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725	Life in the Ocean With over 70 percent of the Earth covered in water, oceans, bays and tributaries make up the largest biome – or ecosystem – on the planet. Scientists have found over 1 million species of animals and plants living here and they believe there may be as many as 9 million species yet undiscovered. - The ocean is divided into three zones – the euphotic zone, the disphotic zone and the aphotic zone. - The euphotic zone is closest to the water’s surface. Here, sunlight warms the water. Phytoplankton and algae convert the sun’s light to food. Zooplankton eat the phytoplankton. Fish eat the zooplankton and larger fish and water mammals eat the fish. Most of the animals in the sea live in this zone. - The disphotic zone is sometimes called the twilight zone because its light resembles twilight on land. The aphotic zone is very dark and very cold. Although this zone is the largest habitat on Earth, few oceanic animals can live here. - Algae and phytoplankton take in carbon dioxide and produce more oxygen than any other organism on Earth. Without them, we probably could not survive. - Shallow seafloors near the coasts support a wide variety of oceanic life. Bottom-feeding fish, such as cod live here, as well as mollusks, anemones, sea urchins, crustaceans and starfish. They feed on plankton and other fish. - Some deep-ocean creatures glow in the dark. This is called bioluminescence. The glow can help them attract a mate, confuse a predator or even find food. - Sperm whales dive into the twilight zone in search of food, such as giant squids. These mammals are specially equipped to dive to great depths and can remain underwater for up to 90 minutes. - Euphotic zone: area closest to the ocean’s surface - Disphotic zone: known as the twilight zone, this area of the ocean is colder and darker than the euphotic zone. - Aphotic zone: the deep ocean – completely dark and very cold Visit the University of California Santa Barbara to learn more about the marine world. Question: Are all oceans salty? Answer: In general, yes. Dissolved salt runs from lakes and rivers into the ocean where it becomes concentrated. One gallon of ocean water usually contains about one cup of salt. The Dead Sea is so salty that nothing can live there. Salt makes the water heavier so things float more easily. Many fish and marine animals have soft or non-existent skeletons. The water supports them instead. Jellyfish, for example, float easily in the water, but they can’t survive on land. Cite This Page You may cut-and-paste the below MLA and APA citation examples: MLA Style Citation Declan, Tobin. " Fun Ocean Life Facts for Kids ." Easy Science for Kids, May 2019. Web. 26 May 2019. < https://easyscienceforkids.com/life-in-the-ocean/ >. APA Style Citation Tobin, Declan. (2019). Fun Ocean Life Facts for Kids. Easy Science for Kids. Retrieved from https://easyscienceforkids.com/life-in-the-ocean/ We've recently added Sponsored Links :<\|endoftext\|>	3.734375
356	Persepolis: The Story of a Childhood Family in 1984 and Persepolis 12th Grade In the two texts, the notion of family is greatly influenced by an external factor, which is the political party in control of the population. In Persepolis, this would be the Iranian government in power during the post Cultural Revolution, while in 1984 it is the totalitarian party, referred to as ‘Ingsoc’ or simply ‘the party’. In Orwell’s novel, the party’s concept of family is defined as the people whom you share a household with, where every member is stripped of affection and comfort, and restricted of every possible aspect of freedom. Orwell depicts this through the melancholic relationship between Winston and Catherine, in which the sexual relations were referred to by Catherine as “their duty to the party”. During the coupling, Winston describes his wife as “cold and rigid” to his touch, mirroring her inability to express pleasure or any emotional response. The relationship reflects the extreme control that the party have manifested over people, who are ultimately reduced to detached machines. Furthermore, it appears like the party are ‘breeding’ the lower class together, which dehumanizes them, making them comparable to caged animals that are experimented on. Therefore it is seen that Orwell manipulates the theme of... Join Now to View Premium Content GradeSaver provides access to 1153 study guide PDFs and quizzes, 8921 literature essays, 2367 sample college application essays, 392 lesson plans, and ad-free surfing in this premium content, “Members Only” section of the site! Membership includes a 10% discount on all editing orders. Already a member? Log in<\|endoftext\|>	3.671875
302	# What is the trigonometric form of -8-i? Feb 23, 2018 $- \left(8 + i\right) \approx - \sqrt{58} \left(\cos \left(0.12\right) + i \sin \left(0.12\right)\right)$ #### Explanation: $- 8 - i = - \left(8 + i\right)$ For a given complex number, $z = a + b i$, $z = r \left(\cos \theta + i \sin \theta\right)$ $r = \sqrt{{a}^{2} + {b}^{2}}$ $\theta = {\tan}^{-} 1 \left(\frac{b}{a}\right)$ Let's deal with $8 + i$ $z = 8 + i = r \left(\cos \theta + i \sin \theta\right)$ $r = \sqrt{{8}^{2} + {1}^{2}} = \sqrt{65}$ $\theta = {\tan}^{-} 1 \left(\frac{1}{8}\right) \approx {0.12}^{c}$ $- \left(8 + i\right) \approx - \sqrt{58} \left(\cos \left(0.12\right) + i \sin \left(0.12\right)\right)$<\|endoftext\|>	4.4375
522	[In memoriam: William Shaw, 1917 - 2012] Have modern funerary practices always been in place? Were there different methods (and reasons) for disposing of the deceased over the ages? \|The 9th century Oseberg ship\| The Judaic tradition was clearly for burial. Deuteronomy 34:6 tells us, of Moses, that "God buried him in the depression in the land of Moab opposite Beth Peor. No man knows the place that he was buried, even to this day." Early Christians favored burial over cremation or any other disposal. Tertullian (160-225 CE) discusses Christian funeral practices, and Christ's placement in the tomb reinforces the idea of keeping the body intact in preparation for resurrection. The Viking image of the funeral pyre on land, or the ship ablaze and pushed out to sea, was another medieval attitude to death. The Viking cultures believed in an afterlife, but they knew it could not be a corporeal life—that was over. They (like the Egyptians) honored their dead by surrounding them with accoutrements that would accompany them into that afterlife. Because they were a sea-faring people, using a ship as a bier was appropriate. When those cultures began to adopt Christianity, they changed their funerary practice but did not give up their cultural symbols: they buried their nobles, but chose to bury them in a boat—like the Oseberg ship pictured above—or a boat-shaped grave-mound. \|Bound body being carried, from the Bayeux Tapestry\| The image of bodily resurrection had taken such a strong hold on Christian doctrine that interfering with the body deliberately seemed sacrilegious. Cremation was likewise considered inappropriate. Which leads me to a personal observation: if resurrection of a body that has decayed for centuries is possible, I do not see how resurrection of a body turned into ashes would be significantly more difficult. Still, this distinction in how bodies should be treated provided a strong visual image for cases when the Church wanted to make a point: it became common practice to throw the corpse of a heretic into the river to be washed away. You may remember the case of Jan Hus, who was burned at the stake and had his ashes thrown into the nearest river, and Jan's inspiration, John Wycliffe, who, although he died in 1384, was declared a heretic in 1415, and whose body was dug up in 1428 so that it could be burned and then thrown into the nearest river!<\|endoftext\|>	3.875
430	# 1 1 Patterns and Inductive Reasoning Inductive Reasoning • Slides: 13 1. 1 Patterns and Inductive Reasoning Inductive Reasoning • Watching weather patterns develop help forcasters… • Predict weather. . • They recognize and… • Describe patterns. • They then try to make accurate predictions based on the patterns they discover. • Inductive reasoning – reasoning from detailed facts to general principles Patterns & Inductive Reasoning • In Geometry, we will • Study many patterns… • Some discovered by others…. • Some we will discover… • And use those patterns to make accurate predictions Visual Patterns • Can you predict and sketch the next figure in these patterns? Number Patterns • Describe a pattern in the number sequence and predict the next number. Using Inductive Reasoning • Look for a Pattern • (Looks at several examples…use pictures and tables to help discover a pattern) • Make a conjecture. • (A conjecture is an unproven “statement” based on observation…it might be right or wrong…discuss it with others…make a new conjecture if necessary) How do you know your conjecture is True or False? • To prove a conjecture is TRUE, you need to prove it is ALWAYS true (not always so easy!) • To prove a conjecture is FALSE, you need only provide a SINGLE counterexample. • A counterexample is an example that shows a conjecture is false. Decide if this conjecture is TRUE or FALSE. • All people over 6 feet tall are good basketball players. • This conjecture is false (there are plenty of counterexamples…even here at CHS) • A full moon occurs every 29 or 30 days. • This conjecture is true. The moon revolves around Earth once approximately every 29. 5 days. Sketch the next figure in the pattern…. How many squares are in the next figure? Mixed Review Mixed Review (-2, 1) (4, 0) (3, -2) (-1, -3) Assignment 1. 1<\|endoftext\|>	4.4375
365	When women’s rights advocate Jeannette Rankin, a Montana Republican, was elected to the House of Representatives a century ago, she noted, “I may be the first woman member of Congress, but I won’t be the last.” Rankin took office in 1917 — a member of the 65th Congress. Since that time, 281 women have been elected full voting members of the House and 50 have become senators. Most women in Congress over the years have been Democrats. For much of the century, the gains made by women in the House were slow, but steady. In the Senate, representation came in fits and starts. In the 1992 election, the combined membership of women in both chambers jumped from 32 to 54 members. The election came to be known as the Year of the Woman. Every state but Mississippi and Vermont has been represented by a woman. Five states have not sent a woman to represent them in the House. Since becoming a state in 1959, Hawaii has had the largest share of female House members with women accounting for nearly 40 percent of its historical representation in the chamber. In the Senate, women currently make up a slightly higher percentage of the membership, at 21 percent to the House’s 19 percent. But there has yet to be a woman at the helm of either party in the world’s greatest deliberative body, and 21 states have never been represented by a female senator. Since the first woman was elected to the Senate in 1932, Maine has led, with 37 percent of its representation by women. Sixty-two combined years of Senate service came from Maine’s prolific trio of Republicans Margaret Chase Smith (1949-1973), Olympia Snowe (1995-2013) and Susan Collins (1997-present).<\|endoftext\|>	3.78125
3,209	# Prove that: Question: Prove that: (i) cos 3A + cos 5A + cos 7A + cos 15A = 4 cos 4A cos 5A cos 6A (ii) cos A + cos 3A + cos 5A + cos 7A = 4 cos A cos 2A cos 4A (iii) $\sin A+\sin 2 A+\sin 4 A+\sin 5 A=4 \cos \frac{A}{2} \cos \frac{3 A}{2} \sin 3 A$ (iv) $\sin 3 A+\sin 2 A-\sin A=4 \sin A \cos \frac{A}{2} \cos \frac{3 A}{2}$ (v) $\cos 20^{\circ} \cos 100^{\circ}+\cos 100^{\circ} \cos 140^{\circ}-140^{\circ} \cos 200^{\circ}=-\frac{3}{4}$ (vi) $\sin \frac{x}{2} \sin \frac{7 x}{2}+\sin \frac{3 x}{2} \sin \frac{11 x}{2}=\sin 2 x \sin 5 x$. (vii) $\cos x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 7 x \sin 8 x$ Solution: (i) Consider LHS : $\cos 3 A+c \cos 5 A+\cos 7 A+\cos 15 A$ $=2 \cos \left(\frac{3 A+5 A}{2}\right) \cos \left(\frac{3 A-5 A}{2}\right)+2 \cos \left(\frac{7 A+15 A}{2}\right) \cos \left(\frac{7 A-15 A}{2}\right)$ $\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$ $=2 \cos 4 A \cos (-A)+2 \cos 11 A \cos (-4 A)$ $=2 \cos 4 A \cos A+2 \cos 11 A \cos 4 A$ $=2 \cos 4 A\{\cos A+\cos 11 A\}$ $=2 \cos 4 A \times\left\{2 \cos \left(\frac{A+11 A}{2}\right) \cos \left(\frac{A-11 A}{2}\right)\right\}$ $=4 \cos 4 A \cos 6 A \cos (-5 A)$ $=4 \cos 4 A \cos 5 A \cos 6 A$ = RHS Hence, LHS = RHS (ii) Consider LHS : $\cos A+\cos 3 A+\cos 5 A+\cos 7 A$ $=2 \cos \left(\frac{A+3 A}{2}\right) \cos \left(\frac{A-3 A}{2}\right)+2 \cos \left(\frac{5 A+7 A}{2}\right) \cos \left(\frac{5 A-7 A}{2}\right)$ $\left\{\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$ $=2 \cos 2 A \cos (-A)+2 \cos 6 A \cos (-A)$ $=2 \cos 2 A \cos A+2 \cos 6 A \cos A$ $=2 \cos A(\cos 2 A+\cos 6 A)$ $=2 \cos A \times 2 \cos \left(\frac{2 A+6 A}{2}\right) \cos \left(\frac{2 A-6 A}{2}\right)$ $=4 \cos A \cos 4 A \cos (-2 A)$ $=4 \cos A \cos 2 A \cos 4 A$ = RHS Hence, LHS = RHS (iii) Consider LHS : $\sin A+\sin 2 A+\sin 4 A+\sin 5 A$ $=2 \sin \left(\frac{A+2 A}{2}\right) \cos \left(\frac{A-2 A}{2}\right)+2 \sin \left(\frac{4 A+5 A}{2}\right) \cos \left(\frac{4 A-5 A}{2}\right)$ $\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$ $=2 \sin \left(\frac{3}{2} A\right) \cos \left(-\frac{A}{2}\right)+2 \sin \left(\frac{9}{2} A\right) \cos \left(-\frac{A}{2}\right)$ $=2 \sin \left(\frac{3}{2} A\right) \cos \left(\frac{A}{2}\right)+2 \sin \left(\frac{9}{2} A\right) \cos \left(\frac{A}{2}\right)$ $=2 \cos \left(\frac{A}{2}\right)\left\{\sin \frac{3}{2} A+\sin \frac{9}{2} A\right\}$ $=2 \cos \left(\frac{A}{2}\right) \times 2 \sin \left(\frac{\frac{3}{2} A+\frac{9}{2} A}{2}\right) \cos \left(\frac{\frac{3}{2} A-\frac{9}{2} A}{2}\right)$ $=4 \cos \left(\frac{A}{2}\right) \cos 3 A \cos \left(-\frac{3}{2} A\right)$ $=4 \cos \frac{A}{2} \cos \left(\frac{3}{2} A\right) \cos 3 A$ = RHS Hence, LHS = RHS (iv) Consider LHS : $=\sin 3 A+\sin 2 A-\sin A$ $=2 \sin \left(\frac{3 A+2 A}{2}\right) \cos \left(\frac{3 A-2 A}{2}\right)-\sin A \quad\left\{\because \sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right\}$ $=2 \sin \left(\frac{5}{2} A\right) \cos \left(\frac{A}{2}\right)-\sin A$ $=2 \sin \left(\frac{5}{2} A\right) \cos \left(\frac{A}{2}\right)-2 \sin \frac{A}{2} \cos \frac{A}{2}$ $=2 \cos \left(\frac{A}{2}\right)\left\{\sin \frac{5}{2} A-\sin \frac{A}{2}\right\}$ $=2 \cos \left(\frac{A}{2}\right) \times 2 \sin \left(\frac{\frac{5}{2} A-\frac{A}{2}}{2}\right) \cos \left(\frac{\frac{5}{2} A+\frac{A}{2}}{2}\right)$ $=4 \cos \left(\frac{A}{2}\right) \sin A \cos \left(\frac{3}{2} A\right)$ $=4 \sin A \cos \left(\frac{A}{2}\right) \cos \left(\frac{3}{2} A\right)$ = RHS Hence, LHS = RHS (v) Consider LHS : $\cos 20^{\circ} \cos 100^{\circ}+\cos 100^{\circ} \cos 140^{\circ}-\cos 140^{\circ} \cos 200^{\circ}$ $=\frac{1}{2}\left(2 \cos 20^{\circ} \cos 100^{\circ}+2 \cos 100^{\circ} \cos 140^{\circ}-2 \cos 140^{\circ} \cos 200^{\circ}\right)$ $=\frac{1}{2}\left[\cos \left(100^{\circ}+20^{\circ}\right) \cos \left(100^{\circ}-20^{\circ}\right)+\cos \left(140^{\circ}+100^{\circ}\right) \cos \left(140^{\circ}-100^{\circ}\right)-\cos \left(200^{\circ}+140^{\circ}\right) \cos \left(200^{\circ}-140^{\circ}\right)\right]$ $=\frac{1}{0}\left[\cos 120^{\circ}+\cos 80^{\circ}+\cos 240^{\circ}+\cos 40^{\circ}-\cos 340^{\circ}-\cos 60^{\circ}\right]$ $=\frac{1}{2}\left[\cos 120^{\circ}+\cos 240^{\circ}-\cos 60^{\circ}+\cos 80^{\circ}+\cos 40^{\circ}-\cos 340^{\circ}\right]$ $=\frac{1}{2}\left[\left(-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right)+\cos 80^{\circ}+\cos 40^{\circ}-\cos 340^{\circ}\right]$ $=\frac{1}{2}\left[-\frac{3}{2}+\left\{2 \cos \left(\frac{80^{\circ}+40^{\circ}}{2}\right) \cos \left(\frac{80^{\circ}-40^{\circ}}{2}\right)-\cos \left(360^{\circ}-20^{\circ}\right)\right\}\right]$ $=\frac{1}{2}\left[-\frac{3}{2}+\left\{2 \cos 60^{\circ} \cos 20^{\circ}-\cos 20^{\circ}\right\}\right]$ $=\frac{1}{2}\left[-\frac{3}{2}+\cos 20^{\circ}-\cos 20^{\circ}\right]$ $=\frac{1}{2}\left[-\frac{3}{2}\right]$ $=-\frac{3}{4}=\mathrm{RHS}$ Hence, LHS = RHS (vi) Consider LHS : $\mathrm{LHS}=\sin \frac{x}{2} \sin \frac{7 x}{2}+\sin \frac{3 x}{2} \sin \frac{11 x}{2}$ $=\frac{1}{2}\left[2 \sin \frac{x}{2} \sin \frac{7 x}{2}+2 \sin \frac{3 x}{2} \sin \frac{11 x}{2}\right]$ $=\frac{1}{2}\left[\cos \left(\frac{7 x}{2}-\frac{x}{2}\right)-\cos \left(\frac{7 x}{2}+\frac{x}{2}\right)+\cos \left(\frac{11 x}{2}-\frac{3 x}{2}\right)-\cos \left(\frac{11 x}{2}+\frac{3 x}{2}\right)\right]$ $=\frac{1}{2}[\cos 3 x-\cos 4 x+\cos 4 x-\cos 7 x]$ $=\frac{1}{2}[\cos 3 x-\cos 7 x]$ $=\frac{1}{2}\left[-2 \sin \left(\frac{3 x+7 x}{2}\right) \sin \left(\frac{3 x-7 x}{2}\right)\right]$ $=\frac{1}{2}[-2 \sin (5 x) \sin (-2 x)]$ $=\sin (5 x) \sin (2 x)=\mathrm{RHS}$ Hence, LHS = RHS (vii) Consider LHS : LHS $=\cos x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}$ $=\frac{1}{2}\left[2 \cos x \cos \frac{x}{2}-2 \cos 3 x \cos \frac{9 x}{2}\right]$ $=\frac{1}{2}\left[\cos \left(x+\frac{x}{2}\right)+\cos \left(x-\frac{x}{2}\right)-\cos \left(3 x+\frac{9 x}{2}\right)-\cos \left(3 x-\frac{9 x}{2}\right)\right]$ $=\frac{1}{2}\left[\cos \frac{3 x}{2}+\cos \frac{x}{2}-\cos \frac{15 x}{2}-\cos \frac{3 x}{2}\right]$ $=\frac{1}{2}\left[\cos \frac{x}{2}-\cos \frac{15 x}{2}\right]$ $=\frac{1}{2}\left[-2 \sin \left(\frac{x+15 x}{4}\right) \sin \left(\frac{x-15 x}{4}\right)\right]$ $=\frac{1}{2}\left[-2 \sin (4 x) \sin \left(-\frac{7 x}{2}\right)\right]$ $=\sin (4 x) \sin \left(\frac{7 x}{2}\right)=\mathrm{RHS}$ Hence, LHS = RHS Disclaimer: The given question is incorrect. The correct question should be $\cos x \cos \frac{x}{2}-\cos 3 x \cos \frac{9 x}{2}=\sin 4 x \sin \frac{7 x}{2}$. Leave a comment Click here to get exam-ready with eSaral<\|endoftext\|>	4.46875
367	The activity of movement and control of water resources to minimize the damage to property and life and also to maximize the efficient beneficial use is known as water management. If the management of water is good in dams and levees it reduces the risk of harm caused due to flooding. Water management is a process of developing, optimizing and planning of water resources via many practices which are defined by many policies and regulations. With the increase in the population which has been doubled to over 6 billion people from 1900, the use of water has popped up to 600%. According to the statistics, the health of people is threatened by inadequate access to clean water for drinking and sanitation. Different methods to Conserving Water With a well-planned system, water is supplied to many places regularly in a city. This is generally planned by civic authorities in a city. But many times we observe that some amount of water is wasted through leakage of pipe and many other reasons. As we know that proper water management is necessary for the conservation of water. Thus civic authorities must take care of these reasons while supplying water to our homes. We usually observe that most of the rainwater gets wasted although it is the most precious natural resource. This rainwater can be used to recharge the groundwater by a technique known as rainwater harvesting. Farmers can play an important role in water management by using a technique called drip irrigation. In this technique, plants are watered using narrow tubes and this water is directly delivered at the base of the plant. We can also play an important role in minimizing the wastage of the water we use. Some of those habits can be turning off the taps while brushing, mopping the floor instead of washing. To learn more about Importance of management and some methods to conserve water download Byju’s- The Learning app.<\|endoftext\|>	3.75
1,013	# Thread: Integratin by Partial Fractions 1. ## Integratin by Partial Fractions Here is what I've done so far... $\displaystyle (x^3-4x^2-16)/((x^2)(x-4))=A/x+B/x^2+C/(x-4)$ so $\displaystyle x^3-4x^2-16=A(x^2)(x-4)+B(x)(x-4)+C(x)(x^2)$ =$\displaystyle A(x^3-4x^2)+B(x^2-4x)+C(x^3)$ =$\displaystyle (A+C)(x^3)+(-4A+B)(x^2)+(-4B)(x)$ Now we have to find what the coefficients are equal to, so A+C=1 -4A+B=-4 B=0 A=1 B=0 C=0 So now I should be able to take the integral of $\displaystyle 1/x+0/x^2+0/(x-4)$ This equals $\displaystyle ln(x)$ on the interval [7,8], so my answer should be $\displaystyle ln(8)-ln(7)$. This is apparently the wrong answer, so I must have made a mistake somewhere, but I'm not seeing it. 2. Originally Posted by kevin100 Here is what I've done so far... $\displaystyle (x^3-4x^2-16)/((x^2)(x-4))=A/x+B/x^2+C/(x-4)$ so $\displaystyle x^3-4x^2-16=A(x^2)(x-4)+B(x)(x-4)+C(x)(x^2)$ =$\displaystyle A(x^3-4x^2)+B(x^2-4x)+C(x^3)$ =$\displaystyle (A+C)(x^3)+(-4A+B)(x^2)+(-4B)(x)$ Now we have to find what the coefficients are equal to, so A+C=1 -4A+B=-4 B=0 A=1 B=0 C=0 So now I should be able to take the integral of $\displaystyle 1/x+0/x^2+0/(x-4)$ This equals $\displaystyle ln(x)$ on the interval [7,8], so my answer should be $\displaystyle ln(8)-ln(7)$. This is apparently the wrong answer, so I must have made a mistake somewhere, but I'm not seeing it. if the degree of the numerator is greater than or equal to the degree of the denominator, we must divide before doing partial fractions. the highest power in the top is the same as that of the bottom. thus we must first divide the numerator by the denominator by long division of polynomials, or alternatively, note that $\displaystyle \frac {x^3 - 4x^2 - 16}{x^3 - 4x^2} = \frac {x^3 - 4x^2}{x^3 - 4x^2} - \frac {16}{x^3 - 4x^2} = 1 - \frac {16}{x^3 - 4x^2}$ so your integral is $\displaystyle \int_7^8 \bigg( 1 - \frac {16}{x^3 - 4x^2} \bigg)~dx$ integrating 1 is easy, use partial fractions for the latter part 3. I'm still not getting it. My work doesn't seem to make much sense so I wont bother typing it out. 4. Originally Posted by kevin100 I'm still not getting it. My work doesn't seem to make much sense so I wont bother typing it out. $\displaystyle \frac {16}{x^2(x - 4)} = \frac Ax + \frac B{x^2} + \frac C{x - 4}$ $\displaystyle \Rightarrow 16 = Ax(x - 4) + B(x - 4) + Cx^2$ ...............(1) plugging in $\displaystyle x = 0$ in (1) we get $\displaystyle \boxed{B = -4}$ plugging in $\displaystyle x = 4$ in (1) we get $\displaystyle \boxed{C = 1}$ plugging in $\displaystyle x=1,~B=-4,$ and $\displaystyle C = 1$ in (1) we get $\displaystyle \boxed{A = -1}$ 5. Here.<\|endoftext\|>	4.40625
477	This term we will be investigating the question ‘How does what we eat affect the world?’ and finding out about Cornish produce. In our topic lessons, we will be finding out about Cornish food and how it has travelled the world. We will then investigate what happens to our food after we’ve eaten it! We will also be tasting, designing and making our own pasties. For this, we welcome donations of £3 per child to go towards the cost of the ingredients. Please let me know if your child has any dietary requirements that the school are not aware of. Dance specialist, Sarah Fairhall will be teaching dance on a Wednesday afternoon and the children will be playing Dodgeball on Mondays. Please ensure that the children have their PE kits in school on those days. Home Learning Opportunities - Investigate your food cupboard. Look on your food packets and see if you can find out where your food was produced. Can you find anything that was made in Cornwall? - Celebrate St Piran’s day on 5th March by drawing your own St. Piran’s flag. - When you are out, look in the fields and see if you can identify what the farmer is producing. You could carry out your own survey to find out what is being produced locally (e.g. dairy cows, cattle, chicken (for eggs), chicken (for meat), pigs, sheep, wheat, potatoes, barley, etc). - Cook a meal for you family to digest! There are some simple recipes on this website: https://www.foodafactoflife.org.uk/recipes/ Don’t forget to tell them what happens to their food after they’ve eaten it! - If you lose a tooth this term, ask the tooth fairy if you can keep it for a science experiment. Place your tooth in a glass of fizzy drink and check it every day for 2 weeks to see what happens. - Keep reading to anyone that will listen! - Keep learning those times tables! Dates for the diary Year 4 Class Assembly: Wednesday 6th March at 9:15am Year 4 Parent Consultations: Tuesday 19th March Please feel free to pop in after school if there is anything you would like to discuss.<\|endoftext\|>	3.84375
13,147	Transmission Control Protocol The Transmission Control Protocol (TCP) is one of the main protocols of the Internet protocol suite. It originated in the initial network implementation in which it complemented the Internet Protocol (IP). Therefore, the entire suite is commonly referred to as TCP/IP. TCP provides reliable, ordered, and error-checked delivery of a stream of octets (bytes) between applications running on hosts communicating via an IP network. Major internet applications such as the World Wide Web, email, remote administration, and file transfer rely on TCP. Applications that do not require reliable data stream service may use the User Datagram Protocol (UDP), which provides a connectionless datagram service that emphasizes reduced latency over reliability. In May 1974, Vint Cerf and Bob Kahn described an internetworking protocol for sharing resources using packet switching among network nodes. The authors had been working with Gérard Le Lann to incorporate concepts from the French CYCLADES project into the new network. A central control component of this model was the Transmission Control Program that incorporated both connection-oriented links and datagram services between hosts. The monolithic Transmission Control Program was later divided into a modular architecture consisting of the Transmission Control Protocol and the Internet Protocol. This resulted in a networking model that became known informally as TCP/IP, although formally it was variously referred to as the Department of Defense (DOD) model, and ARPANET model, and eventually also as the Internet Protocol Suite. The Transmission Control Protocol provides a communication service at an intermediate level between an application program and the Internet Protocol. It provides host-to-host connectivity at the transport layer of the Internet model. An application does not need to know the particular mechanisms for sending data via a link to another host, such as the required IP fragmentation to accommodate the maximum transmission unit of the transmission medium. At the transport layer, TCP handles all handshaking and transmission details and presents an abstraction of the network connection to the application typically through a network socket interface. At the lower levels of the protocol stack, due to network congestion, traffic load balancing, or unpredictable network behaviour, IP packets may be lost, duplicated, or delivered out of order. TCP detects these problems, requests re-transmission of lost data, rearranges out-of-order data and even helps minimize network congestion to reduce the occurrence of the other problems. If the data still remains undelivered, the source is notified of this failure. Once the TCP receiver has reassembled the sequence of octets originally transmitted, it passes them to the receiving application. Thus, TCP abstracts the application's communication from the underlying networking details. TCP is optimized for accurate delivery rather than timely delivery and can incur relatively long delays (on the order of seconds) while waiting for out-of-order messages or re-transmissions of lost messages. Therefore, it is not particularly suitable for real-time applications such as voice over IP. For such applications, protocols like the Real-time Transport Protocol (RTP) operating over the User Datagram Protocol (UDP) are usually recommended instead. TCP is a reliable stream delivery service which guarantees that all bytes received will be identical and in the same order as those sent. Since packet transfer by many networks is not reliable, TCP achieves this using a technique known as positive acknowledgement with re-transmission. This requires the receiver to respond with an acknowledgement message as it receives the data. The sender keeps a record of each packet it sends and maintains a timer from when the packet was sent. The sender re-transmits a packet if the timer expires before receiving the acknowledgement. The timer is needed in case a packet gets lost or corrupted. While IP handles actual delivery of the data, TCP keeps track of segments - the individual units of data transmission that a message is divided into for efficient routing through the network. For example, when an HTML file is sent from a web server, the TCP software layer of that server divides the file into segments and forwards them individually to the internet layer in the network stack. The internet layer software encapsulates each TCP segment into an IP packet by adding a header that includes (among other data) the destination IP address. When the client program on the destination computer receives them, the TCP software in the transport layer re-assembles the segments and ensures they are correctly ordered and error-free as it streams the file contents to the receiving application. TCP segment structureEdit Transmission Control Protocol accepts data from a data stream, divides it into chunks, and adds a TCP header creating a TCP segment. The TCP segment is then encapsulated into an Internet Protocol (IP) datagram, and exchanged with peers. The term TCP packet appears in both informal and formal usage, whereas in more precise terminology segment refers to the TCP protocol data unit (PDU), datagram to the IP PDU, and frame to the data link layer PDU: Processes transmit data by calling on the TCP and passing buffers of data as arguments. The TCP packages the data from these buffers into segments and calls on the internet module [e.g. IP] to transmit each segment to the destination TCP. A TCP segment consists of a segment header and a data section. The TCP header contains 10 mandatory fields, and an optional extension field (Options, pink background in table). The data section follows the header. Its contents are the payload data carried for the application. The length of the data section is not specified in the TCP segment header. It can be calculated by subtracting the combined length of the TCP header and the encapsulating IP header from the total IP datagram length (specified in the IP header). \|0\|\|0\|\|Source port\|\|Destination port\| \|8\|\|64\|\|Acknowledgment number (if ACK set)\| 0 0 0 \|16\|\|128\|\|Checksum\|\|Urgent pointer (if URG set)\| \|Options (if data offset > 5. Padded at the end with "0" bytes if necessary.)\| - Source port (16 bits) - Identifies the sending port. - Destination port (16 bits) - Identifies the receiving port. - Sequence number (32 bits) - Has a dual role: - If the SYN flag is set (1), then this is the initial sequence number. The sequence number of the actual first data byte and the acknowledged number in the corresponding ACK are then this sequence number plus 1. - If the SYN flag is clear (0), then this is the accumulated sequence number of the first data byte of this segment for the current session. - Acknowledgment number (32 bits) - If the ACK flag is set then the value of this field is the next sequence number that the sender of the ACK is expecting. This acknowledges receipt of all prior bytes (if any). The first ACK sent by each end acknowledges the other end's initial sequence number itself, but no data. - Data offset (4 bits) - Specifies the size of the TCP header in 32-bit words. The minimum size header is 5 words and the maximum is 15 words thus giving the minimum size of 20 bytes and maximum of 60 bytes, allowing for up to 40 bytes of options in the header. This field gets its name from the fact that it is also the offset from the start of the TCP segment to the actual data. - Reserved (3 bits) - For future use and should be set to zero. - Flags (9 bits) (aka Control bits) - Contains 9 1-bit flags - NS (1 bit): ECN-nonce - concealment protection (experimental: see RFC 3540). - CWR (1 bit): Congestion Window Reduced (CWR) flag is set by the sending host to indicate that it received a TCP segment with the ECE flag set and had responded in congestion control mechanism (added to header by RFC 3168). - ECE (1 bit): ECN-Echo has a dual role, depending on the value of the SYN flag. It indicates: - If the SYN flag is set (1), that the TCP peer is ECN capable. - If the SYN flag is clear (0), that a packet with Congestion Experienced flag set (ECN=11) in the IP header was received during normal transmission (added to header by RFC 3168). This serves as an indication of network congestion (or impending congestion) to the TCP sender. - URG (1 bit): indicates that the Urgent pointer field is significant - ACK (1 bit): indicates that the Acknowledgment field is significant. All packets after the initial SYN packet sent by the client should have this flag set. - PSH (1 bit): Push function. Asks to push the buffered data to the receiving application. - RST (1 bit): Reset the connection - SYN (1 bit): Synchronize sequence numbers. Only the first packet sent from each end should have this flag set. Some other flags and fields change meaning based on this flag, and some are only valid when it is set, and others when it is clear. - FIN (1 bit): Last packet from sender. - Window size (16 bits) - The size of the receive window, which specifies the number of window size units (by default, bytes) (beyond the segment identified by the sequence number in the acknowledgment field) that the sender of this segment is currently willing to receive (see Flow control and Window Scaling). - Checksum (16 bits) - The 16-bit checksum field is used for error-checking of the header, the Payload and a Pseudo-Header. The Pseudo-Header consists of the Source IP Address, the Destination IP Address, the protocol number for the TCP-Protocol (0x0006) and the length of the TCP-Headers including Payload (in Bytes). - Urgent pointer (16 bits) - if the URG flag is set, then this 16-bit field is an offset from the sequence number indicating the last urgent data byte. - Options (Variable 0–320 bits, divisible by 32) - The length of this field is determined by the data offset field. Options have up to three fields: Option-Kind (1 byte), Option-Length (1 byte), Option-Data (variable). The Option-Kind field indicates the type of option, and is the only field that is not optional. Depending on what kind of option we are dealing with, the next two fields may be set: the Option-Length field indicates the total length of the option, and the Option-Data field contains the value of the option, if applicable. For example, an Option-Kind byte of 0x01 indicates that this is a No-Op option used only for padding, and does not have an Option-Length or Option-Data byte following it. An Option-Kind byte of 0 is the End Of Options option, and is also only one byte. An Option-Kind byte of 0x02 indicates that this is the Maximum Segment Size option, and will be followed by a byte specifying the length of the MSS field (should be 0x04). This length is the total length of the given options field, including Option-Kind and Option-Length bytes. So while the MSS value is typically expressed in two bytes, the length of the field will be 4 bytes (+2 bytes of kind and length). In short, an MSS option field with a value of 0x05B4 will show up as (0x02 0x04 0x05B4) in the TCP options section. - Some options may only be sent when SYN is set; they are indicated below as [SYN]. Option-Kind and standard lengths given as (Option-Kind,Option-Length). - 0 (8 bits): End of options list - 1 (8 bits): No operation (NOP, Padding) This may be used to align option fields on 32-bit boundaries for better performance. - 2,4,SS (32 bits): Maximum segment size (see maximum segment size) [SYN] - 3,3,S (24 bits): Window scale (see window scaling for details) [SYN] - 4,2 (16 bits): Selective Acknowledgement permitted. [SYN] (See selective acknowledgments for details) - 5,N,BBBB,EEEE,... (variable bits, N is either 10, 18, 26, or 34)- Selective ACKnowledgement (SACK) These first two bytes are followed by a list of 1–4 blocks being selectively acknowledged, specified as 32-bit begin/end pointers. - 8,10,TTTT,EEEE (80 bits)- Timestamp and echo of previous timestamp (see TCP timestamps for details) - The remaining options are historical, obsolete, experimental, not yet standardized, or unassigned. Option number assignments are maintained by the IANA. - The TCP header padding is used to ensure that the TCP header ends, and data begins, on a 32 bit boundary. The padding is composed of zeros. TCP protocol operations may be divided into three phases. Connections must be properly established in a multi-step handshake process (connection establishment) before entering the data transfer phase. After data transmission is completed, the connection termination closes established virtual circuits and releases all allocated resources. A TCP connection is managed by an operating system through a programming interface that represents the local end-point for communications, the Internet socket. During the lifetime of a TCP connection the local end-point undergoes a series of state changes: - (server) represents waiting for a connection request from any remote TCP and port. - (client) represents waiting for a matching connection request after having sent a connection request. - (server) represents waiting for a confirming connection request acknowledgment after having both received and sent a connection request. - (both server and client) represents an open connection, data received can be delivered to the user. The normal state for the data transfer phase of the connection. - (both server and client) represents waiting for a connection termination request from the remote TCP, or an acknowledgment of the connection termination request previously sent. - (both server and client) represents waiting for a connection termination request from the remote TCP. - (both server and client) represents waiting for a connection termination request from the local user. - (both server and client) represents waiting for a connection termination request acknowledgment from the remote TCP. - (both server and client) represents waiting for an acknowledgment of the connection termination request previously sent to the remote TCP (which includes an acknowledgment of its connection termination request). - (either server or client) represents waiting for enough time to pass to be sure the remote TCP received the acknowledgment of its connection termination request. [According to RFC 793 a connection can stay in TIME-WAIT for a maximum of four minutes known as two MSL (maximum segment lifetime).] - (both server and client) represents no connection state at all. To establish a connection, TCP uses a three-way handshake. Before a client attempts to connect with a server, the server must first bind to and listen at a port to open it up for connections: this is called a passive open. Once the passive open is established, a client may initiate an active open. To establish a connection, the three-way (or 3-step) handshake occurs: - SYN: The active open is performed by the client sending a SYN to the server. The client sets the segment's sequence number to a random value A. - SYN-ACK: In response, the server replies with a SYN-ACK. The acknowledgment number is set to one more than the received sequence number i.e. A+1, and the sequence number that the server chooses for the packet is another random number, B. - ACK: Finally, the client sends an ACK back to the server. The sequence number is set to the received acknowledgement value i.e. A+1, and the acknowledgement number is set to one more than the received sequence number i.e. B+1. At this point, both the client and server have received an acknowledgment of the connection. The steps 1, 2 establish the connection parameter (sequence number) for one direction and it is acknowledged. The steps 2, 3 establish the connection parameter (sequence number) for the other direction and it is acknowledged. With these, a full-duplex communication is established. The connection termination phase uses a four-way handshake, with each side of the connection terminating independently. When an endpoint wishes to stop its half of the connection, it transmits a FIN packet, which the other end acknowledges with an ACK. Therefore, a typical tear-down requires a pair of FIN and ACK segments from each TCP endpoint. After the side that sent the first FIN has responded with the final ACK, it waits for a timeout before finally closing the connection, during which time the local port is unavailable for new connections; this prevents confusion due to delayed packets being delivered during subsequent connections. A connection can be "half-open", in which case one side has terminated its end, but the other has not. The side that has terminated can no longer send any data into the connection, but the other side can. The terminating side should continue reading the data until the other side terminates as well. It is also possible to terminate the connection by a 3-way handshake, when host A sends a FIN and host B replies with a FIN & ACK (merely combines 2 steps into one) and host A replies with an ACK. Some host TCP stacks may implement a half-duplex close sequence, as Linux or HP-UX do. If such a host actively closes a connection but still has not read all the incoming data the stack already received from the link, this host sends a RST instead of a FIN (Section 184.108.40.206 in RFC 1122). This allows a TCP application to be sure the remote application has read all the data the former sent—waiting the FIN from the remote side, when it actively closes the connection. But the remote TCP stack cannot distinguish between a Connection Aborting RST and Data Loss RST. Both cause the remote stack to lose all the data received. Some application protocols using the TCP open/close handshaking for the application protocol open/close handshaking may find the RST problem on active close. As an example: s = connect(remote); send(s, data); close(s); For a program flow like above, a TCP/IP stack like that described above does not guarantee that all the data arrives to the other application if unread data has arrived at this end. Most implementations allocate an entry in a table that maps a session to a running operating system process. Because TCP packets do not include a session identifier, both endpoints identify the session using the client's address and port. Whenever a packet is received, the TCP implementation must perform a lookup on this table to find the destination process. Each entry in the table is known as a Transmission Control Block or TCB. It contains information about the endpoints (IP and port), status of the connection, running data about the packets that are being exchanged and buffers for sending and receiving data. The number of sessions in the server side is limited only by memory and can grow as new connections arrive, but the client must allocate a random port before sending the first SYN to the server. This port remains allocated during the whole conversation, and effectively limits the number of outgoing connections from each of the client's IP addresses. If an application fails to properly close unrequired connections, a client can run out of resources and become unable to establish new TCP connections, even from other applications. Both endpoints must also allocate space for unacknowledged packets and received (but unread) data. There are a few key features that set TCP apart from User Datagram Protocol: - Ordered data transfer: the destination host rearranges according to sequence number - Retransmission of lost packets: any cumulative stream not acknowledged is retransmitted - Error-free data transfer - Flow control: limits the rate a sender transfers data to guarantee reliable delivery. The receiver continually hints the sender on how much data can be received (controlled by the sliding window). When the receiving host's buffer fills, the next acknowledgment contains a 0 in the window size, to stop transfer and allow the data in the buffer to be processed. - Congestion control TCP uses a sequence number to identify each byte of data. The sequence number identifies the order of the bytes sent from each computer so that the data can be reconstructed in order, regardless of any packet reordering, or packet loss that may occur during transmission. The sequence number of the first byte is chosen by the transmitter for the first packet, which is flagged SYN. This number can be arbitrary, and should, in fact, be unpredictable to defend against TCP sequence prediction attacks. Acknowledgements (ACKs) are sent with a sequence number by the receiver of data to tell the sender that data has been received to the specified byte. ACKs do not imply that the data has been delivered to the application. They merely signify that it is now the receiver's responsibility to deliver the data. Reliability is achieved by the sender detecting lost data and retransmitting it. TCP uses two primary techniques to identify loss. Retransmission timeout (abbreviated as RTO) and duplicate cumulative acknowledgements (DupAcks). If a single packet (say packet 100) in a stream is lost, then the receiver cannot acknowledge packets above 100 because it uses cumulative ACKs. Hence the receiver acknowledges packet 99 again on the receipt of another data packet. This duplicate acknowledgement is used as a signal for packet loss. That is, if the sender receives three duplicate acknowledgements, it retransmits the last unacknowledged packet. A threshold of three is used because the network may reorder packets causing duplicate acknowledgements. This threshold has been demonstrated to avoid spurious retransmissions due to reordering. Sometimes selective acknowledgements (SACKs) are used to give more explicit feedback on which packets have been received. This greatly improves TCP's ability to retransmit the right packets. Whenever a packet is sent, the sender sets a timer that is a conservative estimate of when that packet will be acked. If the sender does not receive an ACK by then, it transmits that packet again. The timer is reset every time the sender receives an acknowledgement. This means that the retransmit timer fires only when the sender has received no acknowledgement for a long time. Typically the timer value is set to where is the clock granularity. Further, in case a retransmit timer has fired and still no acknowledgement is received, the next timer is set to twice the previous value (up to a certain threshold). Among other things, this helps defend against a man-in-the-middle denial of service attack that tries to fool the sender into making so many retransmissions that the receiver is overwhelmed. If the sender infers that data has been lost in the network using one of the two techniques described above, it retransmits the data. Sequence numbers allow receivers to discard duplicate packets and properly sequence reordered packets. Acknowledgments allow senders to determine when to retransmit lost packets. To assure correctness a checksum field is included; see checksum computation section for details on checksumming. The TCP checksum is a weak check by modern standards. Data Link Layers with high bit error rates may require additional link error correction/detection capabilities. The weak checksum is partially compensated for by the common use of a CRC or better integrity check at layer 2, below both TCP and IP, such as is used in PPP or the Ethernet frame. However, this does not mean that the 16-bit TCP checksum is redundant: remarkably, introduction of errors in packets between CRC-protected hops is common, but the end-to-end 16-bit TCP checksum catches most of these simple errors. This is the end-to-end principle at work. TCP uses an end-to-end flow control protocol to avoid having the sender send data too fast for the TCP receiver to receive and process it reliably. Having a mechanism for flow control is essential in an environment where machines of diverse network speeds communicate. For example, if a PC sends data to a smartphone that is slowly processing received data, the smartphone must regulate the data flow so as not to be overwhelmed. TCP uses a sliding window flow control protocol. In each TCP segment, the receiver specifies in the receive window field the amount of additionally received data (in bytes) that it is willing to buffer for the connection. The sending host can send only up to that amount of data before it must wait for an acknowledgement and window update from the receiving host. When a receiver advertises a window size of 0, the sender stops sending data and starts the persist timer. The persist timer is used to protect TCP from a deadlock situation that could arise if a subsequent window size update from the receiver is lost, and the sender cannot send more data until receiving a new window size update from the receiver. When the persist timer expires, the TCP sender attempts recovery by sending a small packet so that the receiver responds by sending another acknowledgement containing the new window size. If a receiver is processing incoming data in small increments, it may repeatedly advertise a small receive window. This is referred to as the silly window syndrome, since it is inefficient to send only a few bytes of data in a TCP segment, given the relatively large overhead of the TCP header. The final main aspect of TCP is congestion control. TCP uses a number of mechanisms to achieve high performance and avoid congestion collapse, where network performance can fall by several orders of magnitude. These mechanisms control the rate of data entering the network, keeping the data flow below a rate that would trigger collapse. They also yield an approximately max-min fair allocation between flows. Acknowledgments for data sent, or lack of acknowledgments, are used by senders to infer network conditions between the TCP sender and receiver. Coupled with timers, TCP senders and receivers can alter the behavior of the flow of data. This is more generally referred to as congestion control and/or network congestion avoidance. In addition, senders employ a retransmission timeout (RTO) that is based on the estimated round-trip time (or RTT) between the sender and receiver, as well as the variance in this round trip time. The behavior of this timer is specified in RFC 6298. There are subtleties in the estimation of RTT. For example, senders must be careful when calculating RTT samples for retransmitted packets; typically they use Karn's Algorithm or TCP timestamps (see RFC 1323). These individual RTT samples are then averaged over time to create a Smoothed Round Trip Time (SRTT) using Jacobson's algorithm. This SRTT value is what is finally used as the round-trip time estimate. Enhancing TCP to reliably handle loss, minimize errors, manage congestion and go fast in very high-speed environments are ongoing areas of research and standards development. As a result, there are a number of TCP congestion avoidance algorithm variations. Maximum segment sizeEdit The maximum segment size (MSS) is the largest amount of data, specified in bytes, that TCP is willing to receive in a single segment. For best performance, the MSS should be set small enough to avoid IP fragmentation, which can lead to packet loss and excessive retransmissions. To try to accomplish this, typically the MSS is announced by each side using the MSS option when the TCP connection is established, in which case it is derived from the maximum transmission unit (MTU) size of the data link layer of the networks to which the sender and receiver are directly attached. Furthermore, TCP senders can use path MTU discovery to infer the minimum MTU along the network path between the sender and receiver, and use this to dynamically adjust the MSS to avoid IP fragmentation within the network. MSS announcement is also often called "MSS negotiation". Strictly speaking, the MSS is not "negotiated" between the originator and the receiver, because that would imply that both originator and receiver will negotiate and agree upon a single, unified MSS that applies to all communication in both directions of the connection. In fact, two completely independent values of MSS are permitted for the two directions of data flow in a TCP connection. This situation may arise, for example, if one of the devices participating in a connection has an extremely limited amount of memory reserved (perhaps even smaller than the overall discovered Path MTU) for processing incoming TCP segments. Relying purely on the cumulative acknowledgment scheme employed by the original TCP protocol can lead to inefficiencies when packets are lost. For example, suppose bytes with sequence number 1,000 to 10,999 are sent in 10 different TCP segments of equal size, and the second segment (sequence numbers 2,000 to 2,999) is lost during transmission. In a pure cumulative acknowledgment protocol, the receiver can only send a cumulative ACK value of 2,000 (the sequence number immediately following the last sequence number of the received data) and cannot say that it received bytes 3,000 to 10,999 successfully. Thus the sender may then have to resend all data starting with sequence number 2,000. To alleviate this issue TCP employs the selective acknowledgment (SACK) option, defined in 1996 in RFC 2018, which allows the receiver to acknowledge discontinuous blocks of packets which were received correctly, in addition to the sequence number immediately following the last sequence number of the last contiguous byte received successively, as in the basic TCP acknowledgment. The acknowledgement can specify a number of SACK blocks, where each SACK block is conveyed by the Left Edge of Block (the first sequence number of the block) and the Right Edge of Block (the sequence number immediately following the last sequence number of the block), with a Block being a contiguous range that the receiver correctly received. In the example above, the receiver would send an ACK segment with a cumulative ACK value of 2,000 and a SACK option header with sequence numbers 3,000 and 11,000. The sender would accordingly retransmit only the second segment with sequence numbers 2,000 to 2,999. A TCP sender can interpret an out-of-order segment delivery as a lost segment. If it does so, the TCP sender will retransmit the segment previous to the out-of-order packet and slow its data delivery rate for that connection. The duplicate-SACK option, an extension to the SACK option that was defined in May 2000 in RFC 2883, solves this problem. The TCP receiver sends a D-ACK to indicate that no segments were lost, and the TCP sender can then reinstate the higher transmission-rate. The SACK option is not mandatory, and comes into operation only if both parties support it. This is negotiated when a connection is established. SACK uses a TCP header option (see TCP segment structure for details). The use of SACK has become widespread—all popular TCP stacks support it. Selective acknowledgment is also used in Stream Control Transmission Protocol (SCTP). For more efficient use of high-bandwidth networks, a larger TCP window size may be used. The TCP window size field controls the flow of data and its value is limited to between 2 and 65,535 bytes. Since the size field cannot be expanded, a scaling factor is used. The TCP window scale option, as defined in RFC 1323, is an option used to increase the maximum window size from 65,535 bytes to 1 gigabyte. Scaling up to larger window sizes is a part of what is necessary for TCP tuning. The window scale option is used only during the TCP 3-way handshake. The window scale value represents the number of bits to left-shift the 16-bit window size field. The window scale value can be set from 0 (no shift) to 14 for each direction independently. Both sides must send the option in their SYN segments to enable window scaling in either direction. Some routers and packet firewalls rewrite the window scaling factor during a transmission. This causes sending and receiving sides to assume different TCP window sizes. The result is non-stable traffic that may be very slow. The problem is visible on some sites behind a defective router. TCP timestamps, defined in RFC 1323 in 1992, can help TCP determine in which order packets were sent. TCP timestamps are not normally aligned to the system clock and start at some random value. Many operating systems will increment the timestamp for every elapsed millisecond; however the RFC only states that the ticks should be proportional. There are two timestamp fields: - a 4-byte sender timestamp value (my timestamp) - a 4-byte echo reply timestamp value (the most recent timestamp received from you). TCP timestamps are used in an algorithm known as Protection Against Wrapped Sequence numbers, or PAWS (see RFC 1323 for details). PAWS is used when the receive window crosses the sequence number wraparound boundary. In the case where a packet was potentially retransmitted it answers the question: "Is this sequence number in the first 4 GB or the second?" And the timestamp is used to break the tie. Also, the Eifel detection algorithm (RFC 3522) uses TCP timestamps to determine if retransmissions are occurring because packets are lost or simply out of order. Recent Statistics show that the level of Timestamp adoption has stagnated, at ~40%, owing to Windows server dropping support since Windows Server 2008 . It is possible to interrupt or abort the queued stream instead of waiting for the stream to finish. This is done by specifying the data as urgent. This tells the receiving program to process it immediately, along with the rest of the urgent data. When finished, TCP informs the application and resumes back to the stream queue. An example is when TCP is used for a remote login session, the user can send a keyboard sequence that interrupts or aborts the program at the other end. These signals are most often needed when a program on the remote machine fails to operate correctly. The signals must be sent without waiting for the program to finish its current transfer. TCP OOB data was not designed for the modern Internet. The urgent pointer only alters the processing on the remote host and doesn't expedite any processing on the network itself. When it gets to the remote host there are two slightly different interpretations of the protocol, which means only single bytes of OOB data are reliable. This is assuming it is reliable at all as it is one of the least commonly used protocol elements and tends to be poorly implemented. Forcing data deliveryEdit Normally, TCP waits for 200 ms for a full packet of data to send (Nagle's Algorithm tries to group small messages into a single packet). This wait creates small, but potentially serious delays if repeated constantly during a file transfer. For example, a typical send block would be 4 KB, a typical MSS is 1460, so 2 packets go out on a 10 Mbit/s ethernet taking ~1.2 ms each followed by a third carrying the remaining 1176 after a 197 ms pause because TCP is waiting for a full buffer. In the case of telnet, each user keystroke is echoed back by the server before the user can see it on the screen. This delay would become very annoying. Setting the socket option TCP_NODELAY overrides the default 200 ms send delay. Application programs use this socket option to force output to be sent after writing a character or line of characters. The RFC defines the PSH push bit as "a message to the receiving TCP stack to send this data immediately up to the receiving application". There is no way to indicate or control it in user space using Berkeley sockets and it is controlled by protocol stack only. TCP may be attacked in a variety of ways. The results of a thorough security assessment of TCP, along with possible mitigations for the identified issues, were published in 2009, and is currently being pursued within the IETF. Denial of serviceEdit By using a spoofed IP address and repeatedly sending purposely assembled SYN packets, followed by many ACK packets, attackers can cause the server to consume large amounts of resources keeping track of the bogus connections. This is known as a SYN flood attack. Proposed solutions to this problem include SYN cookies and cryptographic puzzles, though SYN cookies come with their own set of vulnerabilities. Sockstress is a similar attack, that might be mitigated with system resource management. An advanced DoS attack involving the exploitation of the TCP Persist Timer was analyzed in Phrack #66. PUSH and ACK floods are other variants. An attacker who is able to eavesdrop a TCP session and redirect packets can hijack a TCP connection. To do so, the attacker learns the sequence number from the ongoing communication and forges a false segment that looks like the next segment in the stream. Such a simple hijack can result in one packet being erroneously accepted at one end. When the receiving host acknowledges the extra segment to the other side of the connection, synchronization is lost. Hijacking might be combined with Address Resolution Protocol (ARP) or routing attacks that allow taking control of the packet flow, so as to get permanent control of the hijacked TCP connection. Impersonating a different IP address was not difficult prior to RFC 1948, when the initial sequence number was easily guessable. That allowed an attacker to blindly send a sequence of packets that the receiver would believe to come from a different IP address, without the need to deploy ARP or routing attacks: it is enough to ensure that the legitimate host of the impersonated IP address is down, or bring it to that condition using denial-of-service attacks. This is why the initial sequence number is now chosen at random. An attacker who can eavesdrop and predict the size of the next packet to be sent can cause the receiver to accept a malicious payload without disrupting the existing connection. The attacker injects a malicious packet with the sequence number and a payload size of the next expected packet. When the legitimate packet is ultimately received, it is found to have the same sequence number and length as a packet already received and is silently dropped as a normal duplicate packet—the legitimate packet is "vetoed" by the malicious packet. Unlike in connection hijacking, the connection is never desynchronized and communication continues as normal after the malicious payload is accepted. TCP veto gives the attacker less control over the communication, but makes the attack particularly resistant to detection. The large increase in network traffic from the ACK storm is avoided. The only evidence to the receiver that something is amiss is a single duplicate packet, a normal occurrence in an IP network. The sender of the vetoed packet never sees any evidence of an attack. Another vulnerability is TCP reset attack. TCP and UDP use port numbers to identify sending and receiving application end-points on a host, often called Internet sockets. Each side of a TCP connection has an associated 16-bit unsigned port number (0-65535) reserved by the sending or receiving application. Arriving TCP packets are identified as belonging to a specific TCP connection by its sockets, that is, the combination of source host address, source port, destination host address, and destination port. This means that a server computer can provide several clients with several services simultaneously, as long as a client takes care of initiating any simultaneous connections to one destination port from different source ports. Port numbers are categorized into three basic categories: well-known, registered, and dynamic/private. The well-known ports are assigned by the Internet Assigned Numbers Authority (IANA) and are typically used by system-level or root processes. Well-known applications running as servers and passively listening for connections typically use these ports. Some examples include: FTP (20 and 21), SSH (22), TELNET (23), SMTP (25), HTTP over SSL/TLS (443), and HTTP (80). Registered ports are typically used by end user applications as ephemeral source ports when contacting servers, but they can also identify named services that have been registered by a third party. Dynamic/private ports can also be used by end user applications, but are less commonly so. Dynamic/private ports do not contain any meaning outside of any particular TCP connection. Network Address Translation (NAT), typically uses dynamic port numbers, on the ("Internet-facing") public side, to disambiguate the flow of traffic that is passing between a public network and a private subnetwork, thereby allowing many IP addresses (and their ports) on the subnet to be serviced by a single public-facing address. TCP is a complex protocol. However, while significant enhancements have been made and proposed over the years, its most basic operation has not changed significantly since its first specification RFC 675 in 1974, and the v4 specification RFC 793, published in September 1981. RFC 1122, Host Requirements for Internet Hosts, clarified a number of TCP protocol implementation requirements. A list of the 8 required specifications and over 20 strongly encouraged enhancements is available in RFC 7414. Among this list is RFC 2581, TCP Congestion Control, one of the most important TCP-related RFCs in recent years, describes updated algorithms that avoid undue congestion. In 2001, RFC 3168 was written to describe Explicit Congestion Notification (ECN), a congestion avoidance signaling mechanism. The original TCP congestion avoidance algorithm was known as "TCP Tahoe", but many alternative algorithms have since been proposed (including TCP Reno, TCP Vegas, FAST TCP, TCP New Reno, and TCP Hybla). TCP Interactive (iTCP) is a research effort into TCP extensions that allows applications to subscribe to TCP events and register handler components that can launch applications for various purposes, including application-assisted congestion control. Multipath TCP (MPTCP) is an ongoing effort within the IETF that aims at allowing a TCP connection to use multiple paths to maximize resource usage and increase redundancy. The redundancy offered by Multipath TCP in the context of wireless networks enables the simultaneous utilization of different networks, which brings higher throughput and better handover capabilities. Multipath TCP also brings performance benefits in datacenter environments. The reference implementation of Multipath TCP is being developed in the Linux kernel. Multipath TCP is used to support the Siri voice recognition application on iPhones, iPads and Macs TCP Cookie Transactions (TCPCT) is an extension proposed in December 2009 to secure servers against denial-of-service attacks. Unlike SYN cookies, TCPCT does not conflict with other TCP extensions such as window scaling. TCPCT was designed due to necessities of DNSSEC, where servers have to handle large numbers of short-lived TCP connections. tcpcrypt is an extension proposed in July 2010 to provide transport-level encryption directly in TCP itself. It is designed to work transparently and not require any configuration. Unlike TLS (SSL), tcpcrypt itself does not provide authentication, but provides simple primitives down to the application to do that. As of 2010[update], the first tcpcrypt IETF draft has been published and implementations exist for several major platforms. TCP Fast Open is an extension to speed up the opening of successive TCP connections between two endpoints. It works by skipping the three-way handshake using a cryptographic "cookie". It is similar to an earlier proposal called T/TCP, which was not widely adopted due to security issues. TCP Fast Open was published as RFC 7413 in 2014. Proposed in May 2013, Proportional Rate Reduction (PRR) is a TCP extension developed by Google engineers. PRR ensures that the TCP window size after recovery is as close to the Slow-start threshold as possible. The algorithm is designed to improve the speed of recovery and is the default congestion control algorithm in Linux 3.2+ kernels. TCP over wireless networksEdit TCP was originally designed for wired networks. Packet loss is considered to be the result of network congestion and the congestion window size is reduced dramatically as a precaution. However, wireless links are known to experience sporadic and usually temporary losses due to fading, shadowing, hand off, interference, and other radio effects, that are not strictly congestion. After the (erroneous) back-off of the congestion window size, due to wireless packet loss, there may be a congestion avoidance phase with a conservative decrease in window size. This causes the radio link to be underutilized. Extensive research on combating these harmful effects has been conducted. Suggested solutions can be categorized as end-to-end solutions, which require modifications at the client or server, link layer solutions, such as Radio Link Protocol (RLP) in cellular networks, or proxy-based solutions which require some changes in the network without modifying end nodes. One way to overcome the processing power requirements of TCP is to build hardware implementations of it, widely known as TCP offload engines (TOE). The main problem of TOEs is that they are hard to integrate into computing systems, requiring extensive changes in the operating system of the computer or device. One company to develop such a device was Alacritech. A packet sniffer, which intercepts TCP traffic on a network link, can be useful in debugging networks, network stacks, and applications that use TCP by showing the user what packets are passing through a link. Some networking stacks support the SO_DEBUG socket option, which can be enabled on the socket using setsockopt. That option dumps all the packets, TCP states, and events on that socket, which is helpful in debugging. Netstat is another utility that can be used for debugging. For many applications TCP is not appropriate. One problem (at least with normal implementations) is that the application cannot access the packets coming after a lost packet until the retransmitted copy of the lost packet is received. This causes problems for real-time applications such as streaming media, real-time multiplayer games and voice over IP (VoIP) where it is generally more useful to get most of the data in a timely fashion than it is to get all of the data in order. Also, for embedded systems, network booting, and servers that serve simple requests from huge numbers of clients (e.g. DNS servers) the complexity of TCP can be a problem. Finally, some tricks such as transmitting data between two hosts that are both behind NAT (using STUN or similar systems) are far simpler without a relatively complex protocol like TCP in the way. Generally, where TCP is unsuitable, the User Datagram Protocol (UDP) is used. This provides the application multiplexing and checksums that TCP does, but does not handle streams or retransmission, giving the application developer the ability to code them in a way suitable for the situation, or to replace them with other methods like forward error correction or interpolation. Stream Control Transmission Protocol (SCTP) is another protocol that provides reliable stream oriented services similar to TCP. It is newer and considerably more complex than TCP, and has not yet seen widespread deployment. However, it is especially designed to be used in situations where reliability and near-real-time considerations are important. TCP also has issues in high-bandwidth environments. The TCP congestion avoidance algorithm works very well for ad-hoc environments where the data sender is not known in advance. If the environment is predictable, a timing based protocol such as Asynchronous Transfer Mode (ATM) can avoid TCP's retransmits overhead. Multipurpose Transaction Protocol (MTP/IP) is patented proprietary software that is designed to adaptively achieve high throughput and transaction performance in a wide variety of network conditions, particularly those where TCP is perceived to be inefficient. TCP checksum for IPv4Edit The checksum field is the 16 bit one's complement of the one's complement sum of all 16-bit words in the header and text. If a segment contains an odd number of header and text octets to be checksummed, the last octet is padded on the right with zeros to form a 16-bit word for checksum purposes. The pad is not transmitted as part of the segment. While computing the checksum, the checksum field itself is replaced with zeros. In other words, after appropriate padding, all 16-bit words are added using one's complement arithmetic. The sum is then bitwise complemented and inserted as the checksum field. A pseudo-header that mimics the IPv4 packet header used in the checksum computation is shown in the table below. \|96\|\|Source port\|\|Destination port\| The source and destination addresses are those of the IPv4 header. The protocol value is 6 for TCP (cf. List of IP protocol numbers). The TCP length field is the length of the TCP header and data (measured in octets). TCP checksum for IPv6Edit - Any transport or other upper-layer protocol that includes the addresses from the IP header in its checksum computation must be modified for use over IPv6, to include the 128-bit IPv6 addresses instead of 32-bit IPv4 addresses. A pseudo-header that mimics the IPv6 header for computation of the checksum is shown below. \|320\|\|Source port\|\|Destination port\| - Source address: the one in the IPv6 header - Destination address: the final destination; if the IPv6 packet doesn't contain a Routing header, TCP uses the destination address in the IPv6 header, otherwise, at the originating node, it uses the address in the last element of the Routing header, and, at the receiving node, it uses the destination address in the IPv6 header. - TCP length: the length of the TCP header and data - Next Header: the protocol value for TCP Checksum offload Edit Many TCP/IP software stack implementations provide options to use hardware assistance to automatically compute the checksum in the network adapter prior to transmission onto the network or upon reception from the network for validation. This may relieve the OS from using precious CPU cycles calculating the checksum. Hence, overall network performance is increased. This feature may cause packet analyzers that are unaware or uncertain about the use of checksum offload to report invalid checksums in outbound packets that have not yet reached the network adapter. This will only occur for packets that are intercepted before being transmitted by the network adapter; all packets transmitted by the network adaptor on the wire will have valid checksums. This issue can also occur when monitoring packets being transmitted between virtual machines on the same host, where a virtual device driver may omit the checksum calculation (as an optimization), knowing that the checksum will be calculated later by the VM host kernel or its physical hardware. - Connection-oriented communication - Karn's algorithm - List of TCP and UDP port numbers (a long list of ports and services) - Maximum segment lifetime - Maximum transmission unit - Micro-bursting (networking) - Nagle's algorithm - Port (computer networking) - T/TCP variant of TCP - TCP congestion avoidance algorithms - TCP global synchronization - TCP pacing - TCP segment - TCP sequence prediction attack - TCP tuning for high performance networks - WTCP a proxy-based modification of TCP for wireless networks - Transport Layer § Comparison of transport layer protocols - Vinton G. Cerf; Robert E. Kahn (May 1974). "A Protocol for Packet Network Intercommunication". IEEE Transactions on Communications. 22 (5): 637–648. doi:10.1109/tcom.1974.1092259. Archived from the original \|url=(help) on March 4, 2016. - Bennett, Richard (September 2009). "Designed for Change: End-to-End Arguments, Internet Innovation, and the Net Neutrality Debate" (PDF). Information Technology and Innovation Foundation. p. 11. Retrieved 11 September 2017. - Comer, Douglas E. (2006). Internetworking with TCP/IP: Principles, Protocols, and Architecture. 1 (5th ed.). Prentice Hall. ISBN 978-0-13-187671-2. - "TCP (Linktionary term)". - "RFC 791 – section 2.1". - "RFC 793". - "RFC 1323, TCP Extensions for High Performance, Section 2.2". - "RFC 2018, TCP Selective Acknowledgement Options, Section 2". - "RFC 2018, TCP Selective Acknowledgement Options, Section 3". - "RFC 1323, TCP Extensions for High Performance, Section 3.2". - "Transmission Control Protocol (TCP) Parameters: TCP Option Kind Numbers". IANA. - RFC 793 section 3.1 - RFC 793 Section 3.2 - Tanenbaum, Andrew S. (2003-03-17). Computer Networks (Fourth ed.). Prentice Hall. ISBN 978-0-13-066102-9. - "TCP Definition". Retrieved 2011-03-12. - Mathis; Mathew; Semke; Mahdavi; Ott (1997). "The macroscopic behavior of the TCP congestion avoidance algorithm". ACM SIGCOMM Computer Communication Review. 27 (3): 67–82. CiteSeerX 10.1.1.40.7002. doi:10.1145/263932.264023. - Paxson, V.; Allman, M.; Chu, J.; Sargent, M. (June 2011). "The Basic Algorithm". Computing TCP's Retransmission Timer. IETF. p. 2. sec. 2. doi:10.17487/RFC6298. RFC 6298. Retrieved October 24, 2015. - Stone; Partridge (2000). "When The CRC and TCP Checksum Disagree". ACM SIGCOMM Computer Communication Review: 309–319. CiteSeerX 10.1.1.27.7611. doi:10.1145/347059.347561. ISBN 978-1581132236. - "RFC 879". - "TCP window scaling and broken routers [LWN.net]". - David Murray; Terry Koziniec; Sebastian Zander; Michael Dixon; Polychronis Koutsakis (2017). "An Analysis of Changing Enterprise Network Traffic Characteristics" (PDF). The 23rd Asia-Pacific Conference on Communications (APCC 2017). Retrieved 3 October 2017. - "IP sysctl". Linux Kernel Documentation. Retrieved 15 December 2018. - Wang, Eve. "TCP timestamp is disabled". Technet - Windows Server 2012 Essentials. Microsoft. - Gont, Fernando (November 2008). "On the implementation of TCP urgent data". 73rd IETF meeting. Retrieved 2009-01-04. - Peterson, Larry (2003). Computer Networks. Morgan Kaufmann. p. 401. ISBN 978-1-55860-832-0. - Richard W. Stevens (November 2011). TCP/IP Illustrated. Vol. 1, The protocols. Addison-Wesley. pp. Chapter 20. ISBN 978-0-201-63346-7. - "Security Assessment of the Transmission Control Protocol (TCP)" (PDF). Archived from the original on March 6, 2009. Retrieved 2010-12-23.CS1 maint: BOT: original-url status unknown (link) - Security Assessment of the Transmission Control Protocol (TCP) - Jakob Lell. "Quick Blind TCP Connection Spoofing with SYN Cookies". Retrieved 2014-02-05. - "Some insights about the recent TCP DoS (Denial of Service) vulnerabilities" (PDF). - "Exploiting TCP and the Persist Timer Infiniteness". - "PUSH and ACK Flood". f5.com. - "Laurent Joncheray, Simple Active Attack Against TCP, 1995". - John T. Hagen; Barry E. Mullins (2013). TCP veto: A novel network attack and its application to SCADA protocols. Innovative Smart Grid Technologies (ISGT), 2013 IEEE PES. pp. 1–6. doi:10.1109/ISGT.2013.6497785. ISBN 978-1-4673-4896-6. - "TCP Interactive". www.medianet.kent.edu. - RFC 6182 - RFC 6824 - Raiciu; Barre; Pluntke; Greenhalgh; Wischik; Handley (2011). "Improving datacenter performance and robustness with multipath TCP". ACM SIGCOMM Computer Communication Review. 41 (4): 266. CiteSeerX 10.1.1.306.3863. doi:10.1145/2043164.2018467. - "MultiPath TCP - Linux Kernel implementation". - Raiciu; Paasch; Barre; Ford; Honda; Duchene; Bonaventure; Handley (2012). "How Hard Can It Be? Designing and Implementing a Deployable Multipath TCP". Usenix Nsdi: 399–412. - Bonaventure; Seo (2016). "Multipath TCP Deployments". IETF Journal. - Michael Kerrisk (2012-08-01). "TCP Fast Open: expediting web services". LWN.net. - Yuchung Cheng; Jerry Chu; Sivasankar Radhakrishnan & Arvind Jain (December 2014). "TCP Fast Open". IETF. Retrieved 10 January 2015. - "RFC 6937 - Proportional Rate Reduction for TCP". Retrieved 6 June 2014. - Grigorik, Ilya (2013). High-performance browser networking (1. ed.). Beijing: O'Reilly. ISBN 978-1449344764. - "TCP performance over CDMA2000 RLP". Archived from the original on 2011-05-03. Retrieved 2010-08-30 - Muhammad Adeel & Ahmad Ali Iqbal (2004). TCP Congestion Window Optimization for CDMA2000 Packet Data Networks. International Conference on Information Technology (ITNG'07). pp. 31–35. doi:10.1109/ITNG.2007.190. ISBN 978-0-7695-2776-5. - Yunhong Gu, Xinwei Hong, and Robert L. Grossman. "An Analysis of AIMD Algorithm with Decreasing Increases". 2004. - "Wireshark: Offloading". Wireshark captures packets before they are sent to the network adapter. It won't see the correct checksum because it has not been calculated yet. Even worse, most OSes don't bother initialize this data so you're probably seeing little chunks of memory that you shouldn't. New installations of Wireshark 1.2 and above disable IP, TCP, and UDP checksum validation by default. You can disable checksum validation in each of those dissectors by hand if needed. - "Wireshark: Checksums". Checksum offloading often causes confusion as the network packets to be transmitted are handed over to Wireshark before the checksums are actually calculated. Wireshark gets these “empty” checksums and displays them as invalid, even though the packets will contain valid checksums when they leave the network hardware later. - Stevens, W. Richard (1994-01-10). TCP/IP Illustrated, Volume 1: The Protocols. ISBN 978-0-201-63346-7. - Stevens, W. Richard; Wright, Gary R (1994). TCP/IP Illustrated, Volume 2: The Implementation. ISBN 978-0-201-63354-2. - Stevens, W. Richard (1996). TCP/IP Illustrated, Volume 3: TCP for Transactions, HTTP, NNTP, and the UNIX Domain Protocols. ISBN 978-0-201-63495-2.** \|Wikiversity has learning resources about Transmission Control Protocol\| \|Wikimedia Commons has media related to Transmission Control Protocol.\| - RFC 675 – Specification of Internet Transmission Control Program, December 1974 Version - RFC 793 – TCP v4 - STD 7 – Transmission Control Protocol, Protocol specification - RFC 1122 – includes some error corrections for TCP - RFC 1323 – TCP Extensions for High Performance [Obsoleted by RFC 7323] - RFC 1379 – Extending TCP for Transactions—Concepts [Obsoleted by RFC 6247] - RFC 1948 – Defending Against Sequence Number Attacks - RFC 2018 – TCP Selective Acknowledgment Options - RFC 5681 – TCP Congestion Control - RFC 6247 – Moving the Undeployed TCP Extensions RFC 1072, RFC 1106, RFC 1110, RFC 1145, RFC 1146, RFC 1379, RFC 1644, and RFC 1693 to Historic Status - RFC 6298 – Computing TCP's Retransmission Timer - RFC 6824 – TCP Extensions for Multipath Operation with Multiple Addresses - RFC 7323 – TCP Extensions for High Performance - RFC 7414 – A Roadmap for TCP Specification Documents - Oral history interview with Robert E. Kahn, Charles Babbage Institute, University of Minnesota, Minneapolis. Focuses on Kahn's role in the development of computer networking from 1967 through the early 1980s. Beginning with his work at Bolt Beranek and Newman (BBN), Kahn discusses his involvement as the ARPANET proposal was being written, his decision to become active in its implementation, and his role in the public demonstration of the ARPANET. The interview continues into Kahn's involvement with networking when he moves to IPTO in 1972, where he was responsible for the administrative and technical evolution of the ARPANET, including programs in packet radio, the development of a new network protocol (TCP/IP), and the switch to TCP/IP to connect multiple networks. - IANA Port Assignments - IANA TCP Parameters - John Kristoff's Overview of TCP (Fundamental concepts behind TCP and how it is used to transport data between two endpoints) - TCP fast retransmit simulation animated: slow start, sliding window, duplicated Ack, congestion window[permanent dead link] - TCP, Transmission Control Protocol - Checksum example - Engineer Francesco Buffa's page about Transmission Control Protocol - TCP tutorial - Linktionary on TCP segments - TCP Sliding Window simulation animated (ns2)[permanent dead link] - Multipath TCP<\|endoftext\|>	3.828125
444	# Cumulative Tables and Graphs ## Cumulative Cumulative means "how much so far". Think of the word "accumulate" which means to gather together. To have cumulative totals, just add up the values as you go. ### Example: Jamie has earned this much in the last 6 months: Month Earned March \$120 April \$50 May \$110 June \$100 July \$50 August \$20 To work out the cumulative totals, just add up as you go. The first line is easy, the total earned so far is the same as Jamie earned that month: Month Earned Cumulative March \$120 \$120 But for April, the total earned so far is \$120 + \$50 = \$170 : Month Earned Cumulative March \$120 \$120 April \$50 \$170 And for May we continue to add up: \$170 + \$110 = \$280 Month Earned Cumulative March \$120 \$120 April \$50 \$170 May \$110 \$280 Do you see how we add the previous month's cumulative total to this month's earnings? Here is the calculation for the rest: • June is \$280 + \$100 = \$380 • July is \$380 + \$50 = \$430 • August is \$430 + \$20 = \$450 And this is the result Month Earned Cumulative March \$120 \$120 April \$50 \$170 May \$110 \$280 June \$100 \$380 July \$50 \$430 August \$20 \$450 The last cumulative total should match the total of all earnings: \$450 is the last cumulative total ... ... it is also the total of all earnings: \$120+\$50+\$110+\$100+\$50+\$20 = \$450 So we got it right. So that's how to do it, add up as you go down the list and you will have cumulative totals. We could also call it a "Running Total" ## Graphs We can make cumulative graphs, too. Just plot each cumulative total: Cumulative Bar Graph Cumulative Line Graph<\|endoftext\|>	4.8125
668	Undergraduate Mathematics ← Proof by contradiction Proof by exhaustion Taking the Contrapositive → Proof by exhaustion, also known as proof by cases, perfect induction, or the brute force method, is a method of mathematical proof in which the statement to be proved is split into a finite number of cases and each case is checked to see if the proposition in question holds.[1] A proof by exhaustion contains two stages: 1. A proof that the cases are exhaustive; i.e., that each instance of the statement to be proved matches the conditions of (at least) one of the cases. 2. A proof of each of the cases. In the Curry–Howard isomorphism, proof by exhaustion and case analysis are related to ML-style pattern matching. Example To prove that every integer that is a perfect cube is a multiple of 9, or is 1 more than a multiple of 9, or is 1 less than a multiple of 9. Proof: Each cube number is the cube of some integer n. Every integer n is either a multiple of 3, or 1 more or 1 less than a multiple of 3. So these 3 cases are exhaustive: • Case 1: If n = 3p, then n3 = 27p3, which is a multiple of 9. • Case 2: If n = 3p + 1, then n3 = 27p3 + 27p2 + 9p + 1, which is 1 more than a multiple of 9. For instance, if n = 4 then n3 = 64 = 9x7 + 1. • Case 3: If n = 3p − 1, then n3 = 27p3 − 27p2 + 9p − 1, which is 1 less than a multiple of 9. For instance, if n = 5 then n3 = 125 = 9×14 − 1.∎ Number of cases There is no upper limit to the number of cases allowed in a proof by exhaustion. Sometimes there are only two or three cases. Sometimes there may be thousands or even millions. For example, rigorously solving an endgame puzzle in chess might involve considering a very large number of possible positions in the game tree of that problem. The first proof of the four colour theorem was a proof by exhaustion with 1,936 cases. This proof was controversial because the majority of the cases were checked by a computer program, not by hand. The shortest known proof of the four colour theorem today still has over 600 cases. Mathematicians prefer to avoid proofs with large numbers of cases, as they seem inelegant, and in general the probability of an error in the whole proof increases in proportion to the number of cases. A proof with a large number of cases leaves an impression that the theorem is only true by coincidence, and not because of some underlying principle or connection. Other types of proofs—such as proof by induction (mathematical induction)—are considered more elegant. However, there are some important theorems for which no other method of proof has been found, such as<\|endoftext\|>	4.6875
1,294	Point of View really sets the tone of a story and influences the feel of the story for the reader. If a story is told from a character’s Point of View, you get to be in their head and experience the story from their perspective. When a story is told from a narrator’s Point of View, you get to see the whole picture and experience it from an outsider perspective. There are a few layers to Point of View that can be taught in steps. The first objective is for students to understand that Point of View is the voice that is telling the story, which is usually either a character or a narrator. The second layer is identifying the Point of View as First Person (character) or Third Person (narrator). It’s important for students to understand that the Point of View heavily influences their reading experience so identifying the Point of View will give them a deeper understanding of the story being told. I like to spend a sufficient amount of time on each strategy to allow for an introduction, modeling, scaffolding, independent practice, assessment, and reflection. Therefore, I spend approximately 1 week on each strategy and follow a similar instructional routine. This is day 4 of Point of View Week – Assessment. Connection: I always start by connecting today’s lesson to something kids have previously learned so that it triggers their schema and background knowledge. Since this is the fourth day they are learning about Point of View, I make a connection to the independent practice lesson we did yesterday. I remind them that yesterday, they applied the Point of View strategy to their own books. And now that they’ve practiced Point of View in different ways throughout the week, it is time to prove that they understand it. Teaching Point: This is when I tell kids explicitly what we will be working on. I say that today they will be filling out an Point of View Guide (see resource) while they are reading a book of their choice. They will turn in the Guide as an assessment of their understanding. I will use those to provide feedback to the students and parents. The Guides also help me put together flexible strategy groups for small group instruction. Active Engagement: This is where students get to try out the strategy that I just taught them. I place a copy of the Guide on the projector and show students our current Read Aloud book. I fill in the name, date, and title of the book in the appropriate spots. First I want to know who is telling the story. Kids will often say that it’s a character but if so, I want them to be more specific; what is the character’s name? Next I ask students to use that information to decide whether the Point of View is First Person or Third Person. After a minute or two of thinking time, I tell them to turn and talk to their partner to share their ideas with evidence. I give the students a few minutes then call on some to share. I circle the correct choice, which in this case is First Person. After a brief discussion, I include evidence in the section that asks, “How do you know?” We know because the voice of the story is using words like I, me, my, we, and us and talking about himself. Link to Ongoing Work: During this portion of the mini-lesson, I give the students a task that they will focus on during Independent Reading time. I tell them that during Independent Reading, their job is to complete their own Point of View Guide with their own book. I remind them that this will be turned in for a grade/feedback at the end of Independent Reading time so it should be their best quality work. When they finish their task, they should continue reading books from their browsing box. After asking if there are any questions, I send them off for Prep Time. Transition Time: Every day after the mini-lesson, students get 5 minutes of Prep Time to choose new books (if needed), find a comfy spot, use the bathroom, and anything else they might need to do to prepare for 40 minutes of uninterrupted Independent Reading. I set it up that way so that students have no reason to get out of their spots. They are expected to have 5 books in their browsing box at all times so if they finish a book they have others to choose from without moving around the room. They are also expected to have a pencil and sticky notes in their browsing boxes in case they need them for the day’s task. I strongly encourage them to use the bathroom so they do not need to go during reading time. At the end of the 5 minute Prep Time, I do a countdown, 5 4 3 2 1, Level 0 (referring to volume level). By the end of countdown, students must be in their spots and silent with all of the materials they need to sustain their reading. They must follow the distance rule of arm’s length apart from any other student. They are not to get out of their spots for any reason so that they can focus on their book and their task. Because I use Independent Reading time to work with students one-on-one or in small groups, I really stress to the students that the teacher is not available to everyone during this time. I encourage them to problem solve on their own and hold all questions or comments until the end of Independent Reading time. All of this takes practice but once it is all in place, Independent Reading becomes a magical time when students are engrossed in their books and the teacher is free to meet individual needs of students through conferencing, strategy groups, or guided reading. Guided Practice: Today, I will be conferencing with students right at their comfy spots and helping some with their Guide. This is also when I could pull students for assessments, one-on-one reading, strategy groups, or guided reading groups. Closing: At the end of 40 minutes, I remind students that their job during reading time was to complete their Guide. They should make sure all parts are filled and place them in the Finished Basket. I then tell the class that we will wrap up our Point of View work tomorrow. Reader’s Workshop has come to an end so students put their browsing boxes away and make sure the library is neat and organized.<\|endoftext\|>	4.21875
967	NCERT Solutions: Understanding Elementary Shapes (Exercise 5.7 & 5.8) NCERT Solutions for Class 8 Maths Chapter 5 - Understanding Elementary Shapes (Exercise 5.7 and 5.8) Exercise 5.7 Q1. Say true or false: (a) Each angle of a rectangle is a right angle. (b) The opposite sides of a rectangle are equal in length. (c) The diagonals of a square are perpendicular to one another. (d) All the sides of a rhombus are of equal length. (e) All the sides of a parallelogram are of equal length. (f) The opposite sides of a trapezium are parallel. Ans: (a) True (b) True (c) True (d) True (e) False (f) False Q2. Give reasons for the following: (a) A square can be thought of as a special rectangle. (b) A rectangle can be thought of as a special parallelogram. (c) A square can be thought of as a special rhombus. (d) Squares, rectangles, parallelograms are all quadrilateral. (e) Square is also a parallelogram. Ans: (a) Because its all angles are right angle and opposite sides are equal. (b) Because its opposite sides are equal and parallel. (c) Because its four sides are equal and diagonals are perpendicular to each other. (d) Because all of them have four sides. (e) Because its opposite sides are equal and parallel. Q3. A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral? Ans: A square is a regular quadrilateral because all the interior angles are 90o, and all sides are of the same length. Exercise 5.8 Q1. Examine whether the following are polygons. If anyone among these is not, say why? Ans: (a) As it is not a closed figure, therefore, it is not a polygon. (b) It is a polygon because it is closed by line segments. (c) It is not a polygon because it is not made by line segments. (d) It is not a polygon because it not made only by line segments, it has curved surface also. Q2. Name each polygon: Make two more examples of each of these. (b) Triangle (c) Pentagon (d) Octagon Q3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn. Ans: ABCDEF is a regular hexagon and triangle thus formed by joining AEF is an isosceles triangle. Q4. Draw a rough sketch of a regular octagon. (Use squared paper if you wish). Draw a rectangle by joining exactly four of the vertices of the octagon. Ans: ABCDEFGH is a regular octagon and CDGH is a rectangle. Q5. A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals. Ans: ABCDE is the required pentagon and its diagonals are AD, AC, BE and BD. Exercise 5.9 (Old NCERT) Ques 1: Match the following: (a) Cone (b) Sphere (c) Cylinder (d) Cuboid (e) Pyramid Give two example of each shape. Ans: (a) Cone (b) Sphere (c) Cylinder (d) Cuboid (e) Pyramid Ques 2: What shape is: (a) Your instrument box? (b) A brick? (c) A match box? (e) A sweet laddu? Ans: (a) Cuboid (b) Cuboid (c) Cuboid (d) Cylinder (e) Sphere The document NCERT Solutions for Class 8 Maths Chapter 5 - Understanding Elementary Shapes (Exercise 5.7 and 5.8) is a part of the Class 6 Course NCERT Textbooks & Solutions for Class 6. All you need of Class 6 at this link: Class 6 494 docs NCERT Textbooks & Solutions for Class 6 494 docs Up next Explore Courses for Class 6 exam Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , ;<\|endoftext\|>	4.75
281	Taxation is a cost or levy imposed on individuals and businesses to finance the services and policies provided by governments. Taxation can be one of the most controversial aspects of government policy – in how tax money is collected and spent and whether various groups of taxpayers are treated fairly. Although taxation systems across countries and zones vary, they typically use a direct tax – levied on individuals and businesses that generate income – or indirect taxes, such as sales and value added taxes that are mostly collected for government by business. Governments can and do use taxation to direct activity, for example with tobacco taxes designed to discourage smoking by making it more expensive, or with rebates to companies who train employees, to improve the skills of the workforce. The major uses for taxation revenues in developed countries tend to be for defence, health and education. Nations vary widely in attitudes to taxation, with some seeking to encourage commerce by promoting themselves as low-tax regimes, while others see taxation as a means to enable a more equitable society by redistribution of taxes. This redistribution can provide opportunities for low socioeconomic groups to access universal education, health care or other government run programs such as subsidizing public universities. When starting a new business in a foreign market it is important to understand not only the laws of taxation in that economy but also how taxation works within the culture, how it is collected and what rebates and deductions are available.<\|endoftext\|>	3.71875
276	The Rosetta is a space probe launched by the European Space Agency in March 2004. Back in August 2014, the space probe was able to reach comet 67P and was able to land on it a month after – marking the first time we’ve ever landed on a comet. Yesterday, Rosetta announced that they have discovered molecular oxygen in the gas cloud surrounding the comet – which came as a surprise to the scientists as they initially thought that oxygen reacts with other elements as planets form. This new finding challenges our current ideas on how the solar system was formed. Free oxygen was found to be the fourth most common gas around the comet, with the first three being water vapor, carbon monoxide, and carbon dioxide. The data was so surprising that the scientists had to go everything again to make sure it all checked out. Oxygen reacts very easily with other elements to form compounds, rather than stay in its unique form. The researchers suggest that oxygen must have been frozen very quickly and was trapped in clumps of material early on in the formation of the Solar System. Several theories about how the planets and comets formed around the Sun suggest a violent process that would have heated up the frozen oxygen – which would by then, have reacted with other elements. This new information contradicts that idea, and this indicates that the formation of the Solar System formation may have been calmer.<\|endoftext\|>	4.0625
482	Imagine if someone built two giant coal-fired power plants right next to the last livable place in your country. How terrible would this be? That is what the Bangladesh government is planning to do near the Sundarbans, the largest intact mangrove forest on Earth -- and the only sizeable mangrove area left for the globally endangered Bengal Tiger (Panthera tigris tigris) -- which likely numbers less than 2,500 animals alive today. The Bengal Tigers and their mangrove habitats in the Sundarbans are now under imminent threat. The Bangladesh government has recently signed a $1.7 billion agreement with an Indian corporation, Bharat Heavy Electrics, to build the massive Rampal coal-fired power plant. Comprising two giant coal-fired generators just a dozen or so kilometers from the Sundarbans, the power plant will degrade the environment via air pollution and fly ash, unregulated increased resource extraction, road and infrastructure expansion, and the risk of spillage of coal and its by-products on site and during transportation. This proposed development has raised the ire of civil society, many scientists, and global conservation groups (see here and here). Both UNESCO and the IUCN have urged the Bangladesh government to move the power plant to another suitable area to limit its environmental threats to the Sundarbans. So great are such perils that the government of Sri Lanka cancelled an agreement with the same Indian corporation to build a similar but smaller coal-fired power plant in its eastern port city of Trincomalee, largely because of concerns about the environmental threats it posed. A Sanctuary for Nature and People In addition to providing critical habitat for Bengal Tigers, the Sundarbans harbor many other rare or endangered species, such as the Estuarine Crocodile, Indian Python, and the Ganges and Irrawaddy Dolphins. Moreover, more than one million local people depend on natural resources from the Sundarbans to sustain their livelihoods -- and rely on this natural barrier for protection against calamities such as destructive tropical monsoons and tsunamis. The Sundarbans is already facing threats from climate change and local human-caused disasters. But one thing is for certain, the proposed coal-fired power plant would further imperil Earth's last remaining mangrove habitat for Bengal Tigers and other rare wildlife.<\|endoftext\|>	3.6875
2,256	M01 #11 : Retired Discussions [Locked] - Page 2 Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack It is currently 22 Jan 2017, 17:59 ### 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 # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # M01 #11 Author Message Math Expert Joined: 02 Sep 2009 Posts: 36601 Followers: 7097 Kudos [?]: 93477 [0], given: 10563 ### Show Tags 25 Mar 2013, 04:07 Expert's post 2 This post was BOOKMARKED elmagnifico wrote: A 4 cm cube is cut into 1 cm cubes. What is the percentage increase in the surface area after such cutting? (A) 4% (B) 166% (C) 266% (D) 300% (E) 400% [Reveal] Spoiler: OA D Source: GMAT Club Tests - hardest GMAT questions BELOW IS REVISED VERSION OF THIS QUESTION: If a cube with the length of the side of 4 cm is cut into smaller cubes with the length of the side of 1 cm, then what is the percentage increase in the surface area of the resulting cubes? A. 4% B. 166% C. 266% D. 300% E. 400% A cube has 6 faces. The surface area of a cube with the length of the side of 4 cm is 64^2=616 cm^2. Now, since the volume of the big cube is 4^3=64 cm^3 and the volume of the smaller cubes is 1^3=1 cm^3, then when the big cube is cut into the smaller cubes we'll get 64/1=64 little cubes. Each of those little cubes will have the surface area equal to 61^2=6 cm^2, so total surface are of those 64 little cubes will be 664 cm^2. 664 is 4 times more than 616 which corresponds to 300% increase. Or: general formula for percent increase or decrease, (percent change): $$Percent=\frac{Change}{Original}100$$ So the percent increase will be: $$Percent=\frac{Change}{Original}100=\frac{664-616}{616}100=300%$$. _________________ Intern Joined: 24 Mar 2013 Posts: 30 Followers: 0 Kudos [?]: 4 [0], given: 0 ### Show Tags 25 Mar 2013, 14:13 Explanation : Original SA = 644 = 96 After cutting; SA = n611 = 6n ; where n is the number of pieces of cubes of 1cm side. to find the value of n; original volume = volume after cutting the cube.. which means 4^3 = n1^3 which gives n is equal to 64. so, After cutting; SA =664=384. Percentage increase = ((384-96)/96)100=(288/96)100=300% (Option D) _________________ On the way .. - Alok K G. Intern Joined: 08 Mar 2013 Posts: 19 Followers: 0 Kudos [?]: 3 [0], given: 7 Is increase from 1 to 4 considered 400% or 300%? [#permalink] ### Show Tags 30 Apr 2013, 21:50 This is going to this question: a-4-cm-cube-is-cut-into-1-cm-cube-what-is-the-percentage-37027.html?fl=similar Don't want to repost it, since it seems to have been discussed many many times. I understand the calculations, don't understand the reasoning for calling a 4-fold increase as an increase of 300% rather than 400. As I read that question, I take it to mean "how many fold did this increase" rather than "by how much did it increase". Am I just not reading the question correctly? Misinterpreting the keywords? Senior Manager Joined: 16 Dec 2011 Posts: 452 Followers: 11 Kudos [?]: 197 [0], given: 70 Re: Is increase from 1 to 4 considered 400% or 300%? [#permalink] ### Show Tags 30 Apr 2013, 23:46 When the value increased to 4 from 1, increase is 3 ==> increase is 300% 100% of 1 is 1 200% of 1 is 2. Similarly, 400% of 1 is 4. The theory is: • when you calculate % increase, divide increase by original value. • when you calculate % for new value wrt original value, divide new value by original value. Math Expert Joined: 02 Sep 2009 Posts: 36601 Followers: 7097 Kudos [?]: 93477 [0], given: 10563 Re: Is increase from 1 to 4 considered 400% or 300%? [#permalink] ### Show Tags 01 May 2013, 01:14 Dixon wrote: This is going to this question: a-4-cm-cube-is-cut-into-1-cm-cube-what-is-the-percentage-37027.html?fl=similar Don't want to repost it, since it seems to have been discussed many many times. I understand the calculations, don't understand the reasoning for calling a 4-fold increase as an increase of 300% rather than 400. As I read that question, I take it to mean "how many fold did this increase" rather than "by how much did it increase". Am I just not reading the question correctly? Misinterpreting the keywords? Merging similar topics. Revised version of this question is here: m01-70731-20.html#p1202018 Hope it helps. _________________ Director Joined: 14 Dec 2012 Posts: 842 Location: India Concentration: General Management, Operations GMAT 1: 700 Q50 V34 GPA: 3.6 Followers: 59 Kudos [?]: 1289 [0], given: 197 Re: Is increase from 1 to 4 considered 400% or 300%? [#permalink] ### Show Tags 02 May 2013, 23:37 Dixon wrote: This is going to this question: a-4-cm-cube-is-cut-into-1-cm-cube-what-is-the-percentage-37027.html?fl=similar Don't want to repost it, since it seems to have been discussed many many times. I understand the calculations, don't understand the reasoning for calling a 4-fold increase as an increase of 300% rather than 400. As I read that question, I take it to mean "how many fold did this increase" rather than "by how much did it increase". Am I just not reading the question correctly? Misinterpreting the keywords? Hi, two fold of something ===> twice of something three fold of something====> thrice of something four fold of somthing ===> four times of something lets take any value X....==> FOUR FOLD OF X will be four times of X= 4X Now if calculate percent increase then ((4X-X)/X)100= 300 % Hope it helps... SKM _________________ When you want to succeed as bad as you want to breathe ...then you will be successfull.... GIVE VALUE TO OFFICIAL QUESTIONS... learn AWA writing techniques while watching video : http://www.gmatprepnow.com/module/gmat-analytical-writing-assessment Intern Joined: 25 Apr 2013 Posts: 5 Location: India Concentration: Strategy, Marketing GPA: 2.7 WE: Information Technology (Computer Software) Followers: 0 Kudos [?]: 5 [0], given: 5 ### Show Tags 03 May 2013, 08:53 Total Surface Area of bigger cube = 6a^2 = 6(4)^2 = 96 Total Surface Area of 1 smaller cube = 6a^2 = 6(1)^2 = 6 Let 'x' be total number of small cubes formed after cutting Volume of bigger cube = (x) Volume of smaller cube => 4^3 = (x)1^3 => x = 64 Hence,total surface area of 64 smaller cubes = 64 6 = 384 Now % increase in Total Surface Area = [(Final Area - Initial Area)/Initial Area]100=> [(384-96/96 )]100 = 300% Hence Ans (D) Re: M01 #11 [#permalink] 03 May 2013, 08:53 Go to page Previous 1 2 [ 27 posts ] Similar topics Replies Last post Similar Topics: m01 1 27 May 2009, 10:20 30 M01 #10 24 29 Sep 2008, 23:43 19 m01 Q17 18 18 Sep 2008, 02:14 5 M01 27 11 21 Aug 2008, 09:29 19 M01-Q15 20 16 Jul 2008, 20:25 Display posts from previous: Sort by # M01 #11 Moderator: Bunuel Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.<\|endoftext\|>	4.40625
520	Chemistry - Introduction Chemistry is a branch of Natural Science that studies about the structure, composition, and changing properties of matters. Chemistry studies the smallest part of a matter i.e. atom (along with its all properties) to the large materials (e.g. gold, silver, iron, etc.) and their properties. Chemistry also studies the intermolecular forces (that provide matter the general properties) and the interactions between substances through the chemical reactions. In 1998, Professor Raymond Chang defined Chemistry as − "Chemistry" to mean the study of matter and the changes it undergoes. It is believed that the study of chemistry started with the theory of four elements propounded by Aristotle. The four theory of elements states that “fire, air, earth, and water were the fundamental elements from which everything is formed as combination.” Because of his classical work namely “The Sceptical Chymist,” Robert Boyle, is known as the founding father of chemistry. Boyle formulated a law, became popular as ‘Boyle’s Law.’ Boyle’s law is an experimental gas law that analyzes the relationship between the pressure of a gas and volume of the respective container. By advocating his law, Boyle rejected the classical ‘four elements’ theory. The American scientists Linus Pauling and Gilbert N. Lewis collectively propounded the electronic theory of chemical bonds and molecular orbitals. The United Nations declared 2011 as the ‘International Year of Chemistry.’ The matter is defined in chemistry as anything that has rest mass and volume and also takes space. The matter is made up of particles. The atom is the fundamental unit of chemistry. The atom consists of a dense core known as the atomic nucleus and it is surrounded by a space known as the electron cloud. The nucleus (of an atom) is composed of protons (+ve charged particles) and neutrons (neutral or uncharged particles); collectively, these two are known as nucleons (as shown in the image given below). A chemical element is a pure form of a substance; it consists of single type of atom. The periodic table is the standardized representation of all the available chemical elements. A compound is a pure form of a substance; it composed of more than one elements. A molecule is the smallest indivisible part of a pure chemical substance; molecule has distinctive set of chemical properties (see the image given below).<\|endoftext\|>	3.921875
848	# Special case of linear equations: Horizontal lines #### Everything You Need in One Place Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered. #### Learn and Practice With Ease Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals. #### Instant and Unlimited Help Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now! 0/0 0/3 ##### Examples ###### Lessons 1. Determine the line of equations from the following points. i) $(-5,2), (3,2)$ ii) $(-3, 4), (4,4)$ iii) (b,m), (c,m) 1. Write the equation of the line with the given information. 1. Horizontal, passes through (4,7) 2. Horizontal, passes through ($- {2 \over 5}, -3)$ ###### Topic Notes This lesson will focus on the horizontal lines special case of linear equations. We will learn how to determine the equations for graphs where all the y points have the same value. ## Equation of a horizontal line A horizontal line runs from left to right and lies parallel to the x-axis. It's also a linear line, much like many that you've encountered so far (e.g. slope intercept form, general form). But what makes it so special? The equation of a horizontal line is unique and you'll soon see why. ## What is the slope of a horizontal line When we deal with horizontal lines, what is its slope? You can probably answer this question without too much mathematics. Look at a perfectly horizontal line on the graph below. A slope tells us how steep a line is. In fact, the slope can be found via rise/run, which helps us determine steepness. Looking at the red line above, how steep is it? The slope of a horizontal line is actually 0. It never rises (or falls), so if we were to find the rise/run, we'd get 0/run. 0 divided by anything is still equal to 0, and therefore, our horizontal line's slope is 0. Backtracking a bit, what is the equation of a horizontal line? Since the line is horizontal and runs parallel to the x-axis, the equation is simply: y = y-intercept of the line or, more commonly y = b (where b = the y-intercept) ## Example problems Question 1: Determine the line of equations from the following points. i) (-5, 2), (3, 2) Solution: Once again, take a look at either points or the graph itself. The y axis coordinates are both 4. The line runs along y =4. Your answer will be that the equation is y=4. iii) (b,m), (c,m) Solution: Even without numbers, you now understand that the y-coordinates can tell you the equation of the line. In this case, the equation = y=m. Question 2: Write the equation of the line with the given information: Horizontal, passes through (4,7) Solution: While we aren't given two points to observe, the information given tells us all we need to know. It states that the line we're dealing with is horizontal. Therefore, we can take a look at just the y coordinate of the given point to help us determine the horizontal line's equation. You should find that the equation = y=7. Looking on what to tackle next? Learn about parallel and perpendicular lines in linear functions, and what they should look like. You can also learn how to graph linear inequalities in two variables or graph systems of linear inequalities. There's still a lot to learn in terms of linear functions, so make sure you've got this lesson on horizontal line slope cemented before moving on! Horizontal lines : A graph where all the y points has the same value. Ex. y = 5<\|endoftext\|>	4.53125
1,952	Medically reviewed by Drugs.com. Last updated on May 13, 2019. What Is It? A peptic ulcer is a sore or hole that forms in the lining of the stomach or intestine. The word "peptic" refers to the digestive tract. An ulcer in the lining of the stomach is a gastric ulcer. An ulcer in the first part of the small intestine is a duodenal ulcer. The lining of the stomach is a layer of special cells and mucous. Mucous prevents the stomach and duodenum from being damaged by acid and digestive enzymes. If there is a break in the lining (such as an ulcer), the tissue under the lining can be damaged by the enzymes and corrosive acid. If the ulcer is small, there may be few symptoms. The wound can heal on its own. If the ulcer is deep, it can cause serious pain or bleeding. Rarely, acids in the digestive juices may eat completely through the stomach or duodenum wall. Peptic ulcers are very common. They become more common as people age. The bacterium Helicobacter pylori is believed to cause most peptic ulcers. This bacteria causes inflammation in the stomach lining. This probably makes the lining vulnerable. But only a minority of people infected with H. pylori develop ulcers. NSAIDs block the formation of some prostaglandins. Prostaglandins are chemicals that normally help protect against ulcers. With less prostaglandins, ulcers are more likely to form. Several factors can increase your risk of developing a peptic ulcer. Risk factors include: Smoking (particularly if you are infected with H. pylori) Excessive alcohol use However, contrary to popular belief, stress and spicy foods do not seem to increase the risk of ulcers. Most people with ulcers complain of a burning or gnawing pain in the upper abdomen. This typically occurs when the stomach is empty. These symptoms may be worse at night or upon waking. However, some people's pain may worsen when they eat. Other symptoms include: Loss of appetite Some of these symptoms may be relieved by taking over-the-counter antacids or avoiding spicy or acidic foods. In general, symptoms worsen as an ulcer grows, or if more than one ulcer develops. Some people with mild disease don't have any symptoms. In more severe cases, ulcers may bleed or extend deep into the wall of the stomach or intestine. Bleeding from large ulcers can be life threatening. Blood may appear in the vomit. It may appear either red or black, or resemble coffee grounds. Blood also may appear in the stools, which would look tarry-black or maroon. Peritonitis is a very serious abdominal infection. It may develop if the ulcer eats completely through the wall of the stomach or intestine. If your health care professional suspects that you have a peptic ulcer, he or she may recommend one of the following tests: A blood antibody test for evidence of H. pylori infection. This test is widely available and simple to do. If the test is positive, treatment might be given without more invasive tests. However, the H. pylori blood test is not always accurate. The test results may remain positive for years after an H. pylori infection has been treated. Also, the test cannot tell whether an H. pylori infection has caused an ulcer. A stool test for the presence of H. pylori antigen. This test is more specific than the blood antibody test. An esophagogastroduodenoscopy (EGD or endoscopy). A flexible, lighted tube with a tiny camera on the end is passed through your throat into your stomach and intestines. This allows your doctor to examine the walls of the stomach and duodenum. The doctor may snip off a small piece of the lining of the stomach for a biopsy. A biopsy is a close examination of the tissue in a laboratory. A biopsy can show whether there is an ongoing infection with H. pylori. It can also check to make sure an ulcer did not form because of cancer. An upper-gastrointestinal (GI) series. This test is rarely done today because endoscopy is generally a better test. It involves a series of X-rays. They are taken after you drink a chalky liquid that coats the esophagus, stomach and upper part of the intestine. Other tests for H. pylori. Another test to detect the bacteria is called a urea breath test. You swallow a substance containing carbon (in many cases, a small amount of radioactivity is present). If H. pylori are present in your stomach, you will have a positive breath test. Stool samples can be tested for proteins that are associated with the bacteria. Sometimes, more than one test is needed to diagnose your condition. Ulcers caused by a medication should begin healing shortly after you stop taking the drug. Anti-acid medicine may be used for two to six weeks to help healing and relieve pain. Ulcers caused by H. pylori can heal after the bacteria are killed. Typically, you will take antibiotics along with acid-suppressing medicine for two weeks. Then you may take acid-suppressing medication for another four to eight weeks. Gastric ulcers tend to heal more slowly than duodenal ulcers. Uncomplicated gastric ulcers take up to two or three months to heal completely. Duodenal ulcers take about six weeks to heal. An ulcer can temporarily heal without antibiotics. But it is common for an ulcer to recur or for another ulcer to form nearby, if the bacteria are not killed. Peptic ulcers are not usually preventable the first time around. Infection with H. pylori is extremely common. It is probably spread from person to person. Crowded living space appears to be a risk factor. Good hygiene may limit the spread of H. pylori somewhat. This includes washing your hands thoroughly before eating and after using the bathroom. Recurrent ulcers from H. pylori can usually be prevented if you get appropriate treatment for your first ulcer. This should include antibiotics that kill the bacteria. You may help to prevent peptic ulcers by: Avoiding excessive alcohol use Limiting the use of NSAIDs for pain For ulcers caused by H. pylori, treatment requires a combination of medications. The goals of treatment are to: Kill H. pylori bacteria in the body Reduce the amount of acid in the stomach Protect the lining of the stomach and intestines Most patients are treated with "triple therapy." This requires taking two antibiotics and one acid-suppressing medication for one to two weeks. Your doctor will prescribe a specific regimen based on convenience, cost and any allergies you have. If your ulcer occurred while you were using a non-steroidal anti-inflammatory drug (NSAID), you will need to stop taking it. Healing will begin almost immediately. Your doctor also will recommend medications to reduce acid damage during healing. These may include antacids to neutralize gastric acids. Medications that decrease the amount of acid produced by the stomach may also be used. Examples include H2 blockers or proton pump inhibitors. Emergency treatment may be needed if an ulcer causes serious bleeding. Usually, this treatment is done through an endoscope. Acid-blocking medications may be given intravenously (injected into a vein). Blood transfusions may be necessary if the bleeding is severe. In rare circumstances, surgery may be needed to treat a perforated or bleeding peptic ulcer. Surgery for peptic ulcer disease may involve closing a bleeding artery. Surgery is rarely needed for peptic ulcer treatment these days. That is because treatments for H. pylori infections and other causes of peptic ulcer disease are so successful. When To Call a Professional Call for medical advice if you have continuing abdominal pain or indigestion. Also call if you need to take antacids frequently to prevent these symptoms. Seek emergency care if you experience: A sudden sharp pain in your abdomen Bloody or black vomit Maroon or black stools With proper treatment, the outlook for peptic ulcers is excellent. To prevent another ulcer, people who have had a peptic ulcer should avoid: Aspirin (unless a low dose is needed to prevent a heart attack or stroke) Learn more about Peptic Ulcer IBM Watson Micromedex Symptoms and treatments Mayo Clinic Reference National Institute of Diabetes & Digestive & Kidney Disorders Office of Communications and Public Liaison Building 31, Room 9A04 31 Center Drive, MSC 2560 Bethesda, MD 20892-2560 American College of Gastroenterology (ACG) P.O. Box 3099 Arlington, VA 22302 American Gastroenterological Association 4930 Del Ray Ave. Bethesda, MD 20814 Always consult your healthcare provider to ensure the information displayed on this page applies to your personal circumstances.<\|endoftext\|>	3.859375
3,599	Libra constellation lies in the southern sky. It is one of the zodiac constellations, first catalogued by Ptolemy in the 2nd century. The constellation’s name means “the weighing scales” in Latin, and Libra is usually depicted as the scales held by the Greek goddess of justice Dike (or Astraea), represented by the neighbouring Virgo constellation. Libra is the only zodiac constellation that represents an object, not an animal or a character from mythology. The four brightest stars in the constellation form a quadrangle. Alpha and Beta Librae mark the scales’ balance beam, and Gamma and Sigma Librae represent the weighing pans. Libra is also home to HD 140283, popularly known as Methuselah, currently the oldest known star in the universe. The constellation is represented by the symbol ♎. It does not contain any first magnitude stars. FACTS, LOCATION & MAP Libra is the 29th constellation in size, occupying an area of 538 square degrees. It lies in the third quadrant of the southern hemisphere (SQ3) and can be seen at latitudes between +65° and -90°. The neighboring constellations are Centaurus, Hydra, Lupus, Ophiuchus, Scorpius, Serpens Caput and Virgo. Libra contains three stars with known planets and does not have any Messier objects. The brightest star in the constellation is Zubeneschamali, Beta Librae, with an apparent magnitude of 2.61. There is one meteor shower associated with the constellation, the May Librids. Ancient Greeks knew the part of the sky occupied by the Libra constellation as Chelae, or “claws,” and considered it part of Scorpio constellation. Chelae represented the scorpion’s claws. The association of this region of the sky with scales was established among the Romans in the first century BC. It is said that Moon was located in Libra when Rome was founded. The Romans considered Libra to be a favoured constellation, one associated with balanced seasons and equal length of night and day. The Sun was at the autumnal equinox in Libra until the year 729, when the precession of the equinoxes shifted the equinox to Virgo. The autumnal equinox will move to constellation Leo in the year 2439. The Romans were not the first to associate Libra with the idea of balance. The Babylonians called it ZIB.BA.AN.NA, which means “the balance of heaven,” about a thousand years before Christ. Once Libra became associated with balance, its association with Scorpio’s claws faded and the one with the goddess of justice, the Greek Dike or Astraeia, represented by the constellation Virgo, became stronger. As a reminder that Libra was once considered a part of Scorpio constellation, the brightest star in Libra, Beta Librae, has the name Zubeneschamali, which means “the northern claw” in Arabic, while Alpha Librae, Zubenelgenubi, is “the southern claw.” MAJOR STARS IN LIBRA Zubeneschamali – β Librae (Beta Librae) Beta Librae is the brightest star in the constellation. It has an apparent magnitude of 2.61 and is approximately 185 light years distant from the solar system. It has the stellar classification of B8 V, which means that it is a blue-white dwarf. Beta Librae is a very fast spinner, with a projected rotational velocity of 250 km/s. It has 4.9 times the solar radius and is approximately 130 times more luminous than the Sun. The star’s proper name, Zubeneschamali, comes from the Arabic phrase al-zuban al-šamāliyya, which means “the northern claw.” The star’s Latin name is Lanx Borealis, or “the northern scale.” Beta Librae is classified as a single star, but it shows small periodic variations in luminosity (0.03 of a magnitude), which indicate the presence of a companion star. Zubenelgenubi – α Librae (Alpha Librae) Alpha Librae is the second brightest star in Libra. It is a multiple star system whose two brightest components form a binary star and share a common proper motion through space. They are suspected members of the Castor Moving Group of stars, which share a common origin about 200 million years ago. The Alpha Librae system lies close to the ecliptic and can be occulted by the Moon and, much less frequently, by planets. It will next be occulted by a planet (Mercury) on November 10, 2052. The name Zubenelgenubi is derived from the Arabic phrase al-zuban al-janūbiyy, which means “the southern claw.” The system is also sometimes known as Kiffa Australis or Elkhiffa Australis. Both names are partial Latin translations and come from the Arabic phrase al-kiffah al-janūbiyy, which means “the southern pan (of the scales).” An older Latin name for the star is Lanx Australis, or “the southern scale.” Alpha-1 Librae is the dimmer of the two main components in the system. It has an apparent magnitude of 5.153. Alpha-1 Librae is a spectroscopic binary system with an orbital period of 5,870 days, with components separated by 0.383 seconds of arc (10 astronomical units). The system has the stellar classification F4 and an apparent magnitude of 5.153. It is 74.9 light years distant from Earth. Alpha-2 Librae has an apparent magnitude of 2.741. It is also a spectroscopic binary system, separated from Alpha-1 Librae by about 5,400 astronomical units. It has the stellar classification A3 and an apparent magnitude of 2.741. It is 75.8 light years distant from the Sun. The star KU Librae might be a fifth component in the Alpha Librae system, lying at a separation of 2.6 degrees. It shares a similar proper motion with the other components, but is a parsec away. This is just close enough for KU Librae to be gravitationally bound to the other stars. Brachium – σ Librae (Sigma Librae) Sigma Librae is a red giant star with the stellar classification of M3/M4 III. It has an apparent magnitude of 3.29 and is approximately 288 light years distant from the Sun. The star’s traditional name, Brachium, means “arm” in Latin. It is also sometimes known as Cornu (Latin for “horn”) and Zubenalgubi (“southern claw” in Arabic). Sigma Librae used to have the Bayer designation Gamma Scorpii even though it lies quite far from the border with Scorpius constellation. It only became Sigma Librae in the 19th century, and the designation was confirmed by the International Astronomical Union on July 31, 1930. Brachium is a semi-regular variable star with a single pulsation period of 20 days. It exhibits small variations in magnitude of 0.10 to 0.15 over short periods of 15 to 20 minutes every 2.5 to 3 hours or so. Methuselah – HD 140283 HD 140283 is the oldest known star in the Universe, believed to have been created shortly after the Big Bang. It is a subgiant star that is very metal poor and consists almost entirely of hydrogen and helium. Its iron content is less than 1 percent that of the Sun’s. The star’s age is estimated to be 14.46 billion years old, while the Universe is believed to be 13.77 billion years old. The star’s age does not really conflict with the age of the Universe because the values are uncertain. HD 140283 has an apparent magnitude of 7.223 and is 190.1 light years distant. υ Librae (Upsilon Librae) Upsilon Librae has the stellar classification of K3III, which makes the star an orange giant. It is a multiple star system with an apparent magnitude of 3.60. It is approximately 195 light years distant from the Sun. τ Librae (Tau Librae) Tau Librae is a blue-white dwarf with the stellar classification of B2.5V. It has an apparent magnitude of 3.66 and is approximately 445 light years distant. The star has 3.2 times the Sun’s radius. Zubenelakrab – γ Librae (Gamma Librae) Gamma Librae is an orange giant belonging to the stellar class K0 III. It has an apparent magnitude of 3.91 and is approximately 152 light years distant from the solar system. It has 2.15 solar masses and is about 71 times more luminous than the Sun. The star’s traditional name, Zubenelakrab, or Zuben-al-Akrab, is derived from the Arabic phrase al-Zuban al-Aqrab, which means “the shears of the scorpion.” θ Librae (Theta Librae) Theta Librae is an orange giant with the stellar classification of K0 III. It has a visual magnitude of 4.136 and is approximately 163 light years distant from Earth. It has a mass about 84 percent greater than the Sun’s and it is approximately 35 times more luminous. ι Librae (Iota Librae) Iota Librae has the stellar classification of B9IVpSi and an apparent magnitude of 4.54. It is composed of Iota-1 Librae, a pair consisting of a B9 subgiant and a dwarf star, approximately 377 light years distant, and Iota-2 Librae, a class A3 dwarf star about 240 light years from the Sun. The components in the Iota-1 Librae system orbit each other with a period of 23.469 years and are separated by only 0.13 seconds of arc. Their combined mass is 6.05 times solar and they are 149 and 94 times more luminous than the Sun respectively. There is another pair in the Iota-1 system, two 10th and 11th magnitude class G dwarfs. Zuben Elakribi – δ Librae (Delta Librae) Delta Librae belongs to the spectral class B9.5V. It is a blue-white main sequence star with an apparent magnitude of 4.43, approximately 300 light years from Earth. It is classified as an eclipsing variable star. It has a period of 2.3272 days and its luminosity varies from 4.43 to 5.81 magnitudes. The star’s traditional name, Zuben Elakribi, or Zuben-el-Akribi, comes from the Arabic az-zubānā al-ʿaqrab, which means “the claws of the scorpion.” 48 Librae, or FX Librae, is a shell star, a blue supergiant exhibiting irregular variations in luminosity as a result of its abnormally high rotational velocity, which results in gas being ejected from the star’s equator and forming a gaseous equatorial disk around the star. 48 Librae is one of the most rapid rotators known, with a projected rotational velocity of 400 km/s. 48 Librae has the stellar classification B8Ia/Iab. It has an apparent magnitude of 4.94 and is approximately 515 light years distant from Earth. It has 5.8 solar masses, 3.3 times the solar radius, and is about 965 times more luminous than the Sun. Gliese 581 (HO Librae) Gliese 581, or HO Librae, is a red dwarf star with the stellar classification of M3V. It has an apparent magnitude that varies between 10.56 and 10.58, and is 20.3 light years distant from the Sun. It is the 89th closest star to the Sun, and has only a third of the Sun’s mass and 0.2 percent of its visual luminosity. The star lies about two degrees north of the constellation’s brightest star, Beta Librae. Gliese 581 is classified as a variable star of the BY Draconis type and is sometimes known by its variable designation, HO Librae. BY Draconis variables are typically K or M class main sequence stars that exhibit variations in luminosity as a result of rotation coupled with star spots. Gliese 581 has a planetary system with at least three and possibly up to six planets. The first extrasolar planet in the system, Gliese 581 c, was discovered in April 2007. It is likely too hot to be a habitable zone, not unlike Venus. An unconfirmed planet in the system, Gliese 581 d, might be within or just outside the system’s habitable zone. Gliese 581 e, discovered in April 2009, was the least massive planet known orbiting a normal star at the time of discovery. Another unconfirmed planet was claimed to have been discovered in September 2010, and if its presence is confirmed, it will be the planet most suitable for liquid water as it lies in the middle of Gliese 581’s habitable zone. In November 2012, the European Space Agency found a comet belt in the system, one with at least ten times as many comets as the solar system. 23 Librae is a yellow dwarf with the stellar classification of G5 V. It has an apparent magnitude of 6.45 and is approximately 85 light years distant from Earth. It is another star in Libra constellation that has a planetary system with two confirmed planets. The first extrasolar planet, 23 Librae b, was discovered in 1999, and the second one was detected in 2009. 23 Librae is a much older star than the Sun, with an estimated age between 8.4 and 11.1 billion years. It has 107 percent of the Sun’s mass and 125 percent of the solar radius. HD 141937 is another yellow dwarf (spectral class G2/G3 V) with a confirmed planet in its orbit. The planet is a massive gas giant, discovered in 2001. The star has exactly the same mass as the Sun and a slightly larger radius, 1.06 times that of the Sun’s. It has an apparent magnitude of 7.25 and is approximately 109 light years distant from the solar system. Gliese 570 (33 G. Librae) Gliese 570 is a ternary star system, one consisting of at least three stars, located approximately 19 light years from Earth, in the southwestern part of the constellation. It lies southwest of Alpha Librae and northwest of Sigma Librae. The primary component is an orange dwarf with the stellar classification K4V and an apparent magnitude of 6.79. The star is smaller and less massive than the Sun and has only 15.6 percent of the Sun’s luminosity. It is a known X-ray source. The system also contains a binary star system consisting of two red dwarfs (spectral classes M1V and M3V) that orbit each other. Both stars also emit X-rays. A brown dwarf belonging to the spectral class T7Vwas discovered orbiting in the system in January 2001. At the time, it was one of the coolest brown dwarfs known. The dwarf star has a mass 50 times that of Jupiter. In 1998, an extrasolar planet was believed to orbit the brightest star in the system, but its presence was discounted in 2000. DEEP SKY OBJECTS IN LIBRA NGC 5792 is another barred spiral galaxy in Libra. It has an apparent magnitude of 12.1 and is approximately 83 million light years distant from the Sun. NGC 5890 is an unbarred lenticular galaxy in Libra. It was discovered by the American astronomer Ormond Stone in April 1785. It has an apparent magnitude of 14. NGC 5897 is a relatively large globular cluster in Libra. It has an integrated magnitude of 9 and is approximately 40,000 light years distant from the solar system. NGC 5885 is a barred spiral galaxy in Libra. It has an apparent magnitude of 11.8. The galaxy was discovered by William Herschel on May 9, 1784.<\|endoftext\|>	3.703125
597	Protein chips – or ‘protein arrays’ as they are more commonly known – are objects such as slides that have proteins attached to them and allow important scientific data about the behaviour of proteins to be gathered. Functional protein arrays could give scientists the ability to run tests on tens of thousands of different proteins simultaneously, observing how they interact with cells, other proteins, DNA and drugs. As proteins can be placed and located precisely on a ‘chip’, it would be possible to scan large numbers of them at the same time but then isolate the data relating to individual proteins. These chips would allow large amounts of data to be generated with the minimum use of materials – especially rare proteins that are only available in very small amounts. Researchers working at the Manchester Interdisciplinary Biocentre (MIB) and The School of Chemistry have unveiled a new technique for producing functional ‘protein chips’ in a paper in the Journal of the American Chemical Society (JACS), published online (22 August 2008). The Manchester team of Dr Lu Shin Wong, Dr Jenny Thirlway and Prof Jason Micklefield say the technical challenges of attaching proteins in a reliable way have previously held back the widespread application and development of protein chips. Existing techniques for attaching proteins often results in them becoming fixed in random orientations, which can cause them to become damaged and inactive. Current methods also require proteins to be purified first – and this means that creating large and powerful protein arrays would be hugely costly in terms of time, manpower and money. Now researchers at The University of Manchester say they have found a reliable new way of attaching active proteins to a chip. Biological chemists have engineered modified proteins with a special tag, which makes the protein attach to a surface in a highly specified way and ensures it remains functional. The attachment occurs in a single step in just a few hours – unlike with existing techniques – and requires no prior chemical modification of the protein of interest or additional chemical steps. Prof Jason Micklefield from the School of Chemistry, said: “DNA chips have revolutionised biological and medical science. For many years scientists have tried to develop similar protein chips but technical difficulties associated with attaching large numbers of proteins to surfaces have prevented their widespread application. “The method we have developed could have profound applications in the diagnosis of disease, screening of new drugs and in the detection of bacteria, pollutants, toxins and other molecules.” Researchers from The University of Manchester are currently working as part of a consortium of several universities on a £3.1 million project which is aiming to develop so-called ‘nanoarrays’. These would be much smaller than existing ‘micro arrays’ and would allow thousands more protein samples to be placed on a single ‘chip’, reducing cost and vastly increasing the volume of data that could be simultaneously collected. Source: University of Manchester<\|endoftext\|>	4
312	\|Photo by Cowboytoast\| “Real world” and “authentic” are two of many educational buzzwords overused right now. What if instead of making sure that everything has a truly “real” context we give students a creative opportunity to explore the “unreal.” The inspiration for this post comes from a new blog by Randall Munroe, author of xkcd, called What If?. In this blog he answers hypothetical questions by doing the actual math to answer them. So far he has shown things such as how much force does Yoda have? and what would happen if you gathered a mole (unit of measurement) of moles (the small furry creature) in one place? These questions are not real or authentic but the math and science is. But these questions are fun and interesting! Students love to talk about fantasy and science fiction such as zombies and vampires. So why not expose your students to a few of these kind of questions and have them try to “prove” their answer. Afterwards show them what Randall Munroe came up with. Then have students come up with their own questions and write out their reasoning and solutions. This activity would tap into their creativity but also demonstrate their mathematical computations and more importantly their mathematical reasoning. It also would be a literacy task in math. Finally and most important in my opinion it may also be an avenue to engage a student’s passions in math class that Jeff de Varona has been asking about.<\|endoftext\|>	4.0625
291	You will learn to represent functions in different forms and use your graphing calculator to find equations that match different graphs. TEKS Standards and Student Expectations A(2) Linear functions, equations, and inequalities. The student applies the mathematical process standards when using properties of linear functions to write and represent in multiple ways, with and without technology, linear equations, inequalities, and systems of equations. The student is expected to: A(2)(C) write linear equations in two variables given a table of values, a graph, and a verbal description A(7) Quadratic functions and equations. The student applies the mathematical process standards when using graphs of quadratic functions and their related transformations to represent in multiple ways and determine, with and without technology, the solutions to equations. The student is expected to: A(7)(A) graph quadratic functions on the coordinate plane and use the graph to identify key attributes, if possible, including x-intercept, y-intercept, zeros, maximum value, minimum values, vertex, and the equation of the axis of symmetry Given the graph of a linear or quadratic function, the student will write the symbolic representation of the function. What are the different forms of linear functions? How can a graphing calculator be used to match a graph to a linear equation? How is a quadratic equation different from a linear equation?<\|endoftext\|>	4.0625
1,482	# How does adding a nonzero multiple of one equation to another yield a an equivalent equation in Gauss elimination? When using Gaussian elimination to solve a system of linear equations , we can add a multiple of one equation to another and have an equivalent equation In other words , If we have a system of three linear equations L1 , L2 and L3 we can multiply L1 by a nonzero multiple and add it to L2 to get an equivalent equation to L2 why is that valid ? - Proof is easy, show soloution of one system is also soloution of other. Have you tried it ? – pritam Sep 23 '12 at 13:22 You do not generally get an equation that is equivalent to $L_2$. The system of equations, however, is equivalent to the original system. – David Mitra Sep 23 '12 at 13:26 @Akram Please do not write comments as answers. – Bill Dubuque Sep 24 '12 at 2:27 ## 4 Answers If we have two equations $A$ and $B$, multiplying $A$ by $k$ and adding it to $B$ gives the equation $kA+B$. Conversely, if we have $A$ and $kA+B$, then subtracting $kA$ from $kA+B$ gives $B$. This means that the system with equations $A$ and $B$ and the system with equations $A$ and $kA+B$ are equivalent. Note that this argument works even when $k=0$, in which case there is no change. For example, if $A$ is $a_1x_1+...+a_nx_n=x$ and $B$ is $b_1x_1+...+b_nx_n=y$, then $A+kB$ is simply $(a_1+kb_1)x_1+...+(a_n+kb_n)x_n=x+ky$. - [[If we have two equations A and B, multiplying A by k and adding it to B gives the equation kA+B. Conversely, if we have A and kA+B, then subtracting kA from kA+B gives B. This means that the system with equations A and B and the system with equations A and kA+B are equivalent ]] WHY ??? – Akram Hassan Sep 23 '12 at 23:11 Say $x$ is such that $f(x) = g(x) = 0$ i.e. $x$ solves the equations $f$ and $g$. Let $h = f + \lambda g$ be a linear combination of the two equations. Notice that $h(x) = 0 + \lambda0 = 0$. This proves that every solution of $f, g$ is also a solution of $h, g$. Finally notice that $f = h - \lambda g$ and thus, by the same argument, every solution of $h, g$ is also a solution of $f, g$. - can you explain more ? – Akram Hassan Sep 23 '12 at 22:52 By definition, systems of equations are equivalent if they have the same roots. That is what I prove. You'll have to point out what exactly is unclear. – Karolis Juodelė Sep 24 '12 at 5:03 If $f(x) = a$ and $g(x) = b$, then $c g(x) = c \cdot b$, and \begin{align}f(x) + c g(x) &= a + c g(x) \\&= a + c\cdot b,\end{align} by nothing more than substitution. As to equivalence, this follows from the fact that you can just "re-subtract" the equation $cg(x) = cb$ again. - by nothing more than substitution. As to equivalence, this follows from the fact that you can just "re-subtract" the equation cg(x) = cb ... what does that mean ? – Akram Hassan Sep 23 '12 at 22:52 Instead of adding $cg(x) = cb$, simply subtract $cg(x)=cb$. The result will once again be the equation $f(x)=a$. – Niel de Beaudrap Sep 23 '12 at 23:22 Let's consider the system $$\begin{pmatrix} 1 & 1 \\ 2 & 3\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 5 \\ 8\end{pmatrix}.$$ The shortcut step in Gaussian elimination tells you to subtract 2 times the first row from the second, yielding $$\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 5 \\ -2\end{pmatrix}.$$ It's not so hard to see that this system is equivalent, but what are we actually doing behind the handwaving? What's really happening is that we're left multiplying both sides of the equation by a new matrix: $$\begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & 3\end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ -2 & 1 \end{pmatrix} \begin{pmatrix} 5 \\ 8\end{pmatrix}.$$ You might notice that in this case, the new matrix is lower triangular, and when we carry out the Gaussian elimination step, the resulting matrix is upper triangular. In the $2 \times 2$ case, there is a single sub-diagonal element in any matrix. You may also notice that the Gaussian elimination requires only a single "step". If we extend this to $3 \times 3$ matrices, you will notice that you need to do 3 computations -- and there happen to be three sub-diagonal elements in any $3\times 3$ matrix. This is not a coincidence. It turns out, in fact, that Gaussian elimination is innately linked to LU decomposition -- the transformation of a full-rank square matrix into the product of lower and upper triangular matrices. This is very useful, because triangular matrices are spectacularly easy to invert. The shorthand method for Gaussian elimination gives you the following system: $$LAx = Lb.$$ But we could just as easily obtain $$Ax=b \Longrightarrow LUx = b \Longrightarrow Ux = L^{-1}b.$$ The key difference as to which operation to choose is whether you're trying to invert $A$ or solve $Ax=b$. For solving $Ax=b$, the shorthand Gaussian elimination turns out to be easiest. -<\|endoftext\|>	4.40625
465	Children use crayons, markers, and other writing tools in different ways as they progress through the stages of artistic development from scribbling to detailed drawings. Along the way children experiment with colors, shapes, space, and ideas. Working with crayons, markers and other drawing tools enhances creativity and self-expression while the movement of the tool across paper develops small and large muscle control. Stages in Children’s Drawing There are four main stages you can expect to see in your child’s drawings: 1. Random Scribbling - Large-muscle, whole arm movements. - Scribble may extend beyond the edges of the paper. - Drawing tools are usually held in the fist with minimal use of fingers. - Lots of exploration. 2. Controlled Scribbling - Smaller marks are made. - Certain marking motions are repeated. - Increased wrist control. - More intricate loops and spirals. 3. Named Scribbling - Names scribbles/drawings. - Holds drawing tools with fingers. - Makes a variety of lines. - Increased concentration. - Places marks intentionally. 4. Symbolic Representation - Draws recognizable geometric shapes. - Draws head to toe representation of the human figure. - Includes details like fingers, clothes, hair, etc. - Draws things that are personally meaningful. You can encourage your child’s creativity by: - Put on music and encourage your child to turn the paper in a different direction during the silence between songs. - Provide textured surfaces for crayon rubbings like bark, concrete, combs, sandpaper, etc. - Bundle several crayons or markers together and secure with a rubber band for a new color experience. - Demonstrate how to use the point for making dots or thin lines, and the side for wide lines. - Provide tissues for “polishing” a drawing made with crayon. - Spray water on a drawing made with water-based marker to see what happens. - Have your child try drawing with both hands at once. - Try coloring heavily then lightly, quickly then slowly. Compare the marks made with each technique.<\|endoftext\|>	3.859375
1,120	# Linear Relationships Assignment Help Linear Relationships Assignment Help Sport scientists at XXX have been testing elite football players and want to know if there is a relationship between the back squat exercise and vertical jumping or forward sprinting. They want to do this to understand if they need to modify their training program to meet the specific needs of the elite footballers. Explore the relationships between maximum back squat and the vertical jump, and 10m sprint exercises. Include in your exploration a discussion of the correlation between the two exercises. ## Steps to help answer the question • Graph the data in Table 2.1 for each relationship in separate graphs • Describe each relationship using an equation for a straight line. (Excel may be used but equations must be confirmed using hand calculations) • Determine the correlation coefficient associated with each relationship • Explain the differences between the relationships and any meaning you can draw from your data ## Useful information Table 2.1: Relationship between weight lifted in a back squat and height (cm) of a vertical jump and time (seconds) for a 10 m sprint for 20 athletes. Weight lifted in one back squat (kg) Height in vertical jump (cm) Time (sec) for one 10m Sprint 110 48 2.14 127 54 1.90 115 53 2.19 125 55 1.88 150 58 1.58 180 65 1.43 145 59 1.56 135 62 1.76 130 56 1.90 176 56 1.48 130 55 1.86 176 62 1.44 155 57 1.51 130 54 1.79 118 52 2.01 112 50 2.11 135 57 1.88 120 55 1.95 146 63 1.66 127 54 1.88 The task asks you to explore the relationships between maximum back squat and the vertical jump, and 10m sprint exercises. What you need to do here is produce scatterplots, plotting points for one variable against another, then describe the relationship apparent using a straight line, which you can fit to the scatterplot using Excel. Question 2 (statistics) What is the optimal age for a national leader? Two political science students were discussing the age of one of Canada’s Prime Ministers, Joseph Clark, who happened to be less than 40 years of age. They discussed if there was a typical age for a national leader and if this varied between countries. Compare the ages of the leaders of Australia and Canada. Comment on any similarities or differences from your findings. ## Steps to help answer the question • Calculate the five number summary for each sample and display graphically. Excel can be used, but values should be confirmed by hand calculations. • Use Excel to calculate a different measure of central tendency and its associated measure of spread for each sample. • Compare the two samples using all measures calculated above and where possible comment on a typical age. ## Linear Relationships Assignment Help ##### Useful information Table 3.1 Age (to the nearest year) of the past fifteen leaders of two countries. Australia Canada 55 54 60 40 57 57 68 45 56 46 63 47 56 48 45 47 53 51 48 47 57 46 50 55 49 43 56 60 61 52 Responses for questions 1, 2 and 3 below must each be written up in 3 sections, using the headings: Getting started • Analyse the problem to show your understanding of it. • Interpret and give meaning to any numerical data given. • Select a method of solution. Calculations • Perform the necessary calculations. • Communicate the solution in a logically sequenced way Conclusion • Evaluate and analyse your results and method. • Comment on any assumptions or limitations that affected your results. • Justify the practicality of the method of solution • Discuss alternative methods that could have been used • Discuss the relevance to the real world • All written descriptions must be in correctly structured sentences. • Descriptions should be in formal language (e.g. third person -no use of I, me or us throughout). • Figures and tables must be fully labelled and include headings. Getting started Analyse the problem to show your understanding of it. (In your own words, describe the problem, one or two sentences). Interpret and give meaning to any numerical data given. (Describe any numbers, graphs/tables/charts etc. in the question). Select a method of solution. (How will you calculate the answer). Conclusion Evaluate and analyse your results and method. (Answer the question in one or two sentences). Comment on any assumptions or limitations that affected your results. (What are we assuming about the problem/situation? What limits our understanding of the question?). Justify the practicality of the method of solution (Was the method you used useful? Easy to use? Practical? Why/Why not?). Discuss alternative methods that could have been used (How else could you calculate the answer? Is there another way to find the answer?). Discuss the relevance to the real world (How does this relate to everyday life?).<\|endoftext\|>	4.5
2,095	First in a series of stories about carbon sinks and sources. Read part two here and part three here. BOULDER, Colo. — The main tracking center for greenhouse gases in the Earth's atmosphere is housed in a laboratory here, filled with thousands of large glass flasks and the equipment to test the air samples in them. It has been run by the National Oceanic and Atmospheric Administration for almost 40 years and has given birth to a growing body of science that can assess the failures and the triumphs of mankind's battle against the dangers of global warming. It is a vast and unfinished business. Imagine, if you can, the life-giving blanket of air that surrounds our planet. It weighs 5 quadrillion metric tons. Ninety-nine percent of it consists of nitrogen and oxygen, and most of the blanket's thickness is in the bottom three-quarters of it. The top begins to taper off about 6.8 miles above the ground. What frightens, baffles and sometimes pits scientists against each other is the behavior of the gases they find in the remaining 1 percent. They are parts of a gaseous soup that includes carbon dioxide, methane and other accumulating gases, some of whose sources can be traced to human activities. They absorb the heat from sunlight and then radiate the heat back to Earth, raising its temperature in a process experts call "radiative forcing." What these gases do to the planet as they bake in the sunlight is only partly understood. CO2, which comes from fossil fuel combustion, is responsible for 65 percent of climate warming. Methane produces 15 percent. But when it is released, methane is 28 times more powerful as a warmer than carbon dioxide. Sixty percent of it is estimated to come from human activities. The methane content of the atmosphere has more than doubled since 1750, when the industrial age began. After a brief and still dimly understood plateau of about a decade, methane resumed its rapid rise again around 2007. At the moment, this is among the bigger and more ominous mysteries lingering in the atmosphere. There are disputes among multiple scientists over the causes of the new rise, and there are conflicting studies about how to measure methane and trace it back to its sources. That could result in delays among policymakers trying to understand how to rank the activities they might regulate to reduce the mounting threat of global warming. The result could be tragic. On the flip side, there are gases like HFC-134a. Scientists know it is man-made because it is a refrigerant used around the world in automobile air conditioners. It was designed and marketed before the sources of climate change were well understood and has a global warming potential that, measured pound for pound, is 1,340 times more powerful than CO2. The stage is now set for international action this fall to replace it and its close relatives with a new refrigerant that should be relatively harmless. The result of that would be a triumph. The Boulder lab where these questions are being sorted out is run by NOAA's Global Monitoring Division, and it most closely watches the sources for the "greenhouse gases" generated by human activity. What makes the lab's work assessing the danger of CO2 sources more difficult are the "sinks." These are areas in the world that collectively absorb about half of the CO2 we emit every year and store them in places — including the depths of the oceans and in the billions of trees growing in forested areas. By 2003, NOAA's system to take regular air samples covered much of the globe in an effort to sniff out the behavior of the sinks, but there were still places on the Earth's continents where the contents and behavior of the atmosphere were mostly established by guesswork. As John Miller, a NOAA scientist working on global monitoring data here, put it, they were "voids." Some of them were huge, including the air swirling over South America's Amazon and over the jungles of Africa. The laborious business of collecting air samples Brazil's Amazon, four times the size of Alaska, was the biggest question mark in Miller's puzzle. While some climate research was being done on the ground, trying to benchmark what was happening in the atmosphere was a challenge. Computer models predicting the impacts of climate change on the Amazon conflicted with each other, mainly because of a lack of prior observations. The question hanging in the air was: What was normal? In the United States, NOAA technicians obtained enough atmospheric samples to establish a benchmark by driving to remote areas and sometimes installing their air-sampling devices on television towers to get away from point sources of air contamination near the ground or in busy cities. They filled and tested as many as 20,000 flasks a year in an attempt to calculate the average power and accumulation of sources as they moved and mixed with the atmosphere. But the lack of roads in the Amazon made collecting a regular stream of samples more difficult, and there were relatively few towers that Miller and his team of researchers could use for a proposed experiment to fill the void. When they did have samples, they tried to set up regular shipments to test them in the Boulder laboratory, "but logistics were a nightmare," Miller recalled. For the kind of testing Miller wanted to do, a regular schedule of sampling and testing was essential, so NOAA collaborated with Brazilian scientists to build a replica of Boulder's laboratory of glass flasks and testing equipment in São Paulo. The researchers, including a British team, decided to use aircraft to take samples, flying regular routes over four parts of Brazil's Amazon. They hired air taxis that connect cities and towns in the Amazon in much the same way that bush pilots connect places in Alaska. The air taxi pilots were trained to use automated sampling equipment, often used in remote places in the United States. Twelve of the flask-like bottles were loaded in black plastic suitcases with battery-powered pumps that would fill them. Computers controlled the pumps and outlined the routes and the places where the pilots should fly. "We had a small interface with the pilot that says, 'Now please go to 12,000 feet and press the button,'" explained Miller, who said the goal was to get representative air samples from fixed places along the routes. The researchers booked air taxi flights every two weeks between 1 and 2 p.m., a time when their calculations showed that the atmosphere would be thoroughly mixed with few temperature inversions to trap areas of stagnant air. "Then you're measuring something that's homogeneous, so it's more representative," added Miller. "You need to have a baseline." To enhance the accuracy of the baseline — data that could be used to compare samples of future Amazon air to see the ratio of gases over it as the climate changed — the researchers decided they had to find a way to subtract exterior contaminants in the winds blowing westward into Brazil from the Atlantic Ocean and Africa. They did that by using atmospheric data from Barbados, the Ascension Island and South Africa. The systematic measuring began on one Amazon route in 2001 but expanded to four routes beginning in 2010. This was sheer dumb luck from a scientific point of view because that was the year when Brazil's Amazon was hit by an extreme drought, and by then Miller's team had set up a laboratory and a baseline to measure what happened next. A drought with a dangerous 'legacy' They were prepared to take 1,500 air samples a year for the next four years. Up to that point, the orthodoxy of climate studies, both from scientists and environmental activists, was that the Earth's sinks, particularly from oceans and forested areas, provided what Miller called "a 50 percent discount on our climate emissions." They kept about half of the greenhouse gases from further accumulating in the atmosphere. "We were getting a hell of a lot of help, and we didn't know exactly from where or why, and we don't know the extent to which it will last," explained Miller, who said that without the contribution from the sink, the world's leaders would now be wrestling with a buildup of atmospheric CO2 sources at around 550 parts per million, rather than the current 400 ppm. What the 2010 data indicated from the Amazon, however, was that the sinks that are helping to cool the Earth may not survive more extreme weather. From January to May in 2010, half of the days saw temperatures well above long-term averages. Air samples showed the forests were giving off CO2, not absorbing it, and the flow of CO2 into the air was accelerated by smoke from many forest fires that spread quickly as the forests dried. In August, during the Amazon's dry season, the CO2 emissions intensified. Relatively suddenly during 2010, the Amazon had shifted from becoming a sink to becoming one of the globe's CO2 sources. Scientists believe the smoke, the added heat stress and the drought slowed the photosynthesis that normally stored CO2 in trees, sending substantial quantities of it back into the atmosphere instead. A later Stanford University study of the 2010 Amazon drought, using the Brazilian data, found the shifts from sink to source were "very large" and "surprisingly fast." Although the drought and the heat eased the following year, air samples continued to show "large pulses" of CO2 were still being emitted into the air above the eastern Amazon throughout 2011. Caroline Alden, a University of Colorado scientist who worked with Miller on the measuring project, warns that this could be a "legacy" effect that means that tropical jungles could take several years to recover from more severe droughts. To be sure, jungles also resume their plant growth very quickly, beginning with grass fields that can start to become wooded areas in three to four years. These are sometimes called "secondary forests," but the losses from a more extreme weather event like the 2010 drought — as it was measured in the glass flasks filled in 300 flights over the jungles in three years — showed that the renewed growth of a traditional carbon sink may not start quickly enough to offset losses from its sudden and severe collapse. Or, as Miller puts it: "It can return to being a sink, but to recapture all of the carbon that was taken away in the initial step, that takes a long time." Tomorrow: The enduring mysteries of methane.<\|endoftext\|>	3.921875
2,710	Vitamin C is a popular vitamin best known for its immune strengthening properties. In fact, this water soluble micronutrient plays a critical role in sustaining life. Our body will simply fall apart without it. Vitamin C helps to maintain the integrity of our skeletal system, is required for growth and repair of connective tissues and as a co-factor for biosynthesis of collagen and other important metabolic reactions. It also helps in the prevention and treatment of many infections and chronic conditions including arthritis, cardiovascular diseases, eye degenerative disorders, premature aging and cancer. Vitamin C is a potent anti-oxidant and much of its health imparting properties are drawn from its powerful ability to fight and neutralize free radicals – highly reactive oxygen species that can cause excessive oxidative damage to DNA, cell membranes and other delicate cellular structures. And this property makes Vitamin C indispensable when it comes to minimizing the dangerous side effects of ionic radiations, the kind emitted during Fukushima and other nuclear disasters. How ionizing radiation damages cells As radioactive material decays, it releases high frequency radiations into the environment. These radiations are basically fast moving energy particles – alpha particles, gamma rays and beta particles – that crash into a living cell with a force high enough to knock off electrons free from the molecules of the cell. This results in the formation of highly reactive free radicals and their metabolites that further disrupt the chemical bonds between the surrounding molecules like proteins, DNA, lipids and carbohydrates – damaging their fragile structures and impacting their ability to function normally. The most severe form of radiation damage occurs when DNA is damaged. The ionizing radiations incapacitate the cellular DNA in two major ways: - The water in our body happens to absorb most of the radiation and gets ionized in the process, readily forming free radicals and their metabolites. Almost 80 % of the radiation induced DNA damage results through this mechanism. - Radiations can also collide directly with DNA molecules, thus ionizing and injuring these directly. DNA contains genes that carry unique genetic information or instructions to make new cells. A DNA molecule is made up of two long strands that are entwined around each other like a twisted ladder and held by chemical subunits. Ionizing radiations cause the formation of strand breaks in the DNA . Our body can easily repair single-strand breaks (where only one of the two strands has been cleaved) by employing specialized DNA repair enzymes. On the other hand, correction of double-strand breaks (where both DNA strands are injured) is difficult and complicated, and may result in erroneous re-joining of broken ends. This kind of error in the DNA repair can cause mutations. This change in the structure of DNA alters normal cellular processes, including mechanisms that dictate how and when cells divide. Such misdirection leads to uncontrollable cell division causing cancer and chromosomal abnormalities that can even be passed from one generation to another. With significant damage to DNA, it even loses the ability to replicate further and the cells die. It is very well proven that ionizing radiation can damage DNA, resulting in many types of cancer and inherited mutations. We covered the side effects of radiations in our last blog, ‘Fukushima Disaster Health Penalties’. To sum it up, health risks from the unfortunate Fukushima nuclear accident include increased risk of thyroid cancer, other types of solid cancer, hereditary diseases, fertility issues, impaired immunity, birth defects and cardiovascular diseases. Also, low level exposure to ionizing radiations produces other biological damage besides causing latent cancer. In fact, low level chronic radiations are thought to exert much more long-term impact on the exposed population than short-high doses. Ionizing radiation can cause inheritable genetic mutations that can be expressed in the offspring of exposed organisms, be it plants, animals or human beings. While these genetic mutations result in major abnormalities in the offspring and may be easily noticeable, it is also possible that some inherited mutations may either show up only as subtle effects or have bearings that do not appear until much later generations. In addition, the cells present in the vicinity of an exposed cell can be impacted too. Scientists explain it as a bystander effect phenomenon. Cells that have been irradiated impact the surrounding healthy cells by transmitting signals and end up generating the same kind of responses in the non-irradiated cells that results in faulty gene expression, cell proliferation, apoptosis and cell death. “Most well-known important factors affecting radiation-induced bystander effect include free radicals, immune system factors, expression changes of some genes involved in inflammation pathway and epigenetic factors.” What also deserves the spotlight is the impact of radiations on immunity. Ionizing radiations cause deadly damage to bone marrow stem cells, lymphocytes and natural killer cells. These white blood cells are responsible, in a major way, for our innate and adaptive immunity – protecting our body against opportunistic infections and diseases including cancer. Yes, our immune cells actually have the capability to fight the cancer cells but clearly, damage to these fighter cells is only likely to make people more prone to cancer and opportunistic pathogens. Government authorities and health experts all over the world may try hard to assure us that the radiation risks from Fukushima are overstated. But we have taken our lessons from the studies on survivors of Chernobyl, Hiroshima and Nagasaki nuclear disasters. And the data points out that on-going radiations indeed have far-reaching consequences for us as well as our entire biological ecosystem, putting the health of present as well as future generations in total jeopardy. The potential risks are much higher for small children and pregnant women as the radiations have the most impact on fast reproducing cells in their bodies making them more vulnerable to the toxic side effects. What is troubling is that the Fukushima nuclear disaster site is still a source of toxic radionuclides entering the adjoining sea water, making the situation graver. Research from Woods Hole Oceanographic Institution shows that Fukushima radiation is making its way along the U.S. West Coast. “Scientists monitoring the spread of radiation in the ocean from the Fukushima nuclear accident report finding an increased number of sites off the US West Coast showing signs of contamination from Fukushima. This includes the highest detected level to date from a sample collected about 1,600 miles west of San Francisco.” Can we do something to negate the short as well as long-term damaging effects of these radiations? Fortunately, anti-oxidants, also called free radicals scavengers, can destroy free radicals and minimize radiation induced toxicity. And Vitamin C clearly takes a lead. Role of Vitamin C: Free radical scavenger Vitamin C is a popular anti-oxidant that is known for its powerful ability to mop up damaging free radicals, even those created by radiation exposure. Many scientific studies demonstrate the protective role of high doses of Vitamin C in preventing radiation induced cellular damage. And so much so that a large body of research actually points out that Vitamin C can actually reverse the DNA damage caused by ionizing radiations. - A 2011 study by Dr. Atsuo Yanagisawa and team showed that Fukushima workers who took high-dose Vitamin C before going on to clean up the radioactive wreckage were protected against damage to DNA and overall cancer risk. “Workers with severe radiation exposure at the Fukushima nuclear plant had major reduction in cancer risk when supplemented with vitamin C and other anti-oxidative nutrients……. Four workers who took intravenous vitamin C (25,000 mg) therapy before they went in, and continuously took anti-oxidative supplements during the working period, had no significant change in both free DNA and overall cancer risk. Three workers that did not have preventive intravenous vitamin C had an increase in calculated cancer risk. After 2 months of intervention with intravenous vitamin C and oral anti-oxidative nutritional supplements, free DNA returned to normal level and cancer risk score was significantly decreased.” This significant clinical evidence confirms the result of a 1993 study that showed “Vitamin C may have an important role as a radio-protector against accidental or medical radiation exposures, especially when radionuclides are incorporated in the body and deliver the dose in a chronic fashion.” - The results from a 2004 study showed that administration of ascorbic acid did protect mice against radiation-induced sickness, mortality and improved healing of wounds after exposure to whole-body gamma-radiation – adding to the evidence for the radio-protective nature of Vitamin C. - Bone marrow failure is believed to be one the most common complications after acute radiation exposure, causing severe pancytopenia – considerable loss of white blood cells, red blood cells as well as platelets. The reduction in all these blood cells leads to fatal immune dysfunction. While bone marrow transplantation (BMT) or stem cell transplantation is significantly helpful in saving patients from radiation-caused bone marrow failure, the condition is followed by another serious complication – radiation-induced gastrointestinal (GI) syndrome. A 2010 study showed that ascorbic acid treatment may effectively thwart GI syndrome caused by acute radiation exposure. The study paper explains that “ascorbic acid acts as a hydrogen donator to scavenge radiation-induced free radicals in both GI tract and bone marrow. Hydrogen donation is an important step toward chemical repair of damaged DNA, and ascorbic acid is believed to strongly contribute to this process.” - Recently, a team of researchers investigated the effects of administration of ascorbic acid on mouse survival after the mice received whole body irradiation. The results of this 2015 study published in PLoS One concluded that the “administration of high-dose ascorbic acid might reduce radiation lethality in mice even after exposure.” The research further discusses that after the radiation exposure, secondary reactive oxygen species and inflammatory cytokines are released. This explains why post-radiation therapy with ascorbic acid may be advantageous as it reduces “radiation-induced elevation of inflammatory cytokines and also elevation of free radical metabolites.” What experts recommend? In the statement of “Environmental Radioactivity and Health” issued in 2011, The Japanese College of Intravenous Therapy (JCIT) strongly recommended that “People living in the affected areas should regularly take antioxidant supplements such as vitamin C to counteract the negative consequences of long-term low dose radiation exposure as well as to protect the health of coming generations.” Vitamin C and Cancer What makes Vitamin C therapy even more useful is that it also supports immune system functions, thus working along with the body’s inherent ability to fight cancerous cells. A study published in the Journal of Angiogenesis Research showed that high doses of Vitamin C restricts the formation of new blood vessels, that serve to carry increased flow of blood and nutrients to the tumor site required for growth and proliferation . In addition, intravenous Vitamin C improves the quality of life in cancer patients by reducing the severity of devastating side effects that typically accompany conventional cancer treatment like chemotherapy and radiation therapy . - Scholes, G. (1983) Radiation effects on DNA. The Silvanus Thompson Memorial Lecture, April 1982. Br. J. Radiol. 56: 221–231. - Najafi eat al. The Mechanisms of Radiation-Induced Bystander Effect. J Biomed Phys Eng. 2014 - Woods Hole Oceanographic Institution. Higher Levels of Fukushima Cesium Detected Offshore. 2015 - Yanagisawa A. Orthomolecular approaches against radiation exposure. Presentation Orthomolecular Medicine Today Conference. Toronto 2011 http://www.doctoryourself.com/Radiation_VitC.pptx.pdf ) - Narra VR, Howell RW, Sastry KS, Rao DV. Vitamin C as a radioprotector against iodine-131 in vivo. J Nucl Med 1993; 34(4):637-40 - Jagetia GC, Rajanikant GK, Baliga MS, Rao KV, Kumar P. Augmentation of wound healing by ascorbic acid treatment in mice exposed to gamma-radiation. Int J Radiat Biol. 2004 - Pretreatment with Ascorbic Acid Prevents Lethal Gastrointestinal Syndrome in Mice Receiving a Massive Amount of Radiation J. Radiat. Res., 51, 145–156 (2010) - Tomohito Sato et al. Treatment of Irradiated Mice with High-Dose Ascorbic Acid Reduced Lethality. PLoS One. 2015; 10(2): e0117020. - Nina A Mikirova,Joseph J Casciari,and Neil H Riordan. Ascorbate inhibition of angiogenesis in aortic rings ex vivo and subcutaneous Matrigel plugs in vivo. Journal of Angiogenesis Research. 2010; 2: 2. - Anitra C. Carr, Margreet C. M. Vissers, and John S. Cook. The Effect of Intravenous Vitamin C on Cancer- and Chemotherapy-Related Fatigue and Quality of Life. Frontiers in Oncology. 2014; 4: 283.<\|endoftext\|>	3.71875
1,611	Useful maps to explore the cultural history of Grand Teton National Park: Learn more about the following subjects: For a driving tour and audio descriptions of some historic locations in the park. Enjoy our historic audio tour page. Humans and the Teton Landscape: 11,000 years of history in 1,100 words The human history of Jackson Hole and the Teton Range dates back thousands of years. The stunning beauty and abundant wildlife and plants found here has drawn humans to this place for more than 11,000 years. Nomadic paleo-Indians first entered the Jackson Hole valley shortly after Pleistocene Ice Age glaciers retreated. They left behind tipi rings, fire pits and stone tools. Summers were a time of abundance, and modern-day Indian tribes came to harvest bulbs and berries, fish the lakes and streams, and hunt wildlife. With the approach of the harsh winter, indigenous people followed their prey out of the valley in search of milder weather. The first euro-American explorer who may have entered Jackson Hole was John Colter. He served as a member of the Lewis and Clark 'Corps of Discovery' expedition, but he left the expedition in the fall of 1806 and traveled through this region in the winter of 1807-1808. Unfortunately, Colter left no written record of his journeys. People also came here for wealth. Fur trappers, known as "mountain men," trekked west in search of beaver for fur top hats that were fashionable in the early 1800s. Many trappers including David Edward (Davey) Jackson based their operations in this area. The valley we know today as Jackson Hole was dubbed Davey Jackson's Hole in 1829 by William Sublette, Jackson's trapping partner. The beaver population declined rapidly with over-trapping, and when fashions turned from fur to silk hats, the era of the mountain men faded away by the 1840s. As America expanded westward, survey expeditions mapped the landscape, documented natural resources and scouted for future railroad access. Parties lead by Captain W.F. Raynolds in 1860, Ferdinand V. Hayden in 1872, and Gustavus C. Doane in 1876 traveled to the Teton region and expanded America's knowledge of the land and its wealth. Even though the Homestead Act of 1862 encouraged settlement of the West, homesteaders did not arrive in Jackson Hole until 1884, becoming the valley's first year-round residents. Over the next decade, many settlers established homesteads. Conditions were difficult, however. The soil was sandy and rocky, the winters were long and cold, and the summers were dry. Homesteaders struggled hard to raise crops and ranch cattle. They became desperate as an agricultural depression swept the country around 1920. Wealthy Easterners enchanted with by the West visited Jackson Hole to have an authentic "cowboy" experience. Homesteaders began to shift their operations in 1908 to accommodate these priviledged visitors. Eastern 'dudes' (men) and 'dudenes' (women) paid handsomely for lodging, food, the use of a horse and other outdoor activities. Local ranchers quickly realized that 'dude ranching' was more profitable and easier than traditional cattle ranching, which lead to the golden age of dude ranching in the 1920s. Development began to crowd Jackson Hole: cabins, gas stations, dancehalls, billboards and racetracks sprang up in front of the majestic Tetons. Local ranchers and businessmen wanted to preserve the valley as a "museum on the hoof" and eliminate the unwanted devlopment of the open spaces. They held a meeting in 1923 at Maud Noble's cabin, setting in motion the conservation and preservation of this mountain valley. In 1926, John D. Rockefeller, Jr. toured the area with Yellowstone Superintendent Horace Albright. Rockefeller fell in love with the majestic mountain scenery and began purchasing private land throughout the valley. Over the next two decades, he amassed 35,000 acres through the Snake River Land Company with the intent of donating the land to the federal government to be part of Grand Teton National Park. Local residents became concerned when they discovered Rockefeller's involvement. Transferring control of the land to the federal government meant a loss in local tax revenue, an issue finally resolved by a Congressional hearing. Grand Teton National Park took decades to establish. Congress created the original park in 1929 to protect the Teton Range and several lakes at the foot of the mountains. In 1943, Franklin Delano Roosevelt declared the remaining federal land in the valley as Jackson Hole National Monument. In 1949, John D. Rockefeller, Jr. donated the land he purchased to the government to be included in the national park. Finally in 1950, Congress combined the original park, the national monument, and the Rockefeller lands to establish present-day Grand Teton National Park. In 1972, Congress established the John D. Rockefeller, Jr. Memorial Parkway to honor Rockefeller's philanthropy and commitment to the National Park System. The Parkway connects Yellowstone and Grand Teton National Parks. After World War II, more people owned cars and began to explore America. Taking shorter vacations compared to dudes, these visitors would only spend a night or two at one location before moving on. In response to this new demand, auto camps became common. Small cabins clustered around a central parking area which allowed visitors easy access to the park's attractions. John D. Rockefeller, Jr. also saw a need to accommodate travelers. He set out to develop an assortment of lodging - from small rustic cabins at Colter Bay, to a stately lodge on a bluff overlooking Jackson Lake, to the elegant lodge near Jenny Lake. With increased visitation, the park also saw the need to expand visitor services. As the National Park Service approached its 50th anniversary in 1966, visitor centers were built at Colter Bay and Moose in the late 1950s to provide information and offer ranger activities as part of the "Mission 66" program. Adventure has always drawn people to this area. No one knows who first climbed the Grand Teton. American Indians built an "enclosure" at 13,280' on a sub-peak of the 13,770' tall Grand Teton. Although members of the 1872 Hayden Expedition claimed to have reached the summit, William Owen, Franklin Spalding, John Shive and Frank Peterson made the first documented summit in 1898. Many followed in the footsteps. Paul Petzoldt and Glen Exum established the first guide service in 1931, which still operates today as Exum Mountain Guides. Today over 90 different routes and variations lead to the summit of the Grand Teton. Today, all of the Teton peaks lure climbers with the skills necessary for a summit. Olaus Murie first visited the valley in 1927 when he conducted a study on the local elk herd. Years later in 1945, the Murie families - Olaus and Mardy with Adolph and Louise - purchased the STS Dude Ranch in Moose. Early conservation leaders met at the Murie Ranch in support of wilderness preservation. The Wilderness Act of 1964 was conceived and written by those conservationists (Howard Zahniser, Olaus Murie, Bob Marshall and others) who gathered at the Murie Ranch in the shadow of the Tetons.Together they created a legacy that benefits all to this day. This majestic place inspired and sustained people for thousands of years. Learn how humans shaped the Teton landscape through their settlement and occupations. Follow the links below to learn more about some of the historic buildings and stories of this rugged western landscape. See the historic map for locations of buildings and take time to visit an historic structure during your next visit to Grand Teton National Park. For additional information about historic sites within Grand Teton National Park visit: Last updated: November 8, 2016<\|endoftext\|>	3.796875
320	Two page reading comprehension on how to plant and care for a vegetable garden with reading comp, discussion, short response-functional text questions, and vocabulary word search. Use with Healthy Gardeners Club forms. A one page reading comprehension on the history and nutritional value of the potato, with completion questions, critical thinking exercises, a word search puzzle of related vocabulary, and a research writing prompt. This How to: Growing Potatoes clipart is great to illustrate your teaching materials. As an abcteach member you have unlimited access to our 22,000+ clipart illustrations and can use them for commercial use. This How to: Growing Potatoes clipart is provided in jpeg format. PowerPoint with Audio: Germs in Food and Water - Cleaning. Part 2 of 2 food and water preparation presentations from the Living on Planet Earth Series. This program describes how to limit the ingestion of harmful microbes by properly separating, cleaning, cooking and chilling food. Use this 'Comprehension: Heart Health (elem)' printable worksheet in the classroom or at home. Your students will love this 'Comprehension: Heart Health (elem)'. A one page discussion of ways to keep hearts healthy, with multiple choice and short answer questions, plus a crossword puzzle and a word search. Do you want healthy, happy kids? Join NourishMD.com. Learn how REAL food heals. Find natural solutions for children with medical, emotional, and/or learning difficulties, and easy resources to help you get started - articles, recipes, coupons<\|endoftext\|>	3.90625
311	Anomaly detection refers to the technique of identifying unusual patterns and finding outliers in a set of observations. Outliers are data points that differ considerably from the remainders of the dataset. Usually, extreme values that diverge from the normal or expected behavior. Historically statistics was applied to find and remove outliers, for example from the tails of a Gaussian distribution. The idea was that outliers which result from errors (noise, human, etc) may arise in misleading interpretations. In addition, by filtering them out, modern algorithms in supervised learning can gain in accuracy. On the other hand, the anomalies are nowadays also the object of interest, as it is the case of “rare events” in physics, medicine, business or cybersecurity. Datasets vary in their nature, but the most typical ones are time series and spatial data. There are three main types of outliers - Points are single occurrences anomalous with respect to the complete dataset. For example, the stock values marked as red stars in the plot of next section. - Contextual outliers are anomalous for specific circumstances. - Finally, collective outliers are a collection of occurrences which are not anomalous by themselves, but as a subgroup or subsequence of the entire dataset. Next figure exemplifies it in red, for a plotted electrocardiographic signal. Example of collective outlier. The red portion of the signal is an outlier because it was at that value for a significantly longer duration than normal.<\|endoftext\|>	3.734375
545	Exponential and logarithmic equations Solving exponential and logarithmic equations Exponential equations Exponential and logarithmic equations Exponential equations In exponential equations the variable that has to be solved for is in the exponent. To solve an exponential equation, rewrite the given equation to get all powers (exponentials) with the same base, or use logarithms when solving the exponential equation. Example: Solve 3x - 1 = 81. Solution: 3x - 1 = 81 3x - 1 = 34 x - 1 = 4 => x = 5. Example: Solve 0.25x = 43x - 2. Solution: Example: Solve 43x + 2 = 64 · 22x - 4. Solution: 42 · 43x = 43 · 22(x - 2) \| ¸ 42 43x = 41 + x - 2 3x = x - 1 2x = - 1 => x = - 1/2. Example: Solve 4x + 2 + 4x + 1 + 4x -1 = 3x + 3 + 3x + 2. Solution: Example: Solve 4x - 2 = 5x. Solution: Example: Solve Solution: Example: Solve 3 · 4x + 2 · 9x = 5 · 6x. Solution: Logarithmic equations As logarithmic equations contain a logarithm of variable quantity, we use rules and properties of logarithms to solve a logarithm equation. Example: Solve log x = -2. Solution: log x = log10-2 => x = 10-2 = 0.01. Example: Solve log2 (log3 x) = 1. Solution: log3 = 21 => x = 32 = 9. Example: Solve log (x + 5) - log (2x - 3) = 2 · log 2. Solution: Example: Solve log (log x) + log (log x3 - 2) = 0. Solution: Example: Solve x log x - 2 = 1000. Solution: Functions contents D<\|endoftext\|>	4.53125
304	(a) Explain the purpose of the two accounts: Depreciation Expense and Accumulated Depreciation. (b) Is it customary for the balances of the two accounts to be equal? (c) In what financial statements, if any, will each account appear? Depreciation expense is the amount of depletion, as expressed in dollars that an asset had during the fiscal year. It is incredibly difficult to determine the physical decline in a fixed asset’s ability to provide service. Thus depreciation is estimated based on the asset’s useful life and is measured by taking the value of the asset based on it cost of acquisition, subtracting the value the asset will retain when it is totally depleted, and divide that by the useful life of the asset in years. That amount equals the annual depreciation expense for the asset. A record of the initial cost of a fixed asset must be maintained for tax and other purposes. Hence, for this reason the fixed asset account is not directly reduced for depreciation. Instead an offsetting or contra asset account called accumulated depreciation account is used to record the depreciation. Accumulated depreciation is a contra-asset equal to the total of all depreciation expense incurred relating to fixed asset. On the balance sheet, an asset is carried at its acquisition value. The accumulated depreciation account is a negative balance on the balance sheet which allows for the asset account to reflect the acquisition cost of the asset while ensuring that the totals on the balance sheet reflect the value of the company.<\|endoftext\|>	3.796875
849	Parametric Equations Parametric equations definition The parametric equations of a line The direction of motion of a parametric curve Evaluation of parametric equations for given values of the parameter Sketching parametric curve Eliminating the parameter from parametric equations Parametric and rectangular forms of equations conversions Parametric equations definition When Cartesian coordinates of a curve or a surface are represented as functions of the same variable (usually written t), they are called the parametric equations. Thus, parametric equations in the xy-plane x = x (t and y = y (t) denote the x and y coordinate of the graph of a curve in the plane. The parametric equations of a line If in a coordinate plane a line is defined by the point P1(x1, y1) and the direction vector s then, the position or (radius) vector r of any point P(x, y) of the line r = r1 + t · s, - oo < t < + oo and where, r1 = x1i + y1 j and s = xsi + ys j, represents the vector equation of the line. Therefore, any point of the line can be reached by the r = xi + y j = (x1 + xst) i + (y1 + yst) j since the scalar quantity t (called the parameter) can take any real value from - oo to + oo. By writing the scalar components of the above vector equation obtained is x = x1 + xs · t y = y1 + ys · t the parametric equations of the line. To convert the parametric equations into the Cartesian coordinates solve given equations for t. So by equating Therefore, the parametric equations of a line passing through two points P1(x1, y1) and P2(x2, y2) x = x1 + (x2 - x1) t y = y1 + (y2 - y1) t Parametric curves have a direction of motion When plotting the points of a parametric curve by increasing t, the graph of the function is traced out in the direction of motion Example: Write the parametric equations of the line y = (-1/2)x + 3 and sketch its graph. Solution: Since Let take the x-intercept as the given point P1, so for y = 0 => 0 = (-1/2)x + 3, x = 6 therefore, P1(6, 0). Substitute the values, x1 = 6y1 = 0, xs = 2, and ys = -1 into the parametric equations of a line x = x1 + xs · t, x = 6 + 2t y = y1 + ys · t, y = -t The direction of motion (denoted by red arrows) is given by increasing t. Example: Write the parametric equations of the line through points, A(-2, 0) and B(2, 2) and sketch the graph. Solution: Plug the coordinates x1 = -2y1 = 0, x2 = 2, and y2 = 2 into the parametric equations of a line x = x1 + (x2 - x1) t, x = -2 + (2 + 2) t = -2 + 4t, x = -2 + 4t, y = y1 + (y2 - y1) t, y = 0 + (2 - 0) t = 2t, y = 2t. To convert the parametric equations into the Cartesian coordinates solve x = -2 + 4t for t and plug into y = 2t therefore, Functions contents B<\|endoftext\|>	4.5
788	Solving Linear Systems Using Algebraic Method Worksheet \| Problems & Solutions # Solving Linear Systems Using Algebraic Method Worksheet Solving Linear Systems Using Algebraic Method Worksheet • Page 1 1. Which of the following solutions satisfies the linear system? 2$x$ + 3$y$ = 3 - 15$y$ = - 14 + 8$x$ a. ($\frac{2}{3}$, $\frac{1}{2}$) b. (- $\frac{1}{2}$, $\frac{2}{3}$) c. ($\frac{1}{2}$, $\frac{2}{3}$) d. ($\frac{1}{2}$, - $\frac{2}{3}$) 2. Solve the linear system: 4$x$ + $y$ = - 66 5$x$ - 4$y$ = 12 a. (- 12, -17) b. (14, -16) c. (- 12, -18) d. (12, -17) 3. Which of the following solutions satisfies the linear system? 3$x$ - $y$ = - 3 - 2$x$ + 3$y$ = 2 a. (- 2, 0) b. (0, 1) c. (- 2, 1) d. (- 1, 0) 4. Which of the following ordered pairs satisfies the linear system? - 4$x$ + 3$y$ = - 3 3$y$ = 3 + 2$x$ a. (-4, - 3) b. (2, 3) c. (- 3, 4) d. (3, 3) 5. Solve the linear system. 3$x$ + $y$ = 10 2$x$ + $y$ = 6 a. (4, - 2) b. (4, 2) c. (1, 2) d. (- 2, 4) 6. Which of the following ordered pairs satisfies the linear system? 2$x$ - 3$y$ = 8 - 6$x$ + 3$y$ = 8 a. (- $\frac{16}{3}$, - 4) b. (- 4, - $\frac{16}{3}$) c. (- 4, $\frac{16}{3}$) d. (4, $\frac{16}{3}$) 7. Which of the following ordered pairs satisfies the linear system? 5$x$ + 3$y$ = -7 5$x$ - 3$y$ = -13 a. (2, - 1) b. (2, 1) c. (- 2, 1) d. (- 2, - 1) 8. Solve the linear system. $x$ - 4$y$ = -26 - $x$ + 8$y$ = 58 a. (6, 11) b. (6, 8) c. (6, 7) d. (7, 7) 9. Solve the linear system: $w$ + $z$ = 5 2$w$ = 6 a. $w$ = 2, $z$ = 3 b. $w$ = 2, $z$ = - 3 c. $w$ = - 3, $z$ = 2 d. $w$ = 3, $z$ = 2 $a$ = $b$ - 4 2$a$ + $b$ = 16<\|endoftext\|>	4.65625
195	Let’s subitize! Do your students enjoy a variety of math activities? There are times that students need lots of practice with certain skills. Students do not want to do the same thing again and again. However, I know my students can complete the same activity over and over again if it is in different formats. Subitizing is a great example of a skill that requires lots of practice. To help my students practice the same skill over and over again I created Flip Flop Subitizing is a memory-type game to reinforce my students’ number sense. The game can be played as a memory-type activity. Flip-Flop Subitizing is a wonderful small group activity – flash the dot cards while students increase their subitizing ability to instantly recognize quantities. Subitizing is the skill of instantly recognizing a quantity without having to count each object. This skill aids in students’ number sense. Until next time,<\|endoftext\|>	4.21875
539	# Number Bowling This is a student favorite, simple to start, differentiated, with various levels of success. I’ve done this with grades 2 and up, but my 5th-6th graders have gotten the most out of it. Most importantly students learn to be very clear in writing their expressions. This activity naturally leads to a discussion about order of operations and mathematical “grammar.” Also, in their pursuit of the strike, students often ask to be introduced to new operations. Instructions: Begin by rolling a die three times and recording these numbers as your “1st bowl.” You may cross out (knock down) any number that you can write an equation for using those three numbers each only once. For example, if I rolled a 6, 6 and 5, I could knock down the number four by writing: 5=6-6+5, or knock down 1 by writing 1=(6÷6)^5. Notice I used all three numbers, but each only once. The goal is to knock as many “pins” (numbers) as you can. Knocking all the pins down on your first bowl is called a “Strike.” If you can’t think of any other equations, you may bowl again and try for a “Spare.” Notes on Implementation: I’ve created this Number Bowling handout for students to keep track of their games. I’ve also experimented with keeping score, and I think three frames is a good length time for a game. This can get complicated though, since scoring bowling is foreign and not straightforward to many students. This year, after a few games, I had students write down their “favorite” equations on a notecard. We used these equations during a “strategy session” where we came up with tricks to help knock down more pins. The “tricks” can all be described as using an operation to change a number, or two numbers, into another. For example, 3 can be changed into 6 by using factorial, (3!=6). Some more advanced tricks include using square root, and/or the floor and ceiling function (rounding up or down to nearest integer). Extensions: Last year students wondered whether a strike was possible for every combinations. We chose the brute force method of proof :). First students had to figure out how many unique outcomes were possible with 3 dice. For many I assisted them by having them look at this pdf. Obviously rolling 1,6,6 is the same as rolling 6,6,1. Then whenever a student achieved a strike, we crossed that off the list.<\|endoftext\|>	4.71875
191	Arc Flash Overview Test your knowledge by answering the following questions. The light and heat energy generated from an arcing fault event is known as an __________. a. arc hazard b. arc blast c. arc flash d. electrical shock All flame resistant materials used in Personal Protective Equipment (PPE) are also arc rated. Any material that has a low level of resistance, allowing the flow of electrons is called a/an __________. An electrical ground fault is the same as an electrical overload. One who has demonstrated skills and knowledge related to the construction and operation of the electrical equipment and installations and has received safety training to identify and avoid the hazards involved. c. Knowledgeable person d. Qualified person return to top \| previous page \| next page Content ©2015. All Rights Reserved. Date last modified: October 29, 2015. Created with SoftChalk<\|endoftext\|>	3.796875
486	Dividing by 7 Start Practice How to Divide by 7 Division means breaking a number into smaller, equal groups. In the last lesson, you learned how to divide numbers by 6. Now, let’s learn how to divide numbers by 7. 😎 Division by 7 👉 There are 3 ways to divide a number by 7. 1. Division by grouping. 2. Division by repeated subtraction. 3. Division by multiplication. (the fastest) You can use any of these methods! 🤗 But in this lesson, let's focus on using your multiplication skills to make division easy. Division Using Multiplication 👉 Let's look at an example. 21 ÷ 7 = ? Turn any division problem into a related multiplication problem in your head quickly! First, think about it's whole-part diagram: Can you write a multiplication equation for this same diagram? Yes! 7 x ? = 21 Do you know what number times 7 equals 21? Great job! It's 3. So you figured out that: 21 ÷ 7 = 3 Multiplication and division are opposites of each other. Once you know multiplication, you also know your division! Dividing by 7 Facts Your goal is to memorize these. Complete the practice after you review to help. 7 ÷ 7 = 1 14 ÷ 7 = 2 21 ÷ 7 = 3 28 ÷ 7 = 4 35 ÷ 7 = 5 42 ÷ 7 = 6 49 ÷ 7 = 7 56 ÷ 7 = 8 63 ÷ 7 = 9 70 ÷ 7 = 10 Great job! Now, try the practice! Lesson Streak 0 days Complete the practice every day to build your streak S S M T W T F Start Practice Complete the practice to earn 1 Create Credit 1,000 Create Credits is worth \$1 in real AI compute time. 1 Create Credit is enough to get 1 question answered, or to generate 1 image from text, in the tools tab. Teachers: Assign to students Duplicate Edit Questions Edit Article Assign Preview Questions Edit Duplicate<\|endoftext\|>	4.625
612	# Interquartile Range (IQR) Calculator You can use this interquartile range calculator to determine the interquartile range of a set of numbers, including the first quartile, third quartile, and median. Interquartile Range (IQR) Calculator Results Interquartile Range (IQR): 1st Quartile (Q 1 ): 2nd Quartile (Q 2 ): 3rd Quartile (Q 3 ): Related ## How to use the Interquartile Range Calculator: 1) Enter each of the numbers in your set separated by a comma (e.g., 1,9,11,59,77), space (e.g., 1 9 11 59 77) or line break. 2) Click on the "Calculate" button to calculate the interquartile range. ## What is an Interquartile Range? The interquartile range (IQR) is the range from the 25th percentile to the 75th percentile, or middle 50 percent, of a set of numbers. It is frequently calculated as a means of identifying what the range of an average performance should be. For example, how students will typically perform on an exam or the salary levels of a set of employees working in a given industry. Many people argue that the interquartile range represents a more effective measurement than the median or mean because it provides insights into how the data is dispersed as opposed to giving a single number. ## An Example of Calculating IQR Using an IQR Formula To identify the interquartile range of a set of data, simply subtract the first quartile from the third quartile as follows: IQR = Q3 - Q1 Where Q1 is the first, or lower quartile, and Q3 is the third, or upper quartile. For example, let's say we need to determine the IQR of the following set of data 1, 4, 2, 6, 8, 10, 11, 5. The set of numbers of interest is as follows: 1, 4, 2, 6, 8, 10, 11, 5. First, place the numbers in ascending order: 1, 2, 4, 5, 6, 8, 10, 11. Then, identify the 1st and 3rd quartiles as follows: 1st Quartile = (2 + 4) / 2 = 6 / 2 = 3 3rd Quartile = (8 + 10) / 2 = 18 / 2 = 9 Median = 5.5 The interquartile range (IQR) = 3rd Quartile - 1st Quartile IQR = 9 - 3 = 6 You may also be interested in our Percentile Calculator<\|endoftext\|>	4.40625
1,772	Tangent half-angle formula {{ safesubst:#invoke:Unsubst\|\|$N=Unreferenced \|date=__DATE__ \|$B= {{#invoke:Message box\|ambox}} }} In trigonometry, the tangent half-angle formulas relate the tangent of one half of an angle to trigonometric functions of the entire angle. They are as follows: ${\displaystyle \sin \alpha ={\frac {2\tan {\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}}$ ${\displaystyle \cos \alpha ={\frac {1-\tan ^{2}{\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}}$ ${\displaystyle \tan \alpha ={\frac {2\tan {\frac {\alpha }{2}}}{1-\tan ^{2}{\frac {\alpha }{2}}}}}$ There are also a number of different forms: {\displaystyle {\begin{aligned}\tan \left({\frac {\eta }{2}}\pm {\frac {\theta }{2}}\right)&={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}=-{\frac {\cos \eta -\cos \theta }{\sin \eta \mp \sin \theta }},\\[10pt]\tan \left(\pm {\frac {\theta }{2}}\right)&={\frac {\pm \sin \theta }{1+\cos \theta }}={\frac {\pm \tan \theta }{\sec \theta +1}}={\frac {\pm 1}{\csc \theta +\cot \theta }},~~~~(\eta =0)\\[10pt]\tan \left(\pm {\frac {\theta }{2}}\right)&={\frac {1-\cos \theta }{\pm \sin \theta }}={\frac {\sec \theta -1}{\pm \tan \theta }}=\pm (\csc \theta -\cot \theta ),~~~~(\eta =0)\\[10pt]\tan \left({\frac {\pi }{4}}\pm {\frac {\theta }{2}}\right)&={\frac {1\pm \sin \theta }{\cos \theta }}=\sec \theta \pm \tan \theta ={\frac {\csc \theta \pm 1}{\cot \theta }},~~~~(\eta ={\frac {\pi }{2}})\\[10pt]\tan \left({\frac {\pi }{4}}\pm {\frac {\theta }{2}}\right)&={\frac {\cos \theta }{1\mp \sin \theta }}={\frac {1}{\sec \theta \mp \tan \theta }}={\frac {\cot \theta }{\csc \theta \mp 1}},~~~~(\eta ={\frac {\pi }{2}})\\[10pt]{\frac {1-\tan(\theta /2)}{1+\tan(\theta /2)}}&={\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}.\end{aligned}}} Proofs Algebraic proofs Use double-angle formulae and sin2α + cos2α = 1, then Q.E.D. Geometric proofs A geometric proof of the tangent half-angle formula The sides of this rhombus have length 1. The angle between the horizontal line and the shown diagonal is (a + b)/2. This is a geometric way to prove a tangent half-angle formula. Note that sin((a+b)/2) and cos((a+b)/2) just show their relation to the diagonal, not the real value. The tangent half-angle substitution in integral calculus {{#invoke:main\|main}} In various applications of trigonometry, it is useful to rewrite the trigonometric functions (such as sine and cosine) in terms of rational functions of a new variable t. These identities are known collectively as the tangent half-angle formulae because of the definition of t. These identities can be useful in calculus for converting rational functions in sine and cosine to functions of t in order to find their antiderivatives. Technically, the existence of the tangent half-angle formulae stems from the fact that the circle is an algebraic curve of genus 0. One then expects that the circular functions should be reducible to rational functions. Geometrically, the construction goes like this: for any point (cos φ, sin φ) on the unit circle, draw the line passing through it and the point (−1,0). This point crosses the y-axis at some point y = t. One can show using simple geometry that t = tan(φ/2). The equation for the drawn line is y = (1 + x)t. The equation for the intersection of the line and circle is then a quadratic equation involving t. The two solutions to this equation are (−1, 0) and (cos φ, sin φ). This allows us to write the latter as rational functions of t (solutions are given below). Note also that the parameter t represents the stereographic projection of the point (cos φ, sin φ) onto the y-axis with the center of projection at (−1,0). Thus, the tangent half-angle formulae give conversions between the stereographic coordinate t on the unit circle and the standard angular coordinate φ. Then we have and By eliminating phi between the directly above and the initial definition of t, one arrives at the following useful relationship for the arctangent in terms of the natural logarithm ${\displaystyle \arctan t={\frac {1}{2i}}\ln {\frac {1+it}{1-it}}.}$ In calculus, the Weierstrass substitution is used to find antiderivatives of rational functions of sin(φ) and cos(φ). After setting ${\displaystyle t=\tan {\tfrac {1}{2}}\varphi .}$ This implies that ${\displaystyle \varphi =2\arctan t,\,}$ and therefore ${\displaystyle d\varphi ={{2\,dt} \over {1+t^{2}}}.}$ Hyperbolic identities One can play an entirely analogous game with the hyperbolic functions. A point on (the right branch of) a hyperbola is given by (cosh θ, sinh θ). Projecting this onto y-axis from the center (−1, 0) gives the following: ${\displaystyle t=\tanh {\tfrac {1}{2}}\theta ={\frac {\sinh \theta }{\cosh \theta +1}}={\frac {\cosh \theta -1}{\sinh \theta }}}$ with the identities and The use of this substitution for finding antiderivatives was introduced by Karl Weierstrass. Finding θ in terms of t leads to following relationship between the hyperbolic arctangent and the natural logarithm: ${\displaystyle \operatorname {artanh} t={\frac {1}{2}}\ln {\frac {1+t}{1-t}}.}$ The Gudermannian function {{#invoke:main\|main}} Comparing the hyperbolic identities to the circular ones, one notices that they involve the same functions of t, just permuted. If we identify the parameter t in both cases we arrive at a relationship between the circular functions and the hyperbolic ones. That is, if ${\displaystyle t=\tan {\tfrac {1}{2}}\varphi =\tanh {\tfrac {1}{2}}\theta }$ then ${\displaystyle \varphi =2\tan ^{-1}\tanh {\tfrac {1}{2}}\theta \equiv {\mathrm {gd} }\,\theta .}$ The function gd(θ) is called the Gudermannian function. The Gudermannian function gives a direct relationship between the circular functions and the hyperbolic ones that does not involve complex numbers. The above descriptions of the tangent half-angle formulae (projection the unit circle and standard hyperbola onto the y-axis) give a geometric interpretation of this function.<\|endoftext\|>	4.4375
520	# 1732 Is a Thousand Times More Than… Contents ### Today’s Puzzle: If you know the common factor of 50 and 35 that will put only numbers from 1 to 12 in the first column and the top row of this multiplication table puzzle, then you will have completed the first step in solving the puzzle. Afterward, just work from the top of the puzzle to the bottom filling in factors as you go. That’s how you solve these level-three puzzles. ### Factors of 1732: • 1732 is a composite number. • Prime factorization: 1732 = 2 × 2 × 433, which can be written 1732 = 2² × 433. • 1732 has at least one exponent greater than 1 in its prime factorization so √1732 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1732 = (√4)(√433) = 2√433. • The exponents in the prime factorization are 2 and 1. Adding one to each exponent and multiplying we get (2 + 1)(1 + 1) = 3 × 2 = 6. Therefore 1732 has exactly 6 factors. • The factors of 1732 are outlined with their factor pair partners in the graphic below. ### More About the Number 1732: 1732 is the sum of two squares: 34² + 24² = 1732. 1732 is the hypotenuse of a Pythagorean triple: 580-1632-1732, calculated from 34² – 24², 2(34)(24), 34² + 24². It is also 4 times (145-408-433). 1732 is a palindrome in base 15 because 7(15²) + 10(15) + 7(1) = 1732. I remember one of my college professors telling his class that √2 is about 1.4, and Valentine’s day is February 14, √3 is about 1.7, and Saint Patrick’s day is March 17. That’s how I remember those values, but this tweet reminded me that 1732 is a thousand times more than the square root of three rounded to three decimal places. It also makes a reference to the square root of two. This site uses Akismet to reduce spam. Learn how your comment data is processed.<\|endoftext\|>	4.46875
175	The Iowa Watershed Approach Download Full Text What is a wetland?Wetlands are strategically placed to capture surface runoff and drainage water. They provide temporary storage for floodwater which lowers downstream flood peaks and reduces streamflow by 10-20%. Additionally, they reduce nitrate-N concentration in water by 52% and filter out chemicals from pesticides and insecticides. Wetlands also serve as wildlife habitat for 190 amphibian species and 5,000 plant species. Benning, Jamie and Craft, Kristina, "Wetlands" (2018). Extension and Outreach Publications. 442. Iowa State University Extension and Outreach publications in the Iowa State University Digital Repository are made available for historical purposes only. The information contained in these publications may be out of date. For current publications and information from Iowa State University Extension and Outreach, please visit<\|endoftext\|>	3.71875
538	The Grand Canyon in northern Arizona is a favorite for astronauts shooting photos from the International Space Station, as well as one of the best-known tourist attractions in the world. The steep walls of the Colorado River canyon and its many side canyons make an intricate landscape that contrasts with the dark green, forested plateau to the north and south. The Colorado River has done all the erosional work of carving away cubic kilometers of rock in a geologically short period of time. Visible as a darker line snaking along the bottom of the canyon, the river lies at an altitude of 715 meters (2,345 feet), thousands of meters below the North and South Rims. Temperatures are furnace-like on the river banks in the summer. But Grand Canyon Village, the classic outlook point for visitors, enjoys a milder climate at an altitude of 2,100 meters (6,890 feet). The Grand Canyon has become a geologic icon—a place where you can almost sense the invisible tectonic forces within the Earth. The North and South Rims are part of the Kaibab Plateau, a gentle tectonic swell in the landscape. The uplift of the plateau had two pronounced effects on the landscape that show up in this image. First, in drier parts of the world, forests usually indicate higher places; higher altitudes are cooler and wetter, conditions that allow trees to grow. The other geologic lesson on view is the canyon itself. Geologists now know that a river can cut a canyon only if the Earth surface rises vertically. If such uplift is not rapid, a river can maintain its course by eroding huge quantities of rock and forming a canyon. A more detailed view shows the canyon in an astronaut photo that looks west. Astronaut photograph ISS039-E-5258 was acquired on March 25, 2014, with a Nikon D3S digital camera using a 180 millimeter lens, and is provided by the ISS Crew Earth Observations Facility and the Earth Science and Remote Sensing Unit, Johnson Space Center. The image was taken by the Expedition 39 crew. It has been cropped and enhanced to improve contrast, and lens artifacts have been removed. The International Space Station Program supports the laboratory as part of the ISS National Lab to help astronauts take pictures of Earth that will be of the greatest value to scientists and the public, and to make those images freely available on the Internet. Additional images taken by astronauts and cosmonauts can be viewed at the NASA/JSC Gateway to Astronaut Photography of Earth. Caption by M. Justin Wilkinson, Jacobs at NASA-JSC.<\|endoftext\|>	3.765625
878	I: Stating the problem Suppose you took a picture, and you want to print it out. The color might be accurate on your computer screen, but not so right when you print it out. It might be too dark or maybe not saturated enough. This is the most basic color management problem. Things look different on the screen than on a printout. Got it? Good. :-) So the screen and the printer each have a different 'color profile,' which is why the image appears different when you print it out. That is, a given value for, say, green, might appear one way on the screen, and a different way on the paper. Or perhaps the printout looks washed out in comparison to the screen image, or maybe darker. Now, in either case, you're starting with the same data: your original picture. So the example of a picture that's different on screen from the printed page brings us to the real problem: Different representations of the same color data. Meaning: The screen takes the image data and displays it in one way, while the printer takes the same image data and displays it another way. And that's the most basic way to state the problem. Got it? Good. :-) II: Stating a solution We've got an image, which a screen will display one way, and a printer will display in another. That's our problem. To solve it, we can consider: If these differences are consistent, and we know what the differences are, then we can alter the color data depending on how we want to show the image. So if we want to display the image on a screen, we change the data to be more accurate on a screen. If we want to print the image, we change the data to be more accurate on paper. This means we have three things: 1) The image, 2) a way to change the image for display on the screen, and 3) a way to change the image for printing. Since we have these three things, we can be reasonably sure that colors are accurate no matter whether we show the image on the screen or print it out. Got it? Good. :-) Now, since we're talking about computers, we can get the computer to do this for us automatically. Software that does this is called a 'color management system.' On Mac OS X (and previous versions going back to System 7), there's a color management system called ColorSync. In Windows Vista, there is the Windows Color System. Previous Windows systems have Windows Image Color Management. Linux has a few CMS, including Little CMS, or lcms. So the software can change the color data to be more appropriate for whatever output you're using. It can take the original image and make it look right on the screen, or the printer, because it knows the difference between the original color data, and the way things should look on either device. How does it know? That's a very good question. It uses profiles. A profile is a file that lives on your computer and serves the function of describing how to change the color data for output on a particular device. Meaning: When ColorSync (or whatever other CMS you're using) wants to know how to change the color data in order to print an image on your printer, it consults the printer's profile. The profile tells it what to do. The more accurate the profile, the more accurate the picture will be. It just so happens that in the mid '90s, the International Color Consortium was formed to address this issue of color matching in digital media. And they created a standard for these profiles. That is, if two different CMSs both support the ICC standard, then a given ICC profile should work well with either one. So, to sum up, we've got: 1) An image, 2) a profile for your computer's screen, 3) a profile for your printer, and 4) a color management system to do the math and make everything look right. And that's the most basic thing you need to know about color management. Of course there are more complex issues surrounding this, but the goal in this article is to be as basic as possible. Got a question? Ask it.<\|endoftext\|>	3.6875
1,390	# Difference between revisions of "2019 AMC 12A Problems/Problem 20" ## Problem Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $\|x-y\| > \tfrac{1}{2}$? $\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}$ ## Solution 1 There are several cases depending on what the first coin flip is when determining $x$ and what the first coin flip is when determining $y$. The four cases are: Case 1: $x$ is either $0$ or $1$, and $y$ is either $0$ or $1$. Case 2: $x$ is either $0$ or $1$, and $y$ is chosen from the interval $[0,1]$. Case 3: $x$ is is chosen from the interval $[0,1]$, and $y$ is either $0$ or $1$. Case 4: $x$ is is chosen from the interval $[0,1]$, and $y$ is also chosen from the interval $[0,1]$. Each case has a $\frac{1}{4}$ chance of occurring (as it requires two coin flips). For Case 1, we need $x$ and $y$ to be different. Therefore, the probability for success in Case 1 is $\frac{1}{2}$. For Case 2, if $x$ is 0, we need $y$ to be in the interval $\left(\frac{1}{2}, 1\right]$. If $x$ is 1, we need $y$ to be in the interval $\left[0, \frac{1}{2}\right)$. Regardless of what $x$ is, the probability for success for Case 2 is $\frac{1}{2}$. By symmetry, Case 3 has the same success rate as Case 2. For Case 4, we must use geometric probability because there are an infinite number of pairs $(x, y)$ that can be selected, whether they satisfy the inequality or not. Graphing $\|x-y\| > \tfrac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality: $[asy] filldraw((0,0)--(0,1)--(1/2,1)--(0,1/2)--cycle,black+linewidth(1)); filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1)); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); [/asy]$ The shaded area is $\frac{1}{4}$, which means the probability for success for case 4 is $\frac{1}{4}$ (since the total area of the bounding square, containing all possible pairs, is $1$). Adding up the success rates from each case, we get: $\left(\frac{1}{4}\right) \cdot \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}\right) = \boxed{\textbf{(B) }\frac{7}{16}}$. ## Solution 2 Taking into account that there are two options for the result of the first coin flip, there are four possible combinations with equal possibility of initial coins flips. (4) x: tails, y: tails (1) Because x and y both must be integers, so it's simple to see that the probability that $\|x-y\| > \tfrac{1}{2}$ is $\frac{1}{2}$. (2) Since x has to be an integer it is also easy to see that the probability is again $\frac{1}{2}$. (3) This is the same as case 2 so the probability is $\frac{1}{2}$. (4) This one is a little bit tricky. You could use geometric probability shown above, but if you don't understand it or happened to not think of it, there is another solution that involves using the multiple choice answers. First, the probability we have so far is $\frac{3}{8}$.This is greater than $\frac{1}{3}$ so $A$ is the not the answer. It is either $B, C, D,$ or $E$. Let's draw out the probability with two parallel lines on a paper with represent a number line from $0-1$. The concept is to compare one point and find the fraction of the other line that contains points that are at least $\frac{1}{2}$ away from it. If the point we choose is on the endpoints, then the fraction of the other line that works is exactly $\frac{1}{2}$. But as we move this point closer to the middle, the deadzone (area where the other point cannot be) grows, diminishing the probability. Finally, when it is directly in the middle, there are no points that pass the requirements except at $1$ and $0$. So, looking at the choices again, we have $\frac{7}{16}$, $\frac{1}{2}$, $\frac{9}{16}$, and $\frac{2}{3}$. $\frac{1}{2}$ is exactly $\frac{1}{8}$ more than the probability we had before. Notice that this is impossible because we proved that the average probability of the fourth case is lower than $\frac{1}{2}$, so the answer is $\boxed{B}$. -jackshi2006 ## Video Solution 1 Education, the Study of Everything ~ pi_is_3.14<\|endoftext\|>	4.78125
1,107	How good are you with figures and formulas? Find out with these Mathematics past questions and answers. This Test is useful for both job aptitude test candidates and students preparing for JAMB, WAEC, NECO or Post UTME. 1. Find the mean deviation of a set of numbers: 14, 15, 16, 17, 18, and 19. A. 2.5 B. 1.7 C. 1.5 D. 3.5 x̄ = $$\frac{ 14 + 15 + 16 +17 + 18 + 19}{6} = \frac{99}{6}$$ = 16.5 M . D = $$\frac{ (14 - 16.5) + (15 - 16.5) + (16 - 16.5) + (17 - 16.5) + (18 - 16. 5) + (19 - 16.5) }{ 6}$$ M.D = $$\frac{(-2.5) + (-1.5) + (-0.5) + (0.5) + (1.5) + (2.5)}{6}$$ Taking the absolute value of the deviations M.D = $$\frac{2.5 + 1.5 + 0.5 + 0.5 + 1.5 + 2.5}{6}$$ M.D = $$\frac{9}{6}$$ = 1.5 2. M varies jointly as the square of n and square root of q. If M = 24 when n = 2 and q = 4, find M when n = 5, q = 9. A. 288 B. 400 C. 300 D. 225 $$M ∝ n^2\sqrt{q}$$ $$M = Kn^2\sqrt{q}$$ K = $$\frac{M}{n^2\sqrt{q}}$$ K = $$\frac{24}{2^2\sqrt4}$$ k = $$\frac{24}{8} = 3$$ Now, let's find M when n = 5 and q = 9 M = $$Kn^2\sqrt{q}$$ M = $$3\times5^2\sqrt9$$ $$M = 3\times25\times3$$ Therefore, M = 225. 3. The diagonals of a rhombus are 16 cm and 12 cm find the length of the side. A. 20cm B. 8cm C. 14cm D. 10cm In a rhombus, the diagonals are perpendicular bisectors of each other, and they bisect the angles of the rhombus. This means that a rhombus is essentially made up of four congruent right-angled triangles. We can use the Pythagorean theorem to find the length of one side of the rhombus (s) $$s^2 = 8^2 + 6^2$$ $$s^2 = 64 + 36$$ $$s^2 = 100$$ s = $$\sqrt{100}$$ s =10 cm So, the length of each side of the rhombus is 10 cm. 4. If 2x - 3y = -11 and 3x + 2y = 3, evaluate $$(y - x)^2$$ A. 16 B. 25 C. 9 D. 4 2x - 3y = -11 --- (i) 3x + 2y = 3 --- (ii) Multiply equation (i) by 3 and equation (ii) by 2 6x - 9y = -33 --- (iii) 6x + 4y = 6 --- (iv) Subtract equation (iii) from (iv) 13y = 39 y = $$\frac{39}{13}$$ = 3 substitute (3) for y in equation (ii) 3x + 2(3) = 3 3x + 6 = 3 3x = 3 - 6 3x = -3 x = $$\frac{-3}{3}$$ = - 1 Now, $$(y - x)^2 = (3 - (-1))^2$$ = $$(3 + 1)^2$$ = $$4^2$$ = 16 5. A notebook of length 15 cm was measured to be 16.8 cm, calculate, correct to two d.p, the percentage error in the measurement. A. 12.00% B. 11.71% C. 10.71% D. 11.21% Error = 16.8 - 15 = 1.8cm % error = $$\frac{error}{ Actual value}\times 100%$$ = $$\frac{1.8}{ 15}\times100%$$ = $$0.12\times100%$$ = 12.00% WAEC<\|endoftext\|>	4.53125
1,304	What Is an X-Ray? X-ray imaging became a valuable diagnostic tool before the 20th Century. The process has evolved over time and remains integral to emergency care in which prompt evaluation is necessary. This modality is one of the most accessible, affordable, and expedient methods of obtaining important data. In fact, the process has become even more efficient thanks to digital software and transmission to the referring physician’s office. What Do X-Ray Results Show? X-ray images are often used to observe: - Bone fractures or other abnormalities of bony structures. - The extent of joint space, indicating the severity of osteoarthritis. - The shape and size of the heart, used to diagnose heart conditions. - The presence of fluid in the lungs or other hollow structure. - The presence of denser soft tissue, indicating a potential tumor. Where Are X-Rays Performed? - Bones and teeth: Physicians frequently rely on x-ray images to assess the condition of bones because hard tissue such as bone appears white on an x-ray. Clinicians are able to identify areas of decay, infection, and fracture. X-rays can reveal tumors on bone, as well as arthritic conditions in joints. Physicians may order chest x-rays to evaluate for lung conditions or infection such as pneumonia. Blood vessel blockages may also be revealed if x-rays are performed with contrast material (iodine). X-rays can reveal heart enlargement as well as breast cancer (mammography). X-ray imaging can quickly identify if a foreign object has been swallowed and where along the digestive tract that object lies. Am I a Candidate for an X-Ray? X-ray imaging is a valuable diagnostic tool used to explore potential reasons for chest pain as well as pain in joints and limbs. Chest x-rays may be performed if a physician suspects tuberculosis or if a patient has chest pain, difficulty breathing, persistent or severe allergies or asthma, or a productive cough. X-rays may also be necessary if any bony structure has suffered an injury. For example, falling on an outstretched hand may cause a fracture or other injury to the wrist. An x-ray observes the extent of injury to the wrist, enabling a physician to administer appropriate treatment. X-ray imaging represents a significant advancement in medicine. This form of imaging has been safely used for several years, sharply decreasing the need for exploratory surgery and guesswork in diagnostic care. X-rays are routinely performed to improve the accuracy of medical diagnoses and to monitor various health conditions. X-ray images can support physicians in the development of medical and surgical treatment planning, and also provide assistance during medical procedures such as the removal of blood clots or the insertion of a stent or other device. I tried several other places and I could not do this. I’m extremely claustrophobic. The other places in Humble the technician made me feel very scared and I was totally unable to continue. (The technologists) here at GO Imaging were so great. I didn’t think I could go thru this and be totally eased. He made it where I was able to finally complete. The staff up front was also so great and helpful. God bless you and thank you. To read more, visit our testimonial page. X-Ray vs MRI & CT Scan X-rays and CT scans both rely on radiation to capture accurate images. X-rays enable physicians to differentiate between hard matter and soft, or air. CT scans produce cross-sectional, 360-degree images by revolving x-ray beams around the body and capture images. The result is a high-quality 3-dimensional view. MRI imaging requires more time on the table than most X-rays and CT scans. There is no radiation exposure during Magnetic Resonance Imaging. A clear view of normal versus abnormal tissue is obtained with strong magnets and radio waves. We know from research that the value of X-ray imaging outweighs the risks of radiation exposure. How much exposure takes place is dependent on the part of the body being observed. According to experts, the radiation emitted by x-ray machines is low, and should not pose significant risk of cell mutation. Patients who are pregnant or who think they might be, need to disclose this information to their physician and to GO Imaging. Women that are pregnant will not receive an X-ray. The doctor will offer an alternative exam. What to Expect During X-Ray - Prior to X-ray imaging, the patient may have to remove jewelry and certain articles of clothing. - The patient is then positioned in such a manner to facilitate the clearest view of the anatomy to be examined. - For certain X-rays, the patient may need to hold his or her breath. - The technician will then step just outside of the X-ray area to operate the X-ray machine. - X-rays are reviewed before the patient leaves the facility. This is done to ensure quality images have been obtained. - If necessary, the radiologist will then take more images. Otherwise, the patient is released from the facility. “I just wanted to thank you for my “enjoyable” experience at your facility. (Your staff) was wonderful at fitting me in so quickly and was very kind and personable. (The technologist) was gentle and explained everything that was going to happen. When leaving your facility I was given a thank you gift which was very classy & made me feel really good. You followed up with a thank you note which completed my process. How kind of you to treat you patients like important people. You get an A+. Thank you again for making my experience a great one.” – Candice G. After the x-rays, your radiologist will review and interpret the images. The radiologist will send a report to the referring physician, who will contact or schedule an office visit. Following Up After an X-Ray The results of x-ray screening are used by physicians to confirm, monitor, and plan. During your follow-up with your doctor, the images obtained during your x-ray may be reviewed with you to show you the location and extent of a bone fracture or other aspect of your current condition.<\|endoftext\|>	3.828125
647	# Mathematics ### Chapter : Polynomials #### Relation between the zero and the coefficients of a polynomials Consider quadratic polynomial P(x) = 2x2 – 16x + 30. Now, 2x2 – 16x + 30 = (2x – 6) (x – 3) = 2(x – 3) (x – 5) The zeroes of P(x) are 3 and 5. Sum of the zeroes are = 3 + 5 = 8 = Product of the zeroes are = 3 × 5 = 15 = So if ax2 + bx + c, a ≠ 0 is a quadratic polynomial and a, b are two zeroes of polynomial then Example: Find the zeroes of the quadratic polynomial 6x2 – 13x + 6 and verify the relation between the zeroes and its coefficients. Solution: We have, 6x2 – 13x + 6 = 6x2 – 4x – 9x + 6 = 2x (3x – 2) –3 (3x – 2) = (3x – 2) (2x – 3) So, the value of quadratic equation 6x2 – 13x + 6 is 0, when (3x – 2) = 0 or (2x – 3) = 0 i.e., When x = or Therefore, the zeroes of 6x2 – 13x + 6 are and . Sum of the zeroes : + = = Product of the zeroes = × = = Example: Find the zeroes of the quadratic polynomial 4x2 – 9 and verify the relation between the zeroes and its coefficients. Solution: We have, 4x2 – 9 = (2x)2 – 32 = (2x – 3) (2x + 3) So, the value ofa quadratic equation 4x2 – 9 is 0, when 2x – 3 = 0 or 2x + 3 = 0 i.e., when x = or x = . Therefore, the zeroes of 4x2 – 9 are & . Sum of the zeroes = + = 0 = = Product of the zeroes = = Example: Find the zeroes of the quadratic polynomial 9x2 – 5 and verify the relation between the zeroes and its coefficients. Solution: We have, 9x2 – 5 = (3x)2 – (√5)2 = (3x – √5) (3x +√5) So, the value of 9x2 – 5 is 0, when 3x –√5 = 0 or 3x +√5 = 0 i.e., when x = or x = . Sum of the zeroes == 0 = = Product of the zeroes = ×=<\|endoftext\|>	4.71875
1,085	Archaeologists in Britain have uncovered the charred remains of a 3,000-year-old stilted wooden structure that plunged into the river after it caught fire. The remarkably well-preserved roundhouse is offering an unprecedented glimpse into what domestic life was like during the Bronze Age. This remarkable settlement in East Anglia was occupied at the end of the Bronze Age, sometime around 1200-800 BC. Several families lived in a circular building, which was propped up on stilts above the water. The structure collapsed into the river after a fire damaged the posts. The roundhouse would have looked something like this Celtic Crannog (Credit: Christine Westerback/CC BY SA 2.0) But the fire and the roundhouse’s subsequent collapse into the river contributed to its extraordinary preservation. Like the intact structures found at Pompeii, the flames helped to carbonize and maintain the wooden beams. Silt at the bottom of the river prevented air and bacteria from chewing away at the wood. And because the inhabitants were forced to leave everything behind, virtually everything remains where they left it. Archaeologists are describing it as a time capsule. “Everything suggests the site is not a one-off but in fact presents a template of an undiscovered community that thrived 3,000 years ago ‘beneath’ Britain’s largest wetland” — Mark Knight Posts and rafters stick up from the ground, while footprints of the inhabitants can still be seen in the sediment. A charred roof of a roundhouse remains visible, as are tool marks and a perimeter of wooden posts that once enclosed the site. Experts say the excavations are revealing “the best-preserved Bronze Age dwellings ever found” in the country. “A dramatic fire 3,000 years ago combined with subsequent waterlogged preservation has left to us a frozen moment in time, which gives us a graphic picture of life in the Bronze Age,” noted Duncan Wilson, Chief Executive of heritage organization Historic England. He says the site is of “international significance,” and that it’s poised to “transform our understanding of the period.” The site, located at Must Farm near Peterborough, England, has already produced a treasure-trove of artifacts. Though the excavations are only half complete, the archaeologists have uncovered elaborate textiles made from plant fibers, along with small cups, bowls, and jars that still have their meals inside. The researchers also found glass beads that were attached to a necklace—a sophisticated item of jewelry not typically associated with the Bronze Age. This could mean that the inhabitants were at the upper levels of society. These glass beads were once part of an elaborate necklace. “Must Farm is the first large-scale investigation of the deeply buried sediments of the fens and we uncover the perfectly preserved remains of prehistoric settlement,” said Cambridge Archaeological Unit (CAU) Site Director Mark Knight. “Everything suggests the site is not a one-off but in fact presents a template of an undiscovered community that thrived 3,000 years ago ‘beneath’ Britain’s largest wetland.” It’s prehistoric archaeology in 3D with an unsurpassed finds assemblage both in terms of range and quantity.” — David Gibson A human skull was also found at the site, but further excavations are required to determine if it belongs to a person who died in the fire. The team also plans to bring in a fire expert to determine if the fire was deliberate (e.g. the result of a hostile tribe) or accidental. These textiles were made from plant fibers. Archaeologists have known about the site for decades, but it hasn’t been touched since 2006. Efforts to fully uncover the settlement were given an added sense of urgency after fears emerged that falling water levels could cause the remains to degrade quickly. The $1.58 million (£1.1 million), 4-year project is being handled by the CAU, who are being funded by Historic England. All items are being sent to labs for further analysis. A detailed paper is expected in a few years, after which time the items will go on public display. Similar European prehistoric wetland sites have been found before, but nothing quite like this one. Other examples include the ancient loch-side dwellings known as crannogs in Scotland and Ireland, the stilt houses of the Alpine Lakes, and the terps—human-made hill dwellings—in the Netherlands. “Usually at a Later Bronze Age period site you get pits, post-holes and maybe one or two really exciting metal finds. Convincing people that such places were once thriving settlements takes some imagination,” said CAU Archaeological Manager David Gibson. “But this time so much more has been preserved—we can actually see everyday life during the Bronze Age in the round. It’s prehistoric archaeology in 3D with an unsurpassed finds assemblage both in terms of range and quantity.” All images: Cambridge Archaeological Unit Email the author at [email protected] and follow him @dvorsky.<\|endoftext\|>	3.8125
410	It has been Kit-chi’sippi (“the Great River”) to the Algonquin nations, La Grande Rivière du Nord to the French and even nicknamed “the original Trans-Canada Highway” in recognition of its historic role as one of the most vital transportation and trade arteries in North America. The history of human influence on the Ottawa River spans more than 8,500 years, stretching back to the early ancestors of the Algonquin people, who travelled the great river and lived and hunted along its banks. But large-scale change arrived just four centuries ago with explorers such as Étienne Brûlé and Samuel de Champlain and the legions of coureurs de bois, voyageurs and missionaries who were soon pushing upstream in search of new territories, furs and Christian converts. By the early 1800s, lumber barons were using the Ottawa to carry massive pine timber rafts down to the St. Lawrence, and as lumber camps and farming settlements grew into towns and cities, mills and dams were built to harness the power of the river and its tributaries and tame spring floods. Today, the Ottawa River basin is one of the most dammed and regulated catchments in Canada. The 1,271-kilometre main stem and its tributaries drain and sustain a 146,300-square-kilometre area (a region larger than England) that provides drinking water for millions of people, water for recreation, hydroelectricity production and other industries, and habitat for hundreds of fish, bird, mammal, amphibian and reptile species. Conservation efforts and water management are a dizzyingly complex issue in a watershed that’s home to about 200 municipalities and First Nations communities, split between two provinces that share the water, but not policies for its governance. This map was created by Canadian Geographic and the environmental research- and education-focused Chawkers Foundation to put a spotlight on this ecologically, economically and historically critical waterway.<\|endoftext\|>	3.6875
994	By the 1970s amateurs had noticed that all 110 Messier objects could be observed at low northern latitudes over the course of a night in mid-to-late March. You’ve probably heard of Charles Messier’s catalog of celestial deep-sky objects for 18th century comet hunters. When they stumbled upon an unknown faint fuzzy object, they’d consult this list to see if it was a known object. It was first published in 1774 and expanded with help from fellow observers in that century. Astronomers and amateurs in the early 20th Century rediscovered it, added a few new objects, and made corrections. There are now 110 “M” objects in the modern catalog. Every sky chart labels the location of these objects with the letter “M” and a number. For instance, the brightest Messier is M45 — the famous naked-eye Pleiades open star cluster. A third of the way around the sky you’ll find the dimmest: M95 — a galaxy of magnitude 10 or 11 (depending on the source) in Leo. It’s fascinating to compare Messier’s original notes for the objects with notes from the modern list. While about 40 galaxies are noted in the current object descriptions, Messier’s original text never mentioned galaxies or the uncounted stars they contained. Also, the stunning horde of sparkling jewels in globular cluster was lost on Messier and friends. They often noted these objects as being “nebulae without stars”. Problem was: Messier didn’t have a Tele Vue scope or eyepiece 😉 . Seriously, these differences are due to 244-years of optical and astronomical science advancement. Where 18th century astronomers saw gray nebulae we can now see stars! The value of the list in modern times is that it contains many “showpiece” objects, visible to small telescopes, from northern latitudes. By the 1970s amateurs had noticed that all 110 Messier objects could be observed at low northern latitudes over the course of a night in mid-to-late March. This is due to a lack of Messier objects along the Sun’s path at this time of year. Hence, the phrase “Messier Marathon” was invented to describe the all-night attempt at locating and verify each object on the list over the course of a single night. Traditionally the Messier Marathon is done without a “goto” scope, as an automated scope would make the task too easy. This year, on March 17, 2018, the Moon will be new and the Sun in western Pisces. At sunset, it’ll be a good night to try your own Messier Marathon. But if you want company, head out to The All Arizona Messier Marathon (March 17-18, 2018) in Salome, AZ. This long-running event is held by the Saguaro Astronomy Club (SAC). SAC has been hosting the All Arizona Messier Marathons since 1993 and other Messier Marathons before that. They list the top observers on their website every year and hand out awards. There is even an entry category for kids. Our Tele Vue representative John Rhodes will be at the Marathon this year. He may even be able to lend you some equipment to do the Marathon with — or just to look around the sky. Any of Tele Vue’s scopes — from the super-portable TV-60 to the photo / visual flat-field of the TVNP-127is — will have no problem picking up the dimmest Messier object. Our legendary wide-field eyepieces are perfect for quickly sweeping up targets. Unlike equatorial mounts, the Tele-Pod style head on all our mounts allow you to aim a telescope in a perfectly natural way by hand. We wish you good luck with the Marathon, whether done in the back-yard or at an organized event. If you use Tele Vue equipment and post your results to social media, do include the #televue so we can like your comments. - If you’d like to “browse” through the objects for planning purposes, Wikipedia has a table of Messier object that can be sorted by name, position, magnitude, etc. - Information on the All Arizona Messier Marathon - All Arizona Messier Marathon check-off sheet with objects in observation order. - Tele Vue telescopes on our website, Why Choose Tele Vue Scopes article (mobile version) - Tele Vue eyepieces on our website, Why Choose Tele Vue Eyepieces article (mobile version) - Tele Vue mounts on our website (mobile version)<\|endoftext\|>	3.734375
614	Science, Maths & Technology ### Become an OU student Everyday maths 2 Start this free course now. Just create an account and sign in. Enrol and complete the course for a free statement of participation or digital badge if available. Figure _unit2.9.1 Figure 23 Inverse operations An inverse operation is an opposite operation. In a sense, it ‘undoes’ the operation that has just been performed. Let’s look at two simple examples to begin with. ## Case study _unit2.9.1 Example: Check your working 1 6 + 10 = 16 ### Method Since you have done an addition sum, the inverse operation is subtraction. To check this calculation, you can either do: #### Extract _unit2.9.1 16 − 10 = 6 or 16 − 6 = 10 You will notice here that the same 3 numbers (6, 10 and 16) have been used in all the calculations. ## Case study _unit2.9.2 Example: Check your working 2 5 × 3 = 15 ### Method This time, since you have done a multiplication sum, the inverse operation is division. To check this calculation, you can either do: #### Extract _unit2.9.2 15 ÷ 5 = 3 or 15 ÷ 3 = 5 Again, you will notice that the same 3 numbers (3, 5 and 15) have been used in all the calculations. If you have done a more complicated calculation, involving more than one step, you simply ‘undo’ each step. ## Case study _unit2.9.3 Example: Check your working 3 A coat costing £40 has a discount of 15%. How much do you pay? ### Method Firstly, we find out 15% of £40: #### Extract _unit2.9.3 40 ÷ 100 × 15 = £6 discount £40 − £6 = £34 to pay To check this calculation, firstly you would check the subtraction sum by doing the addition: #### Extract _unit2.9.4 £34 + £6 = £40 To check the percentage calculation, you then do: #### Extract _unit2.9.5 £6 ÷ 15 × 100 = £40 You have now completed the number section of the course. Before moving on to the next session, ‘Units of measure’, complete the quiz on the following page to check your knowledge and understanding. ## Summary In this section you have: • learned that each of the four operations has an inverse operation (its opposite) and that these can be used to check your answers • seen examples and practised checking answers using the inverse operation.<\|endoftext\|>	4.8125
1,557	Many agricultural techniques exist today, but in an effort to adjust to the exponential trends of our population without compromising the integrity of the environment it is necessary to have a global transition towards sustainable farming. In a conventional system farmers will designate entire fields to just one crop, which creates uniformity. As communities grow though, less and less land is available for food production and existing crops become easily exhausted. Department of Agriculture. Fortunately humans have been perfecting agricultural methods for thousands of years, which can help to answer this question. Maintenance is made easy for farmers as conventional farming typically involves monocropping, but is also very expensive. Just another SB You: Web Publishing for You site This is important to highlight because conventional agriculture discourages biodiversity and instead relies on synthetic chemicals to maintain crop health. However, these crops are also sprayed with toxic pesticides and herbicides to make up for their uniformity. In addition to higher levels of biodiversity, sustainable farming is typically associated with better soil quality. Agriculture Waste Management for Sustainable Agriculture When humans first discovered the potential of planting seeds, they suddenly had the ability to explore the world and establish infrastructures wherever soils were fertile. This is a desirable characteristic in agricultural land as it allows crops to be more tolerable to changing climate. With breakthroughs in agriculture, populations increased and development spread. Sustainable Lafayette. This result varies though, and in some instances organic crops actually best conventional crops. This method is so efficient compared to conventional agriculture because it requires no input of synthetic chemicals or fertilizers, which accounts for a large amount of the greenhouse gas emissions. - Sustainable Vs. Conventional Agriculture \| Environmental Topics and Essays - Sustainable Agriculture \| National Institute of Food and Agriculture Support research and education intended to help farmers and ranchers mitigate and adapt to climate change Improve production efficiency, productivity, and profitability Address threats from pests and diseases Improve the quality of surface water and groundwater resources Primary Point of Contact: Conventional farming uses synthetic chemicals and fertilizers to maximize the yield of a particular crop or set of crops, which are typically genetically modified. Conventional agriculture was developed to make farming more efficient, but achieves that efficiency at a major cost to the environment. Sustainable agriculture is a natural way to produce food and has a number of social, economic, and environmental benefits. Smith, J. Most research indicates that sustainable crops produce much creative holiday homework for class 7 english than conventional systems. Agronomy for Sustainable Development [Online] ; http: Better management of agricultural land is required to reduce the effects of crop production. Huntley, EE. Locally, however, water is a scarce resource and must be appropriated efficiently. Is Temperate Agroforestry the Answer? This curriculum vitae director marketing retention rate enables sustainable agricultural systems to produce much higher yields than conventional systems during drought conditions Gomiero, Pimentel, and Paoletti Although levels of production are reduced in sustainable agriculture, studies show that higher levels of biodiversity are linked argumentative essay best healthier crops. Instead, farmers will plant a variety of plants together short essay on my favourite bike promote biodiversity and ward off pests and pathogens Nicholls and Altieri In addition, agricultural production has increased tremendously worldwide over the creative holiday homework for class 7 english century. Engage farmers and ranchers in the development and adoption of practices that are profitable and environmentally sound. Selective breeding was first implemented on plants over 10, years ago to produce desired characteristics in crops USDA. Typically conventional crops are genetically modified to perform better under certain conditions than sustainable crops Carpenter Therefore, my goal in writing this paper was to use reliable, long-term research that made specific assessments of the two generalized types of farming and then compare the results. Despite the impacts conventional methods have on agricultural land, not all conventional farms degrade biodiversity. Sustainable agriculture relies solely on natural processes for input and recycles nutrients on-site to eliminate the use of non-renewable resources. Our programs provide services that: Carpenter, J, E. A significant amount of chemical and energy input is required in conventional agriculture to produce the highest possible yield of crops. The science has encouraged people to live and develop rich, permanent settlements all over the world. Department of Agriculture. A uniform crop is ideal because it reduces labor costs and makes harvesting easy, but it can also impact biodiversity and make crops susceptible to pathogens Gabriel, Salt, Kunin, and Benton And once committed to the conventional practices, farmers find themselves sustainable agriculture essay pdf in a perpetual cycle of loans, subsidies, and debt. The environmental impact and production levels of oxford uni creative writing masters method will determine its overall viability as a solution to growing trends. Conclusion Studies point toward sustainable agriculture as the best solution to managing the growing population. Since the goal of conventional agriculture is to maximize yields, environmental health and biodiversity are usually not preserved. Sustainable agriculture essay pdf relies on ecological processes, biodiversity and cycles adapted to local conditions, rather than the use of inputs with adverse effects. Uniformity can determine both the success and failure of conventional systems. Climate Change and Sustainable Food Production. A Comparison of Agriculture In a comparison of conventional and sustainable agriculture there should be several points of focus: Sustainable agriculture profits farmers, economies, and food banks while existing symbiotically with the landscape. Increase profitable farm income Enhance quality of life for farm families and communities Increase production for human food and fiber needs Importance of Sustainable Agriculture Sustainable agriculture frequently encompasses a wide range of production practices, including conventional and organic. This discovery further contributed to the permanence and size of settlements. Sustainable Vs. Conventional Agriculture GM Crops [Online]Volume 2: Advancements in this science have enabled humans to manipulate entire ecosystems to cater to their survival. There are many environmental benefits associated with sustainable agriculture, but its production capacity is limited. Sustainable agriculture is a natural way to produce food and has a number of social, economic, and environmental benefits. Organic agriculture combines tradition, innovation and science to benefit the shared environment and promote fair relationships and a good quality of life for all involved. Sustainable agriculture has sustainable agriculture essay pdf potential to sequester carbon, feed the world, and enrich the environment. Sustainable crops are more permanent than conventional crops because they work in harmony with the landscape rather than drain it of nutrients and biomass. Mueller, ND. To conserve this resource a drastic overhaul of water saving techniques, especially in agriculture, must occur. But as populations continue to grow, resources are becoming limited. Sustainable Vs. However, there are measures that can be sustainable agriculture essay pdf to increase energy efficiency. This is achieved through the application of synthetic chemicals, genetically modified organisms, and a number of other industrial products. Factors such as growing populations, economic instability, climate change, and pressures from companies to produce higher yields have contributed to this shift. However, adopting these conventional methods subjects farmers to the greed of industry, as their crops depend on a high input of energy, synthetic chemicals, and genetically modified organisms. Although biodiversity does not directly determine crop yield, it does play argumentative essay best major role in the health and permanence of sustainable farms. In general, sustainable agriculture essay pdf agriculture fails to match up to conventional agriculture in terms of production. History of Agriculture Agriculture has played a tremendous role in the advancement of human society. Conventional systems are inefficient at capturing carbon because of soil composition, constant production, and how much energy is being used to maintain the crops. Soil management is vital for existing farms because agricultural production is increasing globally and land is becoming less available to accommodate this growth.<\|endoftext\|>	3.8125
519	The term chord progression simply refers to the order in which chords are played in a song/piece of music. Play a few different songs/pieces and you will see that there are various different ways in which composers order chords. However, the good news is that there are a few simple guidelines which will help you massively when writing your own chord progressions. The 5 basic rules of Chord Progressions There are 5 basic rules to follow when writing a chord progression. Follow these and your chord progression will definitely “work”: - Choose a key to write in (if you are just starting out the C major, G major, A minor and E minor are good keys to start with) - Work out the primary chords (I, IV, V). Start to build your progressions with these. Then move on to using secondary chords (II, III, VI) to develop your chord progressions further. - Always start and end your chord progression on chord I - Try using some common progressions (see below) - Try adding some circle progressions (see below) The Common Chord Progressions There are some chord progressions which are used in lots and lots of different songs/pieces – they have been tried and tested in many different styles and will “work”. Learn these and you will be able to play lots of different songs and easily use them in your composing. They will work in both major and minor keys. Start with the following 4 progressions. (I have done an audio example in both a major key and then a minor key for you to listen to) The Circle Progressions Circle Chord Progressions are progressions where the chords seem to naturally follow on from one another. You will find the following 2 circle progressions really useful. Have a listen to the audio examples for each (again, each recording contains an example in a major key followed by an example in a minor key). Circle Progression 1 Circle Progression 2 Quick Reference Chord Progression Sheet I have put together this cheat sheet to give you the basic chord progressions in every key! Feel free to screenshot it/print it out for quick reference. You will see that some of the keys are really complex with lots of sharps and flats. If you are just starting out I suggest you choose one of the following keys to write in: Major – C, G, D Minor – Am, Em, Bm<\|endoftext\|>	3.6875
219	I first introduce the concept of "surface area" and we find the surface areas of rectangular prisms to get started. (This is prior to the actual lesson) I then ask students to predict the surface area of a sphere, and how one might find the surface area of a sphere. I then pass out an inexpensive kickball to each group, and newspaper. I tell the students to cut the several strips of paper into 1" by 4" strips. They are to then paste these strips onto the kickball, counting the number of strips used. By multiplying the number of strips by the area of each strip, the student can approximate the surface area of a sphere. When this is done, i then have them find the radius of the kickball by rolling the ball on a tiled floor and measuring the distance from the starting point to the ending point of one complete revolution. We then get into a discussion and lecture on diameter and radius, and we develop formulas for finding the surface area of shperes. The kids really love it, and they retain this information very well!<\|endoftext\|>	3.9375
1,369	Homework 8 altering product strength part A; potency key.pdf - Homework 8 – Altering product strength part A potency KEY 1 If a gallon of a 30 w/v # Homework 8 altering product strength part A; potency... • Homework Help • 13 This preview shows page 1 - 4 out of 13 pages. 1 Homework 8 – Altering product strength part A; potency - KEY 1. If a gallon of a 30% w/v solution is to be evaporated so that the solution will have a strength of 50% w/v, what will be its volume in milliliters? Answer Q1 x C1 = Q2 x C2 1 gallon = 3,785 mL 3,785 mL x 30% = V mL x 50% V mL = 3,785 mL x 30 mL x 100 mL = 2,271 mL, answer 100 mL 50 mL 2. How many milliliters of a 15% (w/v) ethanol solution should be diluted with water to yield 750 mL of a 1:50 solution? Answer Let’s write down all the information that we have in the form of the QC equation. The ratio strength will be w/v because the percentage is also w/v. V mL x 15% w/v = 750 mL x 1:50 w/v Let’s express the concentrations as fractions using their definitions V mL x 15 g = 750 mL x 1 g 100 mL 50 mL Let’s use algebra to solve this equation V mL = 750 mL x 1 g x 100 mL = 100 mL, answer 50 mL 15 g 3. How many milliliters of a diluent should be added to 25 mL of a 5% w/v drug solution in order to prepare a solution with a final concentration of 1g/100 mL? Answer Since this is a practice problem of dilution, let’s first write the QC equation 25 mL x 5% w/v = V mL x 1 g 100 mL Let’s express the percentage as a fraction 2 25 mL x 5 g = V mL x 1 g 100 mL 100 mL Let’s solve the equation using algebra V = 25 mL x 5 g = 125 mL 1 g Since the final volume is 125 mL and we start with 25 mL, volume of diluent is 125 mL – 25 mL = 100 mL, answer 4. How many milliliters of a 1:2,500 w/v solution of the phenylmercuric acetate can be made from 125mL of a 0.2% solution? Answer The qualifier of the percent will be w/v because this is the qualifier of the ratio strength Let’s write the information in the form of the QC equation 125 mL x 0.2% w/v = V mL x 1:2,500 w/v Let’s express concentrations as fractions 125 mL x 0.2 g = V mL x 1 g 100 mL 2,500 mL Let’s use algebra to solve the equation V mL = 125 mL x 0.2 g x 2,500 mL 100 mL 1 g V = 625 mL, answer 5. How many milliliters of a 50% w/v stock solution of Atropine are needed to prepare 250 mL of a solution containing 25 mg of drug per milliliter? Answer Let’s write the information in the form of the QC equation. V mL x 50% w/v = 250 mL x 25 mg/mL We will now express the percent as a fraction, while changing the mg to g 1 g = W g W = 0.025 g 3 1,000 mg 25 mg V mL x 50 g = 250 mL x 0.025 g 100 mL 1 mL Arranging the equation V = 250 mL x 0.025 g x 100 mL = 12.5 mL, answer 50 g 6. How many milliliters of a diluent should be added to 10 mL of a 5% w/v antibiotic solution in order to prepare a solution with a final concentration of 2.5 g/100 mL? [Please provide the answer in mL and not ml] Answer Since this is a practice problem of dilution, let’s first write the QC equation 10 mL x 5% w/v = V mL x 2.5 g 100 mL Let’s use the definition of w/v as gram in 100 mL, to re-write the equation to 10 mL x 5 g = V mL x 2.5 g 100 mL 100 mL V mL = 10 mL x 5 g = 20 mL 2.5 g Since the final volume is 20 mL and we start with 10 mL 20 mL – 10 mL = 10 mL We will need to add 10 mL of diluent, answer 7. How many milliliters of normal saline should be added to 75 mL of a solution containing 1,500 mg of atropine sulfate to make a 0.05% (w/v) solution? #### You've reached the end of your free preview. Want to read all 13 pages? • Fall '19 ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern Stuck? We have tutors online 24/7 who can help you get unstuck. Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire ) Answers in as fast as 15 minutes<\|endoftext\|>	4.53125
965	# What is the Golden Ratio? Published in -- You know you’re truly geeking out when you’re gushing about how beautiful a number is, but hey this number is pretty special ;) The Golden Ratio has been heralded as the most beautiful ratio in art and architecture for centuries. From the Parthenon to Salvador Dali’s The Sacrament of the Last Supper the Golden Ratio has been found lurking in some of the world’s most celebrated creations. Now whether you believe this divine proportion is truly a mark of beauty or simply selection bias is up to you, but without a doubt it is one of the most intriguing numbers in existence. # Phi is Golden Represented by the greek letter phi (φ), the Golden Ratio is the irrational value: # Euclid and the Golden Ratio In book 6 of The Elements, Euclid gives us the definition of the Golden Ratio. He instructs us to take a line segment and divide it into two smaller segments such that the ratio of the whole line segment (a+b) to segment a is the same as the ratio of segment a to segment b, like this: Or equivalently as a proportion: (You geometry buffs out there probably recognize that a is the geometric mean of a+b and b 🙌🏼 ) # The Golden Rectangle The Golden Ratio is most commonly represented as the Golden Rectangle, a rectangle with side-length ratio of 1.618:1. Golden Rectangles also have the property that if you cut off a square, you’ll be left with another Golden Rectangle. # Solving the Golden Proportion To find where the value 1.618034… comes from we must solve the proportion. For simplicity assume b=1 and a=x so that you may solve for x. ## Step 1 Take the cross products. ## Step 2 Subtract x+1 to set the equation equal to zero. We now have a standard quadratic with a=1, b=-1, and c=-1. ## Step 3 Plug these values into the quadratic formula and solve. Since we’re working with lengths, we need only the positive solution. And there it is! The Golden Ratio, as promised! For good measure, plug in a=1.618 and b=1 to confirm the proportion holds. ## Notice anything interesting about (1.618 + 1)/1.618 = 1.618? We can write the Golden Ratio in terms of itself! Which is totally awesome. Or equivalently, Now let’s get crazy. Substitute φ=1 + 1/φ for φ in the denominator. Whoa, that’s cool! Let’s do it again! We could keep on doing this forever. Which is pretty spectacular. Turns out the Golden Ratio can be written as an infinite continued fraction. # Finding Fibonacci We can use the continued fraction to approximate the Golden Ratio and uncover an interesting relationship with the Fibonacci Sequence. ## Step 1 To start out we’re going to alter our continued fraction a little. Instead of writing the formula nested in itself, we’ll add subscripts to indicate that the next value (φ_n+1) can be generated from the previous value (φ_n). Since this is an infinite continued fraction, as n increases, the approximation gets closer to the true value of φ. ## Step 2 Define φ_0 = 1. To find φ_1 plug in n=0. ## Step 3 Repeat the process to find φ_2 with n=1, since φ_2 = φ_1+1. Use the result from Step 2 for φ_1. ## Step 4 Keep on repeating this process. ## Step 5 Check it out. There’s the Fibonacci Sequence! Each approximation is the ratio of two adjacent Fibonacci numbers. We no longer need to go through the hassle of plugging values into the continued fraction, we can simply divide successive terms of the Fibonacci Sequence. As we move forward with each calculation, we find that our approximation of the Golden Ratio is getting closer and closer to its true value. In fact, the limit of the F(n+1)/F(n) as n → ∞ (where F(n) and F(n+1) represents the nth and nth plus 1 terms in the Fibonacci sequence) converges to φ. Visually, we can see how the Fibonacci Sequence generates rectangles closer and closer to the coveted Golden Rectangle. While the design world may be arguing over whether the Golden Ratio is folklore or not, I think it’s safe to say that the Golden Ratio is mathematically intriguing nonetheless. ## ❤ STAY CONNECTED ❤ Stay up-to-date with everything Math Hacks is up to!<\|endoftext\|>	4.75
572	According to Chinese legend, there was a female warrior named Hua Mulan who joined the army to fight in place of her father. Mulan was reputed to be a very brave woman who disguised herself as a man and fought in combat for 12 years. She has become an iconic heroine in Chinese and western cultures alike. The legend of Mulan is similar to several other female characters that dressed as men to fight in battle. These include Joan of Arc and the princess Eowyn from Lord of the Rings. The story of Mulan was first recounted in a poem called The Battle of Mulan. It was written somewhere around 500 – 600 A.D., before the Tang Dynasty was founded. A copy of the poem’s text was later recorded in a musical collection during the 12th century and has been subsequently passed down through popular culture to today. It is not known whether Mulan was a real or fictional character. Scholars have deliberated her existence for centuries, but no one has been able to determine if she actually lived. Nonetheless, her story has become a parable, as it sets forth many honored aspects of Chinese culture, such as filial piety (devotion to one’s elders), bravery and modesty (shown through Mulan’s character, when she declines rewards from the emperor in favor of returning home to her family). According to the story, the young Mulan sees that her father has been conscripted into the army. Having no older brothers, she decides to dress as a man and goes to war in his place. She spends more than 10 years fighting alongside other male soldiers. Later, the emperor offers her a government post, but she turns it down saying that she would prefer to return home and see her family. At the end of the poem, Mulan finally reveals her female identity to her army comrades, who are shocked to see her as a woman. They set off together as symbolic male and female rabbits, running side by side as equals. The original poem is composed of five-character phrases, though the lines do not always have equal numbers of syllables. Another feature of the ballad is the heavy use of onomatopoeia to describe specific sounds. For instance, the splish-splash of the Yellow River and the “jiu jiu” of the military horses neighing are shown through repetitious words that sound like the things they are describing. Over the centuries, the poem has been adapted in a variety of ways. During the Ming Dynasty, it was expanded into a novel. One Ming scholar also wrote a play based on the poem, entitling Mulan with the surname Hua. Most recently, Disney made an animated film, titled Mulan (1998) that was loosely based on the story of Hua Mulan.<\|endoftext\|>	3.765625
1,164	# Midpoint Theorem and Equal Intercept Theorem Do you know what midpoint theorem is? What is the purpose of using the midpoint theorem? Yes, it is used in solving geometry problems related to triangles. So let us study the midpoint theorem and equal intercept theorem in detail. ## Midpoint Theorem The line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is congruent to one half of the third side. Let us observe the figure given above. We can see the ΔABC, Point D and E are the midpoints of side AB and side AC respectively. Also, the segment DE connects the two sides at the midpoints, then DE \|\| BC and DE is half the length of side BC. If we know the length of BC then it is very convenient to find the length of DE as DE is half of BC. It also allows us to find the length of sides AE, EC, BD, and DA. Since DE is parallel to BC we know that the distance between these two line segment is equal. So, DE \|\| BC ### Proof of the Theorem Given: In triangle ABC, D and E are midpoints of AB and AC respectively. To Prove: 1. DE \|\| BC 2. DE = 1/2 BC Construction: Draw CR \|\| BA to meet DE produced at R. (Refer the above figure) ∠EAD = ∠ECR. (Pair of alternate angles) ———- (1) AE = EC. (∵ E is the mid-point of side AC) ———- (2) ∠AEP = ∠CQR (Vertically opposite angles) ———- (3) Thus, ΔADE ≅ ΔCRE (ASA Congruence rule) DE = 1/2 DR ———- (4) But, AD= BD. (∵ D is the mid-point of the side AB) Also. BD \|\| CR. (by construction) In quadrilateral BCRD, BD = CR and BD \|\| CR Therefore, quadrilateral BCRD is a parallelogram. BC \|\| DR or, BC \|\| DE Also, DR = BC (∵ BCRD is a parallelogram) ⇒ 1/2 DR = 1/2 BC ### The Converse of MidPoint Theorem The line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side. ## Equal Intercept Theorem The theorem states if a transversal makes equal intercepts on three or more parallel lines, then any other line cutting them will also make equal intercepts. It means that given any three mutually perpendicular lines, a line passing through them forms intercepts in the corresponding ratio of the distances between the lines. For example, Suppose there are three lines, l, m and n. Keep the distance between lm twice than the distance between mn. So any line passing through them, the intercept made by l-m on the line is twice the intercept made by m-n. ## Solved Examples Q1. In the adjoining figure, all measurements are indicated in centimetre. Find the length of AO if AX =9.5 cm. 1. 6.75 cm 2. 9.25 cm 3. 4.25 cm 4. 4.75 cm Solution: D. Given in the figure AD = DB = 4 AE =EC = 6 Then D and E are the midpoints of sides AB and AC respectively. Thus, DE bisects AX at point O ∴ AO = 1/2 AX = 1/2 × 9.5 = 4.75 Hence the length of AO = 4.75 Q2. In the given Δ ABC, Z is the mid-point of the median AD. If the area Δ ABC is 18 m², find the area of ΔBZC. 1. 7 m² 2. 9 m² 3. 6 m² 4. 5 m² Solution: B. In Δ ABC, Z is the mid-point of the median AD Join Z to B and C ∴ it will divide Δ ABC into two triangles of equal area. ∴ Area of ΔBZD = Area of ΔBZA A(ΔBZD) = 1/2 A(ΔABD) Also, A((ΔZDC) = A(ΔAZC) = A((ΔZDC) = 1/2 A(ΔAZC) A(ΔBZD) + A((ΔZDC) =1/2 A((ΔABD) + 1/2 A((ΔADC) A(ΔBZC) = 1/2 A(ΔABC) = 1/2 × 18 = 9m² Share with friends ## Customize your course in 30 seconds ##### Which class are you in? 5th 6th 7th 8th 9th 10th 11th 12th Get ready for all-new Live Classes! Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes. Ashhar Firdausi IIT Roorkee Biology Dr. Nazma Shaik VTU Chemistry Gaurav Tiwari APJAKTU Physics Get Started Subscribe Notify of<\|endoftext\|>	4.78125
390	Why does bullying continue? The following exercise will support your efforts to reduce bullying in your school and to fully engage students in the process. Your assignment is the following: First: Draft an agreement with your students. The goal of the agreement is to ensure the safety and wellbeing of ALL students in your class. SUGGESTIONS FOR ENCOURAGING STUDENTS TO SHARE THEIR OPINIONS: - Let students use their own words; try to resist the temptation to correct their grammar, their syntax, their style and their spelling errors. - Ask them why, in their opinion, it is important to have a definition of bullying, and discuss their answers in class. - Give students the opportunity to debate the overall question. It is likely that the subject of bullying will generate a great deal of discussion. - Launch the discussion in an open-ended way, without preconceived ideas (for example, avoid proposing your own or another definition as a starting point). - Try letting students work in small groups for this exercise. At the same time, it is important to remain sensitive to students who have been targeted by bullying and who may be left out or feel vulnerable in such a context. You can try as much as possible to establish trust and an open atmosphere with these students, without drawing attention to them. - Try doing this exercise individually as well. Ask each student to create their own definition. It is best not to draw more attention to some definitions over others, for example, by making comments such as “good answer”. Instead you can display them together so that students can read them all. Second: Encourage (but never force) the students to sign the agreement. Third: Discuss ways to disseminate the agreement (for example, by posting it in the classroom or hallways, online or in the school bulletin). The students may have other ideas.<\|endoftext\|>	4.21875
536	• 0 Guru # D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC. Q.5 • 0 What is the best way to solve the problem of exercise 6.4 of class 10th, this is very important question for class 10th D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC. Share 1. Given, D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. In ΔABC, F is the mid-point of AB (Already given) E is the mid-point of AC (Already given) So, by the mid-point theorem, we have, FE \|\| BC and FE = 1/2BC ⇒ FE \|\| BC and FE \|\| BD [BD = 1/2BC] Since, opposite sides of parallelogram are equal and parallel ∴ BDEF is parallelogram. Similarly, in ΔFBD and ΔDEF, we have FB = DE (Opposite sides of parallelogram BDEF) FD = FD (Common sides) BD = FE (Opposite sides of parallelogram BDEF) ∴ ΔFBD ≅ ΔDEF Similarly, we can prove that ΔAFE ≅ ΔDEF ΔEDC ≅ ΔDEF As we know, if triangles are congruent, then they are equal in area. So, Area(ΔFBD) = Area(ΔDEF) ……………………………(i) Area(ΔAFE) = Area(ΔDEF) ……………………………….(ii) and, Area(ΔEDC) = Area(ΔDEF) ………………………….(iii) Now, Area(ΔABC) = Area(ΔFBD) + Area(ΔDEF) + Area(ΔAFE) + Area(ΔEDC) ………(iv) Area(ΔABC) = Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) From equation (i)(ii) and (iii), ⇒ Area(ΔDEF) = (1/4)Area(ΔABC) ⇒ Area(ΔDEF)/Area(ΔABC) = 1/4 Hence, Area(ΔDEF): Area(ΔABC) = 1:4 • 0<\|endoftext\|>	4.75
995	# document.write (document.title) Math Help > Basic Math > Mental Math > Estimating It's important to know how to do math problems in your head, because that way you can double-check the results you get from your computer. I was astonished the other day when one of my kids did a problem on the calculator: find the square root of 80, got an answer of 28.28, and just wrote it down and went on to the next problem. "NO!" I hollered. "It must be less than NINE because the square root of 81 is NINE!" # Mental Math and Estimating Quick! Multiply 24 by 26! (The difference of two squares) What's the decimal expansion of 1/59? ## Quick! Multiply 24 by 26! (The difference of two squares) 24 × 26 = 25² - 1² 25² is 625, so 24 × 26 = 624. What if you didn't know 25² is 625? No problem: 25² = 20 × 30 + 5ï¿½ = 600 + 25 = 625 These are all applications of one simple fact: if you have two numbers, the difference of their squares is equal to the product of their sum and their difference. In algebraic terms, a²-b² = (a+b)(a-b) ## What's the decimal expansion of 1/59? 1/59 is nearly 1/60. Set the division out thus ``` 0.0 1 ... ------------------------ 6 ) 1.0 ... ``` Here we have the decimal for 1/59, obtained by dividing 1 by 60; as we obtain each digit we merely enter it in the dividend, one place later, and continue with the division. We "bring down" the digit we just appended to the dividend rather than bringing down a zero as we normally would. ``` 0.0 1 6 ... ------------------------ 6 ) 1.0 1 ... ``` Continuing in this way, ``` 0.0 1 6 9 4 9 1 5 2 5... ------------------------ 6 ) 1.0 1 6 9 4 9 1 5 2... ``` What we're really calculating when we do this is (1+1/59)/60, which is mathematically equivalent to 1/59. But this form has the distinct advantage that the outermost division operation is done with a simple one-digit divisor. The digits of the numerator (1+1/59) are supplied for us as we divide, so this new formulation of 1/59 really helps us divide quickly. This trick helps us divide quickly any number ending in nines, such as 1/59, as you have seen, or 5/299, etc. For the latter example, you would shift digits by two places as you append them to the dividend. ``` 0.0 1 6 7 2 2 4 0 8 0 2 6... -------------------------- 3 ) 5 0 1 6 7 2 2 4 0 8 0... ``` If you know the factors of numbers ending in nines, you can do even more problems. For example, to find 5/23 you need only find 15/69. A slight variation on this trick is to subtract (rather than append) the digits from the dividend. For example 5/23 is equal to 435/2001; and if we note that 435 is the same as 434.999999999..., we have another method, in which, as we obtain the digits, we subtract them from the dividend, so many places later. Thus in the present case ``` 217 391 304 347 ... ------------------------- 2 ) 434 782 608 695 652 ... ``` For example, 217 from 999 gives 782, which we then divide by 2, obtaining 391; this, subtracted from 999, gives 608; and so on. These clever methods of division are credited to Alexander Craig Aitken, who lived from 1895 to 1967. ### Related pages in this website Word Problems, including percents, and the meaning of the words "of", "per", and "what", and much, much more. Procedures The Expansions page gives methods for calculating values, such as Gauss' method for calculating pi. The webmaster and author of this Math Help site is Graeme McRae.<\|endoftext\|>	4.71875
547	Thursday , July 18 2019 # Matrices In Mathematics, matrices are arrays of numbers arranged in rows and columns. ### Null Matrix (0): Null Matrix is that matrix, that only contains number 0 in it. ### Diagonal Matrix: Also known as square matrix, in which all element zero except the diagonal upper left to lower left. ### Identity (or unit) Matrix (I): The elements in the diagonal are one’s only. ### Writing the order of Matrices: Order = Number of Rows * Number of Columns Example: Order = 32 Order = 2 3 ### Addition and Subtraction of Matrices: Matrices of the same order are added (or subtracted) by adding (or subtracting) the corresponding elements in each matrix. Adding A + B: Subtracting A – B: Rules: A+B = B+A : A and B can change position when adding. (A+B) + C = A + (B+C): order of operation bracket first. A – B ≠ B –A: A and B should not change positions when subtracting. Example: Example: ### Equal Matrix: If two matrices A and B are of the same order and their corresponding elements are equal, then A = B. Example: ### Multiplication of Two or More Matrices: Matrices can only be multiplied only if they are compatible. They are compatible when the number of rows of the second matrix is the same as the number of coloumns of the first matrix. Rules: AB ≠ BA : A and B should not change positions (AB)C = A(BC): If 3 or more matrices you can choose whichever 2 to multiply first. Example: ### Inverse Of Matrix: Determinant A = ad – bc A-1 = Remember: • If Determinant = 0 then the matrix has no inverse. • Multipying by the inverse of the inverse of a matrix gives the same result as dividing by the matrix. E.g. If AB = C A-1AB = A-1C B= A1C Example of Inverse of Matrix: • If Determinant = 0 then matrix has no inverse. • Multipying by the inverse of the inverse of a matrix gives the same result as dividing by the matrix. E.g. Example: ## Properties Of A Circle Circle Definitions: Geometrical Properties of Circle: If 2 chords in a circle area congruent, then … ### 2 comments 1. its good for studying it will make me pass 2. Really good. It is better than my school textbook.<\|endoftext\|>	4.75
218	The perimeter of an equilateral triangle is the length of the three sides of the triangle. Since they are all equal in an equilateral triangle, it is easy to find its area by knowing the … [Read more...] about Finding the Area of an Equilateral Triangle from its Perimeter An equilateral triangle is a special case of isosceles triangles, in which all three sides of the triangle are equal: Since it is a special case of isosceles triangles, all the properties of isosceles triangles apply to an equilateral triangle as well, so the base angles are equal. But since in this case, the third side is also the same, any side can serve as the base, and so we see that all angles must be the same. And since the sum of all angles is 180°, each angle of an equilateral triangle must be 180°/3=60°. Now that we've explained the basic concept of equilateral triangles in geometry, let's scroll down to work on specific geometry problems relating to this topic.<\|endoftext\|>	3.96875
1,719	$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 4.4: Vectors, Lists and Functions- $$\mathbb{R}^{S}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ Suppose you are going shopping. You might jot down something like this on a piece of paper: We could represent this information mathematically as a set, $$S=\{\rm apple, orange, onion, milk, carrot\}\, .$$ There is no information of ordering here and no information about how many carrots you will buy. This set by itself is not a vector; how would we add such sets to one another? If you were a more careful shopper your list might look like this: What you have really done here is assign a number to each element of the set $$S$$. In other words, the second list is a function $$f:S\longrightarrow {\mathbb R}\, .$$ Given two lists like the second one above, we could easily add them -- if you plan to buy 5 apples and I am buying 3 apples, together we will buy 8 apples! In fact, the second list is really a 5-vector in disguise. In general it is helpful to think of an $$n$$-vector as a function whose domain is the set $$\{1,\dots,n\}$$. This is equivalent to thinking of an $$n$$-vector as an ordered list of $$n$$ numbers. These two ideas give us two equivalent notions for the set of all $$n$$-vectors: $${\mathbb{R}}^{n} :=\left\{ \begin{pmatrix}a^{1} \\ \vdots \\ a^{n}\end{pmatrix} \middle\vert \, a^{1},\dots a^{n} \in \mathbb{R} \right\} =\{ a:\{1,\dots,n\}\to \mathbb{R}\} := \mathbb{R}^{ \{1,\cdots,n\} }$$ The notation $$\mathbb{R}^{ \{1,\cdots,n\} }$$ is used to denote functions from $$\{1,\dots,n\}$$ to $$\mathbb{R}$$. Similarly, for any set $$S$$ the notation $$\mathbb{R}^{S}$$ denotes the set of functions from $$S$$ to $$\mathbb{R}$$: $$\mathbb{R}^{S}:=\{ f:S\to \mathbb {R}\}\, .$$ When $$S$$ is an ordered set like $$\{1,\dots,n\}$$, it is natural to write the components in order. When the elements of $$S$$ do not have a natural ordering, doing so might cause confusion. Example 48 Consider the set $$S=\{, \star, \# \}$$ from chapter 1 review problem 9. A particular element of $$\mathbb{R}^{S}$$ is the function $$a$$ explicitly defined by $$a^{\star}=3, a^{\#}=5, a^{}=-2.$$ It is not natural to write $$a=\begin{pmatrix}3 \\ 5 \\ -2\end{pmatrix} ~{\rm or} ~a=\begin{pmatrix}-2\\ 3 \\ 5\end{pmatrix}$$ because the elements of $$S$$ do not have an ordering, since as sets $$\{, \star, \# \}=\{,\star,\#\}$$. In this important way, $$\mathbb{R}^{S}$$ seems different from $$\mathbb{R}^{3}$$. What is more evident are the similarities; since we can add two functions, we can add two elements of $$\mathbb{R}^{S}$$: Example 49 Addition in $$\mathbb{R}^{S}$$ If $$a^{\star}=3, a^{\#}=5, a^{}=-2$$ and $$b^{\star}=-2, b^{\#}=4, b^{}=13$$ then $$a+b$$ is the function $$(a+b)^{\star}=3-2=1, (a+b)^{\#}=5+4=9, (a+b)^{}=-2+13=11\, .$$ Also, since we can multiply functions by numbers, there is a notion of scalar multiplication on $$\mathbb{R}^{S}$$: Example 50 Scalar Multiplication in $$\mathbb{R}^{S}$$ If $$a^{\star}=3, a^{\#}=5, a^{}=-2$$, then $$3a$$ is the function $$(3a)^{\star}=3\cdot3=9, (3a)^{\#}=3\cdot5=15, (3a)^{*}=3(-2)=-6\, .$$ We visualize $$\mathbb{R}^{2}$$ and $$\mathbb{R}^{3}$$ in terms of axes. We have a more abstract picture of $$\mathbb{R}^{4}$$, $$\mathbb{R}^{5}$$ and $$\mathbb{R}^{n}$$ for larger $$n$$ while $$\mathbb{R}^{S}$$ seems even more abstract. However, when thought of as a simple "shopping list'', you can see that vectors in $$\mathbb{R}^{S}$$ in fact, can describe everyday objects. In chapter 5 we introduce the general definition of a vector space that unifies all these different notions of a vector.<\|endoftext\|>	4.59375
1,406	# finding equations for two lines through the origin • Apr 18th 2013, 07:27 AM ameerulislam finding equations for two lines through the origin Find equations for two lines through the origin that are tangent to the curve $\displaystyle x^2-4x+y^2+3=0$ ,, Not sure what the question is asking.. :( Any help? • Apr 18th 2013, 07:55 AM HallsofIvy Re: finding equations for two lines through the origin $\displaystyle (x- 2)^2= x^2- 4x+ 4$ so that $\displaystyle x^2- 4x+ y^2+ 3= x^2- 4x+ 4- 4+ y^2+ 3= (x-2)^2+ y^2- 1= 0$ so that $\displaystyle (x- 2)^2+ y^2= 1$. That is the equation of a circle with center (2, 0) and radius 1. The problem ask for you to find two equations through the origin (so y= mx for some m) that are tangent to the circle. It should be clear, geometrically, that one is above the x-axis and the other below. One way to do this is to argue that the line must be tangent to the circle and so replacing y by mx, [must](x- 2)^2+ (mx)^2= 1[/tex], must have a double root. That is a quadratic equation so must have zero, one, or two solutions. Since a tangent touches the circle, it cannot have 0 solutions. Since a tangent does not pass through the circle, it does not cross it twice and so that quadratic equation must have only one solution. You can set the discriminant of the equation equal to 0 to determine m. Another way to do this is to use the fact that a tangent to a circle is perpendicular to a radius. So the origin, (0, 0), the center of the circle, (2, 0), and the point, (x, y), where the line is tangent to the circle form a right triangle with the x-axis, from (0, 0) to (3, 0) as hypontenuse. That is, $\displaystyle x^2+y^2= 4+ 1= 5$ since the distance from (0, 0) to (2, 0) is 2 and the circle has radius 1. That, together with the equation of the circle gives you two equations to solve for x and y. Once you know the point, (x, y), where the line is tangent to the circle, as well as (0, 0) you can write down the equation of the line. • Apr 18th 2013, 08:06 AM Plato Re: finding equations for two lines through the origin Quote: Originally Posted by ameerulislam Find equations for two lines through the origin that are tangent to the curve $\displaystyle x^2-4x+y^2+3=0$ If $\displaystyle (a,b)$ is on that curve we know that $\displaystyle a^2-4a+b^2+3=0$. We know that the slope at each point is $\displaystyle y'=\frac{2-x}{y}$. If $\displaystyle a\ne 0$ then any line through $\displaystyle (0,0)~\&~(a,b)$ has slope $\displaystyle \frac{b}{a}$. Now, you want to find point $\displaystyle (a,b)$ on the curve that has slope $\displaystyle y'=\frac{2-a}{b}=\frac{b}{a}$ • Apr 18th 2013, 09:36 PM MINOANMAN Re: finding equations for two lines through the origin Ameer Problems like the one you posted yesterday can be easily solved if you consider the Polar line of the conic. the Polar line is the line that connects the two points of tangency. it has the same form as the equation of the tangent to a circle . In your case the conic is a circle of the form x^2+y^2+Ax +By +C = 0 where A=-4, B= 0 and C = 3. The point you are reffered to is the origin O(x1,y1)=(0,0). In this case the equation of the polar line to the conic has the form : xx1+yy1+A(x+x1)/2+B(y+y1)/2+C=0 and by substitution you get x = 3/2 since x1=y1=0. now solve the system of the conic and the polar line to find the coordinates of the points of tangency . the rest is grade 8 math.... I attach a figure for easy reference. MINOAS Attachment 28019 • Apr 30th 2013, 06:27 PM ameerulislam Re: finding equations for two lines through the origin Quote: Originally Posted by Plato If $\displaystyle (a,b)$ is on that curve we know that $\displaystyle a^2-4a+b^2+3=0$. We know that the slope at each point is $\displaystyle y'=\frac{2-x}{y}$. If $\displaystyle a\ne 0$ then any line through $\displaystyle (0,0)~\&~(a,b)$ has slope $\displaystyle \frac{b}{a}$. Now, you want to find point $\displaystyle (a,b)$ on the curve that has slope $\displaystyle y'=\frac{2-a}{b}=\frac{b}{a}$ now I got this equation $\displaystyle y^2=2x-x^2$ or $\displaystyle b^2=2a-a^2$ if I follow Plato. But I see in my book they went a step ahead, substituted $\displaystyle b^2$ with $\displaystyle 2a-a^2$ and came with the result $\displaystyle b=(\sqrt3/3)a$ and $\displaystyle b=-(\sqrt3/3)a$ Is that even necessary? I thought $\displaystyle b^2=2a-a^2$ is the equation and should be the answer isn't it? Or at bets I could remove the squares $\displaystyle b=\sqrt{2a-a^2}$<\|endoftext\|>	4.40625
1,204	Most cattle producers provide minerals and vitamin supplements during winter and spring, but some may not provide those supplements when pastures are green and growing. This practice might cut production costs in the short term, but it can prove extremely costly in the long run. Minerals and vitamins are a very small but extremely important element of cattle nutrition, playing vital roles in reproduction, immunity and growth. Cattle who do not receive proper mineral and vitamin nutrition will not grow or reproduce as quickly or efficiently as their well-supplemented counterparts. Minerals are loosely grouped into two categories: macro-minerals and trace or micro-minerals. Macro-minerals — including calcium, phosphorus, magnesium, potassium, sulfur and salt — are needed in relatively large amounts in the body. Micro-minerals — such as cobalt, copper, iodine, iron, manganese, selenium and zinc — are needed in very small, or “trace,” amounts. Vitamins A, D and E are among those that are typically supplemented. Details about the key nutrients that are more likely to be deficient in summer forages are included below. Phosphorus is vitally important for growth, milk production and fertility. Cattle on summer pasture are often at least marginally deficient in phosphorus. Cow requirements for spring-calving herds are much higher early in the grazing period through breeding, so supplemental phosphorus is critical during this period. Phosphorus supplementation should also continue after breeding, since forage phosphorus levels decrease steadily as forages mature. Common deficiency symptoms include breeding problems, such as reduced conception rates, and reduced average daily gains. Copper is critical for both fertility and immunity. Many U.S. soil types are marginally to severely deficient in copper, meaning most U.S. cattle require copper supplementation. Inadequate copper levels will result in decreased conception rates, early embryo deaths, decreased ability to respond to immune challenges and faded hair coats. Selenium is also important for reproduction and immunity. Selenium supplementation can help prevent retained placentas, uterine infections and white muscle disease. Most of the soils in the U.S. are marginal to deficient in selenium, so selenium supplementation is often necessary. Zinc plays a role in the maintenance of skin, hooves, gut linings and the lining of the reproductive organs. Deficiencies will result in decreased fertility, skin problems, hoof and joint problems, and decreased average daily gains due to decreased nutrient absorption. Providing these minerals to bulls throughout the year is important — especially during the summer breeding season. Vitamin A is typically abundant in green growing forages but is less plentiful in mature or drought-stricken forages. Cattle under stress (during weaning, lactation, transportation, etc.) have higher vitamin A requirements than normal and benefit from supplementation. Inadequate levels of vitamin A result in stunted growth, reproductive disorders, runny eyes and an increased susceptibility to diseases like pinkeye. While most cattle can survive on the levels of minerals and vitamins in the available forages, the vast majority of cattle are not receiving what is necessary for producing at high levels and improving efficiency. It is important to remember that the mineral content of forages is generally limited by the mineral make-up of the soils in which they grow; if it’s not in the soil, it can’t get into the plant. And, while soil types vary, no one soil type provides optimal levels of all the minerals cattle need. In fact, some soils are severely deficient in some minerals (e.g., selenium or copper) or have an overabundance of one mineral that interferes with the availability of another mineral (e.g., high sulfur levels interfere with copper and selenium uptake and utilization). For this reason, making free-choice mineral and vitamin supplementation available to cattle at all times is often recommended. If you feed salt or trace mineralized salt as a supplement during the summer, you may think that you are giving cattle what they need — but, while cattle do need salt, their nutritional needs are not necessarily met by salt blocks or trace mineralized salt blocks alone. Trace mineralized salt blocks are mostly salt (typically 92 to 98 percent) and contain relatively low levels of trace minerals. Because of their high salt content, consumption of these blocks will be very low, resulting in poor intake levels of the needed trace minerals. Additionally, these blocks do not contain the macro-minerals or vitamins cattle need. A complete mineral/vitamin supplement will provide both the necessary macro- and micro-minerals as well as the vitamins. Mineral and vitamin supplementation should be thought of as an insurance policy. Maintaining high-quality pastures for your cattle will help you meet the majority of their nutritional needs, but also providing free-choice access to a complete mineral and vitamin supplement will help ensure that ALL of your cattle’s nutritional needs are being met. SWEETLIX® offers a wide range of high-quality mineral and vitamin supplements to ensure that your cattle’s nutritional needs are met, not only in the summer, but year-round. Additionally, these minerals come in a variety of forms; convenient pressed blocks, poured blocks, low-moisture blocks or loose minerals are all available from SWEETLIX to best fit your management style and your cattle’s needs. Also, check out our new Blueprint® line of minerals, provided in a highly bioavailable form to meet the advanced demands of today’s modern genetics and to protect the environment from unnecessary over-supplementation through less bioavailable mineral forms. Ask for SWEETLIX by name at your local SWEETLIX dealer or call 1-87-SWEETLIX for more information.<\|endoftext\|>	3.71875
2,079	Dividing the world was just a simple matter of a Papal agreement at the end of the 15th century: After Columbus had returned from his first trip, the King of Spain asked the Vatican to rule over the division of the world between the two naval superpowers at the time, Spain and Portugal. Portugal already had control of the eastern routes, so the Pope draw an imaginary line “370 leagues west of Cape Verde”, splitting the current Brazil in half. Everything west of that line would belong to Spain, and everything east of that line would belong to Portugal. Easy as that. This Treaty of Tordesillas took it for granted that whatever would be found, those lands and riches would belong to these two countries: never mind the locals they might find along the way. All Vatican wanted in return was turning anyone they might find into faithful Christians. The main reason for this treaty was the access to spices from Malacca, what is currently part of Indonesia: gloves, cinnamon, nutmeg and mace were more valuable than gold, and finding a fast and reliable route to these riches was worth some serious investment. As Portugal had control over the eastern route, Spain wanted to find out a westerly route to Malacca and back. Thanks to the miscalculation made by Ptolemy and the maps based on his assumptions, the true size of the Earth was thought to be much smaller, and therefore, based on the distanced to Malacca as measured by the eastern routes, a western passage was seen as perfectly plausible alternative. Thus, wooing the kings and queens of the era to fund such an expedition was not very hard, as it potentially offered a way to vast riches, both for the funding royals and the crews participating in these voyages. It's quite interesting to see how both Columbus and Magellan had both been driven more by ambition than nationality: Columbus was Italian but got his financing from Spain, and Magellan was ex-Portuguese soldier who leveraged the success of Columbus and managed to get a fleet of five ships and a crew of over 260 men funded by Spain, even though he had never been a captain before. Magellan claimed to know of a passage that would allow crossing the South American continent: nobody knows whether this was a fact or just a way to get funding for an expedition, as finding anything valuable on these unchartered lands could still turn up to be worth the risk to all of the participants, and especially to Magellan. It is possible that he had heard references to what is the mouth of River Plata from someone who had sailed south from Rio de Janeiro: River Plata is wide and deep and could have been misinterpreted as a gateway across the continent. There are no known notes on why Magellan was so sure about the existence of such a passage, and the existing maps of the era did not go far enough South to include River Plata. After leaving Rio de Janeiro Magellan spent three months investigating every possible inlet, including River Plata, and found nothing. The armada was now in the chilly and stormy waters of Southern Argentina, and the expedition was taking much longer than anticipated. Moral was low, and when Magellan reduced the daily rations for the crew, a mutiny ensued: the other captains demanded that they return back to Spain immediately. With a couple of loyal supporters Magellan managed to keep enough control to crush the mutiny, and the ringleaders were killed and chopped to pieces as a warning to others. Majority of the crew ended up being pardoned, though, as Magellan still needed men to handle the ships, and his cruelty against mutineers had made it clear who was in charge. The expedition went to the shore waiting for the Southern winter to pass, but when they continued sailing, things started going wrong again: first Magellan lost one of the ships in a storm and the captain of another ship, San Antonio, turned around without permission and slipped back to Spain. The desertion of San Antonio was a big blow to Magellan, as that ship had most of the provisions for the whole fleet, but he was still determined to push forward. What gave him hope was the fact that the new inlet they had found had saltwater in it, indicating that it was not a mouth of a river but had some connection to ocean. After navigating the twisty, 500 km long waterway through Tierra del Fuego for about a month, the three remaining ships finally came out to open sea again: they had finally reached the Pacific Ocean, which actually got its name right there and then by Magellan, simply because it appeared to be much calmer than the Southern Atlantic that they had left behind. The connection between the Atlantic and the Pacific across what is now known as the Magellan Strait had been found, and it would become a major shipping route until the Panama Canal was opened in 1914. After some further sailing Magellan was certain that they were only days away from Malacca, but his assumption was based on the estimate made by Ptolemy, and it was soon obvious that this figure was incorrect: it took them over three months to finally reach Guam, and by the time they reached the Philippines, only 150 of the original crew were still alive, many of them ill or suffering from scurvy. Magellan was given a warm welcome by the local chief of the island of Cebu, Rajah Humabon, thanks in part to Enrique, a Malaccan slave who spoke the local language: Magellan had originally taken Enrique along for the trip exactly for the purpose of using him as an interpreter. This was the high point of the voyage: they were near to their destination, back on the known sea routes, and now had access to good, healthy food again. But the good times did not last—Humabon asked for Magellan's help to overthrow the ruler of a nearby island of Mactan, and he was happy to offer his help: after all, he could then convert yet another group into Christians, as had happened with Humabon's tribe. Unfortunately Magellan severely underestimated the enemy, ending up fighting a force of 1,500 men with a team of only 50: he was killed in the battle that ensued. Enrique, who had been promised his freedom as a payment for the participation, saw his future prospects dim after Magellan's death: the remaining crew did not want to heed Magellan's promise of freedom, as they still had to reach the Spice Islands to make the trip worthwhile and having an interpreter would make the interactions easier. It is likely that Enrique used his language skills and conspired with the locals, causing an ambush that resulted in the death of about 30 sailors and setting him free. There are no more written notes of Enrique at this point, so he presumably jumped the ship and thus became the first ever person to circumnavigate the globe. But the rest of the crew pushed on: as they did not have full crews for all three ships any more, they burned one of the ships and set sail with the two remaining ones to the Spice Islands, where they loaded the ships up to the hilt with spices. One of the ships eventually sank after being captured by the Portuguese, but the last remaining ship, Victoria, finally made it back to Spain under the command of Juan Elcano, one of the mutineers pardoned by Magellan. Out of the original crew of five ships and 260 sailors, only 18 made it back on board the last remaining ship. Naturally also the approximately 40-strong crew of San Antonio that had deserted Magellan just prior to crossing the Magellan Strait had survived. The captain of San Antonio had been jailed, but after the crew of Victoria described the multitude of hardships of the overall journey, the jailed captain was freed: his action was seen as being in the benefit of his crew. Amazingly enough, the value of their spice cargo of just a single boat was enough to cover the whole cost of this expedition, even making some profit on top. But it was not for the benefit of the crew: the crown of Spain confiscated the cargo as a repayment for the loss of the rest of the fleet. It had taken three years and formidable navigation skill to go around the world, and the man who got his name in history books ended up getting killed along the way, but his remaining crew had finally proven that there is a way to Spice Islands via the westerly route, it just happens to be very long and treacherous, and hence less worthwhile than was expected. But for centuries the Magellan Strait was the only available passage for transferring large amounts of goods between the east and west coasts of the USA, as the Transcontinental Railroad was completed only in 1869. Just like any navigators of today, Magellan and his crew were not using any “Flat Earth Navigation Rules” to finalize their circumnavigation: they knew the shape of the Earth, and that helped them to survive this arduous trip and find their way to the Spice Islands by using a route nobody had ever used before. The only error was Ptolemy's wrong estimate for the circumference of the Earth, which added 11,000 km of ocean crossing to their voyage After all twists and turns, Victoria sailed a total of 60,000 km. The final proof that the Earth is a a rotating globe came from the fact that the crew of Victoria realized that their meticulously kept daily log book was off by one day at the arrival back to Spain: they had circumnavigated the globe towards the west, opposite to the Earth's rotation and hence crossed the then non-existent International Date Line, losing one day in the process. Before Magellan's expedition, the assumption that the Earth is a globe was accepted as a fact, but it was still just a theory: Magellan's expedition proved it in practice. All this happened already 500 years ago, yet we still have people claiming that the Earth is flat. Replica of Magellan's only surviving ship, Victoria.<\|endoftext\|>	3.78125
272	The U.S. Drought Monitor (USDM) is a map that is updated each Thursday to show the location and intensity of drought across the country. The USDM uses a five-category system, labeled Abnormally Dry or D0, (a precursor to drought, not actually drought), and Moderate (D1), Severe (D2), Extreme (D3) and Exceptional (D4) Drought. Drought categories show experts' assessments of conditions related to dryness and drought including observations of how much water is available in streams, lakes, and soils compared to usual for the same time of year. U.S. Drought Monitor data go back to 2000. Where does this come from? Each week, drought experts consider how recent precipitation totals across the country compare to their long-term averages. They check variables including temperatures, soil moisture, water levels in streams and lakes, snow cover, and meltwater runoff. Experts also check whether areas are showing drought impacts such as water shortages and business interruptions. Based on dozens of indicators, experts make their best judgments of regional-scale drought conditions, and then check their assessments with experts in the field before publishing weekly drought maps. Associated statistics show what proportion of various geographic areas are in each category of dryness or drought, and how many people are affected.<\|endoftext\|>	3.78125
549	The Sierra Nevada Batholith includes Yosemite Park. It began to develop when subduction zones formed as an oceanic plate subducted beneath the eastern edge of the North American Plate. The oceanic plate began to melt as it descended into the crust and upper mantle. The molten rock (magma) formed hundreds of plutons above the area where the plate was melting. Plutons are teardrop shaped areas where the magma collected and cooled forming a large area of granite. Some of the magma worked its way upward and erupted onto the Earth's surface during volcanic eruptions. The plutons were deep underground when they cooled and hardened. Over a long period of time tectonic forces on the surface of the Earth created the Basin and Range province and forced the Sierra Nevada Batholith upward. The batholith includes all of the hundreds of plutons that formed in the area where the batholith is located. As the batholith was uplifted erosion increased exposing the granite in the plutons as the overlying layers of rock eroded away. The mountains are not a single piece of granite but instead are made up of individual rocks that formed as each pluton cooled. Most of the rocks that make up the batholith have not been uncovered by erosion yet. The Sierra Nevada Mountains are the largest mountain range in the contiguous United States. The mountains are part of the Sierra Nevada Batholith. The Sierra Nevada Mountains have many jagged snow covered peaks. The granite rocks in the upper part of the batholith began to weather as the overlying layers of rock eroded away. Cracks in the granite allowed rainwater to seep beneath the boulders surface. Expansion and contraction of the rainwater due to freezing and thawing caused pieces of rock to break off. This produced the jagged snow covered peaks found at the summit of the Sierra Nevada Mountains. Mount Whitney is the highest peak in the Sierra Nevada Mountains and the continental United States. Yosemite National Park lies within the Sierra Nevada Batholith. Granite lies underneath almost all of the 1,169 square miles of the park. Yosemite has many natural wonders including hanging valleys, waterfalls, U-shaped valleys, polished domes and cirque lakes. The geologic features all formed over eons of time as glaciers covered the area and retreated many times. The glaciers left behind the spectacular scenery found in the park including Yosemite Falls. Yosemite Falls is 2,245 feet high and is the tallest waterfall in North America. Check out Myrna Martin's award winning textbooks, e-books, videos and rock sets. The Ring of Fire Science Bookstore covers a wide range of earth science topics. Click here to browse.<\|endoftext\|>	4.375
729	## Saturday, July 28, 2012 ### CBSE Class 9 - Maths - Ch6 - Lines and Angles (Set-2) Octahedron Lines and Angles: Q & A Q1: Prove that two lines which are both parallel to the same line, are parallel to each other. Given: Three lines l, m, n in a plane such that l \|\| m and m \|\| n. To prove: l \|\| n Proof: Suppose line l is not parallel to line n. Then l, n will intersect at some unique point, say at P. ⇒ P lies on l but does not lie on m, since l \|\| m. ∴ Through point P outside m, there are two lines ( l and n ) and both are parallel to line m. This is not possible (violates Parallel Axiom). ∴ our assumption is wrong. Hence l \|\| n. Q2: Prove that, if P is a point which divides the line segment AB in the ratio m:n internally, then P is unique. Suppose point P divides segment AB in the ratio m:n internally. Let us assume P is not unique. ⇒ There is another point P1 which also divides the segment in ratio m:n internally. ∴ AP/PB = m:n and AP1/P1B = m/n ∴ nAP = mPB and nAP1 =mP1B nAP = m(AB - AP) and nAP1 = m( AB - AP1) ⇒ nAP = mAB - mAP and nAP1 = mAB - mAP1 ⇒ (m+n) AP = mAB and (m + n)AP1 = mAB ⇒ AP = mAB/(m+n) and AP1 = mAB/(m+n) ∴ AP = AP1 ∴ P and P1 are the same point, which contradicts our assumption. Hence P is unique. Q3: How many least number of distinct points determine a unique line? Q4: In how many point two distinct lines can intersect? Q5: In how many points two distinct planes can intersect? Q6: If B lies between A and C and AC = 8, BC = 3. What is AB and AB2 ? Answer: Since B lies between A and C, ∴ AB + BC = AC BC = 3 and AC = 8 AB = AC - BC = 8 - 3 = 5 AB2 = 5 ✕ 5 = 25 Q7: An angle is 14° more than its complement. Find the angle? Answer: Let the angle be p°. Its complementary angle = 90° - p° As per the given information, p° = (90° - p°) + 14° ⇒ p° + p° = 90° + 14° ⇒ 2p° = 104° ⇒ p° = 52° Q8: What angle is equal to its supplement? Let the angle be y°. ∴ The other angle is = 180 - y Since two angle are equal, ⇒ y = 180 - y ⇒ 2y = 180 ⇒ y = 90°<\|endoftext\|>	4.53125
1,425	Preparing for Black History Month, 2015 Black History Month is celebrated every February. The books below, some of the best published in 2014 and 2015, will help your students to understand the passions in the fight for rights and freedoms and to learn about the people that fought against discrimination in the United States. Click here to find a list of additional new books that are excellent resources for Black History Month. ELEMENTARY SCHOOL – PRIMARY Freedom’s School by Lesa Cline-Ransome (ill. by James E. Ransome). 9781423161035. 2015. Gr 1-4. After Lizzie’s family are freed from slavery, her parents must work extra hard in the field so Lizzie and her brother can go to school. They have only received scraps of learning before; now Lizzie loves her new teacher and school. But not everyone does…and one day everything comes crashing down. Will she ever get a chance to go to school again? Gordon Parks : How The Photographer Captured Black and White America by Carole Boston Weatherford (ill. by Jamey Christoph). 9780807530177. February 2015. Gr K-3. Gordon Parks, inspired by a magazine article on migrant workers, bought a used camera, taught himself how to use it, and became a photographer working for the government in Washington, D.C. Parks, appalled by the conditions under which African Americans lived, shot photos exposing the racism in the nation’s capital, including the iconic “American Gothic.” Last Stop on Market Street by Matt de la Pena (ill. by Christian Robinson). 9780399257742. 2015. Gr PK-1. As CJ and his grandma Nana ride the bus to a soup kitchen in the poorer section of town after church on Sunday, CJ asks his grandma why they don’t own a car, and why he doesn’t have an iPod like other boys. Grandma gives an encouraging answer for each question, showing CJ what’s really important, and the beauty that can be found all around them. ELEMENTARY SCHOOL – INTERMEDIATE The Case for Loving : The Fight for Interracial Marriage by Selina Alko (ill. by Sean Qualls). 9780545478533. 2015. Gr 1-4. Richard and Mildred traveled to Washington, D.C. to get married, but when they returned to their home in Virginia, they were arrested! Why? Because Richard was white and Mildred was part African-American, part Cherokee. They were thrown in jail until they agreed to move out of Virginia. Years later, they took their case to the U.S. Supreme Court. New Shoes by Susan Lynn Meyer (ill. by Eric Velasquez). 9780823425280. 2015. Gr 1-4. Ella Mae has always worn her cousin Charlotte’s hand-me-down shoes, but this year they are too small. Ella Mae is thrilled to go to the store to try on new shoes—but African Americans are not allowed to try on new shoes. Instead, the store owner uses a tracing of their foot to find a pair of shoes. Humiliated, Ellie Mae and Charlotte come up with an idea to fight back. Seeds of Freedom : The Peaceful Integration of Huntsville, Alabama by Hester Bass (ill. by E.B. Lewis). 9780763669195. 2015. Gr 1-4. The violence that characterizes the Civil Rights movement has not taken place in Huntsville, Alabama, but that does not mean that blacks are not discriminated against. Read the story of what African American residents did to help integration come to this southern city. Stella by Starlight by Sharon M. Draper. 9781442494978. 2015. Gr 4-6. When a burning cross set by the Klan causes panic and fear in 1932 Bumblebee, North Carolina, fifth-grader Stella tries to find the strength to demand change in her segregated town. Freedom Summer : The 1964 Struggle for Civil Rights in Mississippi by Susan Goldman Rubin. 9780823429202. 2014. Gr 5-8. In the summer of 1964, many young volunteers moved to Mississippi and stayed with local black hosts. Their goals: to open Freedom Schools and inform disenfranchised adults and children about the rights that are denied to them under the Jim Crow laws. The Girl from the Tar Paper School : Barbara Rose Johns and the Advent of the Civil Rights Movement by Teri Kanefield. 9781419707964. 2014. Gr 6-8. Angered by the huge differences between the schools for white children and her own black school—a tar-paper shack, really—Barbara organized a boycott of her school in 1951 in order to gain a new school building for herself and her fellow African American students. Sittin’ Up by Shelia P. Moses. 9780399257230. 2014. Gr 5-8. In the summer of 1940, Mr. Bro. Wiley, the last man around who had been a slave, died. Twelve-year-old Bean is very sad, for he had taken Mr. Bro. Wiley as his adopted grandpa. Bean and his best friend Pole are in for their first “sittin’ up” at the wake for the dead. Voices from the March on Washington by J. Patrick Lewis. 9781620917855. 2014. Gr 5-8. Former U.S. Children’s Poet Laureate J. Patrick Lewis presents a collection of 70 poems written from the point of view of participants in the 1963 civil rights march on Washington, D.C. March : Book 2 by John Lewis & Andrew Aydin (ill. by Nate Powell). 9781603094009. 2015. Gr 9-12. This graphic novel continues the story of Congressman John Lewis, who participated in the civil rights movement of the 1960s, including the march on Washington, D.C. Turning 15 on the Road to Freedom : My Story of the Selma Voting Rights March by Lynda Blackmon Lowery. 9780803741232. 2015. Gr 9-12. The author—the youngest marcher in the 1965 voting rights march from Selma to Montgomery, Alabama—remembers how she turned 15 during the march. She describes how she also participated in nonviolent demonstrations, was jailed numerous times, and worked with Martin Luther King, Jr. for the rights of African Americans. X by Ilyasah Shabazz. 9780763669676. February 2015. Gr 9-12. This book, written by Malcolm X’s daughter, is a fictionalized account of his childhood and coming of age.<\|endoftext\|>	3.90625
384	Place Value 6-Digits Part II Place Value: Learn Numbers, such as 495,786, have six digits. Each digit is a different place value. You already know about the last 3 digits: hundreds, tens, and ones. The first three follow a similar pattern. The first digit is called the hundred thousands' place. It tells you how many sets of one hundred thousand are in the number. The number 495,786 has four hundred thousands. The second digit is the ten thousands' place. It tells you that there are 9 ten thousands in addition to the four hundreds thousands. The third digit is the one thousands' place which is 5 in this example. Therefore, there are 5 sets of one thousand in the number. \|100 Thousands\|\|10 Thousands\|\|Thousands\|\|Hundreds\|\|Tens\|\|Ones\| All together, there are four sets of one hundred thousand, nine sets of ten thousand, five sets of one thousand, seven sets of one hundred, eight sets of ten, and four ones in the number 495,784. Expanded form shows the number expanded into an addition statement. The expanded form of 495,786 is 400,000 + 90,000 + 5,000 + 700 + 80 + 6. Place Value: Practice What is the place value for X in the number X? Press the Start Button To Begin This is 0 percent correct. \|Game Name\|\|Description\|\|Best Score\| \|How many correct answers can you get in 60 seconds?\|\|0\| Extra time is awarded for each correct answer. Play longer by getting more correct. \|How fast can you get 20 more correct answers than wrong answers?\|\|999\| Math Lessons by Grade - Estimation and Mental Math - Naming Numbers - Percents and Ratios - Place Value<\|endoftext\|>	4.09375
956	When you walk outside into sweltering heat or biting cold, your body responds by sweating or shivering to regulate body temperature. It starts with cells in your skin called thermoreceptor neurons, which sense the temperature of your environment and send that information to the brain for processing. But how does the brain process this information to initiate behavioral responses such as sweating, shivering, or pulling your hand from a hot pan? Two studies published together in Nature have mapped the brain’s representation of temperature in fruit flies, and the findings will provide more insights into how our own human bodies tick. Fruit flies are a great model for studying how the brain processes temperature. Because even a small temperature change can be deadly to these tiny insects (they can’t regulate their own body temperature internally), fruit flies need a quick and efficient system for sensing dangerous temperatures and escaping. Although humans have more options for responding to uncomfortable temperatures, the brain’s logic for representing temperature is likely similar in both species. Previously, fly researchers had found that fruit flies have “hot” and “cold” thermoreceptor neurons in their antenna, similar to the ones we have in our skin. The “hot cells” are activated by heat while the “cold cells” are activated by cold. Interestingly, each cell type is also inhibited by the other temperature—heat makes it harder for cold cells to be activated, and vice versa. These thermoreceptor neurons are the first step for temperature sensing in the fruit fly, but what’s the next step? How does this temperature information get processed in the brain? In their recently published papers, Frank et al. and Liu et al. used different approaches to answer these questions. Both groups independently discovered neurons that receive information from the hot and cold cells in the antenna and carry it to the brain. These cells, called projection neurons (PNs), can be separated into three main groups: cold-PNs, which are activated by the cold thermoreceptor neurons; hot-PNs activated by the hot thermoreceptor neurons; and mixed-PNs, which receive information from both types of thermoreceptor neurons and are activated by rapid temperature changes in both directions. Frank and colleagues found that the mixed-PNs were important for fruit flies to recognize and quickly escape from dangerous temperature environments. These results suggest that the fruit flies’ temperature-sensing system is relatively straightforward: hot and cold information each has its own pathway to the brain. In a small group of PNs, this information also overlaps to provide the brain with a quick escape warning, regardless of whether the dangerous temperature is too high or low. But Liu and colleagues obtained more results that remind us that brains are never as simple as we expect. The researchers found that hot-PNs not only receive input from the hot pathway, but they are also affected by the cold pathway. Under cool conditions, the hot-PNs are suppressed by the cold pathway and unable to activate. As the environment gets warmer, however, hot-PNs are activated by the hot thermoreceptor neurons and also released from the cold pathway’s inhibitory influence. Florence and Reiser worded it best in their Nature review on these studies: Liu et al. revealed that hot projection neurons exploit both the excitatory ‘getting hotter’ signal from the hot receptors and the inhibitory ‘getting less cold’ signal from the cold receptors, and suggest that this not-quite-redundant use of both pathways leads to more-sensitive measurements of temperature change. So what happens to the temperature information after the PNs carry it to the brain? Both groups found that the PNs connect with brain regions important for learned and instinctual behaviors. The next research step will therefore be to uncover the processes by which these brain regions initiate behaviors to protect the fly—whether it be a simple quick escape, or storing a memory of a dangerous place. - Liu W.W. & Rachel I. Wilson (2015). Thermosensory processing in the Drosophila brain, Nature, DOI: http://dx.doi.org/10.1038/nature14170 - Frank D.D., Patrick J. Kearney, Lindsey J. Macpherson & Marco Gallio (2015). Temperature representation in the Drosophila brain, Nature, DOI: http://dx.doi.org/10.1038/nature14284 - Florence T. (2015). Neuroscience: Hot on the trail of temperature processing, Nature, DOI: http://dx.doi.org/10.1038/nature14209<\|endoftext\|>	4.1875
2,522	Bivalve, (class Bivalvia), any of more than 15,000 species of clams, oysters, mussels, scallops, and other members of the phylum Mollusca characterized by a shell that is divided from front to back into left and right valves. The valves are connected to one another at a hinge. Primitive bivalves ingest sediment; however, in most species the respiratory gills have become modified into organs of filtration called ctenidia. In keeping with a largely sedentary and deposit-feeding or suspension-feeding lifestyle, bivalves have lost the head and the radular rasping organ typical of most mollusks. Size range and diversity of structure Bivalves range in size from about one millimetre (0.04 inch) in length to the giant clam of South Pacific coral reefs, Tridacna gigas, which may be more than 137 centimetres (54 inches) in length and weigh 264 kilograms (582 pounds). Such an animal may have a life span of about 40 years. The shell morphology and hinge structure are used in classification. In most surface-burrowing species (the hypothetical ancestral habit) the shells are small, spherical or oval, with equal left and right valves. In deeper-burrowing species the shells are laterally compressed, permitting more rapid movement through the sediments. The shells of the most efficient burrowers, the razor clams Ensis and Solen, are laterally compressed, smooth, and elongated. Surface-burrowing species may have an external shell sculpture of radial ribs and concentric lines, with projections that strengthen the shell against predators and damage. A triangular form, ventral flattening, and secure attachment to firm substrates by byssal threads (byssus; proteinaceous threads secreted by a gland on the foot) have allowed certain bivalves to colonize hard surfaces on wave-swept shores. The byssus is a larval feature that is retained by adults of some bivalve groups, such as the true mussels (family Mytilidae) of marine and estuarine shores and the family Dreissenidae of fresh and estuarine waters. Such a shell form and habit evolved first within sediments (endobyssate), where the byssus serves for anchorage and protection when formed into an enclosing nest. Other bivalves have used the byssus to attach securely within crevices and thus to assume a laterally flattened, circular shape. The best example of this is the windowpane shell Placuna. This form has allowed the close attachment of one valve to a hard surface, and although some groups still retain byssal attachment (family Anomiidae), others have forsaken this for cementation, as in the true oysters (family Ostreidae), where the left valve is cemented to estuarine hard surfaces. Some scallops (family Pectinidae) are also cemented, but others lie on soft sediments in coastal waters and at abyssal depths. By limiting shell thickness (which reduces weight), smoothing the shell contours (which reduces drag), and assuming an aerofoil-like leading edge, such scallops can awkwardly swim several metres at a time. In other species, such as the clams, the foot has become modified for rapid and effective digging, and the folds of the mantle tissue have developed into long siphons. Both these features allow the animals to burrow deeply within sand, mud, and other substrates (even into wood and rock). They are protected from predators within such substrata but are still able to feed and breathe using their long siphons. Bivalve shell and body form is thus intimately related to habitat and the relative degree of exposure to predation. From the simple burrowing, equivalve ancestor, the various bivalve groups have repeatedly evolved an elongated, triangular or circular shell; thus, similar body adaptations have been responses to similar modes of life. Distribution and abundance Most bivalves are marine and occur at all depths in or upon virtually all substrates. In shallow seas, bivalves are often dominant on rocky and sandy coasts and are also important in offshore sediments. They occur at abyssal and hadal depths, either burrowing or surface-dwelling, and are important elements of the midoceanic rift fauna. In addition, bivalves bore into soft shales and compacted muds but may be important also in the bioerosion of corals. Bivalves thus occur at all latitudes and depths, although none are planktonic. There are also estuarine bivalves, and two important families, the Unionidae and Corbiculidae, are predominantly freshwater with complicated reproductive cycles. There are no terrestrial bivalves, although some high-intertidal and freshwater species can withstand drought conditions. To be expected within a class comprising more than 15,000 living species, abundance varies considerably. Commensal and parasitic species are small, often highly host-specific, and comprise some of the rarest animals. Others, such as cockles and clams on soft shores and mussels and oysters on rocky coasts, can occur in densities high enough that they dominate entire habitats and assume important roles in nutrient cycles. The total marine catch of mollusks is twice that of crustaceans, and the great majority of this is bivalve. Some three million metric tons (6,615,000,000 pounds) of bivalves are harvested throughout the world each year. Virtually all bivalves, with the possible exception of the thorny oyster Spondylus, are edible and fall into the main categories of oysters, mussels, scallops, and clams. A number of species are raised commercially. The most important edible oysters are representatives of the genus Crassostrea, notably C. gigas in the western Pacific, C. virginica in North America, and C. angulata in Portugal. Most mussels are cultivated on ropes suspended from floats. The European mussel Mytilus edulis has been introduced into the northern Pacific, and the practice now flourishes widely in Japan and China. Most scallops, Pecten, Placopecten, and Amusium, are caught by offshore trawlers, although cultivation is being attempted. A wide variety of clams are cultivated—e.g., Mya arenaria and Mercenaria mercenaria in the North Atlantic and Venerupis japonica and Tapes philippinarum in the Pacific. In some parts of the world, red tides, caused by large numbers of toxic protozoan dinoflagellates, are lethal to fish and certain invertebrates. Bivalves, by virtue of their filter-feeding apparatus, concentrate the toxin and, if eaten by humans, can cause paralysis or death. Bivalves of the genera Pinctada and Pteria have been collected in many tropical seas for the natural pearls they may contain, although in many countries, most notably Japan, pearl oyster fisheries have been developed. The outer shell of the windowpane oyster, Placuna placenta, is called the capiz shell. It is used, primarily in the Philippines, in the manufacture of lampshades, trays, mats, and bowls. In developing countries, many kinds of bivalve shells are used in the manufacture of jewelry and ornaments. Bivalves are important agents in bioerosion, most notably of calcium carbonate rocks and wood in the sea. Piddocks (family Pholadidae) bore into concrete jetties (particularly where the source of obtained lime is coral), timber, and plastics. Shipworms (family Teredinidae) bore softer woods. Date mussels (Lithophaga) bore into rocks and corals. Marine mussels (family Mytilidae) foul ships, buoys, and wharves; they may also block seawater intakes into the cooling systems of power stations. The freshwater zebra mussel (family Dreissenidae) feeds on phytoplankton and proliferates rapidly, clogging water-intake pipes and damaging boats and bridges. A problem in Europe from the 19th century, the zebra mussel arrived in North America, probably in the ballast water of ships, in 1986. It upset the food web of the Great Lakes and threatened many native bivalve species with extinction. Causing millions of dollars in economic losses each year, zebra mussels clog the water intake systems of power plants and industrial cooling systems. Few bivalves are host to human parasitic infections. Industrial and agricultural effluents—notably trace metals, chlorophenothane (DDT), and chlorinated hydrocarbons—have contaminated bivalves, with subsequent concern over human health. Reproduction and life cycles Although most bivalve species are gonochoristic (that is, they are separated into either male or female members) and some species are hermaphroditic (they produce both sperm and eggs), sexual dimorphism is rare. In gonochoristic species there is usually an equal division of the sexes. Simultaneous hermaphroditism occurs when sperm-producing tubules and egg-producing follicles intermingle in the gonads (as in the family Tridacnidae), or the gonads may be developed into a separate ovary and testis, as in all representatives of the subclass Anomalodesmata. In consecutive hermaphroditism, one sex develops first. Typically, this is the male phase (protandry), but in a few cases it is the female (protogyny). This is most clearly seen in the wood-boring family Teredinidae, where young males become females as they age. Rhythmical consecutive hermaphroditism is best known in the European oyster, Ostrea edulis, in which each individual undergoes periodic changes of sex. Alternative hermaphroditism is characteristic of oysters of the genus Crassostrea, in which most young individuals are male. Later the sex ratio becomes about equal, and finally most older individuals become female. Bivalve sperm have two flagellae. Most eggs are small, and synchronized spawning results in the discharge of both types of gametes into the sea for external fertilization. Hermaphrodites usually bring in sperm from another individual through the incurrent siphon. The embryos are then brooded, and brooding typically occurs within the ctenidia. There the fertilized eggs, well endowed with yolk, develop directly (without a larval stage), and the young are released as miniature adults. Although ctenidial incubation is most common, there are other patterns: egg capsules are produced by Turtonia minuta; a brood chamber is plastered to the shell of the palaeotaxodont Nucula delphinodonta; and in members of the Carditidae the female shell is modified into a brood pouch. For most marine species, however, the fertilized egg undergoes indirect development first into a swimming trochophore larva and then into a shelled veliger larva. The veliger has a ciliated velum for swimming and also for trapping minute particles of food. Following a period in the plankton, which varies from hours in some species to months in others, the veliger descends to the seafloor, where it metamorphoses into the adult form: the velum is lost, the foot develops and usually secretes one or two byssal threads for secure attachment, and the ctenidia develop. In the freshwater Unionidae the released larva, called a glochidium, often has sharp spines projecting inward from each valve. The larva attaches to either the gills or fins of passing fish and becomes a temporary parasite. Eventually, it leaves the fish, falls to the lake floor, and metamorphoses into an adult.<\|endoftext\|>	3.875
635	# 7.13. Search Tree Analysis¶ With the implementation of a binary search tree now complete, we will do a quick analysis of the methods we have implemented. Let’s first look at the put method. The limiting factor on its performance is the height of the binary tree. Recall from the vocabulary section that the height of a tree is the number of edges between the root and the deepest leaf node. The height is the limiting factor because when we are searching for the appropriate place to insert a node into the tree, we will need to do at most one comparison at each level of the tree. What is the height of a binary tree likely to be? The answer to this question depends on how the keys are added to the tree. If the keys are added in a random order, the height of the tree is going to be around $$\log_2{n}$$ where $$n$$ is the number of nodes in the tree. This is because if the keys are randomly distributed, about half of them will be less than the root and half will be greater than the root. Remember that in a binary tree there is one node at the root, two nodes in the next level, and four at the next. The number of nodes at any particular level is $$2^d$$ where $$d$$ is the depth of the level. The total number of nodes in a perfectly balanced binary tree is $$2^{h+1}-1$$, where $$h$$ represents the height of the tree. A perfectly balanced tree has the same number of nodes in the left subtree as the right subtree. In a balanced binary tree, the worst-case performance of put is $$O(\log_2{n})$$, where $$n$$ is the number of nodes in the tree. Notice that this is the inverse relationship to the calculation in the previous paragraph. So $$\log_2{n}$$ gives us the height of the tree, and represents the maximum number of comparisons that put will need to do as it searches for the proper place to insert a new node. Unfortunately it is possible to construct a search tree that has height $$n$$ simply by inserting the keys in sorted order! An example of such a tree is shown in Figure 6. In this case the performance of the put method is $$O(n)$$. Figure 6: A skewed binary search tree would give poor performance Now that you understand that the performance of the put method is limited by the height of the tree, you can probably guess that other methods, get, in, and del, are limited as well. Since get searches the tree to find the key, in the worst case the tree is searched all the way to the bottom and no key is found. At first glance del might seem more complicated, since it may need to search for the successor before the deletion operation can complete. But remember that the worst-case scenario to find the successor is also just the height of the tree which means that you would simply double the work. Since doubling is a constant factor it does not change worst case Next Section - 7.14. Balanced Binary Search Trees<\|endoftext\|>	4.53125
734	What is wastewater? You may not realize what happens after you wash something down the drain or flush the toilet. That sewage or wastewater is conveyed in an extensive underground collection system comprised of piping from your house, your place of business and any other place plumbed to the sanitary sewer. The wastewater collection system (totally separate from the storm water collection system that collects storm run-off from streets and parking lots) collects the wastewater and carries it to the wastewater treatment plant. Sewage is the wastewater released by residences, businesses and industries in the community. It is 99.94% water and only 0.06% solid material. Sewage contains nutrients (such as ammonia and phosphorus), minerals and metals. Sewage treatment is a multi-stage process to treat wastewater before it reenters a body of water, is applied to the land or is reused. The goal is to reduce or remove organic matter, solids, nutrients, disease-causing organisms and other pollutants from wastewater. Preliminary treatment to screen out, grind up, or separate debris is the first step in wastewater treatment. Sticks, rags, large food particles, sand, gravel, toys, etc., are removed at this stage to protect the pumping and other equipment in the treatment plant. The collected debris is usually disposed of in a landfill. Primary treatment is the second step in treatment and separates suspended solids and greases from wastewater. Wastewater is held in a quiet tank for several hours allowing the particles to settle to the bottom and the greases to float to the top. The solids drawn off the bottom and skimmed off the top receive further treatment as sludge. The clarified wastewater flows on to the next stage of wastewater treatment. Secondary treatment is a biological treatment process to remove organic matter from wastewater. Microorganisms are cultivated and added to the wastewater. The microorganisms absorb organic matter from sewage as their food supply. Final treatment focuses on removal of disease-causing organisms from wastewater. Treated wastewater can be disinfected by adding chlorine or by using ultraviolet light. High levels of chlorine may be harmful to aquatic life in receiving streams. Treatment systems often add a chlorine-neutralizing chemical to the treated wastewater before stream discharge. Advanced treatment is necessary in some treatment systems to remove nutrients from wastewater. Chemicals are sometimes added during the treatment process to help settle out or strip out phosphorus or nitrogen. Sludges are generated through the sewage treatment process. The goals of sludge treatment are to stabilize the sludge and reduce odors, to remove water to reduce volume, to decompose some of the organic matter to reduce volume, and to disinfect the sludge to kill disease causing organisms. Depending on the contaminant level in the treated sludge, it is either disposed of in a landfill or used as fertilizer for non-food grade crops. Wastewater treatment processes require careful management to ensure the protection of the water body that receives the discharge. Trained and certified treatment plant operators measure and monitor the incoming sewage, the treatment process and the final effluent. The City of Grand Prairie is part of the Central Regional Wastewater System run by the Trinity River Authority (TRA). All of the wastewater generated in the City runs into the TRA’s collection system to the TRA Central Treatment Plant located at 6500 W. Singleton Blvd. on the edge of Grand Prairie at I-30 and Loop 12. The treatment plant takes in wastewater, treats it, and discharges it to the Trinity River. \|TRA Central Regional Wastewater \|Treated Wastewater Leaving \|<\|endoftext\|>	3.78125
860	# How to calculate 37 divided by 42 using long division? 37 ÷ 42 = 0.88095238095238 Division is a fundamental arithmetic operation where we calculate how many times a number (divisor or denominator) can fit into another number (dividend or numerator). In this case, we are dividing 37 (the dividend) by 42 (the divisor). There are three distinct methods to convey the same information: in decimal, fractional, and percentage formats: • 37 divided by 42 in decimal = 0.88095238095238 • 37 divided by 42 in fraction = 37/42 • 37 divided by 42 in percentage = 88.095238095238% ## What is the Quotient and Remainder of 37 divided by 42? The quotient is calculated by dividing the dividend by the divisor, and the remainder is what's left over if the division doesn't result in a whole number. The quotient of 37 divided by 42 is 0, and the remainder is 37. Thus, ### 37 ÷ 42 = 0 R 37 When you divide Thirty Seven by Forty Two, the quotient is Zero, and the remainder is Thirty Seven. ## Verdict The division of 37 by 42 results in a quotient of 0 and a remainder of 37, meaning 42 goes into 37 Zero time with 37 left over. Understanding this division process is crucial in both basic arithmetic and real-life applications where division is used, such as in financial calculations, data analysis, and everyday problem-solving. ## Random Division Problems? No worries, we got your back! Tell us what are you brainstorming with and we will bring correct answers to you. Start Now ### How do we differentiate between divisor and dividend? A dividend is a number we divide, while a divisor is a number by which we divide. Divisor comes on second, followed by the dividend that we write first. For instance, if you have 12 candies and want to distribute them among 3 children, the equation will be 12 ÷ 3. You will put 12 first because this is the number being divided. So here, 12 is a dividend. On the other hand, 3 is written after 12, and it is the number with which we are dividing 12. Hence, 3 is a divisor. ### Which formula is used to find a divisor? There are two formulas used to find a divisor. The first one is: Divisor = Dividend ÷ Quotient. This formula is used to find a divisor when the remainder is 0. Second is: Divisor = (Dividend – Remainder) /Quotient. This formula is used when the remainder is not 0. ### Is there a possibility of a number having the same divisor? Yes, there is. Every number can be divided by itself, leaving 1 as the quotient. So, it would not be wrong to say that all the numbers can have the same divisors. Let’s take the example of 5. If we divide 5 by 5 (5 ÷ 5), then 5 will be the divisor of 5. And ultimately, 1 will be the quotient. ### What is the difference between a divisor and a factor? A divisor is a number with which we can divide any number. However, a factor is different from a divisor. It is the number that can be divided with another number leaving no remainder. All factors are divisors, but not all divisors are factors. ### Is it possible to do division by repeated subtraction? Fortunately yes. You can do division by repeated subtraction. In repeated subtraction, we continuously subtract a number from a bigger number. It continues until we get the 0 or any other number less than the actual number as a remainder. However, it can be a lengthy process, so we can use division as a shortcut. ### Can I check the remainder and the quotient in a division problem? Yes, you can quickly check the remainder and quotient in a division problem by using this relationship: Dividend = Divisor x Quotient + Remainder<\|endoftext\|>	4.71875
212	Rivers were at this time seen as the ideal means of transporting goods and were increasingly engineered by dredging, to accommodate larger loads. The Swan River was remade for transportation purposes by removing sandbars, filling and dredging channels. The first change made to the natural environment after European settlement was in 1831 when a canal was cut making Burswood into an island. (Camfield Henry, Correspondence and legal documents 1829-1865, Acc 1459A). In 1833 a tender was released for the construction of a dyke to block channels between islands with: "duble stake and a wattled fence filled in with clay, three feet above low water and three feet high." When the first canal was relatively unsuccessful, a second canal was cut in 1834. In addition the River was straightened in an effort to correct what were seen as nature’s shortcomings. The arguments used to dredge and straighten the river were utilitarian and health based. Page last updated: Tuesday 23 November 2010<\|endoftext\|>	3.78125
376	A popular implementation of public-key encryption is the Secure Sockets Layer (SSL). Originally developed by Netscape, SSL is an Internet security protocol used by Internet browsers and Web servers to transmit sensitive information. SSL has become part of an overall security protocol known as Transport Layer Security (TLS). In your browser, you can tell when you are using a secure protocol, such as TLS, in a couple of different ways. You will notice that the "http" in the address line is replaced with "https," and you should see a small padlock in the status bar at the bottom of the browser window. When you're accessing sensitive information, such as an online bank account or a payment transfer service like PayPal or Google Checkout, chances are you'll see this type of format change and know your information will most likely pass along securely. TLS and its predecessor SSL make significant use of certificate authorities. Once your browser requests a secure page and adds the "s" onto "http," the browser sends out the public key and the certificate, checking three things: 1) that the certificate comes from a trusted party; 2) that the certificate is currently valid; and 3) that the certificate has a relationship with the site from which it's coming. The browser then uses the public key to encrypt a randomly selected symmetric key. Public-key encryption takes a lot of computing, so most systems use a combination of public-key and symmetric key encryption. When two computers initiate a secure session, one computer creates a symmetric key and sends it to the other computer using public-key encryption. The two computers can then communicate using symmetric-key encryption. Once the session is finished, each computer discards the symmetric key used for that session. Any additional sessions require that a new symmetric key be created, and the process is repeated.<\|endoftext\|>	4.21875
790	Two 2017 NASA missions set to study edge of space Far above Earth’s tenuous upper atmosphere is a layer of charged particles that have been split into positive and negative ions by the Sun’s harsh ultraviolet radiation. This area where the Earth’s atmosphere and terrestrial weather give way to the space environment is called the ionosphere. In 2017, NASA plans to launch two satellite missions to study this region: the Ionospheric Connection Explorer (ICON) and the Global Observations of the Limb and Disk (GOLD). “The ionosphere doesn’t only react to energy input by solar storms,” said Scott England, a space scientist at the University of California, Berkeley, who works on both the ICON and GOLD missions. “Terrestrial weather, like hurricanes and wind patterns, can shape the atmosphere and ionosphere, changing how they react to space weather.” ICON will simultaneously measure the characteristics of charged particles in the ionosphere and neutral particles in the atmosphere to understand how they interact. GOLD will make many of the same measurements but from geostationary orbit, which will provide a global view of changes in the ionosphere. Both missions will use a phenomenon called airglow, the light emitted by gas that is excited or ionized by solar radiation, to study the ionosphere. By measuring the light from airglow, researchers can track the changing composition, density, and temperature of particles in the ionosphere and neutral atmosphere. ICON’s orbit 350 miles (about 563 kilometers) above Earth should allow the satellite to study the atmosphere in profile, giving researchers an unprecedented view of the state of the ionosphere at a range of altitudes. GOLD’s position 22,000 miles (approximately 35,406 kilometers) above Earth should enable it to track changes in the ionosphere as they move across the globe, the same way a weather satellite tracks a storm. “We will be using these two missions together to understand how dynamic weather systems are reflected in the upper atmosphere, and how these changes impact the ionosphere,” said England. England discussed the ICON and Gold missions, both Explorer-class missions managed out of NASA’s Goddard Space Flight Center in Maryland, during the recent fall meeting of the American Geophysical Union (AGU) in San Francisco. UC Berkeley’s Space Sciences Laboratory is developing the ICON mission as well as the two ultraviolet imaging spectrographs. Meanwhile, the Naval Research Laboratory in Washington, D.C., will develop the MIGHTI instrument, and the University of Texas at Dallas will develop the Ion Velocity Meter. The ICON spacecraft itself is being built by Dulles, Virginia-based Orbital ATK. The GOLD mission is led by the University of Central Florida, and the Laboratory for Atmospheric and Space Physics at the University of Colorado Boulder is building the instrument. ICON is scheduled to launch in June of 2017 from the Reagan Test site on Kwajalein Atoll in the Republic of the Marshall Islands aboard a Pegasus XL launch vehicle from Orbital ATK’s “Stargazer” L-1011 aircraft. Video courtesy of NASA Jim Sharkey is a lab assistant, writer and general science enthusiast who grew up in Enid, Oklahoma, the hometown of Skylab and Shuttle astronaut Owen K. Garriott. As a young Star Trek fan he participated in the letter-writing campaign which resulted in the space shuttle prototype being named Enterprise. While his academic studies have ranged from psychology and archaeology to biology, he has never lost his passion for space exploration. Jim began blogging about science, science fiction and futurism in 2004. Jim resides in the San Francisco Bay area and has attended NASA Socials for the Mars Science Laboratory Curiosity rover landing and the NASA LADEE lunar orbiter launch.<\|endoftext\|>	3.765625
235	Because of our decision to use Arabic vocabulary for identifying culturally important terms that relate to Islam and pre-Islamic concepts, we need to address the issue of Arabic being written with the Arabic alphabet. Transliteration is the system for rendering Arabic sounds, which are normally written in the Arabic alphabet, into the Latin alphabet which is used by most European languages (and modern Turkish). There is a lot of local variation in how these are actually pronounced. However, as mentioned above, we have chosen to present Arabic words according to their pronunciation in Classical Arabic for the sake of clarity and consistency. For the system we use to represent Arabic sounds in the Latin alphabet, please see the Romanization chart above, based on the American Library Association’s system. This chart is based on pronunciation of classical Arabic. Despite our choice of classical Arabic pronunciations, there are many classical pronunciations, that in reality are quite rare in the Middle East, even in Arab countries. Dh, for example: in some Arabic-speaking countries the dh pronunciation remains similar to classical, but it is usually pronounced zh. Thus, Ramadhan becomes Ramazhan in many local contexts.<\|endoftext\|>	3.6875
296	What is Equality? Equality is making sure that children are not treated differently due to their nationality, gender, disability or religion. By promoting equality it should get rid of any discrimination in all of these areas. Also bullying and harassment are also seen to be equality and diversity problems. What is Diversity? Diversity takes into consideration the differences between people and puts a positive value on these differences. The role of a Teacher is extremely important when creating equality and diversity in a Nursery. But what exactly can you do to provide equality and diversity in your classroom? Here are some of our top tips! - Make sure policies, procedures and processes do not discriminate children or their families - Treat all your children fairly - Make an inclusive culture for all children - Be aware that everyone has equal access to opportunities to allow children to be fully involved and participated in the learning process. - Allow their students to develop to their potential and help them reach their targets - Equipping children to with the skills to challenge inequality or discrimination in their environments (when they are old enough) - Be aware that any learning materials do not discriminate against anyone. Here at MTC we have a course that will support practitioners in opening up possibilities for all children, whilst respecting individuality. It will allow practitioners to reflect upon their own practise and support them in promoting inclusive provisions, practise and experiences, that reflect a positive and diverse society. For more information click here.<\|endoftext\|>	4.21875
543	Scientists are finding more and more links between deforestation and global warming. The carbon footprint created by four years of deforestation is equal to the carbon footprint of every single air flight in the history of aviation up to the year 2025 [source: Kristof]. Let's break that down into simple logic: Trees absorb carbon dioxide. So fewer trees means more carbon dioxide is loose in the air. More carbon dioxide means an increased greenhouse effect, which leads to global warming. (You can read more about the greenhouse effect in What is the greenhouse effect?) Reduced biodiversity is another deforestation concern. Rainforests, arguably the biggest victims of deforestation, cover only about 7 percent of the world's surface. However, within this 7 percent live almost half of all plant and animal species on earth. Some of these species only live in small specific areas, which makes them especially vulnerable to extinction. As the landscape changes, some plants and animals are simply unable to survive. Species from the tiniest flower to large orangutans are becoming endangered or even extinct. Biologists believe that the key to curing many diseases resides within the biology of these rare plants and animals, and preservation is crucial [source: Lindsey]. Soil erosion, while a natural process, accelerates with deforestation. Trees and plants act as a natural barrier to slow water as it runs off the land. Roots bind the soil and prevent it from washing away. The absence of vegetation causes the topsoil to erode more quickly. It's difficult for plants to grow in the less nutritious soil that remains. Because trees release water vapor into the atmosphere, fewer trees means less rain, which disrupts the water table (or groundwater level). A lowered water table can be devastating for farmers who can't keep crops alive in such dry soil [source: USA Today]. On the other hand, deforestation can also cause flooding. Coastal vegetation lessens the impact of waves and winds associated with a storm surge. Without this vegetation, coastal villages are susceptible to damaging floods. The 2008 cyclone in Myanmar proved this fact to catastrophic effect. Scientists believe that the removal of coastal mangrove forests over the past decade caused the cyclone to hit with much more force [source: United Nations]. Deforestation also affects indigenous people, both physically and culturally. Because many indigenous people actually have no legal rights to the land on which they live, governments that want to use the forest for profit can actually "evict" them. As these populations leave the rainforest, they also leave their culture behind [source: Plotkin]. On the next page, we'll find out if damage from deforestation is reversible and check out the groups working toward a solution.<\|endoftext\|>	3.921875
6,652	# Maharashtra Board 10th Class Maths Part 2 Problem Set 2 Solutions Chapter 2 Pythagoras Theorem ## Problem Set 2 Geometry 10th Std Maths Part 2 Answers Chapter 2 Pythagoras Theorem Question 1. Some questions and their alternative answers are given. Select the correct alternative. [1 Mark each] i. Out of the following which is the Pythagorean triplet? (A) (1,5,10) (B) (3,4,5) (C) (2,2,2) (D) (5,5,2) Hint: Refer Practice set 2.1 Q.1 (i) ii. In a right angled triangle, if sum of the squares of the sides making right angle is 169, then what is the length of the hypotenuse? (A) 15 (B) 13 (C) 5 (D) 12 Hint: iii. Out of the dates given below which date constitutes a Pythagorean triplet? (A) 15/08/17 (B) 16/08/16 (C) 3/5/17 (D) 4/9/15 Hint: iv. If a, b, c are sides of a triangle and a2 + b2 = c2, name the type of the triangle. (A) Obtuse angled triangle (B) Acute angled triangle (C) Right angled triangle (D) Equilateral triangle v. Find perimeter of a square if its diagonal is 102–√ cm. (A) 10 cm (B) 402–√ cm (C) 20 cm (D) 40 cm Hint: vi. Altitude on the hypotenuse of a right angled triangle divides it in two parts of lengths 4 cm and 9 cm. Find the length of the altitude. (A) 9 cm (B) 4 cm (C) 6 cm (D) 26–√ Hint: vii. Height and base of a right angled triangle are 24 cm and 18 cm find the length of its hypotenuse. (A) 24 cm (B) 30 cm (C) 15 cm (D) 18 cm Hint: viii. In ∆ABC, AB = 63–√ cm, AC = 12 cm, BC = 6 cm. Find measure of ∠A. (A) 30° (B) 60° (C) 90° (D) 45° Hint: Question 2. Solve the following examples. i. Find the height of an equilateral triangle having side 2a. ii. Do sides 7 cm, 24 cm, 25 cm form a right angled triangle? Give reason. iii. Find the length of a diagonal of a rectangle having sides 11 cm and 60 cm. iv. Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm. v. A side of an isosceles right angled triangle is x. Find its hypotenuse. vi. In ∆PQR, PQ = 8–√, QR = 5–√, PR = 3–√ . Is ∆PQR a right angled triangle? If yes, which angle is of 90°? Solution: i. Let ∆ABC be the given equilateral triangle. ∴ ∠B = 60° [Angle of an equilateral triangle] Let AD ⊥BC, B – D – C. In ∆ABD, ∠B = 60°, ∠ADB = 90° ∴ ∠BAD = 30° [Remaining angle of a triangle] ∴ ∆ABD is a 30° – 60° – 90° triangle. ∴ AD = 32 AB [Side opposite to 60°] 32 × 2a = a3–√ units The height of the equilateral triangle is a3–√ units. ii. The sides of the triangle are 7 cm, 24 cm and 25 cm. The longest side of the triangle is 25 cm. ∴ (25)2 = 625 Now, sum of the squares of the remaining sides is, (7)2 + (24)2 = 49 + 576 = 625 ∴ (25)2 = (7)2 + (24)2 ∴ Square of the longest side is equal to the sum of the squares of the remaining two sides. ∴ The given sides will form a right angled triangle. [Converse of Pythagoras theorem] iii. Let ꠸ABCD be the given rectangle. AB = 11 cm, BC = 60 cm In ∆ABC, ∠B = 90° [Angle of a rectangle] ∴ AC2 = AB2 + BC2 [Pythagoras theorem] = 112 + 602 = 121 +3600 = 3721 ∴ AC = 3721−−−−√ [Taking square root of both sides] = 61 cm The length of the diagonal of the rectangle is 61 cm. ∴ The length of the diagonal of the rectangle is 61 cm. iv. Let ∆PQR be the given right angled triangle. In ∆PQR, ∠Q = 90° ∴ PR2 = PQ2 + QR2 [Pythagoras theorem] = 92 + 122 = 81 + 144 = 225 ∴ PR = 225−−−√ [Taking square root of both sides] = 15 cm ∴ The length of the hypotenuse of the right angled triangle is 15 cm. v. Let ∆PQR be the given right angled isosceles triangle. PQ = QR = x. In ∆PQR, ∠Q = 90° [Pythagoras theorem] ∴ PR2 = PQ2 + QR2 = x2 + x2 = 2x2 ∴ PR = 2x2−−−√ [Taking square root of both sides] = x 2–√ units ∴ The hypotenuse of the right angled isosceles triangle is x 2–√ units. ∴ The hypotenuse of the right angled isosceles triangle is x 2–√ units. vi. Longest side of ∆PQR = PQ = 8–√ ∴ PQ2 = (8–√)2 = 8 Now, sum of the squares of the remaining sides is, QR2 + PR2 = (5–√)2 + (3–√)2 = 5 + 3 = 8 ∴ PQ2 = QR2 + PR2 ∴ Square of the longest side is equal to the sum of the squares of the remaining two sides. ∴ ∆PQR is a right angled triangle. [Converse of Pythagoras theorem] Now, PQ is the hypotenuse. ∴∠PRQ = 90° [Angle opposite to hypotenuse] ∴ ∆PQR is a right angled triangle in which ∠PRQ is of 90°. Question 3. In ∆RST, ∠S = 90°, ∠T = 30°, RT = 12 cm, then find RS and ST. Solution: in ∆RST, ∠S = 900, ∠T = 30° [Given] ∴ ∠R = 60° [Remaining angle of a triangle] ∴ ∆RST is a 30° – 60° – 90° triangle. ∴ RS = 12 RT [Side opposite to 30°] 12 × 12 = 6cm Also, ST = 32 RT [Side opposite to 60°] 32 × 12 = 6 3–√ cm ∴ RS = 6 cm and ST = 6 3–√ cm Question 4. Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq. cm. Solution: Let ꠸ABCD be the given rectangle. BC = 16cm Area of rectangle = length × breadth Area of ꠸ABCD = BC × AB ∴ 192 = I6 × AB ∴ AB = 19216 = 12cm Now, in ∆ABC, ∠B = 90° [Angle of a rectangle] ∴ AC2 = AB2 + BC2 [Pythagoras theorem] = 122 + 162 = 144 + 256 =400 ∴ AC = 400−−−√ [Taking square root of both sides] = 20cm ∴ The diagonal of the rectangle is 20 cm. Question 5. Find the length of the side and perimeter of an equilateral triangle whose height is 3–√ cm. Solution: Let ∆ABC be the given equilateral triangle. ∴ ∠B = 60° [Angle of an equilateral triangle] AD ⊥ BC, B – D – C. In ∆ABD, ∠B =60°, ∠ADB = 90° ∴ ∠BAD = 30° [Remaining angle of a triangle] ∴ ∆ABD is a 30° – 60° – 90° triangle. ∴ AD = 32AB [Side opposite to 600] ∴ 3–√ = 32AB ∴ AB = 233 ∴ AB = 2cm ∴ Side of equilateral triangle = 2cm Perimeter of ∆ABC = 3 × side = 3 × AB = 3 × 2 = 6cm ∴ The length of the side and perimeter of the equilateral triangle are 2 cm and 6 cm respectively. Question 6. In ∆ABC, seg AP is a median. If BC = 18, AB2 + AC2 = 260, find AP. Solution: PC = 12 BC [P is the midpoint of side BC] 12 × 18 = 9cm in ∆ABC, seg AP is the median, Now, AB2 + AC2 = 2 A2 + 2 PC2 [Apollonius theorem] ∴ 260 = 2 AP2 + 2 (9)2 ∴ 130 = AP2 + 81 [Dividing both sides by 2] ∴ AP2 = 130 – 81 ∴ AP2 = 49 ∴ AP = 49−−√ [Taking square root of both sides] ∴ AP = 7 units Question 7. ∆ABC is an equilateral triangle. Point P is on base BC such that PC = 13 BC, if AB = 6 cm find AP. Given: ∆ABC is an equilateral triangle. PC = 13 BC, AB = 6cm. To find: AP Consttuction: Draw seg AD ± seg BC, B – D – C. Solution: ∆ABC is an equilateral triangle. ∴ AB = BC = AC = 6cm [Sides of an equilateral triangle] pc = 13 BC [Given] 13 (6) ∴ PC = 2cm ∠D = 90° [Construction] ∠C = 60° [Angle of an equilateral triangle] ∠DAC = 30° [Remaining angle of a triangle] ∴ ∆ ADC is a 30° – 60° – 90° triangle. ∴ AD = 32 AC [Side opposite to 60°] CD = 12 AC [Side opposite to 30°] ∴ CD = 12 (6) ∴ CD = 3cm Now DP + PC = CD [D – P – C] ∴ DP + 2 = 3 ∴ DP = 1cm AP2 = AD2 + DP2 [Pythagoras theorem] ∴ AP2 = (33–√)2 + (1)2 ∴ AP2 = 9 × 3 + 1 = 27 + 1 ∴ AP2 = 28 ∴ AP = 28−−√ ∴ AP = 4×7−−−−√ ∴ AP = 2 7–√cm Question 8. From the information given in the adjoining figure, prove that PM = PN = 3–√ × a Solution: Proof: In ∆PMR, QM = QR = a [Given] ∴ Q is the midpoint of side MR. ∴ seg PQ is the median. ∴ PM2 + PR2 = 2PQ2 + 2QM2 [Apollonius theorem] ∴ PM2 + a2 = 2a2 + 2a2 ∴ PM2 + a2 = 4a2 ∴ PM2 = 3a2 ∴ PM,= 3–√a (i) [Taking square root of both sides] SimlarIy, in ∆PNQ, R is the midpoint of side QN. ∴ seg PR is the median. ∴ PN2 + PQ2 = 2 PR2 + 2 RN2 [Apollonius theorem] ∴ PN2 + a2 = 2a2 + 2a2 ∴ PN2 + a2 = 4a2 ∴ PN2 = 3a2 ∴ PN = 3–√a (ii) [Taking square root of both sides] ∴ PM = PN = 3–√ a [From (i) and (ii)] Question 9. Prove that the sum of the squares of the diagonals of a parallelogram ¡s equal to the sum of the squares of its sides. Given: ꠸ABCD is a parallelogram, diagonals AC and BD intersect at point M. To prove: AC2 + BD2 = AB2 + BC2 + CD2 + AD2 Solution: Proof: ꠸ABCD is a parallelogram. ∴ AB = CD and BC = AD (i) [Opposite sides of a parallelogram] AM = 12 AC and BM = 12 BD (ii) [Diagonals of a parallelogram bisect each other] ∴ M is the midpoint of diagonals AC and BD. (iii) In ∆ABC. seg BM is the median. [From (iii)] AB2 + BC2 = 2AM2 + 2BM2 (iv) [Apollonius theorem] ∴ AB2 + BC2 = 2(12 AC)2 + 2(12 BD)2 [From (ii) and (iv)] ∴ AB2 + BC2 = 2 × BD24+2×AC24 ∴ AB2 + BC2 = BD22+AC22 ∴ 2AB2 + 2BC2 = BD2 + AC2 [Multiplying both sides by 2] ∴ AB2 + AB2 + BC2 + BC2 = BD2 + AC2 ∴ AB2 + CD2 + BC2 + AD2 = BD2 + AC2 [From(i)] i.e. AC2 + BD2 = AB2 + BC2 + CD2 + AD2 Question 10. Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 152–√ km. Find their speed per hour. Solution: Suppose Pranali and Prasad started walking from point A, and reached points B and C respectively after 2 hours. Distance between them = BC = 152–√ km Since, their speed is same, both travel the same distance in the given time. ∴ AB = AC Let AB = AC = x km (i) Now, in ∆ ABC, ∠A = 90° ∴ BC2 = AB2 + AC2 [Pythagoras theorem] ∴ (152–√)2 = x2 + x2 [From (i)] ∴ 225 × 2 = 2 x2 ∴ x2 = 225 ∴ x = 225−−−√ [Taking square root of both sides] ∴ x = 15 km ∴ AB = AC = 15km Now, speed = distance time =152 = 7.5 km/hr ∴ The speed of Pranali and Prasad is 7.5 km/hour. Question 11. In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that 4 (BL2 + CM2) = 5 BC2. Given : ∠BAC = 90° seg BL and seg CM are the medians. To prove: 4(BL2 + CM2) = 5BC2 Solution: Proof: In ∆BAL, ∠BAL 90° [Given] ∴ BL2 = AB2 + AL2 (i) [Pythagoras theorem] In ∆CAM, ∠CAM = 90° [Given] ∴ CM2 = AC2 + AM2 (ii) [Pythagoras theorem] ∴ BL2 + CM2 = AB2 + AC2 + AL2 + AM2 (iii) [Adding (i) and (ii)] Now, AL = 12 AC and AM = 12 AB (iv) [seg BL and seg CM are the medians] ∴ BL2 + CM2 = AB2 + AC2 + (12 AC)2 + (12 AB)2 [From (iii) and (iv)] =AB2+AC2+AC24+AB24 =AB2+AB24+AC2+AC24 =5AB24+5AC24 ∴ BL2 + CM2 = 54 (AB2 + AC2) ∴ 4(BL2 + CM2) = 5(AB2 + AC2) (v) In ∆BAC, ∠BAC = 90° [Given] ∴ BC2 = AB2 + AC2 (vi) [Pythagoras theorem] ∴ 4(BL2 + CM2) = 5BC2 [From (v) and (vi)] Question 12. Sum of the squares of adjacent sides of a parallelogram is 130 cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal. Solution: Let ꠸ABCD be the given parallelogram and its diagonals AC and BD intersect at point M. ∴ AB2 + AD2 = 130cm, BD = 14cm MD = 12 BD (i) [Diagonals of a parallelogram bisect each other] 12 × 14 = 7 cm In ∆ABD, seg AM is the median. [From (i)] ∴ AB2 + = 2AM2 + 2MD2 [Apollonius theorem] ∴ 130 = 2 AM2 + 2(7)2 ∴ 65 = AM2 +49 [Dividing both sides by 2] ∴ AM2 = 65 – 49 ∴ AM2 = 16 [Taking square root of both sides] ∴ AM = 16−−√ = 4cm Now, AC =2 AM [Diagonals of a parallelogram bisect each other] 2 × 4 = 8 cm ∴ The length of the other diagonal of the parallelogram is 8 cm. Question 13. In ∆ABC, seg AD ⊥ seg BC and DB = 3 CD. Prove that: 2 AB2 = 2 AC2 + BC2. Given: seg AD ⊥ seg BC DB = 3CD To prove: 2AB2 = 2AC2 + BC2 Solution: DB = 3CD (i) [Given] ∴ AB2 = AD2 + DB2 [Pythagoras theorem] ∴ AB2 = AD2 + (3CD)2 [From (i)] ∴ AB2 = AD2 + 9CD2 (ii) ∴ AC2 = AD2 + CD2 [Pythagoras theorem] ∴ AD2 = AC2 – CD2 (iii) AB2 = AC2 – CD2 + 9CD2 [From (ii) and(iii)] ∴ AB2 = AC2 + 8CD2 (iv) CD + DB = BC [C – D – B] ∴ CD + 3CD = BC [From (i)] ∴ 4CD = BC ∴ CD = BC4 (v) AB2 = AC+ 8(BC4)2 [From (iv) and (v)] ∴ AB2 = AC2 + 8 × BC216 ∴ AB2 = AC2 + BC22 ∴ 2AB2 = 2AC2 + BC2 [Multiplying both sides by 2] Question 14. In an isosceles triangle, length of the congruent sides is 13 em and its.base is 10 cm. Find the distance between the vertex opposite to the base and the centroid. Given: ∆ABC is an isosceles triangle. G is the centroid. AB = AC = 13 cm, BC = 10 cm. To find: AG Construction: Extend AG to intersect side BC at D, B – D – C. Solution: Centroid G of ∆ABC lies on AD ∴ seg AD is the median. (i) ∴ D is the midpoint of side BC. ∴ DC = 12 BC 12 × 10 = 5 In ∆ABC, seg AD is the median. [From (i)] ∴ AB2 + AC2 = 2 AD2 + 2 DC2 [Apollonius theorem] ∴ 132 + 132 = 2 AD2 + 2 (5)2 ∴ 2 × 132 = 2 AD2 + 2 × 25 ∴ 169 = AD2 + 25 [Dividing both sides by 2] ∴ AD2 = 169 – 25 ∴ AD = 144−−−√ [Taking square root of both sides] = 12 cm We know that, the centroid divides the median in the ratio 2 : 1. ∴ AGGD = 21 ∴ GDAG = 12 [By invertendo] ∴ GD+AGAG = 1+22 [By componendo] ∴ ADAG = 32 [A – G – D] ∴ 12AG = 32 ∴ AG = 12×23 = 8cm ∴ The distance between the vertex opposite to the base and the centroid is 8 cm. Question 15. In a trapezium ABCD, seg AB \|\| seg DC, seg BD ⊥ seg AD, seg AC ⊥ seg BC. If AD = 15, BC = 15 and AB = 25, find A (꠸ABCD). Construction: Draw seg DE ⊥ seg AB, A – E – B and seg CF ⊥ seg AB, A – F- B. Solution: In ∆ ACB, ∠ACB = 90° [Given] ∴ AB2 = AC2 + BC2 [Pythagoras theorem] ∴ 252 = AC2 + 152 ∴ AC2 = 625 – 225 = 400 ∴ AC = 400−−−√ [Taking square root of both sides] = 20 units Now, A(∆ABC) = 12 × BC × AC Also, A(∆ABC) = 12 × AB × CF ∴ 12 × BC × AC = 12 × AB × CF ∴ BC × AC = AB × CF ∴ 15 × 20 = 25 × CF ∴ CF = 15×2025 = 12 units In ∆CFB, ∠CFB 90° [Construction] ∴ BC2 = CF2 + FB2 [Pythagoras theorem] ∴ 152 = 122 + FB2 ∴ FB2 = 225 – 144 ∴ FB2 = 81 ∴ FB = 81−−√ [Taking square root of both sides] = 9 units Similarly, we can show that, AE = 9 units Now, AB = AE + EF + FB [A – E – F, E – F – B] ∴ 25 = 9 + EF + 9 ∴ EF = 25 – 18 = 7 units In ꠸CDEF, seg EF \|\| seg DC [Given, A – E – F, E – F – B] seg ED \|\| seg FC [Perpendiculars to same line are parallel] ∴ ꠸CDEF is a parallelogram. ∴ DC = EF 7 units [Opposite sides of a parallelogram] A(꠸ABCD) = 12 × CF × (AB + CD) 12 × 12 × (25 + 7) 12 × 12 × 32 ∴ A(꠸ABCD) = 192 sq. units Question 16. In the adjoining figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that QS = 13 QR. Prove that: 9 PS2 = 7 PQ2. Given: ∆PQR is an equilateral triangle. QS = 13 QR To prove: 9PS2 = 7PQ2 Solution: Proof: ∆PQR is an equilateral triangle [Given] ∴ ∠P = ∠Q = ∠R = 60° (i) [Angles of an equilateral triangle] PQ = QR = PR (ii) [Sides of an equilateral triangle] In ∆PTS, ∠PTS = 90° [Given] PS2 = PT2 + ST2 (iii) [Pythagoras theorem] In ∆PTQ, ∠PTQ = 90° [Given] ∠PQT = 60° [From (i)] ∴ ∠QPT = 30° [Remaining angle of a triangle] ∴ ∆PTQ is a 30° – 60° – 90° triangle ∴ PT = 32 PQ (iv) [Side opposite to 60°] QT = 12 PQ (v) [Side opposite to 30°] QS + ST = QT [Q – S – T] ∴ 13 QR + ST = 12 PQ [Given and from (v)] ∴ 13 PQ + ST = 12 PQ [From (ii)] ∴ ST = PQ2 – PQ3 ∴ ST = 3PQ2PQ6 ∴ ST = PQ6 (vi) PS2=(32PQ)2+(PQ6)2 [From (iii), (iv) and (vi)] ∴ PS2=3PQ24+PQ236 ∴ PS2=27PQ236+PQ236 PS2=28PQ236 ∴PS2 = 73 PQ2 ∴ 9PS2 = 7 PQ2 Question 17. Seg PM is a median of APQR. If PQ = 40, PR = 42 and PM = 29, find QR. Solution: In ∆PQR, seg PM is the median. [Given] ∴ M is the midpoint of side QR. ∴ PQ2 + PR2 = 2 PM2 + 2 MR2 [Apollonius theorem] ∴ 402 + 422 = 2 (29)2 + 2 MR2 ∴ 1600 + 1764 = 2 (841) + 2 MR2 ∴ 3364 = 2 (841) + 2 MR2 ∴ 1682 = 841 +MR2 [Dividing both sides by 2] ∴ MR2 = 1682 – 841 ∴ MR2 = 841 ∴ MR = 841−−−√ [Taking square root of both sides] = 29 units Now, QR = 2 MR [M is the midpoint of QR] = 2 × 29 ∴ QR = 58 units Question 18. Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM. Solution: In ∆ABC, seg AM is the median. [Given] ∴ M is the midpoint of side BC. ∴ MC = 12 BC 12 × 24 = 12 units Now, AB2 + AC2 = 2 AM2 + 2 MC2 [Apollonius theorem] ∴ 222 + 342 = 2 AM2 + 2 (12)2 ∴ 484 + 1156 = 2 AM2 + 2 (144) ∴ 1640 = 2 AM2 + 2 (144) ∴ 820 = AM2 + 144 [Dividing both sides by 2] ∴ AM2 = 820 – 144 ∴ AM2 = 676 ∴ AM = 676−−−√ [Taking square root of both sides] ∴ AM = 26 units<\|endoftext\|>	4.625
1,959	If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. # Worked example: motion problems (with definite integrals) AP.CALC: CHA‑4 (EU) , CHA‑4.C (LO) , CHA‑4.C.1 (EK) ## Video transcript Let's review a little bit of what we learned in differential calculus. Let's say we have some function S that it gives us as a function of time the position of a particle in one dimension. If we were to take the derivative with respect to time. So if we were to take the derivative with respect to time of this function S. What are we going to get? Well we're going to get ds, dt or the rate in which position changes with respect to time. And what's another word for that, position changes with respect to time? Well, that's just velocity. So that we can write as velocity as a function of time. Now, what if we were to take the derivative of that with respect to time. So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function. Or you could say that we're taking the derivative with respect to time of our velocity function. Well this is going to be, we can write this as, we can write this as dv, dt, the rate at which velocity is changing with respect to time. And what's another word for that? Well, that's also called acceleration. This is going to be our acceleration as a function of time. So, you start with the position function, take it, the position as a function of time. Take its derivative with respected time, you get velocity. Take that derivative with respected time, you get acceleration. Well, you could go the other way around. If you started with acceleration, if you started with acceleration, and you were to take the antiderivative. If you were to take the antiderivative of it, the anti, anti, an antiderivative of it is going to be, actually let me just write it this way. So an antiderivative, I'll just use the interval symbol to show that I'm taking the antiderivative. Is going to be the integral of the anti-derivative of a of t. And this is going to give you some expression with a plus c. And we could say, well, that's a general form of our velocity function. This is going to be equal to our velocity function. And to find the particular velocity function, we would have to know what the velocity is at a particular time. And then, we could solve for our c. Whether then, if we're able to do that and we were to take the anti-derivative again. Then, now we're taking the anti-derivative of our velocity function, which would give us some expression as a function of t. And then, some other constant. And, if we could solve for that constant, then we know, then we know what the position is going to be, the position is a function of time. Just like this, it would have some, plus c here if we know our position at a given time we could solve for that c. So now that we've reviewed it a little bit, but we've rewritten it in. I guess you could say, thinking of it not just from the differential point of view from the derivative point of view. But thinking of it from the anti-derivative point of view. Lets see if we can solve an interesting problem. Lets say that we know that the acceleration of a particle is a function of time is equal to one. So it's always accelerating at one unit per, and you know, I'm not giving you time. Let, let's just say that we're thinking in terms of meters and seconds. So this is one meter per second, one meter per second-squared, right over here. That's our acceleration as a function of time. And, let's say we don't know the velocity expressions, but we know the velocity at a particular time and we don't know the position expressions. But we know the position at a particular time. So, let's say we know that the velocity, at time three. Let's say three seconds is negative three meters per second. And actually I wanna write the units here, just to make it a little, a little bit. So this is meters per second squared, that's going to be our unit for acceleration. This is our unit for velocity. And let's say that we know, let's say that we know that the position at time two, at two seconds is equal to negative ten meters. So, if we're thinking in one dimension, of if this is moving along the number line, this is ten to the left of the origin. So, given this information right over here, and everything that I wrote up here. Can we figure out the actual expressions for velocity as a function of time? So not just velocity at time three, but velocity generally as a function of time. And position as a function of time. And I encourage you to pause this video right now. And try to figure it out on your own. So let's just work through this. What is, we know that velocity, as a function of time, is going to be the anti-derivative. The anti-derivitive of our acceleration is a function of time. Our acceleration is just one. So this is going to be the anti-derivitive of this right over here is going to be t and then we can't forget our constant plus c. And now we can solve for c because we know v of 3 is negative 3. So lets just write that down. So v of 3 is going to be equal to 3, 3 plus c. I just replaced where I saw the, the t or every place where I have the t I replaced it with this 3 right over here. Actually let me make it a little bit clearer. So v of 3, v of 3 is equal to 3 plus c, and they tell us that that's equal to negative 3. So that is equal to, that is equal to negative 3. So what's c going to be? So if we just look at this part of the equation or just this equation right over here. If you subtract 3 from both sides, you get, you get c is equal to negative 6. And so now we know the exact, we know the exact expression that defines velocity as a function of time. V of t, v of t is equal to t, t plus negative 6 or, t minus 6. And we can verify that. The derivative of this with respect to time is just one. And when time is equal to 3, time minus 6 is indeed negative 3. So we've been able to figure out velocity a s a function of time. So now let's do a similar thing to figure out position as a function of time. We know that position is gonna be an anti-derivative of the velocity function, so let's write that down. So, position, as a function of time, is going to be equal to the anti-derivative of v of t, dt. Which is equal to the anti-derivative of t minus 6, dt which is equal to well the anti-derivative of t, is t squared over 2. So, t squared over 2, we've seen that before. The anti-derivative of negative 6 is negative 6 t, and of course, we can't forget our constant, so, plus, plus, c. So this is what s of t is equal to, s of t is equal to all of this business right over here. And now we can try to solve, for our constant. And we do that, using this information right over here. At two seconds were at, our position is negative two meters. So s of 2, or I could just write it this way. Well, let me write it this way. S of 2, at 2 seconds, is going to be equal to 2 squared over 2. That is, let's see, that's 4 over 2, that's going to be 2. Minus 6 times 2. So, minus 12 + c is equal to negative 10, is equal to negative 10. So, let's see, we get 2 minus 12 is negative, is negative 10 plus c equals negative 10. So you add 10 to both sides you get c in this case is equal to 0. So we figured out what our position function is as well. The c right over here is just going to be 0. So our position as a function of time is equal to t squared over 2 minus 6 t. And you can verify. When t is equal to 2, t squared over 2 is 2, minus 12 is negative 10. You take the derivative here, you get t minus 6. And you can see and we already verified that v of 3 is negative 3. And you take the derivative here, you get a of t, just like that. Anyway, hopefully, you found this enjoyable.<\|endoftext\|>	4.375
471	Humans may have contributed to 2003 European heatwave Saturday, December 4, 2004European summer with no human influences (without global warming) have compared it to the same model with global warming included. They have concluded that human influence makes extremely warm summers (i.e. summers as warm as the 2003 heatwave) more than twice as likely.Scientists using a computer model of the typical Commentators have suggested that in the future this may open the way for lawsuits against heavy polluters by persons whose livelihoods have been affected by adverse climate change. However, the report stesses that though it is in fact impossible to pin any specific extreme climate event on either global warming or natural climate variation, human activities increase the frequency of such events. The study is based upon climate activity during the last 50 years, which can only be replicated with computer models which include human forcing (CO2 and other greenhouse gases (GHGs)). However, even within a warmer world the summer of 2003 was anomalously warm in Europe. Most simulations of 2003 climate produce cooler weather in Europe than what actually happened. It is not yet known if flaws in the models cause this difference, if the warmth of 2003 was part of a trend, or 2003 was merely unusually hot. If there will be a warming trend, a model predicts that such summers could become an event of 1-in-2 probability only by 2050. In the Americas, a series of sudden weather changes has been more frequent in the last 15 years since the discovery of the phenomenon knows as El Niño, in which vast regions of North and South America suffered weather extremes ranging from high tides to extreme hail, rain and snowstorms during summer and winter seasons following extremely hot summers. There's still no scientific evidence that El Niño is directly influenced by human factors. - Peter A. Stott, D. A. Stone & M. R. Allen. "Human contribution to the European heatwave of 2003" — , 02 December, 2004 - Catherine Brahic. "Humans raise odds of extreme weather" — , 03 December, 2004 - David R. Legates. "Limitations of Climate Models as Predictors of Climate Change" — , 16 May 2002 - Naomi Oreskes. "Beyond the Ivory Tower:The Scientific Consensus on Climate Change" — , 03 December 2004<\|endoftext\|>	3.828125
237	This newly released Hubble image shows galaxy NGC 2337, which is located 25 million light-years away in the constellation of Lynx. NGC 2337 is an irregular galaxy, meaning that it — along with a quarter of all galaxies in the Universe — lacks a distinct, regular appearance. The galaxy was discovered in 1877 by the French astronomer Édouard Stephan who, in the same year, discovered the galactic group Stephan’s Quintet. Although irregular galaxies may never win a beauty prize when competing with their more symmetrical spiral and elliptical peers, astronomers consider them to be very important. Some irregular galaxies may have once fallen into one of the regular classes of the Hubble sequence, but were warped and deformed by a passing cosmic companion. As such, irregular galaxies provide astronomers with a valuable opportunity to learn more about galactic evolution and interaction. Despite the disruption, gravitational interactions between galaxies can kickstart star formation activity within the affected galaxies, which may explain the pockets of blue light scattered throughout NGC 2337. These patches and knots of blue signal the presence of young, newly formed, hot stars. Credit: ESA/Hubble & NASA<\|endoftext\|>	3.953125
2,077	# How to Solve Algebraic Problems With Exponents Co-authored by wikiHow Staff Updated: March 29, 2019 In algebra, the operations (adding, subtracting, multiplying, and dividing) performed on variables work the same as the operations performed on numbers. When performing these operations on exponents, however, the laws are different. By learning these special rules for exponents, you can easily simplify algebraic expressions that include them. ### Part 1 of 2: Understanding the Laws of Exponents 1. 1 Solve expressions with a positive exponent. An exponent simply tells you how many times you multiply the base (big number) by itself. [1] • For example, ${\displaystyle x^{3}}$ is the same as ${\displaystyle x\times x\times x}$. • Plugging in a number, you would have ${\displaystyle 2^{3}}$ =${\displaystyle 2\times 2\times 2}$ =${\displaystyle 8}$ • Expressions to the first degree (expressions with an exponent of 1) always simplify to the base. [2] It is like saying “x one time.” For example, ${\displaystyle x^{1}=x}$. • Expressions to the zero degree (expressions with an exponent of 0) always simplify to 1. [3] For example, ${\displaystyle x^{0}=1}$. 2. 2 Simplify multiplication expressions with a positive exponent. When you multiply two exponents with the same base, you can simplify the expression by adding the exponents. Do NOT add or multiply the base. [4] • This rule does not apply to numbers that have a different base. For example, you cannot simplify ${\displaystyle 2^{3}\times 3^{2}}$, you simply have to solve the exponents separately and then multiply the two numbers. • For example, ${\displaystyle x^{2}\times x^{4}}$ is the same as ${\displaystyle x^{2+4}}$, which is the same as ${\displaystyle x^{6}}$. • Plugging in a number, you would have ${\displaystyle 2^{2}\times 2^{4}}$ =${\displaystyle 2^{2+4}}$ =${\displaystyle 2^{6}}$ =${\displaystyle 2\times 2\times 2\times 2\times 2\times 2}$ =${\displaystyle 64}$ 3. 3 Simplify division expressions with a positive exponent. When you divide to exponents with the same base, you can simplify the expression by subtracting the exponents.[5] Do NOT divide or subtract the base. • For example, ${\displaystyle {\frac {x^{10}}{x^{5}}}}$ is the same as ${\displaystyle x^{10-5}}$, which is the same as ${\displaystyle x^{5}}$. • Plugging in a number, you would have ${\displaystyle {\frac {2^{10}}{2^{5}}}}$ =${\displaystyle 2^{10-5}}$ =${\displaystyle 2^{5}}$ =${\displaystyle 2\times 2\times 2\times 2\times 2}$ =${\displaystyle 32}$ 4. 4 Simplify exponents with a positive exponent. Sometimes an exponent will have an exponent. In this situation, you would multiply the two exponents.[6] • For example, ${\displaystyle (x^{2})^{3}}$ is the same as ${\displaystyle x^{2\times 3}}$, which is the same as ${\displaystyle x^{6}}$. • Plugging in a number, you would have ${\displaystyle (2^{2})^{3}}$ =${\displaystyle 2^{2\times 3}}$ =${\displaystyle 2^{6}}$ =${\displaystyle 2\times 2\times 2\times 2\times 2\times 2}$ =${\displaystyle 64}$ 5. 5 Simplify expressions with a negative exponent. You can think of a negative exponent as being the opposite of a positive exponent. Since a positive exponent tells you how many times to multiply, a negative exponent tells you how many times to divide.[7] To simplify an expression with a negative exponent, use the formula ${\displaystyle x^{-a}={\frac {1}{x^{a}}}}$. • For example, ${\displaystyle x^{-4}}$ is the same as ${\displaystyle {\frac {1}{x^{4}}}}$. • Plugging in a number, ${\displaystyle 2^{-4}}$ =${\displaystyle {\frac {1}{2^{4}}}}$ =${\displaystyle {\frac {1}{2\times 2\times 2\times 2}}}$ =${\displaystyle {\frac {1}{16}}}$ ### Part 2 of 2: Solving a Problem with Exponents 1. 1 Address the order of operations. Just like any problem in mathematics, an algebraic problem must be completed by the order of operations. You can use the phrase "Please Excuse My Dear Aunt Sally," or the acronym PEMDAS, to help you remember Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.[8] • For example, if the problem is ${\displaystyle 4(x^{10}\div x^{4})\div 2(x^{3}\times x^{2})-3x^{0}}$, you would first complete the calculations inside the parentheses. 2. 2 Simplify the expressions using the laws of exponents. Remember, you can only simplify if the exponents have the same base. • For example, ${\displaystyle x^{10}\div x^{4}}$ can simplify to ${\displaystyle x^{10-4}}$, or ${\displaystyle x^{6}}$. ${\displaystyle x^{3}\times x^{2}}$ can simplify to ${\displaystyle x^{3+2}}$, or ${\displaystyle x^{5}}$. ${\displaystyle x^{0}}$ is 1, since any number to the zero power is 1. So, the simplified problem becomes ${\displaystyle 4(x^{6})\div 2(x^{5})-3(1)}$. 3. 3 Simplify coefficients. Coefficients are the numbers in an algebraic problem. When simplifying coefficients with exponents, you complete the regular operations. • For example, for ${\displaystyle 4(x^{6})\div 2(x^{5})}$, you would first divide the coefficients: ${\displaystyle 4\div 2=2}$. Then, divide the exponents: ${\displaystyle x^{6}\div x^{5}}$ =${\displaystyle x^{6-5}}$ =${\displaystyle x^{1}}$ =${\displaystyle x}$. Since ${\displaystyle 3(1)}$ simplifies to ${\displaystyle 3}$, the final, simplified problem is ${\displaystyle 2x-3}$. ## Community Q&A Search • Question How can I solve an unknown exponent, e.g 2^y? Donagan The expression would have to be part of an equation before you could find the value of y. • Question How do I simplify algebraic expressions? Donagan • Question How do I answer if the problem is in sentence form? Do your best to take key vocabulary and transform it into math, then take it from there. • Question How do I do an algebraic equation with 2 variables, that also has squares? You add up the like terms, different variables do not add up. Also, if you are adding, the exponents have to be the same on the top right corner of the coefficients. 200 characters left ## Tips • An exponent only affects a variable that is to its immediate left, including signs. For example, ${\displaystyle -x{^{2}}=-1(x\times x)=-x}$, while ${\displaystyle (-x){^{2}}=-x\times -x=x}$. Thanks! ## Things You'll Need • Paper • Pencils • Time Co-Authored By: wikiHow Staff Editor This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. Together, they cited information from 8 references. Co-authors: 16 Updated: March 29, 2019 Views: 39,078 Thanks to all authors for creating a page that has been read 39,078 times. • DS Oct 30, 2017 "I appreciate how this article not only described how to simplify the expression, but it actually showed all of the steps visually (including the step involving how one would simplify the exponents through its own micro-equation). This is very useful."..." more • SB Samantha Brizuela Nov 17, 2018 "At my school, geometry goes at a faster pace because so much is expected from us. A lot of the times I find myself going home and relearning most of the lessons. I always come to wikiHow because the answers to my questions are always found here."..." more • SN Sammy Norman Oct 5, 2016 "We had been reviewing this topic in class, but I hadn't understood. This article really went over the rules of exponents and helped me identify what I wasn't doing correctly through visuals and examples."..." more<\|endoftext\|>	4.9375
315	How do you solve x^3 - 10x^2 +24x = 0 ? Apr 7, 2018 $x = 0 \mathmr{and} 4 \mathmr{and} 6$ Explanation: ${x}^{3} - 10 {x}^{2} + 24 x = 0$ Factor, $x \left({x}^{2} - 10 x + 24\right) = 0$ Factor, $x \left(x - 6\right) \left(x - 4\right) = 0$ Hence, $x = 0 \mathmr{and} 4 \mathmr{and} 6$ Apr 7, 2018 $x = 0 , x = 4 , x = 6$ Explanation: $\text{take out a "color(blue)"common factor } x$ $x \left({x}^{2} - 10 x + 24\right) = 0$ $\text{the factors of + 24 which sum to - 10 are - 4 and - 6}$ $\Rightarrow x \left(x - 4\right) \left(x - 6\right) = 0$ $\text{equate each factor to zero and solve for x}$ $\Rightarrow x = 0$ $\Rightarrow x - 4 = 0 \Rightarrow x = 4$ $x - 6 = 0 \Rightarrow x = 6$<\|endoftext\|>	4.46875
704	Paired samples t-test Paired samples t-test The paired t-test is used to test the null hypothesis that the average of the differences between a series of paired observations is zero. Observations are paired when, for example, they are performed on the same samples or subjects. Select the variables for sample 1 and sample 2, and a possible filter for the data pairs. You can use the button to select variables and filters in the variables list. - Logarithmic transformation: if the data require a logarithmic transformation (e.g. when the data are positively skewed), select the Logarithmic transformation option. - Confidence interval: select the required confidence interval for the difference between the means. A 95% confidence interval is the usual selection, select a 90% confidence interval for equivalence testing. - You can select an optional Test for Normal distribution of the differences between the paired observations. The results windows for the paired samples t-test displays the summary statistics of the two samples. Note that the sample size will always be equal because only cases are included with data available for the two variables. Next, the arithmetic mean of the differences (mean difference) between the paired observations is given, the standard deviation of these differences and the standard error of the mean difference followed by the confidence interval for the mean difference. Differences are calculated as sample 2 − sample 1. In the paired samples t-test the null hypothesis is that the average of the differences between the paired observations in the two samples is zero. If the calculated P-value is less than 0.05, the conclusion is that, statistically, the mean difference between the paired observations is significantly different from 0. Note that in MedCalc P-values are always two-sided. If you selected the Logarithmic transformation option, the program performs the calculations on the logarithms of the observations, but reports the back-transformed summary statistics. For the paired samples t-test, the mean difference and confidence interval are given on the log-transformed scale. Next, the results of the t-test are transformed back and the interpretation is as follows: the back-transformed mean difference of the logs is the geometric mean of the ratio of paired values on the original scale (Altman, 1991). Normal distribution of differences For the paired samples t-test, it is assumed that the differences of the data pairs follow a Normal distribution (you do not need to check for normality in the two separate samples). This assumption can be evaluated with a formal test, or by means of graphical methods. The different formal Tests for Normal distribution may not have enough power to detect deviation from the Normal distribution when sample size is small. On the other hand, when sample size is large, the requirement of a Normal distribution is less stringent because of the central limit theorem. To do so, you click the hyperlink "Save differences" in the results window. This will save the differences between the paired observations as a new variable in the spreadsheet. You can then use this new variable in the different distribution plots. - Altman DG (1991) Practical statistics for medical research. London: Chapman and Hall. - To perform different tests in one single procedure, see Comparison of paired samples - Independent samples t-test - Equivalence test - Student's t-test on Wikipedia.<\|endoftext\|>	3.890625
1,303	Etymology of Malayalam The word ‘Malayalam’ may have been a local dialect in the beginning. The first part in the word ‘Malayalam’, ie, ‘Mala’ may refer to hill, and the last part ‘Alam’ to the depths of the ocean (‘Alam’ over the years transformed to Azham’ meaning ‘depths’). So the word ‘Malayalam’ may refer to the land lying between the Western Ghats and the Arabian Sea. Or, ‘Alam’ may be the ‘alam’ in words like Kovalam, Pandalam etc, meaning ‘place’. If that be so, Malayalam refers to a hilly region. It goes with words like ‘Malanad’, ‘Malabar’ etc. It has other names too like Kerala Bhasha, Malayampazha, Malayalim, Malayanma, Malayazhma, and so on. ‘Kerala bhasha’ finds mention in the 14th century grammar text, Leelatilakam (Book of the Sacred Mark). The word ‘Malayampuzha’ as ‘Malayalabasha’ is mentioned in the Latin work Hortus Malabaricus, a treatise on plants brought out by the Dutch of Amsterdam between 1673 and 1703. The word, ‘Malayalam’ appears as ‘Malayalim’ in the grammar text ‘Malayalim bhasha’ written by Pitt and published in 1841. In some of the writings that came out in 1891, the word ‘Malayanma’ (‘Malayalariti’) is used. George Mathan’s (1819- 70) grammar book was titled ‘Malayazhmayude Vyakaranam’ (1863). In the olden days, the Malayalam interpretations of Sanskrit works were known as ‘Tamizh kuthu’ (Tamil book). For e.g., the Sanskrit ‘Amaraghosham’ and its interpretation was ‘Amaram Tamizh Kuthu’. Among the Niranam poets who lived during the 15th century, Rama Panicker claimed that he interpreted ‘Brahmaandapuranam’ in Tamil without mixing it with any other language. Grammarians, linguists, and scholars have put forward their own ideas about the origins of Malayalam. But most of these views are mere ideas. One of the most conservative ideas in that Malayalam originated from Sanskrit. Some scholars believe that Malayalam developed from ancient Prakrit. There is another school of thought who says that tribes living in the forests spoke an independent language mixed with the Dravidian language and got transformed over the years to become Malayalam. Scholars have also expressed surprise on how Malayalam and Tamil having similarity on many themes remained distinct. While some argue that Malayalam is the daughter of Tamil, there are others who contend that Malayalam is the daughter of Dravidian language and sister of Tamil. However, scholars familiar with modern language history and comparative literature developed a new view regarding Malayalam. Today, everybody is in unison on including Malayalam along with Tamil, Kotha, Todak, Kodak, and Kannada as belonging to the Dakshina Dravidian family. It’s true that Malayalam has close affinity with Tamil. It is because at one point in history Tamil and Malayalam had a common root. Malayalam’s evolution as an independent language is found in the records and proclamations of the 9th century. Probably in the course of four or five centuries (9th century to 13th century) Tamil and Malayalam became different languages. Though it was necessary to recreate the spoken language of the period, it never materialised. Among the four major Dravidian languages Malayalam happened to be the last to develop literary works of its own. No doubt, Tamil has proved its existence of being the most ancient and possessing a rich tradition. So, it is only natural that Keralites too would want to get into the bandwagon of being the oldest language. The influence of Tamil on Malayalam language can be discerned in the very first decades of its evolution. Though Malayalam was the language of the masses, Tamil received the state of a scholarly language in the western parts of Kerala. Gradually, with the passage of time, Malayalam rose to dizzy heights, finding place in royal proclamations and documents. Brahmins in South India, and Kerala in particular, had an upperhand in matters of culture, thanks to the overarching influence and understanding of Sanskrit. Thus they could influence Kerala life and language. The influence of the Aryan language can be noticed in the sounds, forms of sentence structure, meanings etc. in use in Malayalam. This influence has acted like a catalyst. Language forms and influence of other languages on Malayalam Different kinds of language forms can be seen in modern Malayalam. Caste, region vocation, style and innumerable language forms constitute Malayalam. Newspaper, radio, study materials and education have helped in fostering a humane language form. Apart from geography, society and culture, caste and religion have also contributed to Malayalam. The language forms used by brahmins, harijans, nairs, ezhavas, Christians, and Muslims have been discovered. While Sanskrit words are commonly used by brahmins in their language to a large extent, it is sparsely used by the marginalized sections. In the language spoken by the Christians, we can find English, Syriac, Latin, and Portuguese words. Muslims use Arabic and Urdu words. The influence of some other languages like Prakrit, Pali, Marathi, Hindi, Persian, Dutch and French can be seen in the course of its evolution and transformation. So, it is little wonder that many foreign words have become part and parcel of Malayalam.<\|endoftext\|>	3.765625
1,816	of 60 /60 1 Math 211 Math 211 Lecture #42 The Pendulum Predator-Prey December 6, 2002 • Author others • Category ## Documents • view 1 0 Embed Size (px) ### Transcript of 1 Math 211 Lecture 42mLθ′′ = −mg sin θ − D θ′, Return 2 mLθ′′ = −mg sin θ − D θ′, We will write this as θ′′ + d θ + b sin θ = 0. Return 2 mLθ′′ = −mg sin θ − D θ′, We will write this as θ′′ + d θ + b sin θ = 0. • Introduce ω = θ′ to get the system θ′ = ω Return 3 AnalysisAnalysis Return 3 AnalysisAnalysis • The equilibrium points are (k π, 0)T where k is any integer. Return 3 AnalysisAnalysis • The equilibrium points are (k π, 0)T where k is any integer. If k is odd the equilibrium point is a saddle. Return 3 AnalysisAnalysis • The equilibrium points are (k π, 0)T where k is any integer. If k is odd the equilibrium point is a saddle. If k is even the equilibrium point is a center if d = 0 or a sink if d > 0. Return Pendulum Return Pendulum • The angle θ measured from straight up satisfies the nonlinear differential equation Return Pendulum • The angle θ measured from straight up satisfies the nonlinear differential equation or • The angle θ measured from straight up satisfies the nonlinear differential equation or θ′′ + d θ − b sin θ = 0. Return Inverted pendulum Pendulum system 5 Return Inverted pendulum Pendulum system 5 θ′ = ω Return Inverted pendulum Pendulum system 5 θ′ = ω ω′ = b sin θ − d ω • The equilibrium point at (0, 0)T is a saddle point and unstable. 5 θ′ = ω ω′ = b sin θ − d ω • The equilibrium point at (0, 0)T is a saddle point and unstable. • Can we find an automatic way of sensing the departure of the system from (0, 0)T and moving the pivot to bring the system back to the unstable point at (0, 0)T ? Return Inverted pendulum Pendulum system 5 θ′ = ω ω′ = b sin θ − d ω • The equilibrium point at (0, 0)T is a saddle point and unstable. • Can we find an automatic way of sensing the departure of the system from (0, 0)T and moving the pivot to bring the system back to the unstable point at (0, 0)T ? Return Inverted pendulum Inverted pendulum system 6 The Control SystemThe Control System • If we apply a force v moving the pivot to the right or left, then θ satisfies Return Inverted pendulum Inverted pendulum system 6 The Control SystemThe Control System • If we apply a force v moving the pivot to the right or left, then θ satisfies • The system becomes where u = v/mL. 6 The Control SystemThe Control System • If we apply a force v moving the pivot to the right or left, then θ satisfies • The system becomes where u = v/mL. • Assume the force is a linear response to the detected value of θ, so u = cθ, where c is a constant. Return Inverted pendulum Inverted pendulum system Controls 7 Return Inverted pendulum Inverted pendulum system Controls 7 • The Jacobian at the origin is J = ( ) 7 • The Jacobian at the origin is J = ( ) • The origin is asymptotically stable if T = −d < 0 and D = c − b > 0. 7 • The Jacobian at the origin is J = ( ) • The origin is asymptotically stable if T = −d < 0 and D = c − b > 0. Therefore require c > b = g Return 8 Predator-PreyPredator-Prey • Equilbrium points: • Equilbrium points: (0, 0) is a saddle Return 8 Predator-PreyPredator-Prey • Equilbrium points: (0, 0) is a saddle, (x0, y0) = (c/d, a/b) is a linear center. Return 8 Predator-PreyPredator-Prey • Equilbrium points: (0, 0) is a saddle, (x0, y0) = (c/d, a/b) is a linear center. • The axes are invariant. • Equilbrium points: (0, 0) is a saddle, (x0, y0) = (c/d, a/b) is a linear center. • The axes are invariant. Return 8 Predator-PreyPredator-Prey • Equilbrium points: (0, 0) is a saddle, (x0, y0) = (c/d, a/b) is a linear center. • The axes are invariant. • The solution curves appear to be closed. Return 8 Predator-PreyPredator-Prey • Equilbrium points: (0, 0) is a saddle, (x0, y0) = (c/d, a/b) is a linear center. • The axes are invariant. • The solution curves appear to be closed. Is this actually true? Return System Along the solution curve y = y(x) we have Return System Along the solution curve y = y(x) we have dy dx = . Along the solution curve y = y(x) we have dy dx = . H(x, y) = by − a ln y + dx − c lnx = C Return System Along the solution curve y = y(x) we have dy dx = . H(x, y) = by − a ln y + dx − c lnx = C • This is an implicit equation for the solution curve. Return System Along the solution curve y = y(x) we have dy dx = . H(x, y) = by − a ln y + dx − c lnx = C • This is an implicit equation for the solution curve. ⇒ All solution curves are closed, and represent periodic solutions. 10 System Return 10 Compute the average of the fish & shark populations. System Return 10 Compute the average of the fish & shark populations. d 10 Compute the average of the fish & shark populations. d 10 Compute the average of the fish & shark populations. d 10 Compute the average of the fish & shark populations. d So y = a/b = y0. 10 Compute the average of the fish & shark populations. d So y = a/b = y0. Similarly x = x0 = c/d. System Averages The effect of fishing that does not distinquish between fish and sharks is the system System Averages The effect of fishing that does not distinquish between fish and sharks is the system x′ = (a − by)x − ex y′ = (−c + dx)y − ey The effect of fishing that does not distinquish between fish and sharks is the system x′ = (a − by)x − ex y′ = (−c + dx)y − ey This is the same system with a replaced by a − e and c replaced by c + e. the average shark population to decrease. Return 13 Ladybird Beetle • Cottony cushion scale insect accidentally introduced from Australia in 1868. Ladybird Beetle • Cottony cushion scale insect accidentally introduced from Australia in 1868. Threatened the citrus industry. Ladybird Beetle • Cottony cushion scale insect accidentally introduced from Australia in 1868. Threatened the citrus industry. Return 13 Ladybird Beetle • Cottony cushion scale insect accidentally introduced from Australia in 1868. Threatened the citrus industry. Natural predator<\|endoftext\|>	4.4375
428	# Lim X → 1 √ 5 X − 4 − √ X X 3 − 1 - Mathematics $\lim_{x \to 1} \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1}$ #### Solution $\lim_{x \to 1} \left[ \frac{\sqrt{5x - 4} - \sqrt{x}}{x^3 - 1} \right]$ It is of the form $\frac{0}{0}$ Rationalising the numerator: $\lim_{x \to 1} \left[ \frac{\left( \sqrt{5x - 4} - \sqrt{x} \right) \left( \sqrt{5x - 4} + \sqrt{x} \right)}{\left( x^3 - 1 \right) \left( \sqrt{5x - 4} + \sqrt{x} \right)} \right]$ = $\lim_{x \to 1} \left[ \frac{5x - 4 - x}{\left( x - 1 \right)\left( x^2 + x + 1 \right)\left( \sqrt{5x - 4} + \sqrt{x} \right)} \right]$ = $\lim_{x \to 1} \left[ \frac{4\left( x - 1 \right)}{\left( x - 1 \right)\left( x^2 + x + 1 \right)\left( \sqrt{5x - 4} + \sqrt{x} \right)} \right]$ = $\frac{4}{3\left( 1 + 1 \right)}$ = $\frac{2}{3}$ Is there an error in this question or solution? #### APPEARS IN RD Sharma Class 11 Mathematics Textbook Chapter 29 Limits Exercise 29.4 \| Q 18 \| Page 28<\|endoftext\|>	4.4375
663	Students in school are rarely given opportunities to rest and reflect on the knowledge they've acquired, but a new study suggests that giving the mind a little targeted downtime could be a highly effective way to boost learning. The brain mechanisms that are engaged when the mind is resting and reflecting on previously acquired information can boost future learning, according to research from the University of Texas at Austin, recently published in the Proceedings of the National Academy of Sciences. To demonstrate this ability, researchers asked 35 adult study participants to memorize pairs of photos in two separate series. In between each series, the participants were given some time to rest and think about anything they wanted. Participants who used the time to reflect on the first series of photos, according to brain scans taken during the break, then outperformed themselves on the subsequent series. This was especially true in cases where minor details of information overlapped between the two tasks. During reflection, the researchers theorized, the participants were making mental connections that helped them to later absorb information that related in some way (even loosely) to the information that they had previously acquired. "Nothing happens in isolation," lead researcher Dr. Alison Preston, associate professor of psychology and neuroscience at the University of Texas, said in a statement. "When you are learning something new, you bring to mind all of the things you know that are related to that new information. In doing so, you embed the new information into your existing knowledge." The findings counter previously held assumptions that older memories are likely to interfere with new learning. In at least some cases, the findings show, prior memories can act as helpful connections when acquiring new knowledge. "We've shown for the first time that how the brain processes information during rest can improve future learning," Preston said. "We think replaying memories during rest makes those earlier memories stronger, not just impacting the original content, but impacting the memories to come." It's important to note, according to Perston, that participants were not necessarily actively reflecting on the previous learning experience. "In fact, our participants did not know they would later be learning related information -- so, they knew of no reason to try to remember what they had just been shown," Preston said in an email to the Huffington Post. "We think that it is more likely the case that memory replay during periods of rest is an automatic process -- the brain automatically reflects on past experiences to make memories for those experiences stronger." Preston's findings are in line with a number of studies which have found that when the mind is at rest (engaging in mind-wandering or daydreaming), parts of the brain that aid in memory storage and consolidation, as well as information retrieval, are highly active. "This study is consistent with an emerging body of research suggesting that the capacity to imagine the future draws on the same mental machinery required to remember our past," Dr. Scott Barry Kaufman, psychology professor at Penn University, who has studied mind-wandering extensively, said in an email to the Huffington Post. "Our deep storehouse of memories is part of the Default Network (or as I like to call it, the Imagination Network), which facilitates not just learning, but also perspective taking, imagination, creativity, future planning, reflection, and morality."<\|endoftext\|>	3.703125
494	St. Paul, Minn. (October 4, 2005)—A number of plant diseases may be haunting the pumpkin patch this Halloween. While plant diseases don’t pose a health risk to humans, plant diseases do affect a pumpkin’s quality, from appearance to taste. “The plant disease that is having the biggest impact on this year’s pumpkin crops is downy mildew,” said Daniel S. Egel, extension plant pathologist at the Southwest Purdue Agricultural Center, Vincennes, IN. “Downy mildew affects the foliage of the plant and, while it doesn’t directly hurt the fruit, it can affect the overall quality of the pumpkin,” Egel said. Pumpkins with downy mildew may be smaller in size and may be lower in quality. For example, because downy mildew causes the plant to lose its leaves, infected pumpkins may have areas that are sunken due to exposure to the sun. Another disease that affects pumpkins is Phytophthora blight. This disease is usually found in regions that have experienced heavy rainfall. Pumpkins with Phytophthora blight will have areas of white mold that appears fuzzy. Another disease, bacterial fruit spot, will cause scabby lesions to appear on the fruit. Plant doctors with The American Phytopathological Society (APS) suggest the following tips to help you select a healthy Halloween pumpkin: - Visually check for moldy areas or soft spots on the fruit (remember to check the bottom). - Check the stem; healthy stems are green in color. A good stem will support the weight of the fruit. - Most pumpkin varieties are a bright orange when mature. A yellow pumpkin may not be completely mature. - If possible, keep your pumpkin in a dry, shady place and try to prevent it from freezing. This should increase the ‘porch life’ of your pumpkin. - Once pumpkins are carved, the process of decay will become more rapid. To help insure that a Jack-o-Lantern lasts through Halloween, don’t carve it until a few days before the event. The American Phytopathological Society (APS) is a nonprofit, professional scientific organization. The research of the organization’s 5,000 worldwide members advances the understanding of the science of plant pathology and its application to plant health.<\|endoftext\|>	3.828125

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