent the development of secondary pneumonia. - In this case, the virus stays in the farm, affecting small groups of pigs, mainly weaned pigs. The virus can be responsible for persistent respiratory diseases, having symptoms similar to the acute disease, but less severe. Causes / Contributing Factors Influenza can be introduced by: - Infected people. - Carrier pigs. - Birds, mainly aquatic birds are reservoirs of the disease. - Secondary bacterial infections. - PRRS co-infections. - Temperature fluctuations. - Humid bedding and flooring. - Inadequate environments and bad ventilation increase the severity of the disease. In the acute disease, a reliable diagnosis can be done based on clinical signs, due to the fact that there are no other diseases with such a dramatically quick transmission and clinical effects. Blood samples taken from sows at the beginning of the disease and 2 to 3 weeks after, show an increase in antibody titters. Influenza virus can be identified from nasal swabs analyzed by PCR. Oral fluids can also be used, with good results. However, the endemic disease can be difficult to differentiate form other viral diseases. PRRS and Aujeszky’s disease must be considered. Sequencing is critical to establish the vaccine strain. - Breeding sows and boars showing symptoms of the acute disease such as increase of temperature and of breathing speed must be treated with wide spectrum antibiotics. - Control secondary bacterial pneumonia with antibiotics. - Inject the most affected pigs. - Medicate water for 3 to 5 days. - Use anti-inflammatory medication to minimize the respiratory system damage. - Several effective vaccines exist, but it is important that the vaccine contains the appropriate strain for each farm.

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The feathertail glider (Acrobates pygmaeus), also known as the pygmy gliding possum, pygmy glider, pygmy phalanger, flying phalanger and flying mouse, is the world’s smallest gliding possum and is named for its long feather-shaped tail. At 6.5–8 cm (2.6–3.1 in) in head-and-body length and 10–14 g (0.35–0.49 oz) in weight, it is only the size of a small mouse, but can leap and glide up to 25 m (82 ft). Like other gliding mammals, the feathertail glider has a skin membrane between the fore and hind legs, thicker than that of the other marsupials like the sugar glider, but smaller in proportion, extending only between the elbows and knees. It is monotypical for its genus. The tail is about the same length as the head and body combined, quite thin, moderately prehensile, and almost hairless except for two obvious rows of long, stiff hairs on either side. The tail, when held straight, looks like a double-sided comb. It is used to grip twigs and small branches, and to control gliding flight: steering and then braking. The coat is a uniform mid-grey, with dark patches around the eyes and often a white patch behind the ears. The underside is lighter; the ears are moderately large and rounded. The natural habitat of the feathertail glider is the eastern seaboard of Australia, and the glider’s distribution is from northern Queensland to Victoria. The feathertail glider’s diet includes nectar, pollen, and arthropods. The New Zoo in Poznań, Poland, was the first European zoo to breed feathertail gliders in 1999 (their animals originated from Sydney’s Taronga Zoo). Some of the feathertail gliders born in Poznań have been sent to other European zoos, meaning that the entire European captive population is of Poznán descent. Australia’s Taronga Zoo was the first zoo to breed feathertail gliders in captivity.

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Guards, Visibility and Polygons prepared by: Joseph Malkevitch Department of Mathematics and Computing York College (CUNY) Jamaica, New York 11451 Email: [email protected] (for additions, suggestions, and corrections) web page: www.york.cuny.edu/~malk Imagine that you are in charge of the security of a military installation in flatland. Buildings in flatland are polygonal such as the one shown below: One way to provide security is to install point surveillance devices, which have the property that they can see unimpeded provided a wall of the polygon does does not interfere with the line of sight. This suggests the following definition: x of polygon P is visible from a point g (for guard) if the segment gx does not include any points of the exterior of polygon P. Look at the figure below. Note our definition means that a line segment joint two interior points of the polygon P such as the points u and v are visible from each other despite the fact that the segment uv may include a vertex such as w which is on the boundary of the polygon P. This suggests that one might want an alternative notion of visibility, sometimes referred to as clear-visibility. We say that x is clearly visible from y if the segment xy is a subset of the polygon and if the intersection of xy with the boundary of the polygon can include no point or points of the polygon other than x and y themselves. 1. In the diagram above u is visible from v and v is visible from u but u is not clearly visible from v and v is not clearly visible from u. Clear visibility is an interesting and more natural concept for security but mathematically it is easier to work with ordinary visibility. How many guard (points) are need to guard all the polygon P above? 2. a. Can you find a formula, f(n), for the number of guards points that are sometimes necessary and always sufficient to guard all of a polygon with n sides? (Hint: look at how many guards are needed for n =3, n = 4, n = 5, n = 6. By now, perhaps you can find a pattern for the number of guards involved.) b. How many guards are sufficient for any n-gonal convex polygon? 3. If a polygon can be guarded with g guards, can one always choose g vertices of the polygon which will still serve as guard points for the polygon? 4. What differences would have to made in your analysis of this problem if instead of dealing with visibility one had formulated the problems using clear visibility? 5. Is the following problem the same as the one discussed above: Given any plane polygon P, determine the minimum number of guards to see all of P? 6. Can you find an algorithm that answers the question above (e.g. 5) quickly? 7. Can you formulate the problem in 2. in a way that might make it possible for a computer program to solve the problem rather than solve it using human eyesight? 8. An interesting variant of the guard problem involves "guarded guards." If a guard might perchance fall asleep, it might be nice to have a collection of guards with the property that a. the guards see all of the polygon and b. for each guard there is some other guard who if that guard appears to be asleep can call him/her on a cell phone! How many guards are needed for our original polygon to solve the "guarded guards" version of the problem? Will you need fewer guards if some of the guards are allowed to be in the interior of the polygon rather than at vertices?

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1963: Birmingham's aggression against African-Americans televised By the spring of 1963, the calls for equal rights were growing louder and louder, and in the decidedly southern city of Birmingham, Ala., those protests quickly turned violent. Martin Luther King Jr. personally considered Birmingham to be one of the most segregated cities in America, with incredible discrepancies in employment, income and opportunity in general. Over a 17-year period, the city was hit with dozens of race-fueled bomb attacks. But in 1963, the city’s black population finally rose up, but were treated with humiliation and brutality from every angle. Black residents held sit-ins at segregated churches, diners and libraries, leading to their arrest. Finally, as spring turned to summer, the authorities became more aggressive. As thousands of students marched through the streets, authorities turned water cannons and police dogs onto the crowd. The images of the savage treatment of the black marchers were broadcast to a nationwide audience, only serving to anger millions about the continuing state of race relations in the South. Their tactics worked, though. On May 8, after the city sat paralyzed for days, the city’s white leaders began desegregating numerous businesses and services. Photo: Firefighters turn their hoses full force on civil rights demonstrators July 15, 1963 in Birmingham, Ala. (AP Photo/Bill Hudson)

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# Question: Is A Squiggly Line On A Graph A Function? ## How do you know if a line on a graph is a function? Inspect the graph to see if any vertical line drawn would intersect the curve more than once. If there is any such line, the graph does not represent a function. If no vertical line can intersect the curve more than once, the graph does represent a function.. ## How do you determine if a graph represents a function? Use the vertical line test to determine whether or not a graph represents a function. If a vertical line is moved across the graph and, at any time, touches the graph at only one point, then the graph is a function. If the vertical line touches the graph at more than one point, then the graph is not a function. ## How do you find the equation of a squiggly line? Use the formula m = (y2 – y1) / (x2 – x1), in which “m” is the slope and (x1, y1) and (x2, y2) are points on the line. In the example, m = (1 – 0) / (2 – 0) = 1/2. ## What is a function on a graph examples? Graphs of functions are graphs of equations that have been solved for y! The graph of f(x) in this example is the graph of y = x2 – 3. It is easy to generate points on the graph. Choose a value for the first coordinate, then evaluate f at that number to find the second coordinate. ## How do you prove that two lines are parallel? We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. Unlike parallel lines, perpendicular lines do intersect. ## Is a straight up and down line a function? Key Takeaways A function is a relation with the property that each input is related to exactly one output. A relation is a set of ordered pairs. The graph of a linear function is a straight line, but a vertical line is not the graph of a function. ## Can a squiggly line be a function? In mathematics, the vertical line test is a visual way to determine if a curve is a graph of a function or not. … If a vertical line intersects a curve on an xy-plane more than once then for one value of x the curve has more than one value of y, and so, the curve does not represent a function. ## What is the squiggly line on a graph called? E. Break. a zigzag on the line of the x- or y-axis in a line or bar graph indicating that the data being displayed do not include all of the values that exist on the number line used. Also called a Squiggle. ## Are two lines on a graph a function? if you can draw any vertical line that intersects more than one point on the relationship, then it is not a function. This is based on the fact that a vertical line is a constant value of x, so if there is one input, x, with more than two outputs, y, then it breaks the function rule. ## How do you tell if something is a function without graphing? If a vertical line crosses the relation on the graph only once in all locations, the relation is a function. However, if a vertical line crosses the relation more than once, the relation is not a function. Using the vertical line test, all lines except for vertical lines are functions. ## Is every line a function? No, every straight line is not a graph of a function. Nearly all linear equations are functions because they pass the vertical line test. The exceptions are relations that fail the vertical line test.

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The wildfires in Australia have burned through more than 38,000 square miles, killing hundreds of millions of animals in the path of the blazes and devastating significant swaths of crucial habitat for the survivors. Numerous animals in the hardest-hit states of New South Wales, Victoria and South Australia face real threats to extinction as they struggle to recover from the destructive fires. The federal government has committed $50 million to a wildlife recovery fund. Treasurer Josh Frydenberg announced half this sum would go to wildlife rescues, hospitals and conservation groups, and the other $25 million would go to an emergency intervention fund advised by a panel of experts. Ecology expert Chris Dickman from the University of Sydney conservatively estimates more than 1 billion animals may have perished across the country, based on mammal, bird and reptile population density estimates multiplied by the area burned. According to a draft of the Victorian state government’s bushfire biodiversity response plan obtained by HuffPost, the blazes in the state had burned through “mostly high biodiversity value areas.” The report listed 54 species for immediate concern based on the extent of habitat burned, with numerous species having lost more than 40% of their habitat and some projected to lose more than 70%. Among the 54 species are 13 amphibians, 2 bats, 8 mammals, 11 birds, 7 reptiles and 13 aquatic fauna. A spokesperson for the NSW National Parks and Wildlife Service said an emergency recovery plan to protect and restore wildlife populations in the state is being developed. “On-ground interventions to protect threatened species and remaining habitat will need to include provision of food, water and shelter as well as pest and weed control,” she said. Here’s a snapshot of the species facing serious threat from the fires. The brush-tailed rock-wallaby was already listed as endangered prior to the bushfires. Mark Eldridge, the principal research scientist at the Australian Museum Research Institute, has been studying the species for more than three decades. He said many of the remaining wallaby colonies had been burned by the “unprecedented large and hot fires.” “The fires have killed some individuals but others have survived as they were able to shelter in their rocky crevices. However, the survivors now face an extremely difficult time as the fire has removed all or most of their food so they face starvation, and with most plant cover gone they are now very exposed to predators ... which are often attracted to burnt areas.” He said that, although he does not usually favor intervention, these were “desperate times,” nodding to the New South Wales government’s current aerial food drops of carrots and sweet potatoes for hungry animals. “The species has certainly taken a hit and will continue to suffer. Once we know more about the impact, we will be able to see if their conservation status needs to be reassessed.” He said there was an urgent need to step up control of non-native predators and herbivores in fire-affected areas to give survivors a chance to recover. Kangaroo Island dunnart The Kangaroo Island dunnart is listed as critically endangered by the International Union for Conservation of Nature. Rosie Hohnen, an ecologist at Charles Darwin University who researches the species, said that the fires have burned “all sites that they have been detected at since 1990, so effectively its entire range.” “Given it was previously considered critically endangered and that it will be very difficult for individuals to survive in burnt areas, it’s clear the species is in real peril, on the very edge of extinction,” Hohnen said. Attempts to recover the species will involve surveying unburned habitat, controlling predators in those areas and fencing off the predator-free areas. Koalas have been hit hard in some of their populations, such as near Port Macquarie, New South Wales, where populations of several hundreds have been reduced to very few, said Mathew Crowther, a University of Sydney expert in wildlife ecology. However, a lot of koala habitat does still remain unaffected by the bushfires, he said. Federal environment minister Sussan Ley said that up to 30% of their habitat in New South Wales had been destroyed and that the fires might move the koala from its “vulnerable” classification to “endangered.” Southern corroboree frog Experts are gravely concerned about the critically endangered southern corroboree frog. The species already faced a grim future due to disease, and climate change has affected its alpine environment in the Kosciuzko National Park, Zoos Victoria says on its site. A mega-blaze comprised of three fires has moved through this region, leaving experts unclear on the frogs’ fate. However, conservation groups have “insurance” populations of captive-bred frogs on site. This critically endangered songbird has lost important breeding habitat, especially in its Capertee Valley stronghold. According to Ross Crates, the lead researcher for the species at Australian National University, at least 20% of its known breeding habitat had been lost. “Probably at least 60% of the areas where they may disperse through to spend the winter has been burnt,” he said. A plan to release captive-bred regent honeyeaters last year has been rescheduled for spring due to the wildfires, however, surveying of burned sites will need to take place to see if this plan is still viable. The spotted-tail quoll was endangered even before the fires and suffered losses to feral predators and habitat destruction from changing fire patterns, land clearing and logging. The fires destroyed key habitat in areas including the Tallaganda National Park, a biodiversity stronghold, which urgently needs rain to regenerate. The long-footed potoroo, found in Victoria’s fire-ravaged East Gippsland region and in southeastern New South Wales, is a forest-dwelling mini kangaroo that feeds almost exclusively on a type of fungi. The animal is likely to have sustained serious loss of food and habitat, and it will be vulnerable and exposed to predators. Glossy black cockatoo Kangaroo Island’s unique subspecies of glossy black cockatoo was, until recently, a success story for Australia’s conservation. Daniella Teixeira, a conservation biologist at the University of Queensland, said that, although it was not yet safe for staff to get on the ground to make assessments, as much as 60% of the habitat may have been lost. “With a population of fewer than 400 birds before this crisis, this loss of habitat will be a major setback to the long-running conservation efforts for this unique bird,” she said. She said the devastating losses may even put the bird back on the “critically endangered” list. However, her team would work hard to recover the species. “By winter, we hope to plants thousands, if not tens of thousands, of food trees. These will provide food within five to 10 years. That probably seems like a long time, but that’s actually really quick for a tree. Our overall objective is to create more habitat so that the birds have a better chance of surviving such events in the future,” she said. You can support organizations saving wildlife by donating to the Nature Foundation’s Wildlife Recovery Fund, the Save the Kangaroo Island Glossy Black-Cockatoo fund, WWF Australia, NSW-based animal rescue group Wildlife Information Rescue and Education Service (WIRES), Zoos Victoria’s bushfire emergency wildlife fund, Australia Zoo’s Wildlife Hospital, or Port Macquarie Koala Hospital.

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Montane and Wadi vegetation To Aucher-Éloy and other early travellers, the mountains of the Arabian Peninsula presented a rugged and difficult terrain for passage, and today the more remote regions have yet to yield their final secrets. These mountain chains, lying on the western, southwestern and eastern periphery of the Peninsula, were uplifted during the period of the great Alpine orogeny from the Mid Tertiary onwards, and they were thereafter eroded, often dramatically, during successive pluvial periods (see Chapter 3). Today these mountains exert a strong influence on the climate of the Peninsula (see Chapter 2), and the soil and vegetation structure of their slopes, valleys and summits undoubtedly play a vital, though as yet largely undocumented, role in the hydrogeology of the region. From the Tertiary onwards the mountains were important phytogeographically, providing a relatively equable climatic ‘corridor’ between Africa and Asia, and vice versa, during the Late Tertiary (see Chapter 4), and today the higher altitudes provide a climatic refuge for species that once had much wider distributions. KeywordsArabian Peninsula Rock Fissure Rocky Habitat Northern Mountain Middle Altitude Unable to display preview. Download preview PDF.

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Dividing Fractions Card Game Dividing Fractions High or Low Card Game Some of my students favorite math games during tutoring involve a deck of cards. They never tire of them. Over the last few years, I've come up with several ways to play with a deck of cards. The above picture is one where my students are dividing fractions. I like using a deck of cards because it makes it fun and saves me the hassle of having to come up or find a worksheet with fractions on it. We just use a white board and make the appropriate boxes and then draw from the deck. On the board you draw two squares over the top of each other.  Then write a divide sign and then another set of squares.The student decides where to put the numbers. The goal of the game is to get the most points.  Before you put cards down on the white board, you need to decide if you are playing for HIGH or LOW numbers. You can make a spinner on the white board with a circle drawn into two equal pieces.  One one side of the circle write HIGH and on the other write LOW.  Using a paper clip in the center of the circle, place a pencil in the paper clip and flick it with your finger. 1.  Spin for HIGH or LOW 2.  Each person draws cards and places them on their own board 3.  Each player solves their set of cards 4.  Whoever has the number that is bigger or smaller (decided by the spin) gets a point 5.  Continue play until the first player reaches 10 points. 1 comment: 1. This Dividing Fractions Card Game is very nice game for childrens to teach math. This game can improve memory of children and he/she can easily understand math. Blogging tips

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# LOGS EQUAL THE The inverse of an exponential function is a logarithmic function. Logarithmic Function x = log a y read: “x equals log base a of y” ## Presentation on theme: "LOGS EQUAL THE The inverse of an exponential function is a logarithmic function. Logarithmic Function x = log a y read: “x equals log base a of y”"— Presentation transcript: LOGS EQUAL THE The inverse of an exponential function is a logarithmic function. Logarithmic Function x = log a y read: “x equals log base a of y” y = b x x = log b y These two equations are equivalent We can convert exponential equations to logarithmic equations and vice versa, using this: Convert to exponential form 1) 2) 3) Convert to logarithmic form 4) 5) 6) Now that we can convert between the two forms we can simplify logarithmic expressions. Without a Calculator! Simplify 1) log 2 32 2) log 3 27 3) log 4 2 4) log 3 1 2 ? = 32 3 ? = 27 4 ? = 2 3 ? = 1 ? = 5 ? = 3 ? = 0.5 ? = 0 “What is the exponent of that gives you 32?” “ What is the exponent of 3 that gives you 27? ” Evaluate We can also use these two forms to help us solve for an inverse. The steps for finding an inverse are the same as before. Easy as 1, 2, 3… 1-Rewrite 2-Switch x and y 3-Solve for y Example: Find the inverse Now you try…….. An inverse you just have to know Ln and are inverses They undo each other 1. 2. Example: Find the inverse Now you try….. Find the inverse: Essential Question: How do I graph & solve exponential and logarithmic functions? Daily Question: How do you expand and condense logs? Change-of-Base Formula Let a, b, and x be positive real numbers such that a  1 and b  1. Ex. 1 a) Evaluate using the change-of-base formula. Round to four decimal places. Ex. 2 You can do the same problem using natural logarithms. a) Evaluate using the natural logarithms. Round to four decimal places. Properties of Logarithms Product Property Quotient Property Power Property Ex. 3 log 10 5x 3 y log 10 5+ log 10 y+ log 10 x 3 log 10 5+ log 10 y+ 3 log 10 x Expand. Ex. 4 Expand. Now you Try….. A.A. B.C. Expand each log 3 2x 6 y Ex. 5 Condense. a) b) Now you Try…..Condense each Ex. 6 Use and to evaluate the logarithm. Ex. 7 Use the properties of logarithms to verify that -ln ½ = ln 2 -ln ½ = -ln (2 -1 ) = -(-1) ln (2) = ln (2) =ln 2 Homework: Page 147  # 1 – 23 odd Page 157  # 1 – 25 odd Download ppt "LOGS EQUAL THE The inverse of an exponential function is a logarithmic function. Logarithmic Function x = log a y read: “x equals log base a of y”" Similar presentations

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