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2,279	One year after Desecheo Island, Puerto Rico was declared free of invasive species, Island Conservation and partners are beginning to see signs of recovery, including Audubon’s Shearwaters sighted on the island for the first time and new Bridled Tern nests discovered. Historians say that you could hardly see the sky for the birds. This was once upon a time on Desecheo Island, a National Wildlife Refuge managed by the U.S. Fish and Wildlife Service (FWS) off the shores of Puerto Rico. The island was previously a major nesting area for thousands of seabirds and was so teeming with life that a perpetual “black cloud” of birds hovered overhead the mountainous terrain. So, what caused that clamorous, lively island atmosphere to fade into calm and quiet? The island was exposed to non-native, damaging (invasive) species in the early 1900s. Feral goats, feral cats, and rats were brought in or floated in on ships and boats and began to alter Desecheo’s ecosystem. Rhesus macaques were brought to the island in the 1970s. Invasive species alone are enough to drive species to extinction and trigger ecosystem collapse. Combined with poaching and egg harvesting by people, as well as military activity, the pressure from invasive species was too much for the fine-tuned, evolutionarily precise ecosystem of Desecheo Island. Little by little, native birds began to disappear from their compromised home. Maybe some birds found other nesting sites—certainly many were lost to predation. Now it was extinction, not a big bustling cloud of birds that was looming over the island. Conservation Turns the Tide A mammoth project to remove the invasive predators was carried out over ten years by multiple conservation groups and agencies. A significant undertaking to remove rats from the island, by the FWS, Island Conservation, and other partners came to fruition in 2017. The project’s success was monumental, making headlines and inspiring celebration. Everybody was excited, hoping to watch the recovery of breeding populations of the ecosystem, including its native large‐bodied seabirds (i.e. Brown Booby, Red‐footed Booby, Magnificent Frigatebird, and possibly the White‐tailed Tropicbird), as well as small‐bodied seabird species (i.e. Bridled Tern and Brown Noddy). Soon after the project’s completion, the remembered storm cloud of seabirds was suddenly displaced by the real storm clouds of hurricane Maria. Despite notable damage to the island’s environment, the ecosystem continued to show signs of recovery. Soon, the region’s endemic (found nowhere else on earth), Federally Threatened Higo Chumbo cactus, free from the damages of invasive species, was surging back. But, what about the birds? Birds sometimes or oftentimes need additional encouragement to come back to a once life-threatening place. Luckily, some dedicated conservationists were scheming up a plan for the seabirds. In February 2018, Island Conservation, FWS, and Effective Environmental Restoration (EER) initiated a seabird “social attraction” project to encourage birds to return to the island. So how exactly do you invite seabirds back to an island they have learned to avoid? Humans already know this one well: you create the impression that others are already there having a good time. Social attraction involves placing visual cues such as decoys—that is, hand-crafted seabird models—and audio recordings of birdcalls in the habitat. Decoys and recordings have proven successful in the past, supporting the establishment of colonies in places where they’ve not been seen for years. The partners set out to specifically encourage Bridled Terns (Onychoprion anaethetus), Audubon’s Shearwaters (Puffinus lherminieri), and Brown Noddies (Anous stolidus) to return to Desecheo Island. Audubon’s Shearwaters are quaint birds with pale undersides and dusky grey wings. They love to nest in rock crevices, though sometimes dig their own burrows. Squid and crustaceans are at the top of their menu. Bridled Terns have a narrow build and a sharp silhouette, along with distinctive orca-like patterns on their faces and bodies. Like the shearwaters, these petite birds, as well as the Brown Noddies, have a taste for squid. Brown Noddies nest on rock ledges, the ground, and rarely on vegetation. Island restoration specialist Cielo Figuerola explained the reasoning behind these choices in bird species for the attraction project: Bridled Terns have been observed nesting on Desecheo for decades but in low numbers. We wanted to increase the number of breeding pairs and also the breeding areas around the island. It has been shown that tern decoys are very successful in attracting tern species, and more so, it is more likely that other seabird species will consider visiting the island again once tern colonies are actively utilizing this habitat – so terns are a great species to start a social attraction project with. As for the Audubon’s Shearwater, although there were no records of this species inhabiting Desecheo Island, the nesting habitat on the island is ideal for this species, with many crevices and burrows along the rocky coast. The species visits nearby islands so we know it is around, it is just a matter of setting up the right scenario on Desecheo for it to visit it as well. These shearwaters are very vulnerable and sensitive to invasive species impacts and presence, so it is very unlikely you will find nests on islands currently with the presence of these predators. Now that Desecheo Island is free of invasive predators, it becomes a perfect place for the shearwaters to visit and nest.” EER placed decoys and sound systems to attract Brown Noddies. Noddies have been observed on Desecheo before, and nests have been documented, but in very low numbers, so EER is looking to attract more individuals and eventually increase the numbers of the Brown Noddies on the island. Seabirds play important roles in island ecosystems—having them present is beneficial to plants and wildlife not only on the island but also in its surrounding waters. Figuerola explains the social attraction strategy that she and partners deployed: For the terns we placed decoys on an area suitable for them to start establishing a colony but where this species had not been observed in years. For the shearwater, we placed solar-powered sound systems with speakers to play back their distinctive breeding call during the night. We placed trail cameras in all the sites to document any seabird activity nearby. Setting up the solar-powered sound systems, the speakers and the decoys, and other equipment pieces was no easy task; we had to carry very heavy batteries and bulky backpacks along steep hills and cliffs. However, all these efforts were worth it.” In June 2018, the team visited the island once again to search for birds, check on the status of the sound systems, collect memory cards, and switch batteries from trail cameras. Checking trail cameras for footage is a time-consuming and tedious task. Imagine clicking through thousands of images in search of just one thing. Each click, however, offers a tiny possibility of a great thrill—and the team was lucky to experience just that. When they checked the photos from the trail cameras they found the kind of footage that only a conservationist working to prevent extinctions on a remote island can truly appreciate: an Audubon’s Shearwater standing on top of one of the speakers. Figuerola exclaimed: We were so excited by this—it marked the first time ever that an Audubon’s Shearwater was recorded on Desecheo Island! This finding is very important because it demonstrates that social attraction with seabirds works; the appearance of the Shearwater on the speaker will inform future conservation management decisions, on Desecheo and undoubtedly other island ecosystems!” There was more good news still to come. When visiting the Bridled Tern decoy colony, the team found two certain and one suspected Bridled Tern nests in an area that Bridled Terns had never before been observed to be present. While Brown Noddies were not observed nesting in the area where decoys, mirrors, and recordings were placed, EER will continue to monitor the island to search for new individuals and nests. EER, in cooperation with FWS, continues to conduct biosecurity monitoring and check cameras, chew tags, and traps for any possible evidence of reinvasion by invasive species. After more than a decade of efforts to remove invasive species, these conservationists will do everything they can to uphold the important work that has been accomplished. Figuerola’s excitement and optimism were palpable. She foreshadowed a promising future: Our next step is to go back to the island in the following months to look for Audubon’s Shearwater nests, conduct further seabird monitoring, and check again on the sound systems and decoy colony on island. It is exciting and rewarding to think about how all of these seabird species are making a comeback to the island, and we are hopeful that one day in the not-too-distant future, we’ll have an island full of activity.” With continued support, the project is expected to make important strides in ecosystem recovery. Thanks to the efforts of conservationists and the supporters that make island restoration possible, there is a chance that one day, looking upward from the scrubby slopes rising from the sea, you will hardly be able to see the sky for the birds—as it once was on Desecheo Island. Featured photo: Audubon’s Shearwater soars over the water. Credit: Dominic Sherony - Seabirds Return to Desecheo Island One Year After Restoration - June 27, 2018 - Children of Palau Design Pledge for Ecological Responsibility - June 11, 2018 - Heightened Aspirations: IUCN Green List Strives for Flourishing Species - April 20, 2018 - Philosophy Talks: Pattern, Practices, and Wisdom - April 16, 2018 - Scientific Study Calls for Holistic Conservation Goals - April 13, 2018 - Goats + Island Ecosystems? Not a Good Match - April 6, 2018 - Wallabies Doing Well on Dirk Hartog Island - April 6, 2018 - National Wildlife Federation: Reversing America’s Wildlife Crisis - April 5, 2018 - Midway: Edge of Tomorrow - April 3, 2018 - Hybrid Iguanas Signal Need for Stricter Biosecurity - March 8, 2018<\|endoftext\|>	3.734375
1,251	This is a popular puzzle, and it appears as problem 35 in chapter 32 of Physics (third edition, part 2) by David Halliday and Robert Resnick. The resistor cube in the photographs above is composed of 12 1,000-ohm resistors. The idea is to calculate the resistance between any two points on the cube, for which there are three possibilities. You can calculate the resistance across an edge of the cube, across a face diagonal or across a body diagonal. The figure below shows a schematic of the cube. Points X and E are the ends of a cube edge, X and F give a face diagonal, and X to B gives a body diagonal. You could draw an arbitrary (DC) voltage source across X and E, X and F or X and B, and then using Kirchoff's rules (that the algebraic sum of the currents at any circuit junction must equal zero, and that the sum of the changes in potential encountered in making a complete loop equals zero), derive six equations in six unknowns for the currents in terms of the voltage and the various resistances, then solve the equations to find the total current and thus the overall resistance. Since all the resistors are identical, however, we can take advantage of the symmetry of the cube and solve the problem in a much simpler way. First, let us start with the resistance along a cube edge. In the photograph below, the cube is oriented to make the symmetrical arrangement easier to see. The currents going into and out of points X and Y must split equally along both resistors attached to each, and then combine equally through the bottom two pairs. So we see that the points marked ‘a’ must be at the same potential, and those marked ‘b’ must be at the same potential. If two points in a circuit are at the same potential, we may think of them as if they are connected by a wire. If we do this, we arrive at the schematic shown on the right. Since resistors add reciprocally in parallel (that is, 1/RTOT= 1/R1+1/R2+. . .+1/Rn), if we call the value of each resistor R, then each parallel pair equals R/2, and our schematic becomes: which by adding the three bottom resistors in series (RTOT= R1 + R2 + . . . + Rn) we can further simplify to: The parallel combination of R/2 and 2R gives 1/RTOT= 2/R+1/(2R) = 5R/2R2, and RTOT= (2/5)R. Added in series with the two resistors of R/2, this gives a parallel combination between X and Y of R and (7/5)R. For the edge, then, 1/RTOT= 1/R + 5/(7R) = 12R/7R2. The total resistance across the edge is thus (7/12)R. With R = 1 kohm (or 1 k), this equals 583 ohms. Next, we’ll do a face diagonal. As for the edge, the photograph below shows the cube in an orientation that helps make the symmetry clear. Again, if we imagine a potential placed across the cube at points X and Y, by symmetry we can see that points a are at the same potential, and so are points b, to give us the schematic shown on the right. Again, all the parallel combinations add to give R/2, and we have: When we add the two pairs of series resistors (R + R/2 = 3R/2), we get: At first, this schematic looks a bit tricky, but we note that the resistances on either side of point a are equal, and so are those on either side of point b. Thus, both point a and point b are at a potential halfway between that at X and that at Y, and so are at the same potential. It therefore does not matter whether or not the resistor of R/2 that connects points a and b is there. If we consider it a dead short, we get the sum of twice the parallel combination of R/2 and 3R/2. This parallel combination is: 1/RTOT= 2/R + 2/(3R) = 8R/3R2; RTOT = (3/8)R. Two such combinations added in series give (3/4)R. If we treat the circuit as if the resistor between a and b is not there, we get the parallel combination of R and 3R: 1/RTOT= 1/R + 1/(3R) = 4R/3R2; RTOT= (3/4)R, which for R = 1 k equals 750 ohms. Last, we look at the body diagonal. Again, the photograph shows the cube in such an orientation as to make the symmetry clear. Since the points a are equidistant from points b and vice versa, the currents must split equally in the resistors going from X to a and from Y to b, and the currents in all the resistors connecting points a and points b must also be equal. Thus, points a are all at the same potential, and points b are also at the same potential. This gives us the schematic shown on the right. This is merely the series sum of three parallel arrangements, or R/3 + R/6 + R/3, which equals (5/6)R. For R = 1 k, this equals 833 ohms. You can measure the resistance across an edge, face diagonal or body diagonal by attaching the ohmmeter probes as shown in the top photo, and use the large display to show the results to the class. The values you measure will, of course, be close to the ones calculated above. (The tolerance on the values of the resistors in this cube is 5%.)<\|endoftext\|>	4.4375
974	Courses Courses for Kids Free study material Offline Centres More Store # A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in the figure. The dimensions of equal strips are 25cm, 25cm, and 14 cm. Find the area of each type of paper needed to make the hand fan.$\left( a \right)280{\text{ c}}{{\text{m}}^2}$$\left( b \right)420{\text{ c}}{{\text{m}}^2}$$\left( c \right)840{\text{ c}}{{\text{m}}^2}$$\left( d \right)384{\text{ c}}{{\text{m}}^2}$ Last updated date: 13th Jun 2024 Total views: 413.1k Views today: 6.13k Verified 413.1k+ views Hint: In this particular question use the concept that if all the sides of the triangle are given then the area (A) of the triangle is given as, $A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$, so use this concept to reach the solution of the question. Given data: The sides of the triangle are 25cm, 25cm, and 14cm, as shown in the given figure. Now we have to find out the area of each type of paper needed to make the hand fan. Now as we see that there are two types of paper used in the given figure each type has 5 similar triangles. As there are a total 10 triangles in a hand fan. So the area $\left( {{A_1}} \right)$ of one type of paper = area $\left( {{A_2}} \right)$ of other types of paper = 5 times the area (A) of any one triangle. Let, a = 25cm, b = 25cm, and c = 14cm. Now as we know that if all the sides of the triangle are given then the area (A) of the triangle is given as, $\Rightarrow A = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$................. (1), where s is the half of the perimeter and a, b and c are the length of the sides of the triangle. Now as we know that the perimeter is the sum of all the sides. So, the perimeter of the triangle = a + b + c = 25 + 25 + 14 = 64cm. So, $s = \dfrac{{a + b + c}}{2} = \dfrac{{64}}{2} = 32$ cm. Now substitute all the values in equation (1) we have, $\Rightarrow A = \sqrt {32\left( {32 - 25} \right)\left( {32 - 25} \right)\left( {32 - 14} \right)}$ Now simplify this we have, $\Rightarrow A = \sqrt {32\left( 7 \right)\left( 7 \right)\left( {18} \right)}$ $\Rightarrow A = \sqrt {{7^2}\left( {16} \right)\left( {36} \right)}$ $\Rightarrow A = \sqrt {\left( {{7^2}} \right)\left( {{4^2}} \right)\left( {{6^2}} \right)}$ $\Rightarrow A = 7\left( 4 \right)\left( 6 \right) = 168$ Sq. cm. So the area of each type of paper needed to make the hand fan is 5 times the above calculated area. $\Rightarrow {A_1} = {A_2} = 5\left( A \right) = 5\left( {168} \right) = 840{\text{ c}}{{\text{m}}^2}$ Hence option (c) is the correct answer. Note:Whenever we face such types of questions the key concept we have to remember is that the perimeter of any shape is the sum of all the sides, so the perimeter of triangle is the sum of all the sides, so calculate it, so s is equal to half of the perimeter so calculate it then simply substitute the values in the area formula as above and simplify we will get the required answer.<\|endoftext\|>	4.65625
5,324	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Real Zeros of Polynomials ## Remainder, Factor, and Rational Zero Theorems and Descartes' Rule of Signs. Estimated19 minsto complete % Progress Practice Real Zeros of Polynomials Progress Estimated19 minsto complete % Real Zeros of Polynomials In the real world, problems do not always easily fit into quadratic or even cubic equations. Financial models, population models, fluid activity, etc., all often require many degrees of the input variable in order to approximate the overall behavior. While it can be challenging to model some of these more complex interactions, the effort can be well worth it. Mathematical models of stocks are used constantly as a way to "look into the future" of finance and make the kinds of educated guesses that are behind some of the largest fortunes in the world. What benefits can you think of to modeling the behavior of large populations? Can you think of other useful applications not mentioned here? ### Watch This PatrickJMT: Finding all Zeroes of a Polynomial Function ### Guidance There are three theorems and a rule that we will be referring to during this lesson in order to help make the discovery of the roots of polynomial functions easier. You should review them and be prepared to refer to them often during the practice problems. The Remainder Theorem If a polynomial f(x)\begin{align}f(x)\end{align} of degree n>0\begin{align}n>0\end{align} is divided by xc\begin{align}x-c\end{align}, then the remainder R\begin{align}R\end{align} is a constant and it is equal to the value of the polynomial when c\begin{align}c\end{align} is substituted for x\begin{align}x\end{align}. That is f(c)=R The Factor Theorem If f(x)\begin{align}f(x)\end{align} is a polynomial of degree n>0\begin{align}n>0\end{align} and f(c)=0\begin{align}f(c)=0\end{align}, then xc\begin{align}x-c\end{align} is a factor of the polynomial f(x)\begin{align}f(x)\end{align}. Further, if xc\begin{align}x-c\end{align} is a factor, then c\begin{align}c\end{align} is a zero of f\begin{align}f\end{align}. The Rational Zero Theorem Given the polynomial f(x)=anxn+an1xn1++a1x+a0 an0\begin{align}a_{n}\ne 0\end{align} and n\begin{align}n\end{align} is a positive integer. If the coefficients are integers and pq\begin{align}\frac{p}{q}\end{align} is a rational zero in lowest terms, then p\begin{align}p\end{align} is a divisor of a0\begin{align}a_{0}\end{align} and q\begin{align}q\end{align} is a divisor of an\begin{align}a_{n}\end{align}. Descartes' Rule of Signs Given any polynomial, p(x)\begin{align}p(x)\end{align}, 1. Write it with the terms in descending order, i.e. from the highest degree term to the lowest degree term. 2. Count the number of sign changes of the terms in p(x)\begin{align}p(x)\end{align}. Call the number of sign changes n\begin{align}n\end{align}. 3. Then the number of positive roots of p(x)\begin{align}p(x)\end{align} is less than or equal to n\begin{align}n\end{align}. 4. Further, the possible number of positive roots is n,n2,n4,\begin{align}n, n-2, n-4, \ldots\end{align} 5. To find the number of negative roots of p(x)\begin{align}p(x)\end{align}, write p(x)\begin{align}p(-x)\end{align} in descending order as above (i.e. change the sign of all terms in p(x)\begin{align}p(x)\end{align} with odd powers), and repeat the process above. Then the maximum number of negative roots is n\begin{align}n\end{align}. #### Example A Use synthetic division and the remainder and factor theorems to find the quotient Q(x)\begin{align}Q(x)\end{align} and the remainder R\begin{align}R\end{align} if f(x)=2x33x2+6\begin{align}f(x)=2x^{3}-3x^{2}+6\end{align} is divided by x5\begin{align}x-5\end{align}. Solution: 5 )23 0 6¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 10 35 175 2 7 35 181 Hence 2x33x2+6=(2x2+7x+35)(x5)+181 Notice that the remainder is 181 and it can also be obtained if we simply substituted x=5\begin{align}x=5\end{align} into f(x)\begin{align}f(x)\end{align}, f(5)=2(5)33(5)2+6=25075+6=181 #### Example B Use the rational zero theorem and synthetic division to find all the possible rational zeros of the polynomial f(x)=x32x2x+2 Solution: From the rational zero theorem, pq\begin{align}\frac{p}{q}\end{align} is a rational zero of the polynomial f\begin{align}f\end{align}. So p\begin{align}p\end{align} is a divisor of 2 and q\begin{align}q\end{align} is a divisor of 1. Hence, p\begin{align}p\end{align} can take the following values: -1, 1, -2, 2 and q\begin{align}q\end{align} can be either -1 or 1. Therefore, the possible values of pq\begin{align}\frac{p}{q}\end{align} are So there are four possible zeros. Of these four, not more than three can be zeros of \begin{align}f\end{align} because \begin{align}f\end{align} is a polynomial with degree 3. To test which of the four possible candidates are zeros of \begin{align}f\end{align}, we use the synthetic division. Recall from the remainder theorem if \begin{align}f(c)=0\end{align}, then \begin{align}c\end{align} is a zero of \begin{align}f\end{align}. We have Hence, 2 is a zero of \begin{align}f\end{align}. Further, by the division algorithm, The remaining zeros of \begin{align}f\end{align} are simply the zeros of \begin{align}Q(x)=x^{2}-1\end{align} which is easier to manipulate, and thus the remaining zeros are -1 and 1. Thus the rational zeros of \begin{align}f\end{align} are -1, 1, and 2. #### Example C Graph the polynomial function \begin{align}h(x)=2x^{3}-9x^{2}+12x-5\end{align}. Solution: Notice that the leading term is \begin{align}2x^{3}\end{align}, where \begin{align}n=3\end{align} odd and \begin{align}a_{n}=2>0\end{align}. This tells us that the end behavior will take the shape of a power function with an odd exponent. Here, as you can see, there is no straight-forward way to find the zeros of \begin{align}h(x)\end{align}. However, with the use of the factor theorem and the synthetic division, we can find the rational roots of \begin{align}h(x)\end{align}. First, we use the rational zero theorem and find that the possible rational zeros are testing all these numbers by the synthetic division, -1 is not a root. Now let's test \begin{align}x=1\end{align}. we find that 1 is a zero of \begin{align}h\end{align} and so we can re-write \begin{align}h(x)\end{align}, and so Thus 1 and \begin{align}\frac{5}{2}\end{align} are the \begin{align}x-\end{align}intercepts of \begin{align}h(x)\end{align}. The \begin{align}y-\end{align}intercept is Further, the synthetic division can be also used to form a table of values for the graph of \begin{align}h(x)\end{align}: We choose test points from each interval and find \begin{align}g(x)\end{align}. Interval Test Value \begin{align}x\end{align} \begin{align}h(x)\end{align} Sign of \begin{align}h(x)\end{align} Location of points on the graph \begin{align}(-\infty,1)\end{align} -1 -28 - below the \begin{align}x-\end{align}axis \begin{align}\left ( 1,\frac{5}{2} \right )\end{align} \begin{align}\frac{3}{2}\end{align} \begin{align}\frac{-11}{4}\end{align} - below the \begin{align}x-\end{align}axis \begin{align}\left ( \frac{5}{2},\infty \right )\end{align} 3 4 + above the \begin{align}x-\end{align}axis From this information, the graph of \begin{align}h(x)\end{align} is shown in the two graphs below. Notice that the second graph is a magnification of \begin{align}h(x)\end{align} in the vicinity of the \begin{align}x-\end{align}axis. Concept question wrap-up: Were you able to identify some valuable real-world uses for modeling higher-degree polynomials? Here are a few possibilities: Identifying what times of the day people are most likely to want coffee (market research like this is key to running your own business) Predicting population growth in a particular neighborhood or area of town (useful for identifying a good location to start a small business) Identifying which stocks are likely to rise or fall based on weather or season Forecasting the weather Calculating the right head start to give a slower car to make a drag race exciting There are many, many more. ### Guided Practice 1) Show that \begin{align}x+3\end{align} is a factor of \begin{align}g(x)=x^{4}+2x^{3}-3x^{2}+4x+12\end{align}. Find the quotient \begin{align}Q(x)\end{align} and express \begin{align}f(x)\end{align} in factored form. 2) Use the 'rational zero' theorem and synthetic division to find all the possible rational zeros of the polynomial 3) Use Descartes' Rule of Signs to identify the possible number of positive and negative roots of \begin{align}f(x)=-2x^{3}+x^{2}-3x^{5}+5x-1\end{align}. 4) Find the root(s) of \begin{align}f(x) = 4x^2 - 3x - 7\end{align} 5) Find the root(s) of \begin{align}f(x) = x ^4 + 1\end{align} 1) By the factor theorem, if \begin{align}f(c)=0\end{align}, then \begin{align}x-c\end{align} is a factor of the polynomial. In other words, if the synthetic division produces a remainder equal to zero, then \begin{align}c\end{align} is a factor of the polynomial.: Using the synthetic division with \begin{align}c=-3\end{align}, Hence, \begin{align}g(-3)=0\end{align}, and the quotient is so that \begin{align}g(x)\end{align} can be written as 2) Assume \begin{align}\frac{p}{q}\end{align} is a rational zero of \begin{align}f\end{align}. By the rational zero theorem, \begin{align}p\end{align} is a divisor of 6 and \begin{align}q\end{align} is a divisor of 1. Thus \begin{align}p\end{align} and \begin{align}q\end{align} can assume the following respective values and Therefore, the possible rational zeros will be Notice that with these choices for \begin{align}p\end{align} and \begin{align}q\end{align} there could be \begin{align}8 \cdot 2=16\end{align} rational zeros. But, eight of them are duplicates. For example \begin{align}\frac{1}{-1}=\frac{-1}{1}=-1\end{align}. The next step is to test all these values by the synthetic division (we'll let you do this on your own for practice) and we finally find that are zeros of \begin{align}f\end{align}. That is 3) First, re-write \begin{align}f(x)\end{align} in descending order The number of sign changes of \begin{align}f(x)\end{align} is 2, so the number of positive roots is either 2 or 0. For the negative roots, write The number of sign changes of \begin{align}f(-x)\end{align} is 2, so the maximum number of negative roots is 2. The graph of \begin{align}f(x)\end{align} below shows that there is one negative root and two positive roots. 4) Solve by factoring and applying the zero product rule: \begin{align}4x^2 - 3x - 7 = 0\end{align} \begin{align} (4x - 7)(x + 1) = 0\end{align} \begin{align}4x - 7 = 0\end{align} or \begin{align}x + 1 = 0\end{align} \begin{align}\therefore x = \frac{7}{4}\end{align} or \begin{align} x = -1\end{align} (each zero has a multiplicity of one) 5) This one is easy: \begin{align}f(x) = x^4 + 1\end{align} \begin{align}x^4 = -1\end{align} Since there are no real roots of even powers, this function has zero real solutions. ### Explore More Problems 1 - 3: Use a) long division and b) synthetic division to perform the divisions. Express each result in the form: \begin{align}f(x)=D(x)\cdot Q(x)+R\end{align} 1. \begin{align}5x^{5}-3x^{4}+2x^{3}+x^{2}-7x+3\end{align} by \begin{align}x-2\end{align} 2. \begin{align}-4x^{6}-5x^{3}+3x^{2}+x+7\end{align} by \begin{align}x-1\end{align} 3. \begin{align}2x^{3}-5x^{2}+5x+11\end{align} by \begin{align}x-\frac{1}{2}\end{align} 4. Use synthetic division to find \begin{align}Q(x)\end{align} and \begin{align}f(c)\end{align} so that \begin{align}f(x)=(x-c)Q(x)+f(c)\end{align} if \begin{align}f(x)=-3x^{4}-3x^{3}+3x^{2}+2x-4\end{align} and \begin{align}c=-2\end{align} 5. If \begin{align}f(x)=x^{3}+2x^{2}-10x+10\end{align}, use synthetic division to determine the following: a) \begin{align}f(-1)\end{align} b) \begin{align}f(-3)\end{align} c) \begin{align}f(0)\end{align} d) \begin{align}f(4)\end{align} e) What are the factors of \begin{align}f(x)\end{align}? 6. Find \begin{align}k\end{align} so that \begin{align}x-2\end{align} is a factor of \begin{align}f(x)=3x^{3}+4x^{2}+kx-20\end{align} 7. Use synthetic division to determine all the zeros of the polynomials: a) \begin{align}f(x)=3x^{3}-7x^{2}+8x-2\end{align} b) \begin{align}g(x)=4x^{4}-4x^{3}-7x^{2}+4x+3\end{align} 8. Graph the polynomial function \begin{align}f(x)=x^{3}-2x^{2}-5x+6\end{align} by using synthetic division to find the \begin{align}x-\end{align}intercepts and locate the \begin{align}y-\end{align}intercepts. 9. Graph the polynomial function \begin{align}h(x)=x^{3}-3x^{2}+4\end{align} by using synthetic division to find the \begin{align}x-\end{align}intercepts and locate the \begin{align}y-\end{align}intercepts. 10. Write a 3rd degree equation of a polynomial function with the zeroes: 0, 2, and -5. 11. Write a 7th degree equation of a polynomial function with the zeroes: 0 (multiplicity 2), 2 (multiplicity 3), and -5 (multiplicity 2) 12. Write a quadratic equation which has 4 (multiplicity 2) as the zero and opens downward. 13. Write a 3rd degree polynomial function with the zeroes: -2, 2, and 6, passing through the point (3, 4) 14. Let \begin{align}f(x)=2x^{3}-5x^{2}-4x+3\end{align} and find the solutions: a) \begin{align}f(x)=0\end{align} b) \begin{align}f(2x)=0\end{align} 15. Graph and find the solution set of the inequality \begin{align}x^{3}-2x^{2}-5x+6 \leq 0\end{align}. 16. Use the graph of \begin{align}f(x)=x (x - 1)(x + 2)\end{align} to find the solution set of the inequality \begin{align}x (x - 1)(x + 2) > 0\end{align}. ### Answers for Explore More Problems To view the Explore More answers, open this PDF file and look for section 2.12. ### Vocabulary Language: English Descartes' Rule of Signs Descartes' Rule of Signs Descartes' rule of signs is a technique for determining the number of positive and negative real roots of a polynomial. factor theorem factor theorem The factor theorem states that if $f(x)$ is a polynomial of degree $n>0$ and $f(c)=0$, then $x-c$ is a factor of the polynomial $f(x)$. factorization theorem factorization theorem The factorization theorem states that If $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$, where $a_{n} \ne 0$, and $n$ is a positive integer, then $f(x)=a_{n}(x-c_{1})(x-c_{2})\cdots(x-c_{0})$ where the numbers $c_{i}$ are complex numbers. Multiplicity Multiplicity The multiplicity of a term describes the number of times the given term acts as a zero of the given function. Polynomial Polynomial A polynomial is an expression with at least one algebraic term, but which does not indicate division by a variable or contain variables with fractional exponents. Rational Zero Theorem Rational Zero Theorem The rational zero theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\pm \frac{p}{q}$. Remainder Theorem Remainder Theorem The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$. Roots Roots The roots of a function are the values of x that make y equal to zero. Synthetic Division Synthetic Division Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used. Zeroes Zeroes The zeroes of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero. Zeros Zeros The zeros of a function $f(x)$ are the values of $x$ that cause $f(x)$ to be equal to zero.<\|endoftext\|>	4.46875
7,849	As a literary device, an allegory is a metaphor in which a character, place or event is used to deliver a broader message about real-world issues and occurrences. Allegory (in the sense of the practice and use of allegorical devices and works) has occurred widely throughout history in all forms of art, largely because it can readily illustrate or convey complex ideas and concepts in ways that are comprehensible or striking to its viewers, readers, or listeners. Writers or speakers typically use allegories as literary devices or as rhetorical devices that convey (semi-)hidden or complex meanings through symbolic figures, actions, imagery, or events, which together create the moral, spiritual, or political meaning the author wishes to convey. Many allegories use personifications of abstract concepts. First attested in English in 1382, the word allegory comes from Latin allegoria, the latinisation of the Greek ἀλληγορία (allegoría), "veiled language, figurative", which in turn comes from both ἄλλος (allos), "another, different" and ἀγορεύω (agoreuo), "to harangue, to speak in the assembly", which originates from ἀγορά (agora), "assembly". Northrop Frye discussed what he termed a "continuum of allegory", a spectrum that ranges from what he termed the "naive allegory" of The Faerie Queene, to the more private allegories of modern paradox literature. In this perspective, the characters in a "naive" allegory are not fully three-dimensional, for each aspect of their individual personalities and the events that befall them embodies some moral quality or other abstraction; the allegory has been selected first, and the details merely flesh it out. The origins of Allegory can be traced at least back to Homer in his "quasi-allegorical" use of personifications of, e.g., Terror (Deimos) and Fear (Phobos) at Il. 115 f. The title of "first allegorist," however, is usually awarded to whoever was the earliest to put forth allegorical interpretations of Homer. This approach leads to two possible answers: Theagenes of Rhegium (whom Porphyry calls the "first allegorist," Porph. Quaest. Hom. 1.240.14-241.12 Schrad.) or Pherecydes of Syros, both of whom are presumed to be active in the 6th century B.C.E., though Pherecydes is earlier and as he is often presumed to be the first writer of prose. The debate is complex, since it demands we observe the distinction between two often conflated uses of the Greek verb "allēgoreīn," which can mean both "to speak allegorically" and "to interpret allegorically." In the case of "interpreting allegorically," Theagenes appears to be our earliest example. Presumably in response to proto-philosophical moral critiques of Homer (e.g. Xenophanes fr. 11 Diels-Kranz ), Theagenes proposed symbolic interpretations whereby the Gods of the Iliad actually stood for physical elements. So, Hephestus represents Fire, for instance (for which see fr. A2 in Diels-Kranz ). Some scholars, however, argue that Pherecydes cosmogonic writings anticipated Theagenes allegorical work, illustrated especially by his early placement of Time (Chronos) in his genealogy of the gods, which is thought to be a reinterpretation of the titan Kronos, from more traditional genealogies. Among the best-known examples of allegory, Plato's Allegory of the Cave, forms a part of his larger work The Republic. In this allegory, Plato describes a group of people who have lived chained in a cave all of their lives, facing a blank wall (514a–b). The people watch shadows projected on the wall by things passing in front of a fire behind them and begin to ascribe forms to these shadows, using language to identify their world (514c–515a). According to the allegory, the shadows are as close as the prisoners get to viewing reality, until one of them finds his way into the outside world where he sees the actual objects that produced the shadows. He tries to tell the people in the cave of his discovery, but they do not believe him and vehemently resist his efforts to free them so they can see for themselves (516e–518a). This allegory is, on a basic level, about a philosopher who upon finding greater knowledge outside the cave of human understanding, seeks to share it as is his duty, and the foolishness of those who would ignore him because they think themselves educated enough. In Late Antiquity Martianus Capella organized all the information a fifth-century upper-class male needed to know into an allegory of the wedding of Mercury and Philologia, with the seven liberal arts the young man needed to know as guests. Other early allegories are found in the Hebrew Bible, such as the extended metaphor in Psalm 80 of the Vine and its impressive spread and growth, representing Israel's conquest and peopling of the Promised Land. Also allegorical is Ezekiel 16 and 17, wherein the capture of that same vine by the mighty Eagle represents Israel's exile to Babylon. Allegorical interpretation of the Bible was a common early Christian practice and continues. For example, the recently re-discovered IVth Commentary on the Gospels by Fortunatianus of Aquileia has a comment by its English translator: The principal characteristic of Fortunatianus’ exegesis is a figurative approach, relying on a set of concepts associated with key terms in order to create an allegorical decoding of the text. (pXIX) Allegory has an ability to freeze the temporality of a story, while infusing it with a spiritual context. Mediaeval thinking accepted allegory as having a reality underlying any rhetorical or fictional uses. The allegory was as true as the facts of surface appearances. Thus, the Papal Bull Unam Sanctam (1302) presents themes of the unity of Christendom with the pope as its head in which the allegorical details of the metaphors are adduced as facts on which is based a demonstration with the vocabulary of logic: "Therefore of this one and only Church there is one body and one head—not two heads as if it were a monster... If, then, the Greeks or others say that they were not committed to the care of Peter and his successors, they necessarily confess that they are not of the sheep of Christ." This text also demonstrates the frequent use of allegory in religious texts during the Mediaeval Period, following the tradition and example of the Bible. In the late 15th century, the enigmatic Hypnerotomachia, with its elaborate woodcut illustrations, shows the influence of themed pageants and masques on contemporary allegorical representation, as humanist dialectic conveyed them. The denial of medieval allegory as found in the 12th-century works of Hugh of St Victor and Edward Topsell's Historie of Foure-footed Beastes (London, 1607, 1653) and its replacement in the study of nature with methods of categorisation and mathematics by such figures as naturalist John Ray and the astronomer Galileo is thought to mark the beginnings of early modern science. Since meaningful stories are nearly always applicable to larger issues, allegories may be read into many stories which the author may not have recognised. This is allegoresis, or the act of reading a story as an allegory. Examples of allegory in popular culture that may or may not have been intended include the works of Bertolt Brecht, and even some works of science fiction and fantasy, such as The Chronicles of Narnia by C.S. Lewis and A Kingdom Far and Clear: The Complete Swan Lake Trilogy by Mark Helprin. The story of the apple falling onto Isaac Newton's head is another famous allegory. It simplified the idea of gravity by depicting a simple way it was supposedly discovered. It also made the scientific revelation well known by condensing the theory into a short tale. It is important to note that while allegoresis may make discovery of allegory in any work, not every resonant work of modern fiction is allegorical, and some are clearly not intended to be viewed this way. According to Henry Littlefield's 1964 article, L. Frank Baum's The Wonderful Wizard of Oz, may be readily understood as a plot-driven fantasy narrative in an extended fable with talking animals and broadly sketched characters, intended to discuss the politics of the time. Yet, George MacDonald emphasised in 1893 that "A fairy tale is not an allegory." J.R.R. Tolkien's The Lord of the Rings is another example of a well-known work mistakenly perceived as allegorical, as the author himself once stated, "...I cordially dislike allegory in all its manifestations, and always have done so since I grew old and wary enough to detect its presence. I much prefer history – true or feigned– with its varied applicability to the thought and experience of readers. I think that many confuse applicability with allegory, but the one resides in the freedom of the reader, and the other in the purposed domination of the author." Tolkien specifically resented the suggestion that the book's One Ring, which gives overwhelming power to those possessing it, was intended as an allegory of nuclear weapons. He noted that, had that been his intention, the book would not have ended with the Ring being destroyed but rather with an arms race in which various powers would try to obtain such a Ring for themselves. Then Tolkien went on to outline an alternative plot for "Lord of The Rings", as it would have been written had such an allegory been intended, and which would have made the book into a dystopia. While all this does not mean Tolkien's works may not be treated as having allegorical themes, especially when reinterpreted through postmodern sensibilities, it at least suggests that none were conscious in his writings. This further reinforces the idea of forced allegoresis, as allegory is often a matter of interpretation and only sometimes of original artistic intention. Like allegorical stories, allegorical poetry has two meanings – a literal meaning and a symbolic meaning. Some unique specimens of allegory can be found in the following works: Some elaborate and successful specimens of allegory are to be found in the following works, arranged in approximate chronological order: Allegory in the Middle Ages was a vital element in the synthesis of biblical and classical traditions into what would become recognizable as medieval culture. People of the Middle Ages consciously drew from the cultural legacies of the ancient world in shaping their institutions and ideas, and so allegory in medieval literature and medieval art was a prime mover for the synthesis and transformational continuity between the ancient world and the "new" Christian world.People of the Middle Ages did not see the same break between themselves and their classical predecessors that modern observers see; rather, they saw continuity with themselves and the ancient world, using allegory as a synthesizing agent that brings together a whole image.Allegory of Hispania The allegory of Hispania is the national personification of Spain. The antecedent of this representation were some coins on which there was a horseman holding a lance and the legend HISPANORVM. These coins corresponded to the first half of the 2nd century BC and were minted in Morgantina (Sicily). These coins were carried out by the Hispanic mercenaries who received the government of this Sicilian city by order of the Roman Senate during the Second Punic War. The first representation of Hispania appeared during the Roman Republic as the head of a woman with the legend HISPAN, and was minted in Rome by the Roman family Postumia (81 B.C.). Since then different coins emerged with allegorical representations of Hispania with different characteristics during the entire Roman era. Like other coins with provincial allegories, it would fall into disuse due to the prevalence of symbols of Rome and Constantinople being minted on coins and would not reappear until the Spanish peseta, which itself was based upon the allegory used during the reign of Hadrian. From then on, the allegory would be made into monuments, statues and reliefs.Allegory of Isabella d'Este's Coronation The Allegory of Isabella d'Este's Coronation is a painting by the Italian Renaissance painter Lorenzo Costa the Elder, dating to about 1505-1506. It is displayed in the Louvre Museum of Paris, France.Allegory of Prudence The Allegory of Prudence (c. 1565–1570) is an oil painting attributed to the Italian artist Titian and his assistants. It is in the National Gallery, London. The painting portrays three human heads, each facing in a different direction, above three animal heads, depicting (from left) a wolf, a lion and a dog. The painting is usually interpreted as operating on a number of levels. At the first level, the different ages of the three human heads represent the "Three Ages of Man" (youth, maturity, old age). The different directions in which they are facing reflect a second, wider concept of Time itself as having a past, present and future. This theme is repeated in the animal heads which, according to some traditions, are associated with those categories of time. The third level, from which the painting has acquired its present name, is suggested by a barely visible inscription, EX PRAETERITO/PRAESENS PRUDENTER AGIT/NE FUTURA ACTIONẼ DETURPET (“From the experience of the past, the present acts prudently, lest it spoil future actions”). It may also be that the human faces are actual portraits of the aged Titian, his son Orazio, and a young cousin, Marco Vecellio, who, like Orazio, lived and worked with Titian. Erwin Panofsky, in his classic exposition, suggests that the painting is specifically associated with the negotiations associated with the passing on of Titian’s property to the younger generations, in the light of his approaching death. So, the painting acts as a visual counsel to the three generations to act prudently in the administration of the inheritance. But Nicholas Penny is highly sceptical of this, and points out discrepancies between the human heads and other evidence of the appearance of the individuals. He doubts it was a personal project of any sort and feels that is "surely more likely that the painting was commissioned".More recently the painting has been explained in quite different ways. Instead of an allegory of prudence, it has been seen as an allegory about sin and penitence. On this view, it amounts to an admission by Titian that his failure to act prudently in his youth and middle age has condemned him to lead a regretful old age.At the other extreme, the painting has been explained as asserting that the prudence which comes with experience and old age is an essential aspect of artistic discrimination and judgement. On this interpretation, the painting therefore acts as a rebuttal of the view that old age is the enemy of artistic achievement. On a more general level, the painting’s depiction of Titian with his assistants Orazio and Marco is also intended as a defence of the prudence of the continuity of the Venetian workshop tradition.Allegory of Vice (Correggio) The Allegory of Vice is an oil on canvas painting by Correggio dating to around 1531 and measuring 149 by 88 cm. It and Allegory of Virtue were painted as a pair for the studiolo of Isabella d'Este, with Vice probably the second of the two to be completed. This hypothesis is since only one (possibly non-autograph) sketch survives for Vice, unlike Virtue, for which several preparatory studies survive, along with a near-complete under-drawing - this suggests Correggio had become more proficient after the difficult gestation of Virtue.Influenced by the Laocoon (as is Corregio's treatment of Saint Roch in his San Sebastiano Madonna and Four Saints), the central male figure is sometimes identified as a personification of Vice but sometimes as Silenus (possibly from Virgil's Eclogues 6, where a sleeping Silenus is tied up by the shepherds Chromi and Marsillo and forced to sing by them and the nymph Egle) or Vulcan. It was even misidentified as Apollo and Marsyas by the writer of the Gonzaga collection inventory of 1542. This misunderstanding may have contributed to an Apollo and Marsyas (actually by the studio or circle of Bronzino) being historically misattributed to Correggio. The putto in the foreground is influenced by Raphael's putti in the Sistine Chapel. In 1542, after Isabella's death, they were both recorded as hanging on either side of the entrance door "in the Corte Vecchia near the grotto", with Vice on the left and Virtue on the right. After the contents of her studiolo were dispersed, it remained in Mantua at least until 1627, but the following year it was sold to Charles I of Great Britain. After his execution it was purchased by cardinal Mazarin in 1661 and later by the banker Everhard Jabach, who later sold it to Louis XIV in Paris, reuniting it with Virtue. They both now hang in the Louvre.Allegory of Virtue (Correggio) The Allegory of Virtue is an oil on canvas painting by Correggio dating to around 1531 and measuring 149 by 88 cm. It and Allegory of Vice were painted as a pair for the studiolo of Isabella d'Este, with Vice probably the second of the two to be completed. This hypothesis is since only one (possibly non-autograph) sketch survives for Vice, unlike Virtue, for which two preparatory studies survive (in the Louvre), along with a near-complete oil sketch (attributed to Correggio in the 1603 inventory of the Aldobrandini collection and now at the Galleria Doria Pamphili) - this suggests Correggio had become more proficient after the difficult gestation of Virtue. As usually interpreted, the central woman is Minerva, holding a read lance and a plumed helmet - the work may even be a continuation of Mantegna's Triumph of the Virtues, painted for the same studiolo and also featuring a Minerva with a red lance. (Others have interpreted the figure as Isabella herself, dressed as Wisdom.) Glory hovers above her holding a crown, whilst a seated female figure to the right is surrounded by symbols of the four cardinal virtues (a snake in her hair for Prudence, a sword for Justice, reins for temperance and Hercules's lion skin for Fortitude). Some interpret the seated black female figure on the right as Astrology, Science or Intellectual Virtue - she points outside the painting's space and thus (like the putto in Vice) draws the viewer's attention from one painting to the other. After the studiolo's contents was dispersed, Virtue and the Mantegna were given to cardinal Richelieu around 1627 and moved to Paris. There they were acquired by Eberhard Jabach in 1671, before being sold by him to Louis XIV - Virtue still hangs in the Louvre.Allegory of Wealth Allegory of Wealth is a circa 1640 painting by the French Baroque artist Simon Vouet. Allegory of Wealth is its traditional title, though Nicolas Milovanovic argues that it should instead be entitled Allegory of Contempt for Wealth and the Louvre (where it now hangs) entitles it Allegory of Faith and of Contempt for Wealth Probably painted for Louis XIII's château at Saint-Germain-en-Laye, it is first mentioned in the French royal collection inventories early in the 18th century as Victory crowned with laurels holding in her arms an infant with a sash and an infant holding bracelets and precious stones. Frédéric Villot entitled it La Richesse in the mid 19th century and this title was not contested until 2015.Allegory of the Cave The Allegory of the Cave, or Plato's Cave, was presented by the Greek philosopher Plato in his work Republic (514a–520a) to compare "the effect of education (παιδεία) and the lack of it on our nature". It is written as a dialogue between Plato's brother Glaucon and his mentor Socrates, narrated by the latter. The allegory is presented after the analogy of the sun (508b–509c) and the analogy of the divided line (509d–511e). All three are characterized in relation to dialectic at the end of Books VII and VIII (531d–534e). Plato has Socrates describe a group of people who have lived chained to the wall of a cave all of their lives, facing a blank wall. The people watch shadows projected on the wall from objects passing in front of a fire behind them, and give names to these shadows. The shadows are the prisoners' reality. Socrates explains how the philosopher is like a prisoner who is freed from the cave and comes to understand that the shadows on the wall are not reality at all, for he can perceive the true form of reality rather than the manufactured reality that is the shadows seen by the prisoners. The inmates of this place do not even desire to leave their prison, for they know no better life. The prisoners manage to break their bonds one day, and discover that their reality was not what they thought it was. They discovered the sun, which Plato uses as an analogy for the fire that man cannot see behind. Like the fire that cast light on the walls of the cave, the human condition is forever bound to the impressions that are received through the senses. Even if these interpretations (or, in Kantian terminology, intuitions) are an absurd misrepresentation of reality, we cannot somehow break free from the bonds of our human condition—we cannot free ourselves from phenomenal state just as the prisoners could not free themselves from their chains. If, however, we were to miraculously escape our bondage, we would find a world that we could not understand—the sun is incomprehensible for someone who has never seen it. In other words, we would encounter another "realm", a place incomprehensible because, theoretically, it is the source of a higher reality than the one we have always known; it is the realm of pure Form, pure fact.Socrates remarks that this allegory can be paired with previous writings, namely the analogy of the sun and the analogy of the divided line.Battle of Lepanto The Battle of Lepanto was a naval engagement that took place on 7 October 1571 when a fleet of the Holy League, led by the Venetian Republic and the Spanish Empire, inflicted a major defeat on the fleet of the Ottoman Empire in the Gulf of Patras. The Ottoman forces were sailing westward from their naval station in Lepanto (the Venetian name of ancient Naupactus Ναύπακτος, Ottoman İnebahtı) when they met the fleet of the Holy League which was sailing east from Messina, Sicily. The Holy League was a coalition of European Catholic maritime states which was arranged by Pope Pius V and led by John of Austria. The league was largely financed by Philip II of Spain, and the Venetian Republic was the main contributor of ships.In the history of naval warfare, Lepanto marks the last major engagement in the Western world to be fought almost entirely between rowing vessels, namely the galleys and galeasses which were the direct descendants of ancient trireme warships. The battle was in essence an "infantry battle on floating platforms". It was the largest naval battle in Western history since classical antiquity, involving more than 400 warships. Over the following decades, the increasing importance of the galleon and the line of battle tactic would displace the galley as the major warship of its era, marking the beginning of the "Age of Sail". The victory of the Holy League is of great importance in the history of Europe and of the Ottoman Empire, marking the turning-point of Ottoman military expansion into the Mediterranean, although the Ottoman wars in Europe would continue for another century. It has long been compared to the Battle of Salamis, both for tactical parallels and for its crucial importance in the defense of Europe against imperial expansion. It was also of great symbolic importance in a period when Europe was torn by its own wars of religion following the Protestant Reformation, strengthening the position of Philip II of Spain as the "Most Catholic King" and defender of Christendom against Muslim incursion. Historian Paul K. Davis writes that, "More than a military victory, Lepanto was a moral one. For decades, the Ottoman Turks had terrified Europe, and the victories of Suleiman the Magnificent caused Christian Europe serious concern. The defeat at Lepanto further exemplified the rapid deterioration of Ottoman might under Selim II, and Christians rejoiced at this setback for the Ottomans. The mystique of Ottoman power was tarnished significantly by this battle, and Christian Europe was heartened."Chariot Allegory See also the chariot allegory in the Indian work Katha Upanishad, and another in the story of Vajira.Plato, in his dialogue Phaedrus (sections 246a–254e), uses the Chariot Allegory to explain his view of the human soul. He does this in the dialogue through the character of Socrates, who uses it in a discussion of the merit of Love as "divine madness".Chinese folklore Chinese folklore encompasses the folklore of China, and includes songs, poetry, dances, puppetry, and tales. It often tells stories of human nature, historical or legendary events, love, and the supernatural. The stories often explain natural phenomena and distinctive landmarks. Along with Chinese mythology, it forms an important element in Chinese folk religion.Columbia (name) Columbia (; kə-LUM-bee-ə) is the personification of the United States. It was also a historical name used to describe the Americas and the New World. It has given rise to the names of many persons, places, objects, institutions and companies; for example: Columbia University, the District of Columbia (the national capital of the United States), and the ship Columbia Rediviva, which would give its name to the Columbia River. Images of the Statue of Liberty largely displaced personified Columbia as the female symbol of the United States by around 1920, although Lady Liberty was seen as an aspect of Columbia. The District of Columbia is named after the personification, as is the traditional patriotic hymn "Hail Columbia", which is the official vice presidential anthem of the United States Vice President. Columbia is a New Latin toponym in use since the 1730s for the Thirteen Colonies. It originated from the name of Italian explorer Christopher Columbus and from the ending -ia, common in Latin names of countries (paralleling Britannia, Gallia, and others).Italia turrita Italia turrita (pronounced [iˈtaːlja turˈriːta]) is the national personification or allegory of Italy, characterised by a mural crown (hence turrita or "with towers" in Italian) typical of Italian civic heraldry of Medieval communal origin. In broader terms, the crown symbolizes its mostly urban history. She often holds in her hands a bunch of corn ears (a symbol of fertility and reference to the agrarian economy); during the fascist era, she held a bundle of fasces.National personification A national personification is an anthropomorphic personification of a nation or its people. It may appear in political cartoons and propaganda. As a personification it cannot be a real person, of the Father of the Nation type, or one from ancient history who is believed to have been real. Some early personifications in the Western world tended to be national manifestations of the majestic wisdom and war goddess Minerva/Athena, and often took the Latin name of the ancient Roman province. Examples of this type include Britannia, Germania, Hibernia, Helvetia and Polonia. Examples of personifications of the Goddess of Liberty include Marianne, the Statue of Liberty (Liberty Enlightening the World), and many examples of United States coinage. Another ancient model was Roma, a female deity who personified the city of Rome and more broadly, the Roman state, and who was revived in the 20th Century as the personification of Mussolini's "New Roman Empire". Examples of representations of the everyman or citizenry in addition to the nation itself are Deutscher Michel, John Bull and Uncle Sam.Parable A parable is a succinct, didactic story, in prose or verse that illustrates one or more instructive lessons or principles. It differs from a fable in that fables employ animals, plants, inanimate objects, or forces of nature as characters, whereas parables have human characters. A parable is a type of analogy.Some scholars of the canonical gospels and the New Testament apply the term "parable" only to the parables of Jesus, though that is not a common restriction of the term. Parables such as "The Prodigal Son" are central to Jesus's teaching method in the canonical narratives and the apocrypha.Parabola Allegory The Parabola Allegory is a Rosicrucian allegory, of unknown authorship, dating from the latter part of the seventeenth century. It is sometimes attributed to German alchemist Henricus Madathanus.Bearing many similarities to The Chymical Wedding of Christian Rosenkreutz, it is steeped in alchemical imagery. It deals with the journey of initiation of an unknown narrator, who, after many trials, enters the Rose Garden and bears witness to the dissolution and reconstitution of a pair of royal lovers into a King and Queen. Like The Chymical Wedding, the Parabola Allegory has the haunting quality of a dream. It was taken as the starting point by Viennese psychologist Herbert Silberer for an analysis of Freudian dream interpretation, in his major work Problems of Mysticism and Its Symbolism, where the Allegory is quoted in full. Silberer interprets the Allegory along Freudian lines then, pointing out the limitations of such an approach, goes on to interpret the narrative along alchemical/mystical lines, placing the story in the context of the Mystery traditions of the world's religions as an allegory of the Unio Mystica.Personification Personification is an anthropomorphic metaphor in which a thing or abstraction is represented as a person. The type of personification discussed here excludes passing literary effects such as "Shadows hold their breath", and covers cases where a personification appears as a character in literature, or a human figure in art. The technical term for this, since ancient Greece, is prosopopoeia. In the arts many things are commonly personified. These include numerous types of places, especially cities, countries and the four continents, elements of the natural world such as the months or Four Seasons, Four Elements, Four Winds, Five Senses, and abstractions such as virtues, especially the four cardinal virtues and sins, the nine Muses, or death. In many polytheistic early religions, deities had a strong element of personification, suggested by descriptions such as "god of". In ancient Greek religion, and the related Ancient Roman religion, this was perhaps especially strong, in particular among the minor deities. Many such deities, such as the tyches or tutelary deities for major cities, survived the arrival of Christianity, now as symbolic personifications stripped of religious significance. An exception was the winged goddess of Victory, Victoria/Nike, who developed into the visualization of the Christian angel.Generally, personifications lack much in the way of narrative myths, although classical myth at least gave many of them parents among the major Olympian deities. The iconography of several personifications "maintained a remarkable degree of continuity from late antiquity until the 18th century". Female personifications tend to outnumber male ones, at least until modern national personifications, many of which are male. Personifications are very common elements in allegory, and historians and theorists of personification complain that the two have been too often confused, or discussion of them dominated by allegory. Single images of personifications tend to be titled as an "allegory", arguably incorrectly. According to Ernst Gombrich, "we tend to take it for granted rather than to ask questions about this extraordinary predominantly feminine population which greets us from the porches of cathedrals, crowds around our public monuments, marks our coins and our banknotes, and turns up in our cartoons and our posters; these females variously attired, of course, came to life on the medieval stage, they greeted the Prince on his entry into a city, they were invoked in innumerable speeches, they quarrelled or embraced in endless epics where they struggled for the soul of the hero or set the action going, and when the medieval versifier went out on one fine spring morning and lay down on a grassy bank, one of these ladies rarely failed to appear to him in his sleep and to explain her own nature to him in any number of lines".Primavera (painting) Primavera (Italian pronunciation: [primaˈveːra], meaning "Spring"), is a large panel painting in tempera paint by the Italian Renaissance painter Sandro Botticelli made in the late 1470s or early 1480s (datings vary). It has been described as "one of the most written about, and most controversial paintings in the world", and also "one of the most popular paintings in Western art".The painting depicts a group of figures from classical mythology in a garden, but no story has been found that brings this particular group together. Most critics agree that the painting is an allegory based on the lush growth of Spring, but accounts of any precise meaning vary, though many involve the Renaissance Neoplatonism which then fascinated intellectual circles in Florence. The subject was first described as Primavera by the art historian Giorgio Vasari who saw it at Villa Castello, just outside Florence, by 1550.Although the two are now known not to be a pair, the painting is inevitably discussed with Botticelli's other very large mythological painting, The Birth of Venus, also in the Uffizi. They are among the most famous paintings in the world, and icons of the Italian Renaissance; of the two, the Birth is even better known than the Primavera. As depictions of subjects from classical mythology on a very large scale they were virtually unprecedented in Western art since classical antiquity. It used to be thought that they were both commissioned by the same member of the Medici family, but this is now uncertain. The history of the painting is not certainly known, though it seems to have been commissioned by one of the Medici family. It draws from a number of classical and Renaissance literary sources, including the works of the Ancient Roman poet Ovid and, less certainly, Lucretius, and may also allude to a poem by Poliziano, the Medici house poet who may have helped Botticelli devise the composition. Since 1919 the painting has been part of the collection of the Uffizi Gallery in Florence, Italy.The Allegory of Love For the group of paintings known by this title, see The Allegory of Love (Veronese). The Allegory of Love: A Study in Medieval Tradition (1936), by C. S. Lewis (ISBN 0192812203), is an exploration of the allegorical treatment of love in the Middle Ages and the Renaissance, which was published on May 21, 1936. In the first chapter, Lewis traces the development of the idea of courtly love from the Provençal troubadours to its full development in the works of Chrétien de Troyes. It is here that he sets forth a famous characterization of "the peculiar form which it [courtly love] first took; the four marks of Humility, Courtesy, Adultery, and the Religion of Love"—the last two of which "marks" have, in particular, been the subject of a good deal of controversy among later scholars. In the second chapter, Lewis discusses the medieval evolution of the allegorical tradition in such writers as Bernard Silvestris and Alain de Lille. The remaining chapters, drawing on the points made in the first two, examine the use of allegory and personification in the depiction of love in a selection of poetic works, beginning with the Roman de la Rose. The focus, however, is on English works: the poems of Chaucer, Gower's Confessio Amantis and Usk's Testament of Love, the works of Chaucer's epigones, and Spenser's Faerie Queene. The book is ornamented with quotations from poems in many languages, including Classical and Medieval Latin, Middle English, and Old French. The piquant English translations of many of these are Lewis's own work.<\|endoftext\|>	3.6875
1,900	So far, our revision prep for punctuation has focused on the comma, which makes sense. Misuse of commas is the single most common mistake made in writing, and SHSAT writers will take advantage of those mistakes and see if you can identify them. But, there are many other punctuation mistakes that are made in writing. You will want to Power Up your revision skills by adding knowledge of these best practices in revising. You might not have known that this common symbol, &, is called the ‘ampersand’ but now you do! The ampersand is a symbol that stands for the word “and”. Although the ampersand is used frequently in texting, there aren’t many good uses in writing for the ampersand. If you see the ampersand in writing, be skeptical! But, there are two appropriate instances in writing when you can revise a text and use the ampersand. First, use the ampersand in the names of some businesses and organizations. Law practices frequently use ampersands to identify partnerships, for example: Smith, Sutherland, Sasseen & Hernandez. You can also use the ampersand to identify when two or more people collaborate on a creative product. Authors, for example, can be referenced using the ampersand: M. Astell, M. Cavendish, P. Lord & W. Grommit. Except for these two cases, however, you will revise your sentences by spelling out the word “and” to join ideas together. You are most familiar with apostrophes being used to show possession—and that is their most frequent use in good writing. (You probably have guessed that the most common mistake in apostrophe use is using “it’s” as a possession of ‘it’. But, “it’s” is the only use of the apostrophe of a pronoun that does not show possession. “It’s” is a contraction, instead, of “it is”. To show possession of “it”, you will just revise to “its”.) But there are other uses of the apostrophe besides showing possession. Master the apostrophe, and you’ll be in fantastic shape to revise for the SHSAT! Contractions. The apostrophe is used to contract two ideas into a single, unified idea. You probably will not see a lot of apostrophes being used for contractions, for one simple reason: academic writing usually does not use contractions. Contractions make writing seem more conversational—more similar to how we talk—and so aren’t usually used on reading passages like the SHSAT. But, if you did run into a contraction on the exam, it probably would be for variations of words will, has, and was, contracted with not (so, won’t, hasn’t, haven’t, wasn’t, etc.). Dates. Many middle schoolers believe that the apostrophe is required when writing about events that occurred over time, written with a year followed by the letter ‘s’. For example, it is common—but wrong—to write the following: “With the advent of Stranger Things and Dark, you can only conclude that we are back in love with 1980’s culture.” But, culture doesn’t belong to the 1980s in this sentence—the 1980s doesn’t possess culture, so we do not use an apostrophe. The only time you would revise a date to include an apostrophe is if you are contracting out the century and focusing on the decade. So, it would be appropriate to write: “With the advent of Stronger Things and Dark, you can only conclude that we are back in love with the ‘80s.” In this example, the apostrophe is not possessive. Instead, it stands in place of ‘19’ as a century-marker. Brackets ([ ]) Brackets are similar to parentheses within a sentence. They set information off, away from the main idea, and move the reader to think about something different for a moment. But, brackets should not be used in place of parentheses in good writing. Only use brackets to stand in for words that have been intentionally changed from a passage or to show that there are words missing. A common use of brackets today is to change texts that rely on a male pronoun to more inclusive pronouns. Take the following original quote, from Shakespeare sentence, “A fool thinks himself to be wise, but a wise man knows himself to be a fool.” If you were asked to revise this to include gender-neutral language, you would use brackets to show where you made a change: “A fool thinks [themselves] to be wise, but a wise [person] knows [themselves] to be a fool.” The brackets show where a slight alteration has been made in the passage, and lets the reader know that you, as someone who has approached this quote, has revised it. (It also allows the reader to disagree with how you changed the text!) Colons, almost always, are used to introduce either a quote or a list of ideas. Colons are placed directly against the word prior to the colon (without spaces), and you won’t capitalize the word that comes after the colon unless it is the beginning of a first word in a whole sentence that is quoted, or a proper noun. You could revise this sentence, which uses a colon incorrectly: I asked Shannon to get Bananas: Apples: Cherries; and Pickles. Your revision would ensure that the list is started with the colon and capitalized correctly: I asked Shannon to get: bananas, apples, cherries, and pickles. Did you know that the three periods in a row (…) used by thought bubbles in graphic novels and cartoons are called ellipses? The ellipsis is used in all types of writing to show that there is a long pause. (Whereas the comma and semicolon indicate a short pause, the ellipsis indicates a longer pause in the writing, and often stands for reflection. If you say, It’s stunning to think what might have happened…if…only…. then there are several ideas communicated. Something else could have happened—if—something else occurred. Something else could have happened if—only—something else occurred! But, also, you are thinking deeply about what could have happened if only something else occurred. The ellipsis at the end of the sentence (followed by the period) means that your idea is concluded by continued thought about this stunning idea! One other way that ellipses are used is to show that there is part of a quotation that has been removed from a passage because that material is not important to the overall meaning of the passage. (Ellipses should not be used to take out text that is not convenient for your view!) Consider this longer passage, from philosopher Rene Descartes (who famously said, “I think, therefore, I am!”: “The fact that I can vividly and clearly think of one thing apart from another assures me that the two things are distinct from one another—that is, that they are two—since they can be separated by God. Never mind how they could be separated; that does not affect that judgement that they are distinct. So, my mind is a distinct thing from my body.” This passage actually tells us where we can use an ellipsis to shorten it! (“Never mind how they could be separate,” etc.). So, we can edit this to say: “The fact that I can vividly and clearly think of one thing apart from another assures me that the two things are distinct from one another—that is, that they are two—since they can be separated by God….So, my mind is a distinct thing from my body.” The ellipsis demonstrates that there has been information (that the reviser believes is not crucial to the argument Descartes makes). It is easy to confuse commas with semicolons, because they both represent pauses in the text. Don’t use semicolons in place of period and instead of commas. If you have two separate ideas, use a period between the ideas. If you have two closely related ideas that could be connected by an “and”, use the comma. But, if you have two or more closely related ideas that are independent of each other, use the semicolon. For example: John worked for the Secretary of State; this job is less dangerous. The writer here wants to strongly connect John’s two different jobs. They could be separated by a sentence, but the meaning would be that the first job (for the Secretary of State) is not connected to this job. The way the sentence is written, they are connected and the fact that this job is less dangerous may be a reason why John wanted this job.<\|endoftext\|>	3.671875
1,392	A mineral's luster is the general appearance of its surface in reflected light. There are two broad types of luster: metallic and nonmetallic. Metallic luster is that of an untarnished metal surface, such as gold, steel, copper, galena, pyrite, and hematite. Minerals with metallic luster can also be described as having a "shiny", "dull", or "iridescent" luster. For example, the pyrite mineral shown in the left photo has mostly a shiny, metallic luster. Minerals of metallic luster are opaque to light, even on thin edges. By contrast, minerals with non-metallic luster are generally lighter in color and show some degree of transparency or translucency, even though this may be on a thin edge. There are a number of terms that describe non-metallic luster, such as vitreous (having the luster of a broken piece of glass); adamantine (having the brilliant luster of a diamond); resinous (having the luster of a piece of resin); pearly (having the luster of a pearl or mother-of-pearl); greasy (appearing to be covered with a thin layer of oil); silky (appearing as the surface of silk or satin); dull (producing little or no reflection); and earthy (having a non-lustrous appearance of raw Earth. The quartz minerals shown in the left photo represent both opaque crystal quartz and a variation of quartz called amethyst. They both have a vitreous appearance. The photo to the left represents rose quartz. Note that this particular one has a "greasy appearance". Rose quartz may also have a vitreous appearance, therefore it takes a little practice to notice the subtle differences between vitreous luster and greasy luster. Calcite may also have a vitreous or greasy luster. In fact, all silicate minerals will generally have one or the other of the two. For example, when comparing the two minerals in their photos, both appear to resemble glass; however, the rose quartz has the shiny-but-blurry appearance as if had been covered with a thin layer of oil. Now this microcline feldspar to the left has a non-reflective appearance, thus displaying a dull luster. More than often, distinguishing dull luster from earthy luster can be difficult for beginners. Earthy luster, represented in the next photo, has a granular appearance; whereas, dull luster does not. Recall that metallic luster can also have a dull appearance and therefore be classified as having a dull metallic luster. However, it is clearly evident that this microcline feldspar does not have a metallic appearance. In some cases, a mineral may have a combination of two non-metallic lusters. If this should be the case, the one most dominant will be classified. This azurite mineral has a granular texture with a freshly broken appearance; therefore, it displays an earthy luster. In some cases, earthy minerals look like dirt or dried mud, while others may be rough and porous in texture. Earthy luster may also look like unglazed pottery. Other minerals which may display earthy luster include malachite, kaolinite, pyrolusite, limonite, and bauxite. In such minerals, relatively little light is reflected from the surface and they lack the shiny appearance of a metallic or glassy appearance. Resinous minerals have the appearance of resin, chewing gum or smooth-surfaced plastic. A common mineral that represents a resinous luster, such as the one in the left photo, is amber. Amber is simply a form of fossilized resin that sometimes contains insects, very small lizards, and pollen/spores. Most often, these type of minerals have a yellow, dark-orange or brown color which resembles honey, but not necessarily the same color. Minerals with resinous luster can show a transparent, translucent, or opaque reflectivity. Other minerals that display resinous luster may include sulfur, anglesite, and wulfenite. Minerals, such as barite (left photo) exhibit a luster similar to the inside of a mollusk shell. Many micas display a pearly luster, and some minerals which have a pearly luster also have an iridescent hue. In some cases, minerals, such as calcite and aragonite may display a pearly luster on cleavage cracks parallel and below the their reflecting surface. Many pearly minerals consist of thin, transparent co-planar sheets. Light reflecting from these layers give them a luster resembling that of a pearl. Pearly luster is a dominant characteristic for opals and moonstones. Silky luster is produced by a mineral's shiny, fibrous body appearance. Minerals with silky luster possess microscopic inclusions (very fine fibrous structures), causing them to display similar optical properties of silk cloth. Such fibrous structures may occur within the mineral and/or aggregates on its surface. This type of luster is best seen on rough specimens of tiger's eye. However, when polished, tiger's eye exhibits a silky-vitreous luster known as "silky sheen" luster. Other examples include asbestos, aragonite, satin-spar gypsum, and actinolite (left photo). Adamantine minerals possess a brilliant, superlative luster which is most notably observed in diamonds. Such minerals are transparent or translucent. Minerals with "true" adamantine luster are uncommon and thus are more valuable. Those minerals exhibiting less luster may still be of value; however, they are referred to as subadamantine minerals. The two crystals in the left photo are known as Herkimer Diamond Quartz. Due to their brilliant appearance and high degree of quality, they are commonly known for displaying adamantine luster. As the name implies, waxy minerals have a luster resembling the surface of a wax candle or beeswax. Examples include jade, carnelian, chalcedony (left photo), opal and turquoise. Such minerals have a translucent to opaque reflectivity and may come in a variety of colors. Waxy luster resembles greasy and resinous luster; however, the surfaces on which it is seen are more irregular. The chalcedony mineral (left photo) has formed in globular aggregates, resembling a bunch of grapes. This type of crystal habit is referred to as botryoidal. Many waxy luster minerals possess such form.<\|endoftext\|>	3.921875
788	Fiber is a substance found in plants. Dietary fiber, which is the type of fiber you can eat, is found in fruits, vegetables, and grains. It is an important part of a healthy diet. There are two forms of fiber: soluble and insoluble. Soluble fiber attracts water and turns to gel during digestion. This slows digestion. Soluble fiber is found in oat bran, barley, nuts, seeds, beans, lentils, peas, and some fruits and vegetables. Research has shown that soluble fiber lowers cholesterol, which can help prevent heart disease. Insoluble fiber is found in foods such as wheat bran, vegetables, and whole grains. It appears to speed the passage of foods through the stomach and intestines and adds bulk to the stool. Eating a large amount of fiber in a short period of time can cause intestinal gas (flatulence), bloating, and abdominal cramps. This problem often goes away once the natural bacteria in the digestive system get used to the increase in fiber. Adding fiber to the diet slowly, instead of all at one time, can help reduce gas or diarrhea. Too much fiber may interfere with the absorption of minerals such as iron, zinc, magnesium, and calcium. In most cases, this is not a cause for too much concern because high-fiber foods tend to be rich in minerals. On average, Americans now eat about 16 grams of fiber per day. The recommendation for older children, adolescents, and adults is to eat 21 to 38 grams of fiber each day. Younger children will not be able to eat enough calories to achieve this amount, but it is a good idea to introduce whole grains, fresh fruits, and other high-fiber foods. To ensure that you get enough fiber, eat a variety of foods, including: - Dried beans and peas - Whole grains Add fiber gradually over a period of a few weeks to avoid stomach distress. Water helps fiber pass through the digestive system. Drink plenty of fluids (about 8 glasses of water or noncaloric fluid a day). Taking the peels off fruits and vegetables reduces the amount of fiber you get from the food. Fiber-rich foods offer health benefits when eaten raw or cooked. Diet - fiber; Roughage; Bulk; Constipation - fiber Hoy MK, Goldman JD. Fiber intake of the US population. What we eat in America, NHANES 2009-2010. Food Surveys Research Group. Dietary Data Brief No. 12. September 2014. www.ars.usda.gov/ARSUserFiles/80400530/pdf/DBrief/12_fiber_intake_0910.pdf. Accessed July 25, 2018. Heimburger DC. Nutrition's interface with health and disease. In: Goldman L, Schafer AI, eds. Goldman-Cecil Medicine. 25th ed. Philadelphia, PA: Elsevier Saunders; 2016:chap 213. Thompson M, Noel MB. Nutrition and family medicine. In: Rakel RE, Rakel DP, eds. Textbook of Family Medicine. 9th ed. Philadelphia, PA: Elsevier; 2016:chap 37. US Department of Health and Human Services. US Department of Agriculture. 2015-2020 Dietary Guidelines for Americans. 8th ed. health.gov/dietaryguidelines/2015/guidelines/. Accessed July 25, 2018. Review Date 6/28/2018 Updated by: Linda J. Vorvick, MD, Clinical Associate Professor, Department of Family Medicine, UW Medicine, School of Medicine, University of Washington, Seattle, WA. Also reviewed by David Zieve, MD, MHA, Medical Director, Brenda Conaway, Editorial Director, and the A.D.A.M. Editorial team.<\|endoftext\|>	3.6875
541	Land-cover changes likely intensified Dust Bowl drought Dramatic human-caused changes in land cover between 1850 and the 1930s had a substantive effect on the 1930s Dust Bowl drought in the Great Plains, a new study by University of Nebraska–Lincoln researchers finds. Atypical summer precipitation patterns in the central United States are influenced by sea surface temperature anomalies in the Pacific Ocean, such as El Niño or La Niña, and by those in the North Atlantic. La Niña has been a typical driver for drier-than-normal conditions leading to drought. "But we show those anomalies are not the full story," said Qi Hu, lead author of the study and agricultural climatologist at Nebraska. "We show that land cover is equally important in magnitude as sea surface temperature anomalies. It can cause and prolong severe drought, even without SST forces at play." The researchers tracked down available data on land cover in the Great Plains from 1850 to 1935. They found 46 percent of native grasslands and 75 percent of savannas in the Great Plains had been plowed under and converted to dryland cropland and pastures by the early 1930s. They then designed climate-model simulations to quantify the effect of such land-cover change on summer precipitation. "The results all showed the same thing: reduced summer precipitation in the Great Plains" during the Dust Bowl years, Hu said. "The '30s land cover was the dominant cause because there wasn't any strong or persistent La Niña in the '30s, and the North Atlantic was fairly quiet, too.” The researchers hypothesize that the change in land cover altered the surface pressure distribution from the central United States to the east, effectively weakening the jet stream that brings moisture to the Plains from the Gulf of Mexico. "It undercut the moisture supply to the region," Hu said. “With no supply, there was no rain." The intensity of the Dust Bowl drought worsened, eventually causing nearly 2.5 million people to flee the drought-stricken states of Texas, New Mexico, Colorado, Nebraska, Kansas and Oklahoma in the largest migration in American history. Hu said further research would be needed to gain more insight into what initiated the drought. The researchers recommend carefully considered policies on land use to prevent future disasters of the Dust Bowl's magnitude. The study was funded by the National Science Foundation and published in the American Meteorological Society’s Journal of Climate. Hu authored the study with Matthew Van den Broeke, associate professor of Earth and atmospheric sciences, and Jose Torres-Alavez, graduate student in Earth and atmospheric sciences.<\|endoftext\|>	3.734375
1,297	Instasolv IIT-JEE NEET CBSE NCERT Q&A 4.5/5 # RD Sharma Class 12 Chapter 28 Solutions (Straight Line In Space) RD Sharma Class 12 Maths Solutions Chapter 28 ‘Straight Line in Space’ teaches you all the concepts about a straight line in space. In this chapter, you’ll learn about vector and cartesian equations of a line, associated theorems, reduction of the cartesian form of a line to vector form and vice-versa, vector to the cartesian, angle between two lines, the intersection of two lines, an algorithm for vector form, perpendicular distance of a line from a point, the shortest distance between two straight lines, skew lines, line of shortest distance, the shortest distance between two skew lines, the shortest distance between two parallel lines, etc. Straight Line in Space Chapter 28 has a total of 5 exercises and 67 questions. You’ll answer questions like finding the vector equation of a line that satisfies the specific conditions and reducing them to a cartesian form, finding the line’s cartesian equation where the line will satisfy the specific condition and then reducing it to vector, checking collinearity of three points, or finding a point on a line, on finding the angle between two lines, equation of a line falling parallel to a specific line and passing through a given point, etc. Instasolv provides best RD Sharma Class 12 Maths Solutions Chapter 28 ‘Straight Line in Space’. These solutions take every single concept that has been taught in the chapter. These solutions are available free of cost. Helping you to prepare well for your CBSE and all other exams like NEET and JEE, Instasolv is known for its years of experience and portfolio in helping students achieve their goals. ## Topics Discussed in RD Sharma Class 12 Maths Solutions Chapter 28 – Straight Line in Space ### Theorem on Vector and Cartesian Equations of a Line 1. A straight line passes through a fixed point having position vector a and is found parallel to a given vector b. The vector equation will be vector r = vector a ∂vector b where ∂ is a scalar. 2. A straight line passes through a specific point (x1, y1, z1). It has direction ratios found to be proportional with a, b, c. Then the cartesian equation of the given straight line will be (x-x1)/a = (y-y1)/b = (z-z1)/c 1. A line passes through points having position vectors a and b. The vector equation will be vector r = vector a + ∂(vector b – vector a) 2. A line passes through the points(x1, y1, z1) and (x2, y2, z2). Then its cartesian equation will be (x-x1)/(x2-x1) = (y-y1)/(y2-y1) = (z-z1)/(z2-z1) The intersection of two lines Here is an algorithm that you can use to check if the given lines meet or not and if they meet, a method to find the intersection point. Algorithm Consider the two lines as (x-x1)/a1 = (y-y1)/b1 = (z-z1)/c1 …(1) (x-x2)/a2 = (y-y2)/b2 = (z-z2)/c2 …(2) • Mention the general points’ coordinates on equation 1 and 2 Let’s call them ∂ and µ respectively. Which gives (a1∂ +x1, b1∂ + y1, c1∂+z1) and (a2µ +x2, b2µ + y2, c2µ + z2) • If the two lines 1 and 2 meet, they will have a point in common. Therefore, (a1∂ + x1) = (a2µ + x2) (b1∂ + y1) = (b2µ + y2) (c1∂ + z1) = (c2µ + z2) 1. You will be required to solve two equations out of three in ∂ and µ provided in the above step. If the values of the two variables satisfy the third equation then the above lines i.e. 1 and 2 will meet or in other words intersect. However, if the value of ∂ and µ do not satisfy the third equation, then the lines 1 and 2 do not meet or in other words do not intersect. 2. To find the coordinates of the intersection point, but the value of ∂ and/or µ in general point(s)’ coordinates that you calculated in the first step. ### Discussion of Exercises in RD Sharma Class 12 Maths Solutions Chapter 28 – Straight Line in Space 1. Exercise 28.1 discusses methods to find vector and cartesian equations of a given line, finding the vector equation of the line, finding the equation of the line in both the cartesian and vector form, finding the points of a given line, etc. 2. Exercise 28.2 asks questions based on the angle between the two lines, the equation of a line that is parallel to a specific line and passes through a given point, or finding the equation of a line that passes through a specific point and is perpendicular to the two lines, etc. 3. Exercise 28.3 talks about the intersection of two lines, the algorithm for vector form, and more. 4. Exercise 28.4 discusses the perpendicular distance of a given line from a specific point, vector form, cartesian form, etc. 5. Exercise 28.5 discusses the shortest distance between two straight lines, two skew lines, two parallel lines, etc. ### Benefits of RD Sharma Class 12 Maths Solutions Chapter 28 – Straight Line in Space By Instasolv Access RD Sharma Solutions for Class 12 Maths Solutions Chapter 28 ‘Straight Line in Space’ any time from anywhere and make your preparations easier. Our stepwise solutions will make you understand the complete methodology of getting the right answers in this chapter. Our experts are well-aware of the CBSE class 12 syllabus and have framed the solutions accordingly. These solutions are error-free and easy. More Chapters from Class 12<\|endoftext\|>	4.625
859	Microarrays provide a way of organizing biological samples for high-throughput analysis. Samples are arranged in columns and rows upon a support surface consisting of a glass slide, a nitrocellulose membrane or a microtiter plate. Credit: Sergei Drozd/ Shutterstock.com Protein microarrays are used to determine the function of proteins, as well as to monitor their interactions. The structure of the array allows for numerous proteins to be tracked in parallel. Protein microarrays were developed by utilizing the technology of DNA microarrays, which are commonly used to analyze gene expression. The requirements of protein microarrays are, however, more complex and necessitate material customization to make them suitable. Even though they are not as commonly seen as DNA microarrays, protein microarrays are increasingly being used for multiple applications in the biosciences. Application for detecting protein interactions One of the earliest applications of protein microarrays was for the detection of protein-binding properties. Test ligands are directly or indirectly labelled with fluorescent dyes to analyze protein-protein interactions. One of the proteins in the pair being tested is arrayed in multiple samples on slides and each sample is probed with a different fluorescently labelled protein. Interactions are identified in the sample spots where the fluorescent label is visible. Protein microarrays have been utilized for monitoring the interactions between: - Different proteins such as enzymes and substrates - Proteins and antibodies - Proteins and lipids - Proteins and peptides - Proteins and low-molecular mass compounds - Proteins and oligosaccharides - Proteins and DNA - Proteins and RNA - Lectin (carbohydrate-binding proteins) and cells Application for clinical diagnostics and prognosis Protein microarrays have also been applied for use in clinical diagnostics and prognosis. Biomarker identification tools can be formed when proteins are treated as potential antigens that can be tested for association with particular diseases. For example, fluorescently labelled anti-human immunoglobulin antibodies are used to detect the pairings of antigens and antibodies. The autoantibody associated with a particular human disease will recognize the human protein sample on the array and the fluorescent label provides a method of identifying the pairing produced. This method creates a profile of autoantibodies associated with the disease. A similar technique has been developed for the detection of infectious diseases. One example is the coronavirus protein microarray produced for diagnosing severe acute respiratory syndrome (SARS). The array included samples of SARS coronavirus proteins and five additional infectious coronaviruses. The presence of the human antibodies against the SARS coronavirus proteins on the microarray led to the ability to distinguish patient serum samples that included the virus with 94% accuracy. In short, protein microarray technology provided a diagnostic method with at least one hundred times the sensitivity than standard diagnostic methods, along with lower sample volume requirements. Application of the reverse phase protein microarray (RPPA) Reverse phase protein microarrays (RPPA) provide a method for application on pure protein lysates isolated from tissue or cultured cells. The technology combines microarrays with laser capture microdissection (LCM). LCM is a method of procuring cells using a laser capable of melting an area between 7.5 and 30 micrometers in diameter. The tissue is stained, placed on a slide then visualized under a microscope in real-time. Cells can be isolated within the diameter of the laser, with more cells obtained by moving the slide. The isolated cells are then lysed in preparation for being placed on the reverse phase protein microarray. The sample spots on the array contain the proteins within a cell that correspond to the given pathological state. While conventional protein arrays immobilize the antibody probe, the reverse phase protein microarray immobilizes the protein being analyzed. Reverse phase protein microarrays have been used for diagnostics, to investigate signal transduction pathways between hormones and the cell membrane, as well as for identifying potential vaccine candidates.<\|endoftext\|>	3.9375
392	123 views In a $10$ km race, $A$, $B$ and $C$, each running at uniform speed, get the gold, silver, and bronze medals, respectively. If $A$ beats $B$ by $1$ km and $B$ beats $C$ by $1$ km, then by how many metres does $A$ beat $C$? 1. $1800$ 2. $1900$ 3. $1500$ 4. $1600$ Let's draw the diagram for a better understanding. We know that, $\text{Speed} = \dfrac{\text{Distance}}{\text{Time}}$ If time is constant, then $\boxed{\text{Speed} \propto \text{Distance}}$ The ratio of the speed of $A$ and $B:$ • $S_{A}:S_{B} =10:9$ The ratio of the speed of $B$ and $C:$ • $S_{B}:S_{C} =10:9$ Now, we can combine the ratios. • $S_{A}:S_{B} =(10:9)\times 10=100:90$ • $S_{B}:S_{C} =(10:9)\times 9=90:81$ $\Rightarrow S_{A}:S_{B}:S_{C} = 100:90:81$ We can write the ratio of Distance, $D_{A}:D_{B}: D_{C} = 10000 : 9000 : 8100.$ $\therefore \; A$ would beat $C$ by $1900$ meters. Correct Answer $:\text{B}$ by 7.7k points 3 8 30 1 145 views 1 vote 2 136 views 1 vote<\|endoftext\|>	4.4375
591	Top Top Quadratic equation is a polynomial equation which is in the form of Ax$^{2}$ + Bx + C = 0 (second degree). Quadratic Equation Calculator (or Quadratic Equation Solver) is an online tool to solve a quadratic equation. If any quadratic equation is given then it can find factors of that equation, i.e. for the equation of the form Ax$^{2}$ + Bx + C = 0, it will find the values of x. This calculator solves the quadratic equation with negative discriminant also, i.e. this calculator can find out imaginary roots (or complex roots) also. ## Steps to Solve the Quadratic Equation Step 1 : Observe the value of a, b and c from the quadratic equation. Step 2 : Find the value of discriminant(D) by applying the formula D = b2 - 4ac. Step 3 : If the value of discriminant is less than zero, write "roots does not exist/ imaginary. And if discriminant is greater or equal to zero, then find the roots of an equation by applying the formula x1= $\frac{-b + \sqrt{D}}{2a}$ and x2= $\frac{-b - \sqrt{D}}{2a}$ ## Examples on Quadratic Equation Calculator 1. ### Find the roots of an equation x2 + 7x + 12 = 0? Step 1 : Here a = 1, b = 7, c = 12 Step 2 : D = b2 - 4ac = 72 - 4 × 1 × 12 = 49 - 48 D = 1 Step 3 : x1 = $\frac{-b + \sqrt{D}}{2a}$ = $\frac{-7 + \sqrt{1}}{2 × 1}$ = $\frac{-6}{2}$ = -3 x2 = $\frac{-b - \sqrt{D}}{2a}$ = $\frac{-7 - \sqrt{1}}{2 × 1}$ = $\frac{-8}{2}$ = -4 x1 = -3 x2 = -4 2. ### Find the root of an equation x2 + 4x + 5 = 0? Step 1 : Here a = 1, b = 4, c = 5 Step 2 : D = b2 - 4ac = 42 - 4 × 1 × 5 = 16 - 20 D = -4 Step 3 : Roots are imaginary, since discriminant is less than zero. Roots are imaginary More Quadratic Equation Calculator Discriminant Calculator *AP and SAT are registered trademarks of the College Board.<\|endoftext\|>	4.375
437	Flash memory is a type of electrically erasable programmable read-only memory [EEPROM] chip that can be used for the transfer and permanent storage of digital data. It is actually a long-term persistent storage computer storage device that can be electrically erased and reprogrammed. In this device, the information is stored as an array of floating gate transistors called cells. Each of these cells can store only one bit of information at a time. The design of each memory cell is somewhat similar to that of a MOSFET. The only difference is that this cell has two gates. The gate on top is called the control gate [CG], and the bottom one is called the floating gate [FG]. FG may be made of conductive materials like poly silicon or can also be non-conductive. These two gates are separated from each other by a thin oxide layer. To know the exact representation, take a look the figure given below. The FG can come in contact with the word line only through CG. When that particular link is closed, the cell will have a value ’1′. In order to change the value to ’0′, a process called tunneling has to be done. This charge cancels the electric field from CG, and thus causes to modify the threshold voltage [V] of the cell. During the read process, a voltage that is below V is applied to CG. This V depends whether the channel should be conducting or insulating, which is in turn controlled by the FG charge. This causes the channel to know the current flow and hence the binary code is formed. This is the method of reproducing the stored data. As told earlier, the flash memory can be read only with one byte at a time. But while erasing it, it must be erased as a block. In a block there are a lot of bytes. If you program any of these bytes and later want to erase it, you will have to erase the whole block. So, in short, the device can offer you random access read and programming operations, it cannot offer you random access rewriting or erasing operations.<\|endoftext\|>	4.125
3,026	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> # Scale and Indirect Measurement Applications ## Multiply by scale factors to measure. Estimated9 minsto complete % Progress Practice Scale and Indirect Measurement Applications MEMORY METER This indicates how strong in your memory this concept is Progress Estimated9 minsto complete % Scale and Indirect Measurement Applications Suppose an architectural firm is constructing a building, and they first build a model that is 4 feet tall. If the scale is 1:125 when compared to the height of the real building, can you figure out how tall the real building is? What proportion would you set up? ### Scale and Indirect Measurements We are occasionally faced with having to make measurements of things that would be difficult to measure directly: the height of a tall tree, the width of a wide river, the height of the moon’s craters, and even the distance between two cities separated by mountainous terrain. In such circumstances, measurements can be made indirectly, using proportions and similar triangles. Such indirect methods link measurement with geometry and numbers. #### Scale A map is a two-dimensional, geometrically accurate representation of a section of the Earth’s surface. Maps are used to show, pictorially, how various geographical features are arranged in a particular area. The scale of the map describes the relationship between distances on the map and the corresponding distances on the Earth's surface. These measurements are expressed as a fraction or a ratio. Recall that there are different ways to write a ratio: using the fraction bar, using a colon, and in words. Outside of mathematics books, ratios are often written as two numbers separated by a colon (:). Here is a table that compares ratios written in two different ways. Ratio Is Read As Equivalent To 1:20 one to twenty (120)\begin{align}\left (\frac{1}{20} \right )\end{align} 2:3 two to three (23)\begin{align}\left (\frac{2}{3} \right )\end{align} 1:1000 one to one-thousand (11000)\begin{align}\left (\frac{1}{1000} \right )\end{align} If a map had a scale of 1:1000 (“one to one-thousand”), one unit of measurement on the map (1 inch or 1 centimeter, for example) would represent 1000 of the same units on the ground. A 1:1 (one to one) map would be a map as large as the area it shows! #### Let's solve the following problems using proportions: 1. Anne is visiting a friend in London and is using the map above to navigate from Fleet Street to Borough Road. She is using a 1:100,000 scale map, where 1 cm on the map represents 1 km in real life. Using a ruler, she measures the distance on the map as 8.8 cm. How far is the real distance from the start of her journey to the end? The scale is the ratio of distance on the map to the corresponding distance in real life and can be written as a proportion. dist.on mapreal dist.=1100,000\begin{align}\frac{\text{dist.on map}}{\text{real dist.}} = \frac{1}{100, 000}\end{align} By substituting known values, the proportion becomes: 8.8 cmreal dist.(x)880000 cmx=1100,000=x=8800 mCross multiply.100 cm=1 m.1000 m=1 km.\begin{align}\frac{8.8 \ cm}{\text{real dist.} (x)} & = \frac{1}{100,000} && \text{Cross multiply}. \\ 880000 \ cm & = x && 100 \ cm = 1\ m. \\ x & = 8800 \ m && 1000 \ m = 1\ km.\end{align} The distance from Fleet Street to Borough Road is 8800 m\begin{align}8800\ m\end{align} or 8.8 km\begin{align}8.8\ km\end{align}. We could, in this case, use our intuition: the 1 cm=1 km\begin{align}1 \ cm = 1 \ km\end{align} scale indicates that we could simply use our reading in centimeters to give us our reading in km. Not all maps have a scale this simple. In general, you will need to refer to the map scale to convert between measurements on the map and distances in real life! 1. Oscar is trying to make a scale drawing of the Titanic, which he knows was 883 feet long. He would like his drawing to be 1:500 scale. How long, in inches, must his sheet of paper be? We can reason that since the scale is 1:500 that the paper must be 883500=1.766 feet\begin{align}\frac{883}{500} = 1.766\ feet\end{align} long. Converting to inches gives the length as 12(1.766) in=21.192 in\begin{align}12 (1.766) \ in = 21.192 \ in\end{align}. The paper should be at least 22 inches long. #### Indirect Measurement Not everything has a scale. Architecture such as the St. Louis Arch, St. Basil’s Cathedral, or the Eiffel Tower does not have a scale written on the side. It may be necessary to measure such buildings. To do so requires knowledge of similar figures and a method called indirect measurement. Similar figures are often used to make indirect measurements. Two shapes are said to be similar if they are the same shape and “in proportion.” The ratio of every measurable length in one figure to the corresponding length in the other is the same. Similar triangles are often used in indirect measurement. #### Let's solve the following problem with indirect measurement: Anatole is visiting Paris, and wants to know the height of the Eiffel Tower. Since he's unable to speak French, he decides to measure it in three steps. 1. He measures out a point 500 meters from the base of the tower, and places a small mirror flat on the ground. 2. He stands behind the mirror in such a spot that standing upright he sees the top of the tower reflected in the mirror. 3. He measures both the distance from the spot where he stands to the mirror (2.75 meters) and the height of his eyes from the ground (1.8 meters). Explain how Anatole is able to determine the height of the Eiffel Tower from these numbers and determine what that height is. First, we will draw and label a scale diagram of the situation. The Law of Reflection states, “The angle at which the light reflects off the mirror is the same as the angle at which it hits the mirror.” Using this principle and the figure above, you can conclude that these triangles are similar with proportional sides. This means that the ratio of the long leg in the large triangle to the length of the long leg in the small triangle is the same ratio as the length of the short leg in the large triangle to the length of the short leg in the small triangle. 500 m2.75 m1.85002.75327.3=x1.8 m=x1.81.8x\begin{align}\frac{500 \ m}{2.75 \ m} & = \frac{x}{1.8 \ m} \\ 1.8 \cdot \frac{500}{2.75} & = \frac{x}{1.8} \cdot 1.8 \\ 327.3 & \approx x\end{align} The Eiffel Tower, according to this calculation, is approximately 327.3 meters high. ### Examples #### Example 1 Earlier, you were told that a model of a building is 4 feet tall. The scale is 1:125 when compared to the height of the real building. How tall is the real building? Let's set the scale up as a ratio: model heightreal height=1125\begin{align}\frac{\text{model height}}{\text{real height}} = \frac{1}{125}\end{align}Now, substitute in known values. model heightreal height=11254real height=1125\begin{align}\frac{\text{model height}}{\text{real height}} = \frac{1}{125}\\ \frac{4}{\text{real height}} = \frac{1}{125}\end{align} To make it easier to read the equation, let's call the real height of the building the variable h\begin{align}h\end{align}. To solve, cross multiply: 4h=11254×125=1×h500=h\begin{align}\frac{4}{h}=\frac{1}{125}\\ 4\times 125 = 1\times h\\ 500 = h\end{align} The real building is 500 feet tall. #### Example 2 Bernard is looking at a lighthouse and wondering how high it is. He notices that it casts a long shadow, which he measures at 200 meters long. At the same time, he measures his own shadow at 3.1 meters long. Bernard is 1.9 meters tall. How tall is the lighthouse? Begin by drawing a scale diagram. You can see there are two right triangles. The angle at which the sun causes the shadow from the lighthouse to fall is the same angle at which Bernard’s shadow falls. We have two similar triangles, so we can again say that the ratio of the corresponding sides is the same. 200 m3.1 m1.9200 m3.1 m122.6=x1.9 m=x1.9 m1.9x\begin{align}\frac{200 \ m}{3.1 \ m} & = \frac{x}{1.9 \ m} \\ 1.9 \cdot \frac{200 \ m}{3.1 \ m} & = \frac{x}{1.9 \ m} \cdot 1.9 \\ 122.6 & \approx x\end{align} The lighthouse is approximately 122.6 meters tall. ### Review 1. Define similar figures. 2. What is true about similar figures? 3. What is the process of indirect measurement? When is indirect measurement particularly useful? 4. State the Law of Reflection. How does this law relate to similar figures? 5. A map has a 1 inch : 20 mile scale. If two cities are 1,214 miles apart, how far apart would they be on this map? 6. What would a scale of 1 mile : 1 mile mean on a map? What problems would the cartographer encounter? 7. A woman 66 inches tall stands beside a tree. The length of her shadow is 34 inches and the length of the tree’s shadow is 98 inches. To the nearest foot, how tall is the tree? 1. Use the scale diagram above to determine: 1. The length of the helicopter (cabin to tail) 2. The height of the helicopter (floor to rotors) 3. The length of one main rotor 4. The width of the cabin 5. The diameter of the rear rotor system 2. On a sunny morning, the shadow of the Empire State Building is 600 feet long. At the same time, the shadow of a yardstick (3 feet long) is 1 foot, 514\begin{align}5 \frac{1}{4}\end{align} inches. How high is the Empire State building? 3. Omar and Fredrickson are 12.4 inches apart on a map with a scale of 1.2 inches : 15 miles. How far apart are their two cities in miles? 4. A man 6 feet tall is standing next to a dog. The man’s shadow is 9 feet and the dog’s shadow is 6 feet. How tall is the dog? 5. A model house is 12 inches wide. It was built with a ratio of 3 inches : 4 meters. How wide is the actual house? 6. Using the diagram below and assuming the two triangles are similar, find DE¯¯¯¯¯¯¯¯\begin{align}\overline{DE}\end{align}, the length of segment DE\begin{align}DE\end{align}. 7. A 42.9-foot flagpole casts a 253.1-foot shadow. What is the length of the shadow of a woman 5 feet 5 inches standing next to the flagpole? Mixed Review 1. Solve for a: (77a)+4a=23+3a\begin{align}a: \ -(7-7a)+4a=-23+3a\end{align}. 3. Simplify 243\begin{align}\sqrt{243}\end{align}. 4. Simplify: 2(8g+2)1+4g(25g)\begin{align}2(8g+2)-1+4g-(2-5g)\end{align}. 5. Draw a graph that is not a function. Explain why your picture is not a function. 6. Jose has 23\begin{align}\frac{2}{3}\end{align} the amount of money that Chloe has. Chloe has four dollars less than Huey. Huey has \$26. How much does Jose have? ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes ### Vocabulary Language: English Spanish indirect measurement Measuring indirectly is a technique that uses similar figures and proportions to find an unknown measure. scale Similar figures are related to each other by a proportion or ratio that defines the size relationships, which is also referred to as the scale .<\|endoftext\|>	4.8125
575	Anteros is an asteroid, a large rock that orbits the Sun mainly between the orbits of Mars and Jupiter. They tend to be an irregular shaped but Ceres asteroid is known to be spherical in shape but because it doesn't clear its path round the Sun, it is only a dwarf planet. It has an alternative name which is 1973 EC. Anteros was discovered on 1973-03-13 by J. Gibson. The absolute magnitude of the object is 15.7 which is the brightness of the object. A higher absolute magnitude means that the object is faint whereas a very low number means it is very bright. The albedo of the object is 1. The Albedo is the amount of radiation that is reflected back into space by the object. The Aphelion of the object is 1.797 A.U. which is the point in the orbit that is furthest from the object that it is orbit. At this point, it will then return back to the orbit target. The Perihelion of the object is 1.064 A.U. which is the point in the orbit that is closest to the object that it is orbit around. The Longitude of Ascending Node of the object is 246.3 degrees. The Argument of Perihelion is 338.4. It is the angle along the orbit of a planet or other Solar System object as measured from the ascending node (analogous to right ascension and longitude) Ref:Hawaii. The mean anomoly is 317.8, is the angular distance of the planet from the perihelion or aphelion. Ref:Dictionary.The Semi-Major Axis of the orbit is 1.430, which is the furthest point from the centre to the edge of an elliptical point. The orbital inclination, the angle at which Anteros orbits in relation to the orbital plane is 8.7 degrees. The orbital eccentricity is 0.256, it is the degree at which Anteros orbits close to a circular (0) orbit as opposed to an elliptical (1) orbit. \|Alternative Name\|\|1973 EC\| \|Date of Discovery\|\|1973-03-13\| \|Aphelion (Furthest)\|\|1.797 A.U.\| \|Perihelion (Nearest)\|\|1.064 A.U.\| \|Longitude Of Ascending Node\|\|246.3\| \|Argument of Perihelion\|\|338.4\| There's no register feature and no need to give an email address if you don't need to. All messages will be reviewed before being displayed. Comments may be merged or altered slightly such as if an email address is given in the main body of the comment.<\|endoftext\|>	3.765625
923	# A General Factoring Strategy 0. Order: Order the terms in descending order according to the powers of the variable of your choice. 1. GCF: Factor out a Greatest Common Factor if possible. Set it aside and be sure to include it in the ﬁnal answer. 2. Special Cases: If there are two terms, then consider the following cases: (a) A2 − B 2 = (A − B)(A + B) (c) A3 − B 3 = (A − B)(A2 + AB + B 2 ) (b) A2 + B 2 = P RIM E (d) A3 + B 3 = (A + B)(A2 − AB + B 2 ) 3. Factor by Grouping, Part I: If there are three terms, (a) multiply the coeﬃcient of the ﬁrst term by the coeﬃcient of the last term (b) factor this result into all possible pairs of factors (c) add or subtract each pair of factors until you ﬁnd a combination that produces the middle term (d) replace the middle term with that sum or diﬀerence so that each piece is multiplied by the variable(s) (Note: you should now have four terms of which the middle two are like each other and the previous middle term.) 4. Factor by Grouping, Part II: If there are four terms, (a) factor the ﬁrst pair (b) then factor the second pair (c) if there is a common binomial left in each part after factoring above, then factor this binomial out, and multiply the remaining terms by it in parentheses (d) if there is not a common binomial, then either you made a mistake or the polynomial cannot be factored this way REMEMBER: If you factored something out in step two, then be sure to include it in your ﬁnal answer. 1 × 72 2 × 36 3 × 24 4 × 18 6 × 12 8×9 Step 3c: combine factors. leaving (x + 2) 3(x + 1)(x + 2) completely factored answer 2. leaving (9x − 4) (x + 2)(9x − 4) completely factored answer A c Formatted with L TEX �Gary Parker 1995 . 1 × 2 combine factors. 1 + 2 = 3 replace 3x with 1x + 2x x2 + 1x + 2x + 2 x(x + 1) + 2x + 2 Step 4a: factor the ﬁrst pair x(x + 1) + 2(x + 1) Step 4b: factor the second pair (x + 1)(x + 2) Step 4c: factor out (x + 1). Factor 3x2 + 9x + 6 completely 3x2 + 9x + 6 3(x2 + 3x + 2) Step 0: already done Step 1: factor out three Step 2: does not apply x2 + 3x + 2 Step Step Step Step 3a: 3b: 3c: 3d: multiply 1 by 2 to get 2 factor 2. Factor 9x2 + 14x − 8 completely 9x2 + 14x − 8 Step 0: already done Step 1: no GCF to factor Step 2: does not apply 9x2 + 14x − 8 Step 3a: multiply 9 by 8 to get 72 Step 3b: factor 72.Examples: 1. 18 − 4 = 14 Step 3d: replace 14x with +18x − 4x 9x2 + 18x − 4x − 8 9x(x + 2) − 4x − 8 Step 4a: factor the ﬁrst pair 9x(x + 2) − 4(x + 2) Step 4b: factor the second pair (x + 2)(9x − 4) Step 4c: factor out (x + 2).<\|endoftext\|>	4.4375
2,543	# Transpose (Redirected from Transposed) Jump to: navigation, search The transpose AT of a matrix A can be obtained by reflecting the elements along its main diagonal. Repeating the process on the transposed matrix returns the elements to their original position. In linear algebra, the transpose of a matrix is an operator which flips a matrix over its diagonal, that is it switches the row and column indices of the matrix by producing another matrix denoted as AT (also written A′, Atr, tA or At). It is achieved by any one of the following equivalent actions: • reflect A over its main diagonal (which runs from top-left to bottom-right) to obtain AT • write the rows of A as the columns of AT • write the columns of A as the rows of AT Formally, the i th row, j th column element of AT is the j th row, i th column element of A: ${\displaystyle \left[\mathbf {A} ^{\mathrm {T} }\right]_{ij}=\left[\mathbf {A} \right]_{ji}}$ If A is an m × n matrix then AT is an n × m matrix. The transpose of a matrix was introduced in 1858 by the British mathematician Arthur Cayley.[1] ## Examples • ${\displaystyle {\begin{bmatrix}1&2\end{bmatrix}}^{\mathrm {T} }=\,{\begin{bmatrix}1\\2\end{bmatrix}}}$ • ${\displaystyle {\begin{bmatrix}1&2\\3&4\end{bmatrix}}^{\mathrm {T} }={\begin{bmatrix}1&3\\2&4\end{bmatrix}}}$ • ${\displaystyle {\begin{bmatrix}1&2\\3&4\\5&6\end{bmatrix}}^{\mathrm {T} }={\begin{bmatrix}1&3&5\\2&4&6\end{bmatrix}}}$ ## Properties For matrices A, B and scalar c we have the following properties of transpose: 1. ${\displaystyle \left(\mathbf {A} ^{\mathrm {T} }\right)^{\mathrm {T} }=\mathbf {A} \quad \,}$ The operation of taking the transpose is an involution (self-inverse). 2. ${\displaystyle \left(\mathbf {A} +\mathbf {B} \right)^{\mathrm {T} }=\mathbf {A} ^{\mathrm {T} }+\mathbf {B} ^{\mathrm {T} }\,}$ The transpose respects addition. 3. ${\displaystyle \left(\mathbf {AB} \right)^{\mathrm {T} }=\mathbf {B} ^{\mathrm {T} }\mathbf {A} ^{\mathrm {T} }\,}$ Note that the order of the factors reverses. From this one can deduce that a square matrix A is invertible if and only if AT is invertible, and in this case we have (A−1)T = (AT)−1. By induction this result extends to the general case of multiple matrices, where we find that (A1A2...Ak−1Ak)T = AkTAk−1TA2TA1T. 4. ${\displaystyle \left(c\mathbf {A} \right)^{\mathrm {T} }=c\mathbf {A} ^{\mathrm {T} }\,}$ The transpose of a scalar is the same scalar. Together with (2), this states that the transpose is a linear map from the space of m × n matrices to the space of all n × m matrices. 5. ${\displaystyle \det \left(\mathbf {A} ^{\mathrm {T} }\right)=\det \left(\mathbf {A} \right)\,}$ The determinant of a square matrix is the same as the determinant of its transpose. 6. The dot product of two column vectors a and b can be computed as the single entry of the matrix product ${\displaystyle \left[\mathbf {a} \cdot \mathbf {b} \right]=\mathbf {a} ^{\mathrm {T} }\mathbf {b} ,}$ which is written as aibi in Einstein summation convention. 7. If A has only real entries, then ATA is a positive-semidefinite matrix. 8. ${\displaystyle \left(\mathbf {A} ^{\mathrm {T} }\right)^{-1}=\left(\mathbf {A} ^{-1}\right)^{\mathrm {T} }\,}$ The transpose of an invertible matrix is also invertible, and its inverse is the transpose of the inverse of the original matrix. The notation A−T is sometimes used to represent either of these equivalent expressions. 9. If A is a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose since they share the same characteristic polynomial. ## Special transpose matrices A square matrix whose transpose is equal to itself is called a symmetric matrix; that is, A is symmetric if ${\displaystyle \mathbf {A} ^{\mathrm {T} }=\mathbf {A} }$ A square matrix whose transpose is equal to its negative is called a skew-symmetric matrix; that is, A is skew-symmetric if ${\displaystyle \mathbf {A} ^{\mathrm {T} }=-\mathbf {A} }$ A square complex matrix whose transpose is equal to the matrix with every entry replaced by its complex conjugate (denoted here with an overline) is called a Hermitian matrix (equivalent to the matrix being equal to its conjugate transpose); that is, A is Hermitian if ${\displaystyle \mathbf {A} ^{\mathrm {T} }={\overline {\mathbf {A} }}}$ A square complex matrix whose transpose is equal to the negation of its complex conjugate is called a skew-Hermitian matrix; that is, A is skew-Hermitian if ${\displaystyle \mathbf {A} ^{\mathrm {T} }=-{\overline {\mathbf {A} }}}$ A square matrix whose transpose is equal to its inverse is called an orthogonal matrix; that is, A is orthogonal if ${\displaystyle \mathbf {A} ^{\mathrm {T} }=\mathbf {A} ^{-1}}$ A square complex matrix whose transpose is equal to its conjugate inverse is called a unitary matrix; that is, A is unitary if ${\displaystyle \mathbf {A} ^{\mathrm {T} }={\overline {\mathbf {A} ^{-1}}}}$ ## Transpose of a linear map The transpose may be defined more generally: If f : VW is a linear map between right R-modules V and W with respective dual modules V and W, the transpose of f is the linear map ${\displaystyle {}^{\mathrm {t} }f:W^{}\to V^{}:\varphi \mapsto \varphi \circ f.}$ Equivalently, the transpose tf is defined by the relation ${\displaystyle \left\langle \varphi ,f(v)\right\rangle =\left\langle {}^{\mathrm {t} }f(\varphi ),v\right\rangle \quad \forall \varphi \in W^{*},v\in V,}$ where ⟨·,·⟩ is the natural pairing of each dual space with its respective vector space. This definition also applies unchanged to left modules and to vector spaces.[2] The definition of the transpose may be seen to be independent of any bilinear form on the vector spaces, unlike the adjoint (below). If the matrix A describes a linear map with respect to bases of V and W, then the matrix AT describes the transpose of that linear map with respect to the dual bases. ### Transpose of a bilinear form Every linear map to the dual space f : VV defines a bilinear form B : V × VF, with the relation B(v, w) = f(v)(w). By defining the transpose of this bilinear form as the bilinear form tB defined by the transpose tf : V∗∗V i.e. tB(w, v) = tf(Ψ(w))(v), we find that B(v, w) = tB(w, v). Here, Ψ is the natural homomorphism VV∗∗ into the double dual. ### Adjoint If the vector spaces V and W have respectively nondegenerate bilinear forms BV and BW, a concept known as the adjoint, which is closely related to the transpose, may be defined: If f : VW is a linear map between vector spaces V and W, we define g as the adjoint of f if g : WV satisfies ${\displaystyle B_{V}(v,g(w))=B_{W}(f(v),w)\quad \forall v\in V,w\in W.}$ These bilinear forms define an isomorphism between V and V, and between W and W, resulting in an isomorphism between the transpose and adjoint of f. The matrix of the adjoint of a map is the transposed matrix only if the bases are orthonormal with respect to their bilinear forms. In this context, many authors use the term transpose to refer to the adjoint as defined here. The adjoint allows us to consider whether g : WV is equal to f −1 : WV. In particular, this allows the orthogonal group over a vector space V with a quadratic form to be defined without reference to matrices (nor the components thereof) as the set of all linear maps VV for which the adjoint equals the inverse. Over a complex vector space, one often works with sesquilinear forms (conjugate-linear in one argument) instead of bilinear forms. The Hermitian adjoint of a map between such spaces is defined similarly, and the matrix of the Hermitian adjoint is given by the conjugate transpose matrix if the bases are orthonormal. ## Implementation of matrix transposition on computers Illustration of row- and column-major order On a computer, one can often avoid explicitly transposing a matrix in memory by simply accessing the same data in a different order. For example, software libraries for linear algebra, such as BLAS, typically provide options to specify that certain matrices are to be interpreted in transposed order to avoid the necessity of data movement. However, there remain a number of circumstances in which it is necessary or desirable to physically reorder a matrix in memory to its transposed ordering. For example, with a matrix stored in row-major order, the rows of the matrix are contiguous in memory and the columns are discontiguous. If repeated operations need to be performed on the columns, for example in a fast Fourier transform algorithm, transposing the matrix in memory (to make the columns contiguous) may improve performance by increasing memory locality. Ideally, one might hope to transpose a matrix with minimal additional storage. This leads to the problem of transposing an n × m matrix in-place, with O(1) additional storage or at most storage much less than mn. For n ≠ m, this involves a complicated permutation of the data elements that is non-trivial to implement in-place. Therefore, efficient in-place matrix transposition has been the subject of numerous research publications in computer science, starting in the late 1950s, and several algorithms have been developed. ## References 1. ^ Arthur Cayley (1858) "A memoir on the theory of matrices," Philosophical Transactions of the Royal Society of London, 148 : 17-37. The transpose (or "transposition") is defined on page 31. 2. ^ Bourbaki, "II §2.5", Algebra I<\|endoftext\|>	4.46875
670	Great Expectations is Charles Dickens's thirteenth novel. It is his second novel, after David Copperfield, to be fully narrated in the first person. Great Expectations is a bildungsroman, or a coming-of-age novel, and it is a classic work of Victorian literature. It depicts the growth and personal development of an orphan named Pip. The novel was first published in serial form in Dickens's weekly periodical All the Year Round, from 1 December 1860 to August 1861. Dickens originally intended Great Expectations to be twice as long, but constraints imposed by the management of All the Year Round limited the novel's length. Collected and dense, with a conciseness unusual for Dickens, the novel represents Dickens's peak and maturity as an author. According to G. K. Chesterton, Dickens penned Great Expectations in "the afternoon of (his) life and fame." It is set among the marshes of Kent and in London in the early to mid-1800s. From the outset, the reader is "treated" by the terrifying encounter between Pip, the protagonist, and the escaped convict, Abel Magwitch. Great Expectations is a graphic book, full of extreme imagery, poverty, prison ships, "the hulks," barriers and chains, and fights to the death. It therefore combines intrigue and unexpected twists of autobiographical detail in different tones. Regardless of its narrative technique, the novel reflects the events of the time, Dickens's concerns, and the relationship between society and man. Dickens felt Great Expectations was his best work. Great Expectations has a colourful cast that has entered popular culture: the capricious Miss Havisham, the cold and beautiful Estella, Joe the kind and generous blacksmith, the dry and sycophantic Uncle Pumblechook, Mr Jaggers, Wemmick with his dual personality, and the eloquent and wise friend, Herbert Pocket. Throughout the narrative, typical Dickensian themes emerge: wealth and poverty, love and rejection, and the eventual triumph of good over evil. Great Expectations has become very popular and is now taught as a classic in many English classes. It has been translated into many languages and adapted many times in film and other media. Unabridged version - 185,258 words - 424 pages in the printed edition. Charles John Huffam Dickens (1812-1870) was an English writer and social critic. He created some of the world's most memorable fictional characters and is generally regarded as the greatest novelist of the Victorian period. During his life, his works enjoyed unprecedented fame, and by the twentieth century his literary genius was broadly acknowledged by critics and scholars. His novels and short stories continue to be widely popular. Dickens was regarded as the literary colossus of his age. His 1843 novella, A Christmas Carol, is one of the most influential works ever written, and it remains popular and continues to inspire adaptations in every artistic genre. Set in London and Paris, his 1859 novel, A Tale of Two Cities, is the best selling novel of all time. His creative genius has been praised by fellow writers, from Leo Tolstoy to George Orwell and G. K. Chesterton, for its realism, comedy, prose style, unique characterisations, and social criticism.<\|endoftext\|>	3.78125
822	In this book the author has looked at a variety of issues relating to schools in Britain today. Some of those issues are behaviour management in the classroom, disciplining pupils, bullying in schools, exclusion, language and communication, diversity, and how children can be engaged in their lessons for a successful outcome. The 1989 Children’s Act placed the interest of children first by giving them priority where their welfare is concerned, but these changes brought about a different perspective in classroom management and pupil/teacher relationship. However, changes are inevitable and when they do occur, solutions must be found to deal with them effectively. This book contains some very useful information to help in solving such misunderstandings even before or when they do occur. The aim of the author in writing this book stems from her concern about the issue of black children underachieving in British schools. We have come a far way since the first arrivals of Caribbean migrants who landed at Tilbury docks in Essex on 22nd June 1948. How has things changed since then? Considering the hardships and difficulties they faced as new arrivals in a country distanced by thousands of miles from their original country of abode, was it all worth it? With a time span of seventy years, let’s assess the situation. The immigrants came to work in Britain by invitation from employers and the government, because being British Commonwealth citizens they were entitled to come to Britain without any restrictions. On arrival, many were met with unforeseen hostility especially in housing accommodation as landlords refused to let their premises to them. When they did find accommodation, many had to share rooms and other facilities which were not very comfortable, but there were no other immediate options. By reading this book, you can learn about the effects of race and diversity in Britain today, and how this impacts on people’s lives daily. Where there are different races there is diversity, and where is diversity, there is bound to be discrimination, because people do not always cooperate where equality is concerned. Therefore inequality is always going to present problems where jobs, housing, education, services, gender, disability and religion are concerned. How do we solve this problem. This book sets out how some of those problems can be solved or even avoided by applying the right knowledge to certain situations. Neither race, racism, inequality nor hostile behaviour needs to create a problem. We need to be aware of the situation which causes those problems, and deal with them accordingly, rather than to allow things to get out of hand, then look for a solution. Certain issues such as race and racism which people find difficult to talk about, should be discussed more openly, and clarify any misconceptions which some people may find terrifying to discuss. Some individuals find it uneasy to use the term ‘black’ because of the connotation which it carries. This ties in deeply with race and racism, and because of that, some people will go to great lengths to avoid using the term ‘black’. If used in the right context, there should not be any problems whatsoever. So how do we overcome such problems which seem to dominate our daily lives, and causing misery and profound hurt in the lives of so many people? We should first look at our own attitudes towards others, and modify any attitude which does not fit in with good behaviour. In a modern and civilized society people should not be behaving in ways which upset and destroy other people’s lives and thinking it is fun to do so, because it certainly is not the right thing to do. Look at what is happening in the field of football today where black players still have banana skin thrown at them on the pitch with monkey chants, while the players are trying to concentrate on their game. Have they ever thought how disgraceful this type of behaviour looks to the rest of the world. Those fans ought to be given special lessons on behaviourism, or modify their behaviour. But it is not just in football, institutions also need to change the way they deal with applicants and employees, in terms or recruitment, promotion, retention and other factors relating to ethnicity.<\|endoftext\|>	3.78125
5,966	Artificial gene synthesis Artificial gene synthesis, sometimes known as DNA printing is a method in synthetic biology that is used to create artificial genes in the laboratory. Based on solid-phase DNA synthesis, it differs from molecular cloning and polymerase chain reaction (PCR) in that it does not have to begin with preexisting DNA sequences. Therefore, it is possible to make a completely synthetic double-stranded DNA molecule with no apparent limits on either nucleotide sequence or size. The method has been used to generate functional bacterial or yeast chromosomes containing approximately one million base pairs. Creating novel nucleobase pairs in addition to the two base pairs in nature could greatly expand the genetic code. Synthesis of the first complete gene, a yeast tRNA, was demonstrated by Har Gobind Khorana and coworkers in 1972. Synthesis of the first peptide- and protein-coding genes was performed in the laboratories of Herbert Boyer and Alexander Markham, respectively. Commercial gene synthesis services are now available. Approaches are most often based on a combination of organic chemistry and molecular biology techniques and entire genes may be synthesized "de novo", without the need for template DNA. Gene synthesis is an important tool in many fields of recombinant DNA technology including heterologous gene expression, vaccine development, gene therapy and molecular engineering. The synthesis of nucleic acid sequences can be more economical than classical cloning and mutagenesis procedures. It is also a powerful and flexible engineering tool for creating and designing new DNA sequences and protein functions. - 1 Gene optimization - 2 Standard methods - 3 Applications - 4 Entire bacterial genomes - 5 Yeast chromosome - 6 Unnatural base pair (UBP) - 7 See also - 8 Notes While the ability to make increasingly long stretches of DNA efficiently and at lower prices is a technological driver of this field, increasingly attention is being focused on improving the design of genes for specific purposes. Early in the genome sequencing era, gene synthesis was used as an (expensive) source of cDNAs that were predicted by genomic or partial cDNA information but were difficult to clone. As higher quality sources of sequence verified cloned cDNA have become available, this practice has become less urgent. Producing large amounts of protein from gene sequences (or at least the protein coding regions of genes, the open reading frame) found in nature can sometimes prove difficult and is a problem of sufficient impact that scientific conferences have been devoted to the topic. Many of the most interesting proteins sought by molecular biologists are normally regulated to be expressed in very low amounts in wild type cells. Redesigning these genes offers a means to improve gene expression in many cases. Rewriting the open reading frame is possible because of the degeneracy of the genetic code. Thus it is possible to change up to about a third of the nucleotides in an open reading frame and still produce the same protein. The available number of alternate designs possible for a given protein is astronomical. For a typical protein sequence of 300 amino acids there are over 10150 codon combinations that will encode an identical protein. Codon optimization, or replacing rarely used codons with more common codons sometimes have dramatic effects. Further optimizations such as removing RNA secondary structures can also be included. At least in the case of E. coli, protein expression is maximized by predominantly using codons corresponding to tRNA that retain amino acid charging during starvation. Computer programs written to perform these, and other simultaneous optimizations are used to handle the enormous complexity of the task. A well optimized gene can improve protein expression 2 to 10 fold, and in some cases more than 100 fold improvements have been reported. Because of the large numbers of nucleotide changes made to the original DNA sequence, the only practical way to create the newly designed genes is to use gene synthesis. Oligonucleotides are chemically synthesized using building blocks called nucleoside phosphoramidites. These can be normal or modified nucleosides which have protecting groups to prevent their amines, hydroxyl groups and phosphate groups from interacting incorrectly. One phosphoramidite is added at a time, the 5' hydroxyl group is deprotected and a new base is added and so on. The chain grows in the 3' to 5' direction, which is backwards relative to biosynthesis. At the end, all the protecting groups are removed. Nevertheless, being a chemical process, several incorrect interactions occur leading to some defective products. The longer the oligonucleotide sequence that is being synthesized, the more defects there are, thus this process is only practical for producing short sequences of nucleotides. The current practical limit is about 200 bp (base pairs) for an oligonucleotide with sufficient quality to be used directly for a biological application. HPLC can be used to isolate products with the proper sequence. Meanwhile, a large number of oligos can be synthesized in parallel on gene chips. For optimal performance in subsequent gene synthesis procedures they should be prepared individually and in larger scales. Annealing based connection of oligonucleotides Usually, a set of individually designed oligonucleotides is made on automated solid-phase synthesizers, purified and then connected by specific annealing and standard ligation or polymerase reactions. To improve specificity of oligonucleotide annealing, the synthesis step relies on a set of thermostable DNA ligase and polymerase enzymes. To date, several methods for gene synthesis have been described, such as the ligation of phosphorylated overlapping oligonucleotides, the Fok I method and a modified form of ligase chain reaction for gene synthesis. Additionally, several PCR assembly approaches have been described. They usually employ oligonucleotides of 40-50 nucleotides long that overlap each other. These oligonucleotides are designed to cover most of the sequence of both strands, and the full-length molecule is generated progressively by overlap extension (OE) PCR, thermodynamically balanced inside-out (TBIO) PCR or combined approaches. The most commonly synthesized genes range in size from 600 to 1,200 bp although much longer genes have been made by connecting previously assembled fragments of under 1,000 bp. In this size range it is necessary to test several candidate clones confirming the sequence of the cloned synthetic gene by automated sequencing methods. Moreover, because the assembly of the full-length gene product relies on the efficient and specific alignment of long single stranded oligonucleotides, critical parameters for synthesis success include extended sequence regions comprising secondary structures caused by inverted repeats, extraordinary high or low GC-content, or repetitive structures. Usually these segments of a particular gene can only be synthesized by splitting the procedure into several consecutive steps and a final assembly of shorter sub-sequences, which in turn leads to a significant increase in time and labor needed for its production. The result of a gene synthesis experiment depends strongly on the quality of the oligonucleotides used. For these annealing based gene synthesis protocols, the quality of the product is directly and exponentially dependent on the correctness of the employed oligonucleotides. Alternatively, after performing gene synthesis with oligos of lower quality, more effort must be made in downstream quality assurance during clone analysis, which is usually done by time-consuming standard cloning and sequencing procedures. Another problem associated with all current gene synthesis methods is the high frequency of sequence errors because of the usage of chemically synthesized oligonucleotides. The error frequency increases with longer oligonucleotides, and as a consequence the percentage of correct product decreases dramatically as more oligonucleotides are used. The mutation problem could be solved by shorter oligonucleotides used to assemble the gene. However, all annealing based assembly methods require the primers to be mixed together in one tube. In this case, shorter overlaps do not always allow precise and specific annealing of complementary primers, resulting in the inhibition of full length product formation. Manual design of oligonucleotides is a laborious procedure and does not guarantee the successful synthesis of the desired gene. For optimal performance of almost all annealing based methods, the melting temperatures of the overlapping regions are supposed to be similar for all oligonucleotides. The necessary primer optimization should be performed using specialized oligonucleotide design programs. Several solutions for automated primer design for gene synthesis have been presented so far. Error correction procedures To overcome problems associated with oligonucleotide quality several elaborate strategies have been developed, employing either separately prepared fishing oligonucleotides, mismatch binding enzymes of the mutS family or specific endonucleases from bacteria or phages. Nevertheless, all these strategies increase time and costs for gene synthesis based on the annealing of chemically synthesized oligonucleotides. Massively parallel sequencing has also been used as a tool to screen complex oligonucleotide libraries and enable the retrieval of accurate molecules. In one approach, oligonucleotides are sequenced on the 454 pyrosequencing platform and a robotic system images and picks individual beads corresponding to accurate sequence. In another approach, a complex oligonucleotide library is modified with unique flanking tags before massively parallel sequencing. Tag-directed primers then enable the retrieval of molecules with desired sequences by dial-out PCR. Increasingly, genes are ordered in sets including functionally related genes or multiple sequence variants on a single gene. Virtually all of the therapeutic proteins in development, such as monoclonal antibodies, are optimized by testing many gene variants for improved function or expression. Major applications of synthetic genes include synthesis of DNA sequences identified by high throughput sequencing but never cloned into plasmids and the ability to safely obtain genes for vaccine research without the need to grow the full pathogens. Digital manipulation of digital genetic code before synthesis into DNA can be used to optimize protein expression in a particular host, or remove non-functional segments in order to facilitate further replication of the DNA. Synthesis of DNA allows DNA digital data storage. DNA synthesis and synthetic biology The significant drop in cost of gene synthesis in recent years due to increasing competition of companies providing this service has led to the ability to produce entire bacterial plasmids that have never existed in nature. The field of synthetic biology utilizes the technology to produce synthetic biological circuits, which are stretches of DNA manipulated to change gene expression within cells and cause the cell to produce a desired product. Entire bacterial genomes Synthia and Mycoplasma laboratorium On June 28, 2007, a team at the J. Craig Venter Institute published an article in Science Express, saying that they had successfully transplanted the natural DNA from a Mycoplasma mycoides bacterium into a Mycoplasma capricolum cell, creating a bacterium which behaved like a M. mycoides. On Oct 6, 2007, Craig Venter announced in an interview with UK's The Guardian newspaper that the same team had synthesized a modified version of the single chromosome of Mycoplasma genitalium using chemicals. The chromosome was modified to eliminate all genes which tests in live bacteria had shown to be unnecessary. The next planned step in this minimal genome project is to transplant the synthesized minimal genome into a bacterial cell with its old DNA removed; the resulting bacterium will be called Mycoplasma laboratorium. The next day the Canadian bioethics group, ETC Group issued a statement through their representative, Pat Mooney, saying Venter's "creation" was "a chassis on which you could build almost anything". The synthesized genome had not yet been transplanted into a working cell. On May 21, 2010, Science reported that the Venter group had successfully synthesized the genome of the bacterium Mycoplasma mycoides from a computer record, and transplanted the synthesized genome into the existing cell of a Mycoplasma capricolum bacterium that had its DNA removed. The "synthetic" bacterium was viable, i.e. capable of replicating billions of times. The team had originally planned to use the M. genitalium bacterium they had previously been working with, but switched to M. mycoides because the latter bacterium grows much faster, which translated into quicker experiments. Venter describes it as "the first species.... to have its parents be a computer". The transformed bacterium is dubbed "Synthia" by ETC. A Venter spokesperson has declined to confirm any breakthrough at the time of this writing. In March 2014, Jef Boeke of the Langone Medical Centre at New York University, published that his team has synthesized one of the S. cerevisiae 16 yeast chromosomes, the chromosome III, that he named synIII. The procedure involved replacing the genes in the original chromosome with synthetic versions and the finished human made chromosome was then integrated into a yeast cell. It required designing and creating 273,871 base pairs of DNA – fewer than the 316,667 pairs in the original chromosome. In March 2017 6 yeast chromosomes were reported to have been synthesized. Unnatural base pair (UBP) DNA sequences have been described which use newly created nucleobases to form a third base pair, in addition to the two base pairs found in nature, A-T (adenine – thymine) and G-C (guanine – cytosine). Multiple research groups have been searching for a third base pair for DNA, including teams led by Steven A. Benner, Philippe Marliere, and Ichiro Hirao. Some new base pairs have been reported. In 2012, a group of American scientists led by Floyd Romesberg, a chemical biologist at the Scripps Research Institute in San Diego, California, published that his team designed an unnatural base pair (UBP). The two new artificial nucleotides or Unnatural Base Pair (UBP) were named d5SICS and dNaM. More technically, these artificial nucleotides bearing hydrophobic nucleobases, feature two fused aromatic rings that form a (d5SICS–dNaM) complex or base pair in DNA. In 2014 the same team from the Scripps Research Institute reported that they synthesized a stretch of circular DNA known as a plasmid containing natural T-A and C-G base pairs along with the best-performing UBP Romesberg's laboratory had designed, and inserted it into cells of the common bacterium E. coli that successfully replicated the unnatural base pairs through multiple generations. This is the first known example of a living organism passing along an expanded genetic code to subsequent generations. This was in part achieved by the addition of a supportive algal gene that expresses a nucleotide triphosphate transporter which efficiently imports the triphosphates of both d5SICSTP and dNaMTP into E. coli bacteria. Then, the natural bacterial replication pathways use them to accurately replicate the plasmid containing d5SICS–dNaM. The successful incorporation of a third base pair is a significant breakthrough toward the goal of greatly expanding the number of amino acids which can be encoded by DNA, from the existing 20 amino acids to a theoretically possible 172, thereby expanding the potential for living organisms to produce novel proteins. The artificial strings of DNA do not encode for anything yet, but scientists speculate they could be designed to manufacture new proteins which could have industrial or pharmaceutical uses. - "DNA 'Printing' A Big Boon To Research, But Some Raise Concerns". - Kimoto, M.; et al. (2013). 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"Efficient and sequence-independent replication of DNA containing a third base pair establishes a functional six-letter genetic alphabet". Proc. Natl. Acad. Sci. USA. 109 (30): 12005–12010. Bibcode:2012PNAS..10912005M. doi:10.1073/pnas.1205176109. PMC 3409741. PMID 22773812. - Callaway, Ewan (May 7, 2014). "Scientists Create First Living Organism With 'Artificial' DNA". Nature News. Huffington Post. Retrieved 8 May 2014. - Sample, Ian (May 7, 2014). "First life forms to pass on artificial DNA engineered by US scientists". The Guardian. Retrieved 8 May 2014. - Pollack, Andrew (May 7, 2014). "Scientists Add Letters to DNA's Alphabet, Raising Hope and Fear". New York Times. Retrieved 8 May 2014.<\|endoftext\|>	3.875
330	“Diversity represents beauty, strength, and pride. In the multicultural aspect, consider more deeply the observation that we are all much more alike than we are unalike.” Dealing with Diversity TWW Inc. help students, and people recognize that all the people are unique in their own way. Their differences could consist of their reading level, athletic ability, cultural background, personality, religious beliefs. To embrace diversity, make positive use of it, value diversity, and the need to model this attitude to their students. Value diversity, they recognize and respect the fact that people are different and that these differences is generally a good thing. Our art instructors provide students with an environment that is conducive to learning. TWW Inc. multicultural education is a field of study and an emerging discipline whose major aim is to create equal educational opportunities for students from diverse racial, ethnic, social-class, and cultural groups. One of its important goals is to help all students to acquire the knowledge, attitudes, and skills needed to function effectively in a pluralistic democratic society and to interact, negotiate, and communicate with peoples from diverse groups in order to create a civic and moral community that works for the common good. Multicultural education not only draws content, concepts, paradigms, and theories from specialized interdisciplinary fields such as ethnic studies and women studies (and from history and the social and behavioral sciences), it also interrogates, challenges, and reinterprets content, concepts, and paradigms from the established disciplines. Multicultural education applies content from these fields and disciplines to pedagogy and curriculum development in educational settings.<\|endoftext\|>	3.6875
228	GROUND FLOOR BEAM LAYOUT A beam is a structural element that primarily resists loads applied laterally to the beam's axis. its mode of deflection is primarily by bending. The loads applied to the beam result in reaction forces at the beam's support points. The total effect of all all the forces acting on the beam is to produce shear forces and bending moments within the beam, that in turn induce internal stresses, strains and deflections of the beam. In engineering, beams are of several types: - Simply supported – a beam supported on the ends which are free to rotate and have no moment resistance. - Fixed – a beam supported on both ends and restrained from rotation. - Over hanging – a simple beam extending beyond its support on one end. - Double overhanging – a simple beam with both ends extending beyond its supports on both ends. - Continuous – a beam extending over more than two supports. - Cantilever – a projecting beam fixed only at one end. - Trussed – a beam strengthened by adding a cable or rod to form a truss.<\|endoftext\|>	4.15625
1,227	Taxation without representation was the call to action during the Revolutionary War, and justly so. While the hefty taxes imposed by England on the colonists were widely unfair due to the lack of commoner representation in parliament. However, it is an inescapable fact that taxes are an essential part of a country’s economy. While George Washington wanted to try his best to avoid taxes, several states had incurred debt from the war so Alexander Hamilton felt taxation was a necessary course of action. Washington then took it upon himself to travel and listen to various voters opinions on the matter to gage his support of the bill; the result was unexpectedly pleasant. With reassurance in his mind, Washington gave the okay to Hamilton to push a federal tax to assist with debt and add an additional tax to whiskey as a backup fund if anything went wrong. This maneuver did not go well, however. Protests started almost immediately after the law was put into place in 1792. It seemed the offended parties were farmers, who had to pay cash only for taxes, and smaller distillers, who had to pay nine cents a batch while larger distillers received several tax breaks. From the beginning sporadic violence arose across the countryside- mostly in Pennsylvania- against tax collectors and other authority figures. By 1794 the conflict had reached a point of do or die for the angry mobs who had been warned several times by the government to stand down or they will be facing a militia. This threat only came after the destructive attack on John Neville’s Pennsylvania estate after he had served writs of summons to several people who did not pay their taxes during the summer of 1794. The pot was quite literally near boiling over as the seasons changed and the simplest thing would set either side off. The crux of the conflict came after flames had just been put out on another in Pittsburg after an angry tax payer had intercepted mail that expressed the distaste officials had for the attack on Neville’s home. David Bradford, the man who read the letters, threatened to destructively riot through Pittsburg the next day, but this was quickly deterred with some whiskey barrels and an apology. Soon, new signs of a reignited conflict had arisen so Hamilton voted for a militia to be sent, but Washington opted for a peace envoy instead- which ultimately failed. Finally, Washington marched with an army of twelve thousand men into Western Pennsylvania to end this rebellion. Oddly enough, however, the opposition failed to show up so suspected rebels were arrested and brought back for trial instead. It turns out that most of the rebel leaders had fled; among the twenty men who were arrested all but two, who were charged with treason, were acquitted. After the conflict mellowed from its rolling boil in 1794, there was little conflict in the following years. Eventually in 1802 the tax was repealed and all was well that end well. Sometimes starting something up does not always prove to be as easy as it was on paper and there are often several hiccoughs along the way once the ball gets rolling. The same could certainly be said for the start of a new country and all its ins and outs. After the end of the Revolutionary War, the newly establish United States of America had to learn to stand on its own two feet. Essentially everything had to be built from the ground up- this included government, law, and currency. While many of these tasks were daunting, it seems perhaps that the most difficult obstacle they had to overcome was currency. In fact not only were conversations regarding the Nations money delayed, but so was the establishment of the mint, coin value, materials, and officials. Thankfully by 1792 congress had finalized the currency system and mint establishment in the Coinage Act. Still, actual mintage was a ways away since the mint had difficulties with copper coinage, which began two years after the Act. Once it was time to move on to precious metal coinage, silver was the first to come, but there was still significant obstacles. To begin, the reserves of precious metals for coinage was absolutely laughable, so they mint had to rely on wealthy depositors to contribute metals to produce coins. The mint planned to pay these men back through coins equal to their deposits as soon as possible. Once the metal was in, the government filled the production with red tape and decided both the Chief Coiner and the Assayer had to put up a bond of $10,000 each, to offset malfeasance. This, by no means, was not easy money to come by; in fact it was nearly impossible for these men. Congress reluctantly relented and lowered the bonds to $5,000 for the coiner and $1,000 for the assayer; both men were assisted by two other wealthy men to post their bonds. Finally coins could be minted! The first three silver coins struck were the dollar, half dollar, and half dime- all of which bore the Flowing Hair Liberty designed by engraver Robert Scot. Unfortunately, the struggle was not over for these precious coins. The mint began with the dollar since it was heftier, carried more value, and was meant to be the “showoff” coin for the nation. However, the dies could not keep up with demand and most of them broke after a thousand or so strikes. Repairs would take a while, but depositors pressured the mint for their return, so they began the half dollars. Finally, after a successful production of 5,300 coins the mint was ready to produce more, but this time the metal roller system broke down so blanks could not be rolled to their proper thickness. This issue was not resolved until the beginning of 1795; despite the date change, the mint still utilized the 1794 dies well into 1795 since there was no use in trashing something that was still useful. Mintage ran more smoothly the following year as reflected in the numbers, regardless the design saw its final few strikes in the remaining months of 1795 and was exchanged for the capped bust liberty in 1796.<\|endoftext\|>	3.84375
964	# High School Math : How to find the perimeter of an equilateral triangle ## Example Questions ### Example Question #1 : How To Find The Perimeter Of An Equilateral Triangle A square rug border consists of a continuous pattern of equilateral triangles, with isosceles triangles as corners, one of which is shown above. If the length of each equilateral triangle side is 5 inches, and there are 40 triangles in total, what is the total perimeter of the rug? The inner angles of the corner triangles is 30°. 188 208 124 200 180 188 Explanation: There are 2 components to this problem. The first, and easier one, is recognizing how much of the perimeter the equilateral triangles take up—since there are 40 triangles in total, there must be 40 – 4 = 36 of these triangles. By observation, each contributes only 1 side to the overall perimeter, thus we can simply multiply 36(5) = 180" contribution. The second component is the corner triangles—recognizing that the congruent sides are adjacent to the 5-inch equilateral triangles, and the congruent angles can be found by 180 = 30+2x → x = 75° We can use ratios to find the unknown side: 75/5 = 30/y → 75y = 150 → y = 2''. Since there are 4 corners to the square rug, 2(4) = 8'' contribution to the total perimeter. Adding the 2 components, we get 180+8 = 188 inch perimeter. ### Example Question #1 : How To Find The Perimeter Of An Equilateral Triangle The height of an equilateral triangle is What is the triangle's perimeter? 24 12 6 8 12 Explanation: An altitude drawn in an equilateral triangle will form two 30-60-90 triangles. The height of equilateral triangle is the length of the longer leg of the 30-60-90 triangle. The length of the equilateral triangle's side is the length of the hypotenuse of the 30-60-90. The ratio of the length of the hypotenuse to the length of the longer leg of a 30-60-90 triangle is The length of the longer leg of the 30-60-90 triangle in this problem is Using this ratio, we find that the length of this triangle's hypotenuse is 4. Thus the perimeter of the equilateral triangle will be 4 multiplied by 3, which is 12. ### Example Question #1 : How To Find The Perimeter Of An Equilateral Triangle An equilateral triangle has a side length of . What is its perimeter? Not enough information to solve. Explanation: An equilateral triangle possesses three sides of equal lengths. Therefore, we can easily calculate its perimeter by tripling the given side length. ### Example Question #2 : How To Find The Perimeter Of An Equilateral Triangle An equilateral triangle has an altitude length of . What is its perimeter? Not enough information to solve Explanation: An altitude slices an equilateral triangle into two triangles. These triangles follow a side length pattern. The smallest of the two legs equals and the hypotenuse equals . By way of the Pythagorean Theorem, the longest leg or . We have the length of the altitude of the triangle . We can solve for the smallest side via substitution and simple algebra. Note, this is only the smallest side of one of the triangles. It needs to be doubled to equal a complete side of the equilateral triangle. Now, the side length can be tripled to calculate the perimeter. ### Example Question #1 : How To Find The Perimeter Of An Equilateral Triangle Find the perimeter of the following equilateral triangle: Explanation: The formula for the perimeter of an equilateral triangle is: Where is the length of the side Plugging in our values, we get: ### Example Question #24 : Equilateral Triangles Determine the perimeter of the following equilateral triangle: Explanation: The formula for the perimeter of an equilateral triangle is: , where is the length of the side. Plugging in our value, we get: ### Example Question #3 : How To Find The Perimeter Of An Equilateral Triangle An equilateral triangle has a side of length feet. What is the perimeter of the triangle? feet foot feet It cannot be determined from the information given. feet<\|endoftext\|>	4.84375
100	Geo 3.3 - Volume of Prisms G.5.2 Given the formula, determine the lateral area, surface area, and volume of prisms, pyramids, spheres, cylinders, and cones; 1. A prism is a solid object with: 2. What are some different types of prisms? 9. Find the volume formulas based on the unique area of the base. 9. Use the volume formula to find the volume of the unique prisms.<\|endoftext\|>	4.0625
1,019	# How To Multiply Any Two Integers In Just 4 Steps? 114 0 The current mathematical exercise targets a legitimate system showing multiplication of integers. It shows techniques for integer number duplication from an exceptionally fundamental issue to tackle any difficult issues. What are the possibilities to show rules of integers? Showing it in four stages makes it less demanding for students to get a handle on the topic. It is critical to keep up with these four particular steps. A brief explanation of every possibility is as per the following: 1. To begin with is multiplying two one-digit numbers. In parallel, demonstrate the same expansion arrangement. Kids must have a complete comprehension of the first step, because this will be the beginning base of individual child who finds multiplication to be a big deal. 2. Second is multiplying one-digit number with a two-digit number. 3. Third is multiplying 2 two-digit numbers. 4. Fourth is multiplying two or more numbers with every number having variable number of digits; towards the finish of this stage, students can take care of any multiplication problem containing n numbers. The principal step i.e. the first step is very critical. It is the kids first face off with multiplication. The procedural intuition is altogether different from addition; inability to move from adding to multiplying will leave children confounded. Why are four stages imperative to show multiplication of numbers? Give us a chance to audit the initial three stages as a gathering and step four later. Here are the points of interest of the initial three stages: 1. Multiplying two – one-digit numbers. Close by the multiplying issue, demonstrate the relating adding issue. The addition issue accommodates visual examination. 2. Multiplying a one-digit number with a two-digit number. 3. Multiplying 2 two-digit numbers. 4. Notice at every step, we are including another digit to the increase learning procedure. Give us a chance to dissect showing step one in more detail; it is more unpredictable than steps two and three. There are three reasons why this is valid. 1. The main step is moving from expansion intuition to augmentation. We urge showing augmentation close by the same expansion issue. This places kids in a recognizable safe place. 2. The second step obliges eliminating the parallel expansion issues. The rate of eliminating relies on the expectation to learn and adapt of the class. 3. The third step shows only augmentation issues – no expansion. 4. The fourth and last step: tackle numerous increase issues with any check of numbers and their digits. The key reason of this last step is building up a safe place to tackle any numbers augmentation issues through practice. What to learn? Whether you have hired an online math tutor for your math homework help, or doing it all on your own, multiplication is not as difficult as it seems. It has a striking resemblance with addition, and if grasped properly through online math tutoring, you can tackle same as you do with addition. Just few things to keep in mind before winding up: • Multiplying any integer with 0 will result in 0 • Multiplying any integer with 1 will result in the integer itself • Multiplying any integer with infinity will result in infinity • Multiplying two positive integers will give a positive number as answer • Multiplying two negative integers will give a positive number as answer • Multiplying one positive and one negative integer will give a negative number as the answer #### Sunil Kumar Mr. Sunil Kumar is the Co-Founder and CEO of a leading education technology company Tutorpace.com , which is based in Texas, USA. He works towards the empowerment of education through providing quality tutoring services via his online tutoring portal. Mr. Sunil Kumar started his tutoring company in 2012 and is interested in helping students who struggle in their studies. He likes to write on various educational topics to create awareness among students about the ongoing educational trends. ## Want to Seek Best Assignment Help? Posted by - February 23, 2016 0 Assignments are big chunks of work that take away much of time from students. It is in their attempts to… ## How To Regain The Lost Confidence In Math Homework Help Posted by - September 29, 2014 0 The way the internet has changed the entire tutoring scenario seems to be incredible sometimes. Now, better and systematic studies… ## The 6 Best Things about Online Tutoring Posted by - August 16, 2013 0 Tutoring online has brought an upward mark in the phase of educating the struggling minds with its stamp of technological… ## What are The Various Methods to Learn Math Effectively Posted by - September 17, 2013 0 Each subject has unique methods that make learner proficient in that particular field. These learning methods give student that way… ## Get Math Homework Help From Online Tutors Posted by - December 7, 2013 0 Does math homework create big fuss for you? Tutor Pace’s Math homework help is your savior. Math homework help: Why…<\|endoftext\|>	4.8125
870	Species: Greenland shark (Somniosus microcephalus) Habitat: deep in the North Atlantic and the cold surface waters of the Arctic Fish that were alive during the Age of Enlightenment are still swimming strong. A Greenland shark has lived at least 272 years, making the species the longest-lived vertebrate in the world – smashing the previous record held by a 211-year-old bowhead whale. But it may have been as old as 500 years. “We definitely expected the sharks to be old, but we didn’t expect that it would be the longest-living vertebrate animal,” says Julius Nielsen of the University of Copenhagen, Denmark. Living deep in the North Atlantic and the frigid surface waters of the Arctic, Greenland sharks have a stable environment and grow just a few centimetres per year. Despite their slow growth, though, they reach more than 5 metres in length and are often the apex predator in their ecosystem. Old blue eyes It was once thought to be impossible to age Greenland sharks. Their skeletons, made of cartilage, lack the calcified growth rings of hard-boned vertebrates. And other fish are aged by measuring calcareous bodies that grow in their ears, but this doesn’t work for sharks. Instead Nielsen and his colleagues focused on radiation in the sharks’ eyes. Nuclear bomb tests in the 1950s and ’60s blasted radioactive particles into the atmosphere. Those particles entered food webs all over the world and show up in the form of radioactive forms of carbon in organisms that lived through that period. Because Greenland sharks’ eye lens tissue doesn’t change during its lifetime, it preserves the historic radiation. After catching a 2.2-metre shark that showed radiation levels indicating it was born in the 1960s and was about 50 years old, the team calculated how fast the sharks grew. 150-year dry spell The team estimated that one 5-metre animal was at least 272 years old – but could be more than 500 years old (392 +/- 120 years). Another was at least 260 years old, and could be more than 400 years old. And the female sharks don’t seem to reach breeding age until they are about 150 years old. “They have to wait more than 100 years to get laid – I’m sure they’re not happy about that,” says Nielsen. Despite the uncertainty in estimating birthdays deep in the past, it’s clear these sharks are centuries-old, says Aaron MacNeil at the Australian Institute of Marine Science. “This is our best estimate of how old these things are, but I don’t think it’s the final word,” he says. “The key message here is that these things are living a very, very long time.” Pining for the fjords Many aspects of the Greenland shark’s life remain unknown, says MacNeil. We don’t know for sure where shark pups are born, but one hypothesis is that females give birth to live young in Arctic fjords, MacNeil says. It’s also unclear how much climate change will affect the species through changing its cold-water habitats, he adds. Finding out more would help scientists determine whether the species has a population healthy enough for a fishing industry, says Nielsen. While there seem to be many young Greenland sharks around, Nielsen believes many breeding-age sharks were harvested for oil around the time of the second world war, and because of the time it takes them to mature sexually, the population will be recovering for another 100 years. The long generation times could also make them vulnerable to habitat disturbances, he says. Other mysterious deep-sea shark species could also have surprising life spans, says Neil Hammerschlag at the University of Miami in Florida. “Every single time you study a deep-sea shark, you make a new discovery,” he says. “This might be just the tip of the iceberg.” Journal reference: Science, DOI: 10.1126/science.aaf1703 More on these topics:<\|endoftext\|>	3.828125
938	# How to write an absolute value equation from a graph Examples of Student Work at this Level The student: What are these two values? If you already know the solution, you can tell immediately whether the number inside the absolute value brackets is positive or negative, and you can drop the absolute value brackets. Examples of Student Work at this Level The student correctly writes and solves the first equation: A difference is described between two values. This is the solution for equation 2. Sciencing Video Vault 1. When you take the absolute value of a number, the result is always positive, even if the number itself is negative. Equation 2 is the correct one. Do you know whether or not the temperature on the first day of the month is greater or less than 74 degrees? If needed, clarify the difference between an absolute value equation and the statement of its solutions. Writes the solutions of the first equation using absolute value symbols. Why is it necessary to use absolute value symbols to represent the difference that is described in the second problem? Evaluate the expression x β 12 for a sample of values some of which are less than 12 and some of which are greater than 12 to demonstrate how the expression represents the difference between a particular value and Emphasize that each expression simply means the difference between x and Ask the student to solve the equation and provide feedback. What is the difference? For a random number x, both the following equations are true: Then explain why the equation the student originally wrote does not model the relationship described in the problem. If you plot the above two equations on a graph, they will both be straight lines that intersect the origin. For example, represent the difference between x and 12 as x β 12 or 12 β x. Instructional Implications Model using absolute value to represent differences between two numbers. To solve this, you have to set up two equalities and solve each separately. Ask the student to consider these two solutions in the context of the problem to see if each fits the condition given in the problem i. Guide the student to write an equation to represent the relationship described in the second problem. Should you use absolute value symbols to show the solutions? Finds only one of the solutions of the first equation. Got It The student provides complete and correct responses to all components of the task. Set Up Two Equations Set up two separate and unrelated equations for x in terms of y, being careful not to treat them as two equations in two variables: Instructional Implications Provide feedback to the student concerning any errors made. You can now drop the absolute value brackets from the original equation and write instead: This means that any equation that has an absolute value in it has two possible solutions. Writing an Equation with a Known Solution If you have values for x and y for the above example, you can determine which of the two possible relationships between x and y is true, and this tells you whether the expression in the absolute value brackets is positive or negative. Provide additional opportunities for the student to write and solve absolute value equations.The other case for absolute value inequalities is the "greater than" case. Let's first return to the number line, and consider the inequality \| x \| > The solution will be all. For absolute value equations multiplied by a constant (for example, y = a \| x \|),if 0 1, it is stretched. Also, if a is negative, then the graph opens downward, instead of upwards as usual. Absolute Value Functions To graph an absolute value function you may find it helpful to plot the vertex and one other point. Use symmetry to plot a third point and then complete the graph. Writing an Absolute Value Function Write an equation of the graph shown. SOLUTION The vertex of the graph is (0, ΒΊ3), so the equation has the form. The general form of an absolute value function is f(x)=a\|x-h\|+k. From this form, we can draw graphs. This article reviews how to draw the graphs of absolute value functions. While absolute-value graphs tend to look like the one above, with an "elbow" in the middle, this is not always the case. However, if you see a graph with an elbow like this, you should expect that the equation is probably an absolute value. You can denote absolute value by a pair of vertical lines bracketing the number in question. When you take the absolute value of a number, the result is always positive, even if the number itself is negative. How to write an absolute value equation from a graph Rated 0/5 based on 73 review<\|endoftext\|>	4.6875
10,553	$$\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}}$$ # 7.6: Calculating Centers of Mass and Moments of Inertia $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ We have already discussed a few applications of multiple integrals, such as finding areas, volumes, and the average value of a function over a bounded region. In this section we develop computational techniques for finding the center of mass and moments of inertia of several types of physical objects, using double integrals for a lamina (flat plate) and triple integrals for a three-dimensional object with variable density. The density is usually considered to be a constant number when the lamina or the object is homogeneous; that is, the object has uniform density. ## Center of Mass in Two Dimensions The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. Figure $$\PageIndex{1}$$ shows a point $$P$$ as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass. Figure $$\PageIndex{1}$$: A lamina is perfectly balanced on a spindle if the lamina’s center of mass sits on the spindle. To find the coordinates of the center of mass $$P(\bar{x},\bar{y})$$ of a lamina, we need to find the moment $$M_x$$ of the lamina about the $$x$$-axis and the moment $$M_y$$ about the $$y$$-axis. We also need to find the mass $$m$$ of the lamina. Then $\bar{x} = \dfrac{M_y}{m}$ and $\bar{y} = \dfrac{M_x}{m}.$ Refer to Moments and Centers of Mass for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral. If we allow a constant density function, then $$\bar{x} = \dfrac{M_y}{m}$$ and $$\bar{y} = \dfrac{M_x}{m}$$ give the centroid of the lamina. Suppose that the lamina occupies a region $$R$$ in the $$xy$$-plane and let $$\rho (x,y)$$ be its density (in units of mass per unit area) at any point $$(x,y)$$. Hence, $\rho(x,y) = \lim_{\Delta A \rightarrow 0} \dfrac{\Delta m}{\Delta A}$ where $$\Delta m$$ and $$\Delta A$$ are the mass and area of a small rectangle containing the point $$(x,y)$$ and the limit is taken as the dimensions of the rectangle go to $$0$$ (see the following figure). Figure $$\PageIndex{2}$$: The density of a lamina at a point is the limit of its mass per area in a small rectangle about the point as the area goes to zero. Just as before, we divide the region $$R$$ into tiny rectangles $$R_{ij}$$ with area $$\Delta A$$ and choose $$(x_{ij}^, y_{ij}^)$$ as sample points. Then the mass $$m_{ij}$$ of each $$R_{ij}$$ is equal to $$\rho (x_{ij}^, y_{ij}^) \Delta A$$ (Figure $$\PageIndex{2}$$). Let $$k$$ and $$l$$ be the number of subintervals in $$x$$ and $$y$$ respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections. Figure $$\PageIndex{3}$$: Subdividing the lamina into tiny rectangles $$R_{ij}$$ each containing a sample point $$(x_{ij}^,y_{ij}^)$$. Hence, the mass of the lamina is $m =\lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l \rho(x_{ij}^,y_{ij}^) \Delta A = \iint_R \rho(x,y) dA.$ Let’s see an example now of finding the total mass of a triangular lamina. Example $$\PageIndex{1}$$: Finding the Total Mass of a Lamina Consider a triangular lamina $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density $$\rho(x,y) = xy \, kg/m^2$$. Find the total mass. Solution A sketch of the region $$R$$ is always helpful, as shown in the following figure. Figure $$\PageIndex{4}$$: A lamina in the $$xy$$-plane with density $$\rho (x,y) = xy$$. Using the expression developed for mass, we see that $m = \iint_R \, dm = \iint_R \rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} xy \, dy \, dx = \int_{x=0}^{x=3} \left[ \left. x \dfrac{y^2}{2} \right\|_{y=0}^{y=3} \right] \, dx = \int_{x=0}^{x=3} \dfrac{1}{2} x (3 - x)^2 dx = \left.\left[ \dfrac{9x^2}{4} - x^3 + \dfrac{x^4}{8} \right]\right\|_{x=0}^{x=3} = \dfrac{27}{8}.$ The computation is straightforward, giving the answer $$m = \dfrac{27}{8} \, kg$$. Exercise $$\PageIndex{1}$$ Consider the same region $$R$$ as in the previous example, and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the total mass. $$\dfrac{9\pi}{8} \, kg$$ Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment $$M_z$$ about the $$x$$-axis for $$R$$ is the limit of the sums of moments of the regions $$R_{ij}$$ about the $$x$$-axis. Hence $M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^) \rho(x_{ij}^,y_{ij}^) \Delta A = \iint_R y\rho (x,y) dA$ Similarly, the moment $$M_y$$ about the $$y$$-axis for $$R$$ is the limit of the sums of moments of the regions $$R_{ij}$$ about the $$y$$-axis. Hence $M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^) \rho(x_{ij}^,y_{ij}^) \Delta A = \iint_R x\rho (x,y) dA$ Example $$\PageIndex{2}$$: Finding Moments Consider the same triangular lamina $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density $$\rho (x,y) = xy$$. Find the moments $$M_x$$ and $$M_y$$. Solution Use double integrals for each moment and compute their values: $M_x = \iint_R y\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x y^2 \, dy \, dx = \dfrac{81}{20},$ $M_y = \iint_R x\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x^2 y \, dy \, dx = \dfrac{81}{20},$ The computation is quite straightforward. Exercise $$\PageIndex{2}$$ Consider the same lamina $$R$$ as above and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the moments $$M_x$$ and $$M_y$$. $$M_x = \dfrac{81\pi}{64}$$ and $$M_y = \dfrac{81\pi}{64}$$ Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by $$\bar{x}$$ and the y-coordinate by $$\bar{y}$$. Specifically, $\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA}$ and $\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA}$ Example $$\PageIndex{3}$$: center of mass Again consider the same triangular region $$R$$ with vertices $$(0,0), \, (0,3), \, (3,0)$$ and with density function $$\rho (x,y) = xy$$. Find the center of mass. Solution Using the formulas we developed, we have $\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5},$ $\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5}.$ Therefore, the center of mass is the point $$\left(\dfrac{6}{5},\dfrac{6}{5}\right).$$ Analysis If we choose the density $$\rho(x,y)$$ instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid, $x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1,$ $y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1.$ Notice that the center of mass $$\left(\dfrac{6}{5},\dfrac{6}{5}\right)$$ is not exactly the same as the centroid $$(1,1)$$ of the triangular region. This is due to the variable density of $$R$$ If the density is constant, then we just use $$\rho(x,y) = c$$ (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina. Exercise $$\PageIndex{3}$$ Again use the same region $$R$$ as above and use the density function $$\rho (x,y) = \sqrt{xy}$$. Find the center of mass. $$\bar{x} = \dfrac{M_y}{m} = \dfrac{81\pi/64}{9\pi/8} = \dfrac{9}{8}$$ and $$\bar{y} = \dfrac{M_x}{m} = \dfrac{81\pi}{9\pi/8} = \dfrac{0}{8}$$. Once again, based on the comments at the end of Example $$\PageIndex{3}$$, we have expressions for the centroid of a region on the plane: $x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} \, \text{and} \, y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA}.$ We should use these formulas and verify the centroid of the triangular region referred to in the last three examples. Example $$\PageIndex{4}$$: Finding Mass, Moments, and Center of Mass Find the mass, moments, and the center of mass of the lamina of density $$\rho(x,y) = x + y$$ occupying the region $$R$$ under the curve $$y = x^2$$ in the interval $$0 \leq x \leq 2$$ (see the following figure). Figure $$\PageIndex{5}$$: Locating the center of mass of a lamina $$R$$ with density $$\rho(x,y) = x+y$$. Solution First we compute the mass $$m$$. We need to describe the region between the graph of $$y = x^2$$ and the vertical lines $$x = 0$$ and $$x = 2$$: $m = \iint_R dm = \iint_R \rho (x,y) dA = \int_{x=0}{x=2} \int_{y=0}^{y=x^2} (x + y) dy \, dx = \int_{x=0}^{x=2} \left[\left. xy + \dfrac{y^2}{2}\right\|_{y=0}^{y=x^2} \right] dx$ $= \int_{x=0}^{x=2} \left[ x^3 + \dfrac{x^4}{2} \right] dx = \left.\left[ \dfrac{x^4}{4} + \dfrac{x^5}{10}\right] \right\|_{x=0}^{x=2} = \dfrac{36}{5}.$ Now compute the moments $$M_x$$ and $$M_y$$: $M_x = \iint_R y \rho (x,y) dA = \int_{x=0}^{x=2} \int_{y=0}^{y=x^2} y(x + y) dy \, dx = \dfrac{80}{7},$ $M_y = \iint_R x \rho (x,y) dA = \int_{x=0}^{x=2} \int_{y=0}^{y=x^2} x(x + y) dy \, dx = \dfrac{176}{15}.$ Finally, evaluate the center of mass, $\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x \rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{176/15}{36/5} = \dfrac{44}{27},$ $\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y \rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{80/7}{36/5} = \dfrac{100}{63}.$ Hence the center of mass is $$(\bar{x},\bar{y}) = \left(\dfrac{44}{27}, \dfrac{100}{63} \right)$$. Exercise $$\PageIndex{4}$$ Calculate the mass, moments, and the center of mass of the region between the curves $$y = x$$ and $$y = x^2$$ with the density function $$\rho(x,y) = x$$ in the interval $$0 \leq x \leq 1$$. $$\bar{x} = \dfrac{M_y}{m} = \dfrac{1/20}{1/12} = \dfrac{3}{5}$$ and $$\bar{y} = \dfrac{M_x}{m} = \dfrac{1/24}{1/12} = \dfrac{1}{2}$$ Example $$\PageIndex{5}$$: Finding a Centroid Find the centroid of the region under the curve $$y = e^x$$ over the interval $$1 \leq x \leq 3$$ (Figure $$\PageIndex{6}$$). Figure $$\PageIndex{6}$$: Finding a centroid of a region below the curve $$y = e^x$$. Solution To compute the centroid, we assume that the density function is constant and hence it cancels out: $x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} \, and \, y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA},$ $x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} = \dfrac{\int_{x=1}^{x=3} \int_{y=0}^{y=e^x} x \, dy \, dx}{\int_{x=1}^{x=3} \int_{y=0}^{y=e^x} dy \, dx} = \dfrac{\int_{x=1}^{x=3} xe^x dx}{\int_{x=1}^{x=3} e^x dx} = \dfrac{2e^3}{e^3 - e} = \dfrac{2e^2}{e^2 - 1},$ $y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA} = \dfrac{\int_{x=1}^{x=3} \int_{y=0}^{y=e^x} y \, dy \, dx}{\int_{x=1}^{x=3} \int_{y=0}^{y=e^x} dy \, dx} = \dfrac{\int_{x=1}^{x=3} \dfrac{e^{2x}}{2} dx}{\int_{x=1}^{x=3} e^x dx} =\dfrac{\dfrac{1}{4} e^2 (e^4 - 1)}{e(e^2 - 1)} = \dfrac{1}{4} e(e^2 + 1).$ Thus the centroid of the region is $(x_c,y_c) = \left( \dfrac{2e^2}{e^2 - 1}, \dfrac{1}{4} e(e^2 + 1)\right).$ Exercise $$\PageIndex{5}$$ Calculate the centroid of the region between the curves $$y = x$$ and $$y = \sqrt{x}$$ with uniform density in the interval $$0 \leq x \leq 1$$. $$x_c = \dfrac{M_y}{m} = \dfrac{1/15}{1/6} = \dfrac{2}{5}$$ and $$y_c = \dfrac{M_x}{m} = \dfrac{1/12}{1/6} = \dfrac{1}{2}$$ ## Moments of Inertia For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section $$6.6$$. The moment of inertia of a particle of mass $$m$$ about an axis is $$mr^2$$ where $$r$$ is the distance of the particle from the axis. We can see from Figure $$PageIndex{3}$$ that the moment of inertia of the subrectangle $$R_{ij}$$ about the $$x$$-axis is $$(y_{ij}^)^2 \rho(x_{ij}^,y_{ij}^) \Delta A$$. Similarly, the moment of inertia of the subrectangle $$R_{ij}$$ about the $$y$$-axis is $$(x_{ij}^)^2 \rho(x_{ij}^,y_{ij}^) \Delta A$$. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis. The moment of inertia $$I_x$$ about the $$x$$-axis for the region $$R$$ is the limit of the sum of moments of inertia of the regions $$R_{ij}$$ about the $$x$$-axis. Hence $I_x = \lim_{k,l\rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^)^2 m_{ij} = \lim_{k,l\rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^)^2 \rho (x_{ij}^, y_{ij}^) \Delta A = \iint_R y^2 \rho(x,y)dA.$ Similarly, the moment of inertia $$I_y$$ about the $$y$$-axis for $$R$$ is the limit of the sum of moments of inertia of the regions $$R_{ij}$$ about the $$y$$-axis. Hence $I_y = \lim_{k,l\rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^)^2 m_{ij} = \lim_{k,l\rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^)^2 \rho (x_{ij}^, y_{ij}^) \Delta A = \iint_R x^2 \rho(x,y)dA.$ Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by $$I_0$$ and obtain it by adding the moments of inertia $$I_x$$ and $$I_y$$. Hence $I_0 = I_x + I_y = \iint_R (x^2 + y^2) \rho (x,y) dA.$ All these expressions can be written in polar coordinates by substituting $$x = r \, \cos \, \theta, \, y = r \, \sin \, \theta$$, and $$dA = r \, dr \, d\theta$$. For example, $$I_0 = \iint_R r^2 \rho (r \, \cos \, \theta, \, r \, \sin \, \theta)dA$$. Example $$\PageIndex{6}$$: Finding Moments of Inertia for a Triangular Lamina Use the triangular region $$R$$ with vertices $$(0,0), \, (2,2)$$, and $$(2,0)$$ and with density $$\rho (x,y) = xy$$ as in previous examples. Find the moments of inertia. Solution Using the expressions established above for the moments of inertia, we have $I_x = \iint_R y^2 \rho(x,y) dA = \int_{x=0}^{x=2} \int_{y=0}^{y=x} xy^3 dy \, dx = \dfrac{8}{3},$ $I_y = \iint_R x^2 \rho(x,y) dA = \int_{x=0}^{x=2} \int_{y=0}^{y=x} x^3y dy \, dx = \dfrac{16}{3},$ $I_0 = \iint_R (x^2 + y^2) \rho(x,y) dA = \int_0^2 \int_0^x (x^2 + y^2) xy \, dy \, dx = I_x + I_y = 8$ Exercise $$\PageIndex{6}$$ Again use the same region $$R$$ as above and the density function $$\rho (x,y) = \sqrt{xy}$$. Find the moments of inertia. $I_x = \int_{x=0}^{x=2} \int_{y=0}^{y=x} y^2 \sqrt{xy} \, dy \, dx = \dfrac{64}{35}$ and $I_y = \int_{x=0}^{x=2} \int_{y=0}^{y=x} x^2 \sqrt{xy} \, dy \, dx = \dfrac{64}{35}.$ Also, $I_0 = \int_{x=0}^{x=2} \int_{y=0}^{y=x} (x^2 + y^2) \sqrt{xy} \, dy \, dx = \dfrac{128}{21}$ As mentioned earlier, the moment of inertia of a particle of mass $$m$$ about an axis is $$mr^2$$ where $$r$$ is the distance of the particle from the axis, also known as the radius of gyration. Hence the radii of gyration with respect to the $$x$$-axis, the $$y$$-axis and the origin are $R_x = \sqrt{\dfrac{I_x}{m}}, \, R_y = \sqrt{\dfrac{I_y}{m}}, \, and \, R_0 = \sqrt{\dfrac{I_0}{m}},$ respectively. In each case, the radius of gyration tells us how far (perpendicular distance) from the axis of rotation the entire mass of an object might be concentrated. The moments of an object are useful for finding information on the balance and torque of the object about an axis, but radii of gyration are used to describe the distribution of mass around its centroidal axis. There are many applications in engineering and physics. Sometimes it is necessary to find the radius of gyration, as in the next example. Example $$\PageIndex{7}$$: Finding the Radius of Gyration for a Triangular Lamina Consider the same triangular lamina $$R$$ with vertices $$(0,0), \, (2,2)$$, and $$(2,0)$$ and with density $$\rho(x,y) = xy$$ as in previous examples. Find the radii of gyration with respect to the $$x$$-axis the $$y$$-axis and the origin. Solution If we compute the mass of this region we find that $$m = 2$$. We found the moments of inertia of this lamina in Example $$\PageIndex{4}$$. From these data, the radii of gyration with respect to the $$x$$-axis, $$y$$-axis and the origin are, respectively, \begin{align} R_x &= \sqrt{\dfrac{I_x}{m}} = \sqrt{\dfrac{8/3}{2}} = \sqrt{\dfrac{8}{6}} = \dfrac{2\sqrt{3}}{3},\\R_y &= \sqrt{\dfrac{I_y}{m}} = \sqrt{\dfrac{16/3}{2}} = \sqrt{\dfrac{8}{3}} = \dfrac{2\sqrt{6}}{3}, \\R_0 &= \sqrt{\dfrac{I_0}{m}} = \sqrt{\dfrac{8}{2}} = \sqrt{4} = 2.\end{align} Exercise $$\PageIndex{7}$$ Use the same region $$R$$ from Example $$\PageIndex{7}$$ and the density function $$\rho (x,y) = \sqrt{xy}$$. Find the radii of gyration with respect to the $$x$$-axis, the $$y$$-axis, and the origin. Hint Follow the steps shown in the previous example. $$R_x = \dfrac{6\sqrt{35}}{35}, \, R_y = \dfrac{6\sqrt{15}}{15},$$ and $$R_0 = \dfrac{4\sqrt{42}}{7}$$. ## Center of Mass and Moments of Inertia in Three Dimensions All the expressions of double integrals discussed so far can be modified to become triple integrals. Definition If we have a solid object $$Q$$ with a density function $$\rho(x,y,z)$$ at any point $$(x,y,z)$$ in space, then its mass is $m = \iiint_Q \rho(x,y,z) dV.$ Its moments about the $$xy$$-plane the $$xz$$-plane and the $$yz$$-plane are $M_{xy} = \iiint_Q z\rho (x,y,z) dV, \, M_{xz} = \iiint_Q y\rho(x,y,z) dV, \, M_{yz} = \iiint_Q x\rho(x,y,z) dV.$ If the center of mass of the object is the point $$(\bar{x}, \bar{y}, \bar{z})$$, then $\bar{x} = \dfrac{M_{yz}}{m}, \, \bar{y} = \dfrac{M_{xz}}{m}, \, \bar{z} = \dfrac{M_{xy}}{m}.$ Also, if the solid object is homogeneous (with constant density), then the center of mass becomes the centroid of the solid. Finally, the moments of inertia about the $$yz$$-plane, $$xz$$-plane, and the $$xy$$-plane are $I_x = \iiint_Q (y^2 + z^2) \, \rho (x,y,z) \, dV,$ $I_y = \iiint_Q (x^2 + z^2) \, \rho (x,y,z) \, dV,$ $I_z = \iiint_Q (x^2 + y^2) \, \rho (x,y,z) \, dV.$ Example $$\PageIndex{8}$$: Finding the Mass of a Solid Suppose that $$Q$$ is a solid region bounded by $$x + 2y + 3z = 6$$ and the coordinate planes and has density $$\rho (x,y,z) = x^2 yz$$. Find the total mass. Solution The region $$Q$$ is a tetrahedron (Figure $$\PageIndex{7}$$) meeting the axes at the points $$(6,0,0), \, (0,3,0),$$ and $$(0,0,2)$$. To find the limits of integration, let $$z = 0$$ in the slanted plane $$z = \dfrac{1}{3} (6 - x - 2y)$$. Then for $$x$$ and $$y$$ find the projection of $$Q$$ onto the $$xy$$-plane, which is bounded by the axes and the line $$x + 2y = 6$$. Hence the mass is $m = \iiint_Q \rho (x,y,z) dV = \int_{x=0}^{x=6} \int_{y=0}^{y=1/2(6-x)} \int_{z=0}^{z=1/3(6-x-2y)} x^2 yz \, dz \, dy \, dx = \dfrac{108}{35}$ Figure $$\PageIndex{7}$$: Finding the mass of a three-dimensional solid $$Q$$. Exercise $$\PageIndex{8}$$ Consider the same region $$Q$$ (Figure $$\PageIndex{7}$$), and use the density function $$\rho (x,y,z) = xy^2z$$. Find the mass. Hint Follow the steps in the previous example. $$\dfrac{54}{35} = 1.543$$ Example $$\PageIndex{9}$$: Finding the Center of Mass of a Solid Suppose $$Q$$ is a solid region bounded by the plane $$x + 2y + 3z = 6$$ and the coordinate planes with density $$\rho (x,y,z) = x^2yz$$ (see Figure $$\PageIndex{7}$$). Find the center of mass using decimal approximation. Solution We have used this tetrahedron before and know the limits of integration, so we can proceed to the computations right away. First, we need to find the moments about the $$xy$$-plane, the $$xz$$-plane, and the $$yz$$-plane: $M_{xy} = \iiint_Q z\rho (x,y,z) dV = \int_{x=0}^{x=6} \int_{y=0}^{y=1/2(6-x)} \int_{z=0}^{z=1/3(6-x-2y)} x^2 yz^2 dz \, dy \, dx = \dfrac{54}{35} \approx 1.543,$ $M_{xz} = \iiint_Q y\rho (x,y,z) dV = \int_{x=0}^{x=6} \int_{y=0}^{y=1/2(6-x)} \int_{z=0}^{z=1/3(6-x-2y)} x^2 y^2z \, dz \, dy \, dx = \dfrac{81}{35} \approx 2.314,$ $M_{yz} = \iiint_Q x\rho (x,y,z) dV = \int_{x=0}^{x=6} \int_{y=0}^{y=1/2(6-x)} \int_{z=0}^{z=1/3(6-x-2y)} x^3 yz \, dz \, dy \, dx = \dfrac{243}{35} \approx 6.943.$ Hence the center of mass is $\bar{x} = \dfrac{M_{yz}}{m}, \, \bar{y} = \dfrac{M_{xz}}{m}, \, \bar{z} = \dfrac{M_{xy}}{m},$ $\bar{x} = \dfrac{M_{yz}}{m} = \dfrac{243/35}{108/35} = \dfrac{243}{108} = 2.25,$ $\bar{y} = \dfrac{M_{xz}}{m} = \dfrac{81/35}{108/35} = \dfrac{81}{108} = 0.75,$ $\bar{z} = \dfrac{M_{xy}}{m} = \dfrac{54/35}{108/35} = \dfrac{54}{108} = 0.5$ The center of mass for the tetrahedron $$Q$$ is the point $$(2.25, 0.75, 0.5)$$. Exercise $$\PageIndex{9}$$ Consider the same region $$Q$$ (Figure $$\PageIndex{7}$$) and use the density function $$\rho (x,y,z) = xy^2z$$. Find the center of mass. Hint Check that $$M_{xy} = \dfrac{27}{35}, \, M_{xz} = \dfrac{243}{140},$$ and $$M_{yz} = \dfrac{81}{35}$$. Then use $$m$$ from a previous checkpoint question. $$\left(\dfrac{3}{2}, \dfrac{9}{8}, \dfrac{1}{2}\right)$$ We conclude this section with an example of finding moments of inertia $$I_x, \, I_y$$, and $$I_z$$. Example $$\PageIndex{10}$$: Finding the Moments of Inertia of a Solid Suppose that $$Q$$ is a solid region and is bounded by $$x + 2y + 3z = 6$$ and the coordinate planes with density $$\rho (x,y,z) = x^2 yz$$ (see Figure $$\PageIndex{7}$$). Find the moments of inertia of the tetrahedron $$Q$$ about the $$yz$$-plane, the $$xz$$-plane, and the $$xy$$-plane. Solution Once again, we can almost immediately write the limits of integration and hence we can quickly proceed to evaluating the moments of inertia. Using the formula stated before, the moments of inertia of the tetrahedron $$Q$$ about the $$yz$$-plane, the $$xz$$-plane, and the $$xy$$-plane are $I_x = \iiint_Q (y^2 + z^2) \rho(x,y,z) dV,$ $I_y = \iiint_Q (x^2 + z^2) \rho(x,y,z) dV,$ and $I_z = \iiint_Q (x^2 + y^2) \rho(x,y,z) dV \, with \, \rho(x,y,z) = x^2yz.$ Proceeding with the computations, we have \begin{align} I_x &= \iiint_Q (y^2 + z^2) x^2 \rho(x,y,z) dV \\[5pt] &= \int_{x=0}^{x=6} \int_{y=0}^{y=\dfrac{1}{2}(6-x)} \int_{z=0}^{z=\dfrac{1}{3}(6-x-2y)} (y^2 + z^2) x^2 yz \, dz \, dy \, dx = \dfrac{117}{35} \approx 3.343,\end{align} \begin{align} I_y &= \iiint_Q (x^2 + z^2) x^2 \rho(x,y,z) dV \\[5pt] &= \int_{x=0}^{x=6} \int_{y=0}^{y=\dfrac{1}{2}(6-x)} \int_{z=0}^{z=\dfrac{1}{3}(6-x-2y)} (x^2 + z^2) x^2 yz \, dz \, dy \, dx = \dfrac{684}{35} \approx 19.543, \end{align} \begin{align} I_z &= \iiint_Q (x^2 + y^2) x^2 \rho(x,y,z) dV \\[5pt] &= \int_{x=0}^{x=6} \int_{y=0}^{y=\dfrac{1}{2}(6-x)} \int_{z=0}^{z=\dfrac{1}{3}(6-x-2y)} (x^2 + y^2) x^2 yz \, dz \, dy \, dx = \dfrac{729}{35} \approx 20.829. \end{align} Thus, the moments of inertia of the tetrahedron $$Q$$ about the $$yz$$-plane, the $$xz$$-plane, and the $$xy$$-plane are $$117/35, \, 684/35$$, and $$729/35$$, respectively. Exercise $$\PageIndex{10}$$ Consider the same region $$Q$$ (Figure $$\PageIndex{7}$$), and use the density function $$\rho(x,y,z) = xy^2z$$. Find the moments of inertia about the three coordinate planes. The moments of inertia of the tetrahedron $$Q$$ about the $$yz$$-plane, the $$xz$$-plane, and the $$xy$$-plane are $$99/35, \, 36/7$$ and $$243/35$$, respectively. ## Key Concepts Finding the mass, center of mass, moments, and moments of inertia in double integrals: • For a lamina $$R$$ with a density function $$\rho (x,y)$$ at any point $$(x,y)$$ in the plane, the mass is $m = \iint_R \rho (x,y)dA.$ • The moments about the $$x$$-axis and $$y$$-axis are $M_x = \iint_R y\rho(x,y) dA \, and \, M_y = \iint_R x\rho(x,y)dA.$ • The center of mass is given by $$\bar{x} = \dfrac{M_y}{m}, \, \bar{y} = \dfrac{M_x}{m}$$. • The center of mass becomes the centroid of the plane when the density is constant. • The moments of inertia about the $$x$$-axis, $$y$$-axis, and the origin are $I_x = \iint_R y^2 \rho(x,y) dA, \, I_y = \iint_R x^2 \rho(x,y) dA, \, and \, I_0 = I_x + I_y = \iint_R (x^2 + y^2) \rho(x,y) dA.$ Finding the mass, center of mass, moments, and moments of inertia in triple integrals: • For a solid object $$Q$$ with a density function $$\rho(x,y,z)$$ at any point $$(x,y,z)$$ in space, the mass is $m = \iiint_Q \rho (x,y,z) dV.$ • The moments about the $$xy$$-plane, the $$xz$$-plane, and the $$yz$$-plane are $M_{xy} = \iiint_Q z\rho (x,y,z)dV, \, M_{xz} = \iiint_Q y\rho (x,y,z)dV, \, M_{yz} = \iiint_Q x\rho (x,y,z)dV$ • The center of mass is given by $$\bar{x} = \dfrac{M_{yz}}{m}, \, \bar{y} = \dfrac{M_{xz}}{m}, \, \bar{z} = \dfrac{M_{xy}}{m}.$$ • The center of mass becomes the centroid of the solid when the density is constant. • The moments of inertia about the $$yz$$-plane, the $$xz$$-plane, and the $$xy$$-plane are $I_x = \iiint_Q (y^2 + z^2) \, \rho (x,y,z) \, dV, \, I_y = \iiint_Q (x^2 + z^2) \, \rho (x,y,z) \, dV, \, I_z = \iiint_Q (x^2 + y^2) \, \rho (x,y,z) \, dV.$ ## Key Equations • Mass of a lamina $m = \lim_{k,l \rightarrow\infty} \sum_{i=1}^k \sum_{j=1}^l m_{ij} = \lim_{k,l \rightarrow\infty} \sum_{i=1}^k \sum_{j=1}^l \rho (x_{ij}^8, y_{ij}^) \Delta A = \iint_R \rho(x,y) dA$ • Moment about the x-axis $M_x = \lim_{k,l \rightarrow\infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^) m_{ij} = \lim_{k,l \rightarrow\infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^) \rho (x_{ij}^8, y_{ij}^) \Delta A = \iint_R y\rho(x,y)dA$ • Moment about the y-axis $M_y = \lim_{k,l \rightarrow\infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^) m_{ij} = \lim_{k,l \rightarrow\infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^) \rho (x_{ij}^8, y_{ij}^*) \Delta A = \iint_R x\rho(x,y)dA$ • Center of mass of a lamina $\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y) dA} \, and \, \bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y) dA}$<\|endoftext\|>	4.59375
1,128	# Math posted by . Determine which, if any, of the three statements are equivalent. I) If the carpet is not clean, then Sheila will run the vacuum. II) Either the carpet is not clean or Sheila will not run the vacuum. III) If the carpet is clean, Sheila will not run the vacuum. • Math - • Math - I think I and III are equivalent?? • Math - Right! :-) • Math - great!! thank you :) • Math - You're welcome. • Math - I don't agree with this answer. I think you are studying on how to solve such problems using truth tables, so you should write down the three truth tables for the three statements and see which ones match. E.g., put: A = carpet is not clean. B = Sheila will run the vacuum. For statement 1 (S1) we have: A=False, B=True, S1 = True A=False, B=False, S1 = True A=True, B=True, S1 = True A=True, B=False, S1 = False If you do this for the two others you see that the table for statement 2 and 3 are the same. You can also see this from the fact that A implies B is equivalent to Not(A) OR B A implies B means that if A is true, then B has to be true. Terefore, only if A is true, there is a requirement for B to be true. The only way to violate this condition is thus when B is false and A is True. So, for the statement to be true requires that either A is false or that B is true. Now A being false making A implies B true can be counterintuitive, and this makes it dangerous to rely on intuition to analyze such problems. Statement 3 is equivalent to: Not(A) implies Not(B) = Not(Not A) Or Not(B) = A Or Not(B) which is exactly what statement 2 says. Another thing to remember is how to write the statement: A implies B in terms of Not(A) and Not(B), as you see statments 1 and 3 are not equivalent. We have: A implies B = Not(A) OR B = B OR Not(A) = Not(Not(B)) OR Not(A) If you put Not(B) = X and Not(A) = Y, you see that the last line is: Not(X) OR Y and that is the same as: X implies Y So, we find that: A implies B = Not(B) implies Not(A) • Math - Simply put, p --> q is not equivalent to ~p --> ~q. Example: If I am in Paris, then I am in France. Therefore, if I am not in Paris, then I am not in France. The conclusion is invalid. Sure, I can be in some other country such as England, but I can be in another city in France. • Math - Okay.... so I and III are not equivalent?? ## Similar Questions 1. ### CRT 205 Crt 205 Vacuum Sales Digital Story Text Axia College Material Text of the Vacuum Sales Digital Story This text provides the script of the Vacuum Sales digital story. Use this text as an alternative or supplement to viewing the digital … 2. ### college mathematics 8. Determine which, if any, of the three statements are equivalent. Give a reason for your conclusion. Show complete work. I.) If the dog wags its tail, then the dog is not calm. II.) Either the dog does not wag its tail or the dog … 3. ### Math Carpet plus installs carpet for \$100 plus \$8 per square yard of carpet. Carpet World charges \$75 for installation and \$10 per square yard of carpet. Find the number of square yards of carpet for which the cost including carpet and … 4. ### Pre-Algebra Carpet Masters charges \$9.50 per square yard to clean a carpet. If you have two rooms with an area of six square yards each, how much will it cost to have the carpets cleaned? 5. ### geometry you plan to carpet 468ft of space. carpet costs \$26.75 per square yard and the tax is 7%. the carpet is sold in 12 ft rolls. what is the cost of the carpet? 6. ### algebra CLEAN CARPET COMPANYS CLEAN 3 ROOMS OF CARPETING FOR 144 DOLLARS. WHICH OF THE FOLLOWING IS AN EQUIVALENT RATE ? 7. ### Math If the cost is 3.00 doars per sq ft to clean a carpet, how much will it cost to clean a carpet 12.5 ft x 23ft? 8. ### English What are you going to do for your home this weekend? 9. ### Math A contractor buys 14 yd of nylon carpet and 20 yd of wool carpet for \$1738. A second purchase, at the same prices, includes 16 yd of nylon carpet and 25 yd of wool carpet for \$2102. Find the cost per yard of the wool carpet. 10. ### math Sheila needs to hire a babysitter for the evening. The babysitter charges \$10 for each hour and also charges for the time she spends driving to and from the house. She thinks it will take 1.5 hours of total driving time. Sheila is … More Similar Questions<\|endoftext\|>	4.46875
924	Blank verse is a type of unrhymed poetry with a regular pattern of rhythm, or meter. It is written in iambic pentameter, meaning that each line is made of five pairs of unstressed/stressed syllables, for a total of 10 syllables. Blank verse is a highly regarded dramatic and narrative verse form in English, and it is the standard form for dramatic verse in Italian and German. Blank verse was adapted from unrhymed Greek and Latin heroic verse. It was introduced in 16th-century Italy along with other classical meters. The Italian humanist Francesco Maria Molza attempted to write unrhymed verse in 1514 in his translation of Virgil’s Aeneid. About the same time, Giovanni Rucellai experimented with the poem Le api (1539). He was the first to use the term versi sciolti, which became translated into English as “blank verse.” It soon became the standard meter of Italian Renaissance drama, and poets such as Ludovico Ariosto and Torquato Tasso used it in their works. Henry Howard, Earl of Surrey, introduced the meter, along with the sonnet and other Italian verse forms, to England in the early 16th century. Thomas Sackville, 1st earl of Dorset, and Thomas Norton used blank verse for the first English tragic drama, Gorboduc (first performed 1561). Shortly thereafter Christopher Marlowe developed the form’s musical qualities and emotional power in his plays Tamburlaine the Great, Doctor Faustus, and Edward II. This blank-verse passage is from Tamburlaine (performed 1587): Nature, that fram’d us of four elements Warring within our breasts for regiment, Doth teach us all to have aspiring minds: Our souls, whose faculties can comprehend The wondrous architecture of the world, And measure every wandering planet’s course, Still climbing after knowledge infinite, And always moving as the restless spheres, Wills us to wear ourselves and never rest, Until we reach the ripest fruit of all, That perfect bliss and sole felicity, The sweet fruition of an earthly crown. Likewise, William Shakespeare used blank verse to create some of the greatest poetry in English drama. In his early plays, he combined it with prose and a 10-syllable rhymed couplet; he later employed a blank verse dependent on stress rather than syllable length. Here is an example of blank verse from Shakespeare’s The Merchant of Venice (written about 1596–97): How sweet the moonlight sleeps upon this bank! Here will we sit and let the sounds of music Creep in our ears. Soft stillness and the night Become the touches of sweet harmony. Sit, Jessica. Look how the floor of heaven Is thick inlaid with patines of bright gold. There’s not the smallest orb which thou behold’st But in his motion like an angel sings, Still quiring to the young-ey’d cherubims; Such harmony is in immortal souls; But whilst this muddy vesture of decay Doth grossly close it in, we cannot hear it. The next great blank-verse practitioner was John Milton, who wrote the epic poem Paradise Lost (1667). In the 18th century, James Thomson used blank verse in his long descriptive poem The Seasons, and Edward Young used it in The Complaint; or, Night Thoughts, a long poem on death. Later authors to use blank verse include William Wordsworth in his autobiography of the poetic spirit, The Prelude (completed 1805–06; published 1850). Percy Bysshe Shelley used blank verse in his drama The Cenci (1819), and John Keats employed it in his epic poem Hyperion (1820). The extreme flexibility of blank verse can be seen in its range from the high tragedy of Shakespeare to the low-key conversational tone of Robert Frost in A Masque of Reason (1945). Blank verse was established in German drama by Gotthold Ephraim Lessing’s Nathan der Weise (1779). Examples of its use are found in the writings of Johann Wolfgang von Goethe, Friedrich Schiller, and Gerhart Hauptmann. It was also used extensively in Swedish, Russian, and Polish dramatic verse.<\|endoftext\|>	3.96875
846	# Class 7 Maths Chapter 2 Exercise 2.5 Pdf Notes NCERT Solutions Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.5 pdf notes:- Exercise 2.5 Class 7 maths Chapter 2 Pdf Notes:- ## Ncert Solution for Class 6 Maths Chapter 2 Fractions and decimals Exercise 2.5 Tips:- John has  15.50 and Salma has  15.75. Who has more money? To find this we need to compare the decimal numbers 15.50 and 15.75. To do this, we first compare the digits on the left of the decimal point, starting from the leftmost digit. Here both the digits 1 and 5, to the left of the decimal point, are same. So we compare the digits on the right of the decimal point starting from the tenths place. We find that 5 < 7, so we say 15.50 < 15.75. Thus, Salma has more money than John. If the digits at the tenths place are also same then compare the digits at the hundredths place and so on. Now compare quickly, 35.63 and 35.67; 20.1 and 20.01; 19.36 and 29.36. While converting lower units of money, length and weight, to their higher units, we are required to use decimals. For example, 3 paise =  3 100 =  0.03, 5g = 5 1000 kg = 0.005 kg, 7 cm = 0.07 m. Write 75 paise =  ______, 250 g = _____ kg, 85 cm = _____m. We also know how to add and subtract decimals. Thus, 21.36 + 37.35 is bserve the shift of the decimal point of the products in the table. Here the numbers are multiplied by 10,100 and 1000. In 1.76 × 10 = 17.6, the digits are same i.e., 1, 7 and 6. Do you observe this in other products also? Observe 1.76 and 17.6. To which side has the decimal point shifted, right or left? The decimal point has shifted to the right by one place. Note that 10 has one zero over 1. In 1.76×100 = 176.0, observe 1.76 and 176.0. To which side and by how many digits has the decimal point shifted? The decimal point has shifted to the right by two places. Note that 100 has two zeros over one. Do you observe similar shifting of decimal point in other products also? So we say, when a decimal number is multiplied by 10, 100 or 1000, the digits in the product are same as in the decimal number but the decimal point in the product is shifted to the right by as, many of places as there are zeros over one. Based on these observations we can now say 0.07 × 10 = 0.7, 0.07 × 100 = 7 and 0.07 × 1000 = 70. Can you now tell 2.97 × 10 = ? 2.97 × 100 = ? 2.97 × 1000 = ? Can you now help Reshma to find the total amount i.e.,  8.50 × 150, that she has to pay? #### Test Paper Of Class 8th • Maths 8th Class • Science 8th class • Sst 8th Class • #### Test Paper Of Class 7th • Maths 7th Class • Science 7th class • #### Test Paper Of Class 6th • Maths 6th Class • Science 6th class<\|endoftext\|>	4.90625
1,240	Respiratory syncytial virus What is respiratory syncytial virus? Respiratory syncytial (sin-sye-shul) virus, or "RSV," is a virus that causes respiratory infections in people of all ages. In Minnesota most RSV infections occur between October and April. Usually older children and adults will have only a cold, but in infants and young children it can cause other respiratory illnesses. An RSV illness can be especially severe in very young or premature infants, those with heart or lung disease, and those with compromised immune systems. In infants and young children, RSV is a common cause of pneumonia and bronchiolitis—inflamed airways in the lungs. (See the education sheets "Pneumonia" and "Bronchiolitis." Older children and adults may get bronchitis from RSV. RSV infects most children during their first 3 years of life. Immunity to RSV is short-lived, and it is common to get it again. What are the signs of RSV? Newborns and premature babies - irritable (crabby) - poor feeding - may have slight cold symptoms such as a stuffy nose - lethargic (very sleepy, hard to arouse) - may have episodes of apnea (breathing stops) Infants and young children - fast breathing - hard time breathing Older children and adults - watery eyes - runny nose How is RSV spread? RSV is easily spread from person to person, especially during the first few days. A person will get the illness 2 to 8 days after being exposed to it. People catch RSV by direct contact with respiratory secretions (saliva and mucus), or things that have been in contact with it. The virus can live on surfaces for many hours, and for a half-hour or more on the hands. Frequent hand hygiene is the most important way to prevent the spread of RSV. All family members should wash hands well with soap and water for at least 15 seconds, or use an alcohol hand sanitizer, such as Purell®, and avoid touching their nose and eyes. Teach children not to share food or beverages, and to cover their mouth with their forearm when they sneeze or cough. How is RSV treated? If an infant or young child is very ill, mucus from the nose will be tested to see if RSV is the cause. Check with your doctor about the results of this test. Most children with RSV do not need a hospital stay. The children who do need to be in the hospital are those with other health problems or very young infants who are prone to develop pneumonia and bronchiolitis. Most children with RSV do not need prescribed medicines. Because RSV is a virus, antibiotics do not help. When caring for your child at home, symptoms may be treated as follows: Give acetaminophen (Tylenol® or another brand) according to package directions. Do not give aspirin. - Raise the head of the crib. - Buy saline nose drops at a drug store, or stir ¼ teaspoon salt into ½ cup water. - Lay the child on the back and put 2 or 3 drops of saline into the nose. - Wait about ½ minute and suction the nose with a bulb syringe. - Do this before feeding and sleeping and as often as needed. Do not give any non-prescription medicine without checking with your doctor. If your child is wheezing or has an ear infection, your doctor may prescribe medicine. If my child is admitted to the hospital, what should I expect during the hospital stay? - Your child may receive oxygen. - Your child may have an intravenous line (IV) to provide liquids until your child can eat and drink well. - Your child will be closely monitored. - Your child may be given medicine in the form of a mist to breathe into his or her lungs. - Your child will be placed in "contact" precautions. This means all staff will do hand hygiene and wear a gown and gloves each time they enter your child's room. You will need to do hand hygiene each time you leave and enter your child's room. If your child is coughing, he or she will be placed in contact and droplet precautions. All staff will do hand hygiene, and wear a mask, eye protection, gown and gloves each time they enter your child's room. When should I call the doctor? - a child younger than 1 year old with heart or lung disease develops a cold - refusing to drink - vomiting, not able to keep liquids down - pulling at the ears - hard time breathing - wheezing or gasping for air - breathing fast (more than 60 times per minute) - hard time staying awake - sucking in on the spaces between the ribs with each breath - has a bluish or grayish tinge around or inside the lips - call 911 When can my child return to day care? Children may return to day care and other group activities when they have no fever, feed normally, and feel well—even if they still have a cough or a runny nose. This sheet is not specific to your child, but provides general information. If you have any questions, please call the clinic. Children's Hospitals and Clinics of Minnesota 2525 Chicago Avenue South Minneapolis, MN 55404 Last Reviewed 7/2015 © Copyright This page is not specific to your child, but provides general information on the topic above. If you have any questions, please call your clinic. For more reading material about this and other health topics, please call or visit Children's Family Resource Center library, or visit www.childrensmn.org/educationmaterials. © 2019 Children's Hospitals and Clinics of Minnesota<\|endoftext\|>	3.953125
543	The hiatus of global warming in the Northern Hemisphere during the mid-20th century may have been due to an abrupt cooling event centered over the North Atlantic around 1970, rather than the cooling effects of tropospheric pollution. Global temps 1850-2010 David W. J. Thompson, an atmospheric science professor at Colorado State University, is the lead author on the paper. Other authors are John M. Wallace at the University of Washington, and John J. Kennedy at the Met Office and Phil D. Jones of the University of East Anglia, both in the United Kingdom. David W.J. Thompson, professor of atmospheric science at Colorado State University, is the lead author of a Nature paper that shows sudden ocean cooling contributed to a global warming hiatus in the middle 20th century in the Northern Hemisphere. Credit: Colorado State University The international team of scientists discovered an unexpectedly abrupt cooling event that occurred between roughly 1968 and 1972 in Northern Hemisphere ocean temperatures. The research indicates that the cooling played a key role in the different rates of warming seen in the Northern and Southern Hemispheres in the middle 20th century. "We knew that the Northern Hemisphere oceans cooled during the mid-20th century, but the sudden nature of that cooling surprised us," Thompson said. While the temperature drop was evident in data from all Northern Hemisphere oceans, it was most pronounced in the northern North Atlantic, a region of the world ocean thought to be climatically dynamic. "Accounting for the effects of some forms of natural variability - such as El Nino and volcanic eruptions - helped us to identify the suddenness of the event," Jones said. The different rates of warming in the Northern and Southern Hemispheres in the middle 20th century are frequently attributed to the larger buildup of tropospheric aerosol pollution in the rapidly industrializing Northern Hemisphere. Aerosol pollution contributes to cooling of the Earth's surface and thus can attenuate the warming due to increasing greenhouse gases. But the new paper offers an alternative interpretation of the difference in mid-century temperature trends. "The suddenness of the drop in Northern Hemisphere ocean temperatures relative to the Southern Hemisphere is difficult to reconcile with the relatively slow buildup of tropospheric aerosols," Thompson said. "We don't know why the Northern Hemisphere ocean areas cooled so rapidly around 1970. But the cooling appears to be largest in a climatically important region of the ocean," Wallace said. Contacts and sources:Vince Stricherz University of Washington Simon Dunford University of East Anglia "An abrupt drop in Northern Hemisphere sea surface temperature around 1970" "When the North Atlantic caught a chill"<\|endoftext\|>	3.734375
198	Your screen resolution is expressed in pixels. One pixel is just one tiny dot on your screen that has a color. If you put a bunch of pixels together, you can, for example, make a drawing. All screens, including the screen of your phone and computer, consists of thousands to millions of pixels. A blue rectangle on your screen is just a collection of blue pixels. On most modern screens you can't see the individual pixels/dots, but if you would zoom in, you would definitely see them. Your screen resolution tells about the number of pixels your current screen is using. The more pixels it uses, the more items can fit on it. Let's say you have a resolution of 1280x720. This means your screen has 1280 pixels horizontally and 720 vertically. It's like a raster. So in the case of 1280x720, your screen consists of 1280*720 = 921600 pixels. There are various situations where people change their screen resolutions:<\|endoftext\|>	3.90625
1,791	# Formula for Circumference of Circle The word circumference means the boundary of any shape. The circumference covers the round shapes body distance. Circumference is a Latin word in which circum means around and ferre means carry. In Latin ferre is a verb. In mathematics, the circumference is used to find out the distance around the circle. So, the circumference formula is used to find out the distance of the circle or round shape things. Round shape things like hat, boundary of ground etc. The distance of such kind of things is measured with the help of circumference formula. ## What is the Formula for Circumference? As, we all get to know about what circumference is and in this section we are going to discuss about what is the formula of circumference? The formula of circumference is used to find out the distance of the circle. The distance of the circle is finding from the centre of the circle. And it covers the whole boundary of the round shape thing or body. Like if someone has covered the round shape distance then to find the distance covered by the person is measured with the help of circumference. Now, to find the circumference we need to know about the diameter of the circle. It is not necessary that the circumference can only be find out of circle, but can find out measurement of other circle shapes as well like we can also find out the circumference of ellipse, cylinder, sphere as well. ### Formula for Circumference The formula of circumference consists of diameter, pie, and radius. Sometimes when the radius is given then first you need to find out the diameter which is double of the radius of the circle. The circle shaped bodies have same distance from the centre to the whole boundary. Here, the term pie is used which is mathematical constant and its value is 3.14 or 22/7 this value of pie is multiplied with the diameter. Diameter is defined by the distance of the across the boundary. And as I said above sometimes you get the value of radius then this radius is the distance of the boundary from the centre of the circle. That’s why when we double this value of radius we get the distance across the circle which is diameter of the circle. Formula: Where, C is the circumference of the circle d represents diameter and “this symbol is used to represent the pie. ### Circumference of a Circle Formula To find out the circumference of the circle we use the above written formula and if the radius is given then you can use And if you know the diameter of the circle then you can use above formula i.e. Here, you need to put the value of pie as well to find out the accurate answer of the circumference of the circle that is 22/7 or 3.14 this is the standard constant value of the pie which is used in mathematics. ### Area and Circumference of a Circle Formula You can find out area of a circle as well if you know the diameter of the circle. As the area of the circle deals with the value of the radius which can be find out from the diameter of the circle. As the radius of the circle from the diameter can be find out by half the value of diameter i.e. r = d/2 Now, area of the circle is And we can also find out the circumference of the circle from the radius as well. I.e. if the area of the circle is given to you and you have to find out the circumference of the circle then in this case you first need to find out the radius of the circle from the area of the circle. You can find out the radius with the help of square root of area of the circle divide by pie. I.e. r= √A/ After finding out the radius you can easily find out the circumference of the circle by putting the value of the radius in the circumference formula. And vice-versa can be used to find out the area of the circle from the circumference of the circle then you need to find out the radius of the circle by dividing the circumference of the circle by 2i.e. r= C/2 And then put the value of r in the area of circle formula. Hence, these formula are correlated to each other and you can find out the values of missing terms from these formulas. ### Circumference to Diameter Formula If the diameter is provided to you then you can simply find out the circumference of the circle from the diameter formula in which the term pie is multiplied with the diameter of the circle. I.e. This formula does not need any complicated calculation if you know the diameter form this you can find out circumference of the circle very easily. You can take example as well for the diameter formula of the circumference. Let us consider a circle shaped ground has diameter of 30 meter and you need to find out the circumference then you will just need to put the value of pie and diameter in the formula i.e. C = 3.14 * 30 meter C = 94.2 meter (Solution) Hope this example has helped you to understand the concept of diameter formula. Circumference of circle can also be find out if the diameter is not provided to you and radius is given. As I mentioned above also, the radius is twice the diameter of the circle so, if you have to find out the circumference of the circle using radius formula then you can use the below formula i.e. And you can help of the example as well to know how to calculate the circumference of the circle using radius formula. If the radius of the circle is 17 meter and you are asked to find out the circumference then put the value of radius in r of the above formula i.e. C = 23.14 17 meter C = 34 * 3.14 meter C = 106.67 meter ### Pi Formula Circumference Pie formula can be used with the diameter or radius as well. Well, Pie is the Greek letter which has very long value when we divide 22/7 but round off value 3.14 is used to find out the accuracy of the measurement. This value provides the approximate value of the measurement. The value of pie always remains constant. That is in any formula when you find out the value of pie it always comes to the value 3.14 or 22/7 and if the value not come out to be this than it means you have done some wrong calculation. In case of circumference the pie can be find out from: C/d = And the pie formula of circumference is ### Circumference of a Sphere Formula Sphere has same distance from the centre to all the sides. That is the diameter of the sphere does not vary horizontally or vertically it remain same from all the angles of the sphere. In this case the circumference formula of the circle is used as in case of sphere. Both possess same formula i.e. ### Circumference of the Sphere With the help of this you can find out the radius of the circle which helps you to find out the area, volume, surface area or other measurements of the sphere. ### Circumference of Cylinder Formula If you see the cylinder carefully then you will find the image of cylinder consists of two circles in the bottom and top and to find out the area of the cylinder you need to know the circumference of the circles and as you all know the formula to find out the circle is 2* pier. And after finding the circumference of one circle and with the help of circumference of the circle you can find out the area of outer edge of the circle by simply multiplying the circumference with the height of the cylinder. In this way, the circumference of the circle helps you to find out the surface area of the cylinder. I.e. 2 And then surface area of cylinder is twice the area of circle and outer edge i.e. A = 2 ( * (2 Ellipse Circumference Formula Circumference of the ellipse can also be known as perimeter of the ellipse and it is much difficult to find out the circumference of ellipse as it varies in the shape from the sphere. It has different diameter with respect to orientation vertical and horizontal. Here, you need to take the both values of horizontal radius as well as vertical radius. And the formula for finding circumference of ellipse is: P = 4 * (a + b) * Where, “a” and “b” is the horizontal and vertical radii of the ellipse. And P stands for the perimeter f the ellipse. In Geometry, the shapes and there measurement can only be find out with the help of these formulas so you need to understand the concepts behind these formulas and remember these formulas as well so that you will be able to calculate the values and measurements easily.<\|endoftext\|>	4.59375
223	# How do you write the equation for "one more than three times a number is 7"? Nov 10, 2017 The equation is: $3 N + 1 = 7$ The solution is: $2$ #### Explanation: You say: "one more than three times a number is 7". So we start with a number $N$. Then we multiply it by $3$ to get: $3 N$ Then we increase it by $1$ to get: $3 N + 1$ And we equal all of that to $7$: $3 N + 1 = 7 \to$ [ANS} Solving: $3 N + 1 = 7$ $3 N + 1 - 1 = 7 - 1$ $3 N = 6$ $N = \frac{6}{3}$ N=2 "one more than three times a number is 7". "one more than three times N is 7". "one more than three times $2$ is 7".<\|endoftext\|>	4.46875
615	In geometry, a square is a regular quadrilateral. This means that it has four equal sides and four equal angles (90-degree angles, or right angles). It can also be defined as a rectangle in which two adjacent sides have equal length. A square with vertices ABCD would be denoted ABCD. The square is the n=2 case of the families of n-hypercubes and n-orthoplexes. Other articles related to "square, squares": ... Bureau, the county has a total area of 906 square miles (2,346.5 km2), of which 903 square miles (2,338.8 km2) is land and 3 square miles (7.8 km2) (0.33%) is water ... ... and two additional polyhedra called square cupolae, which count among the Johnson solids it is thus an elongated square orthobicupola ... These pieces can be reassembled to give a new solid called the elongated square gyrobicupola or pseudorhombicuboctahedron, with the symmetry of a square ... are all locally the same as those of a rhombicuboctahedron, with one triangle and three squares meeting at each, but are not all identical with respect ... ... The heating surface of the boiler was 254.8 square metres (2,743 square feet), of which 98.5 square metres (1,060 square feet) were superheated, while the grate area was 4.72 square ... ... The K4 complete graph is often drawn as a square with all 6 edges connected ... This graph also represents an orthographic projection of the 4 vertices and 6 edges of the regular 3-simplex (tetrahedron) ... ... According to the 2010 census, the village has a total area of 0.21 square miles (0.54 km2), of which 0.20 square miles (0.52 km2) (or 95.24%) is land and ... Famous quotes containing the word square: “The square dance fiddlers first concern is to carry a tune, but he must carry it loud enough to be heard over the noise of stamping feet, the cries of the caller, and the shouts of the dancers. When he fiddles, he fiddles all over; feet, hands, knees, head, and eyes are all busy.” —State of Oklahoma, U.S. public relief program (1935-1943) “Rationalists, wearing square hats, Think, in square rooms, Looking at the floor, Looking at the ceiling. They confine themselves To right-angled triangles.” —Wallace Stevens (18791955) “Mark you the floore? that square & speckled stone, Which looks so firm and strong, —George Herbert (15931633)<\|endoftext\|>	3.65625
635	An innovative use of instruments that measure the ocean near Antarctica has helped Australian scientists to get a clearer picture of how the ocean is melting the Antarctic ice sheet. Until now, most measurements in Antarctica were made during summer, leaving winter conditions, when the sea freezes over with ice, largely unknown. But scientists from IMAS and the CSIRO, supported by ACE CRC, the ARC-funded Antarctic Gateway Partnership and the Centre for Southern Hemisphere Oceans Research (CSHOR), developed a novel mission that allowed year-round measurements to be collected near the Totten Glacier, a fast-melting glacier in East Antarctica. They used instruments known as ARGO floats that are typically designed to drift with ocean currents and measure ocean temperature and salinity profiles. For this mission, however, the floats were designed to "park" on the sea floor between profiles so they stayed in the region and did not drift away, and vital data collected during ice-covered winter months were stored and uploaded via satellite later in ice-free conditions. The study published in the journal JGR Oceans revealed for the first time that deep water driving melting at the base of the Totten Glacier is warmer and in a thicker layer during winter and autumn than during spring and summer. Lead author Alessandro Silvano, from IMAS, said this means the Totten Glacier might melt more rapidly in winter than summer, and that summer measurements might under-estimate the flow of warm water to the ice shelf. "We had a nervous wait during the first winter, wondering if the floats would survive the icy winter conditions after being parked on the rough sea floor for long periods," he said. "When spring arrived and the sea ice started to melt we were very excited to see that the floats returned and transmitted the winter data. "We immediately noticed that the ocean was warmer in autumn and winter than found in our previous summer measurements. "The new measurements confirm that this part of East Antarctica is exposed to warm ocean waters that can drive rapid melt, with the potential to make a large contribution to future sea level rise. "The floats also provided new measurements of ocean depth in the region, revealing a deep trough that allows warm water to approach the glacier year-round," Mr Silvano said. CSIRO co-author Dr Steve Rintoul from CSHOR said the new measurements of ocean depth, temperature and salinity will help improve models used to predict the Antarctic's contribution to sea level rise. "Crashing sensitive oceanographic instruments into the sea floor isn't generally recommended," he said. "But these results show that profiling floats can be used in novel ways to measure the ocean near Antarctica, a critical blind-spot in the global ocean observing system. "Much work remains to be done and more measurements are needed to assess the vulnerability of the ice shelf to changes in the ocean, including in the ocean beneath the floating Totten Glacier. "New technologies, like the autonomous underwater vehicle (AUV) recently acquired by the University of Tasmania, will be needed to fill this gap," Dr Rintoul said.<\|endoftext\|>	3.734375
647	1. ## ellipse 3 Given the ellipse: $7x^2+16y^2=112$. Finding the points of the ellipse whose distances to the focus left is $\frac{5}{2}$ $(-2,+or-\frac{\sqrt{21}}{2})$ 2. Originally Posted by Apprentice123 Given the ellipse: $7x^2+16y^2=112$. Finding the points of the ellipse whose distances to the focus left is $\frac{5}{2}$ $(-2,+or-\frac{\sqrt{21}}{2})$ So we need to find the coordinates of the left focus. Then with (x,y) as any arbitrary point on the ellipse, we equate the distance between the left focus and point (x,y) to 5/2. 7x^2 +16y^2 = 112 ----------(i) Divide both sides by 112 to make the RHS equal to 1, (x^2)/16 +(y^2)/7 = 1 Or, (x^2)/(4^2) +(y^2)/[(sqrt(7))^2] = 1 ---the same ellipse as (i). a = 4 and b= sqrt(7). Center of ellipce is at (0,0) The leftmost end of the ellipse is at (-4,0). The left focus is at (-c,0). c^2 = a^2 -b^2 c^2 = 4^2 -(sqrt(7))^2 = 9 c = 3 Hence, the left focus is at (-3,0). Then, (5/2)^2 = (x-(-3))^2 +(y-0)62 25/4 = x^2 +6x +9 +y^2 From (i), y^2 = (112 -7x^2)/16 so, 25/4 = x^2 +6x +9 +(112 -7x^2)/16 Clear the fractions, multiply both sides by 16, 100 = 16x^2 +96x +144 +112 -7x^2 0 = 9x^2 +96x +156 0 = 3x^2 +32x +52 0 = (3x +26)(x+2) x = -26/3, or -2 When x = -26/3 = - 8.667, this point is not on the ellipse because the leftmost point on the ellipse is at x = -4 only When x = -2, y^2 = (112 -7x^2)/16 y^2 = (112 -7(4))/16 = 84/16 = 21/4 y = +,-sqrt(21) /2 Therefore, two points on the ellipse, (-2, sqrt(21) /2) and (-2, -sqrt(21) /2), are 5/2 units away from the left focus. ----answer. 3. thanks<\|endoftext\|>	4.53125
1,741	The effects that can occur are based on whether the drug is an agonist or an antagonist. Agonists work by chemically imitating the neurotransmitter as it connects to the receptor, therefore strengthening its effects (OpenStax, 2016). One type of drug that has the characteristics of an agonist is a depressant. All depressants affect the gamma-Aminobutyric acid neurotransmitter system (OpenStax, 2016). Because the gamma-aminobutyric acid suppresses the brain, depressants also suppress the brain because they act as an agonist by mimicking the neurotransmitter. Those who continuously take depressants become dependent and therefore addicted to the drug due to its ability to increasingly calm the nerves of an individual. Another example of an agonist is opioids. Opioid agonists attach to the endogenous opioid neurotransmitter system and aid in decreasing pain (OpenStax, 2016). Because humans already have a system in their body that functions to reduce pain, i... ... middle of paper ... ...usal or lack thereof one may receive from using a drug, the drug still negatively affects the entire human body. The effect drugs have on the body is like The Domino Effect. First, the neurotransmitters that bind to matching receptors in order to carry out the correct function of the neurotransmitter system are either blocked or enhanced. The incorrect attachment of the neurotransmitter to the receptor has an effect on the behavior of an individual. As this incorrect bond continuously occurs, it disrupts several other functions of not just the nervous system, but the many other systems that make up the human body. The more systems that stop working correctly, the higher risk a person has for health issues and death. Although one may enjoy the feeling that goes along with using drugs, it is not worth the potentially life-changing effects that drugs have on the body. 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889	# Planning the Trek - 4th Grade Math Lesson ### Overall Rating Subject: Math Lesson Duration: 60 Minutes State Standards: MS Objectives: 3, 3a, 3b, 3c, 3d, 3f, 3h, 5, 5c Thinking Skills: Understanding: Understand the main idea of material heard, viewed, or read. Interpret or summarize the ideas in own words. Applying: Apply an abstract idea in a concrete situation to solve a problem or relate it to a prior experience. Creating: Bring together parts (elements, compounds) of knowledge to form a whole and build relationships for NEW situations. ### Objective Essential Question: What factors make it quicker to hike the Natchez Trace today rather than in the early 1800's? The students will be able to: 1) Locate various places on Natchez Trace Parkway 2) Identify landmarks on a map 3) Measure distances between points on a map using using the four operations 4) Use critical thinking and math operations to plan a successful trip. ### Background Students will use maps to locate specific points and measure the distances between those points. Students they will pretend they are mapping out a trip along the Natchez Trace. They will determine how long it would take to drive (easy), walk (hard), bike, and ride the Natchez Trace Parkway today. They will compare their answers to the average time it took historic boatmen to travel the Trace. The old Natchez Trace was a hard place to walk. A very fast journey would take about two weeks. Most people took much longer; some people died or were murdered before they completed their journey. A healthy person in good shape can walk 15 or 20 miles per day. Today, an experienced bicyclist might travel 50 miles per day. The speed limit on the Natchez Trace Parkway is 50 miles per hour. Horses can average 35 miles on good trails. ### Preparation Materials 1.) Instructions 2.) Natchez Trace Map 3.) Bicycling Tracking Sheet 4.) Car or Motorcycle Tracking Sheet 5.) Hiking Trip Tracking Sheet 6.) Horseback Riding Tracking Sheet 7.) Scissors and tape to put the 4 sections of map in proper sequence ### Materials A math lesson to help student learn to plan a trip in a national park. ### Lesson Hook/Preview Math can help you plan a fun trip on the Natchez Trace Parkway. ### Procedure Student Task: The students will determine the number of stops (nights) or the number of hours it takes to travel the Trace. The students may be assigned only the hiking option, or all of the options. Student Instruction: 1.) Map a trip along the Natchez Trace. 2.) Depending on the teacher assignment, you will hike, bike, ride or drive. 3.) Hikers may hike no more than 20 miles per day. You may hike less than 20 miles. 4.) Bikers may bike no more than 50 miles but may bike less. 5.) Riders may ride no more than 35 miles but may ride less and drives must travel no more than 50 miles per hour. 6.) You will need to stop and spend the night at a town or campground. 7.) On your tracking sheet, write the name of the stop and the mileage you traveled that day. When you are done, count the number of stops you made. This is how many nights you spent on the Natchez Trace. Teacher Closure: Lead the students in a discussion about what factors make it quicker to hike the Natchez Trace today rather than in the early 1800's. ### Vocabulary Legend, map key, boatmen, Kaintuck, math, trip planning, mapping, measuring, distance, map skills ### Supports for Struggling Learners Reminders on board or on top of paper of days per week, hours in day, minutes in an hour. Have students help teacher with writing these important prior knowledge notes on the board. ### Enrichment Activities Extenatnnsions 1.) Play the Trekking the Trace Classroom Game 2.) Visit the Natchez Trace Parkway and see how long it takes your school bus to travel from one point to the next.<\|endoftext\|>	4.40625
1,060	When the goblin shark species was first discovered off the coast of Japan in 1898, it set off a case of scientific déjà vu. Researchers realized that they had seen goblin sharks before -- not swimming in the seas, but in fossil remains. Shark fossils from the Scapanorhynchus species dating back more than 100 million years bore striking resemblance to the goblin shark [source: Bright]. Since the species was thought to be extinct, goblin sharks are now classified as living fossils. A comparison between the current incarnation and fossils also revealed that little about the fish has changed over time [source: Carwadine]. Despite its ugliness, the goblin shark's body structure seems to have been perfected for survival. For starters, its flat snout is lined with openings called ampullae of Lorenzini that serve as electrical sensors that track down food. Sharks have a unique sense called electroreception. Ampullae pick up weak electrical impulses living fish and animals give off whenever a muscle contracts. This built-in homing device allows goblin sharks to seek out food. Once it's tracked down a meal, the goblin shark's retractable jaw provides the quick-draw action to snatch up food. In spite of the goblin shark's large size, it can sneak up on prey effectively thanks to its liver (of all things). In relation to body size, the goblin shark's oily liver takes up enough room on its insides to make it almost as dense as water [source: Ebert]. Since it floats so easily, the fish requires little movement to get around, and it can discretely float up to its desired food source. Research conducted on the small number of goblin sharks that have come out of the water suggest that their main food sources include fish, shrimp and squid [source: Food and Agriculture Organization of the United Nations]. Deciphering more concrete dietary information has been a challenge since many of the goblin sharks examined have empty stomachs. For example, in one large-scale study on goblin sharks performed in 2006, 29 percent of the 148 individual fish studied had no food inside of them [source: Yano et al]. The last goblin shark sighting occurred in 2007 in Tokyo Bay, Japan [source: Mehta]. After one day in captivity, the living fossil fish died. Since opportunities like that one are few and far between, it may be a long time before we decode all the mysteries of the goblin shark. For more shark-related information, visit the links below. Related HowStuffWorks Articles More Great Links - Bright, Michael. "The Private Life of the Sharks: The Truth Behind the Myth." Stackpole Books. 2000. (May 14, 2008)http://books.google.com/books?id=w31fF5IlqdoC - Carwadine, Mark. "Shark." Firefly Books. 2004. (May 14, 2008)http://books.google.com/books?id=Qh44RNa5yh0C - Ebert, David A. "Sharks, Rays and Chimaeras of California." University of California Press. 2003. (May 14, 2008)http://books.google.com/books?id=1SjtuAs702kC - Food and Agriculture Organization of the United Nations. "Mitsukurina owstoni." Fisheries and Aquaculture Department. 2001. (May 14, 2008)http://www.fao.org/fishery/species/13494 - Jordan, Vanessa. "Goblin Shark." Florida Museum of Natural History Ichthyology Department. (May 14, 2008)http://www.flmnh.ufl.edu/FISH/Gallery/Descript/GoblinShark/GoblinShark.html - Krock, Lexi. "Other Fish in the Sea." NOVA Online. January 2003. (May 14, 2008)http://www.pbs.org/wgbh/nova/fish/other.html - Mehta, Aalok. "Rare 'Prehistoric' Goblin Shark Caught in Japan." National Geographic. Feb. 9, 2007. (May 14, 2008)http://news.nationalgeographic.com/news/2007/02/070209-goblin-shark.html - ReefQuest Centre for Shark Research. "Biology of the Goblin Shark." Biology of Sharks and Rays. (May 14, 2008)http://elasmo-research.org/education/shark_profiles/m_owstoni.htm - Yano, Kazunari; Miya, Masaki; Aizawa, Masahiro; Noichi, Tetsuhisa. "Some aspects of the biology of the goblin shark, Mitsukurina owstoni, collected from the Tokyo Submarine Canyon and adjacent waters, Japan." Ichthyological Research. May 22, 2007. (May 14, 2008)http://www.springerlink.com/content/p430623g30188246/fulltext.pdf<\|endoftext\|>	3.671875
92	# How do you solve 4x+7 - 5x = 12 ? Mar 22, 2018 $- 5$ is the answer #### Explanation: 4x+7−5x=12 Step 1: Combine like terms. $- x + 7 = 12$ Step 2: Subtract $- x = 5$ Step 3: Divide $x = - 5$<\|endoftext\|>	4.375
296	## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition) $\frac{ta}{a-t}$=b $\frac{t}{a}+\frac{t}{b}= 1 ; b$ What this problem is asking for is what does the variable b equal. This is indicated because of the use of the semicolon. The first thing we will do is get rid of the fractions. This will be done by multiplying each number times the LCD of the denominators. In this case, the least common denominator is (ab). $\frac{t}{a}\times(ab)+\frac{t}{b}\times(ab)=1 \times(ab)$ By multiplying $\frac{t}{a}$ with (ab) the variable a from the fraction will cancel out with the variable a from (ab). You then multiply t $\times$b giving you tb. tb+ta=ab Next, we subtract tb from both sides. This is done to get the variable b on one side of the equation. tb-(tb)+ta=ab-(tb) ta=ab-tb We will now factor out b. ta=b(a-t) Lastly, you divide (a-t) from both sides. $\frac{ta}{a-t}=b\frac{a-t}{a-t}$ $\frac{a-t}{a-t}$ will cancel itself out. Your final answer should look like this. $\frac{ta}{a-t}=b$<\|endoftext\|>	4.65625
1,101	Views: Category: Education ## Presentation Description No description available. ## Presentation Transcript ### PowerPoint Presentation: Quadratic Equation By 10 ‘C’ Students ### PowerPoint Presentation: What is A Quadratic Equation A Quadratic Equation in Standard Form (a, b, and c can have any value, except that a can't be 0.) The letters a, b and c are coefficients. The letter "x" is the variable or unknown ### PowerPoint Presentation: For Example this makes it a quadratic The name Quadratic comes from "quad" meaning square, because of x 2 (in other words x squared ). It can also be called an equation of degree 2. ### PowerPoint Presentation: Some othe r Example Where a=2, b=5 and c=3 This one is a little more tricky: Where is a ? In fact a=1 , because we don't usually write "1x 2 " b=-3 And where is c ? Well, c=0 , so is not shown. ### PowerPoint Presentation: We can solve Quadratic equation by 3 methods: 1:BY FACTORISATION METHOD 2: BY COMPLETING THE SQUARE METHOD 3:BY QUADRATIC FORMULA ### PowerPoint Presentation: Solving Quadratic Equations by Factorisation method ### PowerPoint Presentation: Solving Quadratic Equations by Factorisation This presentation is an introduction to solving quadratic equations by factorisation. The following idea is used when solving quadratics by factorisation. If the product of two numbers is 0 then one (or both) of the numbers must be 0. So if xy = 0 either x = 0 or y = 0 Considering some specific numbers: If 8 x x = 0 then x = 0 If y x 15 = 0 then y = 0 ### PowerPoint Presentation: Solving Quadratic Equations by Factorisation a x 2 + b x + c = 0 , a  0 Some quadratic equations can be solved by factorising and it is normal to try this method first before resorting to the other two methods discussed. The first step in solving is to rearrange them (if necessary) into the form shown above. x 2 = 4 x Example 1: Solve x 2 – 4 x = 0 x ( x – 4) = 0 either x = 0 or x – 4 = 0 if x – 4 = 0 then x = 4 Solutions (roots) are x = 0 , x = 4 rearrange factorise ### PowerPoint Presentation: 6 x 2 = – 9 x Example 2: Solve 6 x 2 + 9 x = 0 3 x (2 x + 3) = 0 either 3 x = 0 or 2 x + 3 = 0  x = 0 or x = – 1½ rearrange factorise ### PowerPoint Presentation: Solving Quadratic Equations by Completing the Square ### Creating a Perfect Square Trinomial: Creating a Perfect Square Trinomial In the following perfect square trinomial, the constant term is missing. x 2 + 14x + ____ Find the constant term by squaring half the coefficient of the linear term. (14/2) 2 x 2 + 14x + 49 ### Solving Quadratic Equations by Completing the Square: Solving Quadratic Equations by Completing the Square Solve the following equation by completing the square: Step 1: Move quadratic term, and linear term to left side of the equation ### Solving Quadratic Equations by Completing the Square: Solving Quadratic Equations by Completing the Square Step 2: Find the term that completes the square on the left side of the equation. Add that term to both sides. ### PowerPoint Presentation: Solving Quadratic Equations by Completing the Square Step 3: Factor the perfect square trinomial on the left side of the equation. Simplify the right side of the equation. ### Solving Quadratic Equations by Completing the Square: Solving Quadratic Equations by Completing the Square Step 4: Take the square root of each side ### Solving Quadratic Equations by Completing the Square: Solving Quadratic Equations by Completing the Square Step 5: Set up the two possibilities and solve ### PowerPoint Presentation: What Does The Formula Do ? The Quadratic formula allows you to find the roots of a quadratic equation (if they exist) even if the quadratic equation does not factorise. The formula states that for a quadratic equation of the form : ax 2 + bx + c = 0 The roots of the quadratic equation are given by : ### PowerPoint Presentation: Example 1 Use the quadratic formula to solve the equation : x 2 + 5x + 6= 0 Solution: x 2 + 5x + 6= 0 a = 1 b = 5 c = 6 ### PowerPoint Presentation: x = - 2 or x = - 3 These are the roots of the equation. ### PowerPoint Presentation: Thank you Made And Presented By Amar Abhishek Rawat Abhishek Solanki Abhishek Sharma Ajay Anirudh<\|endoftext\|>	4.6875
738	Volcanic Rock(redirected from Igneous volcanic rock) Also found in: Dictionary, Thesaurus. volcanic rock[väl′kan·ik ′räk] vulcanites, rock formed as a result of volcanic eruptions. Depending on the type of explosion (a lava flow or explosive eruption), two types of rock form—effusive and volcanogenic-detrital (pyroclastic). The latter type is divided into friable (volcanic ash, sand, bombs, and so on), compacted, and cemented rock (tuff, tuff breccia, and others). In addition, there are intermediate types of volcanic rocks—tuff lavas, which occur as a result of eruptions of gas-rich foaming lava flows, and ignimbrites, which occur as a result of violent eruptions when pieces of lava are carried into the air and fall on to the surface, forming masses of melted matter that occasionally occupy wide areas measuring hundreds and thousands of square kilometers. The viscosity of the lava and the terrain of the volcanic area determine the body forms of effusive rock. Sheets and flows are formed from low-viscosity basaltic lavas. Cupolas and needles occur during eruptions of viscous lava (dacites and liparites). Dikes and necks are caused by the filling of cracks and underlying channels by lava. Volcanic rock is distinguished by its chemical composition and structural and textural peculiarities and by the degree of disintegration of the rock substance. According to chemical composition, effusive volcanic rock is divided into alkaline-earth and alkaline rock and, in addition, into basic (not saturated with silicic acid), medium (saturated with silicic acid), and acidic (supersaturated with silicic acid) rock. The degree of crystallization of the lava, as well as its structures and textures, depends on its viscosity. The internal parts of effusive bodies are usually crystallized, whereas the external parts are slaglike, porous, and glassy. Porphyritic, microlitic, semi vitreous, and vitreous structures and fluidal, striped, massive, and porous textures are characteristic of effusive rock. Profoundly changed, usually older effusive rock is called a paleotype, and unchanged rock is called a cenotype. The most widely distributed cenotypic rocks are basalts, andesites, trachytes, and liparites; their paleotypic analogues are called diabases, basaltic and andesitic porphyrites, and trachytic and liparitic porphyry. Volcanogenic sedimentary rock with pyroclastic rocks (tuff and breccia), also belongs to the group of detrital volcanic rocks. Many volcanic rocks (for example, basalt, tuff, and pumice) are widely used in industry. REFERENCESZavaritskii, A. N. Izverzhennye gornye porody. Moscow, 1955. Maleev, E. F. Vulkanoklasticheskie gornye porody. Moscow, 1963. Koptev-Dvornikov, V. S., E. B. Iakovleva, and M. A. Petrova. Vulkanogennye porody i melody ikh izucheniia. Moscow, 1967. V. I. VLODAVETS<\|endoftext\|>	3.75
354	What is sedimentation? Sedimentation is very important. Without it we wouldn’t have any dinosaur fossils. It is the building up of layers of small particles like sand or mud. The easiest place to see this is the beach. A beach is made up of lots of sand which have been deposited, or left behind, by the sea. Sand and mud come from inland. Rivers erode them from the land and bring them towards the sea. As the water slows, it can’t carry as much and so sand and mud are dropped. The bigger the grain of sand, the sooner it is dropped. If you look at a cliff, you will often see layers which make the cliff look like a layer cake. These layers are caused by sedimentation. Over a long period of time, the grains of sand and mud build up and up, forming the layers. Fossils are found in these layers. The quicker bones are buried, the more chance they will be saved from scavenging animals and damage by weather. The sea, rivers and lakes are the best depositors of sand and mud and dinosaurs are found where there used to be a sea, lake or river. But big glaciers also carry grains and the air can also carry very small grains. A land-slide, where mud and rock fall down a mountain of a sand dune can also save the bones of a dinosaur. One famous fossil called ‘The Fighting Dinosaurs’ is two dinosaurs entwined as though they are fighting. Palaeontologists think that the Velociraptor was hunting the other dinosaur, Protoceratops when a sand dune collapsed on both, killing them and preserving their bones.<\|endoftext\|>	4.1875
769	While the area now known as Scotland lay on the periphery of the Roman Empire, and only episodically was incorporated within it, there was a relationship with Rome for over 300 years which can only be properly understood within the framework of wider Roman Empire studies. The Roman invasions of Scotland can best be understood within the world of Roman politics, known primarily from textual sources, which saw the ebb and flow of Roman arms related to the interest of successive emperors. Vespasian, who took part in the invasion of AD 43, sent governors to Britain with specific instructions which appear to have included orders to impose the will of Rome over the whole island. These injunctions ultimately brought Agricola to Mons Graupius in the year 83. The personal disposition of Antoninus Pius, who lacked military experience, has been related to the decision to abandon Hadrian's Wall and re-occupy southern Scotland in 139. Cassius Dio and Herodian, in their respective Histories both offer reasons for the campaigns of Septimius Severus from 208 to 211, including that the Emperor enjoyed campaigning and that he wished to take his sons away from the flesh- pots of Rome. Events elsewhere also might have an effect on activities in north Britain. Roman defeats on the Danube in the 80s led to the withdrawal of about a quarter of the army of Britain and the abandonment of the Flavian conquests of Agricola. A requirement to send reinforcements to the Mauretanean War of the late 140s may have resulted in apparent delays in the building of the Antonine Wall, and it is possible that a general overstretching of resources may have lain behind the decision to abandon this frontier in the 180s. The death of Severus at York in 211 led to the abandonment of his conquests and the return of his sons to Rome. It was, it would appear, always the advent of trouble on the northern frontier which brought the Emperor to Britain. Hadrian's visit in 122 followed unrest in Britain, though whether that led directly to the building of Hadrian's Wall is another matter. Severus came following warfare and, arguably, with the intention of completing the conquest of the island. Constantius I with his son Constantine came to fight the Picts in 305; Constantine possibly visited again later; and his grandson Constans came to Britain in 342/3 probably because of trouble on the northern frontier. Roman Scotland was also part of a wider trading network. Pottery came to the northern frontier from Gaul as well as southern Britain. Arms, armour and other items of equipment were imported over long distances to the northern frontier. A good deal of food might have been grown locally but, together with wine, much was also imported from various places, including the Mediterranean. Table 1: Table of Roman Dynasties, Emperors and notable events \|Dynasty\|\|Emperor\|\|Events / people\| \|Flavian 69-96AD\|\|Vespasian 69-79AD \|Flavian invasion c.78-86 (Agricola as governor c.77-84) \|Adopted emperors 96-138 AD\|\|Nerva 96-98AD \|Unrest on northern frontier; building of Hadrian's Wall 119AD onwards\| \|Antonine 138-192 AD\|\|Antoninus Pius 138-161AD Marcus Aurelius 161-180AD Lucius Verus Commodus 180-192AD \|Antonine invasion c.139-165 Antonine wall 140-141AD Wars under Commodus Abandonment of Antonine Wall during 170s ? \|Severan 138-192 AD\|\|Severus 193-211AD \|Severan campaigns 208-211\|<\|endoftext\|>	3.875
761	Child Development is a very important Section for CTET & State TET exam. Learning Disabilities is one of the important topics in the unit "Concept of Inclusive education and understanding children with special needs" Every year some questions are based on Learning Disabilities. Let’s Understand the different type of Learning Disabilities in brief. Please read carefully because once you read this, you will be able to solve all the questions in the exam related to this topic. Learning Disabilities are sometimes also called Learning Disorders. Types of Learning Disabilities are given below. 1. Aphasia: Aphasia is an impairment of language, affecting the production or comprehension of speech and the ability to read or write. 2. Dyslexia: Difficulty in reading. These children might see reversing letters or write backward. Some see Dyslexia as a visual issue. 3. Hyperlexia: Hyperlexia is a syndrome that is characterized by a child's precocious ability to read (far above what would be expected at their age), significant difficulty in understanding and using verbal language (or a profound nonverbal learning disability), and significant problems during social interactions. 4. Dyscalculia: It is difficulty in learning or comprehending arithmetic, such as difficulty in understanding numbers, learning how to manipulate numbers, and learning facts in mathematics. It is generally seen as a specific developmental disorder. 5. Dysgraphia: Difficulty in writing is called Dysgraphia. 6. Dyspraxia: Dyspraxia can affect a child's ability to do a wide range of everyday physical tasks which involve motor skills. These can include things like jumping, speaking clearly and gripping a pencil 7. Body dysmorphic disorder(BDD), occasionally still called dysmorphia phobia, is a mental disorder characterized by the obsessive idea that some aspect of one's own appearance is severely flawed and warrants exceptional measures to hide or fix it 8. Attention deficit hyperactivity disorder (ADHD) is a mental disorder of the neurodevelopmental It is characterized by problems paying attention, excessive activity, or difficulty controlling behaviour which is not appropriate for a person's age. The symptoms appear before a person is twelve years old, are present for more than six months, and cause problems in at least two settings (such as school, home, or recreational activities). In children, problems paying attention may result in poor school performance. Although it causes impairment, particularly in modern society, many children with ADHD have a good attention span for tasks they find interesting. 9. Autism is a complex neurobehavioral condition that includes impairments in social interaction and developmental language and communication skills combined with rigid, repetitive behaviours. 10. Cerebral Palsy is considered a neurological disorder caused by a non-progressive brain injury or malformation that occurs while the child’s brain is under development. Cerebral Palsy primarily affects body movement and muscle coordination. Cerebral Palsy affects body movement, muscle control, muscle coordination, muscle tone, reflex, posture, and balance. It can also impact fine motor skills, gross motor skills, and oral motor functioning Past Year solved questions to check that have we understood the concept: 1. Dyslexia is associated mainly with difficulties in - Speaking and hearing Explanation: We have read that Dyslexia is difficulty in reading. These children might see reversing letters or write backwards. 2. Learning Disability in motor skills is called Explanation: We have read that Dyspraxia can affect a child's ability to do a wide range of everyday physical tasks which involve motor skills. These can include things like jumping, speaking clearly and gripping a pencil. CTET Mock Test 2019:<\|endoftext\|>	3.71875
373	Mountains are natural landforms shaped by moving tectonic plates and volcanic activity, which cause the earth’s crust to rise and form a peak. These geological forces have built mountains around the world, such as the Himalayas, the Andes, the Alps, the Sierra Nevadas, and the Canadian Rockies. Moving continents, rising and falling mountain barriers, vast volcanic eruptions and continental ice sheets all played an essential role in creating the diversity of life in British Columbia today. The cirque glaciers left over from the last ‘little ice age’, which ended 10,000 years ago, carved the knife-edge peaks called arêtes, the u-shaped valleys, and the deep fjords along the coastline. Over the course of 650 million years, British Columbia was shaped by colliding tectonic plates, volcanic eruptions, and glaciations, which resulted in the uplift necessary to create the Rocky Mountains to the east, the Interior Plateau, and the Coast Mountains. The Coast Mountains stretch over 1,600 kilometres from the Fraser River Valley north to the southwestern edge of the Yukon and Alaskan Panhandle. From Mount Waddington (4019m), the highest peak in British Columbia, to the banks of the Fraser River, the geomorphology of the Coast Mountains has long been studied and admired by geologists and tourists alike. Unlike the sedimentary rock and dinosaur fossils discovered in the Rockies, the Coast Mountains have one of the world’s largest granite fields, and are home to one of Canada’s most popular rock climbing areas, the Stawamus Chief. Cannings, Sydney, Richard Cannings, JoAnne Nelson. Geology of British Columbia: a journey through time. Vancouver: Greystone Books, cc2011. Pg. 9.<\|endoftext\|>	4.15625
416	Environmental disasters in the U.S. often hit minority groups the hardest. When Hurricane Katrina slammed New Orleans in 2005, the city’s black residents were disproportionately affected. Their neighborhoods were located in the low-lying, less-protected areas of the city, and many people lacked the resources to evacuate safely. Similar patterns have played out during hurricanes and tropical storms ever since. Massive wildfires, which may be getting more intense due to climate change and a long history of fire-suppression policies, also have strikingly unequal effects on minority communities, a new study shows. Researchers at the University of Washington and The Nature Conservancy used census data to develop a “vulnerability index” to assess wildfire risk in communities across the U.S. Their results, appearing Nov. 2 in the journal PLOS ONE, show that racial and ethnic minorities face greater vulnerability to wildfires compared with primarily white communities. In particular, Native Americans are six times more likely than other groups to live in areas most prone to wildfires. “A general perception is that communities most affected by wildfires are affluent people living in rural and suburban communities near forested areas,” said lead author Ian Davies, a graduate student in the UW School of Environmental and Forest Sciences. “But there are actually millions of people who live in areas that have a high wildfire potential and are very poor, or don’t have access to vehicles or other resources, which makes it difficult to adapt or recover from a wildfire disaster.” This study is one of the first to integrate both the physical risk of wildfire with the social and economic resilience of communities to see which areas across the country are most vulnerable to large wildfires. The approach takes 13 socioeconomic measures from the U.S. census — including income, housing type, English fluency and health — for more than 71,000 census tracts across the country and overlays them with wildfire potential based on weather, historical fire activity and burnable fuels on the landscape. Read the rest of the story at the UW News.<\|endoftext\|>	3.84375
825	Malnutrition is a disparity between the amount of food and other nutrients that the body needs and the amount that it is receiving. This imbalance is most frequently associated with under nutrition, the primary focus of this article, but it may also be due to over nutrition. Chronic over nutrition can lead to obesity and to the metabolic syndrome, a set of cardiovascular risk factors characterised by abdominal obesity, a decreased ability to process glucose (insulin resistance), dyslipidaemia, and hypertension. Those with metabolic syndrome have been shown to be at a greater risk of developing type 2 diabetes and cardiovascular disease. Another relatively uncommon form of overnutrition is vitamin or mineral toxicity. This is usually due to excessive supplementation, for instance, high doses of fat-soluble vitamins such as vitamin A rather than the ingestion of food. Toxicity symptoms depend on the substance(s) ingested, the severity of the overdose, and whether it is acute or chronic. Under nutrition occurs when one or more vital nutrients are not present in the quantity that is needed for the body to develop and function normally. This may be due to insufficient intake, increased loss, increased demand, or a condition or disease that decreases the body’s ability to digest and absorb nutrients from available food. While the need for adequate nutrition is a constant, the demands of the body will vary, both on a daily and yearly basis. - During infancy, adolescence, and pregnancy additional nutrition is crucial for normal growth and development. A severe shortage of food will lead to a condition in children called marasmus that is characterised by a thin body and stunted growth. If enough calories are given, but the food is lacking in protein, a child may develop kwashiorkor – a condition characterised by oedema (fluid retention), an enlarged liver, apathy, and delayed development. Deficiencies of specific vitamins can affect bone and tissue formation. A lack of vitamin D, for instance, can affect bone formation – causing rickets in children and osteomalacia in adults, while a deficiency in folic acid during pregnancy can cause birth defects. - Acute conditions such as surgery, severe burns, infections, and trauma can drastically increase short-term nutritional requirements. People who have been malnourished for some time may have under functioning immune system and a poorer prognosis. They frequently take longer to recover from surgical procedures and spend longer in hospital. Malnutrition is common in patients admitted to hospital but it is easily overlooked by hospital staff. For this reason, many hospitals screen and monitor the nutritional status of their patients. This approach is recommended by the National Institute for Health and Care Excellence (NICE). Patients having surgery are frequently evaluated both prior to surgery and during their recovery process. - Chronic diseases may be associated with nutrient loss, increased nutrient demand, and with malabsorption (the inability of the body to absorb one or more available nutrients). Malabsorption may occur with chronic diseases such as coeliac disease, cystic fibrosis, pancreatic insufficiency, and pernicious anaemia. An increased loss of nutrients occur in chronic kidney disease, diarrhoea, and haemorrhaging. Sometimes conditions and their treatments can both cause malnutrition through decreased intake. Examples of this are the decreased appetite, difficulty swallowing, and nausea associated both with cancer (and chemotherapy), and with HIV/AIDS (and its drug therapies). Increased loss, malabsorption, and decreased intake may also occur in patients who chronically abuse drugs and/or alcohol. - Elderly patients are often less able to absorb nutrients due in part to decreased stomach acid production and are more likely to have one or more chronic ailments that may affect their nutritional status. At the same time, they may have more difficulty preparing meals and may have less access to a variety of nutritious foods. Older patients also frequently eat less due to a decreased appetite, decreased sense of smell, and/or mechanical difficulties with chewing or swallowing. For these reasons, elderly patients are often malnourished and require nutrutional support.<\|endoftext\|>	4.125
598	Practice 7A: \| 1 \| 2 \| 3 \| 4 \| Go up Angular Displacement - by Matt Henderson, 2003 1. A girl sitting on a merry-go -round moves counterclockwise through an arc length of 2.5 m. If the girl's angular displacement is 1.67 rad, how far is she from the center of the merry - go- round. Here's what you know, Dq = 1.67 rad and l = 2.5 use the formula Dq = Ds/r to find the radius 1.67 = 2.5/r r = 1.5 m 2. A beetle sits at the top of a bicycle wheel and flies away just before it would be squashed. Assuming that the wheel turns clockwise, the beetle's angular displacement is (p) rad, which corresponds to an arc length of 1.2 m. What is the wheel's raduis? We know that Dq = (p) rad and l = 1.2 m so now we use the formula Dq = Ds/r to find the radius (p) = 1.2/r r = .38 m 3. A car on a Ferris wheel has an angular displacement of (p/4) rad, which corresponds to an arc length of 29.8. What is the Ferris wheel's radius? We know that Dq = (p/4) rad and l = 29.8 m so now we use the formula Dq = Ds/r to find the radius (p/4) = 29.8/r r = 37.94 m 4. Fill in the unknown quantities in the following table: Dq Ds r a. ?rad .25m .1m Dq = Ds/r which is Dq = .25/.1 = 2.5 rad b. .75 rad ? 8.5m Dq = Ds/r which is .75 = Ds/8.5 so Ds = 6.375m c. ? degr -4.2m .75m Dq = Ds/r which is Dq = -4.2/.75 = -5.6 rad and 1 rad = 57.3 degrees so (57.3)(-5.6) = -320.88 degre d. 135degr 2.6m ? 135/57.3 since 1 rad = 57.3 degrees that = 2.356 rad now plug into Dq = Ds/r so 2.356 = 2.6/r and r = 1.1 m<\|endoftext\|>	4.6875
3,458	Terra Cimmeria is a large Martian region, centered at Coordinates: and covering 5,400 km (3,400 mi) at its broadest extent. It covers latitudes 15 N to 75 S and longitudes 170 to 260 W. It lies in the Eridania quadrangle. Terra Cimmeria is one part of the heavily cratered, southern highland region of the planet. The Spirit rover landed near the area. The word Cimmerium comes from an ancient Thracian seafaring people. The land was always covered in clouds and mist. A high altitude visual phenomena, probably a condensation cloud, was seen above this region in late March 2012. NASA tried to observe it with some of its Mars orbiters, including the THEMIS instrument on the 2001 Mars Odyssey spacecraft and MARCI on the Mars Reconnaissance Orbiter. Terra Cimmeria is the location of gullies that may be due to recent flowing water. Gullies occur on steep slopes, especially on the walls of craters. Gullies are believed to be relatively young because they have few, if any craters. Moreover, they lie on top of sand dunes which themselves are considered to be quite young. Usually, each gully has an alcove, channel, and apron. Some studies have found that gullies occur on slopes that face all directions, others have found that the greater number of gullies are found on poleward facing slopes, especially from 30-44 S. Although many ideas have been put forward to explain them, the most popular involve liquid water coming from an aquifer, from melting at the base of old glaciers, or from the melting of ice in the ground when the climate was warmer. Because of the good possibility that liquid water was involved with their formation and that they could be very young, scientists are excited. Maybe the gullies are where we should go to find life. There is evidence for all three theories. Most of the gully alcove heads occur at the same level, just as one would expect of an aquifer. Various measurements and calculations show that liquid water could exist in aquifers at the usual depths where gullies begin. One variation of this model is that rising hot magma could have melted ice in the ground and caused water to flow in aquifers. Aquifers are layer that allow water to flow. They may consist of porous sandstone. The aquifer layer would be perched on top of another layer that prevents water from going down (in geological terms it would be called impermeable). Because water in an aquifer is prevented from going down, the only direction the trapped water can flow is horizontally. Eventually, water could flow out onto the surface when the aquifer reaches a break—like a crater wall. The resulting flow of water could erode the wall to create gullies. Aquifers are quite common on Earth. A good example is "Weeping Rock" in Zion National Park Utah. As for the next theory, much of the surface of Mars is covered by a thick smooth mantle that is thought to be a mixture of ice and dust. This ice-rich mantle, a few yards thick, smoothes the land, but in places it has a bumpy texture, resembling the surface of a basketball. The mantle may be like a glacier and under certain conditions the ice that is mixed in the mantle could melt and flow down the slopes and make gullies. Because there are few craters on this mantle, the mantle is relatively young. An excellent view of this mantle is shown below in the picture of the Ptolemaeus Crater Rim, as seen by HiRISE. The ice-rich mantle may be the result of climate changes. Changes in Mars's orbit and tilt cause significant changes in the distribution of water ice from polar regions down to latitudes equivalent to Texas. During certain climate periods water vapor leaves polar ice and enters the atmosphere. The water comes back to ground at lower latitudes as deposits of frost or snow mixed generously with dust. The atmosphere of Mars contains a great deal of fine dust particles. Water vapor will condense on the particles, then fall down to the ground due to the additional weight of the water coating. When Mars is at its greatest tilt or obliquity, up to 2 cm (0.79 in) of ice could be removed from the summer ice cap and deposited at midlatitudes. This movement of water could last for several thousand years and create a snow layer of up to around 10 m (33 ft) thick. When ice at the top of the mantling layer goes back into the atmosphere, it leaves behind dust, which insulating the remaining ice. Measurements of altitudes and slopes of gullies support the idea that snowpacks or glaciers are associated with gullies. Steeper slopes have more shade which would preserve snow. Higher elevations have far fewer gullies because ice would tend to sublimate more in the thin air of the higher altitude. The third theory might be possible since climate changes may be enough to simply allow ice in the ground to melt and thus form the gullies. During a warmer climate, the first few meters of ground could thaw and produce a "debris flow" similar to those on the dry and cold Greenland east coast. Since the gullies occur on steep slopes only a small decrease of the shear strength of the soil particles is needed to begin the flow. Small amounts of liquid water from melted ground ice could be enough. Calculations show that a third of a mm of runoff can be produced each day for 50 days of each Martian year, even under current conditions. Magnetic Stripes and Plate TectonicsEdit The Mars Global Surveyor (MGS) discovered magnetic stripes in the crust of Mars, especially in the Phaethontis and Eridania quadrangles (Terra Cimmeria and Terra Sirenum). The magnetometer on MGS discovered 100 km (62 mi) wide stripes of magnetized crust running roughly parallel for up to 2,000 kilometres (1,200 mi). These stripes alternate in polarity with the north magnetic pole of one pointing up from the surface and the north magnetic pole of the next pointing down. When similar stripes were discovered on Earth in the 1960s, they were taken as evidence of plate tectonics. Researchers believe these magnetic stripes on Mars are evidence for a short, early period of plate tectonic activity. When the rocks became solid they retained the magnetism that existed at the time. A magnetic field of a planet is believed to be caused by fluid motions under the surface. The initial data was obtained when MGS traveled close to the planet during aerobraking. However, later measurements, collected over a 2-year period from an altitude of 400 km (250 mi), revealed that the magnetic features even matched up with known features on the surface. However, there are some differences, between the magnetic stripes on Earth and those on Mars. The Martian stripes are wider, much more strongly magnetized, and do not appear to spread out from a middle crustal spreading zone. Because the area containing the magnetic stripes is about 4 billion years old, it is believed that the global magnetic field probably lasted for only the first few hundred million years of Mars' life, when the temperature of the molten iron in the planet's core might have been high enough to mix it into a magnetic dynamo. There are no magnetic fields near large impact basins like Hellas. The shock of the impact may have erased the remnant magnetization in the rock. So, magnetism produced by early fluid motion in the core would not have existed after the impacts. When molten rock containing magnetic material, such as hematite (Fe2O3), cools and solidifies in the presence of a magnetic field, it becomes magnetized and takes on the polarity of the background field. This magnetism is lost only if the rock is subsequently heated above a particular temperature (the Curie point which is 770 °C for iron). The magnetism left in rocks is a record of the magnetic field when the rock solidified. Many features on Mars are believed to be glaciers with a relatively thin coating of debris that keeps the ice from melting. Some of these features are shown in the pictures below. A detailed description of them can be found in the article Glaciers on Mars. Crater floor, as seen by HiRISE under HiWish program. Rough surface was produced by ice leaving the ground. The crater has accumulated much ice that is covered by rocks and dirt. Arrhenius Crater, as seen by CTX camera (on Mars Reconnaissance Orbiter). Glacial features in Arrhenius Crater, as seen by HiRISE under the HiWish program. Arrows point to old glaciers. When there are perfect conditions for producing sand dunes, steady wind in one direction and just enough sand, a barchan sand dune forms. Barchans have a gentle slope on the wind side and a much steeper slope on the lee side where horns or a notch often forms. The whole dune may appear to move with the wind. Observing dunes on Mars can tell us how strong the winds are, as well as their direction. If pictures are taken at regular intervals, one may see changes in the dunes or possibly in ripples on the dune’s surface. On Mars dunes are often dark in color because they were formed from the common, volcanic rock basalt. In the dry environment, dark minerals in basalt, like olivine and pyroxene, do not break down as they do on Earth. Although rare, some dark sand is found on Hawaii which also has many volcanoes discharging basalt. Barchan is a Russian term because this type of dune was first seen in the desert regions of Turkistan. Some of the wind on Mars is created when the dry ice at the poles is heated in the spring. At that time, the solid carbon dioxide (dry ice) sublimates or changes directly to a gas and rushes away at high speeds. Each Martian year 30% of the carbon dioxide in the atmosphere freezes out and covers the pole that is experiencing winter, so there is a great potential for strong winds. Close up of dark dunes, as seen by HiRISE under HiWish program. The image is a little more than 1 km in its longest dimension. The location of this image is shown in the previous image. Dunes, as seen by HiRISE under HiWish program. Location is Eridania quadrangle. Interactive Mars mapEdit - http://planetarynames.wr.usgs.gov/Features/5930[permanent dead link] - Blunck, J. 1982. Mars and its Satellites. Exposition Press. Smithtown, N.Y. - Alan Boyle - Mars' mystery cloud explained (2012) - MSNBC - Alan Boyle -Mysterious cloud spotted on Mars (2012) - MSNBC - Edgett, K. et al. 2003. Polar-and middle-latitude martian gullies: A view from MGS MOC after 2 Mars years in the mapping orbit. Lunar Planet. Sci. 34. Abstract 1038. - Dickson, J. et al. 2007. Martian gullies in the southern mid-latitudes of Mars Evidence for climate-controlled formation of young fluvial features based upon local and global topography. Icarus: 188. 315-323 - Heldmann, J. and M. Mellon. Observations of martian gullies and constraints on potential formation mechanisms. 2004. Icarus. 168: 285-304. - Forget, F. et al. 2006. Planet Mars Story of Another World. Praxis Publishing. Chichester, UK. - Heldmann, J. and M. Mellon. 2004. Observations of martian gullies and constraints on potential formation mechanisms. Icarus. 168:285-304 - Harris, A and E. Tuttle. 1990. Geology of National Parks. Kendall/Hunt Publishing Company. Dubuque, Iowa - Malin, M. and K. Edgett. 2001. Mars Global Surveyor Mars Orbiter Camera: Interplanetary cruise through primary mission. J. Geophys. Res.: 106> 23429-23570 - Mustard, J. et al. 2001. Evidence for recent climate change on Mars from the identification of youthful near-surface ground ice. Nature: 412. 411-414. - Carr, M. 2001. Mars Global Surveyor observations of fretted terrain. J. Geophys. Res.: 106. 23571-23595. - Head, J. et al. 2008. Formation of gullies on Mars: Link to recent climate history and insolation microenvironments implicate surface water flow origin. PNAS: 105. 13258-13263. - Christensen, P. 2003. Formation of recent martian gullies through melting of extensive water-rich snow deposits. Nature: 422. 45-48. - Jakosky B. and M. Carr. 1985. Possible precipitation of ice at low latitudes of Mars during periods of high obliquity. Nature: 315. 559-561. - Jakosky, B. et al. 1995. Chaotic obliquity and the nature of the Martian climate. J. Geophys. Res.: 100. 1579-1584. - MLA NASA/Jet Propulsion Laboratory (2003, December 18). Mars May Be Emerging From An Ice Age. ScienceDaily. Retrieved February 19, 2009, from https://www.sciencedaily.com/releases/2003/12/031218075443.htmAds[permanent dead link] by GoogleAdvertise - Dickson, J. et al. 2007. Martian gullies in the southern mid-latitudes of Mars Evidence for climate-controlled formation of young fluvial features based upon local and global topography. Icarus: 188. 315-323. - Hecht, M. 2002. Metastability of liquid water on Mars. Icarus: 156. 373-386. - Peulvast, J. Physio-Geo. 18. 87-105. - Costard, F. et al. 2001. Debris Flows on Mars: Analogy with Terrestrial Periglacial Environment and Climatic Implications. Lunar and Planetary Science XXXII (2001). 1534.pdf - http://www.spaceref.com:16090/news/viewpr.html?pid=7124[permanent dead link], - Clow, G. 1987. Generation of liquid water on Mars through the melting of a dusty snowpack. Icarus: 72. 93-127. - Barlow, N. 2008. Mars: An Introduction to its Interior, Surface and Atmosphere. Cambridge University Press - ISBN 978-0-387-48925-4 - ISBN 978-0-521-82956-4 - ISBN 978-0-521-85226-5 - Connerney, J. et al. 1999. Magnetic lineations in the ancient crust of Mars. Science: 284. 794-798. - Langlais, B. et al. 2004. Crustal magnetic field of Mars. Journal of Geophysical Research. 109: EO2008 - Sprenke, K. and L. Baker. 2000. Magnetization, palemagnetic poles, and polar wander on Mars. Icarus. 147: 26-34. - Connerney, J. et al. 2005. Tectonic implications of Mars crustal magnetism. Proceedings of the National Academy of Sciences of the USA. 102: 14970-14975 - Acuna, M. et al. 1999. Global distribution of crustal magnetization discovered by the Mars Global Surveyor MAG/ER Experiment. Science. 284: 790-793. - Pye, Kenneth; Haim Tsoar (2008). Aeolian Sand and Sand Dunes. Springer. p. 138. ISBN 9783540859109. - Mellon, J. T.; Feldman, W. C.; Prettyman, T. H. (2003). "The presence and stability of ground ice in the southern hemisphere of Mars". Icarus. 169 (2): 324–340. Bibcode:2004Icar..169..324M. doi:10.1016/j.icarus.2003.10.022. \|Wikimedia Commons has media related to Terra Cimmeria.\|<\|endoftext\|>	3.90625
3,499	By: Clarence Collison Honey bees take repeated orientation flights before becoming foragers at about three weeks of age (Winston 1987). Foragers must learn to navigate between the hive and floral locations that may be miles away. Young pre-foragers prepare for this task by performing orientation flights near the hive, during which they begin to learn navigational cues such as the appearance of the hive, the position of landmarks, and the movement of the sun (Lutz and Robinson 2013). Bees with prior foraging experience will also perform re-orientation flights after the relocation of a colony (Winston 1987). The reproductive castes also take orientation flights prior to mating flights. During the Spring and Summer month’s drones begin to take orientation flights when about eight days old, while still sexually immature (Ruttner 1966). These orientation flights generally take place in the afternoon and last only a few minutes. Most queens also take one or two short orientation flights before leaving on their mating flight(s), most often in mid-afternoon (Winston 1987). Capaldi et al. (2000) initiated their investigations of orientation flights by introducing one-day-old adult worker bees tagged with numbered disks into a colony of about 20,000 workers and a queen, and simply watching them as they exited and re-entered the hive. All flight activities of the tagged bees were recorded. They found that no bee became a forager (defined as a bee that returns to the hive with nectar or pollen) without taking at least one orientation flight and that almost all bees took multiple orientation flights before they began to forage. The number of orientation flights taken by bees before foraging was highly variable, ranging from one to 18 (mean 5.6 ± 0.29 flights). Equally variable was the age at which these flights began, ranging from three to 14 days (mean 6.2 ± 0.18 days). In this study the mean age at onset of pollen foraging was 14 ± 2 days. In a second study, they used harmonic radar to extend the range of their observations. Tagged one-day-old bees were introduced into a standard hive and all of their subsequent flights were visually observed and recorded. They were thus able to select bees of known age and flight history for radar tracking. Transponders were attached to bees as they departed from the hive and orientation flight tracks were recorded by the radar. Each bee was tracked only once. They analyzed 29 complete (out and back) orientation flights taken by bees of different ages (3-27, median = six days), with differing degrees of experience (1-17 orientation flights). For comparison, they also tracked flights taken by experienced foragers of unknown age from the same hive. All flights were naturally occurring; that is, all tracked bees had moved spontaneously from inside the hive to the hive entrance before attachment of the transponder. Most bees fitted with transponders began their orientation flights with a brief period of hovering facing the colony entrance before departure, just as those without transponders did and has been previously reported. The duration of the orientation flights of bees with and without transponders was not significantly different (with transponders 331.6 ± 59.2 seconds; without transponders 340.1 ± 26.4 seconds). They therefore concluded that the attachment of transponders did not significantly alter flight behavior. Significant positive correlations between flight number and speed relative to the ground (averaged over the journey), round trip journey distance, maximum range from the hive and the area covered by the convex polygon circumscribing each flight were found. There was no significant positive correlation between flight number and flight duration; as the bees gained experience they apparently travelled further by moving faster rather than by staying out longer. Capaldi et al. (2000) findings suggest that bees take multiple orientation flights before becoming foragers in order to visit different and larger portions of the landscape around the hive. These flights provide them with repeated opportunities to view the hive and its surroundings from different positions, suggesting that bees learn the local landscape in a progressive fashion. Bees navigate using a combination of cues, including the position of the sun and the location of salient landscape features (Collett 1992; Dyer and Dickinson 1994), but it is not known how or whether information about these cues, obtained during sequential flights is integrated. Scouting behavior is performed in two distinct contexts: scouting for new food sources or new nest sites. There are striking individual differences in scouting behavior- some bees act as scouts and others never do so. Food scouts, who make up five to 35% of a colony’s foraging force, search independently for new food sources and continue to do so even when plentiful sources have been found. Non-scouts do not search for novel food sources and instead rely on information from scouts (communicated via “dance language”) to guide their foraging. By constantly discovering new flower patches, food scouts help ensure a high influx of food to their colony, despite the ephemeral nature of each patch. Experienced foragers scout more than do novice foragers. The cost of finding a forage patch is greater for recruits than scouts, but the patches found by recruits are evidently superior to those found by scouts. The honey bees combined system of recruitment communication, scout-recruit division of labor, and selectivity in recruitment, apparently enhances a colony’s overall foraging efficiency by guiding a large majority of a colony’s foragers to good forage patches (Seeley 1983). Nest scouts make up <5% of the population of a swarm, which is a fragment of a colony that has left its natal nest to start a new colony. Nest scouts search independently for potential nesting cavities and collectively choose the best one, whereas non-scout swarm members rely on information from scouts to guide them to their new home (Seeley 2010). Nest scouting is a crucial behavior; a colony’s survival depends on its nest scouts finding suitable protective living quarters. To determine the consistency of novelty seeking in individual bees across the two behavioral contexts, Liang et al. (2012) determined whether nest scouts are prone to also act as food scouts. They identified and marked nest scouts in both artificial and natural swarms. They then identified food scouts with the standard “hive moving” assay (Seeley 1983; Dreller 1998), after installing each swarm in a beehive and moving it at night (when bees don’t forage) to a new location outside the bees’ original home range. This assay identifies food scouts as the first bees to return to their hive in the morning; under these circumstances, each successful forager must have located a food source on her own. There was a robust tendency of nest scouts to seek novel resources across different contexts, but it did not translate into every nest scout showing food-scouting behavior. In nine trials involving eight different colonies over two years, nest scouts were on average 3.4 times more likely to become food scouts than were bees that did not search for nest sites during swarming. These results demonstrate that some bees show consistent novelty seeking across diverse behavioral contexts. Liang et al. (2012) also investigated the neurochemical basis of scouting behavior and they revealed some of the molecular underpinnings of this behavior relative to foragers that do not scout. Food scouts showed extensive differences in brain gene expression relative to other foragers, including differences related to catecholamine, glutamate, and g-aminobutyric acid signaling. Octopamine and glutamate treatments increased the likelihood of scouting, whereas dopamine antagonist treatment decreased it. Within a swarm cluster, the bees are divided between two behavioral states. The vast majority of the bees in the swarm are inactive. Engorged with honey, these bees function as the swarm’s food reservoir (Combs 1972). A small minority of the bees, however, are active. They serve as the nest-site scouts, flying from the swarm cluster to discover and inspect potential nest cavities. They then return to report their discoveries by performing waggle dances on the surface of the swarm cluster. Gilley (1998) identified the nest-site scouts by examining the age distribution of the bees who engaged in scouting activities for both prime swarms and afterswarms. Statistical differences were found between the age distributions of the swarm and parental colony, the scouts and the swarm and the scouts and the foragers. The median age of the swarm bees was lower than that of the colony bees, that of the scouts was higher than that of the swarm bees and that of the scouts was slightly less than that of the foragers. These results suggest that the nest-site scouts are primarily middle aged bees which have foraging or flight experience. Viewing the nest-site scouting process at the group level, Seeley and Buhrman (1999) found: (1) the scout bees in a swarm find potential nest sites in all directions and at distances of up to several kilometers; (2) initially, the scouts advertise a dozen or more sites with their dances on the swarm, but eventually they advertise just one site; (3) within about an hour of the appearance of unanimity among the dancers, the swarm lifts off to fly to the chosen site; (4) there is a crescendo of dancing just before liftoff, and (5) the chosen site is not necessarily the one that is first advertised on the swarm. Viewing the process at the individual level, they found: (1) the dances of individual scout bees tend to taper off and eventually cease, so that many dancers drop out each day; (2) some scout bees switch their allegiance from one site to another, and (3) the principal means of consensus building among the dancing bees is for bees that dance initially for a non-chosen site to cease their dancing altogether, not to switch their dancing to the chosen site. They hypothesized that scout bees are programmed to gradually quit dancing and that this reduces the possibility of the decision-making process coming to a standstill with groups of unyielding dancers deadlocked over two or more sites. When a scout returns to the swarm cluster after inspecting a high-quality cavity, she performs waggle dances which encode the distance and direction to the site. Since a swarm has many scouts, many potential nest sites may be discovered, and dances for several sites may be performed simultaneously on the surface of the swarm cluster. However, in time, the bees reach a consensus, and only dances for a single site are seen. Shortly thereafter the cluster abruptly breaks and the bees fly to the chosen site (Camazine et al. 1999). Nest-site choice by a honey bee swarm is an impressive example of group decision making. Seeley and Visscher (2003) considered the mystery of how the scout bees in a honey bee swarm know when they have completed their group decision making regarding the swarm’s new nest site. More specifically, they investigated how the scouts sense when it is appropriate for them to begin producing the worker piping signals that stimulate their swarm-mates to prepare for the flight to their new home. Two hypotheses were tested: “consensus sensing,” the scouts noting when all the bees performing waggle dances are advertising just one site; and “quorum sensing,” the scouts noting when one site is being visited by a sufficiently large number of scouts. Their test involved monitoring four swarms as they discovered, recruited to, and chose between two nest boxes and their scouts started producing piping signals. They found that a consensus among the dancers was neither necessary nor sufficient for the start of worker piping, which indicates that the consensus sensing hypothesis is false. They also found that a buildup of 10-15 or more bees at one of the nest boxes was consistently associated with the start of worker piping, which indicates that the quorum sensing hypothesis may be true. In considering why the scout bees rely on reaching a quorum rather than a consensus as their cue of when to start preparing for liftoff, they suggest that quorum sensing may provide a better balance between accuracy and speed in decision making. In short, the bees appear to begin preparations for liftoff as soon as enough of the scout bees, but not all of them, have approved of one of the potential nest sites. Honey bee scouting, where individual bees search the environment without prior information about the possible location of food sources or nest sites is notoriously difficult to study. The use of simulation models is one way to investigate the possible mechanisms behind the regulation of scouting at the group level as well as the ways in which the swarm searches its environment. Janson et al. (2007) used an individual based simulation model to study the scouting behavior of honey bee swarms. In their model they implemented a simple decision rule that regulated the number of scouts: individual bees’ first attempt to find a dance to follow but become scouts if they fail to do so. They showed that this rule neatly allows the swarm to adjust the number of scouts depending on the quality of the nest sites known to the swarm. They also explored different search strategies that allowed the swarm to select good-quality nest sites independent of their distance from the swarm. Assuming that it is costly to move to a site that is far away, the best search strategy would be to give precedence to nearby sites while still allowing the discovery of better sites at distances farther away. Camazine, S., P.K. Visscher, J. Finley and R.S. Vitter 1999. House-hunting by honey bee swarms: collective decisions and individual behaviors. Insect. Soc. 46: 348-360. Capaldi, E.A., A.D. Smith, J.L. Osborne, S.E. Fahrbach, S.M Farris, D.R. Reynolds, A.S. Edwards, A. Martin, G.E. Robinson, G.M. Poppy and J.R. Riley 2000. Ontogeny of orientation flight in the honeybee revealed by harmonic radar. Nature 403: 537-540. Collett, T.S. 1992. Landmark learning and guidance of insects. Phil. Trans. R. Soc. Lond. B 337: 295-303. Combs, G.F. 1972. The engorgement of swarming worker honeybees. J. Apic. Res. 11: 121-128. Dreller, C. 1998. Division of labor between scouts and recruits: genetic influence and mechanisms. Behav. Ecol. Sociobiol. 43: 191-196. Dyer, F.C. and J.A. Dickinson 1994. How partially experienced bees estimate the sun’s course. Proc. Ntl. Acad. Sci. USA 91: 4471-4474. Gilley, D.C. 1998. The identity of nest-site scouts in honey bee swarms. Apidologie 29: 229-240. Janson, S., M. Middendorf, and M. Beekman 2007. Searching for a new home – scouting behavior of honeybee swarms. Behav. Ecol. 18: 384-392. Liang, Z.S., T. Nguyen, H.R. Mattila, S.L. Rodriguez-Zas, T.D. Seeley and G.E. Robinson 2012. Molecular determinants of scouting behavior in honey bees. Science 335: 1225-1228. Lutz, C.C. and G.E. Robinson 2013. Activity-dependent gene expression in honey bee mushroom bodies in response to orientation flight. J. Exp. Biol. 216: 2031-2038. Ruttner, F. 1966. The life and flight activity of drones. Bee World 47: 93-100. Seeley, T.D. 1983. Division of labor between scouts and recruits in honeybee foraging. Behav. Ecol. Sociobiol. 12: 253-259. Seeley, T.D. 2010. Honeybee Democracy. Princeton Univ Press, Princeton, NJ. Seeley, T.D. and S.C. Buhrman 1999. Group decision making in swarms of honey bees. Behav. Ecol. Sociobiol. 45: 19-31. Seeley, T.D. and P.K. Visscher 2003. Choosing a home: how the scouts in a honey bee swarm perceive the completion of their group decision making. Behav. Ecol. Sociobiol. 54: 511-520. Winston, M.L. 1987. The Biology Of The Honey Bee. Harvard University Press, Cambridge, MA, 281 pp. Clarence Collison is an Emeritus Professor of Entomology and Department Head Emeritus of Entomology and Plant Pathology at Mississippi State University, Mississippi State, MS.<\|endoftext\|>	3.9375
607	The urge to explore beyond our solar system grows stronger every day. This proves true for the understanding of wormholes and time travel as well. In order to quench our thirst for the unknown, NASA will research unknown physics revolutionizing exploration of space. We first have to improve on our understanding of space-time, the quantum vacuum, gravity and other physical phenomena. This information will help NASA send robots on interstellar space missions. Exactly 15 areas will be studied including human exploration, landing systems, nanotechnology and robots. Propulsion technologies are the major focus with NASA, revealing physics that create revolutionary engines. As of now, propulsion needed is not feasible. For the next 20 years, NASA will be working on propulsion as part of technology roadmaps, enabling ‘robotic interstellar missions’ into space. Chief Technologist at NASA Headquarters in Washington DC, Dr. David Miller, says: “Candidates included in the technology roadmaps provide capabilities which can be improved, leveraged and built upon.” These improvements will be needed to further explore the outer reaches of space. The capabilities will also increase jobs and improve the future of our youth – a future where exploration of space is common. Studies in these areas will involve intricate scientific work. Time Warp and Wormholes Not only is propulsion a popular study. Wormholes and space warp are also a subject of interest to NASA. Demonstrating microscopic instances of both these phenomena means space travel over huge distances in seconds. How is this possible? One of the most recent technologies revolves around the EmDrive. The EmDrive eliminates the use of rocket fuel by converting electrical energy into thrust. This technology would enable space travel at speeds faster than anything possible as of yet. NASA announced that funding would be provided, in early April, for the Vasimir Engine. This engine uses a plasma powered form of propulsion. This means travel to Mars could take a little more than a month’s time -much less than 8 months, which was the previous record time. NASA’s asteroid mission is planned for 2020, which includes technologies enabling humans to work in the space vacuum, and an unmanned mission to Mars may be possible as early as 2026. Missions to Mars would provide resources from the red planet thus reducing resources from the earth. This would, in turn, increase missions into space Dr. Harold White, NASA scientist, revealed sketches of interstellar warships which could travel to distant space. NASA has also been hard at work with alternate forms of propulsion such as solar sails, antimatter drives and ‘beamed’ propulsion. ‘Beamed’ propulsion is shooting a laser at a ship to make it move. These technologies could be implemented by the year 2033. It seems NASA is hard at work on space travel. It’s simply a race against time. Copyright © 2012-2019 Learning Mind. All rights reserved. For permission to reprint, contact us.<\|endoftext\|>	4
681	This Tutorial talks about basics of Linear regression by discussing in depth about the concept of Linearity and Which type of linearity is desirable. What is the meaning of the term Linear ? In Linear Regression the term linear is understood in 2 ways – - Linearity in variables - Linearity in parameters Linear regression however always means linearity in parameters , irrespective of linearity in explanatory variables. A linear regression for 2 variables is represented mathematically as ( u is the error term )- Y = B1 + B2X + u Or Y = B1 + B2X ² + u Here the variable X can be non linear i.e X or X² and still we can consider this as a linear regression. However if our parameters are not linear i.e say the regression equation is Y = B1² + B2²X + u then this can not be said to represent a linear regression equation. Linear Regression Models \|Model linear in parameters? \|Model linear in variables?\| \|Yes\|\|Linear Model\|\|Linear Model\| \|No\|\|Non Linear Model\|\|Non Linear Model\| Linearity in predictor variables – Xi A function Y = f(x) is said to be linear in X if X appears with a power or index of 1 only. i.e the terms such as x2, Γx, and so on are excluded or if x is not multiplied or divided by any other variable. Linearity in parameters – Bi Y is linearly related to X if the rate of change of Y with respect to X (dY/dX) is independent of the value of X. A function is said to be linear in the parameter, say, B1, if B1 appears with a power of 1 only and is not multiplied or divided by any other parameter (for eg B1 x B2 , or B2 / B1) To reiterate again – For purpose of Linear regression we are only concerned about linearity of parameters B1, B2 …. and not the actual variables X1, X2 …. Non Linear Models - Some models may look non linear in the parameters but are inherently or intrinsically linear. - This is because with suitable transformations they can be made linear in parameters. - However, if these cannot be linearized, these are called intrinsically non linear regression models - When we say ‘non linear regression model’ we mean that it is intrinsically non linear. For Log(Yi) = Log(B1) + B2 Log(Xi) + u B2 is Linear but B1 is non-linear but if we transform α = Log(B1) then the model Log(Yi) = α + B2 Log(Xi) + u is linear in α and B2 as parameters. Implying we can make the regression equation linear in parameters using a simple transformation For other cases we may not have an easy way to transform parameters to their linear form and such equations are hence treated as intrinsically non-linear and are NOT modeled using linear regression Next in the series : Reference : Based on Lectures by Dr. Manish Sinha. ( Associate Prof. SCMHRD )<\|endoftext\|>	3.703125
476	# Prove that if 2 divides n and 7 divides n, then 14 divides n Okay so I have to prove this. I can write that if 2 divides n and 7 divides n, then there must be integers k and m such that $2k=n$ and $7m=n$ So $14km=n^2$ But what to do after that? If I say that then 14 divides $n^2$, I get bit of a circular argument, but if I write that n divides $14km$, then I don't know what to do next. Any help/suggestions? • Some general advice: when you are asked to prove something that seems so incredibly obvious that it appears all proofs would go in circles, what you're really being asked to do is to use the explicitly allowed axioms and definitions to prove it. Don't start doubting your intuition about the patently obvious. Just examine the given definitions to establish to your satisfaction that they are sufficient, all by themselves, to formally prove the statement in question. Commented Oct 17, 2017 at 22:54 Following from what you have written, $$n = 2k=7m \implies k=\frac{7m}{2}.$$ Since $k$ is an integer and $\gcd(2,7)=1$, $m/2$ must be an integer; i.e., $m/2=r \implies m=2r$, where $r$ is an integer. Therefore, $$n=7m=7\times 2r = 14 r.$$ Q.E.D. You can say much more: Say $a$ and $b$ are relatively prime. If $a\|n$ and $b\|n$ then $ab\|n$. Proof: Since $a\|n$ we can write $n=ak$. Now since $b\|ak$ we have, by Euclid lemma $b\|k$, so $k=bl$. Thus $n=abl$ and so $ab\|n$. Write $n = 7 m$. If $m$ were odd, $7m$ would also be odd, contradiction. So $m$ is even, $m=2k$, and $n = 14k$.<\|endoftext\|>	4.40625
804	Hawksbill sea turtles, named for their pointed beaks, are teetering on the edge of extinction. With populations down to perhaps 10 percent of what they were a century ago, according to Richard Hamilton, the Nature Conservancy’s Melanesia program director, hawksbills are the most critically endangered of all seven species of sea turtles. Found in the tropical and subtropical waters of the Atlantic, Pacific, and Indian Oceans, they’re also among the smallest. They face seemingly insurmountable threats: a thriving illegal trade in their shells, poaching of their eggs, hunting for their meat, beach erosion and human development at their nesting sites, and degradation of coral reefs where they forage. We don’t know exactly how many hawksbills are left. Counting them is complicated because they come ashore to nest only once every two to seven years, and when they do, they usually gather in small concentrations. The Sea Turtle Conservancy estimates that there are between 20,000 and 23,000 nesting females worldwide. The Arnavons—four forested bumps within the Solomon Islands that support between 300 and 600 nesting females annually and a total nesting population of 2,000 to 4,000—form the largest rookery in the South Pacific. According to Hamilton, poaching in the Solomons has been getting worse, and hawksbill shell products are for sale in the airport in the capital, Honiara. It’s legal for local people to harvest hawksbills for subsistence, but national law bans the sale of any turtle product, and international trade in hawksbills is also banned. In April, during the peak nesting season, the Nature Conservancy and scientists and community leaders from the Arnavon Community Marine Conservation Area tagged 10 hawksbills with GPS trackers. Although two of the females were subsequently killed by poachers, data from the others have already provided insight on hawksbill movement and nesting and feeding habits. Key findings so far: - Sea turtles island-hop. Satellite tagging reveals that individuals nested on beaches in two main areas on Kerehikapa and Sikopo Islands. As a result the number of rangers has been doubled, from three to six, and rangers are present on both islands. - The bulk of the nesting now occurs on Sikopo Island, likely because rising sea levels and storm surges have eroded the beaches on Kerehikapa. Sikopo has been less affected because it has more beaches at higher elevations. - Some good news: While nesting, and during the two-week period between laying successive clutches of eggs, the turtles are spending nearly all their time within the protected area boundaries of the Arnavons. - Migration patterns vary extremely. One of the tagged turtles stayed in the Solomons, but most of the others swam all the way back to Australia’s Great Barrier Reef, a distance of some 1,300 miles (2,092 kilometers), their home base until the next nesting journey. (Those returning to northern and central parts of the reef found a very different habitat from the one they’d left months earlier: Bleaching caused by warmer temperatures has killed more than 35 percent of the corals in those areas.) Hamilton hopes to expand the turtle-tagging program. “We need to get a handle on current harvest rates,” he says. “Nobody has any idea of either the legal or illegal harvest rates, or on what’s sustainable.” With that information, it will be easier to nip illegal exports and get better enforcement. In the long term, he says, the local people need to be informed about the turtles. If the community is involved in everything from naming the tagged turtles to monitoring the beaches, “I’m optimistic about the way forward.”<\|endoftext\|>	3.765625
5,256	$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 7.4: Ellipses In the definition of a circle, Definition \ref{circledefn}, we fixed a point called the center and considered all of the points which were a fixed distance $$r$$ from that one point. For our next conic section, the ellipse, we fix two distinct points and a distance $$d$$ to use in our definition. Definition: Ellipse Given two distinct points $$F_1$$ and $$F_2$$ in the plane and a fixed distance $$d$$, an ellipse is the set of all points $$(x, y)$$ in the plane such that the sum of each of the distances from $$F_1$$ and $$F_2$$ to $$(x, y)$$ is $$d$$. The points $$F_1$$ and $$F_2$$ are called the foci (the plural of focus') of the ellipse. We may imagine taking a length of string and anchoring it to two points on a piece of paper. The curve traced out by taking a pencil and moving it so the string is always taut is an ellipse. The center of the ellipse is the midpoint of the line segment connecting the two foci. The major axis of the ellipse is the line segment connecting two opposite ends of the ellipse which also contains the center and foci. The minor axis of the ellipse is the line segment connecting two opposite ends of the ellipse which contains the center but is perpendicular to the major axis. The vertices of an ellipse are the points of the ellipse which lie on the major axis. Notice that the center is also the midpoint of the major axis, hence it is the midpoint of the vertices. In pictures we have, An ellipse with center $$C$$; foci $$F_1$$, $$F_2$$; and vertices $$V_1$$, $$V_2$$ Note that the major axis is the longer of the two axes through the center, and likewise, the minor axis is the shorter of the two. In order to derive the standard equation of an ellipse, we assume that the ellipse has its center at $$(0,0)$$, its major axis along the $$x$$-axis, and has foci $$(c,0)$$ and $$(-c,0)$$ and vertices $$(-a,0)$$ and $$(a,0)$$. We will label the $$y$$-intercepts of the ellipse as $$(0,b)$$ and $$(0,-b)$$ (We assume $$a$$, $$b$$, and $$c$$ are all positive numbers.) Schematically, Note that since $$(a,0)$$ is on the ellipse, it must satisfy the conditions of Definition \ref{ellipsedefn}. That is, the distance from $$(-c,0)$$ to $$(a,0)$$ plus the distance from $$(c,0)$$ to $$(a,0)$$ must equal the fixed distance $$d$$. Since all of these points lie on the $$x$$-axis, we get $\begin{array}{rclr} {distance from $$(-c,0)$$ to $$(a,0)$$} + {distance from $$(c,0)$$ to $$(a,0)$$} & = & d & \\ (a+c) + (a-c) & = & d & \\ 2a & = & d \\ \end{array}$ In other words, the fixed distance $$d$$ mentioned in the definition of the ellipse is none other than the length of the major axis. We now use that fact $$(0,b)$$ is on the ellipse, along with the fact that $$d=2a$$ to get $\begin{array}{rclr} {distance from $$(-c,0)$$ to $$(0,b)$$} + {distance from $$(c,0)$$ to $$(0,b)$$} & = & 2a & \\ \sqrt{(0-(-c))^2+(b-0)^2} + \sqrt{(0-c)^2+(b-0)^2} & = & 2a & \\ \sqrt{b^2+c^2}+\sqrt{b^2+c^2} & = & 2a \\ 2 \sqrt{b^2+c^2} & = & 2a \\ \sqrt{b^2+c^2} & = & a \end{array}$ From this, we get $$a^2 = b^2 + c^2$$, or $$b^2 = a^2 - c^2$$, which will prove useful later. Now consider a point $$(x,y)$$ on the ellipse. Applying Definition \ref{ellipsedefn}, we get $\begin{array}{rclr} {distance from $$(-c,0)$$ to $$(x,y)$$} + {distance from $$(c,0)$$ to $$(x,y)$$} & = & 2a & \\ \sqrt{(x-(-c))^2+(y-0)^2} + \sqrt{(x-c)^2+(y-0)^2} & = & 2a & \\ \sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2} & = & 2a \\ \end{array}$ In order to make sense of this situation, we need to make good use of Intermediate Algebra. $\begin{array}{rclr} \sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2} & = & 2a & \\ \sqrt{(x+c)^2+y^2} & = & 2a - \sqrt{(x-c)^2+y^2} & \\ \left(\sqrt{(x+c)^2+y^2}\right)^2 & = & \left(2a - \sqrt{(x-c)^2+y^2}\right)^2 & \\ (x+c)^2+y^2 & = & 4a^2 - 4a\sqrt{(x-c)^2+y^2} + (x-c)^2+y^2 & \\ 4a\sqrt{(x-c)^2+y^2} & = & 4a^2 + (x-c)^2 - (x+c)^2 & \\ 4a\sqrt{(x-c)^2+y^2} & = & 4a^2 - 4cx & \\ a\sqrt{(x-c)^2+y^2} & = & a^2 - cx & \\ \left(a\sqrt{(x-c)^2+y^2}\right)^2 & = & \left(a^2 - cx\right)^2 & \\ a^2\left((x-c)^2+y^2\right) & = & a^4 - 2a^2cx +c^2 x^2 & \\ a^2x^2 - 2a^2cx + a^2c^2+a^2 y^2 & = & a^4 - 2a^2cx +c^2 x^2 & \\ a^2x^2 - c^2 x^2 +a^2 y^2 & = & a^4 - a^2c^2 & \\ \left(a^2 - c^2\right) x^2 +a^2 y^2 & = & a^2 \left(a^2 - c^2\right) & \\ \end{array}$ We are nearly finished. Recall that $$b^2 = a^2 - c^2$$ so that $\begin{array}{rclr} \left(a^2 - c^2\right)x^2 +a^2 y^2 & = & a^2\left(a^2 - c^2\right) & \\ b^2 x^2 +a^2 y^2 & = & a^2 b^2 & \\ \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} & = & 1 & \\ \end{array}$ This equation is for an ellipse centered at the origin. To get the formula for the ellipse centered at $$(h,k)$$, we could use the transformations from Section \ref{Transformations} or re-derive the equation using Definition \ref{ellipsedefn} and the distance formula to obtain the formula below. The Standard Equation of an Ellipse For positive unequal numbers $$a$$ and $$b$$, the equation of an ellipse with center $$(h,k)$$ is $\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1$ Some remarks about Equation \ref{standardellipse} are in order. First note that the values $$a$$ and $$b$$ determine how far in the $$x$$ and $$y$$ directions, respectively, one counts from the center to arrive at points on the ellipse. Also take note that if $$a > b$$, then we have an ellipse whose major axis is horizontal, and hence, the foci lie to the left and right of the center. In this case, as we've seen in the derivation, the distance from the center to the focus, $$c$$, can be found by $$c = \sqrt{a^2 - b^2}$$. If $$b > a$$, the roles of the major and minor axes are reversed, and the foci lie above and below the center. In this case, $$c = \sqrt{b^2 - a^2}$$. In either case, $$c$$ is the distance from the center to each focus, and $$c = \sqrt{ {bigger denominator} - {smaller denominator}}$$. Finally, it is worth mentioning that if we take the standard equation of a circle, Equation \ref{standardcircle}, and divide both sides by $$r^2$$, we get The Alternate Standard Equation of a Circle: The equation of a circle with center $$(h,k)$$ and radius $$r >0$$ is $\dfrac{(x-h)^2}{r^2} + \dfrac{(y-k)^2}{r^2} = 1$ Notice the similarity between Equation \ref{standardellipse} and Equation \ref{standardcirclealternate}. Both equations involve a sum of squares equal to $$1$$; the difference is that with a circle, the denominators are the same, and with an ellipse, they are different. If we take a transformational approach, we can consider both Equations \ref{standardellipse} and \ref{standardcirclealternate} as shifts and stretches of the Unit Circle $$x^2 + y^2 = 1$$ in Definition \ref{UnitCircle}. Replacing $$x$$ with $$(x-h)$$ and $$y$$ with $$(y-k)$$ causes the usual horizontal and vertical shifts. Replacing $$x$$ with $$\frac{x}{a}$$ and $$y$$ with $$\frac{y}{b}$$ causes the usual vertical and horizontal stretches. In other words, it is perfectly fine to think of an ellipse as the deformation of a circle in which the circle is stretched farther in one direction than the other.\footnote{This was foreshadowed in Exercise \ref{circletransunitcircleexercise} in Section \ref{Circles}.} Example $$\PageIndex{1}$$: Ellipse Properties Graph $\frac{(x+1)^2}{9} + \frac{(y-2)^2}{25} = 1$ Also find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. Solution We see that this equation is in the standard form of Equation \ref{standardellipse}. Here $$x-h$$ is $$x+1$$ so $$h = -1$$, and $$y-k$$ is $$y-2$$ so $$k = 2$$. Hence, our ellipse is centered at $$(-1,2)$$. We see that $$a^2 = 9$$ so $$a=3$$, and $$b^2 = 25$$ so $$b=5$$. This means that we move $$3$$ units left and right from the center and $$5$$ units up and down from the center to arrive at points on the ellipse. As an aid to sketching, we draw a rectangle matching this description, called a \index{ellipse ! guide rectangle} \index{guide rectangle ! for an ellipse} \textbf{guide rectangle}, and sketch the ellipse inside this rectangle as seen below on the left. Since we moved farther in the $$y$$ direction than in the $$x$$ direction, the major axis will lie along the vertical line $$x=-1$$, which means the minor axis lies along the horizontal line, $$y = 2$$. The vertices are the points on the ellipse which lie along the major axis so in this case, they are the points $$(-1,7)$$ and $$(-1,-3)$$, and the endpoints of the minor axis are $$(-4,2)$$ and $$(2,2)$$. (Notice these points are the four points we used to draw the guide rectangle.) To find the foci, we find $$c = \sqrt{25-9} = \sqrt{16} = 4$$, which means the foci lie $$4$$ units from the center. Since the major axis is vertical, the foci lie $$4$$ units above and below the center, at $$(-1,-2)$$ and $$(-1,6)$$. Plotting all this information gives the graph seen above on the right. \qed Example $$\PageIndex{2}$$: Find the equation of the ellipse with foci $$(2,1)$$ and $$(4,1)$$ and vertex $$(0,1)$$. Solution Plotting the data given to us, we have From this sketch, we know that the major axis is horizontal, meaning $$a > b$$. Since the center is the midpoint of the foci, we know it is $$(3, 1)$$. Since one vertex is $$(0,1)$$ we have that $$a = 3$$, so $$a^2 = 9$$. All that remains is to find $$b^2$$. Since the foci are $$1$$ unit away from the center, we know $$c=1$$. Since $$a > b$$, we have $$c = \sqrt{a^2-b^2}$$, or $$1 = \sqrt{9-b^2}$$, so $$b^2 = 8$$. Substituting all of our findings into the equation $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$, we get our final answer to be $$\frac{(x-3)^2}{9}+\frac{(y-1)^2}{8}=1$$. \qed As with circles and parabolas, an equation may be given which is an ellipse, but isn't in the standard form of Equation \ref{standardellipse}. In those cases, as with circles and parabolas before, we will need to massage the given equation into the standard form. To Write the Equation of an Ellipse in Standard Form 1. Group the same variables together on one side of the equation and position the constant on the other side. 2. Complete the square in both variables as needed. 3. Divide both sides by the constant term so that the constant on the other side of the equation becomes $$1$$. Example $$\PageIndex{3}$$: Graph $$x^2+4y^2-2x+24y+33 = 0$$. Find the center, the lines which contain the major and minor axes, the vertices, the endpoints of the minor axis, and the foci. Solution Since we have a sum of squares and the squared terms have unequal coefficients, it's a good bet we have an ellipse on our hands.\footnote{The equation of a parabola has only one squared variable and the equation of a circle has two squared variables with identical coefficients. We need to complete both squares, and then divide, if necessary, to get the right-hand side equal to $$1$$. $\begin{array}{rclr} x^2+4y^2-2x+24y+33 & = & 0 & \\ x^2-2x+4y^2+24y & = & -33 & \\ x^2 - 2x + 4\left(y^2+6y\right) & = & - 33 & \\ \left(x^2 - 2x +1\right) + 4\left(y^2+6y+9\right) & = & - 33 + 1 + 4(9)& \\ (x-1)^2 + 4(y+3)^2 & = & 4 & \\[5pt] \dfrac{(x-1)^2 + 4(y+3)^2}{4} & = & \dfrac{4}{4} & \\[10pt] \dfrac{(x-1)^2}{4} + (y+3)^2 & = & 1 & \\[10pt] \dfrac{(x-1)^2}{4} + \dfrac{(y+3)^2}{1} & = & 1 & \\ \end{array}$ Now that this equation is in the standard form of Equation \ref{standardellipse}, we see that $$x-h$$ is $$x-1$$ so $$h = 1$$, and $$y-k$$ is $$y+3$$ so $$k = -3$$. Hence, our ellipse is centered at $$(1,-3)$$. We see that $$a^2 = 4$$ so $$a=2$$, and $$b^2 = 1$$ so $$b=1$$. This means we move $$2$$ units left and right from the center and $$1$$ unit up and down from the center to arrive at points on the ellipse. Since we moved farther in the $$x$$ direction than in the $$y$$ direction, the major axis will lie along the horizontal line $$y=-3$$, which means the minor axis lies along the vertical line $$x = 1$$. The vertices are the points on the ellipse which lie along the major axis so in this case, they are the points $$(-1,-3)$$ and $$(3,-3)$$, and the endpoints of the minor axis are $$(1,-2)$$ and $$(1,-4)$$. To find the foci, we find $$c = \sqrt{4-1} = \sqrt{3}$$, which means the foci lie $$\sqrt{3}$$ units from the center. Since the major axis is horizontal, the foci lie $$\sqrt{3}$$ units to the left and right of the center, at $$(1-\sqrt{3},-3)$$ and $$(1+\sqrt{3},-3)$$. Plotting all of this information gives As you come across ellipses in the homework exercises and in the wild, you'll notice they come in all shapes in sizes. Compare the two ellipses below. Certainly, one ellipse is more round than the other. This notion of roundness' is quantified below. Definition: Eccentricity of an Ellipse The eccentricity of an ellipse, denoted $$e$$, is the following ratio: $e = \dfrac{ \text{distance from the center to a focus}}{ \text{distance from the center to a vertex}}$ In an ellipse, the foci are closer to the center than the vertices, so $$0 < e < 1$$. The ellipse above on the left has eccentricity $$e \approx 0.98$$; for the ellipse above on the right, $$e \approx 0.66$$. In general, the closer the eccentricity is to $$0$$, the more circular' the ellipse; the closer the eccentricity is to $$1$$, the more eccentric' the ellipse. Example $$\PageIndex{1}$$: Find the equation of the ellipse whose vertices are $$(\pm 5,0 )$$ with eccentricity $$e = \frac{1}{4}$$. Solution As before, we plot the data given to us From this sketch, we know that the major axis is horizontal, meaning $$a > b$$. With the vertices located at $$(\pm 5,0)$$, we get $$a = 5$$ so $$a^2 = 25$$. We also know that the center is $$(0,0)$$ because the center is the midpoint of the vertices. All that remains is to find $$b^2$$. To that end, we use the fact that the eccentricity $$e = \frac{1}{4}$$ which means $e = \dfrac{ {distance from the center to a focus}}{ {distance from the center to a vertex}} = \dfrac{c}{a} = \dfrac{c}{5} = \dfrac{1}{4}$ from which we get $$c = \frac{5}{4}$$. To get $$b^2$$, we use the fact that $$c = \sqrt{a^2 - b^2}$$, so $$\frac{5}{4} = \sqrt{25-b^2}$$ from which we get $$b^2 = \frac{375}{16}$$. Substituting all of our findings into the equation $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$, yields our final answer $$\frac{x^2}{25} + \frac{16y^2}{375}=1$$. \qed As with parabolas, ellipses have a reflective property. If we imagine the dashed lines below representing sound waves, then the waves emanating from one focus reflect off the top of the ellipse and head towards the other focus. Such geometry is exploited in the construction of so-called `Whispering Galleries'. If a person whispers at one focus, a person standing at the other focus will hear the first person as if they were standing right next to them. We explore the Whispering Galleries in our last example. Example $$\PageIndex{3}$$: Jamie and Jason want to exchange secrets (terrible secrets) from across a crowded whispering gallery. Recall that a whispering gallery is a room which, in cross section, is half of an ellipse. If the room is 40 feet high at the center and 100 feet wide at the floor, how far from the outer wall should each of them stand so that they will be positioned at the foci of the ellipse? Solution Graphing the data yields It's most convenient to imagine this ellipse centered at $$(0,0)$$. Since the ellipse is $$100$$ units wide and $$40$$ units tall, we get $$a=50$$ and $$b=40$$. Hence, our ellipse has the equation $$\frac{x^2}{50^2}+\frac{y^2}{40^2} = 1$$. We're looking for the foci, and we get $$c = \sqrt{50^2-40^2} = \sqrt{900} = 30$$, so that the foci are $$30$$ units from the center. That means they are $$50-30=20$$ units from the vertices. Hence, Jason and Jamie should stand $$20$$ feet from opposite ends of the gallery. \qed ### Contributors • Carl Stitz, Ph.D. (Lakeland Community College) and Jeff Zeager, Ph.D. (Lorain County Community College)<\|endoftext\|>	4.78125
1,294	# chap1 - Chapter 1 Introductory material Last revised 9... This preview shows pages 1–3. Sign up to view the full content. Chapter 1 Introductory material Last revised 9 October 2008 This chapter gives a quick review of some of those parts of the prerequisite courses (Calculus I and II and Geometry I) which we will actually use, adding some extra material. Those parts which are revision will be without examples. 1.1 Trigonometric functions 1.1.1 Values (See Thomas 1.6) We can quickly obtain the value of a trigonometric function for any argument in terms of values for x [ 0 , 1 2 π ] by remembering a few things. First we have the table 0 30 = π 6 radians 45 = π 4 rad. 60 = π 3 rad. 90 = π 2 rad. cos 1 3 2 1 2 1 2 0 sin 0 1 2 1 2 3 2 1 To get the sign for other values we can use the mnemonic table Radians Degrees sin cos tan Positive functions ( 0 , 1 2 π ) ( 0 , 90 ) + + + All ( 1 2 π , π ) ( 90 , 180 ) + Sin ( π , 3 2 π ) ( 180 , 270 ) + Tan ( 3 2 π , 2 π ) ( 270 , 360 ) + Cos sometimes called the ‘Add Sugar To Coffee’ rule – or use Thomas’ variant “All Students Take Calculus”. (Note: to be entirely accurate we should have special rows in this table for the values 1 2 π etc because at those points one or more of the functions will be zero or unbounded.) 1 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document Then we remember what happens when we replace x by x , x + π / 2 or x + π : cos ( x ) = cos x , sin ( x ) = sin x , cos ( x + π 2 ) = sin x , sin ( x + π 2 ) = cos x , (1.1) cos ( x + π ) = cos x , sin ( x + π )= sin x . These are very easy to derive from e ix = cos x + i sin x , remembering that e i π / 2 = i , e i π = 1. Using them in combination we can get cos ( π x ) = cos x , sin ( π x ) = sin x . and so on. More generally cos ( x + ( 2 m + 1 ) π 2 ) = ( 1 ) ( m + 1 ) sin x , sin ( x + ( 2 m + 1 ) π 2 ) = ( 1 ) m cos x , (1.2) cos ( x + n π ) = ( 1 ) n cos x , sin ( x + n π ) = ( 1 ) n sin x . (1.3) where m and n are integers. These identities enable us to relate the value we want to a value in the first quadrant (i.e. the range [ 0 , 1 2 π ] ). Note in particular cos ( n π ) = ( 1 ) n , sin (( 2 m + 1 ) π / 2 )= ( 1 ) m . (1.4) which will turn up later on. 1.1.2 Identities for the trigonometric functions The most important formulae to remember are sin 2 A + cos 2 A = 1 (1.5) cos ( A + B ) = cos A cos B sin A sin B (1.6) sin ( A + B ) = sin A cos B + cos A sin B . (1.7) If you have trouble remembering which of the last two is which, and which has the minus in it, try substituting some special values such as A = 0 or B = 1 2 π and checking the result. For example, taking A = 0 in the last equation gives sin B = sin B whereas if you had tried sin ( A + B ) = sin A cos B cos A sin B you would get sin B = sin B . From these and the earlier results we get cos ( A B ) = cos A cos B + sin A sin B sin ( A B ) = sin A cos B cos A sin B . and we can use these to get cos A cos B = 1 2 ( cos ( A + B )+ cos ( A B )) (1.8) sin A sin B = 1 2 ( cos ( A B ) cos ( A + B )) (1.9) sin A cos B = 1 2 ( sin ( A + B )+ sin ( A B )) , (1.10) which we will find very useful in doing integrations like integraltext cos ( nx ) cos ( mx ) d x . This is the end of the preview. Sign up to access the rest of the document. {[ snackBarMessage ]} ### What students are saying • As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students. Kiran Temple University Fox School of Business ‘17, Course Hero Intern • I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero. Dana University of Pennsylvania ‘17, Course Hero Intern • The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time. Jill Tulane University ‘16, Course Hero Intern<\|endoftext\|>	4.6875
524	# 2.3: Graphing Polynomial Functions Day 2 ```2.3 Graphing Polynomial Functions Objective: -graph polynomial functions -describe the end behavior Recall from yesterday What cubic functions look like… Must have 3 zeros! (real or imaginary) Will cross x-axis 3 times (usually) What quartic functions look like… Must have 4 zeros! (real or imaginary) Will cross x-axis 4 times (usually) Sketching the Graph of a Polynomial What I expect: • The zeros of the polynomial • Appropriate end behavior Sketch the Graph of a Factored Polynomial • Sketch the graph of 𝑓 𝑥 = (𝑥 + 1)(𝑥 − 1)(𝑥 − 2) – Find and plot zeros – Determine the end behavior – Graph it! Sketch a graph: f(x)= (x-3)(x+1)(2x-1) Sketch a graph: f(x)=(x+3)(x+1)(x-1)(x-2) Sketch a graph: f(x) = -x(x+2)(x-1) Sketch a graph: f(x) = x(x+3)(x-4)(1-x) Sketch a graph: f(x) = (x-1)(x-3)2(x-4) Uh oh! We have 2 of the same root. What happens now? The double root is tangent to the x-axis (the curve “bounces off” the x-axis). Analyze the effect of a repeated factor (multiplicity) • Summary: – Even multiplicity: then the graph is tangent to the x-axis at the zero. – Odd multiplicity: then the graph crosses the xaxis at the zero. Sketch a graph: f(x) = (-x-1)(x-3)2 Sketch a graph: f(x) = x4(x − 2)3(x + 1)2 Now, we aren’t just given the roots. How can we sketch a graph from this? 𝑓 𝑥 = 𝑥 3 − 4𝑥 Practice Sess • Source: Textbook • Page: 66 • Problems: #1-9 (odds), #13, #15 Homework • In your textbook complete #2-16 (evens) on pg. 66 ```<\|endoftext\|>	4.59375
956	What is the flu? Learn about flu prevention, get a flu shot, wash your hands often, and follow travel and public health advisories. The flu, or influenza, is a respiratory illness caused by airborne viruses that spread from person-to-person by droplets from coughing or sneezing. The period between becoming infected with the virus and becoming ill is usually 1 to 4 days. The contagious period is 3 to 5 days from the onset of symptoms. Symptoms of the flu, or influenza, are: - Fever (up to 104 degrees) and sweating/chills - Shortness of breath - Vomiting and nausea (in children) A cold and flu are alike in many ways. A stuffy nose, sore throat and sneezing are usually signs of a cold. "Stomach flu" is not really the flu, as there are no respiratory symptoms. Nausea, vomiting and diarrhea without the fever, cough, aching and respiratory symptoms is actually gastroenteritis, but some people call it "stomach flu." This form is caused by other microorganisms and has no relationship to true influenza. How flu spreads Flu viruses spread in respiratory droplets caused by coughing and sneezing. They usually spread from close person-to-person contact, though sometimes people become infected by touching something with flu viruses on it and then touching their mouth, eyes or nose. The virus can live for as long as two hours on surfaces like doorknobs, desks and tables. Healthy adults, infected with the virus, may be able to infect others beginning one day before symptoms develop and up to five days after becoming sick. That means that you can pass on the flu to someone else before you know you are sick, as well as while you are sick. Avoiding Seasonal Flu There are several things you can do to keep from getting seasonal flu: Get a flu shot When you get vaccinated, it greatly reduces your chances of getting seasonal flu. Since the flu season can last through May, even January is not too late to get a flu shot; however, it takes two weeks after the shot to develop adequate immunity. Students may get a flu shot through the Health Center, usually just after the Bates October break. Faculty and staff will be given an opportunity to receive a free flu shot on campus. Look for an announcement from the Office of Human Relations on scheduling. Wash your hands Hand washing is effective in preventing the flu, cold and other infectious diseases. According to the U.S. Centers for Disease Control and Prevention (CDC), rubbing your hands together with soap and water is one of the most important ways to prevent infection. Disease-causing germs can enter your body when your unwashed hands touch your nose, eyes, mouth, and open wounds. Make hand washing a habit and encourage others in your workplace to do the same. When soap and water are not available, use an antibacterial hand cleaner. Choose alcohol hand rubs with 60 - 95 percent alcohol (usually listed as isopropyl, ethanol or propanol). Glycerol or other skin conditioning agents are helpful additives. Read the directions and use the hand rub appropriately. Never wipe the hand rub off; allow your hands to air dry. When used properly, these sanitizers reduce the transmission of disease-causing germs. Other ways to prevent the flu - Avoid touching your eyes, nose or mouth. - When you come back from public places such as a mall, wash your hands. - Cover your mouth with tissue when sneezing. - Stay away from others if you are sick; don't go to class or work. - Avoid close contact with people who are sick. Get help if you are sick If you develop symptoms of the flu, contact your health care provider. There may be medications to relieve your symptoms. Get plenty of rest, drink lots of liquids and avoid using alcohol and tobacco. The flu can be debilitating, causing the person who is ill to be bedridden for extended periods. Be alert to the well being of your friends, relatives and co-workers. Those with the flu may need assistance in getting medical attention and care. If you are at special risk from complications of flu, you should consult your health care provider immediately upon recognizing flu symptoms. Those at risk include people 65 years or older, people with chronic medical conditions, pregnant women or children. With grateful acknowledgement for resource information from the Campus Safety, Health and Environmental Health Association (CSHEMA) University of North Carolina at Chapel Hill (UNC), Bates College and SUNY Canton.<\|endoftext\|>	3.96875
764	Hundreds of tiny worlds circling beyond Neptune have been thought of as frozen in time, almost unchanged since they formed at the birth of the solar system five billion years ago. A discovery now has some astronomers entertaining visions of icy volcanoes roiling on some of them, at least at one time. David J. Stevenson, a professor of planetary science at the California Institute of Technology, speculates that the decay of radioactive elements within some of these worlds, known as Kuiper Belt Objects, would produce enough warmth to melt a mixture of water ice and ammonia, an antifreeze. The liquid could then rise up through cracks and erupt onto the surface. "This would be in exact analogy with what happens inside the Earth with volcanic processes," he said. Dr. Stevenson described these speculations in today's issue of the journal Nature. The discovery that spurred Dr. Stevenson's imaginings is more modest, at least on the surface. In an article also appearing in today's Nature, David Jewitt of the University of Hawaii and Jane Luu of the Lincoln Laboratory at the Massachusetts Institute of Technology, who together discovered the first Kuiper Belt Object in 1992, report the presence of crystalline water ice on the largest known Kuiper Belt Object, Quaoar (pronounced KWAH-war). They also report signs of ammonia hydrate. Ice cubes and snow on Earth all develop in crystalline form, with the water molecules stacking in a neat pattern.Continue reading the main story But what is common on Earth was a surprise for Quaoar, located about four billion miles from Earth and about one billion miles farther out than Pluto. In ultrafrigid conditions like those found on Quaoar -- the surface temperature dips below minus-360 degrees Fahrenheit -- ice generally takes an amorphous form more like glass, where the water molecules stack haphazardly. Laboratory experiments indicate that crystalline ice forms only at temperatures warmer than minus-260 degrees. Furthermore, bombardment by ultraviolet light and cosmic rays turns crystalline ice into amorphous ice, within about 10 million years. That means the crystalline ice seen by Dr. Jewitt and Dr. Luu must have formed or been exposed relatively recently. The questions then are, how did the crystalline ice form and why is it still there? A mixture of ice and ammonia melts at minus-145 degrees, and Dr. Stevenson thinks temperatures deep within Quaoar still exceed that today. However, because erupted ice would not be likely to return to the core, any volcanism is likely to have occurred during Quaoar's youth, Dr. Stevenson said. "I think it makes most sense if it was in the far past, like three billion years ago," he said. In this theory, the crystalline ice spotted by Dr. Jewitt and Dr. Luu would have formed underground, but was exposed within the last few million years by a collision with a smaller Kuiper Belt Object. Crystalline ice has also been spotted on Pluto's moon, Charon, and on the moons of Saturn and Uranus. "It's everywhere in the outer solar system," said Michael E. Brown, a colleague of Dr. Stevenson's at Caltech. Measurements by Dr. Brown, one of the discoverers of the 780-mile-wide Quaoar two years ago, agree on the presence of water ice, but suggest the presence of methane instead of ammonia hydrate. Dr. Brown proposes a less exotic idea -- the impacts of micrometeorites -- to explain how the surface was heated enough to produce crystalline ice. But, he added, "The story has to be more complicated than just that."Continue reading the main story<\|endoftext\|>	4.0625
335	Free Algebra Tutorials! # Multiplying and Dividing Fractions ## Multiplication Multiply straight across: Multiply the numerators to get the new numerator. Multiply the denominators to get the new denominator. ## Division The first fraction is said to be “divided by” the second fraction. The fraction you are “dividing by” is called the “divisior.” To divide, “invert” the divisor (turn it upside down), then multiply. (When a fraction is inverted we call the resulting fraction the “reciprocal” of the original number. Any whole number can be thought of as a fraction with 1 in the denominator. Since , the reciprocal of 2 is .) ## Chain Multiplication and Division There are times (especially in science classes in high school and beyond) when you need to multiply and divide a whole series of fractions and whole numbers. This is actually easy to do. Think of making a shish kabob: a skewer loaded up with chunks of meat, tomatoes, onions, bell pepper, mushrooms, etc. and cooked on a grill. Think of one long fraction bar as a skewer. To multiply a fraction in the list, simply skewer it on. To divide by a fraction, flip it over and skewer it on. Any whole number can be thought of as a fraction with 1 in the denominator, so can treat it like any other fraction. Multiplying by a whole number puts the whole number into the numerator. Dividing by a whole number puts the whole number into the denominator. (Once everything is in place we can ignore the 1’s.)<\|endoftext\|>	4.53125
937	Research suggests children are more likely to achieve higher academic results if they are able to read, write and perform basic computational mathematics tasks by the time they reach Year 3. Now, more than ever, a flipped view of education is being embraced. More importance is being placed on early primary school education to ensure children are developing the skills that will give them the best possible foundations for a lifetime of learning by the beginning of Year 3. According to New America publication, “Third grade marks a critical turning point in children's education, when they shift from ‘learning to read’ to ‘reading to learn.’” Students can develop and practise effective strategies from Kindergarten to Year 2 that will prepare them for the remainder of their primary and secondary schooling. The evolving landscape of Early Learning in Australia The Australian Bureau of Statistics states that the current landscape of enrolments in Early Learning is on the rise. In 2017, 339,243 children aged four or five were enrolled in a preschool program across Australia. Many schools now include a structured Early Learning curriculum that supports child development, play and creative thinking. Essentially, an effective preschool curriculum includes resilience building, unlocking imagination and creativity, and develops social, cognitive and emotional connectedness through play. At Toorak College, we focus on skills that build grit and resilience while showing more inclusiveness by boosting parent engagement. Emphasis is placed on the whole child and their capabilities rather than what a child cannot do. In short, we identify a child’s strengths in order to further develop weaknesses or skills not yet attained. Why do students need to be proficient by Year 3? According to ACER research conducted in 2006 on Early Learning, “literacy development at the end of Year 1 and Year 2 is strongly influenced by literacy skills developed in the early years.” A child’s academic development begins before school, when they are introduced to a new world of learning. A good start to their education will help instil a better outcome for them later in life with the grasp of strong language and maths skills. If these skills are not adequate by the end of Year 2, the likelihood of this is that the student will feel overwhelmed, stressed and may experience a drop in self-esteem. For example, if a class is writing a narrative (story), a student with minimum or below-average literacy and numeracy skills may struggle to maintain focus, generate ideas and get their thoughts down on paper because they are unable to contextualise and apply their learning. The greater the grasp of core skills learned before Year 3, the more confident the child will feel to strive and accept challenges in their learning. Are we hitting education benchmarks? Australian education organisations are developing more information and data about the standards of students in literacy and numeracy. According to NAPLAN 2017, an average of 95 percent of students who participated in the assessment are at or above minimum standards of proficiency in Victoria in Year 3. This national assessment begins at Year 3 as that is the year when students are expected to be equipped with adequate reading and writing skills to begin developing deeper knowledge. On the other hand, NAPLAN 2018 scores suggest writing has deteriorated to its lowest point in a decade, with a record low of 94 percent of students hitting minimum or above standards. Toorak College is proudly above the average threshold, with three key areas receiving the highest results ever for the School: Year 9 Numeracy, Year 5 Reading and Year 3 Spelling. How can we consolidate basic proficiency skills by Year 3? Ensuring basic proficiency skills by Year 3 begins when one chooses the right kind of education for their child. When searching for a quality Early Learning Centre and Junior School, it is important to consider how the school is going to align the curriculum in terms of literacy, numeracy and life skills such as resilience and creative thinking. Our Early Learning and Prep to Year 2 curriculum aims to be nurturing, structured and rich in experiences that enable students to discover themselves as learners. When deciding if a school is right for your child, I recommend that parents understand the school’s curriculum and pedagogy to determine how much attention is focused on reading, creative arts, science and language. Secondly, focus on how much time is dedicated to embracing skills such as critical and creative thinking, resilience and problem solving. With all this at the forefront of your mind, you will be able to make an informed decision of choosing the right place for your child. This will result in your child to enjoy their education and feel confident and competent to achieve excellence in Year 3 and beyond.<\|endoftext\|>	3.84375
369	In 1958, the National Aeronautics and Space Administration was established. Today, 50 years on, NASA has a host of programmers to showcase its achievements. One of the objectives of NASA has been to interact with the community through extensive outreach programmers impacting the thoughts and imagination of generations of children the world over. Encouraging kids to dream big and study hard, through competitions, designed to get children thinking about space — and by extension, math and science — in new ways is what NASA aims at. Each year over 60,000 students get to directly interact with NASA’s educational opportunities. NASA has numerous activities, games, amazing stories, study material, competitions, student conferences and scholarships for different age groups. From providing an opportunity of asking an astronaut on the space station a question, assisting students with homework and projects, assembling paper model spacecraft to interactive websites with every information on space and space related subjects, NASA is a must visit site for the curious child. Importantly October 4 to 10 is celebrated annually in some 50 nations as World Space Week to mark the anniversary of two great milestones of humanity’s exploration of space. The first human-made Earth satellite, SPUTNIK I being launched on October 4, 1957, and the Treaty on Principles Governing the Activities of States in the Exploration and Peaceful Uses of Outer Space, including the Moon and Other Celestial Bodies, coming into force on October 10, 1967. Special World Space Week instructional materials in several languages with activities for all grades are made available to schools at no charge. You can begin your own space odyssey and lucky for you that you need not even venture into space for it — just get to a computer and connect to NASA. This is one place where the sky is certainly not the limit – only a beginning to the limitless!<\|endoftext\|>	3.6875
1,098	# Search by Topic #### Resources tagged with Angle properties of shapes similar to Lighting up Time: Filter by: Content type: Stage: Challenge level: ### There are 27 results Broad Topics > 2D Geometry, Shape and Space > Angle properties of shapes ### Arclets Explained ##### Stage: 3 and 4 This article gives an wonderful insight into students working on the Arclets problem that first appeared in the Sept 2002 edition of the NRICH website. ### Angle A ##### Stage: 3 Challenge Level: The three corners of a triangle are sitting on a circle. The angles are called Angle A, Angle B and Angle C. The dot in the middle of the circle shows the centre. The counter is measuring the size. . . . ### Floored ##### Stage: 3 Challenge Level: A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded? ### Star Polygons ##### Stage: 3 Challenge Level: Draw some stars and measure the angles at their points. Can you find and prove a result about their sum? ### Pie Cuts ##### Stage: 3 Challenge Level: Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters). ### Bisecting Angles in a Triangle ##### Stage: 3 and 4 Challenge Level: Measure the two angles. What do you notice? ### Polygon Pictures ##### Stage: 3 Challenge Level: Can you work out how these polygon pictures were drawn, and use that to figure out their angles? ### Fred the Class Robot ##### Stage: 2 Challenge Level: Billy's class had a robot called Fred who could draw with chalk held underneath him. What shapes did the pupils make Fred draw? ### Convex Polygons ##### Stage: 3 Challenge Level: Show that among the interior angles of a convex polygon there cannot be more than three acute angles. ### Getting an Angle ##### Stage: 3 Challenge Level: How can you make an angle of 60 degrees by folding a sheet of paper twice? ### Can You Explain Why? ##### Stage: 3 Challenge Level: Can you explain why it is impossible to construct this triangle? ### Tessellating Hexagons ##### Stage: 3 Challenge Level: Which hexagons tessellate? ### Angles in Three Squares ##### Stage: 3 and 4 Challenge Level: Drawing the right diagram can help you to prove a result about the angles in a line of squares. ### Cartesian Isometric ##### Stage: 2 Challenge Level: The graph below is an oblique coordinate system based on 60 degree angles. It was drawn on isometric paper. What kinds of triangles do these points form? ##### Stage: 3 and 4 Challenge Level: Draw some quadrilaterals on a 9-point circle and work out the angles. Is there a theorem? ### Always, Sometimes or Never? Shape ##### Stage: 2 Challenge Level: Are these statements always true, sometimes true or never true? ### Logo Challenge 3 - Star Square ##### Stage: 2, 3 and 4 Challenge Level: Creating designs with squares - using the REPEAT command in LOGO. This requires some careful thought on angles ### Which Solids Can We Make? ##### Stage: 3 Challenge Level: Interior angles can help us to work out which polygons will tessellate. Can we use similar ideas to predict which polygons combine to create semi-regular solids? ### First Forward Into Logo 9: Stars ##### Stage: 3, 4 and 5 Challenge Level: Turn through bigger angles and draw stars with Logo. ### First Forward Into Logo 7: Angles of Polygons ##### Stage: 3, 4 and 5 Challenge Level: More Logo for beginners. Learn to calculate exterior angles and draw regular polygons using procedures and variables. ### Subtended Angles ##### Stage: 3 Challenge Level: What is the relationship between the angle at the centre and the angles at the circumference, for angles which stand on the same arc? Can you prove it? ### Semi-regular Tessellations ##### Stage: 3 Challenge Level: Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations? ### Transformations on a Pegboard ##### Stage: 2 Challenge Level: How would you move the bands on the pegboard to alter these shapes? ### Triangles in Circles ##### Stage: 3 Challenge Level: Can you find triangles on a 9-point circle? Can you work out their angles? ### Triangle Pin-down ##### Stage: 2 Challenge Level: Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs. ### Triangles All Around ##### Stage: 2 Challenge Level: Can you find all the different triangles on these peg boards, and find their angles? ### LOGO Challenge 4 - Squares to Procedures ##### Stage: 3 and 4 Challenge Level: This LOGO Challenge emphasises the idea of breaking down a problem into smaller manageable parts. Working on squares and angles.<\|endoftext\|>	4.4375
1,529	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. ### Course: Arithmetic>Unit 15 Lesson 1: Fractions as division # Creating mixed numbers with fraction division Learn how to interpret fractions as division problems. Watch examples of how to break down fractions into whole numbers and remainders. Then see Sal practice turning these into mixed numbers. Created by Sal Khan. ## Want to join the conversation? • Wow. Just because he barley messes up means you can pounce on him when he does? He even states that he did. • That is fact like he makes a mistake at and you just jump him common, folk. • I'm confused about the remainder I thought 23 divided by 6 is 3 r5 or is it the way you are doing right? • Yes, it is the right answer. • thank yo for the vids!! • When Sal creates extra instances of the denominator, for ex. at and , it confuses me on an intuitive level because it looks like extra numbers are being created. By what rule is a denominator allowed to be replicated in that way? Thanks. • Well, the values are the same. 18/6=3, and 6+6+6/6 = 6/6 + 6/6 + 6/6 is an easier way of writing it, because 6+6+6/6, 18/6, and 6/6 + 6/6 + 6/6 all equal 3. • A Mixed number has a whole number part and a fraction part. Rather than simply having a fraction with a bigger numerator than a denominator (an 'improper fraction'), we sometimes write out the full number and the left part of a fraction. For example, we say "1 and 3/4 inches", rather than "7/4" of an inch. • Why did Sal say 23/6 = 5r3 It's 3r5 that's what his work showed. Don't show us the wrong information Sal!! • what's 2 divided by 9 ? • 4.5 in decimal form • How is 23 divided by 6 five remainder 3? Shouldn’t it be 3 remainder 5? • Yea, but he made a mistake and if you saw the little pop up message on the side it said “Sal means to write 3 r 5 but he wrote 5 r 3 instead.” • I understood completely. Thanks for the videos • how do you draw so neatly on the computer? • Prob a drawing tablet connected to the computer ## Video transcript We've already seen that a fraction like 2/9 can be interpreted as 2 divided by 9. Let me do the 2 in the same color. That if we have a fraction, it could be interpreted as the numerator divided by the denominator. And this leads to all sorts of interesting conclusions, some of which we've already seen, and some of which are a little bit new. So, for example, if I had the fraction 7/7, this can be interpreted now as 7 divided by 7, as our numerator divided by our denominator. And 7 divided by 7 is of course equal to 1. And this is consistent with what we've already seen. 7/7 would get us to a whole, and a whole is the exact same thing as one. But we could do things a little bit more interesting as well. We could take something like 18/6 and realize, wow. This is the same thing as 18 divided by 6, which we know is equal to 3. And we should do a little reality check. Does this make sense, that 18/6 should be equal to 3? Well, we could rewrite it. We could rewrite 18/6. Let me make the sixths that same orange color. That's going to be the same thing. 18 is 6 plus 6 plus 6. And then all of that over 6, and then that's the same thing. That's the same thing as 6/6 plus 6/6 plus 6/6. And I could make this right over here in orange. And we've already seen, or we've seen many, many videos ago, that 6/6, just like 7/7, these are each equal to a whole. These are each equal to 1. And we can now view this as 6 divided by 6, which is the same thing as 1. So this is 1 plus 1 plus 1, which is, of course, equal to 3. But this starts to raise an interesting question. This all worked out just fine because 18 is a multiple of 6. 6 divides evenly into 18. But what happens if we start having fractions where the denominator does not divide evenly into the numerator? Let's say we have a fraction like 23 over 6. Well, we know that we can interpret this as 23 divided by 6. And if we actually divide 23 by 6-- let's do that. So we divide 6 into 23. We know 6 goes into 23 three times. 3 times 6 is 18. And then when you subtract, you end up with a remainder of 5. So we might say, hey, 23 divided by 6 is equal to 5 remainder 3. But that's not that satisfying. What do I do with this remainder? This really isn't a number here. This is just saying that we're going five times, and then we have a little bit left over. What we can do now is manipulate this a little bit so that we can realize that this is a number, and in particular, a mixed number. So for example, we could start with a 23 over 6, and we could divide it into-- or we could decompose the numerator into one part that is divisible by 6, evenly divisible by 6, and the remainder. So for example, 23 over 6 we can rewrite as 18 plus 5 over 6. Notice I decomposed the 23 into one part that is a multiple of 6, and it's the largest multiple that essentially fits into 23, or that is less than or equal to 23, and then the remainder. When you divide 6 into 23, you get a remainder of 5. You could view it as, I divided it into the remainder and everything else. And the reason why this is interesting is because we know that this is going to be equal to 18 over 6 plus 5 over 6. Well, we already know that 18/6 is the same thing as 18 divided by 6, or 3. So this is the same thing as 3. So we know that 23 over 6, which is the same thing as 18 plus 5 over 6, is the same thing as 3 plus 5/6. Or, if we want to write it as a mixed number, we could write it as 3 and 5/6.<\|endoftext\|>	4.46875
476	Or download our app "Guided Lessons by Education.com" on your device's app store. Finding the Subject and Predicate Students will be able to understand the subject and predicate of a sentence as well as identify sentence fragments. - Give the definition for Subject, the person or thing being discussed in a sentence, and for Predicate, the part of the sentence containing a verb and discussing the subject. - Write an example sentence on the board and underline the subject once and the predicate twice. - Ask students to write a short sentence in their notebooks and underline the different parts as you did. - Ask students to share their sentences as you write them on the whiteboard. Explicit Instruction/Teacher modeling(20 minutes) - Hand out the Two Parts of a Sentence worksheet. - Complete numbers 2 and 3 together. - Allow students to finish independently. - Use the document camera to go over the answers. Make sure students correct their answers if they answered incorrectly. Guided practise(10 minutes) - Hand out the Subject and Predicate worksheet. - Help struggling students as the class completes the worksheet. - Review answers on the document camera. Independent working time(20 minutes) - Have students write 10-15 sentences of their own. Give an expected length for the sentences depending on the fluency of your class. - Have students trade papers with a partner. - Have students underline the subject once and the predicate twice on their partners' sentences. - Hand out the Complete Sentences and Sentence or Fragment? #1 worksheets. - Have students work on these two sheets independently. - Enrichment:Students who complete their work early may be allowed to complete the Compound Predicate worksheet for a challenge. - Support:Work with struggling students one-on-one to identify the subjects and predicates of sentences. - Go over the answers for the worksheets on the document camera. - Collect student worksheets and check for correctness. Review and closing(5 minutes) - Have students give you sentences of which you will find the subject and predicate on the board. - Allow students to ask any final questions or voice any concerns they may have about the lesson.<\|endoftext\|>	4.21875
2,323	1. Objectives • To introduce the concept of a difference equation • To introduce the concept of an ordinary differential equation 2. Motivation • Most science models have a mathematical representation that has the form of an ordinary differential equation. 3. Pre-questions • You won $100 and want to invest it. Which bank offers the better deal? • Bank A: 12% interest per year compounded yearly • Bank B: 1% interest per month compounded monthly • Bank C: (1/365.2425)% interest compounded daily "Compounded yearly/monthly/daily" means that at the end of one year/month/day the interest is added to your balance. So, after one • year you have$112 dollars • month you have $101 dollars • day you have 100+(1/365.2425) dollars 4. Slides 4.1. Introduction In computational science, we most often want to simulate systems that continuously change. Previously we modeled population change with a discrete equation. In reality, population continuously changes. The mathematical model of this is P(next instant)-P(this instant) = -(time between this instant and the next)aP(this instant) where the time between iterations (time between this instant and the next) is nearly zero. In a calculus course, you will learn that this is related to limits and derivatives. To see where the "above time between iterations" factor comes in, consider the following observation, which does not require an understanding of calculus. 4.2. Observation Population grows at a rate of 20% per year P(next year) = P(this year) + 0.2P(this year) Won't you get the same answer if you divide the growth rate by 12 and change the equation to go month-to-month? P(next month) = P(this month) + (0.2/12)P(this month) this can be rewritten as P(next month) - P(this month) = (time between iterations)(0.2)P(this month) where (time between iterations) = (1/12)th of a year 4.3. Observation cont. Won't you get the same answer if you divide the growth rate by 12 and change the equation to go month-to-month? No. But close. P(1) = 1; for i = [1] P(i+1) = P(i) + 0.2P(i); end P(2) = 1.2000 P(1) = 1; for i = [1:12] P(i+1) = P(i) + (0.2/12)P(i); end P(13) = 1.2194 4.4. Observation cont. Explore the rule 1. Multiply the number of iterations by some factor 2. Divide the growth rate by the same factor P(1) = 1; for i = 1:10 P(i+1) = P(i) + (1000.001)P(i); end P(11) = 2.5937 for i = 1:100 P(i+1) = P(i) + (100.001)P(i); end P(101) = 2.7048 for i = 1:1000 P(i+1) = P(i) + (10.001)P(i); end P(1001) = 2.7169 (Note that 2.7169 is fairly close to e1 = 2.718281828459046.... If you continue the above process, you will get closer and closer to e1.) 4.5. Observation cont. Suppose that you were asked to compute for i = 1:1000 P(i+1) = P(i) + (0.001)P(i); end P(1001) = 2.7169 it appears that you could do 10 times fewer iterations and get about the same answer by 1. Multiplying the number of iterations by some factor 2. Dividing the growth rate by the same factor for i = 1:100 P(i+1) = P(i) + (100.001)P(i); end P(101) = 2.7048 4.6. Summary In computational science, we most often want to simulate systems that have the computational model of for i = 1:HUGE P(i+1) = P(i) + (SMALL)P(i); end • The main problem is that if HUGE is large, many iterations are required, which means that the calculation will take a long time. • In the field of computational mathematics, they study how modify the right-hand-side of the equation for P(i+1) along with the values of HUGE and SMALL in order to minimize the amount of computation and maximize the accuracy of the calculation. 4.7. Definition • This type of program arises from the translation of a mathematical model that has the form of an ordinary differential equations (ODE) to a computational model. ODEs are used to mathematically model many natural systems. for i = 1:HUGE P(i+1) = P(i) + (SMALL)aP(i); end • For example, the science model for a system that experiences continuous growth at a rate that is proportional to the population is an ordinary differential equation which has the form $\frac{\Delta P}{\Delta t} = aP$ or $\frac{}{}{\Delta P} = {\Delta t}aP$ 4.8. Definition cont. $\frac{}{}\Delta P = \Delta taP$ can be written as P(next instant)-P(this instant) = (time between this instant and the next)aP(this instant) because by definition $\frac{}{}\Delta P = \mbox{P(next instant)-P(this instant)}$ and $\frac{}{}\Delta t = \mbox{time between this instant and the next}$ The differential in differential equation comes from the fact that it is a model of a difference in population from one instant to the next for a difference in time that is very small. 4.9. Definition cont. Another situation where an ODE appears is in the mathematical model of how an object cools. The change in temperature, T, of an object as it cools has the mathematical model of $\frac{\Delta T}{\Delta t} = -q(T-T_a)$ where Ta is the temperature of the room where the object is stored (ambient temperature). This can also be written as T(next instant)-T(this instant) = -(time between this instant and the next)qT(this instant) where the time between the next instant and this instant, Δt, is nearly zero. 5. Questions 5.1. Bank Eqn. Consider the equations for a bank account with 1% interest per year compounded yearly and 1/12 percent interest per month compounded monthly. The initial balance is$100, and no additional deposits are made. B(next year) = B(this year) + 0.01B(this year) B(next month) = B(this month) + (1/12)0.01B(this month) For this problem, type format long as the first command in your program or when you first start MATLAB. This tells MATLAB to show more digits when displaying numbers. 1. Write a program that uses the first equation to compute the balance after 10 years. Post the program (or a screenshot) to your wiki page. 2. Write a program that uses the second equation to compute the balance after 10 years. Post the program (or a screenshot) to your wiki page. 3. Compute the balance after 10 years using 1/365 percent interest compounded daily (assume 1 year = exactly 365 days). Enter the value on your wiki page. 4. Compute the balance after 10 years using 1/(36524) percent interest compounded hourly (assume 1 year = exactly 365 days). Enter the value on your wiki page. 5. Compute the balance after 10 years using 1/(3652460) percent interest compounded every minute (assume 1 year = exactly 365 days). Enter the value on your wiki page. 6. Compute the balance after 10 years using 1/(365246060) percent interest compounded every second (assume 1 year = exactly 365 days). Enter the value on your wiki page. 7. (If you did not already) write a program that computes all of the above balances using only two for loops. 6. Other Questions 6.1. Heat Eqn. The equation that describes how the temperature of an object, T changes with time (Newton's Law of Cooling) is $\frac{\Delta T}{\Delta t} = -q(T-T_a)$ 1. Use a spreadsheet to determine how the temperature of an object that starts out at 100 degrees C cools when the ambient temperature, Ta is 0. Assume that q = 0.1 and Δt = 0.1. At approximately what time does the temperature equal one-half of its initial temperature? If Δt = 0.05, at approximately what time does the temperature equal one-half of its initial temperature? 2. Does the time it takes for the temperature to fall by one-half depend on the initial temperature of the object? Describe in words you how you determined this. 6.2. Population Eqn. Consider the following model of population Every year, population increases a value of 10% of the population in the previous year. However, if the predicted population is over 100, a disease outbreak instantly kills 80% of this predicted population value. For example, if the predicted population is 110, then the next year the population is 0.2110. • Use Excel to plot population as a function of time for 40 years. Assume that the initial population is 20. • Try to make a change to the default setting for the plot that makes it clearer or easier to read (in your opinion; there is no "wrong" answer). 6.3. Code Example Consider the following science model: "The starting balance in a bank account is $1,000 and it grows by 5% per year." The following Matlab/Octave code was written to simulate the month-to-month growth of this bank account for 36 months, and, to print out the balance at the end of months 26, 27 and 28: B(1) = 1000; for i = [2:35] B(i) = B(i-1) + 0.05*B(i-1); end B(26) B(27) B(28) When this code is run in Matlab or Octave, we find that B(26) =$3,386.40, B(27) = $3,555.70 and B(28) =$3,733.50, all of which are incorrect. Why?<\|endoftext\|>	4.59375
903	# mathematics The mathematics essay below has been submitted to us by a student in order to help you with your studies. # The central limit theorem ### The Central Limit Theorem The central limit theorem is the second fundamental theorem in probability after the ‘law of large numbers.' The‘law of large numbers'is atheoremthat describes the result of performing the same experiment a large number of times. According to the law, theaverageof the results obtained after a large number of trials should be close to theexpected value, and will tend to become closer to this value as more trials are carried out. For example, a single roll of afair diceproduces one of the numbers {1, 2, 3, 4, 5, 6} each with equalprobability. Therefore, the expected value (E(x)), of a single dice roll is (1+2+3+4+5+6) ÷ 6 = 3.5. If this dice is rolled a large number of times, the law of large numbers states average of the result of all these trials known as the sample mean , will be approximately equal to 3.5. = 1Nk=1Nxk≈Ex=3.5 If the number of trials was to further increase, the average would further approach the expected value. So in general, as N→∞, →Ex This is the main premise of the law of large numbers. The central limit theorem is similar to the law of large numbers in that it involves the behaviour of a distribution as N→∞. The central limit theorem states that given a distribution with a mean (μ) and variance (σ²), the samplingdistribution of the mean approaches anormal distributionwith a mean (μ) and a variance (σ²N) as N, thesample size,increases. In other words, the central limit theorem predicts that regardless of the distribution of the parent population: 1. Themeanof the population of means isalwaysequal to the mean of the parent population from which the population samples were drawn. 2. Thestandard deviationof the population of means is always equal to the standard deviation of the parent population divided by the square root of the sample size (N). 3. Thedistribution of means will increasingly approximate anormal distributionas the size N of samples increases. →X~N(μ, σ2N) (This is the main consequence of the theorem.) The origin of this celebrated theorem is said to have come from Abraham de Moivre, a French born mathematician who used the normal distribution to approximate the distribution of the number of heads resulting from many tosses of a fair coin. This was documented in his book ‘The Doctrine of Chances' published in 1733 which was essentially a handbook for gamblers. This finding was somewhat forgotten until the famous French mathematicianPierre-Simon Laplacerevived it in his monumental work‘Théorie Analytique des Probabilités', which was published in 1812. Laplace was able to expand on de Moivre's findings by approximating the binomial distribution with the normal distribution. De Moivre Laplace But as with de Moivre, Laplace's finding received little attention in his own time. It was not until the nineteenth century was at an end that the importance of the central limit theorem was discerned, when, in 1901, Russian mathematicianAleksandr Lyapunovdefined it in general terms and proved precisely how it worked mathematically.A full proof of the central limit theorem will be given later in this document. One may be familiar with the normal distribution and the famous ‘bell shaped' curve that is associated with it. This curve is often found when presenting data for something like the heights or weights of people in a large population. Where μ is the mean . When the central limit theorem is applied, the distribution will approach something similar to the graph above. However, the amazing implication The central limit theorem explains why many non-normal distributions tend towards the normal distribution as the sample size N increases. This includes uniform, triangular, inverse and even parabolic distributions. The following illustrations show how they tends towards a normal distribution: ### Request Removal If you are the original writer of this essay and no longer wish to have the essay published on the UK Essays website then please click on the link below to request removal:<\|endoftext\|>	4.46875
852	How to help at home How to help at home Talking about the sounds your child has learnt by looking through thier Phonics Book is a great way to help your child progress and apply their phonic knowledge. These sounds can be revisited and they could have a go at writing the letter again, they may well be very proud of how much their letter formation has improved! They could even attempt to write some words, captions or sentences at home. These could be brought in to school to share with the class as their ‘Magic Moment’. A link to the Jolly Phonics songs we sing at school can be found below: The most important thing you can do to help your child at home is to read, read and read! Reading every single day is so important and a great routine to get in to. It can be a lovely time to spend with your child, not only with their reading book but with story books from school or home. Please write in your child’s reading diary each day you read and if you can, include a brief comment. Help your child to be as independent as possible – holding the book, turning the pages, pointing to the text and joining in with familiar phrases. When sharing a story with your child you can model key skills which will support them in their learning. For example, finger pointing to the words, scanning from left to right, spotting tricky words and familiar sounds, picking up information from illustrations and making predictions. Broadly speaking children apply their phonic knowledge in similar ways: - hearing and writing initial letter sounds in words - hearing and writing initial and final sounds - hearing sounds in the order they occur in words - building a memory of words off by heart – often ‘tricky’ words You can support your child in their writing by encouraging them to listen to the sounds in words and write them down, they will hear more sounds in order as they become more confident. You can help at home by: - providing opportunities for independent mark making - valuing and encourage all mark making - avoiding scribing for your child (dotting / writing over the top / copy writing) - practising the phonemes and grapheme correspondence - encouraging the use of phonemes and accept their own spellings e.g. hows (house) - encouraging the correct pencil grip - 'froggy legs' You can support your child with their number skills at home by: - Singing songs that take away or add things e.g. 10 green bottles, 1 man went to mow, 5 current buns - Exploit all counting opportunities – count stairs, count buttons, count lampposts on a walk, count ‘red’ cars on a journey etc. - Commercial games such as snakes and ladders - these help with the counting on strategy. - Throwing beanbags/balls at numbered targets and adding up scores – who scored the most? The least? - Practise counting in 2s, 5s and 10s. - Look for numbers whilst walking or on a journey - Ask questions like ‘if I took one away how many would I have left? ‘ or if I add one how many have I got now? - Use magnetic numbers on the fridge or foam numbers for the bath. Put them in order. Miss one out of a sequence – do they know which one is missing? Shape, Space and Measure You can support your child in developing their shape, space and measure understanding at home by: - Looking for and naming shapes at home and in the environment - Talking about 3D (solid) shape names - packaging for food items is an excellent way. - Junk modelling with 2D and 3D shapes – can you name them all? - Making pictures with different shapes. - Involve your child in cooking. Look at numbers on scales and measuring jugs. - Measure and compare feet sizes and height of other family members. - Shopping activities – real or pretend – use real money to help identify coins and weight.<\|endoftext\|>	3.96875
338	Intermediate Algebra (12th Edition) $y \text{ is NOT a function of }x \\\text{Domain: } [0,\infty)$ $\bf{\text{Solution Outline:}}$ To determine if the given equation, $x=y^4 ,$ is a function, solve first for $y$ in terms of $x.$ Then check if $x$ is unique for every value of $y.$ To find the domain, find the set of all possible values of $x.$ $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} x=y^4 \\\\ y^4=x \\\\ y=\pm\sqrt[4]{x} .\end{array} If $x=16,$ then $y=2$ or $y=-2.$ That is, the ordered pairs, $\{ (16,2),(16,-2) \}$ satisfy the given equation. This means that the value of $x$ is not unique. Hence, $y$ is NOT a function of $x.$ Since $x$ appears in the radicand of a radical with an even index, then \begin{array}{l}\require{cancel} x\ge0 .\end{array} Hence, the domain is the set of all nonnegative numbers. Hence, the given equation has the following characteristics: \begin{array}{l}\require{cancel} y \text{ is NOT a function of }x \\\text{Domain: } [0,\infty) .\end{array}<\|endoftext\|>	4.40625
1,451	# 2.5 Quadratic equations (Page 6/14) Page 6 / 14 ## Finding the length of the missing side of a right triangle Find the length of the missing side of the right triangle in [link] . As we have measurements for side b and the hypotenuse, the missing side is a. $\begin{array}{ccc}\hfill {a}^{2}+{b}^{2}& =& {c}^{2}\hfill \\ {a}^{2}+{\left(4\right)}^{2}\hfill & =& {\left(12\right)}^{2}\hfill \\ \hfill {a}^{2}+16& =& 144\hfill \\ \hfill {a}^{2}& =& 128\hfill \\ \hfill a& =& \sqrt{128}\hfill \\ & =& 8\sqrt{2}\hfill \end{array}$ Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse. $5\text{\hspace{0.17em}}$ units ## Key equations quadratic formula $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$ ## Key concepts • Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-factor property is then used to find solutions. See [link] , [link] , and [link] . • Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method. See [link] and [link] . • Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution. See [link] and [link] . • Completing the square is a method of solving quadratic equations when the equation cannot be factored. See [link] . • A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. See [link] . • The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each. See [link] . • The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. See [link] . ## Verbal How do we recognize when an equation is quadratic? It is a second-degree equation (the highest variable exponent is 2). When we solve a quadratic equation, how many solutions should we always start out seeking? Explain why when solving a quadratic equation in the form $\text{\hspace{0.17em}}a{x}^{2}+bx+c=0\text{\hspace{0.17em}}$ we may graph the equation $\text{\hspace{0.17em}}y=a{x}^{2}+bx+c\text{\hspace{0.17em}}$ and have no zeroes ( x -intercepts). When we solve a quadratic equation by factoring, why do we move all terms to one side, having zero on the other side? We want to take advantage of the zero property of multiplication in the fact that if $\text{\hspace{0.17em}}a\cdot b=0\text{\hspace{0.17em}}$ then it must follow that each factor separately offers a solution to the product being zero: In the quadratic formula, what is the name of the expression under the radical sign $\text{\hspace{0.17em}}{b}^{2}-4ac,$ and how does it determine the number of and nature of our solutions? Describe two scenarios where using the square root property to solve a quadratic equation would be the most efficient method. One, when no linear term is present (no x term), such as $\text{\hspace{0.17em}}{x}^{2}=16.\text{\hspace{0.17em}}$ Two, when the equation is already in the form $\text{\hspace{0.17em}}{\left(ax+b\right)}^{2}=d.$ A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes. The sequence is {1,-1,1-1.....} has how can we solve this problem Sin(A+B) = sinBcosA+cosBsinA Prove it Eseka Eseka hi Joel June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler? 7.5 and 37.5 Nando find the sum of 28th term of the AP 3+10+17+--------- I think you should say "28 terms" instead of "28th term" Vedant the 28th term is 175 Nando 192 Kenneth if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n write down the polynomial function with root 1/3,2,-3 with solution if A and B are subspaces of V prove that (A+B)/B=A/(A-B) write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°) Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4 what is the answer to dividing negative index In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c. give me the waec 2019 questions<\|endoftext\|>	4.875
2,455	Nevada is nicknamed as the Silver State; Battle Born State; Sagebrush State and is located in the Western, Southwestern and Mountain West regions of the United States. Nevada became the 36th state of the United States on 1864 . Carson City is the capital city and Las Vegas is the largest city of the state. Nevada Fast Facts: - Capital city: Carson City - Largest city: Las Vegas - State Animal: Desert bighorn sheep - State Bird: Mountain bluebird - State Tree: Single-leaf pinon - State Flower: Sagebrush - State Gem: Virgin Valley black fire opal History of Nevada Prior to European exploration, many indigenous people settled in Nevada. Native Americans such as that of Shoshone, the Paiute and Washoe tribes were settled in the current Nevada.The first European to arrive in the state was Francisco Garces.Before it was a part of Spanish Empire. In 1861, the Nevada Territory was detached from the Utah Territory. Additionally, it adopted its current name which is a taken from Sierra Nevada. Nevada joined the Union as the 36th state in 1864. In 1867, it attained its current southern boundaries. In early Nevada mining town, unregulated gambling was quite common. In 1909, gambling was outlawed and it was by far the least populated state. Nevada History Timeline - 1609 - Established town of Santa Fe as Spanish-Indian trade center. - 1776 - Father Garces crossed Colorado River. - 1850 - United States Congress formed the Utah Territory. - 1864 - Nevada attained statehood and became the 36th state. - 1866 - Adopted official Nevada State Seal. - 1874 - Opened University of Nevada at Elko. - 1902 - Goldfield discovered - 1910 - Abolished gambling in Nevada - 1914 - State election granted Nevada women voting right Geography of Nevada Total Area: 77,354 sq mi Longitude: 95o 19' W to 104o 03' W Latitude: 40o N to 43o N Highest point: Panorama Point - 5,424 ft (1654 m) Mean point: 2,600 ft (790 m) Lowest point: Missouri River at Kansas border - 840 ft (256 m) Time Zone Mountain: - most of state - Central: UTC −6/−5 - panhandle Mountain: UTC −7/−6 Nevada is the 7th most extensive state, covering a total land area of 110,622 square miles . It is bordered on the north by Oregon and Idaho and on the south by California. Utah and Arizona lie on the east. On the west, it is bound by California. Nevada lies within the Basin and Range Province, and it is segregated by various north-south mountain ranges. The ranges consist of endorheic valleys between them. The state is divided into 2 time zones, namely the Pacific Time zone in most part of the state and the Mountain Time in the state's West Wendover. Nevada’s geographic center is located in Lander County. The highest point in Nevada is Boundary Peak at 13,140 feet. Topography of Nevada The Nevada landscape consists of sandy deserts, snow covered mountains, rugged, forested mountain slopes, and grassy valleys. The state is nestled within the Great Basin and is sectioned into 3 main land regions - the Sierra Nevada, the Columbia Plateau and the Basin and Range Region. Lava bedrock make up the northeastern corner of Nevada. Many rivers and streams interrupt the bedrock, forming deep canyons with high ridges. Open prairie make up the Idaho border, which is a part of the Columbia Plateau. The Sierra Nevada consists of rugged mountain range. The ranges cut across a a region of the state. The rest of Nevada is made up of large parts of Basin and Range Region, which is segregated by over 150 mountain ranges that run from north to south. Some of the major rivers in Nevada are Colorado River, Humboldt River, Columbia River, Truckee RiverContact Nevada Division of Forestry 2478 Fairview Drive (map) Carson City, Nevada 89701 Climate of Nevada Nevada has the driest climate in the nation, predominantly consisting of desert and semiarid climate regions. Long and freezing winters dominate the northern Nevada region while the state's southern region has milder and shorter winter season. The whole state of Nevada receive scarce precipitation, with most precipitation on the lee side of the Sierra Nevada. The highest recorded temperature in Nevada is 125 o F at Laughlin and the lowest recorded temperature at San Jacinto is −50o Demographics of Nevada Nevada had an estimated population of 2,890,845 as of 2015 estimation by the U.S census bureau, which reflected an increase of 7% since the year 2010. The population density of the state is 24.6 persons per square mile . There are many ancestry groups in Nevada. The five largest groups among them include German, English, Mexican, Italian, American and Scandinavian.Nevada Population Quick Facts: Nevada Racial Groups are- - Population, 2015 - 2,890,845 - Population, 2014 - 2,839,099 - Population, percent change, April 1, 2010 to July 1, 2015 - 7% - Persons under 5 years, percent, 2014 - 6.2% - Persons under 18 years, percent, 2014 - 23.4% - Persons 65 years and over, percent, 2014 -14.2% - Female persons, percent, 2014 - 49.7% The major religious groups in Nevada are – - White alone: 76.2% - Black or African American alone: 9.1% - American Indian and Alaska Native alone: 1.6% - Asian alone: 8.3% - Native Hawaiian and Other Pacific Islander alone: 0.7% - Two or More Races: 4% - Hispanic or Latino: 27.8% - White alone, not Hispanic or Latino: 51.5% - Roman Catholic - Latter-day Saint Economy of Nevada The important economic sectors of Nevada include tourism, mining, and cattle ranching. The state's economy is mainly associated with the tourism sector and major revenue is generated from agriculture and its products. Nevada’s per capita income in millions of current dollars of all industry totals in 2014 was $26515. Nevada's unemployment rate, 2014 is 2.8% Agriculture in Nevada - The top 5 agricultural products based on revenue generated include cattle and calves, dairy products, hay, onions, and potatoes. - Nevadaa's most important crop is hay. - The other major crops of the state are alfalfa seed, garlic, barley, mint, potatoes, onions and wheat. - The state's major livestock products are sheep, lambs and hogs. - Sheep and cattle ranching is the state's primary agricultural activity. Industry in Nevada - Leading service industries in Nevada include tourism, business and personal service sectors. Other important sectors include gambling, recreation and amusement services. - In manufacturing sector, the state's top manufacturing products include printed products, concrete and machinery and food products. - Chemicals, construction equipment and plastics are also other important manufacturing products. - The production of chemicals ranked 2nd and production of machinery ranked 3rd in the state's manufacturing sector. - Gold is the Nevada's most important mined product. It also leads in silver production too. - Finance, insurance and real estate are the second ranking service industry. Tourism in Nevada Tourism is also a principal part of Nevada’s economy and is one of the largest employers of the state. Many plans and programs are implemented to improve the transportation network in the state and accordingly enhance the tourism sector.Some of the major tourist attractions of the state are- Health Care in Nevada The Nevada Department of Health and Human Services (DHHS) promotes the health and well-being of its residents through the delivery or facilitation of a multitude of essential services to ensure families are strengthened, public health is protected, and individuals achieve their highest level of self-sufficiency. The Department is the largest in state government comprised of five Divisions along with additional programs and offices overseen by the DHHS’s Director’s Office. The Nevada Department of Health is the main organization working towards promoting health, preventing disease and well-being. The department’s mission is to protect the lives of all citizens. Nevada's Medicaid works effectively for low-income families by giving them access to best medical facilities by paying their medical expenses including foster and adoptive children, pregnant women and women having breast or cervical cancer and for those people whose income is not enough to meet the cost of required medical services.Learn more: Health Care in Nevada Government of Nevada The Constitution of the State of Nevada was created on July 4,1864. The Government of Nevada has three branches of government: Executive, Legislative and Judicial. The legislative branch is personified in the bicameral General Assembly. The Governor is the head of the executive branch. The Judicial branch is headed by the Supreme Court. The Nevada Legislature is the state legislature of the U.S. state of Nevada. The Legislature is a bicameral body, consisting of the lower house Nevada Assembly, with 42 members, and the upper house Nevada Senate, with 21 members. All 63 members of the Legislature are elected from an equal amount of constituent districts across the state. The Judicial Branch of Nevada consists of Supreme Court, District Courts, Justice Courts and Municipal Courts. The court is responsible to make effective rules and policies for the appropriate working of the entire court system. The court system of Nevada is efficiently managed by the Nevada Supreme Court. The court is responsible to make effective rules and policies for the appropriate working of the entire court system. The Chief Justice and four Associate Justices makes the final decisions in the matter of the District Court.Learn more: Government of Nevada Education in Nevada The Nevada Department of Education operates to improve the public education system and also to enhance student academic achievements and outputs. The department’s mission is to provide world class education and best values to the students to ensure success in education, workplace and in every spheres of life. Education in Nevada is an accumulative sectors of public and private schools (elementary, middle, and high), as well as colleges and universities.Colleges and Universities of Nevada - Nevada System of Higher Education - University of Nevada, Las Vegas (UNLV) - University of Nevada, Reno (Nevada) - Nevada State College - Truckee Meadows Community College (TMCC) - Great Basin College - College of Southern Nevada (CSN) - Western Nevada College (WNC) - Sierra Nevada College - Touro University Nevada - Roseman University of Health Sciences Sports in Nevada Nevada does not have many professional sports team but the state takes pride of its college sports. College teams include- - the Nevada Wolf Pack - the Mountain West Conference - the UNLV Rebels Interesting Facts about Nevada: - Nevada state motto is "All for our country." - Silver and blue are Nevada's state color. - Nevada state fossil is Ichthyosaur. - Nevada state soil is Orovada series. - "Home Means Nevada," is the state song. - The origin of Navada's name is Spanish that means "snow-capped." - Nevada is the 7th largest US state in land area. - Nevada Statehood - Demographics and Total Land Area - Economy Facts<\|endoftext\|>	3.6875
869	Assistive technology is any product, device, or equipment that maintains, or improves functional capabilities of a person with a disability. Assistive technology can include anything from a pencil grip to help a student write more accurately, to a speech synthesizer to help those who have difficulty reading. The first step to finding technologies for the disabled is to identify the area where assistance is needed. A person who has difficulty speaking, for instance, may benefit from a communication device to express themselves. Below find a list of some common tools your child or student may use in the classroom. Assistive Learning Technology in the Classroom For those that have difficulty listening, remembering, or processing spoken words, such as a lecture in the classroom there are a variety of listening devices available. - Tape Recorders -Tape recorders help by allowing the recorded audio to be played back at a later time and repeatedly if necessary. - FM Listening Systems – The speaker wears a microphone and the listener wears a headset. This helps students with mild or moderate hearing loss to better understand lectures. Assistive technology for math includes devices to help with aligning, calculating, or copying problems into columns. - Electronic Math Worksheets – For students that have difficulty copying down math problems with a pencil and paper, or aligning these problems on the page, the MathPad can be used. MathPad, by Intellitools, requires a teacher or para-educator to enter the problem into the tool. - Talking Calculators – Each time a number or an operand is pressed on a talking calculator, a built-in speech synthesizer lets the user know what was pressed. This helps users to verify that they have selected the correct buttons. Talking calculators are very affordable and many offer adjustable volume, clock, and alarm features as well. Technology can help in the area of organization by enabling users to track their schedule, due dates, deadlines, manage the days events, or remember important tasks. For those who struggle with reading, assistive technology helps present the written word in a different format. The most commonly used devices are those that synthesize written text into the spoken word. - Optical Character Recognition – For those that cannot read the written word, optical character recognition is a computer aided scanning device that reads written characters. It is often combined with a speech synthesizer, and reads the word aloud through computer generated speech. Scanning the page of a text generates computer synthesized speech allowing the user to hear almost any text in audio format. - Other Alternative Text Formats – Audio books and books available in braille are an example of alternative forms of text. These technologies can be useful by allowing those that cannot identify written text to read and learn in another format. For those who have difficulty writing, there are tools available which allow a student to dictate their answers rather than write them down. Other tools include spelling or grammar checks and word prediction. This type of technology is used regularly by those without disabilities. Assistive learning technologies can help people succeed in the classroom by assisting and improving their capabilities in their areas of weakness. If one of the above areas is holding someone back, one of these devices may help them succeed. The above list may help identify educational devices that can be of assistance in the classroom. If these devices are determined necessary for a child to learn they will be placed in the Individualized Education Plan, or IEP. These AT devices will be paid for by the school. The school retains ownership of the device and when the child leaves the school, they must return the device. If the child’s IEP specifies that the device is needed at home to ensure appropriate education, the device may be transported home from school. Technologies for the Disabled at Home In addition to the technology listed above for school, remotes and electronic control devices may help people with disabilities to operate televisions, radios, cassette recorders, and other electronics more independently at home. Audio prompting devices may help to remind them to complete a task from start to finish. Examples of these tasks include, making the bed, brushing teeth, staying on task with school work, or managing time. Funding for these devices at home may be available through an assistive technology grant.<\|endoftext\|>	3.875
634	The Department for Education state that there is a need “to create and enforce a clear and rigorous expectation on all schools to promote the fundamental British values of democracy, the rule of law, individual liberty and mutual respect and tolerance of those with different faiths and beliefs.” The government set out its definition of British values in the 2011 Prevent Strategy and these values were reiterated by the Prime Minister in 2014. At Leopold School, these values are reinforced regularly and in the following ways 1. The Rule of Law The importance of laws whether they are those that govern the class, the school or the country, are consistently reinforced. Each class discusses and sets its own rules that are clearly understood by all and seen to be necessary to ensure that every class member is able to learn in a safe and ordered environment. Our pupils are taught the value and reasons behind laws, that they govern and protect us, the responsibilities that this involves and the consequences when laws are broken. Democracy is embedded at the school. Pupils are always listened to by adults and are taught to listen carefully and with concern to each other, respecting the right of every individual to have their opinions and voices heard. Pupils also have the opportunity to air their opinions and ideas through our School Council Team and pupil questionnaires. The elections of the School Council members are based solely on pupil votes, reflecting our British electoral system and demonstrating democracy in action. 3. Individual Liberty Within school, pupils are actively encouraged to make choices, knowing that they are in a safe and supportive environment. As a school we educate and provide boundaries for our pupils to make choices safely, through the provision of a safe environment and an empowering education. Our pupils are encouraged to know, understand and exercise their rights and personal freedoms and are advised how to exercise these safely; examples of this can be clearly seen in our e-safety and P.S.H.E. lessons. Whether it is through choice of challenge; of how they record; of participation in our numerous extra- curricular activities; our pupils are given the freedom to make choices. 4. Mutual Respect Respect is one of the core values of our school. Our PSHE curriculum embodies values of mutual respect through many different units of learning. Respect is promoted during our Head teacher and achievers assemblies. Pupils learn that their behaviours have an effect on their own rights and those of others and all members of the school community are expected to treat each other with respect. 5. Tolerance of Those With Different Faiths And Beliefs Our core value of Respect ensures tolerance of those who have different faiths and beliefs. Leopold Primary School enhances pupils understanding of different faiths and beliefs through religious education studies; P.S.H.E. work; visits to other schools in different settings to participate in celebrations such as Diwali; inviting guest speakers from many different cultural backgrounds . Beliefs, traditions and customs will be studied in depth, with visitors being invited in to our school to enrich and extend understanding. Through this our pupils gain an enhanced understanding of their place in a culturally diverse society.<\|endoftext\|>	3.921875
401	Goals of Essential Studies at UND As UND students progress through the Essential Studies Program, preparing for productive and fulfilling public, private, and professional lives, they will take courses through which they will practice and develop key intellectual skills. To accomplish this, every Essential Studies course will very purposefully include content, activities, and assignments whose focus is on one or more of the following: Critical Inquiry and Analysis Inquiry is a systematic process of exploring issues, objects or works through the collection and analysis of evidence that results in informed conclusions or judgments. Analysis is the process of breaking complex topics or issues into parts to gain a better understanding of them. Quantitative Reasoning is competency and comfort in working with numerical data, using it to reason and solve quantitative problems from a wide array of authentic contexts and everyday life situations, and to create and clearly communicate sophisticated arguments supported by quantitative evidence, such as by using words, tables, graphs, mathematical equations, etc., as appropriate. Written Communication is the development and expression of ideas in writing. Written communication involves learning to work in many genres and styles. It can involve working with many different writing technologies, and mixing text, data, and images. Oral communication involves a purposeful presentation designed to increase knowledge, to foster understanding, or to promote change in the listeners' attitudes, values, beliefs, or behaviors. - Information Literacy Information literacy is the set of integrated abilities encompassing the reflective discovery of information, the understanding of how information is produced and valued, and the use of information in creating new knowledge and participating ethically in communities of learning. - Intercultural Knowledge & Skills Intercultural knowledge and skills foster the capacity to meaningfully engage with the perspectives of people whose cultures and identities are different from one's own. To meaningfully engage with others' perspectives, one must be aware of how those perspectives are shaped by larger social structures, by issues of contemporary importance, and by issues that arise in global society.<\|endoftext\|>	3.828125
3,931	SAT Math 1 & 2 Subject Tests Questions about solid geometry frequently test plane geometry techniques. They’re difficult mostly because the added third dimension makes them harder to visualize. You’re likely to run into three or four solid geometry questions on either one of the Math Subject Tests, however, so it’s important to practice. If you’re not the artistic type and have trouble drawing cubes, cylinders, and so on, it’s worthwhile to practice sketching the shapes in the following pages. The ability to make your own drawing is often helpful. Prisms are three-dimensional figures that have two parallel bases that are polygons. Cubes and rectangular solids are examples of prisms that ETS often asks about. In general, the volume of a prism is given by the following formula: Volume of a Prism V = Bh In this formula, B represents the area of either base of the prism (the top or the bottom), and h represents the height of the prism (perpendicular to the base). The formulas for the volume of a rectangular solid, a cube, and a cylinder all come from this basic formula. Area and Volume In general the volume of a shape involves the area of the base, often referred to as B, and the height, or h, of the solid. A rectangular solid is simply a box; ETS also sometimes calls it a rectangular prism. It has three distinct dimensions: length, width, and height. The volume of a rectangular solid (the amount of space it contains) is given by this formula: Volume of a Rectangular Solid V = lwh The surface area (SA) of a rectangular solid is the sum of the areas of all of its faces. A rectangular solid’s surface area is given by the formula on the next page. Surface Area of a Rectangular Solid SA = 2lw + 2wh + 2lh The volume and surface area of a solid make up the most basic information you can have about that solid (volume is tested more often than surface area). You may also be asked about lengths within a rectangular solid—edges and diagonals. The dimensions of the solid give the lengths of its edges, and the diagonal of any face of a rectangular solid can be found using the Pythagorean theorem. There’s one more length you may be asked about—the long diagonal (or space diagonal) that passes from corner to corner through the center of the box. The length of the long diagonal is given by this formula: Long Diagonal of a Rectangular Solid (Super Pythagorean Theorem) a2 + b2 + c2 = d2 This is the Pythagorean theorem with a third dimension added, and it works just the same way. This formula will work in any rectangular box. The long diagonal is the longest straight line that can be drawn inside any rectangular solid. A cube is a rectangular solid that has the same length in all three dimensions. All six of its faces are squares. This simplifies the formulas for volume, surface area, and the long diagonal. Volume of a Cube V = s3 Surface Area of a Cube SA = 6s2 Face Diagonal of a Cube f = s Long Diagonal of a Cube d = s A cylinder is like a prism but with a circular base. It has two important dimensions—radius and height. Remember that volume is the area of the base times the height. In this case, the base is a circle. The area of a circle is πr2. So the volume of a cylinder is πr2h. Volume of a Cylinder V = πr2h The surface area of a cylinder is found by adding the areas of the two circular bases to the area of the rectangle you’d get if you unrolled the side of the cylinder. That boils down to the following formula: Surface Area of a Cylinder SA = 2πr2 + 2πrh The longest line that can be drawn inside a cylinder is the diagonal of the rectangle formed by the diameter and the height of the cylinder. You can find its length with the Pythagorean theorem. d2 = (2r)2 + h 2 If you take a cylinder and shrink one of its circular bases down to a point, then you have a cone. A cone has three significant dimensions which form a right triangle—its radius, its height, and its slant height, which is the straight-line distance from the tip of the cone to a point on the edge of its base. The formulas for the volume and surface area of a cone are given in the information box at the beginning of both of the Math Subject Tests. The formula for the volume of a cone is pretty straightforward: Volume of a Cone V = πr2h Connect the Dots Notice that the volume of a cone is just one-third of the volume of a circular cylinder. Make memorizing But you have to be careful computing surface area for a cone using the formula provided by ETS. The formula at the beginning of the Math Subject Tests is for the lateral area of a cone—the area of the sloping sides—not the complete surface area. It doesn’t include the circular base. Here’s a more useful equation for the surface area of a cone. Surface Area of a Cone SA = πrl + πr2 If you want to calculate only the lateral area of a cone, just use the first half of the above formula—leave the πr2 off. A sphere is simply a hollow ball. It can be defined as all of the points in space at a fixed distance from a central point. The one important measure in a sphere is its radius. The formulas for the volume and surface area of a sphere are given to you at the very beginning of both Math Subject Tests. That means that you don’t need to have them memorized, but here they are anyway: Volume of a Sphere V = πr3 Surface Area of a Sphere SA = 4πr2 The intersection of a plane and a sphere always forms a circle unless the plane is tangent to the sphere, in which case the plane and sphere touch at only one point. A pyramid is a little like a cone, except that its base is a polygon instead of a circle. Pyramids don’t show up often on the Math Subject Tests. When you do run into a pyramid, it will almost always have a rectangular base. Pyramids can be pretty complicated solids, but for the purposes of the Math Subject Tests, a pyramid has just two important measures—the area of its base and its height. The height of a pyramid is the length of a line drawn straight down from the pyramid’s tip to its base. The height is perpendicular to the base. The volume of a pyramid is given by this formula. Connect the Dots Notice that the volume of a pyramid is just one-third of the volume of a prism. Make memorizing easy! Volume of a Pyramid V = Bh (B = area of base) It’s not really possible to give a general formula for the surface area of a pyramid because there are so many different kinds. At any rate, the information is not generally tested on the Math Subject Tests. If you should be called upon to figure out the surface area of a pyramid, just figure out the area of each face using polygon rules, and add them up. TRICKS OF THE TRADE Here are some of the most common solid geometry question types you’re likely to encounter on the Math Subject Tests. They occur much more often on the Math Level 2 Subject Test than on the Math Level 1 Subject Test, but they can appear on either test. Triangles in Rectangular Solids Many questions about rectangular solids are actually testing triangle rules. Such questions generally ask for the lengths of the diagonals of a box’s faces, the long diagonal of a box, or other lengths. These questions are usually solved using the Pythagorean theorem and the Super Pythagorean theorem that finds a box’s long diagonal (see the section on Rectangular Solids). Here are some practice questions using triangle rules in rectangular solids. The answers to these drills can be found in Chapter 12. 32. What is the length of the longest line that can be drawn in a cube of volume 27 ? 36. In the rectangular solid shown, if AB = 4, BC = 3, and BF = 12, what is the perimeter of triangle EDB ? 39. In the cube above, M is the midpoint of BC, and N is the midpoint of GH. If the cube has a volume of 1, what is the length of MN ? Many solid geometry questions test your understanding of the relationship between a solid’s volume and its other dimensions—sometimes including the solid’s surface area. To solve these questions, just plug the numbers you’re given into the solid’s volume formula. Try the following practice questions. The answers to these drills can be found in Chapter 12. 17. The volume and surface area of a cube are equal. What is the length of an edge of this cube? 24. A rectangular solid has a volume of 30, and its edges have integer lengths. What is the greatest possible surface area of this solid? 28. The water in Allegra’s swimming pool has a depth of 7 feet. If the area of the pentagonal base of the pool is 150 square feet, then what is the volume, in cubic feet, of the water in her pool? (E) It cannot be determined from the information given. 43. A sphere of radius 1 is totally submerged in a cylindrical tank of radius 4, as shown. The water level in the tank rises a distance of h. What is the value of h ? 17. A cube has a surface area of 6x. What is the volume of the cube? 36. A sphere has a radius of r. If this radius is increased by b, then the surface area of the sphere is increased by what amount? (C) 8πrb + 4πb2 (D) 8πrb + 2rb + b2 40. If the pyramid shown has a square base with edges of length b, and b = 2h, then which of the following is the volume of the pyramid? (D) 8h2 − h Some questions on the Math Subject Tests will be based on spheres inscribed in cubes or cubes inscribed in spheres (these are the most popular inscribed shapes). Occasionally you may also see a rectangular solid inscribed in a sphere, or a cylinder inscribed in a rectangular box, etc. The trick to these questions is always figuring out how to get from the dimensions of one of the solids to the dimensions of the other. Following are a few basic tips that can speed up your work on inscribed solids questions. · When a cube or rectangular solid is inscribed in a sphere, the long diagonal of that solid is equal to the diameter of the sphere. · When a cylinder is inscribed in a sphere, the sphere’s diameter is equal to the diagonal of the rectangle formed by the cylinder’s heights and diameter. · When a sphere is inscribed in a cube, the diameter of the sphere is equal to the length of the cube’s edge. · If a sphere is inscribed in a cylinder, both solids have the same diameter. Most inscribed solids questions fall into one of the preceding categories. If you run into a situation not covered by these tips, just look for the way to get from the dimensions of the inner shape to those of the external shape, or vice versa. Here are some practice inscribed solids questions. The answers to these drills can be found in Chapter 12. 32. A rectangular solid is inscribed in a sphere as shown. If the dimensions of the solid are 3, 4, and 6, then what is the radius of the sphere? 35. A cylinder is inscribed in a cube with an edge of length 2. What volume of space is enclosed by the cube but not by the cylinder? 38. A cone is inscribed in a cube of volume 1 in such a way that its base is inscribed in one face of the cube. What is the volume of the cone? Solids Produced by Rotation Three types of solids can be produced by the rotation of simple two-dimensional shapes—spheres, cylinders, and cones. Questions about solids produced by rotation are generally fairly simple; they usually test your ability to visualize the solid generated by the rotation of a flat shape. Sometimes, rotated solids questions begin with a shape in the coordinate plane—that is, rotated around one of the axes or some other line. Practice will help you figure out the dimensions of the solid from the dimensions of the original flat shape. A sphere is produced when a circle is rotated around its diameter. This is an easy situation to work with, as the sphere and the original circle will have the same radius. Find the radius of the circle, and you can figure out anything you want to about the sphere. A cylinder is formed by the rotation of a rectangle around a central line or one edge. A cone is formed by rotating a right triangle around one of its legs (think of it as spinning the triangle) or by rotating an isosceles triangle around its axis of symmetry. Another way of thinking about it is if you spun the triangle in the first figure above around the y-axis (so you’re rotating around the leg that’s sitting on the y-axis) you would get the second figure. Likewise, if you spun the third figure above around the x-axis (so you’re rotating around the axis of symmetry), you would end up with the fourth figure. Try these rotated solids questions for practice. The answers to these drills can be found in Chapter 12. 34. What is the volume of the solid generated by rotating rectangle ABCD around AD ? 39. If the triangle created by OAB is rotated around the x-axis, what is the volume of the generated solid? 46. What is the volume generated by rotating square ABCD around the y-axis? Some solid geometry questions will ask you to figure out what happens to the volume of a solid if all of its lengths are increased by a certain factor or if its area doubles, and so on. To answer questions of this type, just remember a basic rule. When the lengths of a solid are increased by a certain factor, the surface area of the solid increases by the square of that factor, and the volume increases by the cube of that factor. This rule is true only when the solid’s shape doesn’t change—its length must increase in every dimension, not just one. For that reason, cubes and spheres are most often used for this type of question because their shapes are constant. In the illustration above, a length is doubled, which means that the corresponding area is 4 times as great, and the volume is 8 times as great. If the length had been tripled, the area would have increased by a factor of 9, and the volume by a factor of 27. Try these practice questions. The answers to these drills can be found in Chapter 12. 13. If the radius of sphere A is one-third as long as the radius of sphere B, then the volume of sphere A is what fraction of the volume of sphere B ? 18. A rectangular solid with length l, width w, and height h has a volume of 24. What is the volume of a rectangular solid with length , width , and height ? 21. If the surface area of a cube is increased by a factor of 2.25, then its volume is increased by what factor? · Solid geometry questions are often plane geometry questions in disguise. · For the purposes of the SAT Math 1 & 2 Subject Tests, prisms are 3-dimensional figures with two parallel, identical bases. The bases can be any shape from plane geometry. · The volume of a prism is the area of the base, often referred to as B, times the height, h. · Let’s talk rectangular prisms: · The formula for the volume of a rectangular prism is V = lwh. · The formula for the surface area of a rectangular solid is SA = 2lw + 2wh + 2lh. Think about painting the outside of the figure. Find the area of each side. · The Super Pythagorean theorem, which is helpful in solving questions about the diagonal of a rectangular prism, is a2 + b2 + c2 = d2. · Let’s talk cubes. Remember that a cube is just a rectangular prism whose length, width, and height are equal. If you forget a formula, just use the rectangular prism formula! · The volume of a cube is V = s3. · The surface area of a rectangular solid is SA = 6s2. · Let’s talk cylinders. A cylinder is a prism whose bases are circles. · The volume of a cylinder is V = πr2h. · The surface area of a rectangular solid is SA = 2πr2 + 2πrh. If you forget this, remember that you’re just painting the outside. So you’ll need the area of two circles and the area of the other piece, which, when rolled out (like a roll of paper towels), is a rectangle whose sides are the circumference of the circle and the height. · A cone is similar to a cylinder except that one of its bases is merely a point. · The formula for the volume of a cone is V = ≠ r2 h, where the height must be perpendicular to the base. · The formula for the surface area of a cone is SA = πrl + πr2, where l is the slant height. · A sphere is a hollow ball. · The formula for the volume of a sphere is V = ≠ r3. · The formula for the surface area of a cone is SA = 4πr2. · Pyramids are like cones, but the base is a plane geometry shape. The formula for the volume of a pyramid is V = Bh. · Inscribed figures always have a line or curve that connects the inner figure to the outer figure. · Questions about solids produced by rotation usually test your ability to visualize the solid created by the rotation of a flat shape.<\|endoftext\|>	4.6875
1,598	Socrates (c. 470 BCE - c. 399 BCE). Socrates was a Greek philosopher and the main source of Western thought. Little is known of his life except what was recorded by his students, including Plato. Socrates was born circa 470 BC, in Athens, Greece. We know of his life through the writings of his students, including Plato and Xenophon. His "Socratic method," laid the groundwork for Western systems of logic and philosophy. When the political climate of Greece turned, Socrates was sentenced to death by hemlock poisoning in 399 BC. He accepted this judgement rather than fleeing into exile. Born circa 470 BC in Athens, Greece, Socrates's life is chronicled through only a few sources—the dialogues of Plato and Xenophon and the plays of Aristophanes. Because these writings had other purposes than reporting his life, it is likely none present a completely accurate picture. However, collectively, they provide a unique and vivid portrayal of Socrates's philosophy and personality. Socrates was the son of Sophroniscus, an Athenian stone mason and sculptor, and Phaenarete, a midwife. Because he wasn't from a noble family, he probably received a basic Greek education and learned his father's craft at a young age. It is believed Socrates worked as mason for many years before he devoted his life to philosophy. Contemporaries differ in their account of how Socrates supported himself as a philosopher. Both Xenophon and Aristophanes state Socrates received payment for teaching, while Plato writes Socrates explicitly denied accepting payment, citing his poverty as proof. Socrates married Xanthippe, a younger woman, who bore him three sons—Lamprocles, Sophroniscus and Menexenus. There is little known about her except for Xenophon's characterization of Xanthippe as "undesirable." He writes she was not happy with Socrates's second profession and complained that he wasn’t supporting family as a philosopher. By his own words, Socrates had little to do with his sons' upbringing and expressed far more interest in the intellectual development of Athens' young boys. Athenian law required all able bodied males serve as citizen soldiers, on call for duty from ages 18 until 60. According to Plato, Socrates served in the armored infantry—known as the hoplite—with shield, long spear and face mask. He participated in three military campaigns during the Peloponnesian War, at Delium, Amphipolis, and Potidaea, where he saved the life of Alcibiades, a popular Athenian general. Socrates was known for his courage in battle and fearlessness, a trait that stayed with him throughout his life. After his trial, he compared his refusal to retreat from his legal troubles to a soldier's refusal to retreat from battle when threatened with death. Plato's Symposium provides the best details of Socrates's physical appearance. He was not the ideal of Athenian masculinity. Short and stocky, with a snub nose and bulging eyes, Socrates always seemed to appear to be staring. However, Plato pointed out that in the eyes of his students, Socrates possessed a different kind of attractiveness, not based on a physical ideal but on his brilliant debates and penetrating thought. Socrates always emphasized the importance of the mind over the relative unimportance of the human body. This credo inspired Plato’s philosophy of dividing reality into two separate realms, the world of the senses and the world of ideas, declaring that the latter was the only important one. Socrates believed that philosophy should achieve practical results for the greater well-being of society. He attempted to establish an ethical system based on human reason rather than theological doctrine. He pointed out that human choice was motivated by the desire for happiness. Ultimate wisdom comes from knowing oneself. The more a person knows, the greater his or her ability to reason and make choices that will bring true happiness. Socrates believed that this translated into politics with the best form of government being neither a tyranny nor a democracy. Instead, government worked best when ruled by individuals who had the greatest ability, knowledge, and virtue and possessed a complete understanding of themselves. For Socrates, Athens was a classroom and he went about asking questions of the elite and common man alike, seeking to arrive at political and ethical truths. Socrates didn’t lecture about what he knew. In fact, he claimed to be ignorant because he had no ideas, but wise because he recognized his own ignorance. He asked questions of his fellow Athenians in a dialectic method (the Socratic Method) which compelled the audience to think through a problem to a logical conclusion. Sometimes the answer seemed so obvious, it made Socrates's opponents look foolish. For this, he was admired by some and vilified by others. During Socrates's life, Athens was going through a dramatic transition from hegemony in the classical world to its decline after a humiliating defeat by Sparta in the Peloponnesian War. Athenians entered a period of instability and doubt about their identity and place in the world. As a result, they clung to past glories, notions of wealth, and a fixation with physical beauty. Socrates attacked these values with his insistent emphasis on the greater importance of the mind. While many Athenians admired Socrates's challenges to Greek conventional wisdom and the humorous way he went about it, an equal number grew angry and felt he threatened their way of life and uncertain future. The jury was not swayed by Socrates's defense and convicted him by a vote of 280 to 221. Possibly the defiant tone of his defense contributed to the verdict and he made things worse during the deliberation over his punishment. Athenian law allowed a convicted citizen to propose an alternative punishment to the one called for by the prosecution and the jury would decide. Instead of proposing he be exiled, Socrates suggested he be honored by the city for his contribution to their enlightenment and be paid for his services. The jury was not amused and sentenced him to death by drinking a mixture of poison hemlock. Before Socrates's execution, friends offered to bribe the guards and rescue him so he could flee into exile. He declined, stating he wasn't afraid of death, felt he would be no better off if in exile and said he was still a loyal citizen of Athens, willing to abide by its laws, even the ones that condemned him to death. Plato described Socrates's execution in his Phaedo dialogue: Socrates drank the hemlock mixture without hesitation. Numbness slowly crept into his body until it reached his heart. Shortly before his final breath, Socrates described his death as a release of the soul from the body. One of the immediate effects of Socrates’ death was the setting up of new philosophical schools of thoughts by his students and followers. They also exercised their perceptions of his teachings in politics. His student Plato founded the “Academy” in 385 BC which later became so famous that it popularized the word “Academy” for educational institutions. Plato’s protégé, Aristotle was also a key figure in the Classical Era and founded his own school,the Lyceum in 335 BC. Aristotle was also the tutor of Alexander the Great. Socrates’ thoughts of stressing a simplistic way of living later led to the origination of Cynicism by one of his older students, Antisthenes. The contribution of Socrates in Western Philosophy became more evident in the Renaissance and the Age of Reason in Europe. A number of paintings and plays from this period depicted his role in the western intellectual process. His scientific method is still being used in classrooms and law school discourses. --> Click "play-button" below to launch video. (Socrates Biography) Our Mobile Application Check out Our Mobile Application "Ancient Greece Reloaded"<\|endoftext\|>	4.03125
314	# How do you graph the line y= -2x+3? Mar 29, 2017 see explanation. #### Explanation: To graph a line we require 2 points. The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is. $\textcolor{red}{\overline{\underline{\| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{2}{2}} \|}}}$ where m represents the slope and b, the y-intercept. $y = - 2 x + 3 \text{ is in this form}$ $\Rightarrow \text{y-intercept "=b=3to(0,3)color(red)" is point on line}$ To find another point on the line, choose any value for x, substitute into the equation and evaluate for y $x = 3 \to y = \left(- 2 \times 3\right) + 3 = - 6 + 3 = - 3$ $\Rightarrow \left(3 , - 3\right) \textcolor{red}{\text{ is a point on the line}}$ Plot the 2 points, draw a straight line through them to obtain the graph. graph{-2x+3 [-10, 10, -5, 5]}<\|endoftext\|>	4.75
153	Fern leaves are called fronds. Ferns are vascular plants that are non-flowering and do not contain seeds. The method in which this plant reproduces is through spore dispersal. Fern leaves grow from a rhizome. Fern leaves are fan-shaped and also play an important role in reproduction. The production of spores occurs in structures called sporangia. These structures are found on the surface of the fronds, usually on the underside of the leaf. When the sporangia or spore casings dry out, the dispersal of spores occurs into the atmosphere. Certain fern species date back nearly 400 million years. There are about 12,000 fern species. Ferns require water and shady areas to grow.<\|endoftext\|>	4.28125
140	Presents experiments and activities that demonstrate the concepts and scientific principles of magnetism. Complete a variety of fun science experiments using common kitchen items. Introduces simple scientific principles involving heat, and provides instructions for experiments that can be done at home to prove them. Simple text explores the principles of liquids, including melting, freezing, surface tension, and density. Presents experiments demonstrating the basic scientific principles of water, including information on erosion, precipitation, and condensation. Describes the physical properties of solids and explains how matter can change to and from a solid into a liquid or gas. Complete a variety of fun science experiments using the items found in your classroom at school.<\|endoftext\|>	4.1875
470	# How do you calculate the specific heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6.75 * 10^4 joules of heat, and its temperature changes from 32°C to 57°C? Jan 29, 2016 1.8"J"/("g" ""^@"C") #### Explanation: A substance's specific heat tells you how much heat much either be added or removed from $\text{1 g}$ of that substance in order to cause a ${1}^{\circ} \text{C}$ change in temperature. The equation that establishes a relationship between specific heat, heat added or removed, and change in temperature looks like this $\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where $q$ - the amount of heat added / removed $m$ - the mass of the sample $c$ - the specific heat of the substance $\Delta T$ - the change in temperature In your case, the $\text{1500.0-g}$ piece of wood is said to absorb a total of $6.75 \cdot {10}^{4} \text{J}$ of heat. This caused its temperature to increase from ${32}^{\circ} \text{C}$ to ${57}^{\circ} \text{C}$. The difference between the final temperature and the initial temperature of the sample will be the value for $\Delta T$. $\Delta T = {57}^{\circ} \text{C" - 32^@"C" = 25^@"C}$ This means that the specific heat of the wood is equal to $q = m \cdot c \cdot \Delta T \implies c = \frac{q}{m \cdot \Delta T}$ Plug in your values to get c = (6.75 * 10^4"J")/("1500.0 g" * 25^@"C") = color(green)(1.8"J"/("g" ""^@"C")) The answer is rounded to two sig figs, the number of sig figs you have for the two temperatures of the sample.<\|endoftext\|>	4.59375
1,489	3 Q: # If Riya's age is four times the age of Siya and if Riya's age is 24 yrs , find the age difference between Riya and Siya ? A) 2 yrs B) 12 yrs C) 18 yrs D) 6 yrs Explanation: Let Siya's age = x Then Riya's age = 4x But given Riya's age = 24 => 4x = 24 => x = 6 Hence Siya's age = 6 yrs => The age difference = 24 - 6 = 18 yrs. Q: Sum of present ages of P and Q is 41. Age of P 2 year hence is equal to age of R, 1 year ago. Age of P, 4 year hence is equal to age of Q 1 year ago and ratio of present age of P and S is 3 : 4. Find the difference of age of R and S. A) 2 B) 3 C) 4 D) 5 Explanation: According to the given data, P + Q = 41 ......(1) R - 1 = P + 2 R = P + 3 and P + 4 = Q - 1 => Q = P + 5 ....(2) From (1)&(2) P = 18 Q = 18 + 5 = 23 R = 18 + 3 = 21 => P/S = 3/4 => S = 4/3 x 18 = 24 Required Difference = S - R = 24 - 21 = 3. 1 321 Q: Present ages of Deepa and Hyma are in the ratio of 5:6 respectively. After four years, this ratio becomes 6: 7. What is the difference between their present ages? A) 6 yrs B) 4 yrs C) 8 yrs D) 2 yrs Explanation: Let the present age of Deepa = 5p Let the present age of Hyma = 6p After four years their ratio = 6 : 7 => => 35p + 28 = 36p + 24 => p = 4 Therefore, the difference between their ages = 24 - 20 = 4 years. 6 459 Q: The ratio between present ages of Renuka and Sony is 5 : 4 . 4 years ago, Sony's age was 24 years. What will be Renuka's age after 5 years ? A) 25 B) 30 C) 35 D) 40 Explanation: Present age of Sony = 24 + 4 = 28 years After 5 years Renuka's age = 7×5 + 5 = 40 years 8 431 Q: Hari Ram’s present age is three times his son’s present age and two fifth of his father’s present age. The average of the present age of all of them is 46 years. What is the difference between Hari Ram’s son’s present age and Hari Ram’s father’s present age? A) 44 yrs B) 56 yrs C) 67 yrs D) 78 yrs Explanation: Let Hari Ram's present age = x Then, his son's age = x/3 Father's age = 5x/2 Now the required difference = 6 415 Q: Sweety is 54 years old and her mother is 80. How many years ago was Sweety's mother three times her age? A) 20 B) 41 C) 26 D) 33 Explanation: Given P = 54 and M = 80 According to the question, Let ‘A’ years ago, M = 3P => M - A/P - A = 3 => 80 - A = 3(54 - A) => 80 - A = 162 - 3A => 2A = 82 => A = 41 Therefore, A = 41 years ago Sweety’s mother was 3 times of Sweety’s age. Hence, at the age S = 13 and M = 39. 6 540 Q: 7 years ago, the ratio of the ages of Anirudh to that of Bhavana, was 7:9. Chandhu is 12 years older than Anirudh and 12 years younger than Bhavana. What is Chandhu's present age? A) 91 years B) 115 years C) 103 years D) Can't be determined Explanation: Let present Anirudh's age be 'A' and Bhavana's age be 'B' Given seven years, their ratio is 7:9 => A-7 : B-7 = 7 : 9 => 9A - 7B = 14 .......(1) And given, Chandhu is 12 years older than A nad 12 years younger than B => C = A + 12....(2) => B = C + 12 => B = A + 12 + 12 .....(From (2)) => B = A + 24 ....(3) Put (3) in (1) 9A - 7(A+24) = 14 9A - 7A - 168 = 14 2A = 182 => A = 91 years => C = A + 12 = 91 +12 = 103 years. 7 472 Q: The average age of A and B, 2 years ago was 26. If the age of A, 5 years hence is 40 yrs, and B is 5 years younger to C, then find the difference between the age of A and C? A) 11 yrs B) 9 yrs C) 7 yrs D) 13 yrs Explanation: Let the present ages of A and B be 'x' and 'y' respectively From the given data, [(x-2) + (y-2)]/2 = 26 => x+y = 56 But given the age of A, 5 years hence is 40 yrs => present age of A = 40 - 5 = 35 yrs => x = 35 => y = 56 - 35 = 21 => Age of B = 21 yrs Given B is 5 years younger to C, => Age of C = 21 + 5 = 26 yrs => Required Difference between ages of A and C = 35 - 26 = 9 yrs. 5 581 Q: The average age of husband, wife and their child 3 years ago was 24 years and that of wife and child 5 years ago was 25 years. The present age of the husband is: A) 21 yrs B) 27 yrs C) 29 yrs D) 31 yrs<\|endoftext\|>	4.65625
329	The disease-causing organisms transferred by the insect may be viruses (termed “arboviruses”, an abbreviation of arthropod-borne viruses), bacteria (including rickettsias), protists, or filarial nematode worms. Replication of these parasites in both vectors and hosts is required and some complex life cycles have developed, notably amongst the protists and filarial nematodes. The presence of a parasite in the vector insect (which can be determined by dissection and microscopy and/or biochemical means) generally appears not to harm the host insect. When the parasite is at an appropriate developmental stage, and following multiplication or replication (amplification and/or concentration in the vector), transmission can occur. Transfer of parasites from vector to host or vice versa takes place when the blood-feeding insect takes a meal from a vertebrate host. The transfer from host to previously uninfected vector is through parasite-infected blood. Transmission to a host by an infected insect usually is by injection along with anticoagulant salivary gland products that keep the wound open during feeding. However, transmission may also be through deposition of infected feces close to the wound site. In the following survey of major arthropod-borne disease, malaria will be dealt with in some detail. Malaria is the most devastating and debilitating disease in the world, and it illustrates a number of general points concerning medical entomology. This is followed by briefer sections reviewing the range of pathogenic diseases involving insects, arranged by phylogenetic sequence of parasite, from virus to filarial worm.<\|endoftext\|>	3.71875
109	In this activity, students graph and analyze data from sediments collected off the coast of Santa Barbara, California to determine whether this information can be used to study historical climate change. Secrets of the Sediments Lesson Plan Overview Lesson PlansSecrets of the Sediments Lesson Plan OLP#1: The Earth has one big ocean with many features. OLP#2: The ocean and life in the ocean shape the features of the Earth. OLP#7: The ocean is largely unexplored.<\|endoftext\|>	3.78125
3,446	# 9.3: Properties of Chords Difficulty Level: At Grade Created by: CK-12 ## Learning Objectives • Find the lengths of chords in a circle. • Discover properties of chords and arcs. ## Review Queue 1. Draw a chord in a circle. 2. Draw a diameter in the circle from #1. Is a diameter a chord? 3. \begin{align}\triangle ABC\end{align} is an equilateral triangle in \begin{align}\bigodot A\end{align}. Find \begin{align}m\widehat{BC}\end{align} and \begin{align}m \widehat{BDC}\end{align}. 4. \begin{align}\triangle ABC\end{align} and \begin{align}\triangle ADE\end{align} are equilateral triangles in \begin{align}\bigodot A\end{align}. List a pair of congruent arcs and chords. Know What? To the right is the Gran Teatro Falla, in Cadiz, Andalucía, Spain. Notice the five windows, \begin{align}A-E\end{align}. \begin{align}\bigodot A \cong \bigodot E\end{align} and \begin{align}\bigodot B \cong \bigodot C \cong \bigodot D\end{align}. Each window is topped with a \begin{align}240^\circ\end{align} arc. The gold chord in each circle connects the rectangular portion of the window to the circle. Which chords are congruent? Recall from the first section, a chord is a line segment whose endpoints are on a circle. A diameter is the longest chord in a circle. ## Congruent Chords & Congruent Arcs From #4 in the Review Queue above, we noticed that \begin{align}\overline{BC} \cong \overline{DE}\end{align} and \begin{align}\widehat{BC} \cong \widehat{DE}\end{align}. Theorem 9-3: In the same circle or congruent circles, minor arcs are congruent if and only if their corresponding chords are congruent. In both of these pictures, \begin{align}\overline{BE} \cong \overline{CD}\end{align} and \begin{align}\widehat{BE} \cong \widehat{CD}\end{align}. In the second circle, \begin{align}\triangle BAE \cong \triangle CAD\end{align} by SAS. Example 1: Use \begin{align}\bigodot A\end{align} to answer the following. a) If \begin{align}m \widehat{BD} = 125^\circ\end{align}, find \begin{align}m \widehat{CD}\end{align}. b) If \begin{align}m \widehat{BC} = 80^\circ\end{align}, find \begin{align}m \widehat{CD}\end{align}. Solution: a) \begin{align}BD = CD\end{align}, which means the arcs are equal too. \begin{align}m \widehat{CD} = 125^\circ\end{align}. b) \begin{align}m \widehat{CD} \cong m \widehat{BD}\end{align} because \begin{align}BD = CD\end{align}. \begin{align}m\widehat{BC} + m \widehat{CD} + m\widehat{BD} & =360^\circ\\ 80^\circ+2m\widehat{CD}& =360^\circ\\ 2m\widehat{CD} & = 280^\circ\\ m\widehat{CD} & = 140^\circ\end{align} Investigation 9-2: Perpendicular Bisector of a Chord Tools Needed: paper, pencil, compass, ruler 1. Draw a circle. Label the center \begin{align}A\end{align}. 2. Draw a chord. Label it \begin{align}\overline{BC}\end{align}. 3. Find the midpoint of \begin{align}\overline{BC}\end{align} using a ruler. Label it \begin{align}D\end{align}. 4. Connect \begin{align}A\end{align} and \begin{align}D\end{align} to form a diameter. How does \begin{align}\overline{AD}\end{align} relate to \begin{align}\overline{BC}\end{align}? Theorem 9-4: The perpendicular bisector of a chord is also a diameter. If \begin{align}\overline{AD} \perp \overline{BC}\end{align} and \begin{align}\overline{BD} \cong \overline{DC}\end{align} then \begin{align}\overline{EF}\end{align} is a diameter. If \begin{align}\overline{EF} \perp \overline{BC}\end{align}, then \begin{align}\overline{BD} \cong \overline{DC}\end{align} and \begin{align}\widehat{BE} \cong \widehat{EC}\end{align}. Theorem 9-5: If a diameter is perpendicular to a chord, then the diameter bisects the chord and its corresponding arc. Example 2: Find the value of \begin{align}x\end{align} and \begin{align}y\end{align}. Solution: The diameter perpendicular to the chord. From Theorem 9-5, \begin{align}x = 6\end{align} and \begin{align}y = 75^\circ\end{align}. Example 3: Is the converse of Theorem 9-4 true? Solution: The converse of Theorem 9-4 would be: A diameter is also the perpendicular bisector of a chord. This is not true, a diameter cannot always be a perpendicular bisector to every chord. See the picture. Example 4: Algebra Connection Find the value of \begin{align}x\end{align} and \begin{align}y\end{align}. Solution: The diameter is perpendicular to the chord, which means it bisects the chord and the arc. Set up an equation for \begin{align}x\end{align} and \begin{align}y\end{align}. \begin{align}(3x-4)^\circ& =(5x-18)^\circ \qquad y+4=2y+1\\ 14^\circ& =2x \qquad \qquad \qquad \ \ \ 3=y\\ 7^\circ& =x \qquad \end{align} ## Equidistant Congruent Chords Investigation 9-3: Properties of Congruent Chords Tools Needed: pencil, paper, compass, ruler 1. Draw a circle with a radius of 2 inches and two chords that are both 3 inches. Label like the picture to the right. This diagram is drawn to scale. 2. From the center, draw the perpendicular segment to \begin{align}\overline{AB}\end{align} and \begin{align}\overline{CD}\end{align}. You can use Investigation 3-2 3. Erase the arc marks and lines beyond the points of intersection, leaving \begin{align}\overline{FE}\end{align} and \begin{align}\overline{EG}\end{align}. Find the measure of these segments. What do you notice? Theorem 9-6: In the same circle or congruent circles, two chords are congruent if and only if they are equidistant from the center. The shortest distance from any point to a line is the perpendicular line between them. If \begin{align}FE = EG\end{align} and \begin{align}\overline{EF} \perp \overline{EG}\end{align}, then \begin{align}\overline{AB}\end{align} and \begin{align}\overline{CD}\end{align} are equidistant to the center and \begin{align}\overline{AB} \cong \overline{CD}\end{align}. Example 5: Algebra Connection Find the value of \begin{align}x\end{align}. Solution: Because the distance from the center to the chords is congruent and perpendicular to the chords, the chords are equal. \begin{align}6x-7& = 35\\ 6x& = 42\\ x& =7\end{align} Example 6: \begin{align}BD = 12\end{align} and \begin{align}AC = 3\end{align} in \begin{align}\bigodot A\end{align}. Find the radius. Solution: First find the radius. \begin{align}\overline{AB}\end{align} is a radius, so we can use the right triangle \begin{align}\triangle ABC\end{align}, so \begin{align}\overline{AB}\end{align} is the hypotenuse. From Theorem 9-5, \begin{align}BC = 6\end{align}. \begin{align}3^2+6^2& =AB^2\\ 9+36&=AB^2\\ AB&=\sqrt{45}=3\sqrt{5}\end{align} Example 7: Find \begin{align}m\widehat{BD}\end{align} from Example 6. Solution: First, find the corresponding central angle, \begin{align}\angle BAD\end{align}. We can find \begin{align}m \angle BAC\end{align} using the tangent ratio. Then, multiply \begin{align}m\angle BAC\end{align} by 2 for \begin{align}m\angle BAD\end{align} and \begin{align}m\widehat{BD}\end{align}. \begin{align}\tan^{-1} \left ( \frac{6}{3} \right ) & = m\angle BAC\\ m\angle BAC & \approx 63.43^\circ\\ m\angle BAD & \approx 2 \cdot 63.43^\circ \approx 126.86^\circ \approx m\widehat{BD}\end{align} Know What? Revisited In the picture, the chords from \begin{align}\bigodot A\end{align} and \begin{align}\bigodot E\end{align} are congruent and the chords from \begin{align}\bigodot B, \ \bigodot C,\end{align} and \begin{align}\bigodot D\end{align} are also congruent. We know this from Theorem 9-3. ## Review Questions • Questions 1-3 use the theorems from this section and similar to Example 3. • Questions 4-10 use the definitions and theorems from this section. • Questions 11-16 are similar to Example 1 and 2. • Questions 17-25 are similar to Examples 2, 4, 5, and 6. • Questions 26 and 27 are similar to Example 7. • Questions 28-30 use the theorems from this section. 1. Two chords in a circle are perpendicular and congruent. Does one of them have to be a diameter? Why or why not? 2. Write the converse of Theorem 9-5. Is it true? If not, draw a counterexample. 3. Write the original and converse statements from Theorem 9-3 and Theorem 9-6. Fill in the blanks. 1. \begin{align}\underline{\;\;\;\;\;\;\;\;\;} \cong \overline{DF}\end{align} 2. \begin{align}\widehat{AC} \cong \underline{\;\;\;\;\;\;\;\;\;}\end{align} 3. \begin{align}\widehat{DJ} \cong \underline{\;\;\;\;\;\;\;\;\;}\end{align} 4. \begin{align}\underline{\;\;\;\;\;\;\;\;\;} \cong \overline{EJ}\end{align} 5. \begin{align}\angle AGH \cong \underline{\;\;\;\;\;\;\;\;\;}\end{align} 6. \begin{align}\angle DGF \cong \underline{\;\;\;\;\;\;\;\;\;}\end{align} 7. List all the congruent radii in \begin{align}\bigodot G\end{align}. Find the value of the indicated arc in \begin{align}\bigodot A\end{align}. 1. \begin{align}m \widehat{BC}\end{align} 2. \begin{align}m\widehat{BD}\end{align} 3. \begin{align}m\widehat{BC}\end{align} 4. \begin{align}m\widehat{BD}\end{align} 5. \begin{align}m\widehat{BD}\end{align} 6. \begin{align}m\widehat{BD}\end{align} Algebra Connection Find the value of \begin{align}x\end{align} and/or \begin{align}y\end{align}. 1. \begin{align}AB = 32\end{align} 2. \begin{align}AB = 20 \end{align} 3. Find \begin{align}m\widehat{AB}\end{align} in Question 20. Round your answer to the nearest tenth of a degree. 4. Find \begin{align}m\widehat{AB}\end{align} in Question 25. Round your answer to the nearest tenth of a degree. In problems 28-30, what can you conclude about the picture? State a theorem that justifies your answer. You may assume that \begin{align}A\end{align} is the center of the circle. 1 & 2. Answers will vary 3. \begin{align}m\widehat{BC}=60^\circ, m\widehat{BDC}=300^\circ\end{align} 4. \begin{align}\overline{BC} \cong \overline{DE}\end{align} and \begin{align}\widehat{BC} \cong \widehat{DE}\end{align} ### Notes/Highlights Having trouble? Report an issue. Color Highlighted Text Notes Show Hide Details Description Tags: Subjects:<\|endoftext\|>	4.59375
3,896	Nature & Wildlife The word biodiversity simply means the variety of life found in a particular place. Peat bogs are home to a unique combination of species, and although they do not have as much biodiversity as a tropical rain forest, they are just as important, and are also under threat. Most plants and animals have specific requirements for their existence. The main building block of bogs, sphagnum moss, is an outstanding example as it will only succeed on waterlogged acid ground. This and many other plants contribute to biodiversity by occupying distinct ecological niches in the bog system. This niche is called a habitat. Human interference with habitats can affect their natural biodiversity by increasing or decreasing the amount of species present. Despite peat cutting in the past, Clara Bog has been much less developed than many other Irish raised bogs and has remained relatively intact. This has resulted in a rich variety of natural habitats and wetland species. Clara Bog contains the typical components of a raised bog; hummocks, hollows, flushes and lawns. There are also very unique topographical features that are not common to all raised bogs. Soak systems are open areas of water that are more nutrient rich as they are fed with groundwater and can support alkaline plant species thereby increasing the overall biodiversity of the area. Shanley’s Lough is found on the West bank and Lough Roe on the East Bank. Native Bog Woodland consisting mainly of Downy Birch (Betula pubescens) grows in a flushed area near Shanley’s Lough. This is a rare and protected priority habitat under EU legislation. The raised bog flora is very unique. The waterlogged and acidic peat is contrary to what most plants usually thrive in yet Clara Bog is covered in plants. That is because many of them are specially adapted to growing in adverse conditions. Sphagnum moss is without doubt the most important plant on the bog. It is also called the bog builder as it is the main peat forming plant of raised bogs. It has amazing absorbency qualities and can hold up to 20 times its own weight in water. This accounts for the high water content of the bog, typically 95% water. It is a curious little plant. It has no roots and lives in acid water with very few nutrients. It grows by absorbing nutrients from the moisture droplets in the air. The top part of the moss is living while the bottom part is dead. The dead part helps the living part by storing water in its cells to help prevent it from drying out during drier spells. There are about 24 species of sphagnum moss in Ireland; 13 of these grow in Clara Bog. They are like living carpet on the bog’s surface and are varied in colour. They occupy different moisture zones on the bog. In the bog pools you will find the bright green Sphagnum cuspidatum, while on the bog’s surface the crimson Sphagnum magellanicum and ochre Sphagnum papillosum can form a wet mosaic carpet. The ginger brown Sphagnum austinii forms tightly packed hummocks raising the water table above the bog’s surface. Besides Sphagnum there are many other moss types present in Clara Bog. A noticeable one, Leucobryum glaucum, forms tightly packed green “pincushions” that can grow as high as 50cm. Ling Heather can often be seen growing on the pincushions. Clara Bog is also the only known Irish site for Narrow Cruet-moss (Tetraplodon angustatus). Other interesting plants on the bog include the collection of carnivorous plants. Sundews, butterworts and bladderworts all trap and digest their insect prey. Attracted by the glistening and sweet-smelling “tentacles” of the sundew, a fly lands. The ultra-sensitive stalks bend inwards immediately, trapping the insect with their sticky glue. The creature struggles to free itself, only enmeshing itself further. In less than 15 minutes it is dead from asphyxiation or exhaustion. Then enzymes in the plant get to work, digesting the creature to create a nutrient soup for the plant to devour. Two types of sundew grow on Clara Bog; round-leaved sundew (Drosera rotundifolia) can be found in drier areas where there are pockets of bare peat and the larger Great Sundew (Drosera angelica) can be seen growing in wetter areas such as bog pools and remnant drains. The purple flower on the slender stem of the Butterwort may look attractive, but its leaves are deadly for insects. They are attracted by the wet appearance of the sticky mucus secreted by the leaves – once they land; there is no escape… the leaf edges can curl over to pour more of this “glue” on to the creature till it quickly expires. The yellow flowers of Bladderwort can be seen above the bog pools in summer. The feathery submerged leaves have many tiny bladders. These are used to trap insects with a unique trap door and trigger hair mechanism. When the prey touches off the trigger hair, the trap door opens and the insect get sucked into the bladder. The door closes then the unfortunate creature becomes trapped and eventually digested by the plant. Bog cotton, perhaps the quintessential plant of bogs, can give the distant appearance of a snow covered landscape even on a sunny day. Two types of bog cotton grow on Clara Bog. The first to appear is Hare’s-tail Cottongrass (Eriphorum vaginatum) with its single fluffy head followed by Common Cottongrass (Eriophorum angustifolium) with its multi-cottony flower heads. Heather, another classical plant of bogs, begins to swathe the drier areas in a deep pink colour from mid-summer onwards. The more dominant and bushier Ling Heather (Calluna vulgaris) can provide food for hungry birds and insects in the form of nectar, leaves and seeds. The more slender Cross-leaved Heath (Erica tetralix) weaves it way amongst the Ling Heather and has beautiful pink to magenta bell- shaped flowers. Few counties can have a small bog plant on their coat of arms, but County Offaly is proud of the Bog Rosemary. It is easily distinguished by its leathery rolled leaf edges and its delicate pink flowers and usually grows with Sphagnum and sometimes Cranberry. Bog Rosemary features as the emblem for the Irish Peat Society (IPS). Due to the destruction of lowland raised bogs to which it is restricted, this small member of the heather family is relatively local in its distribution in Ireland. Heath-spotted orchid (Dactylorhiza maculata) and Common-spotted Orchid (Dactylorhiza fuchsii) are plentiful in the drier areas and they often hybridize. The Lesser-butterfly Orchid (Platanthera bifolia) with its almost luminous flower spike colonise similar areas. Marsh Helleborine (Epicactus palustris) and Common Twayblade (Listera ovata) are found on either side of the boardwalk before it ascends onto high bog. Several bog-loving berry plants thrive on top of the Sphagnum mosses. Cranberry stems wind their way along the top of wet Sphagnum producing red berries that are edible. These can be seen from the board walk. Bilberries (Vaccinium myrtillus) are plentiful in the Bog woodland and provide food for passing mammals such as Fallow Deer. The Bog Woodland is visible from the board walk on the high bank looking in south westerly direction. Crowberry (Empetrum nigrum) also occurs close to this wooded area. Other plants that can be seen on Clara Bog include Deergrass (Trichophorum germanicum), White Beak-Sedge (Rhynchospora alba), Bogbean (Menyanthes trifoliata), Bog Asphodel (Narthecium ossifragum), Bog-myrtle (Myrica gale), Meadow or Bog Thistle (Cirsium dissectum), Purple Moor-grass (Molinia caerulea), Carnation Sedge (Carex panicea), and Heath Milkwort (Polygala serpyllifolia). The laying of limestone gravel beneath the boardwalk on the low bog near the roadside has buffered the acidity of the peat and allowed a community of more alkaline loving plants to flourish. Bird’s-foot Trefoil (Lotus corniculatus), Lady’s Bedstraw (Galium verum), Yarrow (Achillea millefolium), Burnet-saxifrage (Pimpinella saxifraga), Wild Carrot (Daucus carota), Quaking Grass (Briza media) and Columbine (Aquilegia spp.) are some of the plants to be seen. They attract many insects such as moths, butterflies, bees, hoverflies and beetles especially when they are in flower. Their presence enriches the overall biodiversity of the area. Lichens and fungi belong to their own kingdoms separate from the plant kingdom. Few fungi are found but the brown toadstool ……… is often seen growing on top of Sphagnum moss. Noteworthy are the lichen community. Lichens are unusual living organisms. They are an assemblage of two or three living organisms that have a symbiotic relationship. The dominant partner is a fungus and it can co-exist with either or both an algae and a cyanobacteria. The latter two provide the photosynthetic part for the lichen. Cladonia lichens are the main types that grow on the bog surface. They have a grey appearance. The most obvious ones are the matchstick lichen (Cladonia floerkeana) with its bright red tips. It is found on drier more disturbed areas and the reindeer or antler lichen (Cladonia portentosa) which can grow in large sponge-like clumps in both wetter and drier areas. Viewing this lichen using a hand lens, the “antler” look becomes apparent. When wet, the antler lichen has a soft spongy texture but is brittle and rough when dried out. Lichens are amongst the oldest living things on earth and in the past they had many uses ranging from dyes to antibiotics and some were even used as packing material for ancient Egyptian mummies. Clara Bog is a haven for invertebrates. On a fine summer’s day many dragonflies can be seen darting and gliding over the bog pools. Their transparent wings and colourful bodies may look pretty, but they are fearsome ambush attackers. They spot their prey using their 30,000 eye lenses and can fly at speeds of up to 30 miles an hour. They are formidable aerial acrobats, able to fly in every direction as well as hover – better than helicopters! Four-spotted chasers (Libellula quadrimaculata), Common Hawkers (Aeshna juncea) and Common Darters (Sympetrum striolatum) are often seen. The Keeled Skimmer (Orthetrum coerulescens), a powder-blue dragonﬂy has been recorded from a soak area. Damselflies are also commonly seen and the Scarce Blue-tailed Damselfy (Ischnura pumilio) has been recorded on Clara Bog. Many spiders weave their intricate webs amongst the heathers and surrounding vegetation ready to pounce on their prey. The Raft spider (Dolomedes fimbriatus) is Ireland’s largest spider. It is only found in bogs and it does not spin a web. Instead, it waits at the edge of a pool; it can sense any vibration on the water surface from a potential prey through his sensitive forelegs. The unlucky prey may be met by the Raft spider as it quickly runs across the water’s surface in hot pursuit. Raft spiders are also known as the Jesus Christ spider for their ability to walk on water. The bog vegetation provides much needed food and shelter for moths. The bright green caterpillar of the Emperor moth (Saturina pavonia) has many black bristles that help it go along un-noticed. The adult Emperor moth has a trick to avoid being eaten. It rests with its wings open flat to display its four intimidating eye-spots and it can vibrate them to really frighten a predator! A rare moth, the Dark Tussock (Dicallomera fascelina) has been recorded from Clara bog, as has the Large Heath Butterﬂy (Coenonympha tullia), another moorland specialist. Ireland’s only protected insect, the Marsh Fritillary butterfly (Euphydryas aurinia), can also be found. Its specific food-plant, Devil’s-bit Scabious (Succisa pratensis) grows here. There are many other insects including hoverflies, beetles and grasshoppers. Clara Bog is the only known site in Ireland for two rare midge species, and a click beetle. Two amphibious creatures; the Common Frog (Rana temporaria) and the Smooth Newt (Lissotriton vulgaris) complete their life-cycles both in the bog pools and on the bog’s surface. Both are protected species. Another interesting species that is often spotted basking on the boardwalk on a warm sunny day is the Common Lizard (Lacerta vivipara), Ireland’s only native reptile. Atypical of reptiles the Common Lizard gives birth to live young rather than laying eggs. They are about 10 cm to 16 cm in length. When startled they have the uncanny ability to drop their tail, this gives them the chance to escape from a predator who may be distracted by the tail especially as it can twitch after being dropped. The lizard will eventually grow a new one. Besides peat lands they also live in coastal and woodland habitats. Some of the bird life found on bogs is unique. Wading birds such as Curlew (Numenius arquata) and Snipe (Gallinago gallinago) have long legs and so are specially adapted to living in wetlands such as Clara Bog. Resident Curlews have a red-listed status in Ireland; they are of the highest conservation concern due to a rapid decline in their numbers in recent decades. Thankfully the conservation of Clara Bog has provided a habitat for breeding curlew. During the month of May and June they are most vocal and their distinctive “cur lee” “cur lee” call is heard ascending from the bog. The Snipe produces a most distinctive drumming sound when performing display flights. The sound is actually a vibration from the breeze passing through its stiff tail feathers. The relatively rare Merlin (Falco columbarius), Ireland’s smallest bird of prey, has bred on Clara Bog. It used to nest on the ground but now it usually takes up residence in trees. It is a master of the surprise attack, coming in low and snatching its prey, before plucking and decapitating it on a boulder or fence post. Able to catch birds four times its own weight, it could easily deal with the Snipe. Skylarks (Alauda arvensis) rise up from the ground high into the sky (a staggering 50 to 100 metres!) and continuously warble until they descend with paratrooper-like precision back down to the ground. They can sing their song for extended periods of time and have been known to sing for up to half an hour non-stop. Other birds to be seen and heard on the bog include Woodcock (Scolopax rusticola), Mallard duck (Anas platyrhynchos), Meadow pipit (Anthus pratensis), Stonechat (Saxicola torquata), Kestrel (Falco tinnunculus) and during the summer months Swallow (Hirundo rustica) and Swift (Apus apus) can be seen flying over the bog feeding on aerial insects. Winter migrant visitors include Hen Harrier (Circus cyaneus) and Short eared owl (Asio flammeus). Along with the Peregrine Falcon (Falco peregrinus) they are known to hunt over the bog. Red grouse (Lagopus lagopus) and Greenland white-fronted Goose (Anser albifrons flavirostris) used to occur on the site but have become locally extinct. Larger mammals tend to use the bog more as hunting or feeding ground but the Irish Hare (Lepus timidus hibernicus) makes its home called a form, a hollowed out area amongst the bog vegetation. The Irish hare is a distinct sub species of mountain hare endemic to Ireland. It is thought to have been here since before the last Ice Age, and may be the country’s oldest surviving mammal. Unlike its relative in Scotland and elsewhere, its coat remains brown throughout the year. It now appears to be under threat as its usual habitat range is being affected by changes in land use, and its numbers have declined dramatically in the last 30 years. In the wooded areas immediately surrounding the bog there are Fox (Vulpes vulpes), Badger (Meles meles) and Pine Marten (Martes martes). The Pine Marten or Tree Cat as it is also known is an excellent climber. It belongs to the family Mustelidae, which also includes otter, stoat, mink and badger. It has a cat-like appearance with a flatter more pointed head. It has long legs and its tail is long and fluffy, in fact much fluffier than the tail of a red squirrel. The coat is a rich chocolate brown colour with a creamy coloured chin and chest. They are carnivores and mostly hunt small animals and birds. If hungry they supplement their diet with fruit and invertebrates. Fallow deer (Dama dama) also visit the bog as is evident from droppings often seen near bilberry shrubs. Otter tracks and are sometimes seen near old drains. It is thought they are looking for eels that sometimes end up in the drains by accident.<\|endoftext\|>	3.734375
1,177	Radiation therapy is a type of treatment that uses high doses of radiation to eradicate, stop or slow the growth of cancer cells. It is one of the most common treatments for cancer, either by itself or along with other forms of treatment. At low doses, radiation is used as an x-ray to see inside your body and take images, such as your teeth at the dentist office or broken bones. Radiation used in cancer treatment works in much the same way, except that it is given at much higher energy. External beam radiation therapy is a local treatment, meaning that the radiation is aimed only at a specific part of your body. It comes from a machine outside your body that aims the radiation at your cancer cells. For example, if you have prostate cancer, you will get radiation in the pelvis area only and not the rest of your body. We use On-Board Imaging (OBI) to take x-rays of your internal anatomy to see exactly where to aim the radiation beam for daily treatment. Many people with cancer need radiation therapy. In fact, more than half of people with cancer receive radiation therapy. Sometimes, radiation therapy is the only kind of cancer treatment people need. Most people get external beam radiation therapy once a day, five days a week, Monday through Friday. Treatment lasts for eight to nine weeks, depending on the type of cancer you have and the goal of your treatment. Each treatment lasts about 15 minutes. Three-Dimensional Conformal Radiation Therapy (3D CRT) Three dimensional conformal radiation therapy is also called 3D CRT. 3D CRT is a process during which images obtained using CT (computed tomography) and/or MRI (magnetic resonance imaging) are used to create detailed, accurate, three-dimensional representations of a tumor and any surrounding organs. The information taken from these scans is fed directly into the radiotherapy planning computer, so doctors can see the treatment area in 3 dimensions. The dosimetrist then designs radiation beams that follow the shape of the tumor more closely, so the radiation beam spares healthy tissue as much as possible. Of course, doctors have always tried to avoid affecting as much healthy tissue as possible with radiotherapy. But the main benefit with conformal radiotherapy is that it is more precise, as it allows doctors to plan in 3D. Conformal radiotherapy can give a better chance of killing the cancer by delivering a higher dose of radiation straight to the tumor. Less healthy tissue is included in the radiotherapy field and so you are likely to have fewer long term side effects. If doctors took too much of the surrounding area out of the radiotherapy treatment field, there could be a risk of cancer cells being missed. But radiation oncologists are extremely precise in keeping the right balance between keeping all the cancer inside the radiation treatment field, while as much healthy tissue as possible is kept out. Intensity-Modulated Radiation Therapy (IMRT) SmartBeam IMRT is a type of radiation treatment involving varying (or modulating) the intensity of the radiation beam (in this case, X-rays), through the use of a multileaf collimator. It is a form of radiation therapy that uses computer generated images to plan and then deliver more tightly focused radiation beams to cancerous tumors than is possible with conventional radiotherapy. With this capability, clinicians can deliver a precise radiation dose that conforms to the shape of the tumor, while significantly reducing the amount of radiation to surrounding healthy tissues. Consequently, the technique can increase the rate of tumor control while greatly reducing adverse side effects. Intensity-modulated radiation therapy (IMRT) is the most technologically advanced and most precise method of external beam radiation therapy available. IGRT takes this technology one step further. It delivers high doses of radiation directly to the treatment area while sparing surrounding healthy tissue through the use of daily imaging. IGRT (Image Guided Radiation Therapy) is a type of treatment that uses x-rays daily before treatment to make sure you are aligned properly with radiation delivery. We have a Clinac iX medical linear accelerator from Varian Medical Systems, for treating cancer with Dynamic Image Guided Radiation Therapy (IGRT). This new robotic and automated technology combines a state-of-the-art treatment machine with sophisticated digital imaging, and tracking and monitoring tools, enabling doctors to deliver the most accurate treatments possible, even when treating tumors that move as the patient breathes. We are proud to bring this advanced and sophisticated technology to the Greater Austin area. This highly advanced system incorporates a high-quality imaging device that enables clinicians to position patients very precisely and to deliver the radiation dose directly to a targeted area with sub millimeter accuracy, guided by three-dimensional images of the patient’s anatomy. The system’s versatility makes it appropriate for treating a wide range of abnormalities, from small metastases to large tumors, even in cases where the tumor is close to critical structures like the spinal cord or the optic nerve. Altogether, these tools deliver extremely precise treatments with better outcomes and fewer side-effects. The Clinac iX with its On-Board Imager device is visually compelling. Housed in a specially designed vault visible through closed circuit television, the machine stands approximately nine feet tall and rotates 360 degrees around the patient to deliver radiation from several angles. The On-Board Imager is attached to the machine on a pair of robotic arms that extend out on either side of the patient to generate anatomical images that will guide the treatment. Sophisticated image-matching software shows patient anatomy on computer monitors outside the treatment room. At the Austin Center for Radiation Oncology, we use IMRT and IGRT for prostate cancer patients. These give patients the best dose distribution.<\|endoftext\|>	3.75
1,997	# GSEB Solutions Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 Textbook Questions and Answers. ## Gujarat Board Textbook Solutions Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 Question 1. How many 3-digit numbers can be formed by using the digits 1 to 9, if no digit is repeated? Solution: 3-digit numbers are to be formed. i.e., Three places are to be filled with digits 1 to 9. This can be done in 9 Ã— 8 Ã— 7 = 504 ways, Question 2. How many 4-digit numbers are there with no digit repeated? Solution: Let the digits be 0 to 9. 4-digit numbers are 10P4. This includes those numbers which have 0 in the beginning (one thousandâ€™s place) 3-digit numbers out of 9 digits 1-9 are 9P3. âˆ´ 4-digit numbers which do not have 0 in the beginning (on the extreme left) = 10P49P3 = 10 Ã— 9 Ã— 8 Ã— 7 – 9 Ã— 8 Ã— 7 = 9 Ã— 8 Ã— 7(10 – 1) = 9 Ã— 9 Ã— 8 Ã— 7 = 4536. Question 3. How many 3-digit even numbers can be made, using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated? Solution: Let 2 be fixed at unitâ€™s place. Now, we have 5 digits and 3 places are to be filled up. This can be alone in 5P2 ways. When unitâ€™s place is filled up both 4 or 6, again in each case, we have 5P3 numbers. âˆ´ 4-digit even numbers are 3 Ã— 5P2 = 3 Ã— 5 Ã— 4 = 60. Question 4. Find the 4-digit numbers that can be formed, using the digits 1, 2, 3, 4 and 5, if no digit is repeated. How many of these will be even? Solution: (i) Out of 5 digits, 4-digit numbers are to be formed. Such numbers are = 5P4 = 5 Ã— 4 Ã— 3 Ã— 2 = 120. (ii) When 2 is at unitâ€™s place, then remaining three places are filled in 4P3 ways = 4 Ã— 3 Ã— 2 = 24. When 4 is at unitâ€™s place, then 4-digit numbers are again = 24. âˆ´ Even 4-digit numbers = 2 Ã— 24 = 48. Question 5. From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman, assuming one person cannot hold more than one position? Solution: Out of 8 persons, chairman can be chosen in 8 ways. After selecting a chairman, we have to choose a vice chairman out of 7 persons. This can be done in 7 ways. Out of 8 persons, a chairman and a vice chairman can be chosen 8 Ã— 7 = 56 ways. Question 6. Find n, if n-1P3 : nP4 = 1 : 9. Solution: or n = 9. Question 7. Find r, if (i) 5Pr = 26Pr-1 (ii) 5Pr = 6Pr-1 Solution: (i) 5Pr = 26Pr-1 or (7 – r)(6 – r) = 12. â‡’ 42 – 13r + r2 = 12 or r2 – 13r + 30 = 0. â‡’ (r – 10)(r – 3) = 0 â‡’ r = 3 or 10. r â‰ 10, since r cannot be greater than n. âˆ´ r = 3. (ii) 5Pr = 6Pr-1 or (7 – r)(6 – r) = 6 or 42 – 13r + r2 = 6. or r2 – 3r + 36 or (r – 9)(r – 4) = 0. â‡’ r = 9 or 4. So, r = 4, since r â‰ 9, because r cannot be greater than 4. Question 8. How many words can be formed using all letters of the word EQUATION, using each letter exactly once? Solution: Number of letters in equation = 8. âˆ´ Number of letters to be taken at a time = 8. If p is the number of words thus formed, then P = P(8, 8) = $$\frac{8!}{(8 – 8)!}$$ = 8! = 8 Ã— 7 Ã— 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 = 40320. Question 9. How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if (i) 4 letters are used at a time? (ii) all letters are used at a time? (iii) all letters are used but first letter is a vowel? Solution: (i) MONDAY has 6 letters. 4 letters are taken at a time. âˆ´ Number of words = 6P4 = 6 Ã— 5 Ã— 4 Ã— 3 = 360. (ii) All the letters of word MONDAY are taken at a time. Number of words = 6! = 6 Ã— 5 Ã— 4 Ã— 3 Ã— 2 Ã— 1 = 720 (iii) Let the words begin with A. Number of words formed with 5 letters = 5! = 120 Similarly, when words begin with â€˜Oâ€™, number of such words = 120. âˆ´ Number of words beginning with vowel = 2 Ã— 120 = 240. Question 10. In how many of the distinct permutations of the letters in MISSISSIPPI, do the four ‘Iâ€™s not come together? Solution: In the given word their are 4I, 4S, 2P and 1M. Total number of permutations with no restriction = $$\frac{11!}{4!4!2!}$$ If we take 4I as one letter, then total letters become = 11 – 4 + 1 = 8. If p is the permutations, when 41s are not together, then = 34650 – 840 = 33810. Question 11. In how many ways, can the letters of the word PERMUTATIONS be arranged, if the (ii) Vowels are together? (iii) There are always 4 letters between P and S? Solution: (i) Letters between P and S are ERMUTATION. These 10 letters having T two times. These letters can be arranged in $$\frac{10!}{2!}$$ ways = 1814400 ways. (ii) There are 12 letters in the word PERMUTATIONS which have T two times. Now the vowels a, e, i, o, u are together. Let it be considered in one block. The letters of vowels can be arranged in 5! ways. Thus, there are 7 letters and 1 block of vowels with T two times. âˆ´ Number of arrangements = $$\frac{8!}{2!}$$ Ã— 5! = $$\frac{8!5!}{2}$$ = 2419200. (iii) There are 12 letters to be arranged in 12 places. These 12 letters are filled in 12 places as shown above. P may be filled up at place No. 1, 2, 3, 4, 5, 6, 7 and consequently S may be filled up at place No. 6, 7, 8, 9, 10, 11, 12 leaving four places in between. Now P and S may be filled up in 7 ways. Similarly, S and P may be filled of in 7 ways. Thus, P and S or S and P can be filled up is 7 + 7 = 14 ways. Now remaining 10 places can be filled by E, R, M, U, T, A, T, I, O, N in $$\frac{10!}{2!}$$ ways. âˆ´ No. of ways in which 4 letters occur between P and S = $$\frac{10!}{2!}$$ Ã— 14 = $$\frac{3628800}{2}$$ Ã— 14 = 2540100.<\|endoftext\|>	4.78125
2,007	Radiant barrier is a highly reflective material, typically made of aluminum, which blocks the transfer of radiant heat by reflecting it away from its surface offering a permanent way to reduce your monthly utility costs. Radiant barrier foil insulation systems BLOCK radiant heat energy instead of absorbing it like fiberglass insulation. Because this foil insulation is unaffected by humidity, it will continue to perform at a consistent level no matter how humid the seasonal climate. A radiant barrier insulation system is a layer of reinforced aluminum foil facing an airspace inside a building. In order to fully understand how a radiant barrier works and how one can benefit you, the following concepts and radiant barrier information will enhance your understanding. Heat moves from one location to another in three different ways: conduction, convection, and radiation. Conduction: this method of heat transfer occurs when two solid objects touch each other. The heat is transferred between the two objects via molecular motion. For example, if your hand touches the surface of a light bulb, the heat from the glass will be transferred to your hand. Convection: heat that moves through fluids, such as water, rising heated air, or steam is doing so by the convection. In air, convection results in rising heated air causing the cooler air to drop to the floor. This creates what is called a free convection loop. One can create their own convection loop by using a fan to circulate the air in a room. This creates what is called a forced convection loop. Radiation: heat that is transferred by infrared waves is called radiation. These rays are invisible to the naked eye and are not affected by air flow. Stepping out into the sun on a hot day, you will feel your skin heat up from the heat radiating from the sun. And you will still feel these heat rays regardless of how windy it is outside. Radiation is the transfer of heat from a warmer object across an air space (or vacuum) to a colder surface and all objects radiate some of their heat in varying degrees from 0% to 100% whether it be the surface of a car that's been sitting in the sun or an electric stovetop burner. Other examples of radiant heat transfer: Roof shingles which have absorbed heat via radiant heat transfer from the sun. Heat radiating from a furnace. Most people are familiar with traditional insulating materials such as fiberglass, cellulose, Styrofoam, and rock wool. These products absorb or slow down convective and conductive heat transfers to insulate. These types of insulation do not BLOCK heat - only slow it down. Therefore, after a period of time, 100% of the heat absorbed would eventually transfer through the insulation. The rate in which this heat eventually transfers through an insulation material is the material's R-Value. Fiberglass and blown-in cellulose insulation are manufactured with air spaces inside designed to reduce heat conduction through the material. They also restrict heat transfer by convection by trapping air flows and lowering air circulation. Similar to fiberglass and blown-in insulation, some foam insulations, comprised of hydrochlorofluorocarbons (HCF), also absorb conductive and convective heat. However, HCF's found in some foam insulation products have been found to be extremely potent greenhouse gases and are being phased out by the United States over the next 23 years. A radiant barrier is designed to BLOCK (reflect) radiant heat energy unlike traditional insulations that are designed to slow it down by absorbing it. A radiant barrier can also REDUCE heat transfer caused by convection by blocking convective air flow. How does a radiant barrier reflect/BLOCK radiant heat? There are two physical properties of aluminum that a radiant barrier utilizes to reduce the transfer of radiant heat. A Radiant barrier REFLECTs radiant heat that strikes its surface across an air space from a heat source and conversely, it EMITs very little radiant heat from its surface across an air space opposite a heat source. No matter how you plan to install a radiant barrier, it MUST have at least one air space of at least 3/4 of an inch on either side to be effective at BLOCKING radiant heat. It does NOT matter which side of the radiant barrier the air space is located. The purpose of the air space is to prevent conductive heat transfer. If a radiant barrier does not have at least ONE air space on either side of it, heat will conduct from the surface touching the radiant barrier, through the barrier, and then transfer to the next surface touching the radiant barrier on the opposite side therefore, giving you no protection against the heat you intend to block. Therefore, as long as the air space requirement is achieved, a radiant barrier will be effective at BLOCKING radiant heat regardless of your application, i.e. interior/exterior walls, siding, roofing and attic locations, etc. Because a radiant barrier requires an air space on at least one side of itself to be able to BLOCK radiant heat, a radiant barrier CANNOT be installed directly underneath roofing materials where no air space exists. For example, if you install a radiant barrier on top of roof decking between the felt paper and asphalt shingles, it will NOT provide any benefits as the radiant heat would be transferred through the shingles, through the felt to the radiant barrier, and through the roof decking into the attic space (see image below). A radiant barrier can be effective with an asphalt shingle roof ONLY when installed inside the attic either to the underside of the roof decking or to the underside of the roof rafters. In these attic space applications, there is an air space below the radiant barrier. It is the existence of a single air space that eliminates, almost entirely, the pass-through of radiant heat. Our RadiantGUARD® radiant barriers REFLECT 95-97% of the radiant heat that strike their surface across and air space and conversely only EMIT 3% of the radiant heat from their surface facing an air space. The help you understand the more difficult concept of emissivity, imagine of a hot baked potato wrapped in aluminum foil. If you hold your hand close to the wrapped potato (not touching it), you would feel very little radiant heat coming off the aluminum because aluminum doesn't “emit” much heat across an air space (aluminum has a low emissivity factor). If you were then to touch the aluminum wrapped baked potato, you would feel a great deal of heat because the aluminum would then be conducting heat from the potato, through the aluminum, to your hand. Because you have lost the air space, the heat would contact (conduct) to your hand. A radiant barrier is ONLY effective when at least a 3/4" air space is provided on either side of itself regardless of the location of the heat source. If the air space is on the side of the heat source, the REFLECTIVITY property works to REFLECT the radiant heat. If the air space is on the opposite side of the heat source, the low EMISSIVITY property works to reduce the amount of radiant heat that EMITs from its surface. All building surfaces include roofs, ceilings, and even conventional fiberglass and blown-in insulation radiate heat in varying degrees. Radiant heat from the sun strikes the outer surfaces of roofs and walls and is absorbed causing building surfaces to heat up. This absorbed heat moves through the material (via conduction) to the opposite side and is then radiated from itself into attics and living spaces increasing the temperatures inside the building. Installing a radiant barrier is a MUST to combat the major form of heat transfer (radiant) that is currently not being controlled by your conventional insulation. Per the Department of Energy (DOE), a product classified as a "radiant barrier" MUST have a low emittance of 10% or less and a high reflectance of 90% or more. RadiantGUARD radiant barriers have an emittance of only 3-5% and a reflectance of 95-97%; considerable better than the DOE's radiant barrier minimum classification requirements. RadiantGUARD radiant barriers reflect/BLOCK radiant heat; not just absorb or slow it down like other forms of insulation. RadiantGUARD radiant barriers are unaffected by humidity or ambient temperatures, unlike other forms of insulation, and therefore, perform at a consistent level at all times. RadiantGUARD radiant barriers reflect/BLOCK 95-97% of the radiant heat transfer and when installed in an attic space, they can result in a reduction of attic temperature below the radiant barrier of up to 30 degrees. Lowering the temperatures above living space ceilings provides a significant benefit by reducing air conditioning loads and energy usage. Our radiant barriers can: Reduce heat transfer from attic to living spaces by 16-42%, Extend the life of air conditioning unit, Increase the comfort level of a home or building, and Reduce monthly cooling bills up to 17%. RadiantGUARD radiant barriers are safe and easy to install: No breathing apparatus required Non-toxic / non-carcinogenic Clean and lightweight; easy to handle Installation requires no special tools or clothing Don't promote the growth of fungi or bacteria Provides no nest support for rodent or insect pests Class A / Class 1 Fire Rating Meets fire and smoke safety requirements of most federal, state and local building codes Require no maintenance Do not shrink RadiantGUARD Testing and Approvals: United States Testing Company Tennessee Valley Authority Tennessee Technological University State of California Quality Standards Oak Ridge National Laboratory Metro Dade County Texas A&M University<\|endoftext\|>	3.859375
1,288	2-1 Part I: Polynomial Functions The anatomy of a polynomial Here's a typical polynomial: 3x - 2x + 1 Note: exponents in polynomials must be positive integers Some terminology: This polynomial has three terms. The term having highest degree is called the leading term. The degree of the polynomial is the degree of the leading term. Polynomials can have one, two, three or more terms: #terms name examples one term monomial 3x2, 2x, 3 two terms binomial 2x + y, 3x2 - 1 three terms trinomial 3x - 2y + 5 Polynomials can be of degree 0, 1, 2, 3, or more: degree name examples 0 constant 3 1 linear 3x, 2y + 1 2 quadratic 2x2, -3x2 + 1 3 cubic 2x3, -3x3 + 2x We refer to: 3x2 + 2x + 1 as a quadratic trinomial 3x + 1 as a linear binomial, etc. Algebra and geometry of polynomials 1. How a graph "behaves" at its extreme left and right is called tail behavior, and is a geometric (graphical) concept. • for example, a graph may "shoot up" or "shoot down" at its extremes; i.e., • go off-scale to +oo or -oo 2. The number of turning points and number of x-intercepts of a graph are geometric concepts. Here's some geometry: This graph has ·· turning points and ·· x-intercepts. 3. The degree of a polynomial and the sign of the leading coefficient are algebraic concepts: f(x) = 7x2 - 3x4 + 7 (this is algebra) has degree ·· and the sign of the leading coefficient is ··. There are a couple of principles that tell us something about: • the algebra of a polynomial, by examining its graph • the graph of a polynomial, by examining its algebra The Polynomial Tail Principle Concerning the graph of a polynomial: left tail – behavior at extreme left of the graph right tail – behavior at extreme right of graph Polynomial Tail Principle for a polynomial, the tails will always go to +oo or -oo WHICH? Just look at its leading term: (1) its degree - even or odd (2) its sign - plus or minus sign + sign - degree even example: 3x4 left tail => +oo right tail => +oo picture: example: -3x4 left tail => -oo right tail => -oo picture: degree odd example: 3x3 left tail => -oo right tail => +¥ picture: example: -3x3 left tail => +oo right tail => -¥ picture: Summary (polynomial tails) even leading term: tails go same direction (up for +, down for -) odd leading term: tails go opposite (down-up for +, up-down for -) Example: f(x) = -7x2 - 3x4 + 7 leading term is even or odd? ·· tails go same or opposite ·· direction? coefficient of leading term is positive or negative? ·· direction is up or down? ·· The Polynomial Turning Points and Intercepts Principle In general, a polynomial of lower degree cannot create a graph with lots of turning points; lots of turning points implies a polynomial of higher degree. The same kind of statement applies as well to x-intercepts: lots of x-intercepts imply higher degree. It does not work the other way; a very high degree polynomial can produce a graph with few turning points and even no intercepts. The graph of f(x) = x20 looks pretty much like a parabola! If a polynomial is of odd degree, it must have at least one x-intercept (it has to cross the x-axis, by the Polynomial Tail Principle). Every polynomial f intercepts the y-axis somewhere (f(0) is always defined for a polynomial). To be specific, we have the following principles for "going from algebra (degree) to geometry (turning points/intercepts": degree-to-(turning points/intercepts) principle A polynomial of degree n has: n – 1 turning points or fewer n x-intercepts or fewer exactly 1 y-intercept Furthermore, if n is odd, it must have: at least 1 x-intercept Stating this same principle another way, for "going from geometry to algebra", we have: (turning points/intercepts)-to-degree principle A polynomial having n turning points has: degree n + 1 or greater A polynomial having n intercepts has: degree n or greater Hint: remember "turning points is 1 less than degree" (roughly) • think of a parabola (the graph of a polynomial of degree 2) • but it always has only 1 turning point! Note: • a polynomial of degree 6 could have 5,4,3,2, or 1 turning points • a polynomial of degree 9 could have 9, 8, 7, ... x-intercepts • but it must have at least 1 x-intercept . . . • think about the polynomial tail principle! • a polynomial with 3 turning points can only be of degree 4, 5, 6, ... Application: going from the algebra of a polynomial to its geometry: Example: f(x) = -7x2 - 3x4 + 7 The graph of f can have at most ·· turning points, and ·· x-intercepts. Application: going from the geometry of a polynomial to its algebra : The above graph has: ·· turning points ·· x-intercepts So, if it is the graph of a polynomial, it must be the graph of a polynomial of degree: ·· or greater<\|endoftext\|>	4.75
1,950	Pssst… we can write an original essay just for you. Any subject. Any type of essay. We’ll even meet a 3-hour deadline.Get your price 121 writers online Rosa Parks has been known for decades as the African American women who refused to give up her seat to a white passenger in Montgomery, Alabama. This bold move triggered bus boycotts all throughout the city. She is known now as a civil rights activist that was a significant driving factor behind desegregation of public facilities in the South. Parks was not the first black woman to refuse her seat to a white man, but yet she was seen as an inspiration (Klein 2013). There was something about her overwhelming courage, dedication, and pride that made her a leader all throughout the country in a matter of hours. Rosa Parks was an African American woman who was born in Tuskegee, Alabama on February 4, 1923. Her desire to push for civil rights came from her grandparents, whom she lived with as a child. They were former slaves that would constantly preach to Parks about the importance of equality (“Rosa Parks Biography”). She attended a segregated school, but received a fairly good education. By the time she was an adult, she found herself in Montgomery, Alabama working as a line-worker in a textile factory (“Rosa Parks Biography”). She began her life as an outward civil rights activist in 1943 by joining the National Association for the Advancement of Colored People, also known as the NAACP (“Rosa Parks Biography”). She was actively involved in the organization; she was a secretary of the president and was a youth leader throughout the organization. As far as Parks was concerned, if there was a will then there was a way and she was going to change the world by advocating civil rights. Parks was a determined, articulate, and brave young woman that was willing to do whatever it took to be an equal. On December 1, 1955 Parks got on a segregated bus after a long day of hard work. At this time, the social norm was for the whites to sit at the front of the bus while the blacks sat at the back. Parks boarded the bus and sat at one of the seats closest to the front that was still designated for blacks. Once the bus began to fill, the bus driver realized several whites were standing, waiting for seats, while many blacks were comfortably seated. At that point, the bus driver asked Rosa to give up her seat to the white citizens that were standing; Rosa refused. She felt that she should not have to give up her seat (“Rosa Parks Biography”). First and foremost, it was wrong to treat any race inferior to the other. But secondly, Rosa was seated in an “African American section.” She was seated where she was told to be seated, so why should she have to get up? The bus driver called the local police on Parks and she was arrested shortly after the incident, but was released on bail that same night (“Rosa Parks Biography”). Parks has since been treated as a hero, but a lot of the information taught about Parks is incorrect or speaks only the half-truth. The first truth to be told is that Parks was not the first African American woman to not want to give up her seat. There were three other women prior to her: Aurelia Browder, Mary Louise Smith, and Susie McDonald (Klein 2013). The second truth is that this act of civil disobedience was in no way pre-meditated. Parks did not know that she was going to be standing up for a cause she believed in that day. She simply had gotten on the bus and was not willing to be disrespected yet again. The third truth, and perhaps the largest in my opinion, is that Parks did not refuse to give up her seat due to her being tired. She even wrote in her own autobiography, “No, the only tired I was, was tired of giving in. (Klein 2013)” I feel that Parks was not taught in her correct historical context because she was meant to be a one of a kind, a true leader. The people needed Parks to be the spark of a revolution. If people were told that others had done the same as her prior and she had it planned for months, it would not seem as heroic. I believe that over time, the myths continued and began to spiral out of control. Parks was an incredible woman and I am very happy she was able to start a civil rights revolution, but I also believe her acts were greatly exaggerated. People wanted someone to stand up to the mistreatment, someone who feared nothing but fear itself, and someone who did what they always couldn’t; they made Parks this woman. After Parks’ encounter on the Montgomery bus, president of the NAACP, E.D. Nixon, began organizing a boycott of city buses all throughout Montgomery with Martin Luther King Jr. as the leader. He began placing local ads to urge other African Americans to stay off of all city buses on December 5, 1955; the day of Parks’ trial. There was a great success with over 40,000 African American commuters who supported Parks and her actions (“Rosa Parks Biography”). Since the initial boycott was so successful, the boycott continued with great success for many months. The buses sat empty and the city began wasting tax dollars on nothing. There were revolts throughout the city to protest the boycott, but the African Americans pressed on. By June 1956 racial segregation was deemed unconstitutional by the Montgomery district court (“Rosa Parks Biography”). Finally, after legislation was passed, bus segregation was ended all throughout the city so the boycott continued until December 20, 1956; lasting a total of 381 days of protest thanks to Parks and her incredible courage. In later years other campaigns, such as the “Don’t Buy Where You Can’t Work” campaign, used the Montgomery bus boycott as leverage. In the late 1920s and early 1930s this campaign was launched to urge blacks to not shop in white stores throughout their neighborhoods that they were incapable of working at (“Don’t Buy…”). As far as these citizens were concerned, why bother giving a store business when they are ignorant and refuse to let the community work within the business? The NAACP was also behind this campaign; urging the citizens to commit to a mass civil rights protest that could change the country, just like the Montgomery bus boycott. Thankfully for them, it did change the views of the people just like the prior campaign. Whites had no choice but to hire blacks for skilled and white-collar positions if they wanted to stay in business (“Don’t Buy…”). The campaign was a bustling success that once again showed the importance of peaceful protest and black activism. While there were dozens of people involved in the fight against segregated buses in cities all over America the majority seems to focus in on Rosa Parks. I feel this is because Rosa parks just so happened to be in the right place at the right time. Parks did this right in the most oppressed time for African Americans in the South. While slavery was no doubt an absolute tragedy, there is nothing more frustrating than being told you are free and yet still being treated as an inferior being. Parks was sick of it and so was everybody else; the only difference was that she decided to take a stand (or rather a seat). As I had mentioned prior, E.D. Nixon and the NAACP were leaders in the fight against segregation yet they are rarely mentioned by scholars today. Nixon was the president of NAACP and he arranged countless boycotts, rallies, civil disobedience movements, and personally dedicated his life to the cause. They were determined people who wanted to get all of the African American citizens fighting for legitimate freedom. Nixon and the NAACP did not want to sit back and accept this life of turmoil for what it was; they wanted action. As the old saying goes, “If you want something done right then you just have to do it yourself.” Nixon and the NAACP was a genuine example of this expression. Rosa Parks was an incredible woman with the opportunity to help direct the masses in a fight for equality. She did more than just refuse to give up a seat on a bus; she began a revolution. She taught tens of thousands of people what it can mean to stand up for not only a cause, but a dream. While Rosa is a true inspiration she did not do it on her own. She had the help of E.D. Nixon and the entire NAACP. With their help she managed to gain local support and begin the Montgomery bus boycotts that changed segregation legislation forever in the South. The Montgomery bus boycott strategy was used many years after in a variety of situations pertaining to more than just black activism. Parks was a true inspiration and leader with a heart full of courage, faith, and determination. To export a reference to this article please select a referencing style below: Sorry, copying is not allowed on our website. If you’d like this or any other sample, we’ll happily email it to you. Your essay sample has been sent. Want us to write one just for you? We can custom edit this essay into an original, 100% plagiarism free essay.Order now Are you interested in getting a customized paper?Check it out!<\|endoftext\|>	4.5
202	Common Core Connections: Math is the perfect tool for helping second grade students master Common Core math skills.The Common Core Standards for Math in second grade focuses on four critical areas: extending understanding of base ten-notation, building fluency with addition and subtraction, standard units of measure, and shapes. This resource provides focused practice pages for targeting and reinforcing these skills while helping students connect comprehension with knowledge and application. Connecting the standards to content has never been easier with the Common Core Connections series for Math. The Common Core Connections series provides teachers with the skill assessments to help determine individualized instruction needs. Focused, comprehensive practice pages and self-assessments guide students to reflection and exploration for deeper learning! Grade specific coherent content progresses in difficulty to achieve optimum fluency. It is also an ideal resource for differeniation and remediation. Each 96-page book includes an assessment test, test analysis, Common Core State Standards Alignment Matrix, and answer key. - Answer Key<\|endoftext\|>	4.53125
798	## APTITUDE SOLVED PROBLEMS Questions : 1. The ratio between the present ages of Arun and Arvind is 6 : 7. If Arvind is 4 years old than Arun, what will be the ratio of the ages of Arun and Arvind after 4 years ? 2. The volume of a cube is 729 cm3 . Find the surface area of a cube? 3. The sum of three numbers is 123. If the ratio between first and second numbers is 2:5 and that of between second and third is 3:4, then find the difference between second and the third number ? 4. A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. What is the rate of interest per annum ? 1. 7 : 8. 2. 486 cm2. 3. 15. 4. 12 %. 1. Explanation: Let Arun age and Aravind age is 6x years and 7x years. 7x – 6x = 4 x = 4 Required ratio = ( 6x + 4 ) : ( 7x + 4 ) = [ ( 6 * 4 ) + 4 ] : [ ( 7 * 4 ) + 4 ] = ( 24 + 4 ) : ( 28 + 4 ) = 28 : 32 = 7 : 8 ∴ The ratio of the ages will be 7 : 8. 2. Explanation: Volume of a cube = a 3 cm3 729 cm3 = a3 cm3 a3 cm3 = 93 cm3 Taking cube root on both sides a = 9 cm The surface area of a cube = 6 a2 cm2 = 6 * ( 92) cm2 = ( 6 * 81 ) cm2 = 486 cm2 ∴ The surface area of a cube is 486 cm2. 3. Explanation: a : b = 2 : 5 and b : c = 3 : 4 So, a : b : c = 6 : 15 : 20 ( 2 * 3 = 6 ), ( 5 * 3 = 15) or ( 3 * 5 = 15), ( 5 * 4 = 20 ) Total Ratio = 6 + 15 + 20 Total Ratio = 41 41x = 123 x = ( 123 / 41 ) x = 3 Difference between second and the third number = 20x – 15x = 5x = ( 5 * 3 ) = 15 ∴ The difference between second and the third number is 15. 4. Explanation: Simple interest for 3 years = Simple interest for 8 years – 5 years Simple interest for 3 years = ( 12005 – 9800 ) Simple interest for 3 years = 2205 Simple interest for 1 year = ( 2205 / 3 ) Simple interest for 5 years = ( 2205 / 3 ) * 5 = ( 735 * 5 ) = Rs. 3675 Principal = ( 9800 – 3675 ) = 6125 Rate = ( 100 * Simple Interest ) / ( Principal * Time ) Rate = ( 100 * 3675 ) / ( 6125 * 5 ) = ( 20 * 3675 ) / 6125 ( 73500 / 6125 ) = 12 % ∴ Rate of interest per annum is 12 %.<\|endoftext\|>	4.625
330	## 2 Points A circular garden has a diameter of 8 yards. What is the area of the garden? Use 3.14 for TT. O A. 12.56 ya? Question 2 Points A circular garden has a diameter of 8 yards. What is the area of the garden? Use 3.14 for TT. O A. 12.56 ya? O B. 25.12 yd2 O C. 200.96 yd2 D. 50.24 yd2 SUBMIT in progress 0 7 hours 2021-09-12T19:25:12+00:00 1 Answer 0 50.24 yd^2 Step-by-step explanation: The area of a circle can be found using: First, we need to find the radius. We have the diameter. The diameter is twice the radius, or d=2r We know that the diameter is 8, so we can substitute 8 in for d 8=2r Since we are trying to find r, we need to get r by itself. To do this, divide both sides by 2. 8/2=2r/2 4=r Now we know the radius, and can substitute 4 in for r. Also, we can substitute 3.14 in for pi. a=3.144^2 Solve the exponent a=3.1416 Multiply a=50.24 The area is 50.24 yards^2<\|endoftext\|>	4.5

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