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523	The following figure shows the annual average temperature (°C) based on 35 stations spread over the whole country. Sweden only covers a small portion of the earth’s surface, and the figure shows regional variations in comparison with the global values. For example the warm period during the 1930s and 1940s, including the cold war years, can be clearly seen in the Swedish data but are not nearly as obvious in the global data. The cold year in 2010 is another example of a regional variation since this year was globally one of the warmest. Similar variations in the global average temperature In general there are many similarities between the variations in the global and the Swedish temperature, for example the end of the 1800s was colder than the 1900s. Since 1988, all years except 1996 and 2010 have been warmer or much warmer than the average for 1961-1990, which is the normal period currently being used. This is in line with the consequences of an increased greenhouse effect. The annual temperatures in the diagram for Sweden are based on homogenised date since 1860. Homogenised data means that as well as correcting inaccuracies and interpolating missing data, consideration is also given to differences that can occur due to changing instrument (measurement method) or moving the measurement site. The temperature climate can vary by several tenths of a degree over a distance of just a few kilometres. After homogenisation the entire data period should behave as if it was measured at the same location with the same instruments and methods. The average temperature for Winter (December, January and February) together with deviations from the average value for the period 1961-1990. The average temperature for Spring (March, April and May) together with deviations from the average value for the period 1961-1990. The average temperature for Summer (June, July and August) together with deviations from the average value for the period 1961-1990. The average temperature for Autumn (September, October and November) together with deviations from the average value for the period 1961-1990. Temperature in the future The average temperature in Sweden is expected to continue to rise. By the end of the century the annual average temperature is predicted to be 2-6 °C higher than for the period 1961-1990, depending on which scenario is used. The temperature is expected to increase during all seasons, with the biggest increase in northern Sweden during winter. Winter is also the season with the greatest variations from year to year. This means that individual winters in the future can be significantly warmer or colder than the average climate.<\|endoftext\|>	3.78125
1,515	# Dividing with Decimals 18 teachers like this lesson Print Lesson ## Objective SWBAT: â¢ Make estimates of division problems â¢ Divide whole numbers and decimals â¢ Interpret a remainder #### Big Idea How can you split \$27 equally between 4 people? Students apply what they know about division to divide with whole numbers and decimals. ## Do Now 10 minutes Notes: • Before the lesson you will need to copy, cut, and organize the Division Matching cards. I like to label cards in sets (write #1 on all of first set, etc.) so that students know which cards are theirs. • I use the data about division from the pre-test in Unit 1 to Create Homogeneous GroupsFor the do now I have students work in partners or groups of 3. See my Do Now in my Strategy folder that explains my beginning of class routines. Often, I create do nows that have problems that connect to the task that students will be working on that day. Here, I want students to connect what is being described in the word problem with a picture. Students must think about what is going on and what the action will look like. After most students have finished most of the matches, I call the class together. I ask students to share their number sentences and pictures for #4 and #8. There is a subtle difference in these problems, which I want students to notice. A common mistake is that students confuse the situations that represent 24/4= 6 and 24/6= 4 ## Estimation 5 minutes Here I want students to quickly review the vocabulary word quotient and then work on estimating. I give students a couple minutes to make and write explanations for their estimates. I stress to students that estimates need to be relatively quick. I walk around and monitor student work. I’m looking to see what strategies students are using. Some students may round numbers to whole numbers and divide. Other students may make connections to multiplication. I am interested to see what students do with #3. If they struggle I ask, “Do you think the quotient is going to be bigger or smaller than 1.89? Why?” I call one student to explain one of their estimates. These students are students I observed using a particular strategy as I was walking around. I am not giving exact answers at this time. We will work on finding exact quotients later in the lesson. It is important that students are able to use their number sense to make reasonable estimates. ## Practice 15 minutes We do number 1 together. I ask students how they want to find the exact quotient. I also show them the algorithm. I stress that students need to use their number sense and the patterns they noticed earlier to decide if their final answer makes sense (MP1) Students work independently on the rest of the problems. They can check in with their partner if they are stuck. I Post A Key so that students can check their work when they finish a page. I am looking that students are making reasonable estimates and that they are successfully dividing using a strategy of their choice. If students successfully complete the practice problems, they can play the Leftovers From 100 game. If students are struggling, I may intervene in one of the following ways: • Ask them what their estimate is and how they got it. • Let them use a multiplication chart. • Give them a word problem situation that represents the problem. • Have them use the grids to organize their problems. • Pull a small group of students who are struggling to work together. ## Sharing \$\$ 10 minutes I have students participate in a Think Write Pair Share. Working with money forces students to interpret the remainder. I have a few students share and explain their strategies. Some students will use the picture, others will use the algorithm, others will make connections to multiplication. Here students are using MP 6: Attend to precision. We move on to the next page where I ask students to take a couple minutes to analyze the examples. What do they notice? What are the similarities? The differences? I give the students a couple minutes to jot down notes. I want students to see that all of the answers are technically correct, although not as useful as others. Problem A solves the problem but the answer doesn’t help. I also want students to start moving away from the partial quotient method and towards the standard algorithm. Problem B turns the remainder into a fraction, but we then need to figure out how to do that. A common struggle is that students don’t know where to put a decimal point. This can be easily remedied if students are able to make reasonable estimates. For example, offer \$67.50 as an answer. Hopefully students will quickly eliminate this as a possibility, since it is way too big! I want students to notice the differences between C and D. In problem C the person ignored the decimal point until the end, and then used estimation to decide where it goes. Problem D kept the decimal point in the dividend and brought it up to the quotient. I ask students how could these students check their answers. I want students to get into the habit of using multiplication to check their answers. ## More Practice 10 minutes Just like the previous Practice section, we do number 1 together. I ask students how they want to find the exact quotient. I stress that students need to use their number sense and estimates to decide if their final answer makes sense. Students work independently on the rest of the problems. They can check in with their partner if they are stuck. I Post A Key so that students can check their work when they finish a page. I am looking that students are making reasonable estimates and that they are successfully dividing using a strategy of their choice. If students successfully complete the practice problems, they can play the Leftovers From 100 game. If students are struggling, I may intervene in one of the following ways: • Ask them what their estimate is and how they got it. • Let them use a multiplication chart. • Give them a word problem situation that represents the problem. • Have them use the grids to organize their problems. • Pull a small group of students who are struggling to work together. ## Closure and Ticket to Go 10 minutes I begin the Closure by asking students to look at questions 3 and 4. What was your estimate? Why? How did you find the exact quotient. I look for students who used different strategies and I have them show and explain their work. If there is a common mistake I see students making, I will present it and ask students to address it here. I am also interested to see what students estimated with #4. Even if students struggle with estimating, they should be able to see that they answer is going to be smaller than 2.34, since you are dividing it by a number that is larger. With about five minutes left I pass out the Ticket to Go for students to complete independently. Then I pass out the HW Dividing with Decimals. I may also assign one of the “Facts about College” or “Working During College” pages for homework, depending on how much students finished.<\|endoftext\|>	4.53125
1,882	# Few notes on the Binomial and Fibonacci heaps 8 min read • Published: January 23, 2018 Having just implemented and tested a Fibonacci heap on a large dataset I thought I’d write up a bit about it, mostly so that I can reference this post later in the future, and to help me remember things I’ve learned better. Note that this blog post is not a tutorial on how to implement a Binomial/Fibonacci heap. First, let’s begin with a few definitions. Throughout the article we’ll be talking about min-heaps. The heap property of a tree says that the value in each node is less than or equal to the value of its children. It doesn’t say which values go to the left and which go to the right, so it doesn’t help us with searching. It only tells us that the minimum of any subtree is in its root. Also note that we are not restricting ourselves just to binary trees, this property works for any kind of tree. ## Binomial heap Before we can define a binomial heap, we need to define a binomial tree. We’ll use a recursive definition. 1. A binomial tree of rank 0 is a single node without any children. 2. A binomial tree of rank k is a tree where the root has exactly k children, which are, going from left to right, binomial trees of rank 0..k-1. To make things a little more confusing, here’s a picture from wikipedia, which uses uses a reverse order, putting children of lower rank to the right. Both definitions are equivalent. Before we move onto the binomial heap, let us prove a small property which will be useful later. A binomial tree of rank $k$ has $2^k$ nodes. For k=0 we get $2^0 = 1$, which is true. Now taking $k>0$, we know that a binomial tree of rank $k-1$ has $2^{k-1}$ nodes. We also know, that we can use two trees of rank $k-1$ and combine them into a binomial tree of rank $k$. If we do that, we get $2 \cdot 2^{k-1} = 2^k$ nodes in total. This shows that the number of children is logarithmic in the total number of nodes. Since increasing the rank by one increases the depth of the tree by one as well, we get that a binomial tree with $n$ nodes has the depth of $O(\log n)$ and also has $O(\log n)$ children at the root. A small sidenote, when we merge two binomial trees, in order to preserve the heap property, we have to put the one with a higher value in its root under the one with the lower value. Now moving onto a binomial heap, we define it as a list of binomial trees T1,…,Tk, which are sorted by their rank, each rank from 0 to $k$ occurs at most once, and each tree obeys the heap property. The operations we want from the binomial heap are the following: • Min: Finding the minimum, which can be either obtained in $O(\log n)$ by iterating the roots, or in $O(1)$ by keeping a separate minimum pointer and updating it along the other operations. • Merge: Taking two binomial heaps and merging them together, we simply iterate both lists, looking at the same rank at a time … if both heaps contain a tree of the same rank, we merge them together, creating a tree of one rank higher. We keep doing if there’s also a tree of one rank higher, much like we would carry over 1 in binary addition. This whole operating is $O(\log n)$, as it does the same exact operations as binary addition does, which can be shown as $O(\log n)$ using basic amortized analysis. • Insert: Adding a single item into the heap. We do this by creating a singleton heap with just one element and merging it into our heap. By definition this is also $O(\log n)$. • Build: Building a binomial heap from a list of $n$ elements. Unlike in a binary heap, we can simply call Insert for each element. I won’t go into why, but the complexity is just $O(n)$, not $O(n \log n)$ as we could expect. • ExtractMin: Removing a minimum from the heap, we take the tree with the minimum value at its root, remove it from the heap, create a new heap into which we insert all of its children, and merge that heap back into our initial heap. The whole operation is again $O(\log n)$. ## Lazy binomial heap We can go a bit further and make our binomial heap lazy. This will help us improve the amortized time of some of the operations. The only change we’re going to make is allow multiple trees of the same rank to co-exist in our binomial heap. We’ll simplify our Merge operation so that it works in constant time. Instead of doing all of those complicated operations, we simply take the both lists of trees of both heaps, and concatenate them together. This can be done in $O(1)$ when using double linked lists. (Note that the minimum pointer should be updated to the minimum of both heaps when doing this). We also modify our ExtractMin operation so that it performs a new operation called Consolidation. This will fix our heap so that it again looks like a binomial heap. In a nutshell, we’ll do a bucket sort on the lists of trees in our heap, and then merge trees in each bucket until there’s only one left (note that when we merge two trees we create a tree of a higher rank, so we move it one bucket up). Iterating the buckets from lower to higher ranks will result in a bucket lists where each bucket contains zero or one tree. We can convert this back to a binomial heap. The whole trick is that the consolidation itself is $O(\log n)$ (amortized), so we keep the amortized complexity of ExtractMin, but improve the complexity of Insert and Merge to $O(1)$. Also note that the worst case time of ExtractMin is $O(n)$. ## Fibonacci heap Going even further, we want our heap to also support the Decrease operation, which takes a pointer to a node and changes its key to a specific value. As doing this blindly could break the heap property, we have to do some tweaking to our data structure. A regular binomial heap can do a Decrease in $O(\log n)$ by simply propagating the decreased element as far up to the root as needed to maintain the heap property. But our Fibonacci heap will be able to do this in just $O(1)$!. To allow this, we tweak our definition of the binomial heap. We keep the heap ordering on our trees, but we don’t require them to be binomial. All of the above mentioned operations will be identical to the lazy binomial heap, with the exception of Decrease. We’ll also be keeping an additional flag on each node, which says if the node is marked. When Decrease is called on a node, we check if it changes the keys in such the parent node now has a higher value. If not, we stop right here as the tree keeps the heap property. If the parent now has a higher value, we Cut the subtree at the changed note (including the changed node, acting as a root). The Cut operation takes the subtree, removes it from its parent, and Inserts it back into the heap. If the parent was marked, we recursively call Cut on the parent. If the parent wasn’t marked, we mark it and end right there. This means our first Decrease will simply take the subtree at the decreased note, mark its parent, and insert the subtree back into the heap. If we then call a second Decrease under the same parent node, we will end up Cutting the parent as well. This prevents the tree from becoming too degenerate. The most interesting part here (which I’m however not going to prove), is that the amortized cost of Cut is $O(1)$, and the amortized cost of Decrease is also $O(1)$. This makes for a very interesting data structure, in which all operations except for the ExtractMin run in amortized constant time. ## Conclusion I do realize that I’ve skipped most of the amortized analysis, and simplified a few things, but this article mostly serves as a mental refresher for people already somewhat familiar with the Fibonacci heap. For a complete reference, check out the references at the Wikipedia page. ### Discussion of "Few notes on the Binomial and Fibonacci heaps" If you have any questions, feedback, or suggestions, please do share them in the comments! I'll try to answer each and every one. If something in the article wasn't clear don't be afraid to mention it. The goal of these articles is to be as informative as possible.<\|endoftext\|>	4.53125
1,030	After more than a century of Partitions between the Austrian, the Prussian, and the Russian imperial powers, Poland re-emerged as a sovereign state at the end of the First World War in Europe in 1917-1918. The victorious Allies of World War I confirmed the rebirth of Poland in the Treaty of Versailles of June 1919. It was one of the great stories of the 1919 Paris Peace Conference. Poland solidified its independence in a series of border wars fought by the newly formed Polish Army from 1918 to 1921. The extent of the eastern half of the interwar territory of Poland was settled diplomatically in 1922 and internationally recognized by the League of Nations. End of World War I In the course of World War I (1914-1918), Germany gradually gained overall dominance on the Eastern Front as the Imperial Russian Army fell back. German and Austro-Hungarian armies seized the Russian-ruled part of what became Poland. In a failed attempt to resolve the Polish question as quickly as possible, Berlin set up a German puppet state on 5 November 1916, with a governing Provisional Council of State and (from 15 October 1917) a Regency Council (Rada Regencyjna Królestwa Polskiego). The Council administered the country under German auspices (see also Mitteleuropa), pending the election of a king. A month before Germany surrendered on 11 November 1918 and the war ended, the Regency Council had dissolved the Council of State, and announced its intention to restore Polish independence (7 October 1918). With the notable exception of the Marxist-oriented Social Democratic Party of the Kingdom of Poland and Lithuania (SDKPiL), most Polish political parties supported this move. On 23 October the Regency Council appointed a new government under Józef Świeżyński and began conscription into the Polish Army. Formation of the Republic Second Polish Republic between 1921 and 1939 (light beige), including the Eastern Borderlands In 1918–1919, over 100 workers' councils sprang up on Polish territories; on 5 November 1918, in Lublin, the first Soviet of Delegates was established. On 6 November socialists proclaimed the Republic of Tarnobrzeg at Tarnobrzeg in Austrian Galicia. The same day the Socialist, Ignacy Daszyński, set up a Provisional People's Government of the Republic of Poland (Tymczasowy Rząd Ludowy Republiki Polskiej) in Lublin. On Sunday, 10 November at 7 a.m., Józef Piłsudski, newly freed from 16 months in a German prison in Magdeburg, returned by train to Warsaw. Piłsudski, together with Colonel Kazimierz Sosnkowski, was greeted at Warsaw's railway station by Regent Zdzisław Lubomirski and by Colonel Adam Koc. Next day, due to his popularity and support from most political parties, the Regency Council appointed Piłsudski as Commander in Chief of the Polish Armed Forces. On 14 November, the Council dissolved itself and transferred all its authority to Piłsudski as Chief of State (Naczelnik Państwa). After consultation with Piłsudski, Daszyński's government dissolved itself and a new government formed under Jędrzej Moraczewski. In 1918 Italy became the first country in Europe to recognise Poland's renewed sovereignty. Coat of arms of Poland during 1919-1927 Centers of government that formed at that time in Galicia (formerly Austrian-ruled southern Poland) included the National Council of the Principality of Cieszyn (established in November 1918), the Republic of Zakopane and the Polish Liquidation Committee (28 October). Soon afterward, the Polish–Ukrainian War broke out in Lwów (1 November 1918) between forces of the Military Committee of Ukrainians and the Polish irregular units made up of students known as the Lwów Eaglets, who were later supported by the Polish Army (see Battle of Lwów (1918), Battle of Przemyśl (1918)). Meanwhile, in western Poland, another war of national liberation began under the banner of the Greater Poland uprising (1918–19). In January 1919 Czechoslovakian forces attacked Polish units in the area of Zaolzie (see Polish–Czechoslovak War). Soon afterwards the Polish–Lithuanian War (ca 1919-1920) began, and in August 1919 Polish-speaking residents of Upper Silesia initiated a series of three Silesian Uprisings. The most critical military conflict of that period, however, the Polish–Soviet War (1919-1921), ended in a decisive Polish victory. In 1919 the Warsaw government suppressed the Republic of Tarnobrzeg and the workers' councils.<\|endoftext\|>	4.1875
5,339	Mathematics Part II Solutions Solutions for Class 9 Maths Chapter 7 Co Ordinate Geometry are provided here with simple step-by-step explanations. These solutions for Co Ordinate Geometry are extremely popular among Class 9 students for Maths Co Ordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 9 Maths Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate. #### Question 1: State in which quadrant or on which axis do the following points lie. • A($-$3, 2), • B($-$5, $-$2), • K(3.5, 1.5), • D(2, 10), • E(37, 35), F(15, -18), G(3, $-$7), H(0, $-$5), • M(12, 0), N(0, 9), P(0, 2.5), Q($-$7, $-$3) The x co-ordinate of A(−3, 2) is negative and its y coordinate is positive. Therefore, point A(−3, 2) is in the second quadrant. The x co-ordinate of B(−5, −2) is negative and its y coordinate is negative. Therefore, point B(−5, −2) is in the third quadrant. The x co-ordinate of K(3.5, 1.5) is positive and its y coordinate is positive. Therefore, point K(3.5, 1.5) is in the first quadrant. The x co-ordinate of D(2, 10) is positive and its y coordinate is positive. Therefore, point D(2, 10) is in the first quadrant. The x co-ordinate of E(37, 35) is positive and its y coordinate is positive. Therefore, point E(37, 35) is in the first quadrant. The x co-ordinate of F(15, −18) is positive and its y coordinate is negative. Therefore, point F(15, −18) is in the fourth quadrant. The x co-ordinate of G(3, −7) is positive and its y coordinate is negative. Therefore, point G(3, −7) is in the fourth quadrant. The x co-ordinate of H(0, −5) is zero. Therefore, point H(0, −5) is on the Y-axis. The y co-ordinate of M(12, 0) is zero. Therefore, point M(12, 0) is on the X-axis. The x co-ordinate of N(0, 9) is zero. Therefore, point N(0, 9) is on the Y-axis. The x co-ordinate of P(0, 2.5) is zero. Therefore, point P(0, 2.5) is on the Y-axis. The x co-ordinate of Q(−7, −3) is negative and its y coordinate is negative. Therefore, point Q(−7, −3) is in the third quadrant. #### Question 2: In which quadrant are the following points ? (i) whose both co-ordinates are positive. (ii) whose both co-ordinates are negative. (iii) whose x co-ordinate is positive, and the y co-ordinate is negative. (iv) whose x co-ordinate is negative and y co-ordinate is positive. (i) The x co-ordinate and y co-ordinate of a point are both positive in the first quadrant. (ii) The x co-ordinate and y co-ordinate of a point are both negative in the third quadrant. (iii) The x co-ordinate of a point is positive and y co-ordinate of a point is negative in the fourth quadrant. (iv) The x co-ordinate of a point is negative and y co-ordinate of a point is positive in the second quadrant. #### Question 3: Draw the co-ordinate system on a plane and plot the following points. L($-$2, 4), M(5, 6), N($-$3, $-$4), P(2, $-$3), Q(6, $-$5), S(7, 0), T(0,$-$5) The given points are L(−2, 4), M(5, 6), N(−3, −4), P(2, −3), Q(6, −5), S(7, 0) and T(0, −5). These point can be plotted on the co-ordinate system as follows: #### Question 1: On a graph paper plot the points A (3,0), B(3,3), C(0,3). Join A, B and B, C. What is the figure formed? The given points are A(3, 0), B(3, 3) and C(0, 3). These points can be plotted on the co-ordinate plane as follows: The x co-ordinate of point is its distance from the Y-axis and y co-ordinate of point is its distance from the X-axis. Here, OA = AB = BC = OC = 3 units Therefore, the figure formed is a square. #### Question 2: Write the equation of the line parallel to the Y-axis at a distance of 7 units from it to its left. The equation of the line parallel to Y-axis at a distance of 7 units from it to its left is x = −7. #### Question 3: Write the equation of the line parallel to the X-axis at a distance of 5 units from it and below the X-axis. The equation of the line parallel to X-axis at a distance of 5 units from it and below the X-axis is y = −5. #### Question 4: The point Q($-$3, $-$2) lies on a line parallel to the Y-axis. Write the equation of the line and draw its graph. The x co-ordinate of a point is its distance from the Y-axis and y co-ordinate is its distance from the X-axis. The point of intersection of the line parallel to Y-axis at a distance 3 units from it to its left and the line parallel to X-axis at a distance 2 units below it is (−3, −2). The equation of the line parallel to Y-axis at a distance 3 units from it to its left is x = −3. #### Question 5: X-axis and line x$-$4 are parallel lines. What is the distance between them? Disclaimer: The question is incorrect. The question should be "Y-axis and line x = $-$4 are parallel lines. What is the distance between them? or X-axis and line y = $-$4 are parallel lines. What is the distance between them?" If the question is "Y-axis and line x = $-$4 are parallel lines. What is the distance between them?" then the answer is as follows. x = −4 is the equation of the line parallel to the Y-axis at a distance of 4 units and to the left of Y-axis. Thus, the distance between them is 4 units. OR If the question is "X-axis and line y = $-$4 are parallel lines. What is the distance between them?" then the answer is as follows. y = −4 is the equation of the line parallel to the X-axis at a distance of 4 units and below the X-axis. Thus, the distance between them is 4 units. #### Question 6: Which of the equations given below have graphs parallel to the X-axis, and which ones have graphs parallel to the Y-axis ? (i) x = 3 (ii) $-$2 = 0 (iii) x + 6 = 0 (iv) y = $-$5 (i) x = 3 is the equation of the line parallel to Y-axis at a distance of 3 units and to the right of Y-axis. Thus, the graph of the line x = 3 is parallel to Y-axis. (ii) y − 2 = 0 ⇒ y = 2 y = 2 is the equation of the line parallel to X-axis at a distance of 2 units and above the X-axis. Thus, the graph of the line y − 2 = 0 is parallel to X-axis. (iii) x + 6 = 0 ⇒ x = −6 x = −6 is the equation of the line parallel to Y-axis at a distance of 6 units from it to its left. Thus, the graph of the line x + 6 = 0 is parallel to Y-axis. (iv) y = −5 is the equation of the line parallel to X-axis at a distance of 5 units and below the X-axis. Thus, the graph of the line y = −5 is parallel to X-axis. #### Question 7: On a graph paper, plot the points A(2, 3), B(6,$-$1) and C(0, 5). If those points are collinear then draw the line which includes them. Write the co-ordinates of the points at which the line intersects the X-axis and the Y-axis. The given points are A(2, 3), B(6, −1) and C(0, 5). The line which includes these points is shown below: It can be seen from the graph that the line intersects the X-axis at (5, 0) and the Y-axis at (0, 5). #### Question 8: Draw the graphs of the following equations on the same system of co-ordinates. Write the co-ordinates of their points of intersection. x + 4 = 0, y $-$1 = 0, 2x + 3 = 0, 3y $-$ 15 = 0 x + 4 = 0 ⇒ x = −4 x = −4 is the equation of the line parallel to Y-axis at a distance of 4 units from it to its left. y − 1 = 0 ⇒ y = 1 y = 1 is the equation of the line parallel to X-axis at a distance of 1 unit and above the X-axis. 2x + 3 = 0 ⇒ x = $-\frac{3}{2}$ x = $-\frac{3}{2}$ is the equation of the line parallel to Y-axis at a distance of $\frac{3}{2}$ units from it to its left. 3y − 15 = 0 ⇒ y = $\frac{15}{3}$ = 5 y = 5 is the equation of the line parallel to X-axis at a distance of 5 units and above the X-axis. It can be seen from the figure that the co-ordinates of the points of intersection are , (−4, 5) and (−4, 1). #### Question 9: Draw the graphs of the equations given below (i) x + y = 2 (ii) 3x $-$ y = 0 (iii) 2x + y = 1 (i) The equation of the given line is xy = 2. x + y = 2 ⇒ y = 2 − x .....(1) Putting x = 0 in (1), we get y = 2 − 0 = 2 Putting x = 2 in (1), we get y = 2 − 2 = 0 Putting x = −1 in (1), we get y = 2 − (−1) = 2 + 1 = 3 Putting x = 5 in (1), we get y = 2 − 5 = −3 These values can be represented in the table in the form of ordered pairs as follows: x 0 2 −1 5 y 2 0 3 −3 (x, y) (0, 2) (2, 0) (−1, 3) (5, −3) Plot these points on the graph paper. The line is the graph of the equation x + = 2. (ii) The equation of the given line is 3x y = 0. 3x y = 0 ⇒ y = 3x .....(1) Putting x = 0 in (1), we get y = 3 × 0 = 0 Putting x = 1 in (1), we get y = 3 × 1 = 3 Putting x = −1 in (1), we get y = 3 × (−1) = −3 Putting x = 2 in (1), we get y = 3 × 2 = 6 These values can be represented in the table in the form of ordered pairs as follows: x 0 1 −1 2 y 0 3 −3 6 (x, y) (0, 0) (3, 0) (−1, −3) (2, 6) Plot these points on the graph paper. The line is the graph of the equation 3x y = 0. (iii) The equation of the given line is 2x + y = 1. 2x + y = 1 ⇒ y = 1 − 2x .....(1) Putting x = 0 in (1), we get y = 1 − 2 × 0 = 1 − 0 = 1 Putting x = 1 in (1), we get y = 1 − 2 × 1 = 1 − 2 = −1 Putting x = −1 in (1), we get y = 1 − 2 × (−1) = 1 + 2 = 3 Putting x = 2 in (1), we get y = 1 − 2 × 2 = 1 − 4 = −3 These values can be represented in the table in the form of ordered pairs as follows: x 0 1 −1 2 y 1 −1 3 −3 (x, y) (0, 1) (1, −1) (−1, 3) (2, −3) Plot these points on the graph paper. The line is the graph of the equation 2x + y = 1. #### Question 1: Choose the correct alternative answer for the following quesitons. (i) What is the form of co-ordinates of a point on the X-axis ? (A) ( b , b ) (B) ( o , b ) (C) ( a , o ) (D) ( a , a ) (ii) Any point on the line y = x is of the form ..... (A) ( a , a ) (B) ( o , a ) (C) ( a , o ) (D) ( a , $-$a ) (iii) What is the equation of the X-axis ? (A) x = 0 (B) y = 0 (C) x + y = 0 (D) x = y (iv) In which quadrant does the point ($-$4, $-$3) lie ? (A) First (B) Second (C) Third (D) Fourth (v) What is the nature of the line which includes the points ($-$5,5), (6,5), ($-$3,5), (0,5) ? (A) Passes through the origin,,(B) Parallel to Y-axis. (C) Parallel to X-axis (D) None of these (vi) Which of the points P ($-$1,1), Q (3,$-$4), R(1,$-$1), S ($-$2,$-$3), T ($-$4,4) lie in the fourth quadrant ? (A) P and T (B) Q and R (C) only S (D) P and R (i) The y co-ordinate of every point on the X-axis is 0. Thus, the co-ordinates of a point on the X-axis is (a, 0). Hence, the correct answer is option (C). (ii) Putting xa in yx, we get ya Thus, any point on the line yx is of the form (aa). Hence, the correct answer is option (A). (iii) The y co-ordinate of every point on the X-axis is 0. Therefore, the equation of the X-axis is y = 0. Hence, the correct answer is option (B). (iv) The x co-ordinate of (−4, −3) is negative and its y co-ordinate is negative. Therefore, the point (−4, −3) lies in the third quadrant. Hence, the correct answer is option (C). (v) The y co-ordinate of all the points (−5, 5), (6, 5), (−3, 5) and (0, 5) is 5. All these points lies on the line y = 5, which is parallel to the X-axis. Thus, the line which includes the points (−5, 5), (6, 5), (−3, 5) and (0, 5) is parallel to the X-axis. Hence, the correct answer is option (C). (vi) The point whose x co-ordinate is positive and y co-ordinate is negative lie in the fourth quadrant. Thus, the points Q(3, −4) and R(1, −1) lie in the fourth quadrant. Hence, the correct answer is option (B). #### Question 2: Some points are shown in the given figure, With the help of it answer the following questions : (i) Write the co-ordinates of the points Q and R. (ii) Write the co-ordinates of the points T and M. (iii) Which point lies in the third quadrant ? (iv) Which are the points whose x and y co-ordinates are equal? (i) The co-ordinate of a point is its distance from the Y-axis and y co-ordinate of a point is its distance from the X-axis. The co-ordinates of point Q are (−2, 2) and the co-ordinates of point R are (4, −1). (ii) The x co-ordinate of every point on the Y-axis is 0 and the y co-ordinate of every point on the X-axis is 0. The co-ordinates of point T are (0, −1) and the co-ordinates of point M are (3, 0). (iii) The point whose x co-ordinate is negative and y co-ordinate is negative lies in the third quadrant. Thus, the point S(−3, −2) lies in the third quadrant. (iv) The co-ordinates of point O are (0, 0). Thus, O is the point whose x and y co-ordinates are equal. #### Question 3: Without plotting the points on a graph, state in which quadrant or on which axis do the following point lie. (i) (5,$-$3) (ii) ($-$7, $-$12) (iii) ($-$23, 4) (iv) ($-$9, 5) (v) (0, $-$3) (vi) ($-$6, 0) (i) The x co-ordinate of the point (5, −3) is positive and its y co-ordinate is negative. Therefore, the point (5, −3) lie in the fourth quadrant. (ii) The x co-ordinate of the point (−7, −12) is negative and its y co-ordinate is negative. Therefore, the point (−7, −12) lie in the third quadrant. (iii) The x co-ordinate of the point (−23, 4) is negative and its y co-ordinate is positive. Therefore, the point (−23, 4) lie in the second quadrant. (iv) The x co-ordinate of the point (−9, 5) is negative and its y co-ordinate is positive. Therefore, the point (−9, 5) lie in the second quadrant. (v) The x co-ordinate of the point (0, −3) is zero. Therefore, the point (0, −3) lie on the Y-axis. (vi) The y co-ordinate of the point (−6, 0) is zero. Therefore, the point (−6, 0) lie on the X-axis. #### Question 4: Plot the following points on the one and the same co-ordinate system. A(1, 3), B($-$3, $-$1), C(1, $-$4), D($-$2, 3), E(0, $-$8), F(1, 0) The given points are A(1, 3), B(−3, −1), C(1, −4), D(−2, 3), E(0, −8) and F(1, 0). These points can be plotted on the co-ordinate system as follows: #### Question 5: In the graph alongside, line LM is parallel to the Y-axis. (i) What is the distance of line LM from the Y-axis ? (ii) Write the co-ordinates of the points P, Q and R. (iii) What is the difference between the x co-ordinates of the points L and M? (i) The line LM is parallel to the Y-axis and passing through the point (3, 0). The equation of the line LM is x = 3 and at a distance of 3 units to the right of the Y-axis. Thus, the distance of line LM from the Y-axis is 3 units. (ii) The x co-ordinate of a point is its distance from the Y-axis and y co-ordinate is its distance from the X-axis. Thus, the co-ordinates of the points P, Q and R are (3, 2), (3, −1) and (3, 0), respectively. (iii) x co-ordinate of point L = 3 x co-ordinate of point M = 3 ∴ Difference between the x co-ordinates of the points L and M = 3 − 3 = 0 Thus, the difference between the x co-ordinates of the points L and M is 0. #### Question 6: How many lines are there which are parallel to X-axis and having a distance 5 units? The equations of the lines parallel to X-axis and at a distance 5 units from it are y = 5 and y = −5. Thus, there are two lines which are parallel to X-axis and having a distance 5 units from it. #### Question 7: If ‘a’ is a real number, what is the distance between the Y-axis and the line x = a ?<\|endoftext\|>	4.6875
1,606	In 1933-1934, SS chief Heinrich Himmler secured SS control over a centralized concentration camp system. Throughout Germany, various civilian authorities and police agencies had established concentration camps during 1933 to incarcerate political enemies of the Nazi government. Impressed with the Dachau concentration camp established by the SS in March 1933, Hitler authorized Himmler to centralize these camps under SS leadership. Himmler established (in the SS Main Office) an SS Inspectorate of Concentration Camps under the leadership of Dachau camp commandant and SS General Theodor Eicke. After 1934, SS officers commanded all concentration camps in Germany and in German-occupied territory. Units known as SS Death's-Head Units (SS-Totenkopfverbände) guarded and administered the camps. Although the Security Police (Gestapo and Kripo) had exclusive authority to incarcerate, release, and “officially” order the execution of prisoners, the daily life of prisoners lay in the brutal and merciless hands of the camp commandants and these SS Death's-Head Units, which were not part of the police forces. In 1937, there were only four concentration camps in Germany; by 1944, there were approximately 30 main camps and hundreds of subcamps located throughout the Greater German Reich and German-occupied Europe. Before 1938, the vast majority of concentration camp prisoners were political opponents of the Nazi regime with minorities of Roma (Gypsies), Jehovah's Witnesses, homosexuals, repeat criminal offenders and so-called asocials. Roma were often classified as “criminals” and/or “asocials.” Although in reality most Roma in Germany were settled and somewhat integrated into German society, the SS and police saw them in terms of traditional negative stereotypes of Gypsies as petty criminals and persons who engaged in anti-social behavior. Likewise, with the drastic expansion of police enforcement of laws criminalizing real and perceived homosexual acts between consenting adults, homosexuals in the camps were frequently classified as “criminals” and “asocials.” Despite their presence in numbers vastly out of proportion to their percentage in the German population, Jews remained a minority among the prisoners prior to 1938. In most cases, Gestapo or Kripo officials incarcerated Jews in the camps because they had been activists in the Social Democratic, Communist, or liberal democratic parties, had been visible opponents of the Nazi Party or specific Nazi policies, or had been advocates of policies or members of organizations that the Nazis found to be “hostile to the state and the race.” Examples of such were membership in pro-democratic Masonic lodges, activism in organizations dedicated to pacifism or international peace and understanding, or advocacy for rights for national minorities or homosexuals. With German expansion in 1938, the availability of prisoners for forced labor in the concentration camp system took on added significance. The SS was determined that the Thousand-Year Reich would be ruled by its self-selected, “racially pure” elite. To ensure this development, its leaders invested significant financial and human resources in planning for the construction of the German settlements in Poland and the Soviet Union in accordance with their visions of permanent German rule. As early as the mid-1930s, the SS leaders of the concentration camps and the chief of the SS Administration Main Office (SS-Verwaltungshauptamt), SS General Oswald Pohl, recognized the potential of concentration camp prisoners as forced laborers to produce construction materials, and eventually to do the manual labor to build and maintain these settlements. At this time, the SS founded a number of companies, such as the German Earth and Stone Works (Deutsche Erd- und Steinwerke; DESt) and the German Equipment Works (Deutsche Ausrüstungswerke; DAW) to produce construction materials and equipment for the SS. The Sachsenhausen (1936), Buchenwald (1937), Flossenbürg (1938), and Mauthausen (1938) concentration camp sites were chosen precisely because of their proximity to soil suitable for making bricks, to a brickworks factory, or to stone quarries. In 1938, the Criminal Police initiated two massive roundups of so-called “work-shy” and asocial individuals, in part to increase the concentration camp population available for forced labor. Among the so-called work-shy and asocials whom the Kripo incarcerated were Roma (Gypsies). In the first three years of the war, the SS leaders expanded the concentration camp system not merely to detain the tens of thousands of new political prisoners of non-German nationality who chose to resist German occupation policies, but also to increase the pool of forced laborers available for the settlements that the SS planned to construct now that Poland and the western Soviet Union were in German hands. SS leaders reached decisions to construct new gigantic concentration camps at Auschwitz-Birkenau and Lublin-Majdanek (1941-1942) based on expectations of thousands of Soviet prisoners of war who would be available for forced labor (1941) and, later, on tens of thousands of Jewish forced laborers (January-March 1942). After spring 1942, the overwhelming majority of prisoners in both Auschwitz and Majdanek (until November 1943) were Jews. In April 1942, Auschwitz-Birkenau took on the function of a killing center. Until November 1943, Lublin-Majdanek served primarily as a forced-labor camp for Jews within the framework of Operation Reinhard. Majdanek also served the SS from time to time as a killing site, where the SS and police killed Jews under Operation Reinhard. In winter 1942, it became clear that Germany would be engaged in a long war and that labor needs for German war production could not be met even by the deportation of millions of civilian forced laborers from occupied Poland and the occupied Soviet Union. After this time, the leadership of the SS administration and the Inspectorate of Concentration Camps deployed concentration camp labor, at a profit, in accordance with contracts signed by German military and civilian agencies and private firms producing armaments, related war materials (e.g. uniforms), and construction materials to repair or replace facilities destroyed by Allied bombing. To facilitate this process, Himmler incorporated the Inspectorate of Concentration Camps, now led by SS General Richard Glücks, with the SS Economic-Administration Main Office (SS-Wirtschafts-Verwaltungshauptamt-WVHA) under Pohl in March 1942. Henceforth, the concentration camps and their prisoners, the SS-owned companies, and the administrative offices of the SS were all together in one agency. Under the auspices of the WVHA, the number of subcamps multiplied into the hundreds and even thousands. Enterprises using concentration camp labor ranged from large state combines like the Hermann-Göring-Werke and large, private corporate conglomerates like the I.G. Farben chemical combine to small private firms like Oskar Schindler's German Enamelware Factory (Deutsche Emalwarenfabrik) in Krakow, Poland. During World War II, SS camp authorities would kill around two million prisoners—Jews, political prisoners, Roma (Gypsies), so-called asocials, recidivist convicts, homosexuals, Jehovah's Witnesses and others—in the concentration camp system. Series: The SS Critical Thinking Questions - How does the SS embody the systematic nature of the attempted “Final Solution?” - What other institutions and professions collaborated with the SS to operate the camp system? - What other institutions and groups in society benefited from the camp system?<\|endoftext\|>	3.9375
260	In the physical sciences, an interface is the boundary between two spatial regions occupied by different matter, or by matter in different physical states. The interface between matter and air, or matter and vacuum, is called a surface, and studied in surface science. In thermal equilibrium, the regions in contact are called phases, and the interface is called a phase boundary. An example for an interface out of equilibrium is the grain boundary in polycrystalline matter. The importance of the interface depends on the type of system: the bigger the quotient area/volume, the greater the effect the interface will have. Consequently, interfaces are very important in systems with large interface area-to-volume ratios, such as colloids. Surface tension is the physical property which rules interface processes involving liquids. For a liquid film on flat surfaces, the liquid-vapor interface keeps flat to minimize interfacial area and system free energy. For a liquid film on rough surfaces, the surface tension tends to keep the meniscus flat, while the disjoining pressure makes the film conformal to the substrate. The equilibrium meniscus shape is a result of the competition between the capillary pressure and disjoining pressure. One topical interface system is the gas-liquid interface between aerosols and other atmospheric molecules.<\|endoftext\|>	3.796875
425	# A design is made on a rectangular Question: A design is made on a rectangular tile of dimensions 50 cm x 17 cm as shown in figure. The design shows 8 triangle, each of sides 26 cm, 17 cm and 25 cm. Find the total area of the design and the remaining area of the tiles. Solution: Given, the dimension of rectangular tile is $50 \mathrm{~cm} \times 70 \mathrm{~cm}$. $\therefore$ Area of rectangular tile $=50 \times 70=3500 \mathrm{~cm}^{2}$ The sides of a design of one triangle be $a=25 \mathrm{~cm}, b=17 \mathrm{~cm}$ and $c=26 \mathrm{~cm}$ Now, semi-perimeter, $s=\frac{a+b+c}{2}=\frac{25+17+26}{2}=\frac{68}{2}=34$ $\therefore \quad$ Area of one triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula] $=\sqrt{34 \times 9 \times 17 \times 8}$ $=\sqrt{17 \times 2 \times 3 \times 3 \times 17 \times 2 \times 2 \times 2}$ $=17 \times 3 \times 2 \times 2=204 \mathrm{~cm}^{2}$ $\therefore$ Total area of eight triangles $=204 \times 8=1632 \mathrm{~cm}^{2}$ Now, area of the design $=$ Total area of eight triangles $=3500-1632$ $=1868 \mathrm{~cm}^{2}$ Hence, the total area of the design is $1632 \mathrm{~cm}^{2}$ and the remaining area of the tile is $1868 \mathrm{~cm}^{2}$.<\|endoftext\|>	4.84375
976	### How to Find Percent Error? Formula & Explanation How to find percent error? In mathematics and science, a percentage error is a measure of the difference between a measured value and the true value of the quantity being measured. The percentage error is computed as the absolute value of the difference divided by the true value and multiplied by 100. Percentage errors are often reported in scientific papers, where they are used to compare the results of different experiments or to track the progress of a research project. However, they can also be used in everyday life to check whether a number that someone quotes are accurate. For example, if you hear that your favorite band is playing at a venue that holds 1000 people, you can check to see if that number is close to the expected attendance by doing some simple math. Percentage errors can be both positive and negative. ## How to Calculate Percent? In order to calculate a percentage, divide the number of items that match the criteria by the total number of items and multiply by 100. This will give you the percentage that matches the criteria. ## How to Calculate Percent Error? In order to find the percent error, you will need to divide the difference between the two numbers by the average of the two numbers. Once you have that number, you will need to move the decimal point two places to the right. This is your percent error. ## Percentage Error Formula In mathematics and statistics, the percentage error (PE) is a measure of how close a measured value is to the true value. The PE is computed as the absolute value of the difference between the measured and true values, divided by the true value: The percentage error is also called the relative error. Percent error = (Approximate or experimental Value – Exact or known Value/Exact or known Value)∗100 Example 1 For example, if you are working with a set of data that has a difference of 5 and an average of 8, your percent error would be 25%. Example 2 In order to calculate percent error, you need to take the absolute value of the difference between the experimental value and the theoretical value, divide it by the theoretical value, and multiply it by 100%. So, if an experiment yielded a result of 9.7 grams and the theoretical value was supposed to be 10 grams, then the percent error would be (9.7-10)/10100%= -3%. If an experiment yielded a result of 9.7 grams and the theoretical value was supposed to be 9 grams, then the percent error would be (9.7-9)/9100%= 8%. ## Percent Error Versus Absolute and Relative Error The purpose of this paper is to discuss percent error versus absolute and relative error. In scientific and mathematical circles, percent error has long been the standard way to express accuracy. This is partly because it can be easily calculated from the difference between the measured and actual values. As a measure of accuracy, however, percent error is not without its flaws. One major problem with the percent error is that it is always relative to the size of the measurement unit. For example, if you measure the length of a room in feet and calculate a percent error of 2%, your result means very little unless you also know what the original measurement was in feet. In other words, percent error tells you how far off your measurement was from the true value but says nothing about how close it was to the true value. A related issue is that percent error can be misleading when there are large measurements involved. ## FAQs Q: Can you have a Negative Percent Error? A: A percent error is a measure of how far off a measured value is from the true value. Percent error can be positive or negative, but it cannot be a negative percent error. This is because percent error is defined as the absolute value of the difference between the measured and true values, and a negative number cannot be an absolute value. Q: What does a Percent Error tell us? We can use a tool called a caliper to measure the width of a piece of metal with great accuracy. However, when we do this experiment, we may not get the same result every time. This variation is due to random errors in our measurement. A percent error tells us how much our measurement deviates from the actual value. It is important to note that percent error does not tell us the absolute value of the deviation; it only tells us how large the deviation is relative to the size of the measured value. For example, if I measure the width of a metal piece to be 10 cm and my actual width is 9 cm, my percent error would be 10%. This means that my measurement was off by 1 cm (or 10%), relative to the actual width.<\|endoftext\|>	4.625
909	‘Challenging behaviour’ is a very broad term to define when a child acts out of character and causes distress to themselves, parents, carers, teachers and other professionals. It can include temper tantrums, physical harm such as hitting or kicking other people, harming themselves and throwing things. You can classify behaviour as challenging if it is harmful to them or others around them. It may also stop them getting involved in their usual daily activities such as concentrating on homework or at school and playing with friends or creating new friendships. How do you manage challenging behaviour when a child has a learning disability? Whilst challenging behaviour is not a learning disability, children with a disability are more likely to show challenging behaviour, which can be due to them struggling to communicate and expressing frustration at things they cannot do due to their disability. Challenging behaviour may also indicate that something is wrong with your child, especially if they are very young and cannot tell you in any other way, that they may be in pain or discomfort for example. What causes challenging behaviour, and can I do anything to stop it? Every child is different and therefore what causes them to get upset will depend on the child. There is no one cause, but being aware of the environment they are in, the relationships or how they are interacting with other children and adults as well as general discomfort and frustration are all common reasons for them to act out of character. Taking your child to one side and talking to them about why they are behaving in the way they are and allowing them to tell you what is upsetting them is the first step in helping to overcome their challenges and help them to communicate more effectively. If you are worried that your child is demonstrating challenging behaviour on a regular basis and causing harm to themselves or others, you may wish to speak to your GP who may then refer you to another professional, such as a child psychologist if you think that it may be sign of a mental health problem. How I deal with Challenging Behaviour in My Home Promoting positive behaviour is very important for the children in my care as I have several of them to look after at once. Setting rules and guidelines for them to work with is essential in ensuring they get along with each other as much as possible. I do this by: - Giving lots of praise for good behaviour - Giving the children individual attention so they feel valued - Setting a good example, being a good role model - Listening to what the children have to say - Rewarding good behaviour (allowing them to choose the next activity for example). All the children understand the house rules, which are realistic and achievable by all of them and I am consistent in the enforcing the rules so that I do not give out confusing signals. I am aware of the different reasons why children misbehave and will endeavour to keep to routines so that children feel safe and cared for and that they are not tired or hungry which can also cause them to misbehave. How I Work to Diffuse Any Challenging Behaviour in My Home I have strategies in place that I use to calm situations and these also depend on the child’s age and ability. This includes: - Distraction – Remove the child from the situation and give them an alternative activity. - Discussion – If the child can understand, I will discuss their behaviour and try and get them to appreciate the consequences of their actions on others. I will make sure that they understand that it is their behaviour I am unhappy with, not them. - Time Out – Removing the child from the activity and sitting them quietly for a few minutes. If occasionally a child misbehaves I let parents know verbally when they collect their child. Some children can become upset if the incident is retold in front of them, so as much as possible I take them to one side to let them know what has happened. I also inform parents of how the matter was dealt with, which in most cases means that it doesn’t require any further action. Managing challenging behaviour is par for the course when you have a child. They are usually testing boundaries for what they can or can’t get away with, but in most cases there are genuine frustrations that trigger them to act out. It’s important to remember that they also mirror behaviour, so staying calm in difficult situations will help them to learn to do the same.<\|endoftext\|>	3.78125
715	Basic Division Lesson 1: Number 1 Welcome to Basic Division. In this unit, you will learn how to divide numbers from one to twelve and also zero. FoxyMath wishes you success! Here's an easy example: 2 ÷ 1 = 2 ÷ = Peru Box Tan Box divided by Red Box Box equals Peru Box Tan Box We start with two boxes. We divide these two boxes by one. This means: how many groups of one can we make with the two boxes that we have? Answer: 2. Here's another example: 6 ÷ 1 = 6 ÷ = Violet Box Cyan Box Orchid Box Blue Box ForestGreen Box Plum Box divided Red Box Box equals Violet Box Cyan Box Orchid Box Blue Box ForestGreen Box Plum Box We start with six boxes. We divide these six boxes by one. This means: how many groups of one can we make with the six boxes that we have? Answer: 6. Violet Box Cyan Box Orchid Box Blue Box ForestGreen Box Plum Box Try this problem. 12 ÷ 1 = ÷ = MediumAquamarine Box SteelBlue Box DarkSeaGreen Box MidnightBlue Box Aquamarine Box Khaki Box DarkKhaki Box Fuchsia Box DarkOrchid Box DarkViolet Box Gold Box Chartreuse Box divided Red Box Box equals MediumAquamarine Box SteelBlue Box DarkSeaGreen Box MidnightBlue Box Aquamarine Box Khaki Box DarkKhaki Box Fuchsia Box DarkOrchid Box DarkViolet Box Gold Box Chartreuse Box We start with twelve boxes. We divide these twelve boxes by one. This means: how many groups of one can we make with the twelve boxes that we have? Answer: 12. MediumAquamarine Box SteelBlue Box DarkSeaGreen Box MidnightBlue Box Aquamarine Box Khaki Box DarkKhaki Box Fuchsia Box DarkOrchid Box DarkViolet Box Gold Box Chartreuse Box In order to do basic division, you must memorize some equations but fear not: FoxyMath will help you! For this lesson, there is only one equation to learn: 1 ÷ 1 = 1 2 ÷ 1 = 2 3 ÷ 1 = 3 4 ÷ 1 = 4 5 ÷ 1 = 5 6 ÷ 1 = 6 7 ÷ 1 = 7 8 ÷ 1 = 8 9 ÷ 1 = 9 10 ÷ 1 = 10 11 ÷ 1 = 11 12 ÷ 1 = 12 As we see, any number divided by one is the number itself.<\|endoftext\|>	4.75
443	#### Need solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 24.1 question 9 sub question (iii) Answer : $\frac{\sqrt{26}}{2} \text { sq.units }$ Hint : To solve this we use area of parallelogram formula Given : $3 \hat{\imath}+4 \hat{\jmath} \text { and } \hat{\imath}+\hat{\jmath}+\hat{k}$ Solution :Area of parallelogram $=\frac{1}{2}\left(d_{1} \times d_{2}\right)$ $\begin{gathered} d_{1}=3 \hat{\imath}+4 \hat{\jmath} \\ d_{2}=\hat{\imath}+\hat{\jmath}+\hat{k} \end{gathered}$ \begin{aligned} &d_{1} \times d_{2}=\left\|\begin{array}{lll} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 4 & 0 \\ 1 & 1 & 1 \end{array}\right\| \\\\ &=\hat{\imath}(4 \times 1-0 \times 1)-\hat{j}(3 \times 1-0 \times 1)+\hat{k}(3 \times 1-4 \times 1) \\\\ &=4 \hat{\imath}-3 \hat{\jmath}-1 \hat{k} \end{aligned} \begin{aligned} &A \quad=\frac{1}{2}\left(d_{1} \times d_{2}\right) \\\\ &=\frac{1}{2} \sqrt{(-4)^{2}+(-3)^{2}+(-1)^{2}} \\\\ &=\frac{1}{2} \sqrt{16+9+1} \\\\ &=\frac{\sqrt{26}}{2} \text { sq.units } \end{aligned}<\|endoftext\|>	4.46875
4,951	A BRIEF HISTORY OF SCOTLAND By Tim Lambert During the ice age Scotland was uninhabited. However when the ice melted forests spread across Scotland and stone age hunters moved there. By 6,000 BC small groups of people lived in Scotland by hunting animals like red deer and seals and by gathering plants for food. Then about 4,500 BC farming was introduced into Scotland. The early farmers continued to use stone tools and weapons and this period is called the Neolithic (new stone age). The Neolithic people used stone axes or fire to clear forests for farming and they grew wheat, barley and rye. They also bred cattle and sheep. They lived in simple stone huts with roofs of turf or thatch. The finest example of a Neolithic village was found in Orkney after a storm in 1850. The inhabitants lived in stone huts with stone shelves and stone seats inside. They also had stone beds, which were probably covered with straw or heather. The people of Skara Brae used pottery vessels. By 1,800 BC people in Scotland had learned to make bronze. The Bronze Age people continued to live in simple huts but they are famous for their stone monuments. They arranged huge stones in circles. The fact that they were able to do so indicates they lived in an organised society. The Picts and Scots The Picts lived in round huts of wood or stone with thatched roofs. Some Picts lived in crannogs, which were huts erected on artificial platforms in lochs or estuaries. Pictish chieftains built hill forts of stone, wood or earth. Pictish farmers raised cattle, pigs and sheep. They also fished, hunted deer and seals and caught birds. They grew crops of wheat, barley and rye. They also gathered wild fruits such as crab-apples, sloes, raspberries, blackberries and damsons. Although the vast majority of Picts were farmers some worked as craftsmen such as blacksmiths, bronze smiths, goldsmiths and potters. The Picts were very skilled at making jewelry. They also carved pictures on stones. Upper class Picts spent their days hunting on horseback or hunting with falcons. In the evenings they drank and feasted. Scotland's written history begins with the Romans. The Romans invaded Scotland in 80 AD led by Agricola. They advanced into southern Scotland and then marched into the northeast. In 84 the Romans severely defeated the Picts at a place they called Mons Graupius (its exact location is unknown). However in the years after the battle the Romans slowly withdrew and in 123 the Emperor Hadrian began building a wall to keep out the Picts. Later in the Second century the Romans advanced again and in 140 they built the Antonine Wall from the Clyde to the Forth. However the Romans finally abandoned the Antonine Wall in 196 AD. Afterwards Hadrians Wall became the frontier. The Romans advanced into Scotland again in 209 AD but only temporarily. In 367-68 the Picts took part in a great raid upon Roman Britain. In the 6th century a people from Ireland called the Scots invaded what is now Scotland. They settled in what is now Argyll and founded the kingdom of Dalriada. Meanwhile Christian missionaries had begun the work of converting the Picts. Some Picts in southeast Scotland accepted Christianity in the 5th century. Columba who went there in 563 converted southwest Scotland to Christianity. He founded a monastery at Iona, which became very important in the history of Christianity in Britain. During the 6th and 7th centuries Christianity spread across Scotland and by the end of the 7th century all of Scotland was Christian. Further south in the 6th century Angles invaded Northeast England and they created the kingdom of Northumbria. In the early 7th century the Northumbrians expanded into southeast Scotland and as far as Dunbar and Edinburgh. Then, in 843 Kenneth MacAlpin who was king of the Scottish kingdom of Dalriada also became king of the Picts of northern and central Scotland. So the Scots and Picts merged to form single kingdom. However the new kingdom of Scotland only included land north of the Clyde and Forth. The English ruled the southeast of Scotland until 1018 when the Scots conquered it. Furthermore southwest Scotland and Cumbria formed a separate kingdom called Strathclyde. However in 1018 Strathclyde was peacefully absorbed into Scotland. Meanwhile Scotland faced another threat - the Vikings! They raided the monastery at Iona in 795. Then in the early 9th century Vikings settled on the Shetland and Orkney Islands. Later in the 9th century they settled in the Hebrides and in Caithness and Sutherland as well as on the western coast of Scotland. In 1034 Duncan became king of Scotland. He proved to be incompetent and in 1040 Macbeth who then replaced him as king killed him. Unlike the character created by Shakespeare Macbeth was a good king and in 1050 he went on a pilgrimage to Rome. However in 1057 Macbeth was killed at the battle of Lumphanan and Duncan's son became Malcolm III. Scotland in the Middle Ages In 1066 the Normans conquered England. Norman influence was soon felt in Scotland. In 1069 Malcolm married an English woman named Margaret who promoted Norman ways at the Scottish court. Malcolm was killed in a battle against the English at Alnwick in 1093. Nevertheless during the reigns of his three sons Edgar (1097-1107), Alexander I (1107-1124) and David I (1124-1153) Norman influence in Scotland gradually increased. During the reign of David I many Normans came to live in Scotland. Dioceses were organised for bishops and new monasteries were founded. Government was reformed. Also in the 12th century many towns or burghs were founded in Scotland and trade flourished. David I was the first Scottish king to found mints and issue his own coins. However Scottish kings had little power. In the west and north chieftains frequently rebelled against the king during the 12th and 13th centuries. Nevertheless in 1265 the Scottish king conquered the Western Islands, which until then were ruled by Norway. By the Treaty of Perth in 1266 the Norwegian king formally surrendered all his territory in Scotland except for the Orkney and Shetland Islands. One night in 1286 Alexander III's horse fell in the darkness and he was killed. His heir was a little girl called Margaret who lived in Norway. However she died in 1290 on her way to Scotland. There were now many claimants to the throne. In fact there were 13. The Bishop of St Andrews asked Edward I to arbitrate. Edward was happy to oblige and he chose John Balliol who was crowned in 1292. Edward claimed to be overlord of Scotland and he soon made it clear he wanted Balliol to be a puppet. Finally in 1295 Edward tried to force the Scots to join him in a war against France. Balliol rebelled and formed an alliance with France. However in 1296 Edward invaded Scotland. Balliol was captured and forced to surrender the throne. Edward tried to rule Scotland directly, without a puppet king. He forced many Scottish nobles and landowners to submit to him at Berwick. He then installed English officials to govern Scotland and withdrew. However the Scots were not subdued so easily. Many small landowners rose in rebellion led by William Wallace. In 1297 Wallace severely defeated the English at Stirling Bridge. However English won a victory at Falkirk in July 1298. Yet the Scots continued to resist and the English only really controlled the southeast. Yet Wallace was captured in 1305 and executed. From 1306 Robert the Bruce, who was crowned king of Scotland that year, led resistance. Scottish resistance gradually increased and Edward I died in 1307. Then in 1314 the English were utterly defeated at the Battle of Bannockburn. After the battle Scottish independence was assured. However it was another 14 years till the English finally recognized Scottish independence by the Treaty of Northampton in 1328. Nevertheless the treaty did not bring peace. Robert the Bruce died in 1329 and his 5-year-old son became David II. However in England there were some Scottish nobles who had been deprived of their lands in Scotland for supporting the English. They now attempted to make John Balliol's son Edward king of Scotland. They invaded Scotland by sea and defeated an army sent to meet them. They marched to Scone where Edward Balliol was crowned king. He tried to get the support of the English king by promising him Berwick. However Balliol was soon driven out of Scotland. Nevertheless the English took Berwick anyway and invaded southern Scotland. King David was sent to France for safety. However after 1338 the English were at war with France and they were gradually forced to withdraw from Scotland. Then in 1346 the French king appealed to the Scots for help. David invaded England but he was defeated and captured at Neville's Cross. David was released in 1357 when the Scot's paid a ransom. He died in 1371. In the late Middle Ages the Scottish kings still had little power and the barons sometimes acted virtually as independent rulers. Accordingly Scotland suffered from lawlessness. On the other hand the burghs thrived and Scotland's first university St Andrews was founded in 1413. Meanwhile during the late 14th and 15th centuries intermittent warfare between the Scots and the English continued. 16th Century Scotland James IV (1488-1513) restored order. Furthermore his reign was a great age for literature in Scotland. Also the first printing press was set up in Edinburgh in 1507. Meanwhile Aberdeen University was founded in 1495 and in 1496 a law was passed requiring all well off landowners to send their eldest sons to school. Then in 1503 James married Margaret, daughter of Henry VII of England. In 1511 James built a huge warship called the Great Michael. However in 1513 he invaded England. The Scots were badly defeated at the battle of Flodden and James himself was killed. His heir James V was only a child and he did not begin to rule Scotland till 1528. The Scots invaded England in 1542 but were defeated at the battle of Solway Moss in November. The king died in December 1542 while still a young man. The throne passed to Mary Queen of Scots, who was only a baby. Henry VIII of England wanted his son to marry Mary. The Regent of Scotland, the Earl of Arran signed the Treaty of Greenwich in 1543, agreeing to the marriage. However in December 1543 the Scottish parliament repudiated the treaty. So in 1544 and 1545 the English invaded southern Scotland and devastated it. The English invaded Scotland again in 1547 and defeated the Scots at Pinkie. The English invaded again in 1548 so Mary was sent to France. Later she married a French prince. In the 16th century Scotland, like the rest of Europe, was rocked by the Reformation. Early in the century Protestant ideas spread through Scotland and gradually took hold. Finally in 1557 a group of Scottish nobles met and signed a covenant to uphold Protestant teachings. However the leading figure in the Scottish Reformation was John Knox (1505-1572). In 1559 he returned from Geneva where he had learned the teachings of John Calvin. Knox's preaching won many converts and finally in 1560 the Scottish parliament met and severed all links with the Pope. Parliament also banned the Catholic mass or any doctrine or practice contrary to a confession of faith drawn up by Knox. The Scottish Reformation had succeeded and Scotland was now a Protestant country. In 1561 Queen Mary returned from France after the death of her husband. Mary was a Catholic. She was forced to accept the Scottish Reformation but she kept her old religion. In 1565 Mary married her Catholic cousin Henry Stuart, Lord Darnley. However Darnley became jealous of Mary's Italian secretary David Riccio. In March 1566 Darnley and his friends murdered Riccio. Mary never forgave Darnley and she came under the spell of the Earl of Bothwell. In 1567 a house where Darnley was staying was blown up. When Darnley's body was found it was discovered that he had been strangled. Shortly afterwards Mary married Bothwell. Enraged, the Protestant nobles rose and captured Mary. They forced her to abdicate in favor of her baby son, who became James VI. Mary escaped and raised an army but she was defeated at the Battle of Langside and fled to England. Scotland was ruled by regents until James was old enough to rule himself. (In 1587 his mother Mary was beheaded in England). In 1589 James married Anne of Denmark. Then in 1603, on the death of Elizabeth I he became King James I of England as well as King James VI of Scotland. 17th Century Scotland However the Scottish Church was different in some of its doctrines and practices from the English Church. James's son Charles I (1625-1649) foolishly tried to bring the Scottish religion in line with the English religion. In 1637 he tried to impose a prayer book on the Scots. However the Scots rejected it utterly. On 28 February 1638 and the following two days nobles and gentlemen in Edinburgh signed a document promising to uphold the 'true religion'. The document became known as the National Covenant and messengers took copies all over Scotland for people to sign. Charles tried to force the Scots to submit and in 1639 he raised an army in England. However he was desperately short of money and he made a peace treaty to buy time. In 1640 Charles raised another army but the Scots invaded England and they occupied Newcastle and Durham. They withdrew in 1641. Meanwhile Charles managed to alienate his English subjects and in 1642 civil war began in England. At first the Scots remained neutral. However in 1643 the English parliament persuaded the Scots to join their side by promising to make England Presbyterian. In 1644 the Scots sent an army to England. Yet not all Scots agreed with this decision. Some supported the king and in 1644 the Marquis of Montrose raised an army in the Highlands to fight for him. At first Montrose had some success but in 1645 he was defeated at Philiphaugh. Meanwhile the king was defeated in England and in 1646 he surrendered to the Scottish army at Newark. Montrose fled to Norway. However the English now dragged their feet about introducing Presbyterianism. When it became clear they were not going to the Scots made a deal with the king. He promised to introduce Presbyterianism in England for a 3-year trial period. So a Scottish army invaded England in 1648 but it was defeated at Preston. Then in January 1649 the English beheaded Charles I. The Scots immediately proclaimed his son Charles II king. Charles II like his father Charles I and his grandfather James VI was an Episcopalian. He believed bishops should govern the Church. Nevertheless to gain the support of the Scots he agreed to accept Presbyterianism in Scotland. In June 1650 he went to Scotland and he was crowned king at Scone in January 1651. Meanwhile in July 1650 another English army invaded Scotland and occupied Edinburgh. In the summer of 1651 they defeated a Scottish army at Inverkeithing. A Scottish army then invaded England. They hoped English royalists would join them but they did not. The Scots were routed at Worcester in September 1651. Charles II fled abroad. The English army then occupied the whole of Scotland. However the English occupation ended in 1660 when Charles II became king of England and Scotland. Charles II restored bishops to the Church of Scotland. However about a third of ministers resigned. Many Scots, especially in the southwest, held secret religious meetings called conventicles. Gradually the government treated them more harshly. Finally in 1679 the Archbishop of St Andrews was murdered and unrest spread through the west. However the government sent troops to quell it and the Covenanters were defeated at the battle of Bothwell Brig. Nevertheless the Covenanters continued to resist and the government continued to persecute them. The 1680s became known as the killing time. Charles II died in 1685 and his brother James became King James II. However James II was a Roman Catholic and both English and Scots feared he would restore Roman Catholicism. James II was deposed in 1688 and William and Mary became king and queen of Scotland. The Scottish parliament restored Presbyterianism. However not all Scots welcomed the new monarchs. The Highlanders rose under Viscount Dundee. They won a victory at Killiecrankie in 1689 but their leader was killed and the Highlanders dispersed. The government was determined to bring the Highlands to heel and they ordered the chiefs of all the clans to take an oath of loyalty to King William by the last day of 1691. However the chief of the MacDonalds of Glencoe arrived late and only took the oath on 6 January 1692. Even though he was only a few days late the government decided to make an example of him. So troops led by Captain Robert Campbell of Glenlyon were sent to Glencoe and billeted in cottages there. The MacDonalds treated them hospitably. However early in the morning of 13 February Campbell and his men fell on the sleeping guests. They went from house to house killing the inhabitants and then burning the houses. Altogether 38 people were murdered including the clan chief. This appalling massacre became known as the massacre of Glencoe. 18th Century Scotland King William realised the deposed king, James II might go to Scotland and claim the Scottish throne. To try and prevent that he urged a union of England and Scotland. The next monarch, Queen Anne did the same. Scottish merchants saw economic advantages from a union and in 1706 they Scots agreed to open negotiations. The Scots wanted a federal union but the English refused. However in 1706 a treaty was drawn up. The two nations would share a flag and a parliament. Scotland would keep its own church and its own legal system. The Scottish parliament accepted the treaty of Union in 1707. The United Kingdom came into existence on 1 May 1707. However the Act of Union was unpopular with many Scots and it soon became more so. Meanwhile James II, the king who was deposed in 1688 died in 1701 but his son James Edward was keen to regain the throne. His followers were called Jacobites from the Latin for James, Jacobus. James had many supporters in the Highlands and in 1715 the Earl of Mar proclaimed him king. Lord Mar also denounced the Act of Union. Highlanders flocked to join Lord Mar and in September 1715 his forces captured Perth. However towns south of the Tay stayed loyal to the government. On 13 November the Jacobites fought government troops at Sheriffmuir near Dunblane. The battle ended indecisively. However afterwards the government army was reinforced. On 22 December 1715 James Edward landed at Peterhead but the government army advanced and the Jacobites withdrew from Perth. James Edward grew discouraged and on 4 February 1716 he and Lord Mar left Scotland. Afterwards the rebellion petered out. However the Highlanders were by no means defeated and they remained a threat to the government. Still the government took some measures to control the Highlands. Fort Augustus was built in 1716 and in 1725-36 General Wade built a network of roads in the Highlands to make it easier for government troops to march from place to place. Then in August 1745 Charles Stuart, grandson of the king who was deposed in 1688 landed in Scotland hoping to reclaim the throne. 'Bonnie Prince Charlie' persuaded some of the Highlanders to support him and in September 1745 they captured Edinburgh. They then routed a government army at Prestonpans. The Jacobites then marched south and in December they reached Derby. However the English failed to rise to support Charles and some of his Highland troops deserted. So on 6 December 1745 the Jacobites began a retreat. They retreated to Inverness but the government was busy raising reinforcements. On 16th April 1746 the Jacobites were totally defeated by a government army at Culloden. Charles Stuart managed to escape to France. The commander of the government army was William Augustus, Duke of Cumberland, known as 'Butcher Cumberland' because of his cruelty. After Culloden Cumberland ordered that the Jacobites should be given no quarter. Many wounded Jacobites were killed. Furthermore 120 prisoners were executed and more than 1,000 were transported to colonies. Following the defeat of the Jacobite rebellion the government passed laws to destroy the Highlanders way of life forever. In 1746 a law banned the kilt and the bagpipes. Lands owned by Jacobites were confiscated and the 'heritable jurisdictions' (the right of clan chiefs to hold courts and try certain cases) were abolished. Despite the Jacobite rebellions Scotland's economy grew rapidly during the 18th century. Landowners were keen to improve their estates and new methods of farming were introduced. Turnips and potatoes were introduced into Scotland. Unfortunately the Highland Clearances caused much suffering. From the 1760s landowners evicted tenant farmers and turn their land over to sheep farming. Many of the dispossessed migrated to North America. Others moved to the rapidly growing industrial cities. In the late 18th century the industrial revolution began to transform Scotland. The linen industry and the cotton industry boomed. The iron industry also grew rapidly. Meanwhile transport improved. Turnpike roads were built. (Those roads were privately owned and maintained and you had to pay to use them). In the late 18th century canals were built in Scotland. Many Scottish towns grew very rapidly especially Glasgow and Paisley. Meanwhile art, learning and architecture flourished in Scotland and Edinburgh was called the Athens of the north. 19th Century Scotland In the 19th century the history of Scotland merged into the history of Britain. In the early 19th century the Highland clearances continued. Many Highlanders were forced to emigrate. Meanwhile further south Scotland's industries boomed. Coal and iron industries flourished. So did ship building. Scottish cities continued to grow rapidly. However housing conditions in the new industrial towns were often appalling. Disease and overcrowding were common. Still in the late 19th century conditions improved and living standards rose. Furthermore at the end of the 19th century Scottish workers began to form powerful trade unions. Meanwhile in the mid-19th century railways were built across Scotland. In 1842 a railway was built from Glasgow to Edinburgh. 20th Century Scotland Scotland suffered very high unemployment during the 1920s and 1930s. Traditional industries such as shipbuilding, mining, iron and steel were badly affected by depression. The Second World War brought a return to full employment and the 1950s and 1960s were years of prosperity. However recession returned in the early 1980s and early 1990s. Nevertheless new hi-tech and service industries grew up in Scotland in the late 20th century to replace the old manufacturing ones and in 1990 Glasgow was made the Cultural Capital of Europe. During the 20th century there was a growing nationalist movement in Scotland. The National Party of Scotland was formed in 1928. In 1934 it changed its name to the Scottish National Party. The first SNP MP was elected in 1945. In 1974 11 SNP MPs were elected. Finally in 1999 Scotland gained its own parliament. 21st Century Scotland Then in 2011 the Scottish Nationalist Party won a majority in the Scottish Parliament. However in a referendum in 2014 a majority of Scots voted against independence. Today the population of Scotland is 5.3 million. A brief history of Aberdeen A brief history of Ayr A brief history of Dumfries A brief history of Dundee A brief history of Edinburgh A brief history of Inverness A brief history of Perth A brief history of St Andrews A brief history of Stirling Last Revised 2019<\|endoftext\|>	3.71875
1,163	# number of multiples of 4 that are multiples of 4 even if you permute their digits How many 4 digit numbers are multiples of 4 no matter how you permute them? (base 10) - Nothing very useful. factorizing the 4 outside. to make it look like a number a4. – Carry on Smiling Jan 21 '13 at 1:19 Probably useful: A number is divisible by four if and only if the last two digits represent a number that is divisible by four. – Austin Mohr Jan 21 '13 at 1:21 $4044$ is divisible by $4$. Does $0444$ count as a permutation? – Henry Jan 21 '13 at 7:34 yes it does. It does not say the premutation needs to be also 4 digits. – Carry on Smiling Jan 22 '13 at 23:36 A number written in base ten is a multiple of $4$ if and only if the two-digit number formed by its last two digits is a multiple of $4$. If these two digits are $ab$ in that order, clearly $b$ must be even. It’s not hard to verify that if $b$ is a multiple of $4$ (i.e., $0,4$, or $8$), then the two-digit number $ab$ is a multiple of $4$ if and only if $a$ is even, while if $b\equiv2\pmod 4$ (i.e., $2$ or $6$), then $ab$ is a multiple of $4$ if and only if $a$ is odd. Now suppose that you have your four-digit number $n$ whose permutations are all multiples of $4$. Clearly $n$ cannot contain an odd digit, so it also cannot contain a digit $2$ or $6$. We conclude that all four digits must be multiples of $4$, i.e., must be $0,4$, or $8$. Now just count the four-digit numbers that can be formed from these three digits, being careful to exclude leading zeroes. - Let's introduce variables for the digits. It is always useful to be able to talk about the objects that you are dealing with, and the most direct way is to give them names. So, let's say the digits you are using are $a,b,c,d$. The number that in base $10$ is written as $abcd$ is really $1000a+100b+10c+d$. Since $1000$ and $100$ are already multiples of $4$, this number will be a multiple of $4$ precisely when $10c+d$ is a multiple of $4$. In particular, of course, $d$ is even. But you are told that no matter how you permute the digits, the result is again a multiple of $4$. So, in particular, $abdc$ is also a multiple of $4$. (Why did I look at this specific number? Well, I already know that $10c+d$ is a multiple of $4$, I may as well try to learn something more about $c$ and $d$ before bringing the other two variables into the mix.) Ok, so $10d+c$ is a multiple of $4$. Subtracting, we see that $9(c-d)$ is a multiple of $4$, so $c-d$ is a multiple of $4$. Now, since the same happens with all the other permutations, we see that also $c-a,c-b,a-b$ are multiples of $4$. If one of them is $2$ or $6$ (remember, they must be even), the others must also be among $2$ and $6$. But $62$ is not a multiple of $4$, so this cannot be. So the only remaining options are that $a,b,c,d$ are among $0,4,8$. Now we see all the possible numbers formed this way work, and counting should be easy. (You may want to double check whether something like $0484$ is a valid number, since it is not a four digit integer. If not, $0$ cannot be used, unless $a=b=c=d=0$.) - +1 for the 1000a + 100b + 10c + d comment, which convinced me that "A number written in base ten is a multiple of 4 if and only if the two-digit number formed by its last two digits is a multiple of 4." – masonk Jan 21 '13 at 16:14 I could be wrong; but a number is divisible by 4 if and only if the last 2 numbers are. There are of course 25 = 100/4 of these combinations of 2 numbers. Then the same applies to the first two which might be permuted with the last two and we have 25^2 = 625 of these numbers. If we do not accept the leading 0, it would be (88/4)25 = 550 numbers or not far from it. More or less. Permutations between digits 1 and 2 and between 3 and 4 are not counted. - Well, this begs the question of why permutations between digits 1 and 2 and between 3 and 4 are not counted, or any other permutations? $(13)(24)$ generates an awfully small subgroup of $S_4$. – Erick Wong Jan 21 '13 at 6:30<\|endoftext\|>	4.40625
1,870	# Formal Teaching of the DP The various instances of exposure to the DP in earlier grades served to prepare the Chinese students to formally learn the DP in the fourth grade. As stated in Section 5.2.4, the G4 textbook devotes two lessons to teaching the DP, including a lesson that introduces the property (lesson 1) and one that applies the property for computation (lesson 2). Although the project only videotaped the first lesson on the DP’s formal introduction, my prior textbook analysis of the second lesson (Ding & Li, 2010) found that Chinese computational lessons also stressed sense-making for the targeted strategies. For instance, in lesson 2, the worked example was to find the total cost of 102 sets of Chinese chess with a price of ¥32. Students were guided to first find the costs for 100 sets and 2 sets of Chinese chess, respectively, and then to find the total cost. This led to the computational strategy of breaking apart 102 to calculate 32 x 1027. Additionally, deep questions in the textbook (e.g., Is this way effective? What property of operation did you use?) highlighted the DP as the underlying property of the computational strategy (see more descriptions in Ding & Li, 2010). Given that this book is based on videotaped lessons, I will focus only on the initial, formal introduction of the DP (lesson 1). As to be described, this lesson contains many instructional insights that can be incorporated into the U.S. textbook and lessons. While the worked example from the textbook was situated in a word problem with an equal groups structure (see Table 5.3, task 4), the Chinese G4 teacher started the lesson with five different tasks (including Table 5.3 Five Review Tasks Used in the Formal Teaching of the DP. Review Task Discussed Resulting DP instance 1 2 4 x 1 3 (10 + 3) x 24 = 10 x 24 + 3 x24 2 Find the perimeter for this rectangle. 10m (10 + 7) x 2 = 10x2 + 7x2 3 Find the area for this rectangle. (15 + 10) x 8 = 15 x 8 + 10 x 8 4 The fourth grade of a school has 6 classes. The fifth grade has 4 classes. Each class gets 24 jumping ropes. Flow many jumping ropes do the fourth and fifth grade get altogether? (6 + 4) x 24 = 6 x 24 + 4 x 24 5 (19 + 51) x 12 = 19 x 12 + 51 x 12 The teacher purposefully chose each task as a different circumstance that could generate an instance of the DP. For instance, after a student calculated 24 x 13 in task 1, the teacher asked follow-up questions to ensure that the class had a clear understanding that 13 could be viewed as 10 + 3. Consequently, the two partial products 10 x 24 and 3 x 24 were derived. The teacher then recorded this computational process on the board as (10 + 3) x 24 = 10 x 24 + 3 x 24, which generated an instance of the DP. The teacher then encouraged the students to solve the rest of the tasks in different ways. Students reported two solutions for each task and then compared them to generate corresponding instances of the DP (see Table 5.3, right). As has been observed in nearly all the Chinese lessons in this book, the teacher asked the students to explain each of their solution steps. It should be noted that the approach to the perimeter problem (task 2) was similar to the NRC (2001) suggestion that a perimeter problem can be solved in multiple ways [e.g., L + L + W + W, 2L + 2W, or 2(L + W)] , which can be used to illustrate the DP. In addition, the area task (task 3) was similar to the kinds of array/area models that were frequently observed in U.S. classrooms (to be elaborated upon later). Overall, the above tasks used to formally introduce the DP demonstrated vertical connections to many lessons the students encountered across earlier grades. As will be reported below, these review tasks provided a foundation for the fourth graders to discover, make sense of, and formally learn the DP. After the five DP instances were generated (Table 5.3, right), the teacher moved the lesson to the next level through the following activities: • • Observe and compare the equations to identify new patterns. • • Pose one more example of each kind to verify the observations. • • Represent findings with words, letters and shapes. • • Formally reveal the name of the DP. The teacher first asked students to observe the five equations to see if they could identify any patterns that were new or noteworthy: T: We have obtained the five equations on the blackboard. Look here. What do you find? Or what do you feel? Observe. Observe carefully. Why are the left and the right sides equal? Recall the Chinese saying, “Review the old to find the new.” What new information do you find? If you find something new, it indicates that you can teach others! What do you find? When he noticed that some students did not seem able to discover the pattern, he adjusted his instruction by asking the class to pose an example similar to any of the review tasks and to explain why both sides of the equation were equivalent. T: Okay, some students may need some guidance from the others... Please pose an example similar to this. Then, you need to explain why they (both sides) are the same using the situation we just recalled. ... No need to compute. You must explain how they are the same. Students were then given time to first work independently and then discuss in small groups before reporting to the class. Two students shared their examples. Both examples (one about perimeter and the other a story problem) were examined to test whether the generated equations made sense. The teacher asked the class, “Can we exhaust examples like this?” When the class admitted that they could not, they agreed to add to the list of the examples to indicate that it could continue on forever. This is a similar teaching move as the one when teaching the CP of addition (see Section 4.3). Students’ explanations of their examples indicated that they were aware of the embedded property, even if they did not formally know the name of the DP yet. As such, the teacher prompted the class, “Can you represent this pattern? Who can try? Can we use words? Who can use words to do this first?” Several students described their observations, which were then explicitly restated as a summary by the teacher. The teacher further asked students if they could represent this statement in other ways. Students then suggested using shapes and letters to represent this property: (A + Q) x = ^x + )x or (a+b)xc=axc+bxc. Next, the teacher prompted students to explain why (a + b) x c = a x c + b x c was a reasonable suggestion. This lifted students’ reasoning to a higher level of abstraction. Through collective discussion, the students were able to explain this formula based on the meaning of multiplication: The left side represents a + b groups of c, whereas the right side represents a groups of c plus b groups of c. It is worth mentioning that the students were also guided to reason about this formula from both directions (from left to right and vice versa). This serves as another instance where Chinese students were given explicit preparation to apply the opposite direction of the DP, a concept which was often overlooked by U.S. textbooks (Ding 8c Li, 2010) and poorly grasped by the U.S. students (Ding et ah, 2019). At the end of this worked example, the teacher revealed the name of the distributive property and requested students think about the meaning of the word of “distributive.” One may notice that instructional activities described above were common across every formal Chinese lesson on the basic properties. First, concrete example tasks were solved in different ways that generated typical instances of the targeted property. Even though these tasks were often familiar to students (due to informal introduction in earlier grades), teachers still asked deep questions that focused on meaning-making and connections between concrete and abstract representations. Next, Chinese teachers tended to shift students’ attention to the structural features of the generated instances. To do so, they prompted students to discover, verbalize, confirm, and represent the identified patterns. During this process, representations further shifted from concrete to abstract and from specific to general, in alignment with the concreteness fading method. Meanwhile, teachers’ questions tended to facilitate student reasoning such as comparisons, example posing, generalization, and abstraction. The above forms of representation uses and teacher questioning provide insights on how TEPS can be used to teach the basic properties of operations.<\|endoftext\|>	4.4375
232	How to Recognize Parallel and Perpendicular Lines Two lines are parallel if they have the same slope. Two lines are perpendicular if their slopes are negative reciprocals of one another. Numbers that are negative reciprocals have a product –1. Consider the following slopes of some lines or line segments: Here are the slopes of the lines that are parallel: m1 = 1/2 and m4 = 5/10 have the same slope. m2 = –2 and m5 = –14/7 also have the same slope. Here are the slopes of the lines that are perpendicular: m1 = 1/2 and m2 = –2 have slopes that are negative reciprocals. m4 = 5/10 and m5 = –14/7 also have slopes whose product is –1. As a matter of fact, because the lines with slopes of 1/2 and 5/10 are equal to one another, they’re both perpendicular to the lines with slopes of –2 and –14/7, which are also equal in slope. It’s one big, happy family.<\|endoftext\|>	3.875
1,147	# Maharashtra Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.4 Questions and Answers. ## Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4 Question 1. If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e-1 = 0.3678. Solution: ∵ m = 1 ∵ X follows Poisson Distribution = e-m × 1 + e-m × 1 = e-1 + e-1 = 2 × e-1 = 2 × 0.3678 = 0.7356 Question 2. If X ~ P($$\frac{1}{2}$$), then find P(X = 3) given e-0.5 = 0.6065. Solution: Question 3. If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e-3 = 0.0497 Solution: ∵ X follows Poisson Distribution Question 4. The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives (i) only two complaints on a given day, (ii) at most two complaints on a given day. Use e-4 = 0.0183. Solution: ∵ m = 1 ∵ X ~ P(m = 4) ∴ p(x) = $$\frac{e^{-m} \cdot m^{x}}{x !}$$ X = No. of complaints recieved (i) P(Only two complaints on a given day) (ii) P(Atmost two complaints on a given day) P(X ≤ 2) = p(0) + p(1) + p(2) = $$\frac{e^{-4} \times 4^{0}}{0 !}+\frac{e^{-4} \times 4^{1}}{1 !}$$ + 0.1464 = e-4 + e-4 × 4 + 0.1464 = e-4 [1 + 4] + 0.1464 = 0.0183 × 5 + 0.1464 = 0.0915 + 0.1464 = 0.2379 Question 5. A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that (i) no car is used on a given day. (ii) some demand is refused on a given day, given e-1.5 = 0.2231. Solution: Let X = No. of demands for a car on any day ∴ No. of cars hired n = 2 m = 1.5 ∵ X ~ P(m = 1.5) Question 6. Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has (i) no defect, (ii) at least one defect. Use e-1 = 0.3678. Solution: ∵ X = No. of defects on a plywood sheet ∵ m = -1 ∵ X ~ P(m = -1) ∴ p(x) = $$\frac{e^{-m} \cdot m^{x}}{x !}$$ (i) P(No defect) P(X = 0) = $$\frac{e^{-1} \times 1^{0}}{0 !}$$ = e-1 = 0.3678 (ii) P(At least one defect) P(X ≥ 1) = 1 – P(X < 1) = 1 – p(0) = 1 – 0.3678 = 0.6322 Question 7. It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has (i) exactly 5 rats (ii) more than 5 rats (iii) between 5 and 7 rats, inclusive. Given e-5 = 0.0067. Solution: X = No. of rats ∵ m = 5 ∴ X ~ P(m = 5) ∴ p(x) = $$\frac{e^{-m} \cdot m^{x}}{x !}$$ (i) P(Exactly five rats) (ii) P(More than five rats) P(X > 5) = 1 – P(X ≤ 5) (iii) P(between 5 and 7 rats, inclusive) P(5 ≤ x ≤ 7) = p(5) + p(6) + p(7) = 0.0067 × 3125 × 0.02 = 0.0067 × 62.5 = 0.42<\|endoftext\|>	4.65625
655	# Lesson 14 Fractional Lengths in Triangles and Prisms Let’s explore area and volume when fractions are involved. ### Problem 1 Clare is using little wooden cubes with edge length $$\frac12$$ inch to build a larger cube that has edge length 4 inches. How many little cubes does she need? Explain your reasoning. ### Problem 2 The triangle has an area of $$7\frac{7}{8}$$ cm2 and a base of $$5\frac14$$ cm. What is the length of $$h$$? Explain your reasoning. ### Problem 3 1. Which expression can be used to find how many cubes with edge length of $$\frac13$$ unit fit in a prism that is 5 units by 5 units by 8 units? Explain or show your reasoning. • $$(5 \boldcdot \frac 13) \boldcdot (5 \boldcdot \frac 13) \boldcdot (8 \boldcdot \frac 13)$$ • $$5 \boldcdot 5 \boldcdot 8$$ • $$(5 \boldcdot 3) \boldcdot (5 \boldcdot 3) \boldcdot (8 \boldcdot 3)$$ • $$(5 \boldcdot 5 \boldcdot 8) \boldcdot (\frac 13)$$ 2. Mai says that we can also find the answer by multiplying the edge lengths of the prism and then multiplying the result by 27. Do you agree with her? Explain your reasoning. ### Problem 4 A builder is building a fence with $$6\frac14$$-inch-wide wooden boards, arranged side-by-side with no gaps or overlaps. How many boards are needed to build a fence that is 150 inches long? Show your reasoning. (From Unit 4, Lesson 12.) ### Problem 5 1. $$2\frac17 \div \frac27$$ 2. $$\frac {17}{20} \div \frac14$$ (From Unit 4, Lesson 12.) ### Problem 6 Consider the problem: A bucket contains $$11\frac23$$ gallons of water and is $$\frac56$$ full. How many gallons of water would be in a full bucket? Write a multiplication and a division equation to represent the situation. Then, find the answer and show your reasoning. (From Unit 4, Lesson 11.) ### Problem 7 There are 80 kids in a gym. 75% are wearing socks. How many are not wearing socks? If you get stuck, consider using a tape diagram. (From Unit 3, Lesson 12.) ### Problem 8 1. Lin wants to save $75 for a trip to the city. If she has saved$37.50 so far, what percentage of her goal has she saved? What percentage remains? 2. Noah wants to save $60 so that he can purchase a concert ticket. If he has saved$45 so far, what percentage of his goal has he saved? What percentage remains? (From Unit 3, Lesson 11.)<\|endoftext\|>	4.8125
1,305	$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ # 2.8: Theory of Existence and Uniqueness A Sketch of the Proof of the Existence and Uniqueness Theorem Recall the theorem that says that if a first order differential satisfies continuity conditions, then the initial value problem will have a unique solution in some neighborhood of the initial value. More precisely, Theorem (A Result For Nonlinear First Order Differential Equations) Let $y'=f(x,y) \;\;\; y(x_0)=y_0$ be a differential equation such that both partial derivatives $f_x \;\;\; \text{and} \;\;\; f_y$ are continuous in some rectangle containing $$(x_0,y_0)$$/ Then there is a (possibly smaller) rectangle containing $$(x_0,y_0)$$ such that there is a unique solution $$f(x)$$ that satisfies it. Although a rigorous proof of this theorem is outside the scope of the class, we will show how to construct a solution to the initial value problem. First by translating the origin we can change the initial value problem to $y(0) = 0.$ Next we can change the question as follows. $$f(x)$$ is a solution to the initial value problem if and only if $f'(x) = f(x,f(x)) \;\;\; \text{and} \;\;\; f(0) = 0.$ Now integrate both sides to get $\phi (t) = \int _0^t f(s,\phi (s)) \, ds .$ Notice that if such a function exists, then it satisfies $$f(0) = 0$$. The equation above is called the integral equation associated with the differential equation. It is easier to prove that the integral equation has a unique solution, then it is to show that the original differential equation has a unique solution. The strategy to find a solution is the following. First guess at a solution and call the first guess $$f_0(t)$$. Then plug this solution into the integral to get a new function. If the new function is the same as the original guess, then we are done. Otherwise call the new function $$f_1(t)$$. Next plug in $$f_1(t)$$ into the integral to either get the same function or a new function $$f_2(t)$$. Continue this process to get a sequence of functions $$f_n(t)$$. Finally take the limit as $$n$$ approaches infinity. This limit will be the solution to the integral equation. In symbols, define recursively $f_0(t) = 0$ $\phi_{n+1} (t) = \int _0^t f(s,\phi_n (s)) \, ds .$ Example Consider the differential equation $y' = y + 2, \;\;\; y(0) = 0.$ We write the corresponding integral equation $y(t) = \int_0^t \left(y(s)+2 \right) \, ds .$ We choose $f_0(t) = 0$ and calculate $\phi_1(t) = \int_0^t \left(0+2 \right) \, ds = 2t$ and $\phi_2(t) = \int_0^t \left(2s+2 \right) \, ds = t^2 + 2t$ and $\phi_3(t) = \int_0^t \left(s^2+2s+2 \right) \, ds = \frac{t^3}{3}+t^2 + 2t$ and $\phi_4(t) = \int_0^t \left(\frac{s^3}{3}+s^2+2s+2 \right) \, ds = \frac{t^4}{3.4}+ \frac{t^3}{3}+t^2 + 2t.$ Multiplying and dividing by 2 and adding 1 gives $\frac{f_4(t)}{2} + 1 = \frac{t^4}{4.3.2}+\frac{t^3}{3.2}+\frac{t^2}{2}+\frac{t}{1}+\frac{1}{1}.$ The pattern indicates that $\frac{f_n(t)}{2} + 1 = \sum\frac{t^n}{n!}$ or $\frac{f(t)}{2} + 1 = e^t.$ Solving we get $f(t) = 2\left(e^t - 1\right).$ This may seem like a proof of the uniqueness and existence theorem, but we need to be sure of several details for a true proof. 1. Does $$f_n(t)$$ exist for all $$n$$. Although we know that $$f(t,y)$$ is continuous near the initial value, the integral could possible result in a value that lies outside this rectangle of continuity. This is why we may have to get a smaller rectangle. 2. Does the sequence $$f_n(t)$$ converge? The limit may not exist. 3. If the sequence $$f_n(t)$$ does converge, is the limit continuous? 4. Is $$f(t)$$ the only solution to the integral equation? ### Contributors • Integrated by Justin Marshall.<\|endoftext\|>	4.4375
6,503	Derivative of lnSolve Derivatives using our free online calculator. Calculate double and triple integrals and get step by step explanation for each solution.Calculus: Integral with adjustable bounds. example. Calculus: Fundamental Theorem of Calculus Derivatives of Composite Functions. As with any derivative calculation, there are two parts to finding the derivative of a composition: seeing the pattern that tells you what rule to use: for the chain rule...So the derivative of f^-1(y) is 1/ (df/dx) BUT you have to write df/dx in terms of y. The derivative of ln y is 1/ (derivative of f = e^x) = 1/e^x. This is 1/y, a neat slope ! Changing letters is OK : The derivative of ln x is 1/x. Watch this video for GRAPHS Ln is the most common way it is written due to being shorter and easier to write. Example Problems As we can see, taking the derivative of ln requires differentiating the function inside of the natural log and dividing that by the function inside of the natural log.The derivative of ln y is 1/ (derivative of f = e^x) = 1/e^x. This is 1/y, a neat slope ! Changing letters is OK : The derivative of ln x is 1/x. Watch this video for GRAPHS. Professor Strang's Calculus textbook (1st edition, 1991) is freely available here.2.Rewrite the right side lnf(x)g(x) as g(x) ln(f(x)): 3.Di erentiate both sides. 4.Solve the resulting equation for y0. Example 1. Find the derivative of y = xx: Solution. Follow the steps of the logarithmic di erentia The Derivative of the Natural Logarithmic Function. If x > 0 x > 0 and y = lnx y = ln. ⁡. x, then. dy dx = 1 x d y d x = 1 x. More generally, let g(x) g ( x) be a differentiable function. For all values of x x for which g′(x)> 0 g ′ ( x) > 0, the derivative of h(x) =ln(g(x)) h ( x) = ln. ⁡. ( g ( x)) is given by.Answer (1 of 7): Since ln2 is a constant (~0.6931) , its derivative is 0. To ellaborate it a bit more: Derivatives can be considered as the slopes of the tangent lines at any x of the original function. The reason why constants don't modify the value of the derivatives is because all they do is...The second derivative of ln(x) is -1/x 2. This can be derived with the power rule, because 1/x can be rewritten as x-1, allowing you to use the rule. Derivative of ln: Steps. Watch this short (2 min) video to see how the derivative of ln is obtained using implicit differentiation.winston salem craigslistDec 21, 2020 · Use logarithmic differentiation to find this derivative. $$\ln y=\ln (2x^4+1)^{\tan x}$$ Step 1. Take the natural logarithm of both sides. $$\ln y=\tan x\ln (2x^4+1)$$ Step 2. Expand using properties of logarithms. Financial derivatives are contracts to buy or sell underlying assets. They include options, swaps Derivatives have four large risks. The most dangerous is that it's almost impossible to know any...The derivative of $\log_a(x)$: \begin{eqnarray} y & = & \log_a(x) \cr x & = & a^y \cr 1 & = & \frac{d}{dx} \left( a^y\right)\cr 1 & = & a^y \ln(a) \frac{dy}{dx} \cr ...Ln is the most common way it is written due to being shorter and easier to write. Example Problems As we can see, taking the derivative of ln requires differentiating the function inside of the natural log and dividing that by the function inside of the natural log. If you have Telegram, you can view post and join Ищи своих right away.How to take the derivative of ln? Ln is the most common way it is written due to being shorter and easier to write. As we can see, taking the derivative of ln requires differentiating the function inside of the natural log and dividing that by the function inside of the natural log. This definition therefore derives its own principle branch from the principal branch of nth roots. How to establish this derivative of the natural logarithm depends on how it is defined firsthand.In morphology, derivation is the process of creating a new word out of an old word, usually by adding a prefix or a suffix. The word comes from the Latin, "to draw off," and its adjectival form is derivational.derivative of ln (x) \square! \square! . Get step-by-step solutions from expert tutors as fast as 15-30 minutes. Solve Derivatives using our free online calculator. Calculate double and triple integrals and get step by step explanation for each solution.which can be translated as "compute the derivative of the outer function with the inner function as argument, and multiply the derivative of the inner function". To complete our scheme, we need the derivatives: we have. f (x) = ln(x) ⇒ f '(x) = 1 x. g(x) = x2 + 1 ⇒ g'(x) = 2x.ruthless season 2 episode 10Firstly log (ln x) has to be converted to the natural logarithm by the change of base formula as all formulas in calculus only work with logs with the base e and not 10. Hence log ( ln x ) = ln ( ln x ) / ln (10) and then differentiating this gives [1/ln (10)] [d (ln (ln x)) / dx].The derivative of ln y is 1/ (derivative of f = e^x) = 1/e^x. This is 1/y, a neat slope ! Changing letters is OK : The derivative of ln x is 1/x. Watch this video for GRAPHS. Professor Strang's Calculus textbook (1st edition, 1991) is freely available here.Instead, the derivatives have to be calculated manually step by step. The rules of differentiation (product rule, quotient rule, chain rule, …) have been implemented in JavaScript code. There is also a table of derivative functions for the trigonometric functions and the square root, logarithm and exponential function.Proof of Derivative of ln(x) The proof of the derivative of natural logarithm $$\ln(x)$$ is presented using the definition of the derivative. The derivative of a composite function of the form $$\ln(u(x))$$ is also included and several examples with their solutions are presented. Derivative definition, derived. See more. How to use derivative in a sentence. Roni Israelov, the President of investment firm Ndvr and the author of several academic papers on derivatives, says...The Derivative of the Natural Logarithmic Function. If x > 0 x > 0 and y = lnx y = ln. ⁡. x, then. dy dx = 1 x d y d x = 1 x. More generally, let g(x) g ( x) be a differentiable function. For all values of x x for which g′(x)> 0 g ′ ( x) > 0, the derivative of h(x) =ln(g(x)) h ( x) = ln. ⁡. ( g ( x)) is given by.Derivative of y = ln u (where u is a function of x). Unfortunately, we can only use the logarithm laws to help us in a limited number of logarithm differentiation question types. Most often, we need to find the derivative of a logarithm of some function of x.For example, we may need to find the derivative of y = 2 ln (3x 2 − 1).. We need the following formula to solve such problems.3.6 Derivatives of Logarithmic Functions Math 1271, TA: Amy DeCelles 1. Overview Derivatives of logs: The derivative of the natural log is: (lnx)0 = 1 x and the derivative of the log base bis: (log b x) 0 = 1 lnb 1 x Log Laws: Though you probably learned these in high school, you may have forgotten them because you didn't use them very much.Derivative. Notice that we would apply softmax to calculated neural networks scores and probabilities first. Cross entropy is applied to softmax applied probabilities and one hot encoded classes calculated...Derivative f' of the function f(x)=cos x is: f'(x) = - sin x for any value of x. Also in this section. Derivative of ln x.See full list on calculushowto.com 123 fast teesThe f here is the external ln, while the g is the internal ln(x). The derivative of the logarithm is. d dx ln(x) = 1 x. so the f '[g(x)] = 1 ln(x) and the g'(x) = 1 x. The final result is. d dx ln(ln(x)) = 1 ln(x) 1 x = 1 xln(x). Answer link.Free derivative calculator - differentiate functions with all the steps. Type in any function derivative to get the solution, steps and graphDirectional Derivatives We know we can write. The partial derivatives measure the rate of change of If this limit exists, this is called the directional derivative of f at the point (a,b) in the direction of u^.Derivative f' of the function f(x)=cos x is: f'(x) = - sin x for any value of x. Also in this section. Derivative of ln x.Approach Then we need to derive the derivative expression using the derive() function. Below are some examples where we compute the derivative of some expressions using NumPy.Aug 27, 2020 · If f is a real-valued function, differentiable and f(x) ≠ 0, then the logarithmic derivative of f is evaluated with the chain rule: f′(x) / f(x). In notation, that’s: Using similar logic, the log derivative of fg is the log derivative of f + log derivative of g. Example problem: Find the logarithmic derivative of ln(√sin x) is the function itself: f′(x)=ex. . The natural logarithm ln(y). Here we present a version of the derivative of an inverse function page that is specialized to the natural logarithm.The derivative of $\log_a(x)$: \begin{eqnarray} y & = & \log_a(x) \cr x & = & a^y \cr 1 & = & \frac{d}{dx} \left( a^y\right)\cr 1 & = & a^y \ln(a) \frac{dy}{dx} \cr ...cool names for charactersHow to take the derivative of ln? Ln is the most common way it is written due to being shorter and easier to write. As we can see, taking the derivative of ln requires differentiating the function inside of the natural log and dividing that by the function inside of the natural log. Derivatives of Trigonomteric Functions. Because trigonometric functions have periodic oscillating An interesting fact about the derivatives of inverse sine and inverse secant is that their domains are...Derivative of y = ln u (where u is a function of x). Unfortunately, we can only use the logarithm laws to help us in a limited number of logarithm differentiation question types. Most often, we need to find the derivative of a logarithm of some function of x.For example, we may need to find the derivative of y = 2 ln (3x 2 − 1).. We need the following formula to solve such problems.Attached is an image of y=ln(ax) with a = 1,2,3,4,5. It also shows all of their derivatives to be the same (the curve at the top). But they are clearly different curves!!!! There must be some point at which one...Since exponential functions and logarithmic functions are so similar, then it stands to reason that their derivatives will be equal as well. Steps for differentiating an exponential function: Rewrite. Multiply by the natural log of the base. Multiply by the derivative of the exponent.pizza perfect dallas paTheorem 15. When δ is invertible on L, the logarithmic derivative Dδ is a bijection between the set of group-like elements in U (L) and L. Indeed, for h ∈ L the Magnus-type equation in L −adl l = δ−1 ( (h)) exp (−adl ) − 1 has a unique recursive solution l ∈ L such that exp (l) = Dδ−1 (h). Derivative calculator find derivative using limit definition. Our differentiation calculator uses The derivative of a function is a basic concept of mathematics. Derivative occupies a central place in...Chapter. Application of derivatives. With the Calculus as a key, Mathematics can be successfully applied to the explanation of the course of Nature." — WHITEHEAD.2.Rewrite the right side lnf(x)g(x) as g(x) ln(f(x)): 3.Di erentiate both sides. 4.Solve the resulting equation for y0. Example 1. Find the derivative of y = xx: Solution. Follow the steps of the logarithmic di erentia Proof of Derivative of ln(x) The proof of the derivative of natural logarithm $$\ln(x)$$ is presented using the definition of the derivative. The derivative of a composite function of the form $$\ln(u(x))$$ is also included and several examples with their solutions are presented. Derivative of ln(x-4)^(1/2). Simple step by step solution, to learn. Simple, and easy to understand, so dont hesitate to use it as a solution of your homework. The derivative of ln x is 1/x. i.e., d/dx (ln x) = 1/x. In other words, the derivative of the natural logarithm of x is 1/x. But how to prove this? Before proving the derivative of ln x to be 1/x, let us prove this roughly by using its graph.Apr 03, 2015 · How do you find the derivative of ln x4? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e 1 Answer Antoine Apr 4, 2015 If y = lnx4 = 4lnx then dy dx = 4 ⋅ 1 x = 4 x Answer link Also check the Derivative Calculator ! Calculadora de Integrales en español Integralrechner auf In the case of antiderivatives, the entire procedure is repeated with each function's derivative, since...Numberbender. 258 тыс. подписчиков. Подписаться. Calculus - Derivative of Ln Functions. CALCULUS: Derivative of Ln Natural Logarithms and Logarithms in Filipino.The derivative of $$\ln(x)$$ is $$\dfrac{1}{x}$$. In certain situations, you can apply the laws of logarithms to the function first, and then take the derivative. Values like $$\ln(5)$$ and $$\ln(2)$$ are constants; their derivatives are zero. $$\ln(x + y)$$ DOES NOT EQUAL $$\ln(x) + \ln(y)$$; for a function with addition inside the natural log ...The derivative of $\log_a(x)$: \begin{eqnarray} y & = & \log_a(x) \cr x & = & a^y \cr 1 & = & \frac{d}{dx} \left( a^y\right)\cr 1 & = & a^y \ln(a) \frac{dy}{dx} \cr ...Derivatives are the Fundamental tools of Calculus. It is very useful for optimizing a loss function with gradient descent in Machine Learning is possible only because of derivatives.How to take the derivative of ln? Ln is the most common way it is written due to being shorter and easier to write. As we can see, taking the derivative of ln requires differentiating the function inside of the natural log and dividing that by the function inside of the natural log. Financial derivatives are contracts to buy or sell underlying assets. They include options, swaps Derivatives have four large risks. The most dangerous is that it's almost impossible to know any...Derivatives of Matrices, Vectors and Scalar Forms. Keywords: Matrix algebra, matrix relations, matrix identities, derivative of determinant, derivative of inverse matrix, dierentiate a matrix.Derivative of constant multiple. Derivative of sum or difference.The derivative of the natural logarithmic function (ln[x]) is simply 1 divided by x. This derivative can be found using both the definition of the derivative and a calculator. Derivatives of logarithmic functions are simpler than they would seem to be, even though the functions themselves come from an important limit in Calculus. What is the derivative of #ln(2x+1)#? calculus Basic-Differentiation-Rules Chain-Rule 0 0. Add a comment Improve this question. Next > < Previous. Sort answers by oldest. black knight patrolderivative definition: 1. If something is derivative, it is not the result of new ideas, but has been (Definition of derivative from the Cambridge Academic Content Dictionary © Cambridge University...Derivatives of Matrices, Vectors and Scalar Forms. Keywords: Matrix algebra, matrix relations, matrix identities, derivative of determinant, derivative of inverse matrix, dierentiate a matrix.How to take the derivative of ln? Ln is the most common way it is written due to being shorter and easier to write. As we can see, taking the derivative of ln requires differentiating the function inside of the natural log and dividing that by the function inside of the natural log. The f here is the external ln, while the g is the internal ln(x). The derivative of the logarithm is. d dx ln(x) = 1 x. so the f '[g(x)] = 1 ln(x) and the g'(x) = 1 x. The final result is. d dx ln(ln(x)) = 1 ln(x) 1 x = 1 xln(x). Answer link.Aug 27, 2020 · If f is a real-valued function, differentiable and f(x) ≠ 0, then the logarithmic derivative of f is evaluated with the chain rule: f′(x) / f(x). In notation, that’s: Using similar logic, the log derivative of fg is the log derivative of f + log derivative of g. Example problem: Find the logarithmic derivative of ln(√sin x) This definition therefore derives its own principle branch from the principal branch of nth roots. How to establish this derivative of the natural logarithm depends on how it is defined firsthand.Section 4.7 Implicit and Logarithmic Differentiation Subsection 4.7.1 Implicit Differentiation. As we have seen, there is a close relationship between the derivatives of $$\ds e^x$$ and $$\ln x$$ because these functions are inverses. Ln is the most common way it is written due to being shorter and easier to write. Example Problems As we can see, taking the derivative of ln requires differentiating the function inside of the natural log and dividing that by the function inside of the natural log. The derivative of ln(x) with respect to x is (1/x) The derivative of ln(s) with respect to s is (1/s) In a similar way, the derivative of ln(2x) with respect to 2x is (1/2x). We will use this fact as part of the chain rule to find the derivative of ln(2x) with respect to x. How to find the derivative of ln(2x) using the Chain Rule:According to the derivative structure all word fall into two big classes: simplexes (or non-derived words) and complexes (or derived words). Simplexes are words that derivationally cannot be...3.6 Derivatives of Logarithmic Functions Math 1271, TA: Amy DeCelles 1. Overview Derivatives of logs: The derivative of the natural log is: (lnx)0 = 1 x and the derivative of the log base bis: (log b x) 0 = 1 lnb 1 x Log Laws: Though you probably learned these in high school, you may have forgotten them because you didn't use them very much.Dec 21, 2020 · Use logarithmic differentiation to find this derivative. $$\ln y=\ln (2x^4+1)^{\tan x}$$ Step 1. Take the natural logarithm of both sides. $$\ln y=\tan x\ln (2x^4+1)$$ Step 2. Expand using properties of logarithms. Instead, we first simplify with properties of the natural logarithm. We have. ln [ (1 + x) (1 + x 2) 2 (1 + x 3) 3 ] = ln (1 + x) + ln (1 + x 2) 2 + ln (1 + x 3) 3. Now the derivative is not so daunting. We have use the chain rule to get. We define logarithms with other bases by the change of base formula.According to the derivative structure all word fall into two big classes: simplexes (or non-derived words) and complexes (or derived words). Simplexes are words that derivationally cannot be...indian groceries near meDerivative. Notice that we would apply softmax to calculated neural networks scores and probabilities first. Cross entropy is applied to softmax applied probabilities and one hot encoded classes calculated...\left( a^x \right) ' = a^x \cdot \ln a.Derivative Proofs. Derivatives of Inverse Trig Functions. To get the derivative of cos, we can do the exact same thing we did with sin, but we will get an extra negative sign.Theorem 15. When δ is invertible on L, the logarithmic derivative Dδ is a bijection between the set of group-like elements in U (L) and L. Indeed, for h ∈ L the Magnus-type equation in L −adl l = δ−1 ( (h)) exp (−adl ) − 1 has a unique recursive solution l ∈ L such that exp (l) = Dδ−1 (h). Derivatives are the Fundamental tools of Calculus. It is very useful for optimizing a loss function with gradient descent in Machine Learning is possible only because of derivatives.Derivative of constant multiple. Derivative of sum or difference.which can be translated as "compute the derivative of the outer function with the inner function as argument, and multiply the derivative of the inner function". To complete our scheme, we need the derivatives: we have. f (x) = ln(x) ⇒ f '(x) = 1 x. g(x) = x2 + 1 ⇒ g'(x) = 2x.Produced in 2003 by Derivative with architectural visionaries Herzog and de Meuron for the then-brand new Prada Epicenter Store in Tokyo, this was the longest-running TouchDesigner installation as of...Derivative f' of the function f(x)=cos x is: f'(x) = - sin x for any value of x. Also in this section. Derivative of ln x.A derivative is a securitized contract whose value is dependent upon one or more underlying assets. Types of Derivatives. Advantages and Disadvantages. What Is a Derivative?Numberbender. 258 тыс. подписчиков. Подписаться. Calculus - Derivative of Ln Functions. CALCULUS: Derivative of Ln Natural Logarithms and Logarithms in Filipino.texas code of criminal procedureThe derivative of ln x is 1/x. i.e., d/dx (ln x) = 1/x. In other words, the derivative of the natural logarithm of x is 1/x. But how to prove this? Before proving the derivative of ln x to be 1/x, let us prove this roughly by using its graph. Theorem 15. When δ is invertible on L, the logarithmic derivative Dδ is a bijection between the set of group-like elements in U (L) and L. Indeed, for h ∈ L the Magnus-type equation in L −adl l = δ−1 ( (h)) exp (−adl ) − 1 has a unique recursive solution l ∈ L such that exp (l) = Dδ−1 (h). y=ln 4x dxd (ln(u))=u1 ⋅dxdu OR we can use properties of logarithms to rewrite the function. (Note that ln 4 is some constant, hence its derivative is 0.)Ln Derivative Calculator Economic! Analysis economic indicators including growth, development Details: Derivative calculator is able to calculate online all common derivatives: sin, cos, tan, ln, exp...The derivative of ln(x) with respect to x is (1/x) The derivative of ln(s) with respect to s is (1/s) In a similar way, the derivative of ln(2x) with respect to 2x is (1/2x). We will use this fact as part of the chain rule to find the derivative of ln(2x) with respect to x. How to find the derivative of ln(2x) using the Chain Rule:Derivatives are extremely useful. They're one of the most powerful tools we can use to build our Wood is a derivative of a tree. The word herb is a derivative the Latin word, herba, meaning grass.Theorem 15. When δ is invertible on L, the logarithmic derivative Dδ is a bijection between the set of group-like elements in U (L) and L. Indeed, for h ∈ L the Magnus-type equation in L −adl l = δ−1 ( (h)) exp (−adl ) − 1 has a unique recursive solution l ∈ L such that exp (l) = Dδ−1 (h). Ln Derivative Calculator Economic! Analysis economic indicators including growth, development Details: Derivative calculator is able to calculate online all common derivatives: sin, cos, tan, ln, exp...So the derivative of f^-1(y) is 1/ (df/dx) BUT you have to write df/dx in terms of y. The derivative of ln y is 1/ (derivative of f = e^x) = 1/e^x. This is 1/y, a neat slope ! Changing letters is OK : The derivative of ln x is 1/x. Watch this video for GRAPHS Derivative of ln(x-4)^(1/2). Simple step by step solution, to learn. Simple, and easy to understand, so dont hesitate to use it as a solution of your homework. Directional Derivatives We know we can write. The partial derivatives measure the rate of change of If this limit exists, this is called the directional derivative of f at the point (a,b) in the direction of u^.how is force measuredAnswer (1 of 7): Since ln2 is a constant (~0.6931) , its derivative is 0. To ellaborate it a bit more: Derivatives can be considered as the slopes of the tangent lines at any x of the original function. The reason why constants don't modify the value of the derivatives is because all they do is...Formulas and graphs of derivatives and integrals of the trigonometric and hyperbolic functions. For the formulas of the integrals, the +C is omitted. Clicking ↓ shows the according graph.which can be translated as "compute the derivative of the outer function with the inner function as argument, and multiply the derivative of the inner function". To complete our scheme, we need the derivatives: we have. f (x) = ln(x) ⇒ f '(x) = 1 x. g(x) = x2 + 1 ⇒ g'(x) = 2x.How to take the derivative of ln? Ln is the most common way it is written due to being shorter and easier to write. As we can see, taking the derivative of ln requires differentiating the function inside of the natural log and dividing that by the function inside of the natural log. In morphology, derivation is the process of creating a new word out of an old word, usually by adding a prefix or a suffix. The word comes from the Latin, "to draw off," and its adjectival form is derivational.According to the derivative structure all word fall into two big classes: simplexes (or non-derived words) and complexes (or derived words). Simplexes are words that derivationally cannot be...derivative of ln(x), with definition & implicit differentiation, proof of derivative of ln(x), derivative of This video works through the Derivative of ln(sqrt(x 1)). This type of derivative would typically be...Derivative of ln(x-4)^(1/2). Simple step by step solution, to learn. Simple, and easy to understand, so don`t hesitate to use it as a solution of your homework. Finding derivative of ln(x). As one more example, let me show how you can use this technique to help find new derivative formulas. ln(x) can be thought of as an implicit curve; all the points on the. x y xy.Financial derivatives are contracts to buy or sell underlying assets. They include options, swaps Derivatives have four large risks. The most dangerous is that it's almost impossible to know any...which can be translated as "compute the derivative of the outer function with the inner function as argument, and multiply the derivative of the inner function". To complete our scheme, we need the derivatives: we have. f (x) = ln(x) ⇒ f '(x) = 1 x. g(x) = x2 + 1 ⇒ g'(x) = 2x.The derivative of ln(x) is 1/x, and is actually a well-known derivative that most put to memory. However, it's always useful to know where this formula comes from, so let's take a look at the ...The derivative of ln(x) is 1/x, and is actually a well-known derivative that most put to memory. However, it's always useful to know where this formula comes from, so let's take a look at the ...arctic cat atv -fc<\|endoftext\|>	4.5
374	If you cannot find a Mapping Toolbox display function that does what you need, you may be able to use a non-mapping MATLAB function. You can use the MATLAB function to add the graphic objects to the display, using latitude and longitude as x and y, and then project the data afterwards. For example, the Mapping Toolbox does not have a function that displays a triangulated surface from random data, but MATLAB does. Create such a map of the seamount data provided with MATLAB. If the displayed objects are already in the right place and do not need to be projected, you can trim them to the map frame and convert them to mapped objects using trimcart and makemapped. They can then be manipulated as if they had been created with map display functions. This is useful when you want to get back the geographic coordinates. You can also combine Mapping Toolbox and MATLAB commands to mix spherical and Cartesian coordinates. An example would be a quiver plot in which the locations of the vectors are geographic, but the lengths are not. In that case, you can use Mapping Toolbox projection calculations and MATLAB graphics commands. Cylindrical projections are the simplest. North is up and east is right, so just project the locations. An extra step may be required for non-cylindrical projections. In these projections, the directions of north and east vary with location. To make the directions agree with the map grid, vectors should be rotated to bring them into alignment. This can be done with the vector transformation function vfwdtran. Consider the same data displayed on a conic projection. Conformal projections are often the best choice for displays like this. They preserve angles, ensuring that the difference between north and east will always be 90 degrees in projected coordinates. Source.<\|endoftext\|>	3.71875
3,053	Stellar evolution and the problem of the ‘first’ stars According to evolutionists, the first chemical elements heavier than hydrogen, helium and lithium formed in nuclear reactions at the centres of the first stars. Later, when these stars exhausted their fuel of hydrogen and helium, they exploded as supernovas, throwing out the heavier elements. These elements, after being transformed in more generations of stars, eventually formed asteroids, moons and planets. But, how did those first stars of hydrogen and helium form? Star formation is perhaps the weakest link in stellar evolution theory and modern big bang cosmology. Especially problematic is the formation of the first stars—Population III stars as they are called. Evolutionists generally agree that star formation began early after the big bang. Supposedly, the cosmic microwave background originated after some 300,000 years, while the first stars and structure in the universe developed after 1 million to 10 million years. Forming so early, the first stars would be expected to have red shifts or z numbers > 10 today. Some researchers believe the first stars formed at z = 20 to 30 or greater.1 However, it should be pointed out that astronomers have not yet observed objects with such large red shifts. Evolutionists hope that the Next Generation Space Telescope (NGST) may be able to find them or at least find fossils (elements formed by the r and s-process2 for example) left behind by these first stars. The ‘first’ stars The process of star formation is assumed to begin with molecular gas clouds like those that are currently observed in the galaxies. The process is envisaged to be gradual, slow, and inefficient. However, present day molecular gas clouds have no relevance to the origin of the very first Population III stars because conditions soon after the big bang were greatly different from what exists now. ‘The very first stars to form in a galaxy, called first-generation stars, are pure hydrogen and helium, for almost no elements heavier than helium were formed in the Big Bang … . When some of these extremely massive stars explode as supernova, they produce even heavier elements, including uranium, the heaviest of all natural elements. These heavy elements are spewed out into the interstellar void and become part of the clouds of gas and dust that coalesce into the second generation and succeeding generations of stars. These stars will have more heavy elements than do the first-generation stars. The Sun is a star of the second generation or later.’3 Stellar evolution theory considers three populations of stars—Population I, II, and III.4 Population III stars are the most significant to the development of the universe, followed by multiple generations of Population I stars. The idea of stellar populations dates back to research in WWII by Walter Baade at Mount Wilson observatory. He categorised Population II and Population I stars (in terms of their metalicity, distribution and motion) and incorporated them into an evolutionary paradigm. In this paradigm, Population II stars are considered to be the older generation of stars. Thus they lack O and B stars (which burn ‘quickly’) and have a higher proportion of red giants. Population I stars are considered young, and have all spectral classes including the O and B, hot blue stars. The idea of Population III stars is a later addition to the Baade paradigm resulting from the development of big bang cosmology. In big bang cosmology, Population III stars are the first generation of stars. As such, Population III stars would contain no metals (elements heavier than helium) with the possible exception of some primordial Li. This distinctive composition means that their spectra would stand out as sharply different from Population II and I stars—that is, if they could be observed today. Unlike the spectra of Population II or I stars, the C/H and Fe/H ratios in Population III stars would not be detectable. In addition, the stellar spectra of Population III stars would reflect the supposed primordial H/He abundance with possible exception of some primordial Li. How did they form? Because they were first, Population III stars would not have formed by the same mechanisms that evolutionists use to explain the origin of Population I stars, which are observed today. There are a number of significant differences. First, evolutionists cannot invoke a supernova to trigger the gas cloud collapse. Supernovae did not occur until after Population III stars had formed and burned all their nuclear fuel. Second, there were no dust grains or heavy molecules in the primordial gas to assist with cloud condensation and cooling, and form the first stars. (Evolutionists now believe that molecular hydrogen may have played a role, in spite of the fact that molecular H almost certainly requires a surface—i.e. dust grains—to form.) Thus, the story of star formation in stellar evolution theory begins with a process that astronomers cannot observe operating in nature today. Also, evolutionists have modified the equations of state used in the computer models that describe the formation of stars from molecular gas clouds (M42 in Orion is a classic example). Another significant change for Population III stars is the introduction of dark matter to alter the calculations for the minimum Jeans mass.1 The minimum Jeans mass, defined by density, temperature, pressure, and gravitational potential, is critical in stellar evolution theory. It is an attempt to define the minimum stellar mass that could form via gas cloud fragmentation.5 Dark matter assumptions highly influence the calculation for the Jeans mass. Certainly, gas clouds like M42 are missing this in the equations used in computer models. Early reports from the 1970s and 1980s showed that evolutionists commonly believed that Population III stars could form with masses ranging from 0.1 to 100 solar masses (M☉).6 Since low masses (< 1 M☉) were considered possible for Population III stars, and since low-mass stars deplete their nuclear fuel more slowly, astronomers concluded that some Population III stars should still exist. Careful searches, however, failed to find any Population III stars. In more recent times the computer models used to predict the masses of Population III stars have been modified and now favour much larger masses (remember that the equations of state have been altered since the early reports.) Some reports now suggest the lower mass range could be 3 to 16 M☉.7 Thus, evolutionists now do not expect to find any Population III stars today because they were all too massive and burnt their nuclear fuel a long time ago. Some of the massive stars are believed to have evolved into white dwarfs, but most are assumed to have exploded as supernovae, seeding the universe with the heavy elements created from the r and s-process. Do they exist? In the big bang model for star-formation we see a big difference between the story for the first unobservable stars and the stars that are observed today. Keep in mind that about 90% of the stars observed today plot on the main sequence of the H-R star diagram. Of these, the majority (about 70% or more) are less than 0.8 M☉. However, evolutionists could not tolerate this situation for the Population III stars, otherwise the universe would be filled with numerous examples to observe. Yet, none have been found. It seems that evolutionists commonly gloss over this part of the story when they attempt to convince the public that they understand the origin of stars (and by implication the origin of people, i.e. carbon, oxygen, and iron in our bodies forged in the stars.) Astronomy recently published some information on the origin of Population III stars: ‘The problem: If water is crucial in the formation of stars in these clouds, how would the first stars have formed since no water was available?’8 The editors answer: ‘Astronomers don’t know for sure how the universe made its first stars, but they do have a reasonably good guess. (As you can imagine, there’s no way to observe the formation of the first generation of stars, so all the work is based upon theoretical considerations.) The best scenario has molecular hydrogen playing the role of the cooling agent. If the clouds from which stars formed were some four to five times denser in the early universe than they are today, then enough collisions between hydrogen atoms would have taken place to create a lot of molecular hydrogen. The big question is: Were the first galaxies that much denser? Obviously the overall density of the universe was much higher back in the early days, but no one knows whether the star-forming clouds were this much denser. ‘Most astronomers would say that the fact that stars do exist tells us that the density was higher back then, because otherwise there would be no stars … Nowadays, of course, nature has found a simpler, easier way to cool the clouds (with water), so that’s what she uses.’8 It seems there is a major problem with the answer given: ‘The first generation of stars likely formed when the universe was only a few million years old (though these ‘Population III’ stars have not yet been identified).’9 The term, ‘have not yet been identified’ leads to the ‘big question’: Where are examples of these first generation stars or Population III stars? There is no evidence that the universe ever had or does contain Population III stars. There is no evidence that the universe ever contained the primordial star forming clouds that contained no metals. The editors failed to point this out in the ‘theoretical’ answer provided. Their answer assumes that Population III stars are real. Indeed, their answer is tantamount to conjecture and circular reasoning based upon the big bang. The editors also failed to identify the critical role of dark matter in the equations of state used to model the formation of Population III stars, as say compared to molecular gas clouds like M42. Recent reports about possible gas giant planets located in M42 have been drawing attention.10 If confirmed, this shows that evolutionists cannot predict the minimum Jeans mass for gas clouds like M42 or what type of stellar mass distribution may form with any reliability. This makes me wonder about how reliable predictions for the minimum Jeans mass may be that are modelled using unobserved primordial star forming clouds and dark matter. No one has observed or can observe the primordial star forming gas clouds that evolutionists believe existed in the early universe, shortly after the big bang. Their existence remains a matter of conjecture, not fact. The formation of Population III stars in big bang cosmology is very dependent upon assumptions of dark matter used in the equations of state to define the minimum Jeans mass. This again is conjecture, not fact. The existence of Population III stars remains untested. ‘Although the search for Population III stars has proved elusive so far, upcoming CMB anisotropy probes (MAP/Planck) will study their signature, and NGST might be able to directly image them.’1 Star formation in stellar evolution theory is a topic that needs to be critically examined. Some of the mechanisms invoked by evolutionists to explain star formation appear plausible when extrapolated over millions and billions of years. However, current theory based upon observations of molecular gas clouds like M42 breaks down when applied to the origin of Population III stars. Other components of the theory, such as the minimum Jeans mass and stellar mass distribution, indicate that, contrary to the impression we are given, evolutionists are far from solving the origin of the myriad stars we do observe. - Ostriker, J.P. and Gnedin, N.Y., Reheating of the universe and Population III, Astrophysical Journal Letters 472:L63, 1996. Also see Bromm, V., Exploring the physics of primordial star formation, American Astronomical Society Meeting 195, #125.05, 12/1999 as cited at the NASA ADS Astronomy abstract service <adswww. harvard.edu>, 13 July 2000. Return to text. - Elements heavier than iron are believed to have formed via a complex network of nuclear reactions known as r and s processes. These involve a competition between neutron-capture and beta-decay. Return to text. - National Audubon Society Field Guide to the Night Sky, Chanticleer Press, p. 22, 1998. Return to text. - Faulkner, D.R., The role of stellar population types in the discussion of stellar evolution, CRSQ 30(1):8–12, 1993 Return to text. - Rose, W.K, Advanced Stellar Astrophysics, Cambridge University Press, pp. 16–17, 1998. Also see Low, C. and Lynden-Bell, D., The minimum Jeans mass or when fragmentation must stop, Royal Astronomical Society, Monthly Notices 176:367–390, August 1976 as cited at the NASA ADS Astronomy abstract service <adswww. harvard.edu>, 29 June 2000. This abstract shows that for normal molecular gas clouds, it is about 0.007 M☉. Return to text. - Silk, J., On the fragmentation of cosmic gas clouds. I—The formation of galaxies and the first generation of stars, Astrophysical J. 211(1):638–648, 1977 and also Silk, J., On the mass range of the first stars, ESO Workshop on Primordial Helium, Garching, West Germany, 2–3 February 1983, Proceedings (A83-50030 24-90) as cited at the NASA ADS Astronomy abstract service <adswww.harvard.edu>, 6 July 2000. Return to text. - Nakamura, F. and Umemura, M., On the mass of Population III Stars, The Astrophysical J. 515(1):239–248, 1999 as cited at the NASA ADS Astronomy abstract service <adswww. harvard.edu>, 6 March 2000. Return to text. - Talking Back, Water, water (almost) everywhere, Astronomy 27(6):16, 1999. Return to text. - Adams F.C. and Laughlin, G., The future of the universe, Sky & Telescope 96(2):34, 1998. Return to text. - NewsNotes, Free-floating planets in the Orion Nebula? Sky & Telescope 100(1):18–19, 2000. Also see Lucas, P.W. and Roche, P.F., A population of very young brown dwarfs and free-floating planets in Orion, Monthly Notices of the Royal Astronomical Society 314(4):858–864, 2000 as cited at the NASA ADS Astronomy Abstract Service, <adswww. harvard.edu>, 11 July 2000.Return to text.<\|endoftext\|>	3.953125
5,933	Origins of Molecular Genetics The concept of genes as carriers of phenotypic information was introduced in the early 19th century by Gregor Mendel, who later demonstrated the properties of genetic inheritance in peas. Over the next 100 years, many significant discoveries lead to the conclusions that genes encode proteins and reside on chromosomes, which are composed of DNA. These findings culminated in the central dogma of molecular biology, that proteins are translated from RNA, which is transcribed from DNA. DNA is comprised of 4 nucleotides or bases, adenine, thymine, cytosine, and guanine (abbreviated to A, T, C, and G respectively) that are organized into a double stranded helix. The order of these 4 nucleotides makes up the genetic code and provides the instructions to make every protein within an organism. Proteins are made up of amino acids. Each amino acid is encoded for by 3 nucleotides termed a codon. As there are only 20 natural amino acids and 64 codon combinations each amino acid is encoded for by multiple codons. Plasmids and Recombinant DNA Technology Techniques in chemistry enable isolation and purification of cellular components, such as DNA, but practically this isolation is only feasible for relatively short DNA molecules. In order to isolate a particular gene from human chromosomal DNA, it would be necessary to isolate a sequence of a few hundred or few thousand basepairs from the entire human genome. Digesting the human genome with restriction enzymes would yield about two million DNA fragments, which is far too many to separate from each other for the purposes of isolating one specific DNA sequence. This obstacle has been overcome by the field of recombinant DNA technology, which enables the preparation of more managable (i.e., smaller) DNA fragments. In 1952, Joshua Lederberg coined the term plasmid, in reference to any extrachromosomal heritable determinant. Plasmids are fragments of double-stranded DNA that typically carry genes and can replicate independently from chromosomal DNA. Although they can be found in archaea and eukaryotes, they play the most significant biological role in bacteria where they can be passed from one bacterium to another by a type of horizontal gene transfer (conjugation), usually providing a benefit to the host, such as antibiotic resistance. This benefit can be context-dependent, and thus the plasmid exists in a symbiotic relationship with the host cell. Like the bacterial chromosomal DNA, plasmid DNA is replicated upon cell division, and each daughter cell receives at least one copy of the plasmid. By the 1970s the combined discoveries of restriction enzymes, DNA ligase, and gel electrophoresis allowed for the ability to move specific fragments of DNA from one context to another, such as from a chromosome to a plasmid. These tools are essential to the field of recombinant DNA, in which many identical DNA fragments can be generated. The combination of a DNA fragment with a plasmid or vector DNA backbone generates a recombinant DNA molecule, which can be used to study DNA fragments of interest, such as genes. Plasmids that are used most commonly in the field of recombinant DNA technology have been optimized for their use of studying and manipulating genes. For instance, most plasmids are replicated in E. coli and are relatively small (∼3000 - 6000 basepairs) to enable easy manipulation. Typically plasmids contain the minimum essential DNA sequences for this purpose, which includes a DNA replication origin, an antibiotic-resistance gene, and a region in which exogenous DNA fragments can be inserted. When a plasmid exists extrachromosomally in E. coli, it is replicated independently and segregated to the resulting daughter cells. These daughter cells contain the same genetic information as the parental cell, and are thus termed clones of the original cell. The plasmid DNA is similarly referred to as cloned DNA, and this process of generating multiple identical copies of a recombinant DNA molecule is known as DNA or molecular cloning. The process of molecular cloning enabled scientists to break chromosomes down to study their genes, marking the birth of molecular genetics. Today, scientists can easily study and manipulate genes and other genetic elements using specifically engineered plasmids, commonly referred to as vectors, which have become possibly the most ubiquitous tools in the molecular biologist’s toolbox. Plasmids used by scientists today come in many sizes and vary broadly in their functionality. In their simplest form, plasmids require a bacterial origin of replication (ori), an antibiotic-resistance gene, and at least one unique restriction enzyme recognition site. These elements allow for the propagation of the plasmid within bacteria, while allowing for selection against any bacteria not carrying the plasmid. Additionally, the restriction enzyme site(s) allow for the cloning of a fragment of DNA to be studied into the plasmid. Below are some common plasmid elements: \|Origin of Replication (ori)\|\|DNA sequence which directs initiation of plasmid replication (by bacteria) by recruiting DNA replication machinery. The ori is critical for the ability of the plasmid to be copied (amplified) by bacteria, which is an important characteristic of why plasmids are convenient and easy to use.\| \|Antibiotic Resistance Gene\|\|Allows for selection of plasmid-containing bacteria by providing a survival advantage to the bacterial host. Each bacterium can contain multiple copies of an individual plasmid, and ideally would replicate these plasmids upon cell division in addition to their own genomic DNA. Because of this additional replication burden, the rate of bacterial cell division is reduced (i.e., it takes more time to copy this extra DNA). Because of this reduced fitness, bacteria without plasmids can replicate faster and out-populate bacteria with plasmids, thus selecting against the propagation of these plasmids through cell division. To ensure the retention of plasmid DNA in bacterial populations, an antibiotic resistance gene (i.e., a gene whose product confers resistance to ampicillin) is included in the plasmid. These bacteria are then grown in the presence of ampicillin. Under these conditions, there is a selective pressure to retain the plasmid DNA, despite the added replication burden, as bacteria without the plasmid DNA would not survive antibiotic treatment. It is important to distinguish that the antibiotic resistance gene is under the control of a bacterial promoter, and is thus expressed in the bacteria by bacterial transcriptional machinery. \|Multiple Cloning Site (MCS)\|\|Short segment of DNA which contains several restriction enzyme sites, enabling easy insertion of DNA by restriction enzymes digestion and ligation. In expression plasmids, the MCS is often located downstream from a promoter, such that when a gene is inserted within the MCS, its expression will be driven by the promoter. As a general rule, the restriction sites in the MCS are unique and not located elsewhere in the plasmid backbone, which is why they can be used for cloning by restriction enzyme digestion. For more information about restriction enzymes check out NEB's website .\| \|Insert\|\|The insert is the gene, promoter, or other DNA fragment cloned into the MCS. The insert is typically the genetic element one wishes to study using a particular plasmid.\| \|Promoter Region\|\|Drives transcription of the insert. The promoter is designed to recruit transcriptional machinery from a particular organism or group of organisms. Meaning, if a plasmid in intended for use in human cells, the promoter will be a human or mammalian promoter sequence. The promoter can also direct cell-specific expression, which can be achieved by a tissue-specific promoter (e.g., a liver-specific promoter). The strength of the promoter is also important for controlling the level of insert expression (i.e., a strong promoter directs high expression, whereas weaker promoters can direct low/endogenous expression levels).\| \|Selectable Marker\|\|The selectable marker is used to select for cells that have successfully taken up the plasmid for the purpose of expressing the insert. This is different than selecting for bacterial cells that have taken up the plasmid for the purpose of replication. The selectable marker enables selection of a population of cells that have taken up the plasmid and that can be used to study the insert. The selectable marker is typically in the form of another antibiotic resistance gene (this time, under the control of a non-bacterial promoter) or a fluorescent protein (that can be used to select or sort the cells by visualization or FACS).\| \|Primer Binding Site\|\|A short single-stranded DNA sequence used as an initiation point for PCR amplification or DNA sequencing of the plasmid. Primers can be utilized to verify the sequence of the insert or other regions of the plasmid. For commonly used primers check out Addgene's sequencing primer list.\| Working with Plasmids Plasmids have become an essential tool in molecular biology for a variety of reasons, including that they are: Easy to work with - Plasmids are a convenient size (generally 1,000-20,000 basepairs) for physical isolation (purification) and manipulation. With current cloning technology, it is easy to create and modify plasmids containing the genetic element that you are interested in. Self-replicating - Once you have constructed a plasmid, you can easily make an endless number of copies of the plasmid using bacteria, which can uptake plasmids and amplify them during cell division. Because bacteria are easy to grow in a lab, divide relatively quickly, and exhibit exponential growth rates, plasmids can be replicated easily and efficiently in a laboratory setting. Stable - Plasmids are stable long-term either as purified DNA or within bacterial cells that have been preserved as glycerol stocks. Functional in many species and can useful for a diverse set of applications - Plasmids can drive gene expression in a wide variety of organisms, including plants, worms, mice, and even cultured human cells. Although plasmids were originally used to understand protein coding gene function, they are now used for a variety of studies used to investigate promoters, small RNAs, or other genetic elements. Types of Plasmids Plasmids are versitile and can be used in many different ways by scientists. The combination of elements often determines the type of plasmid and dictates how it might be used in the lab. Below are some common plasmid types: Cloning Plasmids - Used to facilitate the cloning of DNA fragments. Cloning vectors tend to be very simple, often containing only a bacterial resistance gene, origin of replication, and MCS. They are small and optimized to help in the initial cloning of a DNA fragment. Commonly used cloning vectors include Gateway entry vectors and TOPO cloning vectors. If you are looking for an empty plasmid backbone for your experiment, see Addgene's empty backbone page for more information. Expression Plasmids - Used for gene expression (for the purposes of gene study). Expression vectors must contain a promoter sequence, a transcription terminator sequence, and the inserted gene. The promoter region is required for the generation of RNA from the insert DNA via transcription. The terminator sequence on the newly synthesized RNA signals for the transcription process to stop. An expression vector can also include an enhancer sequence which increases the amount of protein or RNA produced. Expression vectors can drive expression in various cell types (mammalian, yeast, bacterial, etc.), depending largely on which promoter is used to initiate transcription. Gene Knock-down Plasmids - Used for reducing the expression of an endogenous gene. This is frequently accomplished through expression of an shRNA targeting the mRNA of the gene of interest. These plasmids have promoters that can drive expression of short RNAs. Genome Engineering Plasmids - Used to target and edit genomes. Genome editing is most commonly accomplished using CRISPR technology. CRISPR is composed of a DNA endonuclease and guide RNAs that target specific locations in the genome. For more information on CRISPR check out Addgene’s CRISPR guide. Reporter Plasmids - Used for studying the function of genetic elements. These plasmids contain a reporter gene (for example, luciferase or GFP) that offers a read-out of the activity of the genetic element. For instance, a promoter of interest could be inserted upstream of the luciferase gene to determine the level of transcription driven by that promoter. Regardless of type, plasmids are generally propagated, selected for, and the integrity verified prior to use in an experiment. E. coli strains for propagating plasmids E. coli are gram-negative, rod shaped bacteria naturally found in the intestinal tract of animals. There are many different naturally occurring strains of E. coli, some of which are deadly to humans. The majority of all common, commercial lab strains of E. coli used today are descended from two individual isolates, the K-12 strain and the B strain. K-12 has led to the common lab strains MG1655 and its derivatives DH5alpha and DH10b (also known as TOP10) among others, while the B strain gave rise to BL21 and its derivatives. We've included a small number of E. coli strains below and recommend checking out these two Addgene blog posts relating to common E. coli lab strains and E. coli strains specialized for protein expression for additional strain-related information and a more extensive strain list. \|BL21\|\|Invitrogen; New England BioLabs\|\|E. coli B F dcm ompT hsdS(rB mB) gal\| \|ccdB Survival\|\|Invitrogen\|\|F- mcrA Delta(mrr-hsdRMS-mcrBC) Phi80lacZDeltaM15 Delta-lacX74 recA1 araDelta139 D(ara-leu)7697 galU galK rpsL (StrR) endA1 nupG tonA::Ptrc ccdA\| \|DB3.1\|\|Invitrogen\|\|F- gyrA462 endA Delta(sr1-recA) mcrB mrr hsdS20 (rB- mB-) supE44 ara14 galK2 lacY1 proA2 rpsL20(StrR) xyl5 lambda- leu mtl1\| \|DH5alpha\|\|Invitrogen\|\|F- Phi80lacZDeltaM15 Delta(lacZYA-argF) U169 recA1 endA1 hsdR17(rk-, mk+) phoA supE44 thi-1 gyrA96 relA1 tonA\| \|JM109\|\|Addgene; Promega\|\|e14-(McrA-) recA1 endA1 gyrA96 thi-1 hsdR17(rK- mK+) supE44 relA1 Delta(lac- proAB) [F traDelta36 proAB lacIqZDeltaM15]\| \|NEB Stable\|\|New England Biolabs\|\|F' proA+B+ lacIq ∆(lacZ)M15 zzf::Tn10 (TetR) ∆(ara-leu) 7697 araD139 fhuA ∆lacX74 galK16 galE15 e14- Φ80dlacZ∆M15 recA1 relA1 endA1 nupG rpsL (StrR) rph spoT1 ∆(mrr-hsdRMS-mcrBC)\| \|Stbl3\|\|Invitrogen\|\|F– mcrB mrr hsdS20 (rB–, mB–) recA13 supE44 ara-14 galK2 lacY1 proA2 rpsL20 (StrR ) xyl-5 λ– leu mtl-1\| \|Top10\|\|Invitrogen\|\|F- mcrA Delta(mrr-hsdRMS-mcrBC) Phi80lacZM15 Delta-lacX74 recA1 araD139 Delta(ara-leu)7697 galU galK rpsL (StrR) endA1 nupG\| Antibiotics commonly used for plasmid selection Many plasmids are designed to include an antibiotic resistance gene, which when expressed, allows only plasmid-containing bacteria to grow in or on media containing that antibiotic. These antibiotic resistance genes not only give the scientist with an easy way to detect plasmid-containing bacteria, but also provide those bacteria with a pressure to maintain and replicate your plasmid over multiple generations. More information relating to antibiotic resistance genes as well as additional antibiotics not listed in the table below can be found in this blog post. Below you will find a few antibiotics commonly used in the lab and their recommended concentrations. We suggest checking your plasmid's datasheet or the plasmid map to confirm which antibiotic(s) to add to your LB media or LB agar plates. \|Antibiotic\|\|Recommended Stock Concentration\|\|Recommended Working Concentration\| \|Ampicillin\|\|100 mg/mL\|\|100 µg/mL\| \|Carbenicillin\|\|100 mg/mL\|\|100 µg/mL\| (dissolve in EtOH) \|Hygromycin B\|\|200 mg/mL\|\|200 µg/mL\| \|Kanamycin\|\|50 mg/mL\|\|50 µg/mL\| \|Spectinomycin\|\|50 mg/mL\|\|50 µg/mL\| \|Tetracycline\|\|10 mg/mL\|\|10 µg/mL\| Note: Carbenicillin can be used in place of ampicillin. Create a stock solution of your antibiotic. Unless otherwise indicated, the antibiotic powder can be dissolved in dH20. Addgene recommends making 1000X stock solutions and storing aliquots at -20°C. To use, dilute your antibiotic into your LB medium at 1:1,000. For example, to make 100 mL of LB/ampicillin growth media, add 100 μL of a 100 mg/mL ampicillin stock (1000X stock) to 100 mL of LB. DNA sequencing for plasmid verification DNA is made up of 4 bases, adenine, thymine , cytosine, and guanine. The order of these bases makes up the genetic code and provides all the information needed for cells to make proteins and other molecules essential for life. Scientists often “sequence DNA” to identify the order of these four nucleotide bases in a particular DNA strand. Sequencing DNA and understanding the genetic code allows scientists to study gene function as well as identify changes or mutations that may cause certain diseases. Sequencing DNA is extremely important when verifying plasmids to ensure each plasmid contains the essential elements to function and the correct gene of interest. So how do scientists sequence DNA? In 1975, Frederick Sanger developed the process termed Sanger sequencing, sometimes referred to as chain-termination sequencing or dideoxy sequencing. To understand Sanger sequencing, we first need to understand DNA replication. DNA is a double helix, where a base on one strand pairs with a particular base on the other, complementary, strand. Specifically, A pairs with T and C pairs with G. During replication, DNA unwinds and the DNA polymerase enzyme binds to and migrates down the single stranded DNA adding nucleotides according to the sequence of the complementary strand. The replication process can also be done in a test tube to copy DNA regions of interest. In vitro DNA replication requires the 4 nucleotides, a DNA polymerase enzyme, the template DNA to be copied, and a primer. A primer is a small piece of DNA, approximately 18-22 nucleotides, that binds to complementary DNA and acts as a starting point for the DNA polymerase. Thus to replicate a piece of DNA in vitro one has to know some of its sequence to design a effective primer. Sanger sequencing is modeled after in vitro DNA replication but relies on the random incorporation of modified, fluorescently tagged bases onto the growing DNA strand in addition to the normal A, T, C, or G nucleotide. The 4 standard bases are tagged with a different fluorophore so they can be distinguished from one another. Similar to DNA replication, the Sanger sequencing reaction begins when a primer binds to its complementary DNA and the DNA polymerase adding nucleotides. The major difference in this process occurs when the polymerase incorporates a fluorescently tagged nucleotide. Because these special bases do not have a binding site for adding the next nucleotide, the reaction is halted once the fluorescently tagged base is incorporated. Sanger sequencing requires a lot of DNA because the ultimate goal is to have a fluorescently tagged nucleotide at each position in the DNA sequence. Thus, the final result is a group of newly synthesized DNA strands of varying lengths whose last nucleotide is labeled. Once all the newly synthesized DNA is made, the DNA molecules are then separated by size from shortest to longest and "read" using a sequencing machine that recognizes the different fluorescent labels. The machine detects which fluorescently labeled nucleotide is present at the end of each fragment and assembles that information into the DNA sequence. Sanger sequencing results are presented as a sequencing chromatogram which provides the color and intensity of each fluorescent signal. Sanger can sequence approximately 500-1000 bases downstream of the known primer region with very few errors making it an efficient and reliable sequencing method. Next Generation Sequencing Although Sanger sequencing is quick and efficient, it is low throughput and can only sequence short pieces of DNA. This is not extremely useful when trying to sequence an entire plasmid or an organism’s genome. One Sanger sequencing reaction would give you only 20 pieces of a 2,000 piece puzzle. A scientist would need to run a ton of Sanger sequencing reactions on different pieces of DNA to be able to assemble the whole puzzle. That’s where Next Generation Sequencing (NGS) comes in. NGS is a high-throughput, multi parallel sequencing platform that can generate sequencing data for up to 600 billion bases in one reaction. In other words NGS can give you most of the puzzle pieces in only a few reactions.There are multiple approaches to acquire NGS but one of the most commonly used is the Illumina NGS platform. This is the platform used by Addgene’s sequencing partner, seqWell. The actual process of Illumina NGS is not that different from Sanger sequencing. This process, like Sanger, is based on DNA replication and utilizes modified fluorescently tagged nucleotides. During illumina NGS, a long piece of DNA is first fragmented into small pieces, labeled with a short DNA barcode, and amplified. These DNA fragments are attached to a glass slide so that different fragments of DNA, or templates, are spatially separated from each other. These attached DNA templates are then amplified again producing ~1,000 copies of each template. Each template is then replicated using the modified bases and a microscope captures the fluorescent color that is emitted each time a base is added. Again, each base (A,C,T, or G) is labelled with a different color making it easy to identify the order of the DNA strand. Unlike Sanger however, these modified bases can be converted back to a regular base and thus do not halt the reaction. Illumina NGS, therefore does not require any “normal” bases in the reaction. All the sequenced templates are then aligned to each other to assemble the entire sequence or puzzle. It is important to note that NGS platforms in general do not require a specific primer for your DNA of interest thus a completely unknown piece of DNA can be sequenced. At Addgene all incoming plasmids are sequenced with NGS during our quality control process. NGS allows us to sequence entire plasmids providing scientists with even more information to aid in the reproducibility of scientific research. The genetic code can be defined as a set of rules for translating the information encoded by DNA and RNA into proteins. DNA is comprised of 4 nucleotides Adenine (A), Thymine (T), Cytosine (C) and Guanine (G). In the double helix A always pairs with T and C always pairs with G. RNA on the other hand consists of Adenine, Cytosine, Guanine and Uracil (U). Uracil replaces thymine in RNA molecules. Every 3 nucleotides (codons) in a DNA sequence encodes for an amino acid. The genetic code is degenerate thus multiple codons code for each amino acid. There are 20 amino acids plus a start and stop codon. Below you will find helpful resource tables about the genetic code. This table includes the nucleotide and amino acid code in addition to ambiguous bases and common epitope tags. Ambiguous bases are included in a DNA sequence when sequencing is not 100% efficient and the machine cannot distinguish between the 4 labelled nucleotides. Epitope tags on the other hand are commonly used in molecular cloning to tag a gene within a plasmid. DNA and RNA \|Single Letter Code: Primary bases\|\|Nucleobase\| \|Single Letter Code: Ambiguous bases\|\|Nucleobase\| \|B\|\|C, G, or T\| \|D\|\|A, G, or T\| \|H\|\|A, C, or T\| \|K\|\|G or T\| \|M\|\|A or C\| \|N\|\|A, T, C, or G\| \|R\|\|A or G\| \|S\|\|C or G\| \|V\|\|A, C, or G\| \|W\|\|A or T\| \|Y\|\|C or T\| \|Name\|\|Three Letter Code\|\|Single Letter Code\|\|Codons (RNA)\| \|Alanine\|\|Ala\|\|A\|\|GCU, GCC, GCA, GCG\| \|Arginine\|\|Arg\|\|R\|\|CGU, CGC, CGA, CGG, AGA, AGG\| \|Aspartic acid\|\|Asp\|\|D\|\|GAU, GAC\| \|Glutamic Acid\|\|Glu\|\|E\|\|GAA, GAG\| \|Glycine\|\|Gly\|\|G\|\|GGU, GGC, GGA, GGG\| \|Isoleucine\|\|Ile\|\|I\|\|AUU, AUC, AUA\| \|Leucine\|\|Leu\|\|L\|\|UUA, UUG, CUU, CUC, CUA, CUG\| \|Proline\|\|Pro\|\|P\|\|CCU, CCC, CCA, CCG\| \|Serine\|\|Ser\|\|S\|\|UCU, UCC, UCA, UCG, AGU,AGC\| \|Threonine\|\|Thr\|\|T\|\|ACU, ACC, ACA, ACG\| \|Valine\|\|Val\|\|V\|\|GUU, GUC, GUA, GUG\| \|Stop\|\|UAG (amber), UGA (opal), UAA (ochre)\| *AUG is the most common start codon. Alternative start codons include CUG in eukaryotes and GUG in prokaryotes. Common Epitope Tags \|Tag\|\|Amino Acid Sequence\| \|Xpress\|\|DLDDDDK or DLYDDDDK\| \|BAD (Biotin Acceptor Domain)\|\|GLNDIFEAQKIEWHE\| \|Factor Xa\|\|IEGR or IDGR\| \|SV40 NLS\|\|PKKKRKV or PKKKRKVG\| Webpage and Blog References Addgene's blog, including our popular Plasmids 101 series covers topics ranging from the newest breakthroughs in plasmid technologies and research, to overviews of molecular biology basics and plasmid components. There are many different lab techniques that are important in molecular biology and plasmid cloning. Check out our webpages dedicated to lab protocols and videos for specific techniques and tips!<\|endoftext\|>	3.90625
496	An algorithm is a systematic procedure for carrying out a computation. Algorithms are an important part of the study of mathematics in elementary school. The Principles and Standards for School Mathematics (NCTM, 2000) states that “students must become fluent in arithmetic computation” and that “standard algorithms for arithmetic computation are one means of achieving this fluency.” The standard algorithm for two-digit addition with regrouping is introduced in this chapter. The focus throughout this chapter should not only be on how to carry out the algorithm, but also on why the algorithm works. Addition of two-digit numbers is developed sequentially, beginning by finding the sums of numbers that are multiples of ten. For example, in order to find the sum of 20 and 30, children may think of the basic addition fact 2 + 3 and then append the zero to the answer. The important point to emphasize with children is that they really are not adding 2 and 3, but rather 2 tens and 3 tens. Next, children move on to adding two-digit numbers with sums no greater than two-digits. This means that the sum of the tens must be less than 10 tens. This avoids regrouping. That is, having to exchange, for example, 15 tens for one hundred and 5 tens. The addition algorithm is based on our base-ten positional numeration system, and when adding two-digit numbers, place value must be acknowledged. Consider the example of 34 + 52. The first digit on the left is in the tens place and the second digit from the left is in the ones place. It should be explained that, written in expanded form, 34 means 3 tens 4 ones, or 30 + 4. So an addition such as 34 + 52 can be written in the following ways. 34 + 52 86 30 + 4 + 50 + 2 80 + 6 = 86 Adding 34 and 52 is equivalent to finding the sum of 3 tens (30) and 5 tens (50) and then adding to this result the sum of 4 ones and 2 ones. Of course, in standard practice the numbers are not written in expanded form before finding the sum. The process of finding the sum in expanded form can help students develop an understanding of why the algorithm works. Teaching Model 11.4: Add Three Numbers<\|endoftext\|>	4.78125
680	The Protestant Reformation changed Western culture both broadly and deeply starting in the 16th century, shaping the early modern world in ways that are still being felt. Such a significant and complex cultural development may seem too difficult to tackle in a preschool Sunday school class, but concrete, hands-on lessons can help even young children grasp some of the important concepts of the Reformation and the theologian credited with initiating it, Martin Luther. On October 31, 1517, Luther posted his critique of the Roman Catholic Church, the 95 Theses. The anniversary of this event is now known in some Protestant communities as Reformation Day. Preschoolers can enjoy dressing up in 16th-century costumes to celebrate Reformation Day, especially if your church presents it as an alternative to Halloween. Cut adult T-shirts to make children's tabards or tunics, and cut no-sew cloaks and robes out of fabric. Introduce the concept of selling indulgences, a key point of Luther’s argument, with simple coin-toss games, suggests Christian homeschooler Amy Maze. Coins reflect medieval belief that worshipers could reduce dead loved ones’ tenure in Purgatory by making donations to the church. Let preschoolers receive candy or favors, or keep their coins, to signify grace, which Luther and other reformers argued was a gift freely given, not earned. If your Sunday school has outdoor space, hold a three-legged race or tag game to simulate Luther’s flight from religious persecution by Pope Leo X in 1521. Coloring and Decorating Coloring pages have the advantage of providing children something to take home, encouraging families to continue the conversation. A Sunday school guide by the Church of the Lutheran Confession reproduces Luther’s distinctive seal with a color guide. Other opportunities for coloring include pictures of medieval life or significant people, such as Luther or Pope Leo X. Have students decorate a printed Bible verse -- using large text -- in the style of medieval manuscripts. Serving Special Snacks Food is often a highlight of preschool lessons, so incorporate themed snacks into your Reformation class. Since Luther was German, serve traditional German treats like anise cookies, rye bread or doughnuts. Plan a more substantial meal, perhaps involving whole families as part of a Reformation Day celebration, with the children serving food in their costumes. Sausages are easy to find and prepare, but you can also include fancier fare like sauerbraten, a type of roast. In addition to potatoes and sauerkraut, offer asparagus, noodles and dumplings as side dishes. Desserts can include cookies, gingerbread, tortes and the famous Black Forest cake. Reinforce the connection of the food to the topic by reminding children that the Reformation began in Germany, where these dishes are popular. - New Dictionary of the History of Ideas: Reformation - Christian Post: Churches Remembering Martin Luther with Reformation Sunday Observance - Sandra Dodd: Children’s Costumes at SCA Events - Living and Learning at Home: Reformation Day Celebration - Church of the Lutheran Confession, Sunday School: Reformation, Lesson 2 - Research in Germany: Discover Germany, Food and Drink - Elke Jacob/iStock/Getty Images<\|endoftext\|>	3.78125
1,574	In the mid-1980s, scientists began noticing a curious phenomenon in goose nesting grounds along the western edge of Canada’s Hudson Bay. Once-verdant salt marshes were transforming into barren mud flats. With plant cover gone, evaporation accelerated and the soil quickly became too salty for all but a few species. By August, only the reddish salt-tolerant salicornia plant remained in a landscape often littered with dead, bleached willow branches. Lesser snow goose populations had quadrupled since the 1970s. So many geese were arriving at their summer breeding grounds that they were eating not just plant shoots, but roots as well. The non-interventionist camp argued that goose numbers would naturally return to an equilibrium once the food ran out. But others feared that continued overgrazing would irrevocably damage the ecosystem. Humans had to intervene, they said. The interventionists prevailed and, 10 years ago, wildlife agencies ratcheted up hunting pressure. Now they are unsure if the increased culling is working. Further deterioration of an already degraded landscape appears to have halted, but the marshes haven’t begun to recover. Unsure what else they can do, they have adopted a wait-and-see approach for now. Here’s the problem: Further population control, which would probably mean some sort of mass extermination, is as technically difficult as it is ethically questionable. “Here you’re dealing with things that are ethically at the crossroads.” Saltwater marshes, like those where the geese summer, are among the most productive and diverse ecosystems on earth. They’re also rare, occurring only along coasts. For this reason, scientists are particularly concerned about the northern marshes. The transformation of these productive ecosystems into mud flats hurts not only geese, but also a suite of species that lives there. In degraded marshes scientists find fewer bugs, for example, which other birds feed on. They find fewer other birds, like the normally hardy Savannah Sparrow. “If you can damage or impact a really robust species” like the Savannah Sparrow, says Robert Rockwell, a biology professor at City College of New York who has studied the geese for 40 years, “you can only imagine what’s happening to the most delicate species.” In times past, the lesser snow goose wintered in marshes along the Gulf of Mexico. (The greater snow goose, whose population has also risen in recent decades, winters along the US East Coast and breeds in Canada’s High Arctic.) When scientists looked into the population explosion since the ‘70s, they noted that goose numbers had risen in lockstep with the increased agricultural output of rice, corn, and wheat across the US Midwest. Where their numbers were once limited by the winter food supply, now they weren’t. The birds were reaping the benefits of increased farm production and the government subsidies that had boosted it. For scientists, the shift in feeding behavior revealed how adaptable the geese were. They had moved from marshes, which were disappearing, to expanding rice paddies and cornfields. “We used to characterize these things as narrowly defined niche preferences,” says Ken Abraham, a waterfowl and wetlands scientist with the Ontario Ministry of Natural Resources in Peterborough. Snow geese “have broken every rule.” More farming has meant more geese, and more geese mean more pressure on a delicate northern ecosystem. The fact that humans are indirectly responsible for the destruction bolstered the argument that humans should correct the problem. “We’re manipulating nature at many different levels,” says Mr. Abraham. It’s irresponsible “to stand back and say, because it’s far away, we should let nature take its course.” There was also some worry about the geese themselves. “There could be a tremendous crash in their population,” says James Kelley, the US Fish and Wildlife Service’s (FWS) Mississippi Flyway representative in Fort Snelling, Minn. “We don’t think [that] is a prudent thing to let happen.” In the late 1990s, experts recommended halving the population of 3.2 million snow geese by increased hunting. (Some scientists say the actual population was nearer 6 million.) Over the protests of the Humane Society of the United States, which was denied a court injunction, a culling program began in 1999. Hunting seasons were extended and techniques previously prohibited, like electronic calls, were allowed. Now, between 1 million and 1.5 million geese are taken annually, according to the FWS. And yet, it’s unclear if the increased harvest is working. The salt marsh ecosystem seems to have reached a new equilibrium, says Abraham, meaning that further degradation has halted. But the marshes are not recovering. That could mean that goose numbers are still greater than what the habitat can sustain, or that the habitat takes a long time to recover. For the Humane Society, the so-far inconclusive results are evidence that increased hunting was pointless and unnecessarily cruel. “It’s alarmist to suggest that … this is an ecosystem that’s on the verge of collapse,” says John Grandy, senior vice president for wildlife with the Humane Society in Gaithersburg, Md. “There hasn’t been a catastrophe. The habitat still exists. The arctic still exists.” “We’ve just got to live with the notion that this is a vibrant, functioning ecosystem,” he adds. But the geese will eat through a lot more summer habitat before reaching a natural limit, says biology professor Rockwell. “Density dependent regulation,” in which an ecosystem supports a finite number of individuals, “only works if you stay put,” he says. Snow geese don’t. In fact, with the saltwater marshes exhausted, the ever-adaptable geese are moving into freshwater marshes. “Almost everywhere we go, we find more snow geese,” he says. Without intervention, he foresees a landscape dominated by snow geese and salicornia, and not much else. And that raises the sticky prospect of a Plan B. What else can be done? “We’ve pretty much exhausted many of the hunting-related approaches,” says Mr. Kelley of the FWS. Population control at the nesting grounds is too expensive – they’re too remote. And large-scale government-sponsored goose removal is none too savory. States like Missouri do remove nuisance Canada geese, but fewer than 1,000 of them per year. And whoever wants them gone has to foot the bill. (The meat is donated to charities.) But with snow geese, hundreds of thousands of animals would have to be taken. “I don’t think this would be feasible from a public relations standpoint,” says David Graber, a waterfowl biologist with the Missouri Department of Conservation in Columbia, Mo, who has worked on goose-control contingency plans. Public sentiment aside, how would you do it, where would you do it, who would pay for it, and what would you do with the carcasses? “The logistics are just a nightmare,” he says. “I don’t think that direct control is something that’s on the radar screen.”<\|endoftext\|>	4
184	# If 10 balloons cost $0.15, how much would 50 balloons cost? ##### 1 Answer Nov 30, 2016 50 balloons will cost 75 cents. #### Explanation: This can start by looking at a ration:$0.15 : 10 balloons = c : 50 balloons where c is the cost. Rewriting and solving gives: ($0.15)/(10 ballo ons) = c/(50 ballo ons) (50 ballo ons) * ($0.15)/(10 ballo ons) = (50 ballo ons) * c/(50 ballo ons) (50 cancel(ballo ons)) * ($0.15)/(10 cancel(ballo ons)) = cancel(ballo ons) * c/cancel(ballo ons)) (50 $0.15)/10 = c c = 5 $0.15 c =$0.75 or 75 cents<\|endoftext\|>	4.4375
1,627	While doing some reading on lighthouses, I needed a formula to compute the distance to the horizon as a function of height. This formula would give me an idea of how far away a lighthouse could be seen. As I usually do, I started with the Wikipedia and it listed two interesting equations: - Eq. 1: where s is the distance along the Earth’s surface in kilometers and h is the height of the observer in meters. This formula assumes that light moves strictly in straight lines. You will also see it in the form , where is the radius of the Earth. - Eq. 2: This formula assumes that light is refracted and travels along a circular arc. Allowing for refraction means that light will curve around the geometrical horizon (Eq. 1) because means that we will see objects that are just beyond the will compute a horizon distance that is about 7% further than Eq. 1. Figure 1 illustrates this point. While I frequently see refraction modeled using Eq. 2, I have not seen anyone go into the details of Eq. 2 — I am sure this analysis exists — I just have not seen it. So I will dive into the details of Eq. 2 in this post. I have seen the Eq.1 derived in a number of places, so I will not repeat its derivation here. The material presented in this blog is similar to that presented on mirages. I have included some mirage information in Appendix B. Light bends because of density differences in the atmosphere. These density differences are caused by (1) altitude and (2) temperature. Of course, temperature and pressure vary with time. To get some idea of the effect of refraction on the distance to the horizon, we need to assume a “typical” atmosphere. We can derive Eq. 2 using the typical atmosphere assumption, but we need to understand that those conditions are often not present. Here is my problem solving approach: - Show that the air’s index of refraction is a simple function of density. - Develop an expression for the rate of change of air’s index of refraction. - Develop an expression for the radius of curvature of a refracted light beam. - Develop an expression for the arc length along the Earth’s surface that a refracted light beam will traverse. - Simplify the expression by assuming typical atmospheric conditions and common units. Air’s Refractivity as a function of Density Because air’s index of refraction is very close to 1, physicists usually talk in terms of refractivity. The refractivity of air is defined as , where n is the index of refraction. One common formula for the refractivity of dry air is given by Equation 3 (Source). - P is air pressure [kPa]. - T is the air temperature [K]. - is the wavelength of light [cm]. Using the ideal gas law, we can show that temperature and pressure are closely related to density as shown in Equation 4. - V is the volume of air. - m is the mass of air. - nm is the number of moles of air = , where is the molecular weight of air. - R is the universal gas constant. - is the air’s density. Equation 4 states that the ratio of P/T is proportional to (R and MWAir are constants). This means we can restate Equation 3 in the simpler form shown in Equation 5. - is a function of the wavelength of light. The index of refraction is a function of wavelength, as is shown in Figure 2. Note how the shorter wavelengths have a larger index of refraction than the longer wavelengths. For this post, I will assume that the wavelength of light is fixed and that is a constant. Air’s Index of Refraction Rate of Change For light to refract in the atmosphere, it must encounter air with an index of refraction that varies. The air’s index of refraction is a function of its density, which varies with altitude and temperature. We can express the variation of the atmosphere’s index of refraction using Equation 6. - n is the air index of refraction. - z is the altitude. - g is the acceleration due to gravity. - is the atmospheric lapse rate, which is defined as the rate of temperature decrease with height (i.e. ). Equation 6 has a couple of interesting aspects: This value of is known as the autoconvective lapse rate. This rare condition means that light does not refract. It occurs over surfaces that are easily heated, like open fields and deserts. This condition requires that the rate of temperature decrease with altitude to cause an air density increase that exactly cancels the air density decrease with altitude. The change of index with altitude is normally negative because the air density increase with altitude associated with decreasing temperature is not enough to cancel the air density decrease with altitude. Since light bends towards regions of higher density, light projected straight from a lighthouse will refract toward the Earth’s surface under normal circumstances (as shown in Figure 1). We can compute the autoconvective lapse rate as shown in Figure 3. The value I obtain here can be confirmed at this web page. Radius of Curvature for the Refracted Light Beam Assuming a typical atmosphere, we can model the path of a refracted beam of light in the atmosphere as an arc on a circle. Figure 4 shows a derivation for the radius of curvature for a refracted beam of light. The radius of curvature will be constant (i.e. a circle) when is a constant. In the case of a lighthouse beam, . This means we can write the refraction radius as . Arc Length Traversed By A Refracted Light Beam We are going to model the refracted light beam as moving along the arc of a circle of radius larger than that of the Earth. Figure 5 illustrates this situation. Given the geometry shown in Figure 4, we can derive an expression for the arc length upon the Earth’s surface that the light beam traverses as shown in Figure 6. Equation 7 shows the key result from the derivation in Figure 6. - rE is the radius of the Earth. - rR is the radius of the refraction circle. - h is the height of the lighthouse. Notice how Eq.7 can be viewed as Eq. 1 with an enlarged Earth radius of . I see refraction often modeled by engineers who simply use Eq. 1 with an enlarged Earth radius. The radius used depends on the wavelength of the photons under consideration. Simplified Arc Length Expression Figure 7 shows how I obtained Equation 2 from Equation 7. While the derivation was a bit long, I was able to derive Equation 2 from first principles. The derivation shows that the final result is sensitive to the choice of lapse rate, which varies throughout the day. I should note that the use of a refraction radius that is a multiple of the Earth’s radius is often applied to other types of electromagnetic signals. For example, radar systems frequently use a “4/3 Earth radius” for refraction problems (Source). The refraction radius for radar is different than for optical signals because the index of refraction in the radio band is different from that of optical signals. Appendix A: Derivation of Rate of Change of Atmospheric Index of Refraction with Lapse Rate. Appendix B: Excellent Discussion of Index of Refraction, Lapse Rate, and Mirages. I really like the way this author discusses mirages (Source).<\|endoftext\|>	4.03125
570	How Cheenta works to ensure student success? Explore the Back-Story Smallest Perimeter of Triangle \| AIME I, 2015 \| Question 11 Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Smallest Perimeter of Triangle. Smallest Perimeter of Triangle - AIME 2015 Triangle $$ABC$$ has positive integer side lengths with $$AB=AC$$. Let $$I$$ be the intersection of the bisectors of $$\angle B$$ and $$\angle C$$. Suppose $$BI=8$$. Find the smallest possible perimeter of $$\triangle ABC$$.. • is 107 • is 108 • is 840 • cannot be determined from the given information Key Concepts Inequalities Trigonometry Geometry AIME, 2015, Question 11 Geometry Vol I to IV by Hall and Stevens Try with Hints Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$, so $\angle ADB = \angle ADC = 90^o$.Now let $BD=y$, $AB=x$, and $\angle IBD$ =$\frac{\angle ABD}{2}$ = $\theta$.Then $\mathrm{cos}{(\theta)} = \frac{y}{8}$and $\mathrm{cos}{(2\theta)} = \frac{y}{x} = 2\mathrm{cos^2}{(\theta)} - 1 = \frac{y^2-32}{32}$. Cross-multiplying yields $32y = x(y^2-32)$. Since $x,y>0$, $y^2-32$ must be positive, so $y > 5.5$. Additionally, since $\triangle IBD$ has hypotenuse $\overline{IB}$ of length $8$, $BD=y < 8$. Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$, $6.5$, $7$, and $7.5$. However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\frac{32(6)}{(6)^2-32}=48$. Thus the perimeter of $\triangle ABC$ must be $2(x+y) = {108}$. Subscribe to Cheenta at Youtube This site uses Akismet to reduce spam. Learn how your comment data is processed. Knowledge Partner Cheenta is a knowledge partner of Aditya Birla Education Academy<\|endoftext\|>	4.40625
191	This book takes a fresh look at approaches to teaching reading, writing and communication skills. It presents a wealth of innovative ideas specifically designed to support visual learners, including those with autism spectrum conditions and special educational needs. Some children are more responsive to visual stimulation than spoken words, and this book shows how to engage these children in literacy lessons by using strategies that cover everything from the latest assistive technology to getting creative on a limited budget. There are tips for sharing stories with children who find it hard to sit still, supporting reluctant writers, enabling the pre-verbal child to answer questions and helping the child who never stops talking to develop listening skills. The strategies are supported by practical resources, examples and case studies, to show how to instil in children the confidence to create and share their thoughts. This is a must-have resource for special education teachers and coordinators, as well as speech and language therapists, looking for new strategies for teaching literacy.<\|endoftext\|>	3.65625
1,226	## LetsPlayMaths.Com WELCOME TO THE WORLD OF MATHEMATICS # Class 2 Multiplication Basics of Multiplication Multiplication Tables Multiplication of 2-Digit Number by 1-Digit Number Multiplication of 3-Digit Number by 1-Digit Number Multiplication by 10,20,30 Word Problem Multiplication Test Multiplication Worksheet ## Basics of Multiplication In class-I, we have learnt some basic concept of multiplication. Here we will go little in depth of multiplication.As we know, multiplication is nothing, but the repeated addition and it denoted as ‘X’ sign. The result of multiplication is known as product. We have 3 balls in 3 groups. So total number of balls = 3 + 3 + 3 = 9 ## Multiplication Tables In class-I, we have learnt multiplication table up to 5. Here, we will learn till 15. Try to memorize all the below mentioned tables for quicker problem resolution. ## Multiplication of 2-Digit Number by 1-Digit Number Follow the below mentioned rules for this kind of multiplication where there is no carry over. 1. Write both the numbers according to the place value. 2. Bigger number should occupy the upper row and single digit number should occupy the second row as shown below 3. Multiply the single digit present in 2nd row with the ones place of the 2-digit number. Write the result in ones place 4. Multiply the single digit present in 2nd row with the tens place of the 2-digit number. Write the result in tens place Example 1. Multiply 34 by 2. Solution. 1. Write 34 and 2 according to their place value. 34 should remain in the top row. 2. Multiply 2 with 4 = 2 X 4 = 8.Write 8 in the ones column. 3. Multiply 2 with 3 = 2 X 3 = 6 Write 6 in tens place as shown. So, the answer is 68. Example 2.Multiply 26 by 3. Solution. 1. Write 26 and 3 according to their place value. 26 should remain in the top row. 2. Multiply 3 by 6 = 3 X 6 = 18 = 10 ones + 8 Ones = 1 Ten + 8 Ones Write 8 ones in ones place and 1 ten in tens place. This 1 ten is known as carry over. 3. Multiply 3 by 2 = 3 X 2 = 7 Add 1 (carry over) to 7 = 1 + 7 = 8 Write 8 in tens column. So, the answer is 88. ## Multiplication of 3-Digit Number by 1-Digit Number Process of multiplying 3-digit number by 1-digit number is like the process of multiplying 2-digit number by 1-digit number. Let see some examples. Example 1. Multiply 158 by 4. Solution. Step 1. Multiply ones column 8 X 4 = 32 = 3 tens + 2 ones Write 2 in ones place and carry over 3 to tens place. Step 2. Multiply tens column 5 X 4 = 20 Add 3 tens (Carry over) with 20 tens = 20 + 3 = 23 tens = 20 tens + 3 tens = 2 hundred + 3 tens Write 3 tens in tens column and carry over 2 hundreds to hundred column. Step 3. Multiply hundreds column. 1 x 4 = 4 Add 2 hundreds (Carry over) with 4 hundred = 4 + 2 = 6 hundred Write 6 hundred in hundreds column. So , the answer is 632. ## Multiplication by 10,20,30 When we multiply any number with 10,20,30, etc. then multiply the number with the multiplicand without zero and add zero at the end of result. Example 1. Multiply 36 by 10. Solution. 36 X 1 = 36 Now add one zero to the extreme right of 36 i.e. 360. Example 2. Multiply 36 by 20. Solution. 36 X 2 = 72 Add one zero to the extreme right of 72 i.e. 720. Example 3. Multiply 36 by 30. Solution. 36 X 3 = 108 Add one zero to the extreme right of 108 i.e. 1080. ## Word Problem In our day to day life multiplication is used to solve different problems. Let’s have a look at some examples. Example 1. In a color box there are 12 colors present. How many colors will be there in 5 such color boxes. Solution. Number of colors present in 1 box = 12 Number of colors present in 5 such boxes = No. of colors in 1 box X No. of boxes = 12 X 5 = 60 So, 60 colors will be there in 5 color boxes. Example 2. A jar contains 65 chocolates. How many chocolates do 8 similar jars contain? Solution. No. of chocolates in a jar = 65 No. of chocolates present in 8 jars = No. of chocolates in a jar X No. of jars = 65 X 8 = 520 So, there are 520 chocolates in 8 jars. ## Multiplication Test Multiplication Test - 1 Multiplication Test - 2 ## Class-2 Multiplication Worksheet Multiplication Worksheet - 1 Multiplication Worksheet - 2 Multiplication Worksheet - 3<\|endoftext\|>	4.90625
243	1.1 billion-year-old bright pink pigments, extracted from rocks deep beneath the Sahara desert in Africa, have finally been traced.The team from The Australian National University (ANU) discovered the oldest colours in the geological record. Dr. Nur Gueneli from ANU said that the pigments taken from marine black shales of the Taoudeni Basin in Mauritania, West Africa, were more than half a billion years older than previous pigment discoveries. He added, “The bright pink pigments are the molecular fossils of chlorophyll that were produced by ancient photosynthetic organisms inhabiting an ancient ocean that has long since vanished.” The fossils range from blood red to deep purple in their concentrated form, and bright pink when diluted. The researchers crushed the billion-year-old rocks to powder, before extracting and analysing molecules of ancient organisms from them. Senior lead researcher Associate Professor Jochen Brocks from ANU said that the emergence of large, active organisms was likely to have been restrained by a limited supply of larger food particles, such as algae. The research is published in the journal- Proceedings of the National Academy of Sciences.<\|endoftext\|>	3.796875
560	# Thread: vectors? a+b 100x greater than a-b 1. ## vectors? a+b 100x greater than a-b Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be 100 times larger than the magnitude of A - B, what must be the angle between them? I don't even know where to start. I attempt to solve this visually as any help I look up online includes formulas such as the "dot" formula which we have yet to discuss in class or even hint towards. Any help? 2. Originally Posted by thedoge Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be 100 times larger than the magnitude of A - B, what must be the angle between them? I don't even know where to start. I attempt to solve this visually as any help I look up online includes formulas such as the "dot" formula which we have yet to discuss in class or even hint towards. Any help? Hello, the sum of two vectors can be described as the diagonal in a parallelogram with the vectors as its sides. The difference of two vectors gives the second diagonal in this parallelogram. If $\displaystyle \vec {a}=\vec {b}$ then you are dealing with a rhombus and the diagonals are perpendicular that means you are dealing with a right triangle. (See attachment) The angle between the vectors is $\displaystyle 2 \alpha$. the ratio of the length of the diaogonals is 1 : 100. Thus $\displaystyle \tan(\alpha)=\frac{\frac{1}{2} \cdot (\vec{a}-\vec{b})} {\frac{1}{2} \cdot (\vec{a}+\vec{b})}$. That means: $\displaystyle \tan(\alpha)=\frac{\frac{1}{2} \cdot \left(\frac{1}{100}(\vec{a}+\vec{b})\right)} {\frac{1}{2} \cdot (\vec{a}+\vec{b})}=\frac{1}{100}$ Now you can calculate the angle between the vectors. I've got: $\displaystyle \approx 1.146^\circ$ EB 3. Thank you. I somehow ended up with 11.46 and couldnt figure out why. Whoops I just realized this was in the wrong forum. Glad you found it anyway earboth. , , , , , , , , , , , , , , # magnitude to be 10 times greater Click on a term to search for related topics.<\|endoftext\|>	4.4375
393	Identify the line of symmetry in a two-dimensional design or shape SPI 0306.4.3 Links verified on 12/26/2014 - Ask Hannah - describe symmetry in two-dimensional shapes - Finding Lines of Symmetry - examples, links to interactive activity, and links to handouts for students to use as homework - lesson plan from NCTM - Identify lines of symmetry - You Tube video; explanation of identification of lines of symmetry. - Lines of Symmetry of Plane Shapes - Simple explaination of how to identify lines of symmetry. - Line Symmetry - animated lesson on symmetry - Line Symmetry - explanations and worksheets to use with Smartboards or print off for class discussions. - Line Symmetry - examples followed by two problems for students to try, solutions given - Line Symmetry - this is a PowerPoint show with many good examples - Line Symmetry in Polygons - Video that discusses how to identify lines of symmetry. - Line Symmetry Worksheets - Draw the part of the shape that is mising on these worksheets. Scroll to the bottom of the page for the worksheets. Worksheets come with answers. - Lines of Symmetry - drag the shapes that have lines of symmetry into one bin and those that do not into the other bin - Reflection Symmetry - several examples of this type of line symmetry - Symmetry - worksheet, draw a line of symmetry on each shape - 10 shapes on the page - Symmetry - worksheet; tell whether the dotted line on the shape is a line of symmetry. - Symmetry Activity - practice making patterns that have symmetry site for teachers \| PowerPoint show \| Acrobat document \| Word document \| whiteboard resource \| sound \| video format \| interactive lesson \| a quiz \| lesson plan \| to print<\|endoftext\|>	4.3125
829	{[ promptMessage ]} Bookmark it {[ promptMessage ]} Assignment 3 Solutions # Assignment 3 Solutions - Assignment 3 Solutions Each... This preview shows pages 1–3. Sign up to view the full content. Assignment 3 Solutions Each question is worth 1 point unless otherwise marked. Marked out of 10 Factor each of these polynomials: 1. f ( x ) = 2 x 2 - x - 1 f (1) = 0. By Vieta’s formula, 1/2 is the sum of the roots. r 1 + r 2 = 1 / 2 1 + r 2 = 1 / 2 r 2 = - 1 / 2 f ( x ) = ( x - 1)(2 x + 1) = 2( x - 1)( x - 1 / 2) 2. f ( x ) = 5 x 2 + 5 x - 10 f ( x ) = 5( x 2 + x - 2) f (1) = 0. By Vieta’s formula, -1 is the sum of the roots. r 1 + r 2 = - 1 1 + r 2 = - 1 r 2 = - 2 f ( x ) = 5( x - 1)( x + 2) 3. f ( x ) = 2 x 2 - 5 x - 3 f (1) = - 6, f ( - 1) = 4, f (3) = 0 By Vieta’s formula, 5/2 is the sum of the roots. r 1 + r 2 = 5 / 2 3 + r 2 = 5 / 2 r 2 = - 1 / 2 f ( x ) = ( x - 3)(2 x + 1) = 2( x - 3)( x + 1 / 2) 4. f ( x ) = 1 2 x 2 + 2 x + 3 2 This preview has intentionally blurred sections. Sign up to view the full version. View Full Document f ( x ) = 1 2 ( x 2 + 4 x + 3) Since the coefficients are all positive, there are only negative roots. This means the only possible rational roots are 1 and 3. Since f (1) = 0 and f (3) = 0, we get the polynomial: f ( x ) = 1 2 ( x + 1)( x + 3) 5. A child’s age increased by 5 years gives a number which has a square root (whole number). Decreased by 5 years, the child’s age gives the square root. How old is the child? 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\|>	4.65625
2,307	# IGCSE Mathematics Higher: Inequalities ## Scheme of work: IGCSE Foundation: Year 11: Term 1: Inequalities #### Prerequisite Knowledge 1. Solving Linear Inequalities • Concept: Understanding how to solve linear inequalities and represent their solutions on a number line. • Example: $\text{Solve the inequality } 2x – 3 < 7.$ $2x - 3 < 7 \rightarrow 2x < 10 \rightarrow x < 5.$ $\text{Solution: } x < 5 \text{, represented on a number line as an open circle at } 5 \text{ with a line extending to the left.}$ 2. Plotting Straight Line Graphs • Concept: Knowing how to plot straight line graphs using the slope-intercept form $$y = mx + c$$ or by finding and plotting points. • Example: $\text{Plot the graph of } y = 2x + 3.$ $\text{Find points by substituting } x \text{ values: }$ $x = -1 \rightarrow y = 2(-1) + 3 = 1$ $x = 0 \rightarrow y = 3$ $x = 1 \rightarrow y = 5$ $\text{Plot the points and draw the line.}$ 3. Solving Quadratic Equations by Factorisation • Concept: Understanding how to solve quadratic equations by factorising the quadratic expression. • Example: $\text{Solve } x^2 – 5x + 6 = 0 \text{ by factorisation.}$ $x^2 – 5x + 6 = (x – 2)(x – 3) = 0$ $\text{So, } x – 2 = 0 \text{ or } x – 3 = 0 \rightarrow x = 2 \text{ or } x = 3.$ • Concept: Knowing how to sketch the graph of a quadratic equation by identifying its key features such as the vertex, axis of symmetry, and x-intercepts. • Example: $\text{Sketch the graph of } y = x^2 – 5x + 6.$ $\text{Factorise: } y = (x – 2)(x – 3)$ $\text{X-intercepts at } x = 2 \text{ and } x = 3.$ $\text{Vertex: } x = \frac{-b}{2a} = \frac{5}{2} = 2.5 \rightarrow y = (2.5)^2 – 5(2.5) + 6 = -0.25$ $\text{Plot these points and sketch the parabola.}$ #### Success Criteria 1. Solve Quadratic Inequalities in One Unknown • Objective: Solve quadratic inequalities by finding the critical points and testing the intervals to determine the solution set. • Example: $\text{Solve the inequality } x^2 – 5x + 6 < 0.$ $\text{Factorise: } (x - 2)(x - 3) < 0.$ $\text{Critical points: } x = 2 \text{ and } x = 3.$ $\text{Test intervals: } \begin{cases} x < 2 \rightarrow (1 - 2)(1 - 3) = 1 \cdot (-2) = -2 & \text{(negative)} \\ 2 < x < 3 \rightarrow (2.5 - 2)(2.5 - 3) = 0.5 \cdot (-0.5) = -0.25 & \text{(negative)} \\ x > 3 \rightarrow (4 – 2)(4 – 3) = 2 \cdot 1 = 2 & \text{(positive)} \end{cases}$ $\text{Solution: } 2 < x < 3.$ 2. Represent the Solution Set of a Quadratic Inequality on a Number Line • Objective: Accurately represent the solution set of a quadratic inequality on a number line. • Example: $\text{For } x^2 – 5x + 6 < 0, \text{ the solution set } 2 < x < 3 \text{ is represented on the number line as an open interval between 2 and 3.}$ 3. Identify Harder Examples of Regions Defined by Linear Inequalities • Objective: Identify and represent regions defined by multiple linear inequalities, including more complex cases. • Example: $\text{Find the region defined by } y \geq 2x + 1 \text{ and } y \leq -x + 4.$ $\text{Plot both lines and shade the region where } y \geq 2x + 1 \text{ and } y \leq -x + 4 \text{ intersect.}$ 4. Interpret Solutions Graphically • Objective: Accurately interpret the solutions of quadratic and linear inequalities by examining their graphical representations. • Example: $\text{For } x^2 – 5x + 6 < 0 \text{, plot the parabola and identify the interval where the graph is below the x-axis.}$ $\text{For } y \geq 2x + 1 \text{ and } y \leq -x + 4 \text{, identify the overlapping shaded region.}$ #### Key Concepts 1. Solve Quadratic Inequalities in One Unknown • Concept: Quadratic inequalities can be solved by finding the roots of the corresponding quadratic equation and testing intervals to determine where the inequality holds true. • Example: $\text{For } x^2 – 5x + 6 < 0, \text{ factorise to } (x - 2)(x - 3) < 0.$ $\text{Find the roots } x = 2 \text{ and } x = 3 \text{, then test the intervals around the roots.}$ 2. Represent the Solution Set of a Quadratic Inequality on a Number Line • Concept: The solution set of a quadratic inequality can be represented on a number line by using open or closed circles to indicate whether the endpoints are included or excluded, and shading the interval where the inequality holds. • Example: $\text{For } 2 < x < 3 \text{, use open circles at 2 and 3 and shade the interval between them.}$ 3. Identify Harder Examples of Regions Defined by Linear Inequalities • Concept: Regions defined by multiple linear inequalities can be identified by plotting each inequality on the same graph and finding the overlapping region that satisfies all inequalities. • Example: $\text{For } y \geq 2x + 1 \text{ and } y \leq -x + 4, \text{ plot both lines and shade the region that satisfies both conditions.}$ 4. Interpret Solutions Graphically • Concept: The graphical representation of inequalities can help visually identify the solution set by showing where the graphs of the functions lie above, below, or intersect the relevant lines or curves. • Example: $\text{For } x^2 – 5x + 6 < 0, \text{ plot the parabola and identify where it is below the x-axis.}$ $\text{For } y \geq 2x + 1 \text{ and } y \leq -x + 4, \text{ find the overlapping shaded region on the graph.}$ #### Common Misconceptions 1. Solve Quadratic Inequalities in One Unknown • Common Mistake: Incorrectly finding the critical points or misinterpreting the intervals around the roots. • Example: $\text{Incorrect: For } x^2 – 5x + 6 < 0, \text{ solving the quadratic incorrectly as } (x - 2)(x + 3) \text{ instead of } (x - 2)(x - 3).$ $\text{Correct: Factorise correctly and test intervals properly: }$ $\text{Critical points are } x = 2 \text{ and } x = 3.$ $\text{Test intervals: } \begin{cases} x < 2 \rightarrow (1 - 2)(1 - 3) = 1 \cdot (-2) = -2 \, \text{(negative)} \\ 2 < x < 3 \rightarrow (2.5 - 2)(2.5 - 3) = 0.5 \cdot (-0.5) = -0.25 \, \text{(negative)} \\ x > 3 \rightarrow (4 – 2)(4 – 3) = 2 \cdot 1 = 2 \, \text{(positive)} \end{cases}$ $\text{Solution: } 2 < x < 3.$ 2. Represent the Solution Set of a Quadratic Inequality on a Number Line • Common Mistake: Misrepresenting the solution set on the number line by using closed circles instead of open circles or shading the wrong interval. • Example: $\text{Incorrect: Representing } 2 < x < 3 \text{ with closed circles at 2 and 3.}$ $\text{Correct: Use open circles at 2 and 3 and shade the interval between them.}$ 3. Identify Harder Examples of Regions Defined by Linear Inequalities • Common Mistake: Misidentifying the region that satisfies all inequalities or incorrectly shading the graph. • Example: $\text{Incorrect: For } y \geq 2x + 1 \text{ and } y \leq -x + 4, \text{ shading the wrong region or missing the overlapping part.}$ $\text{Correct: Plot both lines accurately and shade the region where } y \geq 2x + 1 \text{ and } y \leq -x + 4 \text{ intersect.}$ 4. Interpret Solutions Graphically • Common Mistake: Misinterpreting the graphical representation by not correctly identifying where the graph is above or below the axis or other lines. • Example: $\text{Incorrect: For } x^2 – 5x + 6 < 0, \text{ thinking the solution is where the parabola is above the x-axis.}$ $\text{Correct: Identify the interval where the graph is below the x-axis, which represents the solution.}$ ### Mr Mathematics Blog #### Estimating Solutions by Rounding to a Significant Figure Explore key concepts, FAQs, and applications of estimating solutions for Key Stage 3, GCSE and IGCSE mathematics. #### Understanding Equivalent Fractions Explore key concepts, FAQs, and applications of equivalent fractions in Key Stage 3 mathematics. #### Transforming Graphs Using Function Notation Guide for teaching how to transform graphs using function notation for A-Level mathematics.<\|endoftext\|>	4.90625
1,567	# Math posted by . The numerator of a fraction is two less than its denominator. When both numerator and denominator are increased by 3, the fraction is incre ased by 3/20. Find the original fraction. Pls include solution and workings thx :) • Math - first denominator -- x first numerator ---- x-2 new denominator -- x+3 new numerator ---- x+1 (x-2)/x + 3/20 = (x+1)/(x+3) times 20x(x+3) , the LCD 20(x+3)(x+1) + 3x(x+3) = 20x(x+1) expanding and simplifying gave me x^2 + 3x - 40 = 0 (x-5)(x+8) = 0 x = 5 or x=-8 if x=5, the original fraction was 3/5 if x=-8 the original fraction was -10/-8 or 5/4 check for 3/5 , new fraction would be 6/8 or 3/4 3/5 + 3/20 = 12/20 + 3/20 = 15/20 = 3/4 but for 5/4, new fraction would be 8/7 5/4 + 3/20 = 28/20 = 7/5 ≠ 8/7 BUT, if we take the unsimplified fraction -10/-8 , new fraction would be -7/-5 = 7/5 So the original fraction would be 3/5 for sure, but also the unsimplified fraction -10/-8 • Math - Let denominator and numerator be x The numerator is 2 less than its denominator so X-2/x It will increased by 3 and will become 3/20 so • Math - 3/5 • Math - Numerator = x New Numerator = x+3 Denominator = x+2 New Denominator = x+5 x/(x+2) +3/20 = (x+3)/(x+5) (20x+3x+6)/(20x+40) = (x+3)/(x+5) Cross Multiply (x+5)(23x+6) = (x+3)(20x+40) 23x^2+6x+115x+30 = 20x^2+40x+60x+120 23x^2 +121x+30 = 20x^2 +100x+120 Take one of the equations to the other side to simplify: 23x^2 +121x+30 -(20x^2 +100x+120) =0 3x^2 -21x-90 =0 Divide by 3 x^2 -7x-30=0 x^2 +10x-3x-30=0 x(x+10) -3(x+10) =0 (x+10)(x-3) =0 x= -10 or x =3 x =3 (because its positive) so x+2 =3+2 =5 The Original fraction was : x/x+2 3/5 <---- Answer Check: 3/5+3/20 = 3+3/3+5 15/20 = 6/8 3/4 =3/4 Correct! • Math - Numerator = x New Numerator = x+3 Denominator = x+2 New Denominator = x+5 x/(x+2) +3/20 = (x+3)/(x+5) (20x+3x+6)/(20x+40) = (x+3)/(x+5) Cross Multiply (x+5)(23x+6) = (x+3)(20x+40) 23x^2+6x+115x+30 = 20x^2+40x+60x+120 23x^2 +121x+30 = 20x^2 +100x+120 Take one of the equations to the other side to simplify: 23x^2 +121x+30 -(20x^2 +100x+120) =0 3x^2 -21x-90 =0 Divide by 3 x^2 -7x-30=0 x^2 +10x-3x-30=0 x(x+10) -3(x+10) =0 (x+10)(x-3) =0 x= -10 or x =3 x =3 (because its positive) so x+2 =3+2 =5 The Original fraction was : x/x+2 3/5 <---- Answer Check: 3/5+3/20 = 3+3/3+5 15/20 = 6/8 3/4 =3/4 Correct! ## Similar Questions 1. ### math The numerator of a fractionis 5 less than twice the denominator. If the numerator is decreased by 1 and the denominator is increased by 3, the value of the new fraction is 5/4. Find the original fraction. 2. ### Maths When the numerator and denominator of a fraction are each increased by 5, the value of the fraction becomes 3/5 When the numerator and denominator of that same fraction are each decreased by 5, the fraction is then 1/5 Find the original … 3. ### math The denominator of a fraction is 4 more than the numerator. If both the denominator and the numerator are increased by 1, the resulting fraction equals 1/2. What is the original fraction? The denominator of a fraction is 4 more than the numerator. If both the denominator and the numerator are increased by 1, the resulting fraction equals ½. What is the original fraction? 5. ### Mathematics The numerator of a fraction is two less than its denominator. When both numerator and denominator are increased by 3, the fraction is increased by 3/20. Find the original fraction. Explain your solution 6. ### Algebra When nine is added to both the numerator and the denominator of an original fraction, the new fraction equals six sevenths. When seven is subtracted from both the numerator and the denominator of the original fraction, a third fraction … 7. ### MATHS The numerator of a fraction is 4 less than twice the denominator. If both the numerator and the denominator are increased by 1, the fraction becomes 3/2. What is the original fraction? 8. ### algebra The denominator of a fraction is 4 less than the numerator .If the denominator is doubled and the numerator is increased by 6,the sum of the original fraction and the new one is 3 .Find the original fraction. 9. ### math Solve this plxxx The numerator of a fraction is 2less than it's denominator.when both numerator and denominator are increased by 3,the fraction is increased by 3/20.find the original fraction. 10. ### math the numerator of a fraction is 3 less than the denominator. if 4 us added to the numerator and to the denominator, the resulting fraction is equivalent to 3/4. find the original fraction. Show your solution More Similar Questions<\|endoftext\|>	4.375
1,501	# What are the multiples of 8 from 1 to 1000? ## What are the multiples of 8 from 1 to 1000? 0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360, 368, 376, 384, 392, 400, 408, 416, 424, 432, 440, 448, 456, 464, 472, 480, 488, 496, 504, 512, 520, 528, … ## How do you know if a number is a multiple of 8? Just check the last three digits of the number. If they are divisible by 8 , the entire number is divisible by 8 . For example, last three digits of 53927592 are 592 and they are divisible by 8 as 592=74×8 . People also asking: What does it mean to gloat over someone? ## Is the number 8 a multiple of 2? The first 10 multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20. ## What is a factor of 8? A number or an integer that divides 8 exactly without leaving a remainder, then the number is a factor of 8. As the number 8 is an even composite number, it has more than two factors. Thus, the factors of 8 are 1, 2, 4 and 8. ## What are the prime multiples of 8? 8, 16, 32, 40, 48, 80. ## What is multiple of a number? A multiple in math are the numbers you get when you multiply a certain number by an integer. For example, multiples of 5 are: 10, 15, 20, 25, 30…etc. Multiples of 7 are: 14, 21, 28, 35, 42, 49…etc. ## How do you find multiples of a number? To find multiples of a number, multiply the number by any whole number. For example, 5 × 3 = 15 and so, 15 is the third multiple of 5. For example, the first 5 multiples of 4 are 4, 8, 12, 16 and 20. 1 × 4 = 4, therefore the 1st multiple of 4 is 4. ## What is the multiple of 9? Solutions. The first ten multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90. ## What is the least multiple of 8? Multiples of 8 are: 8, 16, 24, 32, 40, …. So 24 is the least common multiple (I can’t find a smaller one!) Hint: We can have smaller lists for the bigger numbers. People also asking: Why was the first steamboat invented? ## What is the multiple of 7? The multiples of 7 between 1 to 100 are 7 , 14 , 21 , 28 , 35 , 42 , 49 , 56 , 63 , 70, 77, 84, 91, 98. ## What is the third multiple of 8? 24 If we look at 24, it is the third multiple of 8, i.e., 8 x 3 = 24. The second option isn’t a multiple of 8, then 56 is the 7th multiple of 8. Lastly, 40 is the 5th multiple of 8, it’s because 8 x 5 = 40. ## What are the divisors of 8? What is the list of divisors from 1 to 100? Number List of Divisors Divisors of 8 1,2,4,8 Divisors of 9 1,3,9 Divisors of 10 1,2,5,10 Divisors of 11 1,11 ## What are the multiples of 2? First 10 Multiples of 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20. In other words, any number that can be denoted in form 2n where n is a natural number is a multiple of 2. ## What is the sum of first multiples of 8? The first 15 multiples of 8 are 8, 16, 24, 32,……. This is an AP in which a = 8,d =( 16-8) =8 and n =15. Hence, the required sum is 960. ## What is a prime factors of 8? Factors of 8 By Prime Factorization The factors of 8 by the prime factorization method are 1, 2, 4, and 8. Here, 2 is the prime factor of 8. ## What are the multiple of 10? The first 10 multiples of 10 are 10, 20, 30, 40, 50, 60, 70, 80, 90 and 100. People also asking: Who is James Berwind? ## What is the trick of table of 8? The Multiplying 2 Trick To Learn The Table Of 8: Here is one of the easiest tricks to master the table of 8. The number 8 is triple of 2, meaning that 8 is the result you get when you multiply the number 2 three times. 2 x 2 x 2 = 8. So, to learn the table of 8, double a number 3 times. ## What’s the 8 times table? 8 Times Table up to 10 8 × 1 = 8 8 × 6 = 48 8 × 2 = 16 8 × 7 = 56 8 × 3 = 24 8 × 8 = 64 8 × 4 = 32 8 × 9 = 72 ## What is the multiplication table of 8? What is the multiplication table of 8? 8, 16, 24, 32, 40, 48, 56, 64, 72, 80. Scroll to Top Scroll to Top<\|endoftext\|>	4.40625
986	Self-Awareness and Self-Regulation What do you do to increase your energy after the midday slump? How do you wind down at the end of a busy day? These are techniques you’ve consciously or unconsciously chosen to regulate your energy. “How Does Your Engine Run?”® program helps Pathfinders students find their metaphorical cup of coffee or bubble bath. “How Does Your Engine Run?”® is one of the foundational programs we use at Pathfinders. This program promotes self-awareness and self-regulation – big words that just mean helping our students check in with themselves and change their energy levels to make it easier to learn and focus. Let’s look at this program more closely then discuss why it is so important. We introduce “How Does Your Engine Run?”® by asking our students to imagine their body as a car engine. A car engine can run on high speed, low speed or somewhere in the middle, which we call “just right.” These speeds are on a spectrum. A high engine can by incredibly high or slightly high. A low engine can be kind of low or very low. And, a just right engine can be heading toward a high speed or low speed. We visually label a low engine as a blue color, a just right engine as green, and a high engine from yellow to red. However, it is very common for a student to have a blue/green engine or a yellow/green engine. To help our students identify where their engine is we ask them to think about how they feel and look at what their body is doing. Does the student’s body feel heavy, like they don’t want to move? Maybe their engine is running low. Does she want to make silly noises and tap her foot under the desk? It sounds like her engine might be high. Does he want to sit quietly and attend to his trainer’s directions? This feels like a just right engine. As their trainer, we obviously can’t feel what they are feeling, but we can observe their behavior. Are they slumped over the desk, head held up by their hand? Are they wiggling their legs in the chair and looking all over the room? Are they sitting up straight with their eyes focused on the present activity? These observations can help a trainer, parent, guardian or teacher guide a student in identifying their engine level. Once our students can identify how their engine is running, they need to decide if that is an appropriate speed for what they are doing. High and low engines are not inherently “bad” engines to have. A low engine will make falling asleep much easier than a just right or high engine. And, it’s quite normal for our engines to be low right after we wake up in the morning or after a long soccer game. A high engine is appropriate for a lively birthday party or when playing that soccer game. A just right engine is best during school and homework, where the student needs to feel calm and focused. If our student has identified her engine and decided it is not appropriate for the current task, she needs to change her engine. If her engine is too low, she needs to increase her energy. Maybe she can do some Cross Crawls, jump on a trampoline or play a quick game. If her engine is too high, she can take some slow deep breaths, have her trainer do relaxing joint compressions on her hands or take slow, deep belly breaths. Each of our students finds which methods work best for them to regulate their engines. This is one of our most important programs because it is all about self-awareness. Being self-aware is the only way we can fully understand and improve ourselves, children and adults alike. All of our programming at Pathfinders requires this self-awareness to create change in the body and brain. This mindfulness is what makes our students able to notice how they are doing an activity and then change the way they do it to improve their body skills or neural pathways. Self-awareness also helps us know when we need to take a break and what that break needs to be to renew our focus. This decreases frustration levels and ultimately promotes more efficient work habits. Checking your engine is a wonderful introduction to self-awareness and mindfulness. This is a skill that is beneficial at school, work, home… everywhere, really. “How Does Your Engine Run?® program is an incredibly useful tool for our Pathfinders students and is easily transferred into daily life. When was the last time you checked your engine? “How Does Your Engine Run?”® The Alert Program® for Self-Regulation by Mary Sue Williams and Sherry Shellenberger (TherapyWorks, Inc.)<\|endoftext\|>	3.765625
1,181	Measurement of Pressure with the Manometer Pressure is defined as a force per unit area - and the most accurate way to measure low air pressure is to balance a column of liquid of known weight against it and measure the height of the liquid column so balanced. The units of measure commonly used are inches of mercury (in. Hg), using mercury as the fluid and inches of water (in. w.c.), using water or oil as the fluid. Fig. 2-1. In its simplest form the manometer is a U-tube about half filled with liquid. With both ends of the tube open, the liquid is at the same height in each leg. Fig. 2-2. When positive pressure is applied to one leg, the liquid is forced down in that leg and up in the other. The difference in height, "h," which is the sum of the readings above and below zero, indicates the pressure. Fig. 2-3. When a vacuum is applied to one leg, the liquid rises in that leg and falls in the other. The difference in height, "h," which is the sum of the readings above and below zero, indicates the amount of vacuum. Instruments employing this principle are called manometers. The simplest form is the basic and well-known U-tube manometer. (Fig. 2-1). This device indicates the difference between two pressures (differential pressure), or between a single pressure and atmosphere (gage pressure), when one side is open to atmosphere. If a U-tube is filled to the half way point with water and air pressure is exerted on one of the columns, the fluid will be displaced. Thus one leg of water column will rise and the other falls. The difference in height "h" which is the sum of the readings above and below the half way point, indicates the pressure in inches of water column. Fig. 2-4. At left, equal pressure is imposed on the fluid in the well and in the indicating tube. Reading is zero. At the right, a positive pressure has been imposed on the liquid in the well causing the level to go down very slightly. Liquid level in indicating tube has risen substantially. Reading is taken directly from scale at liquid level in indicating tube. The scale has been compensated for the drop in level in the well. The U-tube manometer is a primary standard because the difference in height between the two columns is always a true indication of the pressure regardless of variations in the internal diameter of the tubing. This principle makes even the Dwyer Slack Tube® roll-up manometer as accurate as a laboratory instrument. This provides a real convenience to the person who might otherwise have to board an airplane carrying a 60" long rigid glass U-tube manometer. VARIATIONS IN MANOMETER DESIGN Fig. 3-1. At left, equal pressure is imposed on the liquid in the well and the indicating tube. Reading is zero. At the right, a positive pressure has been imposed on the liquid in the indicating tube pushing it down to a point on the scale equal to the pressure. Liquid level in the well rises proportionately. Inclining the indicating tube has opened up the scale to permit more precise reading of the pressure. To improve and expand readability, certain Dwyer U-type and well-type manometers are available with a .826 sp. gr. red oil indicating fluid, and scales compensated to read pressure directly in inches of water. To further increase readability and sensitivity the well-type manometer indicating tube is inclined, as in Fig. 3-1, to cause a greater linear movement along the tube for a given pressure difference. The inclined manometer is frequently called a Draft Gage because it is widely used for determining the over-fired draft in boiler uptakes and flues. For an inclined manometer to be a primary device, the inclined tube must be straight and uniform. Dwyer's precision machined solid plastic construction has been applied to a basic line of rugged manometers, inclined and inclined-vertical, which are industry accepted as primary instruments. Fig. 3-2. At left, with equal pressure on liquid in well and indicating tube, reading is zero. When positive pressure is imposed on liquid in indicating tube, liquid level is depressed in tube and rises slightly in well. Reading is direct since scale is compensated for change of level in well. The combination of an inclined and a vertical manometer is very useful in air movement determination. See Fig. 3-2. For air velocity measurement, an inclined scale, generally up to 1" w.c. is used (1" w.c. velocity pressure = 4000 fpm). In the Dwyer Durablock® inclined-vertical instrument, this scale is combined with a vertical section allowing readings of high pressures, usually 1" w.c. to 5 to 10" w.c., to be taken. The vertical section is used primarily for determining static pressure above the range of the inclined section. Many special purpose types of manometers exist. Examples are the Dwyer Hook Gage and Microtector® gage. These are simply U-tube manometers modified so the liquid level can be read with a micrometer, yet retaining the basic "Physics" of the hydrostatic U-tube primary standard. Readings accurate to ±.001" w.c. in a range of differential pressures from 0-24" w.c. are accomplished with Dwyer Model No. 1425-24 Hook Gage. The Model 1430 Microtector® gage incorporates modern electronics to increase the accuracy of readings to ±.00025" w.c. on a 2" w.c. scale.<\|endoftext\|>	3.875
966	# How to factorise 12x^2-11x+2? Please give steps as well as answer. I am doing this for revision. I can't find anything that would help. The method I remember being taught doesn't seem to work at all, although it worked with expressions with a larger number than x. Update: 'expressions with a larger number than x' example being '12x^2-11x+30' or something. Relevance • 2 years ago Factor by grouping is the method you must use. • 2 years ago (4 x - 1) (3 x - 2) • Como Lv 7 2 years ago ( 4x - 1 ) ( 3x - 2 ) • 2 years ago Multiply 122 = 24 Find two factors of 24 that add up to -11 (12x^2 + -8x )+ ( -3x + 2 ) Now factor out -4 from first bracket • 2 years ago 12x^2-11x+2 = (4x - 1)(3x - 2) answer// Source(s): factor • Mike G Lv 7 2 years ago (4x-1)(3x-2) • 2 years ago When things get this large, I usually don't try to factor, but if I need to factor, I would solve for the roots and then turn the roots into factors: 12x² - 11x + 2 x = [ -b ± √(b² - 4ac)] / (2a) x = [ -(-11) ± √((-11)² - 4(12)(2))] / (2 12) x = [ 11 ± √(121 - 96)] / 24 x = [ 11 ± √(25)] / 24 Note that if what's under the radical isn't a positive rational number, this can't be factored over rational numbers. x = (11 ± 5) / 24 x = 6/24 and 16/24 x = 1/4 and 2/3 So we get: (x - 1/4) (x - 2/3) Recall this would be in an equation equal to 0 if we were factoring to solve for zeroes, so let's do that: (x - 1/4) (x - 2/3) = 0 Now the only problem is in factoring, we wouldn't have fractions. So we can multiply each factor by its denominator (and multiply the right side by the same values), so multiply by 4 and 3: 4(x - 1/4) * 3(x - 2/3) = 0 * 4 * 3 (4x - 1)(3x - 2) = 0 So the factored form of your expression is: (4x - 1)(3x - 2) • khalil Lv 7 2 years ago make it equal zero 12x² - 11x +2 = 0 solve this equation ax² + bx + c = 0 b rule {11 ± √ [ 11² - (4)(12)(2)] } / 24 = (11 ± 5) / 24 16/24 =2/3 and 6/24 = 1/4 now (a) in front and changing the signs 12(x - 2/3 ) ( x - 1/4 ) = (3x-2)(4x-1) • 2 years ago 12x²-11x+2 12x²-(11+d)x+dx+2 Here we want 12:-(11+d) = d:2 in order to get a common factor, so we have d²+11d+24=0. This should be more straightforward to solve by your usual methods for d (-3 or -8). Substitue one of these back into our equation. 12x²-(11-3)x-3x+2 12x²-8x-3x+2 4x(3x-2)-(3x-2) (4x-1)(3x-2) • 2 years ago 12x²-11x+2=(4 x - 1) (3 x - 2) x²-(11/12)x+(2/12)= etc,.<\|endoftext\|>	4.5625
1,169	# 3-Estimation of max and min-moment of inertia for a section. ## Estimation of Max and Min-moment of Inertia For a Section. ### Steps to get the expression for I-max and min-moment of inertia, and tan(2θ). Our calculation for the bending of stress is based on the pure moment assumption as we will estimate the product of inertia Ixy to ensure that it is equal to zero. The neutral axis, for which Ixy = 0, must pass through the Cg. The principal axes for the two perpendicular axes should be number #2, meaning that Ixy should equal zero. We must examine both of the orthogonal axes when we estimate Ixy and discover that it is not equal to zero, as this will result in Ixy=0. Numerous axes can be used to obtain l-polar = Ix+Iy or = Ix1+Iy1 or Ix’+Iy’ because the value of I-polar is constant. However, the axes that will produce the moment of inertia are the ones that we have an interest in, for instance, x’ will produce the most moment of inertia and y’ will provide the smallest, and Ix’y’ should equal zero simultaneously. When it comes to any two axes if Ix is the maximum, Iy is the minimum, and Ixy is zero, then these two axes are classified as principal axes. The terms “principal axes” refer to these two axes. ### Derive the expression for max and min-moment Of Inertia. Using the x and y axes, we will create a general expression and assume that we have two inclined axes, x’ and y’, with an angle of = θ, where θ is positive and pointing counterclockwise. To establish a relationship between (x’ and y’) and x and y, for small strip dA, the coordinates are X and Y equally referring to the new two axes (x’ and y’). The formula for x’ is xcos θ + ysin θ. This is the part of the x’ that is represented by y’=ycosδ -xsinδ. Then, since we know that Ix’ = ∫ dAy’^2, this is the dA, and we wish to make an integration about x’, we continue to calculate the moment of inertia about x’ and y’ while also attempting to obtain an equation for Ix’ and Iy’, for the special case where Max and Min-Moment of Inertia are needed. However, it will be = ∫ dAx’^2 for the moment of inertia surrounding y’. Conversely, for the inertia product Ix’y’,= ∫ dAx’y’. The values of x’=xcos θ+ysin θ and y’=ycosδ -xsinθ will be substituted. As seen in the following graphic, we have Ix’, Iy’, and Ix’y’. These items are written after the values of x’ and y’ are substituted in terms of x and y. To some extent, this calculation is lengthy, but our goal is to obtain three expressions. In terms of x, y, and angle θ—which is the angle closed between axis x and axis x’—the first expression is for Ix’. In terms of x, y, and angle θ—which is the angle closed between axis x and axis x’—the second expression is for Iy’. Regarding x,y, the third expression is for Ix’y’, where θ is the enclosed angle between axis x and axis x’. Since the polar of inertia is equal to Ix + Iy, the first two expressions for Ix’ and Iy are displayed in the next slide. Once we sum up Ix and Iy from Ip, we can obtain the expression for tan(2θp). We will add Ix and Iy together from Ip, we can get an expression for tan(2θp). The angle θp, which produces Ix max and Iy min, can be obtained by differentiating Ix’ concerning θ and setting it equal to zero. ### The expression for max and min-moment of inertia and principal angle θp. The tan(2θp) is expressed as tan(2θp)=2(Ixy)/(Iy-Ix). As we recall this expression, we can change it to produce (-2Ixy/Ix-Iy) as a denominator, and this is the θp. when we set θ as equal to θp, we can plug in the expressions for Ix’ and iy’ and substitute by θp. then we can get the Max and Min-Moment of Inertia expression. Please refer to the next slide image for more information. ### The four cases for the relations between Ix, Iy, and Ixy. We have obtained the enclosed angle (θp) between the X- and X’-axes, as we need it to formulate the tan2θ expression. Four cases are presented, the first of which is Ix>Iy and Ixy is positive. The second case is when Ix>Iy and Ixy are negative. The third case is when Ix < Iy and Ixy is positive. The fourth case is when Ix < Iy and Ixy is negative. This is the pdf file used in the illustration of this post. The next post is an Easy introduction to Mohr’s Circle of Inertia Part 1. This is a link to a useful external resource. Calculator for Cross Section, Mass, Axial & Polar Area Moment of Inertia, and Section Modulus. Scroll to Top<\|endoftext\|>	4.4375
1,412	MCAT Organic Chemistry Review 1.1 IUPAC Naming Conventions Nomenclature is one of the most important prerequisites for answering organic chemistry questions on Test Day; if you don’t know which chemical compound the question is asking about, it’s hard to get the answer right! That’s why it’s so important to understand both IUPAC and common nomenclature. Once you have a handle on these systems, you can easily translate question stems and focus on finding the correct answer. Let’s begin by examining the IUPAC naming conventions before we examine specific compounds and functional groups. Nomenclature is often tested on the MCAT by providing the question stem and the answer choices in different formats. For example, the question stem will give you the IUPAC name of the reactant, and the answer choices will show product structures—leaving you to figure out both the structure of the reactant and the reaction taking place. The primary goal of the International Union of Pure and Applied Chemistry (IUPAC) naming system is to create an unambiguous relationship between the name and structure of a compound. With the conventions established by IUPAC, no two distinct compounds have the same name. The IUPAC naming system greatly simplifies chemical naming. Once we understand the rules, we can match names to structures with ease. 1. Identify the Longest Carbon Chain Containing the Highest-Order Functional Group This will be called the parent chain and will be used to determine the root of the name. Keep in mind that if there are double or triple bonds between carbons, they must be included. If one of the functional groups would provide a suffix for the compound (for example, an alcohol, which will be discussed later), then the parent chain must contain this functional group. We’ll examine priorities of functional groups throughout this chapter, but keep in mind that the highest-priority functional group (with the most oxidized carbon) will provide the suffix. This step may sound easy, but be careful! The molecule may not be drawn so that the longest carbon chain is immediately obvious. If there are two or more chains of equal length, the more substituted chain gets priority as the parent chain. Figure 1.1 shows a hydrocarbon with the longest chain labeled. Figure 1.1. Finding the Longest Carbon Chain 2. Number the Chain In order to appropriately name a compound, we need to number the carbon chain, as shown in Figure 1.2. As a convention, the carbon numbered 1 will be the one closest to the highest-priority functional group. If the functional groups all have the same priority, numbering the chain should make the numbers of the substituted carbons as low as possible. Figure 1.2. Numbering the Longest Carbon Chain The highest-priority functional group should have the lowest possible number; if all substituents have the same priority, make their numbers as low as possible. After we have discussed the functional groups most commonly tested on the MCAT, we’ll review a table of those functional groups in order of priority. For now, keep in mind that the more oxidized the carbon is, the higher priority it has in the molecule. Oxidation state increases with more bonds to heteroatoms (atoms besides carbon and hydrogen, like oxygen, nitrogen, phosphorus, or halogens) and decreases with more bonds to hydrogen. Just like straight chains, rings are numbered starting at the point of greatest substitution, continuing in the direction that gives the lowest numbers to the highest-priority functional groups. If there is a tie between assigning priority in a molecule with double and triple bonds, the double bond takes precedence. 3. Name the Substituents Substituents are functional groups that are not part of the parent chain. A substituent’s name will be placed at the beginning of the compound name as a prefix, followed by the name of the longest chain. Remember that only the highest-priority functional group will determine the suffix for the compound and must be part of the parent chain. Carbon chain substituents are named like alkanes, with the suffix –yl replacing –ane. The prefix n– that we see in Figure 1.3 on n-propyl simply indicates that this is “normal”—in other words, a straight-chain alkane. Because this prefix will not always be present, it is safe to assume that alkane substituents will be normal unless otherwise specified. Figure 1.3. Common Normal Alkyl Substituents The bond on the right side of each substituent connects to the parent molecule. In Figure 1.4, we see some examples of what alternative alkyl substituents may look like. Figure 1.4. Common Alternative Alkyl Substituents The bond on the right side of each substituent connects to the parent molecule. If there are multiple substituents of the same type, we use the prefixes di–, tri–, tetra–, and so on to indicate this fact. The prefix is included directly before the substituent’s name. 4. Assign a Number to Each Substituent Pair the substituents that you have named to the corresponding numbers in the parent chain. Multiple substituents of the same type will get both the di–, tri–, and tetra– prefixes that we have previously noted and also a carbon number designation—even if they are on the same carbon. 5. Complete the Name Names always begin with the names of the substituents in alphabetical order, with each substituent preceded by its number. Note, however, that prefixes like di–, tri–, and tetra– as well as the hyphenated prefixes like n– and tert– are ignored while alphabetizing. Nonhyphenated roots that arepart of the name, however, are included; these are modifiers like iso–, neo–, or cyclo–. Then, the numbers are separated from each other with commas, and from words with hyphens. Finally, we finish the name with the name of the backbone chain, including the suffix for the functional group of highest priority. Figure 1.5 shows an example of an entire hydrocarbon named with IUPAC nomenclature. Figure 1.5. An Example of a Complete IUPAC Name MCAT Concept Check 1.1: Before you move on, assess your understanding of the material with these questions. 1. List the steps of IUPAC nomenclature: 2. Circle or highlight the parent chain in each of the following compounds: 3. How are numbers separated from each other and from words in a compound’s name?<\|endoftext\|>	3.953125
140	In this activity, students set up an experiment to determine the biodegradability of different substances. By the end of this activity, students should be able to: - organise a group of items into categories - carry out a simple experiment - understand that some materials are biodegradable and others aren’t - explain what might affect how quickly something degrades - make predictions about the biodegradability of materials. Download the Word file (see link below) for: - introduction/background notes - what you need - what to do - discussion questions - student handout Find out more about backyard composting on this website.<\|endoftext\|>	4.5625
650	In the brain, the complex circuits behind our cognition largely construct themselves. In a dish or on a chip however, the normal meaningful cues are missing, and any attempts to grow predefined circuits from neurons have largely met with failure. A new technique, dubbed magnetogenetics, has now provided a way to create these neural circuits by combining magnetic manipulation with the cell’s own machinery for stabilizing new growth and extensions. The technique works by making use of the cell’s ability to target a protein, known as Rac-GTPase, which promotes the formation and stabilization of new branches. The researchers used 500nm (0.0005mm) magnetic beads that have been modified so they can hook up to these protein machines soon after they come off the presses. They can then be magnetically acquired, and then moved to the desired locations using a precisely controlled force. In order to see what they are doing, they also attached fluorescent beacons to the beads. Other researchers had previously tried mechanically pulling on the cells with microactuators, and while impressive growth could be obtained, fine control was impossible. Other techniques using laser tweezers to pull on subcellular transparent beads, or even the cell directly also had some successes, but the level of light power necessary to do this is typically damaging to the cells. (See: Manipulating nanoparticles with an electron tractor beam.) To grow neurons on semiconductors, adhesion molecules have typically been pre-patterned along traces for the neurons to follow. When these molecules invariably degraded over time, the neuron was left without any natural supporting structure. By using a machine that nucleates the cellular endoskeleton directly, new processes can be drawn out in a more natural way that the cell can adapt to. In the picture above, the major dynamic skeletal component, actin, has been labelled with red fluorescent molecules, while other more permanent structures known as microtubules (MTs) fluoresce in green. If the actin is nucleated in a persistent fashion, the cell’s secondary crew, the MTs, can infiltrate the new process and make it stronger. The MTs are really the railroad infrastructure of the cell. Once the actin pioneers have laid the desired course, the MTs deliver the goods, including mitochondrial power stations, ribosome factories to build proteins on site, and all other sorts of useful spare parts. In the body, things are a little more difficult for magnetogenetic control because you can not get up close to the cells you want to manipulate. New techniques like Magnetic Resonance Navigation have demonstrated the ability to precisely position engineered nanoparticles within blood vessels. While not quite at the level of the individual cell, for guiding and stabilizing larger axon tracts in cases of neurotrauma, or if you simply need to connect a new memory chip to your hippocampus, magnetogenetics may be just what you are looking for. Now read: The first 3D-printed human stem cells Research paper: doi:10.1038/nnano.2013.23 – “Subcellular control of Rac-GTPase signalling by magnetogenetic manipulation inside living cells”<\|endoftext\|>	3.71875
584	# In figure below, $l \\| \mathrm{m}$ and line segments $\mathrm{AB}, \mathrm{CD}$ and $\mathrm{EF}$ are concurrent at point $\mathrm{P}$.Prove that $\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}}$." Given: $l \\| \mathrm{m}$ and line segments $\mathrm{AB}, \mathrm{CD}$ and $\mathrm{EF}$ are concurrent at point $\mathrm{P}$. To do: We have to prove that $\frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}}$. Solution: In $\triangle A P C$ and $\triangle B P D$ $\angle A P C=\angle B P D$ (Vertically opposite angles) $\angle P A C=\angle P B D$ (Alternate angles) Therefore, by AA similarity, $\triangle A P C \sim \triangle B P D$ This implies, $\frac{A P}{P B}=\frac{A C}{B D}=\frac{P C}{P D}$..........(i) In $\triangle A P E$ and $\triangle B P F$ $\angle A P E=\angle B P F$ (Vertically opposite angles) $\angle P A E=\angle P B F$ (Alternate angles) Therefore, by AA similarity, $\triangle A P E \sim \triangle B P F$ This implies, $\frac{A P}{P B}=\frac{A E}{B F}=\frac{P E}{P F}$........(ii) In $\triangle P E C$ and $\triangle P F D$ $\angle E P C=\angle F P D$ (Vertically opposite angles) $\angle P C E =\angle P D F$ (Alternate angles) Therefore, by AA similarity, $\triangle P E C \sim \triangle P F D$ This implies, $\frac{P E}{P F} =\frac{P C}{P D}=\frac{E C}{F D}$........(iii) From (i), (ii) and (iii), we get, $\frac{A P}{P B}=\frac{A C}{B D}=\frac{A E}{B F}=\frac{P E}{P F}=\frac{E C}{F D}$ $\frac{A E}{B F}=\frac{A C}{B D}=\frac{C E}{F D}$ Hence proved. Tutorialspoint Simply Easy Learning<\|endoftext\|>	4.40625
3,358	# Graphs Higher Revision Card Answers ## HG1 – Gradient and y=mx+c Question: Find the equation of the straight line that passes through the points (-3, -7) and (2, 8) Answer: Remember we just need to find two numbers to describe our line: the gradient ($m$) and the y-axis intercept ($c$). It’s easier if we start with the gradient. $m=\dfrac{\text{change in y}}{\text{change in x}}=\dfrac{y_{2}-y_1}{x_2-x_1}=\dfrac{8--7}{2--3}=\dfrac{15}{5}=3$ So our equation so far is $y=3x+c$. To find c, we substitute any coordinates that the line passes through for x and y in this equation – let’s use the point (2,8). $8=3\times 2 + c$ Now we just rearrange this equation to find $c$. $8=6+c$ $c=2$ So our completed equation is $y=3x+2$ ## HG2 – Coordinates and Midpoints Question: Find the midpoint of the line segment that joins $A$ and $B$ as shown. Answer: Point $A$ has coordinates $(-2, -2)$. Point $B$ has coordinates $(0, 3)$. By taking the average of the $x$ coordinates of $A$ and $B$, the $x$ coordinate of the midpoint is: $\dfrac{-2 + 0}{2} = -1$ By taking the average of the $y$ coordinates of $A$ and $B$, the $y$ coordinate of the midpoint is: $\dfrac{-2 + 3}{2} =\dfrac{1}{2}$ Therefore, the coordinates of the midpoint are $\left(-1, \dfrac{1}{2}\right)$. ## HG3 – Parallel and Perpendicular Lines Question: Lines $𝐿_1$ and $𝐿_2$ intersect at right angles at the point (-2, 3). $𝐿_1$ has the equation $𝑦=−3𝑥−3$. Find the equation of $𝐿_2$. Answer: Given that $L_1$ and $L_2$ are perpendicular, their gradients must follow the relationship $m_1 \times m_2 = -1$. Rearranging this equation tells us that the gradient of $L_2$ is given by $m_2=\dfrac{-1}{m_1}=\dfrac{-1}{-3}=\dfrac{1}{3}$. So our equation for $L_2$ takes the form $y=\dfrac{1}{3}x +c$. To find c, we substitute the coordinates of any point that the line passes through into this equation. We’re told that the lines intersect at $(-2,3)$, so we can substitute $y=3$ and $x=-2$ into this equation and solve for $c$: $3=\dfrac{1}{3}\times -2+c$ $3=-\dfrac{2}{3}+c$ $c=3\dfrac{2}{3}=\dfrac{11}{3}$ So our equation for $L_2$ is $y=\dfrac{1}{3}x +\dfrac{11}{3}$ Question: A curve is given by the function $y=x^2+4x-9$. Describe the shape of this curve. Plot the graph to check. Answer: The coefficient of $x^2$ is positive, so the graph will be a U-shaped curve. To plot the graph, we need to make a table of coordinates. First, we pick some x coordinates (e.g. -5 to 2), then we use the equation of the graph to work out the y-coordinates. For example, when x is -5, the y coordinate is given by $y=(-5)^2+4\times(-5)-9=25-20-9=-4$. We then do this with the other x coordinates until we have the following table: Then we plot the coordinates $(-5,-4), (-4,-9)$ etc. to get the following graph. ## HG5 – Cubic, Reciprocal and Exponential Graphs Question: Sketch the graph of the equation $y=ax^3+bx^2+c$ on the axis shown, where 𝑎, 𝑏 and 𝑐 are integers. What is the name given to this type of function? Answer: The name of this function is cubic. The sketch should be any general cubic sketch, like the one shown below. ## HG6 – Turning Points of Quadratic Graphs Question: Sketch a graph of the function $𝑦=𝑥^2+5𝑥+6$. Label the coordinates of the turning point and any intersections with the x-axis. Answer: Firstly, we must find the roots of this quadratic by factorising it and setting it equal to zero. Observing that $2\times3=6$ and $2+3=5$, we get that $x^2+5x+6=(x+2)(x+3)$ Therefore, to find the roots we set $(x+2)(x+3)=0$ This clearly gives two roots: $x=-2$ and $x=-3$. Then, to find the coordinates of the turning, we need the halfway point between the roots, which is $\dfrac{-2+(-3)}{2}=-2.5$ This is the $x$ coordinate of the turning point. To find the $y$ coordinate, we put this value back into the equation to get $y=(-2.5)^2+5(-2.5)+6=-0.25$ Then, the resulting sketch of the graph should look like Note: because the question asked you for a sketch, it doesn’t have to be a perfect drawing, it just has to have the correct shape and correctly identified & labelled roots and turning points. ## HG7 – Graph Transformations Question: The graph of $𝑓(𝑥)=𝑥^2−𝑥−6$. Sketch, on the same axes, the graphs of $𝑦=𝑓(𝑥)$ and $𝑦=−𝑓(𝑥)+4$. Answer: Firstly, $x^2 - x - 6$ factorises to $(x - 3)(x + 2)$, so it is a positive quadratic with roots at 3 and -2. Now, $-f(x) + 4$ is both a reflection in the x-axis and a translation of 4 in the positive y-direction. We should do the reflection first and the translation second – it often helps to sketch the intermediate step to help you, and you can always rub it out afterwards. Here, the dotted line will be the intermediate step (the reflection before the translation). So, we get ## HG8 – Area under a graph Question: Estimate the area under this graph. (Hint: you can do this by approximating the shape of the graph with straight lines, and subdividing it into simple shapes.) Answer: First, we can draw a straight line that approximately follows the graph. Then we can divide this into simple shapes – in this case, we can split it up into 1 triangle and 3 trapeziums. Next we work out the areas of each shape using the relevant shape area formulas. A: $A=\dfrac{1}{2}a\times b = \dfrac{1}{2}10\times 3=15$ B: $A=\dfrac{1}{2}(a+ b)\times h_v=\dfrac{1}{2}(3+4)\times 20 = 70$ C:$A=\dfrac{1}{2}(a+b)\times h_v= \dfrac{1}{2}(4+6.5)\times 17 = 89.25$ D: $A=\dfrac{1}{2}(a+ b)\times h_v= \dfrac{1}{2}(6.5+2)\times 13= 55.25$ Adding all of these together gives us an estimated total area of $15+70+89.25+55.25=229.5$$\space (\pm 10)$ ## HG9 Equations of Circles & Finding Tangents to Circles Question: What is the equation of a circle with centre (0,0) and radius √5? Find the equation of the tangent to this circle at (−1,2). Answer: Remember the form of the equation of a circle is $x^2+y^2=r^2$, where $r$ is the radius. The radius is $\sqrt(5)$, so $r^2=(\sqrt(5))^2=5$, so the equation of our circle is $x^2+y^2=5$. To find the equation of the tangent, we first think about the radius that meets the tangent at (-1,2): We know that the line perpendicular to the tangent passes through (0,0) and (-1,2), so we can work out the gradient: $(2-0)/(-1-0)=2/(-1)=-2$. We know that the tangent is perpendicular to this, so the gradient of the tangent must be $\dfrac{1}{2}$ (because the tangents of perpendicular lines multiply to make -1, and $\dfrac{1}{2}\times-2=-1$. So the equation of the tangent so far is $y=\dfrac{1}{2}x+c$. To find $c$, we substitute the coordinates (-1,2) into the equation: $2=\dfrac{1}{2}\times -1 +c$ Then solve to get $c=2.5$ So the complete equation of the tangent is $y=\dfrac{1}{2}x+2.5$ ## HG10 – Distance-Time Graphs Question: Neil’s journey is shown in the distance time graph below. At which time interval 12:00 – 12:30, 12:30 – 13:30, 13:30 – 16:30 or 16:30 – 18:30 was Neil’s speed greatest. First we need to work out how far Neil travels between each time period: – 12:00 – 12:30, he travels from 0km away to 12km away; remembering that speed = the gradient at the time. $Gradient = \dfrac{12}{0.5}=24 kmph$ – 12:30 – 13:30, he travels from 12km away to 44km away; $Gradient = \dfrac{44-12}{1}=32kmph$ – 13:30 – 16:30, he stays in one place; $Gradient = \dfrac{0}{1}=0 kmph$ – 16:30 – 18:30, he travels from 44km away to 0km away. $Gradient = \dfrac{-44}{2}=22 kmph$ From this we can see that the greatest speed is between 12:30 – 13:30. ## HG11 – Velocity-Time Graphs Question: The speed-time graph below describes a 50-second car journey. Work out the total distance travelled by the car and work out the maximum acceleration of the car during the 50 seconds. Answer: Before we get stuck in to answering this, let’s go through the journey described by the graph. Firstly, the car accelerated from 0 to 15m/s over the first 10 seconds (because the line is straight, the acceleration is constant). Then, the line is flat, meaning the car’s speed was not changing for 10 seconds – it was moving at constant speed. Next, the car accelerated up to 25m/s over the next 10 seconds, and finally it spent the last 20 seconds decelerating back down to 0m/s. Now, we want to work out the distance travelled. On a speed-time graph, the distance travelled is the area under the graph. To work out the area under this graph, we will break it into 4 shapes: A, B, C, and D – two triangles, a rectangle, and a trapezium are all shapes that we can work out the area of. So, we get $\text{A}=\dfrac{1}{2}\times10 \times15= 75\text{m}$ $\text{B}=10\times15=150\text{m}$ $\text{C}=\dfrac{1}{2}(15+25)\times10=200 \text{m}$ $\text{D}=\dfrac{1}{2}\times20\times25=250\text{m}$ Therefore, the total distance travelled is: $75+150+200+250=675\text{m}$. Next up, the acceleration. To do this, consider that acceleration is a measure of how quickly something’s speed is increasing. Therefore, given that the gradient is a rate of change of $y$ (speed) with respect to $x$ (time), we can work out the acceleration by finding the gradient of the graph. Now, the question asks for “maximum acceleration”, so we can rule out certain parts of the journey. Specifically, the part where the graph is flat – there is no acceleration here – and the part where the graph slopes downward – it is decelerating here, so can’t be the maximum acceleration. It might be obvious to you which of the two sections of the graph is steeper, but it isn’t always, so we’ll work out both just to be sure. Firstly, the first 10 seconds: the car’s speed goes from 0 to 15m/s and it takes 10 seconds, so we get $\text{acceleration between 0s and 10s = gradient}=\dfrac{15-0}{10-0}=1.5\text{m/s}^2$ Then, the other portion we’re interested in is between 20 and 30 seconds. During this period, the speed increases from 15 to 25, so we get $\text{acceleration between 20s and 30s = gradient}=\dfrac{25-15}{30-20}=1\text{m/s}^2$ The first one is larger, so the maximum acceleration is $1.5\text{m/s}^2$. ## HG12 – Inequalities on Graphs Question: Shade the region of a graph that is satisfied by the inequalities y ≥1, 𝑦 ≤3, 𝑥>0 and 𝑦>𝑥, and mark it with an R. Answer: We’re going to pretend that the inequalities are equations and plot them as straight lines. The first one will be the solid plot of the line $y=1$, the second will be a solid plot of the line $y=3$, the third will be a dashed plot of the line $x=0$, and the fourth will be a dashed plot of the line $y=x$. Now, we want to shade the area that is above the line $y=1$, below the line $y=3$, to the right of the line $x=0$, and above the line $y=x$. The resulting graph looks like: Labelling the lines is not necessary but you may find it be a helpful intermediate step. Or you might not, which is totally fine – be your own person.<\|endoftext\|>	4.75
251	Food is important for keeping us alive. Our bodies need nutrients from the foods we eat to keep our bodies working and healthy. And some of those nutrients give our bodies energy to keep working. Those nutrients are called ‘carbohydrates’. But what are they and how do they give us the energy we need? There are three kinds of carbohydrates foods – sugars, starches and fibres: - Sugars make foods taste sweet. They are found in foods like fruits and milk. - Starches make you feel full-up. They are found in root vegetables and cereals. - Fibres help you to make poo. They move undigested food through your body. When we eat foods that have carbohydrates in them, our bodies digest, or break down, those carbohydrates so they can be absorbed by our blood. One of the most important sugars that comes from carbohydrates is glucose. Our bodies need glucose it helps to keep us alive. Glucose is used in a special chemical reaction called ‘respiration‘. When the cells in our bodies respire, they take glucose from the food we have eaten and oxygen from the air we have breathed in and use it to make energy.<\|endoftext\|>	4.03125
4,063	Kalahari Desert, large basinlike plain of the interior plateau of Southern Africa. It occupies almost all of Botswana, the eastern third of Namibia, and the northernmost part of Northern Cape province in South Africa. In the southwest it merges with the Namib, the coastal desert of Namibia. The Kalahari’s longest north–south extent is roughly 1,000 miles (1,600 kilometres), and its greatest east–west distance is about 600 miles; its area has been estimated at some 360,000 square miles (930,000 square kilometres). Physiography and geology The Kalahari Desert is a featureless, gently undulating, sand-covered plain, which everywhere is 3,000 feet (900 metres) or more above sea level. Bedrock is exposed only in the low but vertical-walled hills, called kopjes, that rarely but conspicuously rise above the general surface. Aside from the kopjes, three surfaces characterize virtually all of the Kalahari: sand sheets, longitudinal dunes, and vleis (pans). The sand sheets appear to have been formed during the Pleistocene Epoch (about 2,588,000 to 11,700 years ago), and they have been fixed in place since then. In some areas they appear to have been of fluvial origin, the result of sheet flooding in times of much greater precipitation, but by far the greater part of them were wind-formed. The sheets occupy the eastern part of the Kalahari. Their surface elevation varies only slightly, with relief measured in tens of feet per mile. The depth of the sand there generally exceeds 200 feet. In many areas the sand is red, the result of a thin layer of iron oxide that coats the grains of sand. The entire western Kalahari Desert is characterized by long chains of dunes, oriented roughly to the north or northwest. The dunes measure at least 1 mile in length, several hundred feet in width, and 20 to 200 feet in height. Each dune is separated from its neighbour by a broad parallel depression locally called a straat (“street,” or “lane”), because each constitutes the easy way to travel. Vleis, or pans, are the terminal features of desert drainage systems, the “dry lakes” at the end of ephemeral streams. Many are remnant features from an earlier period of greater precipitation. Very little water ever flowed to the sea from the Kalahari. Rather, each stream ended its course in a slightly lower depression from which there was no outlet. There, as the stream dried up, the fine silt particles carried in suspension by the sluggish stream were deposited along with soluble calcium minerals and salts precipitated out of the evaporating water. The results are pans—flat surfaces devoid of vegetation that are gleaming white when dry, hardened by the cementing action of the soluble minerals, and, on occasion, covered by a shallow layer of standing water. Where the salt content is low, pans may become covered with grasses after a rain. In the southern and central parts of the Kalahari Desert, surface water is found only in small, widely scattered waterholes, and surface drainage is nonexistent. Nearly all of the rain that falls disappears immediately into the deep sand. Some is absorbed by the underlying rock strata; some is drawn to the surface by capillary action and evaporated into the air; and some, lifted from the depths by tree roots, is transpired from leaf surfaces. A small amount, landing on nonsandy surfaces, may flow short distances into pans, but this occurs only immediately after the infrequent rains. In some parts of the central and southern Kalahari, extensive ancient drainage systems have been detected—some on the ground and others by way of aerial photographs. None of these operate today, even in the wettest of years. In the northern Kalahari an extraordinary drainage system prevails. During the summer heavy rains fall on the uplands of central Angola, far to the northwest of the Kalahari. Large amounts of runoff water feed a number of south-flowing streams, which merge to form the Okavango and Cuando (Kwando) rivers. The Okavango flows to the southeast and into the northernmost portion of the Kalahari, eventually breaking up into a number of distributary channels and feeding the vast area of swamps in northern Botswana. After an abnormally wet rainy season in Angola, excess water fills the swamps and overflows, filling Lake Ngami farther to the south, and flows eastward through the Boteti River into Lake Xau and the Makgadikgadi Pans. Similarly, the Cuando River flows south from Angola and partly into a northeastern extension of the same swamps. Thus is created the paradoxical situation of an area with an extensive excess of water in a region chronically short of water. Soils in the Kalahari Desert are largely based on sand, are reddish in color, and are low in organic material. Chemically, they are relatively alkaline, and they are extremely dry. In and near the pans, the soils tend to be highly calcareous or saline, and frequently they are toxic to most vegetation. Traditionally, an area was classed as desert if it received less than 10 inches (250 millimetres) of rain annually. A more accurate definition of a desert is a region in which the potential evaporation rate is twice as great as the precipitation. Both of these criteria are applicable to the southwestern half of the Kalahari. The northeastern portion, however, receives much more rainfall and, climatically, cannot qualify as a desert; and yet, it is totally lacking in surface water. Rain drains instantly through the deep sands of the area, which creates a situation of edaphic drought (i.e., soil completely devoid of moisture). Moisture-bearing air is derived from the Indian Ocean, and precipitation is greatest in the northeast (with a mean annual precipitation of more than 20 inches) and declines toward the southwest (less than 5 inches on the southern fringe of the Kalahari). Precipitation, however, is highly variable. Most of the rain comes as summer thunderstorms, with great variation from place to place and from year to year. Winters are extremely dry: humidity is very low, and no rain falls for six to eight months. Great ranges in both diurnal and seasonal temperatures are the rule, the result of the Kalahari’s relatively high altitude and predominantly clear, dry air (allowing strong insolational heating in daytime and great radiational heat loss at night). As a result, shade temperatures often reach 110–115 °F (43–46 °C) on summer days but drop to 70–80 °F (21–27 °C) on the same nights; temperatures on winter nights commonly drop to freezing and may go as low as 10 °F (−12 °C). The presence of a deep sand cover over most of the area greatly affects the vegetation that grows there. Shallow-rooted plants cannot survive on a perennial basis, although annuals that grow very rapidly after a good rain may be able to sow seeds that will endure until the next good rainy season. Trees with roots deep enough to reach permanently moist sand levels do well. The southwestern Kalahari Desert, with its very low precipitation, has few trees or large bushes—only scattered xerophytic (drought-tolerant) shrubs and short grasses. The central Kalahari, with more rain, has scattered trees (several species of Acacia) and some shrubs and grasses. The northern Kalahari does not have the appearance of a desert at all. It has open woodlands, palm trees growing among thorn brush, and forests of both evergreen and deciduous trees that grow to heights of 50 feet and yield some species suitable for timber; one of the largest and most unusual of these trees is the baobab. The Okavango Swamp supports a dense growth of reeds, papyrus, pond lilies, and other water-loving plants. The animal life of the Kalahari Desert is also richer and more varied in the north than in the south. Yet even in the arid south, many individuals of several species stay for long periods of the year despite the absence of surface water. The principal species found in the south are springbok, gnu (wildebeest), and hartebeest—all of which occasionally are present in great herds—gemsbok (oryx), eland, and many smaller nongregarious species, such as kudu (in areas of denser brush), steenbok, and duiker. The northern Kalahari supports a considerable population of giraffes, zebras, elephants, buffalo, and antelopes (roan, sable, tsessebe, and impala); predators such as lions, cheetahs, leopards, wild hunting dogs, and foxes; other large and medium-sized mammals, such as jackals, hyenas, warthogs, baboons, badgers, anteaters, ant bears, hare, and porcupines; and numerous small rodents, several types of snakes and lizards, and a wealth of birdlife.Richard F. Logan People and economy The Bantu-speaking peoples—the Tswana, the Kgalagadi, and the Herero—are relative newcomers to the Kalahari. In the late 18th century the Tswana spread west from the Limpopo basin into the northern and eastern Kalahari; the Kgalagadi moved north and west into the southern and western Kalahari; and the Herero refugees from the German-Herero conflict of 1904–07 in German South West Africa (now Namibia) fled east into the western and northern Kalahari at the beginning of the 20th century. Those in the remoter parts of the Kalahari who are unaffected by mining or other industry live in villages of between 200 and 5,000 people. Housing is mostly of the traditional type: single-roomed huts with mud walls and thatched roofs. Water is the limiting factor, confining settlement to places situated near wells or boreholes with potable water. Cattle, the basis of the economy, are kept on the outskirts of villages, or at distances of up to 50 miles away. Wells and boreholes are owned by local government councils, syndicates of cattle owners, or private individuals; year-round cattle grazing is limited to their vicinity. In summers of above-average rains, however, pastoralists may trek with their stock to remote pastures, where for a short time water may occur in pools. Cattle and goats feed upon a small range of the available vegetation. Since effective pasture management is little practiced, the grazing of these animals is highly destructive. Pasture loss and subsequent desertification are serious threats to the ecology of the Kalahari Desert. Cattle are prized beyond their economic value, as their ownership is a measure of social status and personal worth. Thus, the desire to possess more cattle puts an increasing load on diminishing pasture, leaving it no chance for recovery. The traditional dangers to livestock—drought, disease, internal parasites, and wild predators—have diminished markedly as more boreholes have been sunk, veterinary care improved, and indigenous fauna have grown scarcer. In addition, wealthier cattle owners have improved their herds by introducing better stock and practicing scientific breeding. Goats furnish most of the meat and milk for home consumption, and nearly all households cultivate crops of corn (maize), sorghum, and pumpkins. Because of the threat of drought, more crops fail than are successful. Wild food plants and the meat of game animals are important components of diet in the smaller and more remote villages. All villages have trading stores or are visited by hawkers who sell foodstuffs and other commodities. All but the smallest villages have state-run primary schools, which are attended by the great majority of children, although few proceed to secondary education. State-run health clinics and hospitals in the larger villages supplement the services of herbalists and diviners. Riding horses and donkeys are the usual means of local travel. Trucks belonging to traders or to the mine labour recruiting agency are used for longer journeys. Large diamond deposits were discovered in Botswana soon after the country’s independence, and the opening of the diamond mine at Orapa in 1971 marked the beginning of the development of mining activities in scattered locations of the Kalahari. In addition, tourism and the sale of handicrafts have become economically important. The San—or Basarwa, as they are called in much of the region—are now either clients of Bantu-speaking pastoralists and work at cattle posts in return for support or they are employees of cattle ranches or are dependents of such employees. Few San still follow their traditional pattern of hunting and gathering. Many have been resettled—often, against their will—by the government of Botswana from their traditional homes in the Central Kalahari Game Reserve to new villages built outside the reserve. Although all San traditionally were hunter-gatherers, there were significant cultural and social differences between groups. For example, a number of groups had long-standing clientships with Bantu-speaking stockowners, while other groups lived—until the 1970s—solely as autonomous foragers. Of these latter peoples, the Kung (!Kung), !xong, and G/wi tribes (the “! ” and “/” representing click sounds) were intensively studied. While each group was distinct, the G/wi of the Central Kalahari Game Reserve can be considered an example of the traditional San hunter-gatherer way of life. The G/wi lived together in bands, each consisting of 5 to 16 households linked by bonds of kinship and friendship. Each band had a recognized territory of 300 to 400 square miles, selected for its resources of food plants (the main part of the diet), wet-season water holes (used during the six to eight weeks when sufficient rainwater gathered in pools), trees (for shade, shelter, firewood, and wood for making artifacts), and areas of grazing to attract and sustain herds of game animals. Subsistence was based on a number of species of edible plants, of which eight were staples in their various seasons. This diet was supplemented by the meat of antelopes and other herbivorous mammals, by tortoises and other reptiles, and by the flesh and eggs of all but raptorial and scavenging birds. Plant-gathering was mostly done by women ranging within five miles of the camp, while men hunted over a much larger area. The main hunting weapon was a light bow shooting flimsy, unfletched, poisoned arrows. The range of these bows was only about 75 feet, and great skill was needed to stalk the quarry within this distance. Antelope leather provided material for clothing, which included cloaks that also served as blankets and carrying bags. From November to some time between late June and early August—a period when there is sufficient food—the band lived as one community, moving from camp to camp every three or four weeks as the local supply of food plants became exhausted. Blighting frosts depleted the available food plants in winter (May to September), and the band would then split into its constituent households, each retreating to a separate part of the territory. Early-fruiting plants increased the food supply just before the approach of the wet season, allowing the band to reunite at a joint campsite. During dry seasons, shelters were little more than open windbreaks made of branches and grass. In rainy periods, domed structures of branches were thatched and made rainproof. Europeans first entered the Kalahari early in the 19th century as travelers, missionaries, ivory hunters, and traders. The only European settlement was in the Ghanzi District, where a number of families were allowed ranching blocks in the 1890s. Until the 1960s they led a life of isolation and poverty, but since then they have been able to gain ownership of the land and improve their living conditions. Most other whites in the Kalahari are government employees or are engaged in private enterprise.George Bertrand Silberbauer Because of its sparsely populated expanse, the Kalahari is served by infrequent roads and tracks, the majority of which are passable only by trucks or four-wheel drive vehicles. Maintained roads connect administrative centres, major habitations, and marginal farming areas in the south, southwest, and northwest. Constructed roads now link eastern Botswana with the Okavango Swamp and with mining developments south of the Makgadikgadi Pans. Study and exploration The Kalahari Desert’s lack of surface water and deep sands constituted a major obstacle to early travelers. The Scottish missionary and explorer David Livingstone, with assistance from local peoples, traversed the Kalahari in 1849 with great effort by utilizing local waterholes. In 1878–79 a party of Boers in the Dorsland (“Thirstland”) Trek crossed the Kalahari from the Transvaal to central Angola by a circuitous route, losing along the way about 250 people and 9,000 cattle, largely from thirst. The introduction of motor vehicles in the 20th century greatly improved transport into the Kalahari, but even as late as the 1950s large areas were virtually inaccessible and were never visited by outsiders. By the mid-1970s, however, vehicle mobility had improved to such a degree that the whole of the Kalahari had been opened to study, hunting, and tourist expeditions, and interest in the desert continued into the 21st century. Learn More in these related Britannica articles: South Africa: Relief…(600 metres) in the sandy Kalahari in the west. The central part of the plateau comprises the Highveld, which reaches between 4,000 and 6,000 feet (1,200 and 1,800 metres) in elevation. South of the Orange River lies the Great Karoo region.… Africa: Pleistocene and Holocene developments…of the Sahara and the Kalahari were alternately subjected first to humid and then to dry and arid phases that expanded the desert surface at the expense of adjacent forested zones.… Botswana: Relief…is the area of deep Kalahari sand covering the rest of the country. The third region consists of ancient lake beds superimposed on the northern sandveld in the lowest part of the Kalahari Basin.… Namibia: Relief…the Central Plateau, and the Kalahari. The Namib is partly rocky and partly (in the central stretch) dunes. While having complex flora and fauna, it is a fragile and sparsely covered environment unsuitable for pastoral or agricultural activities. Diamonds (probably washed down from the Basotho highlands by the Orange River)… duricrust: Effects of climate and time…of aridity in Africa, with Kalahari sand extending 1,600–3,000 kilometres (1,000–1,900 miles) beyond its present limit, is well documented. Similarly, former climates of the current humid tropical type are probably responsible for the presence of fossil crusts outside the tropics and for relict Paleogene and Neogene deep weathering. Such climates…<\|endoftext\|>	3.953125
474	Quiz time: What plant tries to attract flies and digs itself into the ground? Ah February! It’s 5 degrees outside after an ice storm. Nothing green is poking through that layer of snow and ice. But coming soon (not soon enough is what we all think) new shoots will be just visible, then leaves and flowers! A miracle of nature! One of the very earliest plants to look for is skunk cabbage (official name – Symplocarpus foetidus). Look in marshy places like Hathaway Pond or at the edges of brooks that cross the bike trail. It’s called skunk cabbage because it smells skunky when any part of the plant is bruised. Skunk cabbage is an oddity. Unlike most plants, the flowers come up before the leaves, sometime between February and April. Look first for a spotted purple-brown tulip leaf-shaped hood, called a spathe. The hood wraps around the flower stalk, called a spadix, which is a pale green bumpy-looking knob. The actual flowers are very tiny dark-colored dots on the spadix. Skunk cabbage depends on flies, in addition to bees, to pollinate the flowers. The flower stalk and spathe actually make and give off heat which attracts flies during cold weather. They can give off enough heat to melt snow and leave bare soil around them. When the leaves come out, the plants stand 1-2 feet tall. Native Americans and early settlers used skunk cabbage leaves to treat respiratory diseases, nervous disorders, rheumatism, and edema. Here’s something skunk cabbage does that very few other plants do. It has contractile roots. The roots actually can pull against themselves after growing into the earth, which pulls the stem of the plant deeper into the mud. In a way the plant is growing downward, not upward. Each year, the plant grows deeper into the earth, so that older plants are practically impossible to dig up. Keep your eyes open for skunk cabbage while you are enjoying your walk through Marblehead’s conservation lands. It won’t be too long before the flowers are up. (Photo by Ryan Hodnett, Used Under Creative Commons Attribution-Share Alike 4.0 International)<\|endoftext\|>	3.8125
243	When someone has an allergy, they can have many different physical reactions when they are exposed to allergens. The type of reaction and the severity of it depends on the individual and the severity of their allergy. Very small amounts of some allergens, such as nuts, can cause severe adverse reactions including potentially fatal anaphylactic shock. The most common symptoms of an allergic reaction include: \|Body part affected \|\|Sore, red and/or itchy \|\|Runny and/or blocked \|\|Swelling of the lips \|\|Coughing, dry, itchy and swollen throat \|\|Coughing, wheezing and shortness of breath \|\|Nausea and feeling bloated, diarrhoea and/or vomiting \|\|Itchy and/or a rash When someone has a severe reaction to an allergen, this can lead to faintness and/or the person might collapse. Find out more Visit the food allergy and intolerance section of the NHS website: NHS Choices Advice on food allergen labelling (Download this pdf, 0.34mb) Food intolerance and coeliac disease – How to avoid certain foods<\|endoftext\|>	3.84375
2,279	<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" /> You are viewing an older version of this Concept. Go to the latest version. # Graphs of Logarithmic Functions ## Graphing the inverse of exponential functions using transformations Estimated11 minsto complete % Progress Practice Graphs of Logarithmic Functions Progress Estimated11 minsto complete % Graphs of Logarithmic Functions Logarithmic functions are in many ways similar to exponential functions, and can be applied in similar situations. Log functions can be used to model radioactive decay (to estimate the age of fossils, for instance), or estimate the discharge of electricity from a capacitor. Logarithmic functions can be graphed to allow quick and reasonably accurate estimation of the useable area of a WiFi network, or estimate the minimum length of a filtering pipe in order to have a specific purity of kerosine. As you work with the graphs of log functions in this lesson, you will notice that they bear a resemblance to another family of graphs which you have explored already. At the end of the lesson, we will revisit this fact. Can you identify what other family the log functions resemble, and explain in your own words why that is the case before the review? Embedded Video: ### Guidance In a previous lesson, log functions were identified as the inverses of exponential functions, in this lesson we explore that fact visually through the graphs of logarithmic functions. As you can see below, because the function f(x) = log2x is the inverse of the function g(x) = 2x, the graphs of these functions are reflections over the line y = x. We can verify that the functions are inverses by looking at the graph. For example, the graph of g(x) = 2x contains the point (1, 2), while the graph of f(x) = log2x contains the point (2, 1). Also, note that while that the graph of g(x) = 2x is asymptotic to the x-axis, the graph of f(x) = log2x is asymptotic to the y-axis. This behavior of the graphs gives us a visual interpretation of the restricted range of g and the restricted domain of f. When graphing log functions, it is important to consider x- values across the domain of the function. In particular, we should look at the behavior of the graph as it gets closer and closer to the asymptote. Consider f(x) = log2x for values of x between 0 and 1. If x = 1/2, then f(1/2) = log2(1/2) = -1 because 2-1 = 1/2 If x = 1/4, then f(1/4) = log2(1/4) = -2 because 2-2 = 1/4 If x = 1/8, then f(1/8) = log2(1/8) = -3 because 2-3 = 1/8 From these values you can see that if we choose x values that are closer and closer to 0, the y values decrease (heading towards !). In terms of the graph, these values show us that the graph gets closer and closer to the y-axis. Formally we say that the vertical asymptote of the graph is x = 0. Graphing Logarithmic Functions Using Transformations Consider again the log function f(x) = log2x. The table below summarizes how we can use the graph of this function to graph other related functions. Equation Relationship to f(x) = log2x Domain g(x) = log2(x - a), for a > 0 Obtain a graph of g by shifting the graph of f a units to the right. x > a g(x) = log2(x+a) for a > 0 Obtain a graph of g by shifting the graph of f a units to the left. x > -a g(x) = log2(x) + a for a > 0 Obtain a graph of g by shifting the graph of f up a units. x > 0 g(x) = log2(x) - a for a > 0 Obtain a graph of g by shifting the graph of f down a units. x > 0 g(x) = alog2(x) for a > 0 Obtain a graph of g by vertically stretching the graph of f by a factor of a. x > 0 g(x) = -alog2(x) , for a > 0 Obtain a graph of g by vertically stretching the graph of f by a factor of a, and by reflecting the graph over the x-axis. x > 0 g(x) = log2(-x) Obtain a graph of g by reflecting the graph of f over the y-axis. x < 0 #### Example A Graph the function f(x) = log4x and state the domain and range of the function. Solution: The function f(x) = log4x is the inverse of the function g(x) = 4x. We can sketch a graph of f(x) by evaluating the function for several values of x, or by reflecting the graph of g over the line y = x. If we choose to plot points, it is helpful to organize the points in a table: x y = log4x 1/4 1 0 4 1 16 2 The graph is asymptotic to the y-axis, so the domain of f is the set of all real numbers that are greater than 0. We can write this as a set: . While the graph might look as if it has a horizontal asymptote, it does in fact continue to rise. The range is . #### Example B Graph the functions: f(x) = log2(x) g(x) = log2(x) + 3 h(x) = log2(x + 3) Solution: The graph below shows these three functions together: Notice that the location of the 3 in the equation makes a difference! When the 3 is added to log2x , the shift is vertical. When the 3 is added to the x, the shift is horizontal. It is also important to remember that adding 3 to the x is a horizontal shift to the left. This makes sense if you consider the function value when x = -3: h(-3) = log2(-3 + 3) = log20 = undefined This is the vertical asymptote! Note that in order to graph these functions, we evaluated them by investigating specific values of x. If we want to know what the x value is for a particular y value, we need to solve a logarithmic equation. #### Example C Graph the function Solution Start by making a table: x y 3 4 0 5 1 6 1.585 7 2 8 2.32 Since that means and That means that when Since that means and That means that when The graph looks like: Do you recall the question from the introduction to the lesson? Can you identify what other family the log functions resemble, and explain in your own words why that is the case? By now, I am sure you are quite familiar with the answer, it was repeated several times throughout the lesson. Logarithmic function graphs are the inverses of exponential function graphs, because the very definition of a logarithmic function it as the inverse of an exponential function! ### Vocabulary Exponential functions are functions with the input variable (the x term) in the exponent. Logarithmic functions are the inverse of exponential functions. Recall: logbn = a is equivalent to ba=n. log is the shorthand term for 'the logarithm of', as in: "logbn" = "the logarithm, base 'b', of 'n' ". ### Guided Practice 1) Which of the following functions is graphed in the image below? a) b) c) 2) Graph the function 3) Transform the graph of from problem #2 into the graph of 1) All three functions are varieties of The image shows the function reflected across the axis, therefore: c) is correct. 2) To graph we will start with a table of values: x y 3 1 9 2 27 3 81 4 Plotting those points and drawing a smooth curve between them gives: 3) The graph of above ran up the positive side of the y-axis to reach the x-axis. The "+3" inside the parenthesis of means there is a shift of 3 to the left. The image of shifted 3 units to the left looks like this: ### Practice Identify the domain and range, then sketch the graph. Look at the graphs below and identify the function that the graph represents from the functions listed below. a) b) c) d) e) Graph the following logarithms. 1. - 19. The table below shows the x and y values of the points on an exponential curve. Switch them and identify the corresponding coordinates of the points that would appear on the logarithmic curve. Can you identify the function? Point on exponential curve Corresponding point on logarithmic curve (-3, 1/8) (1/8, -3) (-2, 1/4) 14. (__ , __) (-1, 1/2) 15. (__ , __) (0, 1) 16. (__ , __) (1, 2) 17. (__ , __) (2, 4) 18. (__ , __) (3, 8) 19. (__ , __) Graph logarithmic functions, using the inverse of the related exponential function. Then graph the pair of functions on the same axes. ### Vocabulary Language: English Asymptotes Asymptotes An asymptote is a line on the graph of a function representing a value toward which the function may approach, but does not reach (with certain exceptions). Exponential Function Exponential Function An exponential function is a function whose variable is in the exponent. The general form is $y=a \cdot b^{x-h}+k$. log log "log" is the shorthand term for 'the logarithm of', as in "$\log_b n$" means "the logarithm, base $b$, of $n$." Logarithmic function Logarithmic function Logarithmic functions are the inverses of exponential functions. Recall that $\log_b n=a$ is equivalent to $b^a=n$. operation operation Operations are actions performed on variables, constants, or expressions. Common operations are addition, subtraction, multiplication, and division.<\|endoftext\|>	4.53125
278	# How would you solve this: Tickets for a community dinner cost \$4 for adults and \$3 for children. A total of 390 tickets was sold ,earning \$1,380. how many of each type of ticket were sold? 1 by elizabethec123 2014-11-20T20:40:49-05:00 We can make a system of equations to solve for this. We'll use the variable a for adult ticket costs and c for children ticket costs. 4a+3c=1380 (cost of adult plus child tickets) a+c=390 (number of child tickets plus adult tickets) Now we can use substitution or elimination. I'll be using substitution since we already have singular variables in the second equation. First we isolate a variable. a+c=390 a=390-c Now we plug in and solve 4(390-c)+3c=1380 1560-4c+3c=1380 -c=-180 c=180 Now we know that there were 180 children tickets sold. We could do the same thing to find the amount of adult tickets but we already know that there were a total of 390 tickets sold. So if there are 180 child tickets, the rest have to be adult tickets. 390-180=210 180 Children tickets<\|endoftext\|>	4.4375
771	Courses Courses for Kids Free study material Offline Centres More Store # Work done for the process shown in the figure is: (A) $1\,J$(B) $1.5\,J$(C) $4.5\,J$(D) $0.3\,J$ Last updated date: 14th Jun 2024 Total views: 51.9k Views today: 1.51k Verified 51.9k+ views Hint From the given figure, the detailed figure is drawn, then the solution can be determined. In the detailed diagram it forms one rectangle and one right angled triangle. By finding the area of the two shapes and adding the two areas is equal to the work done. Useful formula: The area of the rectangle is given as, ${A_r} = l \times b$ Where, ${A_r}$ is the area of the rectangle, $l$ is the length of the rectangle and $b$ is the breadth of the rectangle. The area of the right-angled triangle is given as, ${A_t} = \dfrac{{ab}}{2}$ Where, ${A_t}$ is the area of the right angle triangle, $a$ is the base of the triangle and $b$ is the height of the triangle. Complete step by step solution Given that, The initial condition of the volume is, ${V_1} = 10\,cc$. The final condition of the volume is, ${V_2} = 25\,cc$. The initial condition of the pressure is, ${P_1} = 10\,kPa$. The final condition of the pressure is, ${P_2} = 30\,kPa$. The work done of the process is given by the following figure. From the above diagram, the area of the rectangle is given by, ${A_r} = l \times b$ By substituting the length and the breadth of the rectangle from the above diagram, then the above equation is written as, ${A_r} = 10 \times 15$ By multiplying the terms in the above equation, then the above equation is written as, ${A_r} = 150\,kPa.cc$ From the above diagram, the area of the triangle is given by, ${A_t} = \dfrac{{ab}}{2}$ By substituting the base and the height of the triangle from the above diagram, then the above equation is written as, ${A_t} = \dfrac{{20 \times 15}}{2}$ By multiplying the terms in the above equation, then the above equation is written as, ${A_t} = \dfrac{{300}}{2}$ By dividing the terms in the above equation, then the above equation is written as, ${A_t} = 150\,kPa.cc$ The work done is the sum of the two areas, then the work done is given as, $W = 150 + 150$ By adding the terms in the above equation, then the above equation is written as, $W = 300\,kPa.cc$ Then the above equation is written as, $W = 0.3\,J$ Hence, the option (D) is the correct answer. Note: The work done of the object is directly proportional to the force of the object and the distance between them. As the force of the object and the distance between them increases, then the work done of the object is also increasing. As the force of the object and the distance between them decreases, then the work done of the object is also decreasing.<\|endoftext\|>	4.5625
843	Create Your Own Maps Students learn about their home state by creating it using Mr. Nussbaum’s Map Builder 2. This site is great for creating just about any land map imaginable! This activity will take several class periods. - Be able to create a map of their state. - Be able to add a compass rose, legend and important attributes of their state. Map: A map is a picture or chart that shows the rivers, mountains, streets, etc., in a particular area a map of the country. Map Key/ Legend: A map key or legend is included with a map to unlock it. It gives you the information needed for the map to make sense. Symbol: Maps often use symbols or colors to represent things, and the map key explains what they mean. Symbols in the key might be pictures or icons that represent different things on the map. Compass rose: A compass rose is a figure on a compass, map, nautical chart, or monument used to display the orientation of the cardinal directions Cardinal directions: Cardinal directions are North, East, South, and West Capital city: A capital city is the most important city or town of a country or region, usually its seat of government and administrative center. To prepare for this lesson: Decide what book/ video you want to use to begin your lesson. NOTE: If you have access to the program KID PIX, in the BACKGROUNDS, you can drag up just about any map and you can find correlating flags in the STICKERS. Students are able to use the paint bucket to color in and they can create their own symbols using stamps. See Accommodations Page and Charts on the 21things4students.net site in the Teacher Resources. Directions for this activity: The teacher will say to the class, "Boys and girls, today you are going to be map makers. You are going to create maps of our state. As a map maker, what do you think would be some important components we would need?" Have the group brainstorm what they would need while you guide them. Suggestions could be a title, capital, places of interest, a body of water. You could use this to add to the rubric. (MAKE A COPY OF THE RUBRIC BEFORE YOU EDIT.) Play or Read: Review with students the Video How To on Mr. Nussbaum. Have students explore Mr. Nussbaum’s map site. Different options for assessing the students: - Check for understanding - Create Your Own Maps rubric MITECS: Michigan adopted the "ISTE Standards for Students" called MITECS (Michigan Integrated Technology Competencies for Students) in 2018. 4a. Students know and use a deliberate design process for generating ideas, testing theories, creating innovative artifacts or solving authentic problems. 4b. Students select and use digital tools to plan and manage a design process that considers design constraints and calculated risks. 4c. Students develop, test and refine prototypes as part of a cyclical design process. 4d. Students exhibit a tolerance for ambiguity, perseverance and the capacity to work with open-ended problems. Devices and Resources CONTENT AREA RESOURCES Have the students create a commercial about their state including the key points on the map. The teacher will add map scales to show students how to calculate distances. The students may Include natural resources on their map. 2 – G1 0 1 Construct maps of the local community that contain symbols, labels, and legends denoting human and natural characteristics of place 2 – G1 0 2 Use maps to describe the spatial organization of the local community by applying concepts including relative location, and using distance, direction, and scale 2 – G1 0 3 Use maps to describe the location of the local community within the state of Michigan in relation to other significant places in the state. This task card was created by Courtney Conley, Utica Community Schools, May 2018.<\|endoftext\|>	4.34375
277	Who is at risk? Approximately half of the world's population is at risk of malaria. Most malaria cases and deaths occur in sub-Saharan Africa. However, Asia, Latin America, and to a lesser extent the Middle East and parts of Europe are also affected. In 2013, 97 countries and territories had ongoing malaria transmission. Specific population risk groups include: \|→\|\|young children in stable transmission areas who have not yet developed protective immunity against the most severe forms of the disease;\| \|→\|\|non-immune pregnant women as malaria causes high rates of miscarriage and can lead to maternal death;\| \|→\|\|semi-immune pregnant women in areas of high transmission. Malaria can result in miscarriage and low birth weight, especially during first and second pregnancies;\| \|→\|\|semi-immune HIV-infected pregnant women in stable transmission areas, during all pregnancies. Women with malaria infection of the placenta also have a higher risk of passing HIV infection to their newborns;\| \|→\|\|people with HIV/AIDS;\| \|→\|\|international travellers from non-endemic areas because they lack immunity;\| \|→\|\|immigrants from endemic areas and their children living in non-endemic areas and returning to their home countries to visit friends and relatives are similarly at risk because of waning or absent immunity\|<\|endoftext\|>	3.71875
389	In 1965, engineer Gordon Moore predicted that the number of transistors on an integrated circuit -- a precursor to the microprocessor -- would double approximately every two years. Today, we call this prediction Moore's Law, though it's not really a scientific law at all. Moore's Law is more of a self-fulfilling prophecy about the computer industry. Microprocessor manufacturers strive to meet the prediction, because if they don't, their competitors will [source: Intel]. To fit more transistors on a chip, engineers have to design smaller transistors. The first chip had about 2,200 transistors on it. Today, hundreds of millions of transistors can fit on a single microprocessor chip. Even so, companies are determined to create increasingly tiny transistors, cramming more into smaller chips. There are already computer chips that have nanoscale transistors (the nanoscale is between 1 and 100 nanometers -- a nanometer is one billionth of a meter). Future transistors will have to be even smaller. Enter the nanowire, a structure that has an amazing length-to-width ratio. Nanowires can be incredibly thin -- it's possible to create a nanowire with the diameter of just one nanometer, though engineers and scientists tend to work with nanowires that are between 30 and 60 nanometers wide. Scientists hope that we will soon be able to use nanowires to create the smallest transistors yet, though there are some pretty tough obstacles in the way. In this article, we'll look at the properties of nanowires. We'll learn how engineers build nanowires and the progress they've made toward creating electronic chips using nanowire transistors. In the last section, we'll look at some of the potential applications for nanowires, including some medical uses. In the next section, we'll examine the properties of nanowires.<\|endoftext\|>	3.953125
1,134	DNA (deoxyribonucleic acid) is a large molecule which carries the genetic information, or blueprint, of all life on Earth. Mutations arising in the DNA code account for the diversity upon which evolution by natural selection can work. Therefore, it is not far-fetched to say that DNA is one of the central, most important molecules in living organisms. For such an important molecule, it sure looks beautiful: DNA is a double helix i.e. two individual strands running along each other in an anti-parallel way, connected to one another by relatively weak hydrogen bonds. DNA’s structure can be learned easily by thinking about the strands and the “stuff in-between” separately. What are the strands made of? The strands are made of repeating units consisting of a deoxyribose (sugar) molecule with a phosphate molecule attached to it; hence, it is called a sugar-phosphate backbone. Phosphodiester bonds between nucleotides (above) create the backbone: DNA is made of 2 of these strands running in an anti-parallel structure. Because the strands have a specific orientation, they are termed differently. It happens that at the end of each strand, the nucleotide with the unattached phosphate group is found at one end, while the deoxyribose sugar is found at the opposite end, with its phosphate group linked to the above nucleotide via the phosphodiester bond. The end with the free phosphate group is called the 5′ (five prime) end because the group is attached to its nucleotide by the fifth carbon in the ring, while the opposite end is called the 3′ (three prime) end because the (hydroxyl) group is attached to the third carbon in the ring. The 5′ to 3′ direction is also the direction DNA is synthesised in living things. What is the centre made of? Attached to the sugar molecules in the backbone are a different type of molecule called nitrogenous base. There are 4 bases in DNA: adenine, thymine, cytosine and guanine. These are abbreviated by their initials: A, T, C and G. The hydrogen bonds are formed between these bases. Due to their complementary shapes, A always pairs with T, and C always pairs with G. A-T is linked by 2 H bonds, while C-G is linked by 3. Here is a diagram of this arrangement: The bases can be sorted into two categories: purines and pyrimidines depending on their ring structure: As you can see, adenine and guanine are bigger and have two rings, while thymine and cytosine only have one ring. Uracil is similar to thymine and also pairs with adenine. The presence of uracil instead of thymine occurs in RNA rather than DNA. It is the sequence of these bases that encodes genes and the sum of an organism’s genetic material, termed genotype. DNA is a very stable molecule, as its purpose of carrying genetic information is very important. Features of this are: 1. DNA is very temperature-resistant, and the H bonds only break at temperatures of about 92 degrees Celsius 2. The sugar-phosphate backbone acts as a shield to the bases, preventing interference from outside chemical reactions 3. The double helix gives stability 4. Many H bonds contribute to the stability 5. The structure of the sugar-phosphate backbone itself confers strength. DNA and chromosomes may seem like completely separate things. Well, they’re not. In fact, all chromosomes are individual DNA molecules coiled and twisted around, because DNA is huge. At least in eukaryotes it is. That’s one of the first differences between eukaryotes and prokaryotes in their DNA – prokaryotes have less DNA. Eukaryotic DNA is stored within the nucleus of each cell (apart from cells without one, e.g. red blood cells). Because of its sheer size, it must be organised well. Proteins called histones help do just that: This is the second difference: eukaryotes have histones around which the DNA coils, while prokaryotes don’t. So what does prokaryotic DNA look like? The DNA above is stored as a small loop (the bacterial chromosome), and as a plasmid. A plasmid is even smaller, and may be copied and transferred to another bacterium of the same or different species by a process called conjugation (or, more colloquially, bacteria sex). Eukaryotic DNA is linear, rather than circular. That means the DNA, despite being coiled numerous times, has a distinguishable start point and end point, while prokaryotic DNA is a continuous circle (see above picture). An exception to this is the DNA of yeast which are also eukaryotic but whose genetic material is instead a circular plasmid. Circular chromosomes are also found in eukaryotic cell organelles like mitochondria and chloroplasts. These organelles have functions in energy production through cellular respiration and photosynthesis. Their different DNA can be explained by the ancestry of these organelles. They used to be standalone prokaryotic cells, and underwent phagocytosis events where they were engulfed by bigger cells.<\|endoftext\|>	4.25
894	# Often asked: What Is Solid Mensuration Math? ## What is a solid figure in math? Solid figures are three-dimensional objects, meaning they have length, width, and height. Because they have three dimensions, they have depth and take up space in our universe. Solid figures are identified according to the features that are unique to each type of solid. ## What are the differences between plane and solid mensuration? Answer: A plane figure is two-dimensional, and a solid figure is three-dimensional. The difference between plane and solid figures is in their dimensions. The same shape takes on extra dimension by adding additional points and lines to give the shape height, width and depth. ## What are the types of mensuration? So let us study the topic mensuration and mensuration formulas in detail. • Cylinder. • Circles. • Polygons. • Rectangles and Squares. • Trapezium, Parallelogram and Rhombus. • Area and Perimeter. • Cube and Cuboid. ## What are the different types of solids in mathematics? The list of solid shapes includes cube, cuboid, sphere, cone, hemisphere, prism, cylinder, and pyramid, etc. You might be interested: Quick Answer: What Does Expression Mean In Math? ## Which object is solid example? A solid has a defined shape and volume. A common example is ice. A liquid has a defined volume, but can change state. An example is liquid water. ## What is a solid object? A solid is characterized by structural rigidity and resistance to a force applied to the surface. Unlike a liquid, a solid object does not flow to take on the shape of its container, nor does it expand to fill the entire available volume like a gas. ## Is a square solid or flat? plane shapes: The flat surfaces of solid figures are plane shapes. rectangle: a rectangle is a plane shape with two sets of opposite sides that are the same length. square: A square is a plane shape. All sides are the same length. ## Is a cylinder solid or flat? Cylinder A cylinder is a solid figure with two parallel congruent circular bases. Plane FigureA plane figure is a flat, two-dimensional figure. PrismA prism is a three-dimensional object with two congruent parallel bases that are polygons. ## What is the formula of mensuration? Volume of Cylinder = π r2 h. Lateral Surface Area (LSA or CSA) = 2π r h. Total Surface Area = TSA = 2 π r (r + h) Volume of hollow cylinder = π r h(R2 – r2) ## What is mensuration explain? Menstruation, or period, is normal vaginal bleeding that occurs as part of a woman’s monthly cycle. Every month, your body prepares for pregnancy. If no pregnancy occurs, the uterus, or womb, sheds its lining. The menstrual blood is partly blood and partly tissue from inside the uterus. ## Who is the father of mensuration? Answer: Leonard Digges is the father of mensuration. Leonard Digges was a well-known English mathematician and surveyor, credited with the invention of the theodolite, and a great populariser of science through his writings in English on surveying, cartography, and military engineering. You might be interested: Question: What Is Lcd In Math Example? ## What is basic mensuration? Mensuration is the skill of measuring the length of lines, areas of surfaces, and volumes of solids from simple data of lines and angles. Mensuration in its literal meaning is to measure. Measuring these quantities is called Mensuration. ## What are the five solids? Platonic solid, any of the five geometric solids whose faces are all identical, regular polygons meeting at the same three-dimensional angles. Also known as the five regular polyhedra, they consist of the tetrahedron (or pyramid), cube, octahedron, dodecahedron, and icosahedron. ## How many types of solid figures are there? Answer: The major types of solid shapes are: cubes, cuboids, prisms, pyramids, platonic solids, torus, cone, cylinder, and sphere. ## What are the 4 types of solids? There are four types of crystalline solids: ionic solids, molecular solids, network covalent solids and metallic solids.<\|endoftext\|>	4.46875
670	Funerary customs, diet, and social behavior in a pre-Roman Italian Celtic community Credit: Laffranchi et al, 2019. Analysis of human remains from a Pre-Roman Celtic cemetery in Italy shows variations in funerary treatment between individuals that could be related to social status, but these variations were not reflected by differences in their living conditions. Zita Laffranchi of Universidad de Granada, Spain, and colleagues present these new findings in the open access journal PLOS ONE on April 17, 2019. Archeological and written evidence have provided scant insight into social organization among the Cenomani Gauls, Celtic people who began to spread into Northern Italy in the 4th century BC. In a preliminary step to address this knowledge gap, Laffranchi and colleagues analyzed remains from Verona – Seminario Vescovile, a Cenomani burial site that was used between the 3rd and 1st centuries BC. Seminario Vescovile features some variety in funerary approaches. Some individuals were buried face up, face down, or on their sides; some with animals; and some with varying amounts and types of objects, such as pottery and decorations. Focusing on 125 individuals buried at the site, the researchers set out to explore links between funerary treatment, age, sex, and diet–the first such study for Cenomani Gauls. Bone analysis revealed different carbon and nitrogen isotope levels between sexes, indicating that men and women at Verona-Seminario Vescovile had different diets. Tooth analysis suggests that the population had higher rates of dental enamel defects than did other Pre-Roman and Celtic groups of the Italian Peninsula, pointing to higher physiological stress levels during childhood. The researchers were surprised to find no differences between sexes in funerary treatment. They also found no links between funerary differences and isotope measurements or enamel defects. Increased age was somewhat related to increased number of grave objects, but age and burial position were not linked. These findings suggest the existence of weak social differentiation that could have influenced funerary treatment, but that, apart from different diets for men and women, individuals in the group shared similar living conditions. The authors conclude that the individuals they analyzed were likely members of a homogeneous and potentially low-status group within the original population. Still, they call for extensive additional research to clarify the implications of their findings. The authors add: “This is the first combined study of diet, developmental stress, and funerary patterns in a Celtic community from the Italian peninsula: the little known Cenomani (North-Eastern Italy).” Citation: Laffranchi Z, Cavalieri Manasse G, Salzani L, Milella M (2019) Patterns of funerary variability, diet, and developmental stress in a Celtic population from NE Italy (3rd-1st c BC). PLoS ONE 14(4): e0214372. https:/ Funding: The authors received no specific funding for this work. Competing Interests: The authors have declared that no competing interests exist. In your coverage please use this URL to provide access to the freely available article in PLOS ONE: http://journals. Related Journal Article<\|endoftext\|>	3.71875
1,347	Special thanks to String for alerting me to this amazing recent discovery: About a week ago, a team of astrophysicists from the University of Helsinki led by Dr. Lauri Jetsu submitted research suggesting that the ancient Egyptians understood the periodicity of the star Algol, and that they incorporated its cycle into their charts for the relative auspiciousness and unluckiness of days throughout the year (in fact, not just of days, but of sections of days -- their calendars were three times more precise than simply lucky and unlucky days, breaking each day into three parts and discussing the merits of each). Algol is the second-brightest star (designated as β-Persei) in the constellation Perseus (almost as bright as the brightest star in Perseus, Mirfak, also known as α-Persei). The constellation Perseus is shown below, with a red arrow pointing to Algol. His position in the sky relative to nearby celestial landmarks is discussed in this previous post and this previous post, although those tend to depict Perseus as seen rising in the east, and during this time of the year he is descending in the west in the hours after sunset and before midnight. Why would the ancient Egyptians (circa 1200 BC according to the texts examined by Dr. Jetsu and colleagues) tie their predictions as to the relative propitiousness of days (and portions of days) to Algol, and how did these modern researchers discover that they were looking at Algol based on texts discussing the merits of those days and times throughout the year? As the article above explains, the researchers analyzed the pattern of predictions preserved in the ancient Egyptian Cairo Calendar and discovered two cycles throughout the year, one corresponding quite accurately to the cycle of the moon, and one corresponding to a period of 2.85 days. The scholars present evidence that the Cairo Calendar's 2.85-day cycle is linked to the cycle of Algol, which dims appreciably every 2.867 days. This phenomenon is visible to the naked eye, and is caused by the fact that Algol is actually a system of three stars, two of which orbit one another rather closely (one of which being the star we see and the other a much larger but less massive and dimmer giant star which causes the dimming phenomenon as it eclipses the smaller brighter star) plus a third star orbiting this pair at a greater distance. The system and the dimming phenomenon (also known as the "minima of Algol") are described here in the very informative website of Professor Jim Kaler. The entire eclipse lasts about ten hours from start to finish, but the appreciable dimmed period is shorter than that, perhaps four to five hours. Some astronomers have long suspected that the ancients may have known about the antics of Algol. For one thing, there is the Arabic name of the star, Al Ghul or "the demon" ("the ghoul"), which gives us the star's modern name. The great H.A. Rey describes this Arabic name as meaning "the prankster" (The Stars: A New Way to See Them, 42). There is also the fact that Ptolemy (c. AD 90 - c. AD 168) in his Almagest (which was preserved by and heavily influenced the Arabic astronomers) designates this star as representing the head of the Gorgon Medusa being carried by Perseus (who slew her with the help of Athena, Hermes, and Pegasus). There have also been those who have denied the ancients the sophistication to have identified the periodicity of Algol, although it seems quite presumptuous to do so, particularly in light of other evidence demonstrating that their knowledge clearly surpassed anything for thousands of years afterwards (such as this evidence that they understood the atmospheric effects which cause a phenomenon called "extinction," to say nothing of the even more ancient evidence that they understood the subtle phenomenon of precession, a fact that most conventional historians continue to dispute). Now, this intrepid team from Finland has added further evidence for the high level of precision attained at very remote dates in mankind's ancient history. They even provide an argument that the periodicity captured in the Cairo Calendar (which is slightly shorter than the periodicity recorded in modern times, since Algol's behavior was "rediscovered" in the 1600s) was not in error, but that the observations provided by the ancient texts can help astronomers today to determine the change in the motions of the Algol system over the 3000-plus intervening years! We should all be grateful to Drs. Jetsu, Porceddu, Lyytinen, Kajatkari, Lehtinen, Markkanen, and Toivari-Viitala for their diligent work in uncovering this amazing new window into human history and knowledge. Left unexamined in many articles discussing this incredible new find is one other big elephant in the room. Remember, this entire discovery by the Finnish team came about because the ancient Egyptians recorded very detailed beliefs about the relative auspiciousness of the segments of each day throughout the year. Nobody seems to ask whether -- if they knew so much about astronomy, as well as about many other subjects (and see here as well) -- they could have actually known something we don't know about the impact of the different "arrangements of the heavens" on affairs here on our little planet. This previous post discusses some of the logic behind the idea that the "harmonies" created by the arrangements of heavenly bodies might have some impact on the way that we feel, just as the relative "harmonies" or "dis-harmonies" of a building's architecture (or even the arrangement of furniture inside) can have an impact on us but from a smaller scale of distance. This post explores the idea as well. Here is one other reference discussing the minima of Algol, and the traditions that these eclipses of the star can lead to unfortunate "losing one's head" with unpleasant results (see Medusa, above). Note that this site contains a list of times and dates for these cycles of Algol, running all the way up to the end of 2012. As this is published, Algol has just emerged a few hours ago from one minima, and another eclipse will take place early in the small hours of the morning just after midnight on May 11 (Greenwich Mean Time, which will still be May 10 in North America and points west, out to the International Date Line).<\|endoftext\|>	3.859375
1,106	Anyone who took electronics classes will be familiar with the circuit elements inductor, capacitor, and resistor. In 1971, a man named Leon Chua speculated that there should be a fourth class of elements for a fundamental passive circuit called a short memory resistor, or memristor. The memristor concept showed invaluable properties pertaining to the augmentation of classical circuits. Until 2008, no one had given evidence of a physical model or pragmatic use of a memristor. Since then many research companies and industry entrepreneurs have been expending massive amounts of resources to realize the untapped potential of the memristor, as it has advantageous prospects for things such as solid state drives and ionic transport components. Memristor layout. Image courtesy of NCBI. As neuroscientists have taken on the task of replicating a human brain in an attempt to mimic human recognition, researchers from the Moscow Institute of Physics and Technology have been attempting to replicate the neural activity of a real brain by using analog components. In a recent paper published in the Journal of Nanoscale Research Letters, the MIPT team reinvented the memristor to get one step closer to creating artificial synapses. The key component to creating a computerized brain is, of course, the memristor. In short, what a memristor does is regulate the flow of electrical current through a circuit and remember the charge value that passed through it. By their very nature, memristors are non-volatile: They can retain information with or without power. An analogy is often made to a water pipe: Water flows through the pipe and expands in diameter, allowing water to pass through the pipe faster. When water flows back through the pipe, the diameter contracts and thus slows down the water flow; even when the water is turned off, the pipe retains the same structure. Similarly in electrical standards, even when a memristor is turned off, it is capable of retaining the same resistance value. For example, if you cut the power of a computer with a memristor in it, the monitor would immediately be able to display the images and documents it displayed before the power was cut— as soon as power passed through it again. It is remarkably difficult to replicate the mind of a human. However, we are currently capable of mimicking the brain of a cat. Neural network simulations such as Google’s DeepMind network have seen strange results, like placing the heads of dogs onto shapes by using image recognition software dubbed Deep Dream. An image before (left) and after (right) being interpreted by Google DeepDream. Image generated via Deep Dream Generator. Original photo by Jay Ruzesky. Unfortunately, the human brain is much faster and more complicated than its electrical counterparts. Biological brains can recognize images in a tenth of a second, so scientists want to imitate the brain, but the architecture of a brain is very different from that of a computer. In some models, neurons are connected and relay information via synapses, but each neuron can have over ten thousand connections with other neurons. Currently, it is postulated that changes in synapse conductivity are responsible for the processing of information, indicating that information essentially needs to pass through a giant chain of synapse changes to be "received". What makes it incredibly hard to replicate the human brain is the fact that with classic systems, in order to replicate a synapse and neuron, one would need a massive amount of circuitry to mimic the billions of neurons and thousands of synapses. This would also consume a lot of power. Image courtesy of the University of Buffalo. Now what the researchers at MIPT did was recreate a memristor more suitable to mimic the brain. MIPT’s reinvention of the memristor made use of hafnium oxide (HfO2), a compound commonly used in CMOS fabrications. By applying voltage to the HfO2 layer, oxygen ions are driven from the lattice to electrodes. This leaves oxygen vacancies which allow for the transport of electrons. The resistance of the memristor is then defined by the number of oxygen vacancies in a channel, and can change over time by biasing the memristor. The MIPT team is currently using their memristor-based electronic brain to simulate the functions of an organic brain such as memorizing and forgetting, noted as "long term potentiation" and "long term depression", respectively. This allows the electronic brain to gain plasticity in that it is capable of forgetting useless information in order to make room for new knowledge. The combination of these elements is called spike-timing-dependent plasticity. In biology, this is the process that adjusts the strength of the connections between neurons in the brain. Connection strengths are adjusted based on the timing of neurons input and output action potentials, referred to as spikes. STDP has been proven to be the process of associated with learning. Thus, many researchers have been attempting to mimic the process using memristors. A synapse spike. Image courtesy of MIPT. Currently, no one has been able to devise a system that can integrate the memristor synapses into a full neural system, but configuring a memristor to become a synapse is the first step in emulating a human brain. You can find the full published research paper here via the Journal of Nanoscale Research Letters.<\|endoftext\|>	4
1,249	Upcoming SlideShare × # 2.7 find square roots and compare real numbers day 1 492 -1 Published on Published in: Technology, Business 0 Likes Statistics Notes • Full Name Comment goes here. Are you sure you want to Yes No Your message goes here • Be the first to comment • Be the first to like this Views Total Views 492 On Slideshare 0 From Embeds 0 Number of Embeds 0 Actions Shares 0 12 0 Likes 0 Embeds 0 No embeds No notes for slide ### 2.7 find square roots and compare real numbers day 1 1. 1. Finding Square Roots and Comparing Real Numbers <br />Section 2.7<br />P. 110<br /> 2. 2. Objectives<br />Evaluate & approximate square roots<br />Looking at the number “sets” again and understanding the members of each set.<br /> 3. 3. Square root: If b2 = a, then b is a square root of a<br />Example:<br /> If 32 = 9, then 3 is a square root of 9.<br /> 4. 4. Vocabulary:<br />Positive square root<br />Negative square root<br />Radicand<br />RadicalSymbols:<br />Undefined – why??<br />±<br /> 5. 5. More vocabulary:<br />Perfect Squares – numbers whose square roots are integers or quotients of integers<br />Irrational Number – a number that cannot be written as the quotient of two integers (it never terminates or repeats)<br />REALS - the set of all rational and irrational numbers. (see P. 112)<br /> 6. 6. Evaluate & approximate square roots<br /> 7. 7. Evaluate & approximate square roots<br />Approximate the square root to the nearest integer<br /> 8. 8. Looking at the number “sets” again and understanding the members of each set<br />Counting<br /> 9. 9. Looking at the number “sets” again and <br />understanding the members of each set<br />Tell whether each of the following numbers is a real number, a rational number, an irrational number, an integer, or a whole number: , , – .<br />Integer?<br />Real Number?<br />Whole Number?<br />Irrational Number?<br />Rational Number?<br />Number<br /><br /><br /><br />24<br />81<br />100<br /><br />No<br />No<br />Yes<br />No<br />Yes<br />24<br />Yes<br />Yes<br />Yes<br />No<br />Yes<br /><br />100<br />No<br />Yes<br />No<br />Yes<br />Yes<br /><br />81<br />EXAMPLE 3<br /> 10. 10. 9<br />9<br />9.<br />Tell whether each of the following numbers. A rational<br />number,an irrational number, an integer, or a whole<br />number: ,5.2, 0, , 4.1,. There order the <br />number from least to greatest.<br /> 2<br /> 2<br />–<br />–<br /><br />7<br /><br />20<br />20 <br /><br />4.4<br />=<br />0<br /><br />7<br />2.6<br />–<br />=<br />4.1<br />–2<br />–8<br />–1<br />4<br />3<br />1<br />2<br />–3<br />–4<br />–5<br />–6<br />–7<br />–9<br />5<br />0<br />5.2<br />Looking at the number “sets” again and understanding the members of each set<br />EXAMPLE 4<br />Graph and order real numbers<br />GUIDED PRACTICE<br />SOLUTION<br />Begin by graphing the numbers on a number line.<br /> 11. 11. Looking at the number “sets” again and<br /> understanding the members of each set<br />,<br />4<br />4<br />,<br />,<br />Order the numbers from least to greatest:<br />–<br /> 3<br /> 3<br />.<br />,<br />–2.5<br />.<br /><br />5<br /><br />13<br /><br />9<br /><br />5<br /><br />9<br /><br />13<br />ANSWER<br />Read the numbers from left to right:<br />–2.5,<br />–<br />,<br />,<br />,<br />EXAMPLE 4<br />SOLUTION<br />Begin by graphing the numbers on a number line.<br /> 12. 12. Assignment: P. 113 (#1-27)<br />May want to use a chart for 24-27<br />Integer?<br />Real Number?<br />Whole Number?<br />Irrational Number?<br />Rational Number?<br />Number<br /><br />No<br />No<br />Yes<br />No<br />Yes<br />24<br />Yes<br />Yes<br />Yes<br />No<br />Yes<br /><br />100<br />No<br />Yes<br />No<br />Yes<br />Yes<br /><br />81<br /> 1. #### A particular slide catching your eye? Clipping is a handy way to collect important slides you want to go back to later.<\|endoftext\|>	4.4375
861	Substitution reaction, any of a class of chemical reactions in which an atom, ion, or group of atoms or ions in a molecule is replaced by another atom, ion, or group. An example is the reaction in which the chlorine atom in the chloromethane molecule is displaced by the hydroxide ion, forming methanol: CH3Cl + −OH→ CH3OH + Cl- If the chlorine atom is displaced by other groups—such as the cyanide ion (−CN), the ethoxide ion (C2H5O−), or the hydrosulfide ion (HS-)—chloromethane is transformed, respectively, to acetonitrile (CH3CN), methyl ethyl ether (CH3OC2H5), or methanethiol (CH3SH). Thus an organic compound such as an alkyl halide can give rise to numerous types of organic compounds by substitution reactions with suitable reagents. Substitution reactions are divided into three general classes, depending on the type of atom or group that acts as the substituent. In one, the substituent is electron-rich and provides the electron pair for bonding with the substrate (the molecule being transformed). This type of reaction is known as nucleophilic substitution. Examples of nucleophilic reagents are the halogen anions (Cl-, Br-, I-), ammonia (NH3), the hydroxyl group, the alkoxy group (RO−), the cyano group, and the hydrosulfide group. In the second type of substitution reaction, the substituent is deficient in electrons, and the electron pair for bonding with the substrate comes from the substrate itself. This reaction is known as electrophilic substitution. Examples of electrophilic species are the hydronium ion (H3O+), the hydrogen halides (HCl, HBr, HI), the nitronium ion (NO2+), and sulfur trioxide (SO3). Substrates of nucleophiles are commonly alkyl halides, while aromatic compounds are among the most important substrates of electrophiles. The third class of substitutions involves the reactions of free radicals with suitable substrates. Examples of radical reagents are the halogen radicals and oxygen-containing species derived from peroxy compounds. Learn More in these related Britannica articles: coordination compound: SubstitutionOne of the most general reactions exhibited by coordination compounds is that of substitution, or replacement, of one ligand by another. This process is depicted in a generalized manner by the equation ML x− 1Y + Z → ML x− 1Z + Y for a metal… chemical reaction: Substitution, elimination, and addition reactionsIn a substitution reaction, an atom or group of atoms in a molecule is replaced by another atom or group of atoms. For example, methane (CH4) reacts with chlorine (Cl2) to produce chloromethane (CH3Cl), a compound used as a topical anesthetic. In this reaction, a chlorine atom… chemical compound: Substitution reactionsThe simple replacement of one atom or group of atoms in a molecule by a second atom or group of atoms is called a substitution reaction. An illustrative example is the conversion of benzyl bromide to benzyl alcohol, using a solution of sodium… amine: SubstitutionAcylation is one of the most important reactions of primary and secondary amines; a hydrogen atom is replaced by an acyl group (a group derived from an acid, such as RCOOH or RSO3H, by removal of ―OH, such as RC(=O)―, RS(O)2―, and so on).… hydrocarbon: Chemical properties…characteristic reaction of alkanes is substitution; that of alkenes and alkynes is addition to the double or triple bond. Hydrogenation is the addition of molecular hydrogen (H2) to a multiple bond, which converts alkenes to alkanes. The reaction occurs at a convenient rate only in the presence of certain finely… More About Substitution reaction5 references found in Britannica articles - major reference - organic compounds<\|endoftext\|>	4.375
318	- For Teachers - For Families - Kids Corner - About the Resource Students will have an understanding of what influences their decisions and impacts their decision making process: In understanding these, students will be better able to make informed decisions about their physical and mental health and wellbeing. The definition of ‘influence’ is displayed on the board, without displaying the actual word “Influence”. The children are asked to guess what word is being defined. Once the definition has been discovered the class discusses various influences that they have in their lives. The students think about people who inspire and influence them in their life. Questions to consider: Students are invited to think, pair and share, giving reasons as to how this person may influence them. As a class discuss the term “role model”. What does it mean to you? Encourage the students to think about who their role model/s may be. Remind the students that the person doesn’t have to be famous. Students are chosen to present their responses to the class, explaining why that person is their role model. Discuss how they, as senior students of the primary school, can be role models to younger students. Give examples of times when they have been a role model, or how they can be a role model. Watch the short video ‘Healthy Eating Matters’ and discuss how the AIS athletes in the video are behaving as role models. Step 1: Select your school Can't see your school or organisation? Click here<\|endoftext\|>	4.4375
1,478	# Which Number Line Represents the Solutions to x + 5 = 1? ## Introduction Mathematics can be a daunting subject for many students, especially when it involves solving equations. One common equation that students struggle with is x + 5 = 1. While it may seem like a simple problem, it requires a clear understanding of number lines and their representations in order to find the correct solution. It’s important to understand this problem and how number lines can help with its solution because it lays the foundation for more complex equations that students will face in the future. This article will explore different types of number lines and how they can be used to represent solutions to the equation x + 5 = 1. ## Visualizing Solutions to x + 5 = 1 on Different Number Lines Before diving into the different types of number lines, it’s important to understand what a number line is. A number line is a visual representation of numbers that shows their relative position to each other. It’s a great tool for understanding mathematical concepts and can help make abstract ideas concrete and easier to grasp. There are different types of number lines, including horizontal and vertical number lines, integer number lines, fraction number lines, and decimal number lines. Each type of number line is useful for different purposes, and it’s important to choose the right one depending on what you’re trying to accomplish. Using number lines to visualize solutions to x + 5 = 1 is particularly helpful because it allows us to see the relationship between the numbers and understand the problem on a deeper level. ## Comparing Four Number Lines: Which One Shows Solutions to x + 5 = 1? Now that we understand number lines and their importance in solving mathematical equations, let’s compare four different types of number lines: horizontal number lines, vertical number lines, integer number lines, and fraction number lines. Horizontal number lines are the most common and are represented by a line that stretches from left to right. They’re great for showing the relationship between positive and negative numbers and are commonly used in basic arithmetic problems. Vertical number lines, on the other hand, are represented by a line that stretches from top to bottom. They’re useful for showing positive and negative numbers, just like horizontal number lines, but are particularly helpful for visualizing graphs and quadrants in coordinate planes. Integer number lines are used to represent whole numbers. They’re helpful when working with problems that require whole numbers, such as counting or measuring discrete objects. Fraction number lines, as the name suggests, are used to represent fractions. They’re useful for understanding the relationship between fractions and can be helpful in solving problems that require the use of fractions. When it comes to solving x + 5 = 1, which number line is the most useful? All four types of number lines can be used, but the horizontal number line is the most commonly used for this problem. We’ll use the horizontal number line in our demonstration on how to solve x + 5 = 1. ## Solving x + 5 = 1: Understanding Number Lines and Their Representations Now that we’ve established that the horizontal number line is the most effective for solving x + 5 = 1, let’s demonstrate how to use it to find the solution. The first step is to plot the numbers on the line. We start with the larger number, which in this case is 5, and plot it on the right-hand side of the line. Next, we plot the smaller number, which is 1, on the left-hand side of the line. Now we need to find the unknown value of x. To do this, we need to move from 5 to 1 by subtracting 5 from both sides of the equation. This gives us x = -4, which is the solution to the equation. By using the horizontal number line, we were able to visualize the relationship between the numbers and find the solution to the equation. This is a great example of how number lines can be used to solve problems and understand mathematical concepts on a deeper level. ## The Search for x: Analyzing Number Lines to Find Solutions Another way number lines can help in solving mathematical equations is by analyzing them to find solutions. Let’s take the example of the equation x + 3 = 7. We can use the horizontal number line to plot the numbers and find the solution. We start by plotting 7 on the right-hand side of the line and 3 on the left-hand side. To find the value of x, we need to move from 3 to 7 by adding 4 (which is the difference between 7 and 3) to the left side of the equation. This gives us x = 4, which is the solution to the equation. By analyzing the number line and understanding the relationship between the numbers, we were able to find the solution to the equation. This is a great example of how number lines can help in problem-solving. ## Different Number Lines, Different Solutions: Decoding x + 5 = 1 It’s important to understand that different types of number lines can lead to different solutions for the same problem. Let’s take a look at the example of x + 5 = 1 and how different number lines can lead to different solutions. If we use a vertical number line instead of a horizontal one, the solution to the equation changes. We plot 5 at the top of the line and 1 at the bottom. To find the value of x, we need to move from 5 to 1 by subtracting 4 (which is the difference between 5 and 1) from the top of the equation. This gives us x = -4, which is a different solution than what we found using the horizontal number line. This example demonstrates that it’s important to be aware of different types of number lines and their representations in order to find the correct solution to a problem. ## Cracking the Code of x + 5 = 1: Navigating the Necessity of Number Lines By now, we’ve seen how number lines can be used to represent and solve equations and how different types of number lines can lead to different solutions. It’s important to understand number lines and their representations in order to navigate the complexity of mathematical equations. Number lines are a great tool for understanding mathematical concepts and solving problems. They make abstract ideas concrete and allow us to see the relationship between numbers. By visualizing the problem on a number line, we gain a better understanding of the problem and can find the correct solution. ## Conclusion The equation x + 5 = 1 may seem simple, but it requires a clear understanding of number lines and their representations in order to find the correct solution. By using a horizontal number line and subtracting 5 from both sides of the equation, we were able to find the solution to the problem -4. We saw how different types of number lines can be used to represent and solve problems and how analyzing number lines can lead to finding solutions. Finally, we learned that different number lines can lead to different solutions and it’s important to be aware of this when solving problems. By understanding the importance of number lines and their representations, we can navigate the complexity of mathematical equations and become more confident problem-solvers.<\|endoftext\|>	4.875
226	Children gain a great sense of accomplishment from learning to say something in a foreign language – it’s like learning to crack a code! Language learning provides frequent opportunities to perform before an audience. This nurtures pupils’ self-esteem and self-confidence and develops strong interpersonal skills. Here at Gooseacre, all children in Key Stage 2 have the benefit of learning French. We believe that foreign language learning increases critical thinking skills, creativity, and flexibility of mind in young children. Pupils who learn a foreign language increase their density of ‘grey matter’ in the brain and synapses, that interconnect parts of the brain. Through studying a foreign language, grammatical concepts and rules in English become clearer. Children use what they have learned in one language to reinforce what they have learned in another. It also equips children with additional life skills and with opportunities to use and apply them in real life situations. Our Modern Foreign Languages lessons are provided by Mrs Vikki Bruff and her MFL company VB Primary Languages. We also offer a residential visit to France for children in Year 6.<\|endoftext\|>	3.703125
661	Copper (Cu), number 29 on the Periodic Table, is a naturally-occurring nonferrous metal that has excellent electrical and thermal conductivity as well as anti-corrosive and antimicrobial properties. It’s also an essential nutrient for human health. Copper is obtained by mining, but it’s also a highly-recycled metal; in 2012 more than 30% of copper consumption came from recycled copper. This post will discuss copper mining; to learn more about copper, its recycling, production, or uses, visit Analyzing Metals. Copper can be found in the earth in pure form, or in many different kinds of mineral deposits including sulfide deposits such as chalcopyrite (most common), bornite, chalcocite, and covellite; carbonate deposits such as azurite and malachite; or silicate deposits such as chrysycolla and dioptase. The U.S. Geological Survey (USGS) Copper for the Ages Fact Sheet describes various copper deposits, which are broadly classified by how the deposits formed: - Porphyry copper deposits: Associated with igneous intrusions, porphyry copper deposits yield about two-thirds of the world’s copper. Large copper deposits of this type are found in mountainous regions of western North and South America. - Sedimentary copper deposits: Copper deposits contained in sedimentary rocks account for approximately one-fourth of the world’s identified copper resources. These deposits occur in such areas as the central African copper belt and the Zechstein basin of Eastern Europe. A 2014 U.S. Geological Survey global assessment of copper deposits indicates that known resources contain about 2.1 billion tons of copper (porphyry deposits accounted for 1.8 billion tons of those resources), and undiscovered resources contain an estimated 3.5 billion tons. In 2014, U.S. mine production of copper was valued at about $9.7 billion. Copper mining usually employs open pit mining techniques although there are also underground copper mining operations. Individual copper deposits may contain hundreds of millions of tons of copper-bearing rock. However, ore deposits are inconsistent, having high concentrations of metals in one area but much lower concentrations in other areas. The grade may be high at the surface, but diminish with depth, or vice versa. Geologists perform ore grade control to pinpoint the most profitable ore deposits and create models or maps of these locations so that they can make the most efficient and economical drilling and excavation decisions. Knowing how and where copper resources are deposited also helps target where undiscovered deposits may lie. Field-portable x-ray fluorescence (FPXRF) instruments provide fast acquisition of geochemical data for mine mapping and ore deposit modeling. FPXRF is an established technique for easily determining elemental constituents for most natural low concentration samples, and it is now recognized as an effective analytical tool for high grade ore concentrates and grade control as well. Editor’s Note: If you want more information about copper, visit our Analyzing Metals blog. We are publishing two months of articles about copper. Check every Tuesday throughout July and August for Copper Compendium Parts 1-8.<\|endoftext\|>	4.0625
479	- Volcano List - Learn More - All About Volcanoes - Kids Only! - Adventures and Fun - Sitemap (Under Construction) Symmetrical Cotopaxi is one of the most prominent volcanoes that line both sides of the Interandean valley along Ecuador's "Avenue of Volcanoes." Cotopaxi, one of Ecuador's most active volcanoes, has produced more than 50 eruptions since the 16th century. Glaciers cover the upper part of the cone from 4700 m altitude on the west flank, seen here, to the 5911-m-high summit. Devastating lahars in historical time swept this valley before turning south and then east into the Amazon basin. Lahars to the NW reached the Pacific Ocean. Photo by Lee Siebert, 1978 (Smithsonian Institution) / Caption SI/USGS. Cotopaxi is a stratovolcano that has erupted 50 times since 1738. The 1877 eruption melted snow and ice on the summit, which produced mudflows that traveled 60 miles (100 km) from the volcano. The most recent eruption of Cotopaxi ended in 1904. Reports of an eruption in 1942 have not been confirmed. The most recent activity was an increase in steam emissions, melting snow, and small earthquakes from 1975-1976. Photo by Chuck Wood - 1979. Photo by J.W. Ewert . The right image above shows a Global Positioning System (GPS) receiver at Cotopaxi Volcano, Ecuador uses data transmitted by orbiting satellites to locate points on the ground. The USGS has made baseline GPS measurements at several volcanoes in the United States and in Latin America. In the event of an awakening of one of these volcanoes, GPS receivers would be set up at these points again to determine whether or not measurable deformation had occurred and to monitor for precursory deformation that might herald an eruption. A view from the summit of Cotopaxi with Antisana to the left and Chimborazo to the right. Copyrighted photo by Timothy Burns . A view of the crater on the summit of Cotopaxi. Copyrighted photo by Timothy Burns . Another view of the crater at the summit of Cotopaxi. Copyrighted photo by Timothy Burns.<\|endoftext\|>	3.703125
532	Bananas (Musa spp.) are relatively strange-looking fruit that grow on even odder plants. Though bushy in appearance, banana plants are not shrubs, but they are also not trees. To make matters more confusing, there is a plant called a banana shrub (Michelia figo). The bananas you eat are grown on plants that grow in U.S. Department of Agriculture plant hardiness zones 9b through 11. Shrubs are generally considered perennial plants that have woody stems or trunks and grow to a final height of less than 13 feet. Stems on shrubs are thin, usually less than 3 inches in diameter, but the main trunk can be thicker than that. There are several reasons why bananas cannot be classified as shrubs. The most obvious is that bananas are not woody. They are leafy with a pseudostem instead of a woody trunk, which a shrub would have. The real stem of a banana grows up through the middle of the pseudostem to later produce the flower and fruit of the plant. The root systems also differ. Unlike the root system commonly seen with shrubs, bananas grow from rhizomes underground that spread in the same way as many running grasses. Banana plants are also too tall to fit the definition of a shrub. While dwarf varieties reach only 6 feet, standard banana plants can tower to a height of 30 feet. A banana plant is classified as a herbaceous perennial. The banana shrub may confuse you into thinking that it produces bananas or is a member of the banana family, but this plant, which grows in USDA zones 8 through 11, is not related to bananas. It is a member of the Magnolia family. The name of this shrub comes from the scent of its magnolialike flowers, which have a strong odor like bananas and a color reminiscent of ripe bananas. Overall, this bush is small compared to banana plants. The shrub grows up to 10 feet tall and wide, but this growth happens slowly. While the banana shrub does produce fruit, these are not bananas. Bananas and the fruit from a banana shrub are completely different. Botanically, bananas are berries that grow in clusters, called hands. These hands have several elongated fruits on them. They are edible when green but taste best if cooked in that state. Yellow bananas are those you're familiar with. Despite the name and scent of the banana shrub, the small brown fruit that measure less than 1/2 inch are not edible. - Photos.com/Photos.com/Getty Images<\|endoftext\|>	3.65625
399	Algebra is chock-full of words that are useful but often misunderstood. To do well on the ACT Math test, you should be able to define these important terms: variable, constant, equation, expression, term, and coefficient. • Variable. A variable is any letter that stands for a number. The most commonly used letters are x and y, but you can use any letter. • Constant. A constant is a number without a variable. For example: The equation 6m + 7 = –m has one constant, the number 7. • Equation. An equation is any string of numbers and symbols that makes sense and includes an equal sign. For example, here are three equations: • Expression. An expression is any string of numbers and symbols that makes sense when placed on one side of an equation. For example, here are four expressions: • Term. A term is any part of an expression that’s separated from the other parts by either a plus sign (+) or a minus sign (–). Important: A term always includes the sign that immediately precedes it. For example, • The expression 3x – 7 has two terms: 3x and –7. • The expression –x2 – 9x + 11 has three terms: –x2, –9x, and 11. • The expression has one term: • Coefficient. A coefficient is the numerical part of a term, including the sign that precedes it (+ or –). Important: Every term has a coefficient. When a term appears to have no coefficient, its coefficient is either 1 or –1, depending on the sign. For example, • The term 3x has a coefficient of 3. • The term –9x has a coefficient of –9. • The term –x2 has a coefficient of –1. Being clear about the meanings of these six words can help you with any other math you study.<\|endoftext\|>	4.5
1,286	# Question Video: Solving a System of Two Equations Using Determinants Mathematics • 10th Grade Use determinants to solve the system −9𝑥 = −8 + 8𝑦, 6𝑦 = 7 + 3𝑥. 05:41 ### Video Transcript Use determinants to solve the system negative nine 𝑥 equals negative eight add eight 𝑦, six 𝑦 equals seven add three 𝑥. There are lots of methods for solving a system of linear equations. But when we’re asked to use determinants, that’s when we use Cramer’s rule. Cramer’s rule involves converting our system of linear equations into a matrix equation. Recall that Cramer’s rule is the following. We can find 𝑥 by calculating Δ sub 𝑥 over Δ and 𝑦 by calculating Δ sub 𝑦 over Δ. Here, Δ is the determinant of the coefficient matrix, and Δ𝑥 and Δ𝑦 are the determinants of the matrices found by substituting elements of the constants matrix with the elements from the columns of the 𝑥- and 𝑦-coefficients. So let’s begin this question by converting this system into a matrix equation. Recall that when we put a system like this into a matrix equation, there are three parts. We have the coefficient matrix, the variable matrix, and the constant matrix. In order to put this into matrix form, the first thing we need to do is rearrange our equations into a form that can easily be converted into a matrix equation. We should try to align the 𝑥’s and the 𝑦’s and the constants. For the first equation, we could add nine 𝑥 to both sides and then add eight to both sides. That gives us nine 𝑥 add eight 𝑦 equals eight. So let’s now try and get the second equation into this similar format. We could do this by subtracting six 𝑦 from both sides and seven from both sides. And that gives us three 𝑥 minus six 𝑦 equals negative seven. Rearranging in this way makes it much easier to put it into a matrix equation. These are the coefficients which go into the coefficient matrix. That’s nine, eight, three, and negative six. Next, we have the variable matrix. This matrix consists of the variables for our system, so that’s going to be 𝑥 and 𝑦. And finally, we have the constant matrix. This just consists of the constants of our system, so that’s going to be eight and negative seven. So now we’ve set up our matrix equation, we can look at using Cramer’s rule. Let’s begin by finding Δ sub 𝑥 and Δ sub 𝑦. Remember, Δ sub 𝑥 and Δ sub 𝑦 are the determinants of the matrices found as a result of substituting the elements of the constants matrix with the elements from the columns of the 𝑥- and 𝑦-coefficients. So to find Δ sub 𝑥, we consider the coefficients matrix. But what we do is swap out the 𝑥-coefficients in the coefficient matrix, that’s nine and three, with the elements of the constant matrix, that’s eight and negative seven. So Δ sub 𝑥 is the determinant of the matrix eight, eight, negative seven, negative six. We now need to actually calculate this determinant. So let’s start by recalling how we find the determinant of a two-by-two matrix. To find the determinant of a matrix 𝑎, 𝑏, 𝑐, 𝑑, we subtract the product of the diagonals, that is, 𝑎𝑑 minus 𝑏𝑐. So the determinant of matrix eight, eight, negative seven, negative six is eight multiplied by negative six minus eight multiplied by negative seven, that is, negative 48 minus negative 56. But that’s just negative 48 add 56. And that gives us eight. So now we need to do the same for Δ sub 𝑦. That’s going to be the determinant of the coefficient matrix, but with the 𝑦-coefficients replaced with the constant matrix, that is, nine, eight, three, negative seven. So we now need to find the determinant of this matrix. Using the same method as we use for the matrix Δ sub 𝑥, this is nine multiplied by negative seven minus eight multiplied by three. And that gives us negative 63 minus 24, which gives us negative 87. So now we’ve found Δ sub 𝑥 and we’ve found Δ sub 𝑦. But we still need to find the value for Δ. Δ is the determinant of the coefficient matrix. That is the determinant of the matrix nine, eight, three, negative six, that is, nine multiplied by negative six minus eight multiplied by three, which is negative 54 minus 24. And that gives us negative 78. So now we have Δ sub 𝑥, Δ sub 𝑦, and Δ, we can now apply Cramer’s rule. This tells us that we can find the value of 𝑥 by doing Δ sub 𝑥 over Δ. We already found Δ sub 𝑥 to be eight. And we found Δ to be negative 78. Therefore, 𝑥 is eight over negative 78. We can actually simplify this fraction by dividing both the numerator and denominator by two, which gives us 𝑥 equals negative four over 39. Cramer’s rule also tells us that 𝑦 is equal to Δ𝑦 over Δ. We found Δ sub 𝑦 to be negative 87 and Δ to be negative 78. So 𝑦 is equal to negative 87 over negative 78. As the highest common factor of 78 and 87 is three, we can divide both the numerator and denominator by three. In fact, we can actually divide our numerator and denominator by negative three, which gives us 29 over 26. So that leads us to our final answer: 𝑥 equals negative four over 39 and 𝑦 equals 29 over 26.<\|endoftext\|>	4.90625
1,022	What Do Rabbits See? Many house rabbit “parents” are curious to know what the world looks like to their lagomorph companion. Why does it seem difficult for my bunny to find food right in front of his face? Why is my bunny so easily startled or frightened if I walk into the room holding a box or a grocery bag? Can my rabbit see colors? The first thing to remember is that a rabbit’s visual system evolved under evolutionary pressures completely different from those which “designed” your eyes. We human primates, like our simian cousins, have forward-placed eyes which confer binocular vision and depth perception. This is essential for an animal originally designed to leap through the trees. Also, we have excellent color vision, a trait which helped our ancestors to find ripe fruit and tasty flowers in the forest canopy. On the other hand, the rabbit visual system is designed–not for foraging and locomotion–but to quickly and effectively detect approaching predators from almost any direction. The eyes are placed high and to the sides of the skull, allowing the rabbit to see nearly 360 degrees, as well as far above her head. Rabbits tend to be farsighted, which explains why they may be frightened by an airplane flying overhead even if their human companion can barely see it. (It could be a hawk! Run!) The price the bunny pays for this remarkable field of vision is a small blind spot directly in front of his face, but forward-placed nostrils and large, spooning ears compensate for that minor loss of predator-detecting space. For an animal to have binocular vision, the field of view of both eyes must overlap to some degree. The central blind spot in the rabbit’s field of view precludes a three-dimensional view of nearby objects. When your bunny cocks her head and seems to be looking at you “sideways,” she is actually looking as straight at you as is possible for a bunny. As far as we know, she does not have a primate’s level of depth perception at such close range. What about color vision? In general, vertebrates have two different types of photoreceptor cells in their retinas: rods and cones. Cones confer high resolution, and, if more than one cone type is present, they also confer the ability to perceive various wavelengths of light as distinct colors. For example, we humans have three different categories of cone–their maximum sensitivities in the red, blue and green regions of the spectrum. The differing sensitivities of each cone type enable us to perceive different (visible) wavelengths of light as the colors of the rainbow. Behavioral studies published in the early 1970’s indicate that rabbits do have a limited ability to discriminate between some wavelengths of light, perceiving them as different colors. Evidently, they can discriminate between the wavelengths we call “green” and “blue.” Although rabbits may not perceive green and blue the way we do, they can tell them apart. This means they have limited color vision, probably conferred by two different categories of cone cells (blue and green). The other type of photoreceptor, the rod cell, confers high visual sensitivity in low light situations, but relatively poor resolution (i.e., a “grainy” picture). The rabbit retina has a much higher ratio of rods to cones than the human retina has. Although a rabbit can see better than a human in low light conditions, his low light image has much poorer resolution (clarity) than the daytime images formed by your cone-rich, primate retina. Now you may wonder: “Can my rabbit see me clearly, or am I just a big blur?” As you read this page, you are focusing on the letters with a very tiny part of your retina called the fovea. This is a minuscule, cone-shaped depression in the retina, lined wall-to-wall with high resolution cone cells. Rabbits, too, have small retinal areas with more cones than rods. However, this (italics) area centralis (close italics) is not indented, and it has far lower cone density than our fovea has. The image formed by the area centralis is relatively “grainy” compared to the one formed by your fovea, but it serves the rabbit well. Using this image, your voice, body movements and scent as cues, your rabbit can recognize you (his favorite human)–as long as you’re not carrying a scary box that completely changes your familiar shape! Knowing a little more about how another creature sees the world allows us to come one step closer to understanding its behavior–and modifying our own to make life happier for everyone. Remember that the next time your rabbit gazes at you with those deep, ancient eyes. copyright 1998 – Dana Krempels<\|endoftext\|>	3.796875
891	C. megalodon Extinct by 2.6 million Years Ago The giant prehistoric shark known as Megalodon (C. megalodon) has certainly attracted a lot of scientific attention in recent years. This predatory shark, believed to be the largest, carnivorous shark to have ever existed, might have reached lengths in excess of sixteen metres. Bodyweight estimates vary, but a number of scientists have calculated weights around the twenty tonnes mark. With a mouth that could gape nearly 3 metres wide, this fish would have been capable of swallowing an extant Great White shark (C. carcharias) whole! Some fishermen, such as those who fish the waters off South Africa have claimed that Megalodon still exists, but a team of researchers from the University of Zurich and Florida University have used a statistical technique to provide the best estimate yet of the extinction of this apex marine predator. Safari Ltd Introduced a Model of the Giant Shark Known as Megalodon this Year Picture Credit: Everything Dinosaur Writing in the on line journal PLOS One (Public Library of Science), the researchers Catalina Pimiento (Department of Biology, University of Florida) and Christopher Clements, (Institute of Evolutionary Biology and Environmental Studies, Zurich University), used a statistical analysis to calculate the approximate time of the extinction of these predators. By developing a better understanding of the time of the extinction, the research team are able to map the consequences of the extinction of an apex predator on the marine ecosystem. Using a mathematical technique called Optimal Linear Estimation (OLE) the scientists have calculated that around 2.6 million years ago C. megalodon became extinct. This is the first time the extinction of C. megalodon has been quantitatively assessed. A total of fifty-three fossils of this ancient shark were used in this study, but this number was reduced to forty-two for the statistical assessment, as eleven of the fossil specimens gave the researchers less confidence over their actual age. Optimal Linear Estimation is used for mapping the extinction of modern species, but with the abundant and widespread fossil record of Megalodon, mainly the large number of fossilised teeth associated with this species, the team were able to apply this mathematical method to calculate the time when the very last of these magnificent creatures lived. Fossilised Tooth from C. megalodon Picture Credit: Everything Dinosaur Although the team admit that their calculations are subject to margins of error, this work represents a refinement on previous estimates of the extinction of this shark species. Why this apex predator became extinct remains a mystery. Most Megalodon fossil material dates from around 16 to 11.6 million years ago (Miocene Epoch) with some further fossils known from the Pleistocene Epoch, although a lot of the fossil material associated with Pleistocene deposits were not included in this particular study. With a more accurate extinction date established, the researchers can then assess the impact of the removal of a large predator from the marine environment on other genera including Cetaceans. The date of 2.6 million years ago marks the border between the Pliocene and Pleistocene Epochs, it was after this point that the baleen whales began to increase in size and to evolve into the many giant forms alive today. As marine mammals are thought to have made up a substantial part of the bus-sized shark’s diet, although there is no conclusive evidence to suggest Megalodon fed on large whales, if Megalodon died out, did this trigger the flourishing of the baleen whales? Commenting on the research, Catalina Pimiento stated: “When we calculated the time of Megalodon’s extinction, we noticed that the modern function and gigantic sizes of filter feeder whales became established around that time. Future research will investigate if Megalodon’s extinction played a part in the evolution of these new classes of whales.” Megalodon is a prime candidate for this sort of statistical analysis, due to the relative abundance and widespread distribution of its fossilised teeth. It is hoped that this technique can be applied to other extinct species to plot more accurately the impact of a single species extinction on the wider ecosystem. Did Megalodon Feed on Large Cetaceans? Picture Credit: Science Photo Library<\|endoftext\|>	3.8125
643	# Lesson 2.3 p. 87 Deductive Reasoning Goals: to use symbolic notation to apply the laws of logic. ## Presentation on theme: "Lesson 2.3 p. 87 Deductive Reasoning Goals: to use symbolic notation to apply the laws of logic."— Presentation transcript: Lesson 2.3 p. 87 Deductive Reasoning Goals: to use symbolic notation to apply the laws of logic Symbolic Notation Conditional Statements Hypothesis “p”Conclusion “q” p Example: If today is Monday, then there is school. q Symbolic Notation Conditional Statement: If p, then qorp q Biconditional : If p q and if q p or p q p if and only if q Converse: If q, then porq p If today is Monday, then there is school. If there is school, then today is Monday. Today is Monday, if and only if there is school. More Symbolic Notation Example Converse or q p: Conditional Statement or p q: Tim will buy a car if he finds a summer job. Inverse or Contrapositive or q: p: Deductive Reasoning vs. Inductive Reasoning Inductive Reasoning: Uses previous examples Uses patterns Can make an educated guess Example: Kate lost her instrument. The only instrument Kate plays is the clarinet. Kate lost her clarinet. Deductive Reasoning: Uses facts Uses definitions Can prove or make a logical argument Example: It rained for the last two days, so it will rain today. Two Laws of Deductive Reasoning 2)Law of Syllogism: If p q and q r are true conditional statements, then p r is true. 1) Law of Detachment: If p q is a true conditional statement and p is true, then q is true. Example: Jamal knows that if he misses practice the day before a game, then he will not be able to pitch. Jamal misses practice on Tuesday. Jamal will not be able to pitch in the game on Wednesday. Example: If it is Friday, then tomorrow is Saturday. If it is Saturday, then we will go swimming. If it is Friday, then tomorrow we will go swimming. Examples of Laws of Deductive Reasoning 2)Law of Syllogism: If p q and q r are true condition statements, then p r is true. 1) Law of Detachment: If p q is a true condition statement and p is true, then q is true. Example: If two angles form a linear pair, then they are supplementary. Angles A and B are supplementary. Conclusion, or can it not be reached? Example: If I break my leg, then I can’t ice skate. If I can’t ice skate, I won’t go to the ice skating party. Conclusion: Example A foot is a measurement of twelve inches. If-then form: Converse: Inverse: Contrapositive: Download ppt "Lesson 2.3 p. 87 Deductive Reasoning Goals: to use symbolic notation to apply the laws of logic." Similar presentations<\|endoftext\|>	4.53125
1,368	# AP Physics C: Mechanics : Electric Circuits ## Example Questions ### Example Question #1 : Electric Circuits A parallel plate capacitor has a capacitance of . If the plates are apart, what is the area of the plates? Explanation: The relationship between capacitance, distance, and area is . We can rearrange this equation to solve for area. Now, we can use the values given in teh question to solve. ### Example Question #1 : Capacitors Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area and are separated by distance . A material of dielectric constant has been placed between the two plates. Which of the following would not result in an increase in the measure of electric potential difference between the two plates? Increase the area, , of the plates Replace the material between the two plates with one of a lower dielectric constant, Increase the value of the surface charge density, , on each plate Increase the distance, , between the plates Increase the area, , of the plates Explanation: The electric potential difference created between the plates of a parallel plate capactor is given by the equation: The charge can be calculated by using the equation: The value of the capacitance is related to the dimensions of the capacitor with the equation: Combining these equations yields: The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference. ### Example Question #1 : Using Capacitor Equations You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of out of those metal sheets, and the sheets needed to be apart, what area do the two metal plates need to be? Explanation: For parallel plate capacitors, the equation is . Solve for . Now we can plug in our given values. ### Example Question #1 : Electric Circuits The plates of a parallel plate capacitor are apart and in area. A potential difference of is applied across the capacitor. Find the capacitance. Explanation: Capacitance is related to plate area and distance by the equation . Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant. ### Example Question #1 : Electric Circuits The plates of a parallel plate capacitor are apart and in area. A potential difference of is applied across the capacitor. Compute the charge on each plate. Explanation: Charge on a capacitor is given by the equation . We know that the voltage is , but we need to determine the capacitance based off of the area and distance between the plates. We can plug this value into the equation for charge. ### Example Question #1 : Capacitors The plates of a parallel plate capacitor are apart. A potential difference of is applied across the capacitor. Compute the magnitude of the generated electric field. Explanation: The electric field given by a capacitor is given by the formula . We do not have these variables, so we will have to adjust the equation. The capacitance can be determined by the area of the plates and the distance between them. This equation simplifies to , allowing us to solve using the values given in the question. ### Example Question #1 : Electric Circuits If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, ? Explanation: Relevant equations: Use the series equation, replacing C1 and C2 with the given constant C: This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors. ### Example Question #1 : Using Capacitor Equations Two parallel conducting plates each have an area of and are separated by a distance, . One plate has a charge evenly distributed across it, and the other has a charge of . A proton (charge ) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process? Explanation: Relevant equations: According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy. 1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other: 2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy): ### Example Question #211 : Ap Physics C and capacitor are connected in series with a battery. What is the magnitude of the charge on one plate of the capacitor? Explanation: Relevant equations: First, find the equivalent capacitance of the whole circuit: Use this equivalent capacitance to find the total charge: Capacitors in series always have the same charge on each unit, so the charge on the capacitor is also . ### Example Question #1 : Electric Circuits Resistors are one of the most important basic components of a circuit. With very few exceptions, all circuits have at least one kind of resistor component. An ammeter is a device that measures current flowing through a circuit. Ammeters are always connected to a circuit in series. Which of the following accurately explains why ammeters must be connected in series within a circuit, and never in parallel? An ammeter connected in parallel would not read any current because the current would have no driving force through the ammeter An ammeter connected in parallel would give readings of half the current flowing through the circuit because current would flow evenly between the ammeter and the circuit's regular pathway An ammeter connected in parallel would give the current flowing through the circuit a pathway with minimal resistance, creating a very large current that could harm the ammeter An ammeter connected in parallel would give high readings of current because the current would be flowing through both the regular current path and the ammeter, but disproportionately through the ammeter An ammeter connected in parallel would give very low readings of current because current would be flowing through both the regular current path and the ammeter, but disproportionately through the regular path<\|endoftext\|>	4.40625
2,306	The curriculum for the Kindergarten at St. James Episcopal School is designed to utilize the Elements of Depth and Complexity in combination with Multiple Intelligences to provide a high cognitive level experience aligned with private school curricular standards. The program provides diverse opportunities for scholars to develop to their fullest through first hand experiences. An important goal of the Kindergarten program is to create a love of learning through relationships, encouragement, and structured play. Kindergarten scholars are taught to look at the world through the Global Theme of Patterns, identifying the following essential, conceptual truths (generalizations): - Patterns can be natural or man-made. - Patterns are subject to change. - Patterns provide structure. - Patterns allow for predictions. - Patterns repeat. - Patterns have rules. - Patterns may have symmetry. - Patterns communicate or tell a story. The language arts approach in Kindergarten is built upon a combined approach offering both balanced literacy and direct phonics instruction.Â The marriage of these two curriculums: Spalding Integrated Language Arts and Houghton-Mifflin Journeyâ€™s, exposes scholars to grammar concepts, writing skills, vocabulary development, and reading comprehension and fluency. Listening and Speaking - Identify and apply spatial and temporal relationships - Use narrative language to describe - Discuss and use common sayings - Follow multi-step, oral directions. - Give simple directions. - Provide simple explanations. - Recite a nursery rhyme, poem, or song independently. - Present a short oral report to peers. - Listen to and understand a variety of texts read aloud. - Describe illustrations. - Sequence four to six pictures illustrating events. - Ask and answer questions requiring literal recall and understanding details. - Ask and answer questions to clarify information. - Use narrative language to describe people, places, things, locations, events, and actions - Compare and contrast similarities and differences. - Make personal connections to events or experiences in a read-aloud - Use pictures accompanying the read-aloud to check and support understanding. - Make predictions prior to and during a read-aloud - Ask and answer questions that require interpretations, judgments, or opinions. - Create and tell an original story. - Understand literary terms. - Discuss important facts and information from a nonfiction read-aloud. - Identify parts and the function of books. - Use correct book orientation. - Recognize that sentences in print are made up of separate words. - Understand words are separated by spaces. - Distinguish letters, words, sentences, and stories. - Use upper and lower-case forms of the alphabet. - Orally segment sentences into individual words. - Understand words are made up of sound sequences. - Recognize initial/medial/final position of phonemes in a spoken word. - Identify whether pairs of phonemes are the same or different. - Orally blend two to three sounds to form a word. - Segment a spoken word into phonemes. - Produce rhyming words. - Identify the number of syllables. - Recognize the relationship between written letters (graphemes) and spoken sounds (phonemes). - Blend individual phonemes to pronounce printed words. - Understand that sometimes two or more printed letters stand for a single sound. - Read consonant-vowel-consonant words. - Read high frequency words. - Recognize, speak, and write phonogram sounds as outlined in Spaldingâ€™s The Writing Road to Reading including single and multi-letter phonograms. - Maintain a spelling notebook with high-frequency words to be written independently. - Recognize and apply phonic rules including the jobs of silent final e. - Use commas and end punctuation while reading orally. - Draw pictures to represent a text. - Draw pictures to represent a preference or opinion. - Write narratives, informative, and explanatory texts. - Add details to writing with assistance. - Create a title or caption to accompany a picture and/or shared writing. - Apply phonetic spelling to write independently. - Use upper and lower-case letters in first and last names and in simple messages. - Use letters, words, phrases, and sentences to communicate thoughts and ideas. - Apply basic spelling conventions. - Capitalize and punctuate sentences. - Write from left to right, leaving spaces between words, and top to bottom, using return sweep. - Offer phonemically plausible spellings for words. - Capitalize first word in a sentence and the pronoun I. - Read and memorize Mother Goose and other traditional poems. - Identify and use rhyming words or words that end with the same sounds. - Identify and use rhythm or a pattern of sound. - Use alliteration. - Identify characters and dialogue. - Maintain a poetry notebook. Sayings and Phrases - Use sayings, phrases, and proverbs appropriately. HISTORY AND GEOGRAPHY In Kindergarten, children study aspects of the world around them including the family, the school, and the community. The history and geography program in Kindergarten is meant to complement and extend that focus. The goal of studying selected topics in world history is to foster curiosity and begin understandings about the larger world outside the childâ€™s immediate community including varied civilizations and ways of life. This is presented through stories, drama, art, music, and discussion. World Geography-Spatial Sense - Geographical awareness - Maps and globes: what they represent, how we use them - Specific areas on maps and globes Overview of the Seven Continents - Identify and locate seven continents on a map and globe. - Differentiate state, city, town, and community. - Locate North America, the continental United States, Alaska, and Hawaii. Native American Peoples, Past, and Present - Study Native American peoples in different regions. - Incorporate in-depth study of at least one specific group of Native Americans-past and present. Early Exploration and Settlement - The Voyage of Columbus in 1492 - Queen Isabella and King Ferdinand of Spain - The NiĂ±a, Pinta, and Santa Maria - Columbusâ€™ mistaken identification of â€śIndiesâ€ť and â€śIndiansâ€ť - The idea of what was, for Europeans, a â€śNew Worldâ€ť - The Pilgrims - The Mayflower - Plymouth RockÂ Â Â Â Â Â Â Â - Thanksgiving Day celebration - July 4, â€śIndependence Dayâ€ť - The â€śbirthdayâ€ť of our nation - Democracy (rule of the people): Americans wanted to rule themselves instead of being ruled by a faraway king. - People who were not free- slavery in early America Presidents Past and Present - Famous Presidents - George Washington - Thomas Jefferson, author of Declaration of Independence - Abraham Lincoln - Â Theodore Roosevelt - The way one becomes a president and what the president does - Current United States President Symbols and Figures - American flag - Statue of Liberty - Mount Rushmore - The White House Integrated themes in Kindergarten science are identified as the focus of instruction for extended periods of time. Â In classroom and out of classroom experiences serve as laboratories for exploring, classifying, making predictions, and recording outcomes of scientific data by word and illustration.Â Â Â Science activities are designed so scholars will discern patterns in the biological and physical world. Plants and Plant Growth - What plants need to live - Basic parts of plants - Two kinds of plants: deciduous and evergreen - Food from plants Animals and Their Needs - Common characteristics and needs of animals - What animals need to live - Offspring of animals - Special needs of animals and pets The Human Body-The Five Senses and Taking Care of Your Body - Sight: eyes - Hearing: ears - Smell: nose - Taste: tongue - Touch: skin - Taking care of your body Introduction to Magnetism - Experiments with magnets - Familiar everyday uses of magnets - Materials attracted by magnets Seasons and Weather - The four seasons - Local weather patterns - The sun - Daily weather changes - Temperature and thermometers - Rain, thunderstorms, and hail - Snow, snowflakes, and blizzards Taking Care of the Earth - Practical measures for conserving energy and resources - George Washington Carver - Jane Goodall - Wilbur and Orville Wright The math program in Kindergarten enables scholars to solve problems using deep number sense and a hands-on approach. Using the Saddler-OxfordProgress in Mathematics text and workbook combined with Singapore Math, scholars solve meaningful problems.Â Math manipulatives are integrated into guided and independent practice. Daily calendar math routines are integrated with math skills. Â Patterns and Classification - Sort by size, color and shape, two ways. - Patterns of color, shape, and size - Grow, transfer, identify, and create patterns. - Positions (above, below, top, middle, bottom, over, under, left, right, inside, outside) - Tell time to the hour. - Identify, order, write numbers 0-100. - Form equal sets. - Use a number line. - Estimate groups. - Count objects in a set. - Ordinal positions to the 10th place - Place value- ones, tens, hundreds - Tally marks - Simple graphs and charts - Fractions-whole, half, fourth - Even and odd numbers - Recognize and add pennies, nickels, dimes, and quarters. - Fair trade for coins - Add and subtract coins - Compare money - Dollar and cent signsÂ Â Â Â Â - Join groups - Count by 5s, 10s, 2s - Single-digit addition - Part to whole relationships - Single-digit subtraction - Addition and subtraction patterns - Ten frames to add and subtract - Problem solving strategies and choosing operations - Compare size, length, and height. - Non-standard measures - Length and distance around - Compare and order by weight and capacity. - Read a thermometer. - Three dimensional shapes - Moving shapes - Plane figures on solids - Combine and separate figures.<\|endoftext\|>	3.875
3,619	\|US Army Indian Scouts - Wikipedia \|Native Americans have made up an integral part of U.S. military conflicts since America's beginning. Colonists recruited Indian allies during such instances as the Pequot War from 1634–1638, the Revolutionary War, as well as in War of 1812. Native Americans also fought on both sides during the American Civil War, as well as military missions abroad including the most notable, the Codetalkers who served in World War II. The Scouts were active in the American West in the late 19th and early 20th centuries. Including those who accompanied General John J. Pershing in 1916 on his expedition to Mexico in pursuit of Pancho Villa. Indian Scouts were officially deactivated in 1947 when their last member retired from the Army at Fort Huachuca, Arizona. For many Indians it was an important form of interaction with white American culture and their first major encounter with the whites’ way of thinking and doing things. Soldiers and Indian scouts take observations before the Battle of Big Dry Wash Recruitment and enlistment of the 3rd cavalry described Apache scouts in Arizona as “almost naked, their only clothing being a muslin loin-cloth, a pair of point toed moccasins and a hat of hawk feather”. In 1876 a description of Crow Scouts reads that they wore, “an old black army hat with top cut out and sides bound round with feathers, fur and scarlet cloth”. With the availability of army clothing some Native scouts took advantage of the availability of the clothing. In 1902 when new regulations were introduced in March the U.S. Scouts received a new more regulated uniform. \|Recruitment of Indian scouts was first authorized on 28 July 1866 by an act of Congress. "The President is authorized to enlist and employ in the Territories and Indian country a force of Indians not to exceed one thousand to act as scouts, who shall receive the pay and allowances of cavalry soldiers, and be discharged whenever the necessity for further employment is abated, at the discretion of the department commander." There were different types of scouts, some enlisted as Indian Scouts for brief terms and there were others who were hired as scouts by the U.S. Army. Some individual may have served at different times as a hired scout and an enlisted scout. Prior to the act in 1866 these scouts were considered employees rather than soldiers. Enlistment records and muster rolls, from 1866 to 1912 were in many instances filed by state, some records were broken down by company or military post providing information such as when, where, and by whom the scout was enlisted; period of enlistment; place of birth; age at time of enlistment; physical description; and possibly additional remarks such as discharge information, including date and place of discharge, rank at the time, and if the scout died in service. Indian scouts who were officially enlisted in the army after 1866 were issued old pattern uniforms from surplus stock legally exempt from sale. Their uniforms were worn with less regulation, sometimes mixed with their native dress. In 1870, Captain Bourke A group of Warm Spring Apache In the Indian wars following the U.S. Civil War, the Indian scouts were a fast-moving, aggressive, and knowledgeable asset to the U.S. army. They often proved to be immune to army notions of discipline and demeanor, but they proved expert in traversing the vast distances of the American West and providing intelligence -- and often a shock force -- to the soldiers who sought hostile Indians. Pawnee Scout leader Luther H. North commented, "Neither the Wild Tribes, nor the Government Indian Scouts ever adopted any of the white soldier's tactics. They thought their own much better." Another chief of scouts, Stanton G. Fisher, emphasized the importance of Indian Scouts by saying of the soldiers, “Uncle Sam's boys are too slow for this business." There existed doubts as to whether Indian Scouts would remain faithful or whether they would betray the white soldiers and turn against them in conflict. The Cibicue Apaches were among the first regular Army Scouts. They are also the only recorded 19th-century incident in which Indian scouts turned against the U.S. Army at Cibicue Creek in Arizona Territory. These Apache scouts were asked to campaign against their own kin, resulting in a mutiny against the army soldiers. Three of the scouts were court-martialed and executed. Reduction of Forces/Pensions The end of hostilities on the frontier meant a reduction in the number of the Indian scouts needed. Army General Order No. 28 issued on March 9, 1891 reduced the number of scouts to 150, distributed among the different departments. This brought the numbers down to; Department of Arizona, 50, Departments of the Dakota, Platte and Missouri, 25 each; Department of Texas, 15, and Departments of the Columbia, 10. Pension files provide information not only on Indian Scouts but also about his family and others with whom he may have served or who knew him or his wife. Indian Scouts and their widows became eligible for pensions with the passage of an act on March 4, 1917, relating to Indian wars from 1859 to 1891. Frontier Scouts included; black, native and mixed blood individuals. Native involvement in military service come from different tribes and regions across the United States including Narragansett, Mohegan, Apache, Navajo and Alaska Natives (who would become involved in the 1940s). One of the most notable U.S. Army Indian Scouts is Curley, a member of the Crow tribe who became a scout in April 1876 under Colonel John Gibbon. He then joined General Custer. Curley is most often identified as the lone survivor of “Custer's Last Stand”. He denied witnessing the battle. The Chicago Tribune published an article claiming that Curly had made statements to them about the battle. John F. Finerty claimed that "Curley said that Custer remained alive throughout the greater part of the engagement, animating his men to determined resistance, but about an hour before the close of the fight received a mortal wound." The official website of the Navy lists the American Indian Medal of Honor recipients, including twelve from the 19th century. In the 20th century, five American Indians have been among those soldiers to be distinguished by receiving the United States' highest military honor: This honor is given for military heroism "above and beyond the call of duty”, exhibiting extraordinary bravery, and for some, making the ultimate sacrifice for their country. The role of Native American women in the U.S. Army is being slowly filled by the efforts of such groups as The Women In Military Service For America Memorial Foundation. It is known of individuals such as Tyonajanegen, an Oneida woman, Sacajawea, a Shoshone, and various female nurses have aided the military as far back as the American Revolution. Little information is currently listed on women's roles as scouts during the 19th century. In 1890 the Scouts were authorized to wear the branch of service insignia of crossed arrows. In 1942 the insignia was authorized to be worn by the 1st Special Service Force. As their traditions passed into the U.S. Army Special Forces, the crossed arrows became part of their insignia being authorized as branch of service insignia in 1984. \|Battle of Big Dry Wash - Wikipedia The Battle of Big Dry Wash was fought on July 17, 1882, between troops of the United States Army's 3rd Cavalry Regiment and 6th Cavalry Regiment and members of the White Mountain Apache tribe. The location of the battle was called "Big Dry Wash" in Major Evans' official report, but later maps called the location "Big Dry Fork", which is how it is cited in the four Medal of Honor citations that resulted from the battle. In the spring of 1882, a party of about 60 White Mountain Apache warriors, coalesced under the leadership of a warrior called Na-tio-tish. In early July some of the warriors ambushed and killed four San Carlos policemen, including the police chief. After the ambush, Na-tio-tish led his band of warriors northwest through the Tonto Basin. Local Arizona settlers were greatly alarmed and demanded protection from the army which immediately sent out fourteen companies of cavalry from forts in the region. In the middle of July, Na-tio-tish led his band up Cherry Creek to the Mogollon Rim, intending to reach General Springs, a well known water hole on the Crook Trail. The Apaches noticed that they were trailed by a single troop of cavalry and decided to lay an ambush seven miles north of General Springs where a fork of East Clear Creek cuts a gorge into the Mogollon Rim. The Apaches hid on the far side and waited. The cavalry company was led by Captain Adna R. Chaffee. However, Chaffee's chief scout, Al Sieber, discovered the Apaches' trap and warned the troops. During the night, Chaffee's lone company was reinforced by four more from Fort Apache under the command of Major Andrew W. Evans. Early in the morning of July 17, one company of cavalry opened fire from the rim facing the Apaches. Meanwhile, Chaffee sent two companies upstream and two downstream to sneak across the canyon and attack the Apaches. Na-tio-tish failed to post lookouts and the troops crossed over undetected. From sixteen to twenty-seven warriors were killed, including Na-tio-tish. On the ridge overlooking the wash, Lieutenant George H. Morgan commanded the first major engagement of the battle. As he exposed himself to enemy fire, a bullet ripped through his arm and into his body. Observation before the battle. About two hours into the battle, Lieutenant Thomas Cruse spotted an encampment site of the Apaches which appeared to be deserted. He took command of four men and dashed across the ravine to capture the camp. Upon reaching the site, several hidden warriors fired upon Cruse and his men, mortally wounding the soldier to Cruse's immediate right, Private Joseph McLernan. Cruse dragged Pvt. McLernan to back to the safety of their previous position. As the battle pitched in intensity, Lieutenant Frank West took command of Chaffee's cavalry troop while Chaffee was engaged with commanding the battle. The first shots were fired around 3:00pm and the battle lasted until nightfall, when a heavy thunderstorm struck, bringing rain and hail. Sieber, together with fellow scout Tom Horn and soldier Lt. George H. Morgan, slipped to the banks opposite of the Apache line, and provided rifle fire for the cavalry. Pressured and outgunned, the remaining Apache warriors, under the cover of darkness and the storm, slipped away on foot and retreated to a nearby Apache reservation, about 20 miles away. The site of the battle is now a historical park, in Coconino County, Arizona. Four men received the Medal of Honor for actions at this battle: Thomas Cruse, George H. Morgan, Charles Taylor, and Frank West. Cruse, Morgan, and West were all Lieutenants and West Point graduates. Taylor was a career soldier and troop First Sergeant at the time of the battle. This engagement was the last major battle between the United States Army and Apache warriors, the wars were not over yet however. The U.S. army would soon begin the Geronimo Campaign which ended with Geronimo's capture in 1886. Even then Apache attacks on white settlers in Arizona continued as late as the year 1900. \|Canyon Diablo, Arizona - from Wikipedia Canyon Diablo (Navajo: Kin ?igaaí) is a ghost town on the Navajo Reservation in Coconino County, Arizona, United States on the edge of the arroyo Canyon Diablo. The town, which is about 12 miles northwest of Meteor Crater, was the closest community to the crater when portions of the meteorite were removed. Consequently, the meteorite that struck the crater is officially called the "Canyon Diablo Meteorite." (33 miles east of Flagstaff on Hwy 40) The town originated about 1882, due to construction delays attributed to the Atlantic and Pacific Railroad ordering the wrong span length railroad bridge across the canyon. The bridge story is that the original bridge when ordered was not long enough to span Canyon Diablo, and this was only discovered when the bridge arrived on site from the manufacturer. Consequently, for six months the transcontinental railroad ended at the lip of Canyon Diablo while another bridge was manufactured and shipped to the work site. The original pillars the bridge was mounted on were excavated from the surrounding Kaibab Limestone and shaped on site by Italian stonemasons. The ruins of the lodgings of the railroad workmen are on the west end of the bridge site. Although the railroad ended at the edge of the canyon, work on the railroad route still progressed. Crews were sent ahead to survey the route, prepare the grade and bed, cut and prestage railroad ties and other supplies in advance of the iron rails that would accompany the trains once the canyon was spanned when the new bridge arrived. Work quickly progressed until the A&P crew linked up with the Southern Pacific Railroad crews at Needles, California on August 9, 1883. Originally a small mobile business community catering to the needs of railroad men, once the railroad stopped at the edge of the canyon this community quickly produced numerous saloons, brothels, dance halls, and gambling houses, all of which remained open 24 hours a day. No lawmen were employed by the community initially, so it quickly became a very dangerous place. Its population was mostly railroad workers, along with passing outlaws, gamblers, and prostitutes. The town was designed with two lines of buildings facing one another across the rock bed main street. The center street, however, was not named Main Street, but "Hell Street". It consisted of fourteen saloons, ten gambling houses, four brothels and two dance halls. Also on this street were two eating counters, one grocery store, and one dry goods store. Scattered about in the vicinity of downtown were large numbers of tents, shotgun houses, and hastily thrown up shacks that served as local residences. Within a short time the town had 2,000 residents. A regular stagecoach route from Flagstaff to Canyon Diablo began running and was often the victim of robberies. Within its first year, the town received its first marshal. He was sworn in at 3:00pm, and was being buried at 8:00pm that same night. Five more town marshals would follow, the longest lasting one month, and all were killed in the line of duty. A "Boot Hill" cemetery sprouted up at the end of town, which in less than a decade had 35 graves, all of whom had been killed by way of violent death. The 36th grave was that of former trading post owner Herman Wolfe, who died in 1899, the only one to have died a nonviolent death. Herman Wolfe's trading post was at "Wolfe's Crossing" on the Little Colorado River about 12 miles north of Leupp, Arizona and near a place called Tolchaco. Herman Wolfe died there and his body was transported to Canyon Diablo for burial. Currently Wolfe's grave is heavily monumented and the story is that after World War II a relative from Germany found his grave and installed the headstone and other improvements on the When the railroad bridge was completed, the town quickly died. By 1903, the only thing remaining in the town was a Navajo trading post. Later in the 20th century, when Route 66 passed within several miles of the town, a gas station and roadhouse called Two Guns sprang up, but it too was short-lived. What remains today at Canyon Diablo are a few building foundations, the grave marker and grave of Herman Wolfe, and the ruins of the trading post. \|All articles submitted to the "Brimstone Gazette" are the property of the author, used with their expressed permission. The Brimstone Pistoleros are not responsible for any accidents which may occur from use of loading data, firearms information, or recommendations published on the Brimstone Pistoleros web site.<\|endoftext\|>	4.09375
1,667	## Engage NY Eureka Math 4th Grade Module 4 Lesson 9 Answer Key ### Eureka Math Grade 4 Module 4 Lesson 9 Problem Set Answer Key Question 1. Complete the table. Refer to the below table. Explanation: In pattern block A, the total number that fit around one vertex is 4 and the interior angle measurement is 360° ÷ 4 which is 90° and the sum of the angles around the vertex is 90°+90°+90°+90°= 360°. In pattern block B, the total number that fit around one vertex is 6 and the interior angle measurement is 360° ÷ 6 which is 60° and the sum of the angles around the vertex is 60°+60°+60°+60°+60°+60°= 360°. In pattern block C, the total number that fit around one vertex is 3 and the interior angle measurement is 360° ÷ 3 which is 120° and the sum of the angles around the vertex is 120°+120°+120°= 360°. In pattern block D, the total number that fit around one vertex is 6 and the interior angle measurement is 360° ÷ 6 which is 60° and the sum of the angles around the vertex is 60°+60°+60°+60°+60°+60°= 360°. In pattern block E, the total number that fit around one vertex is 3 and the interior angle measurement is 360° ÷ 3 which is 120° and the sum of the angles around the vertex is 120°+120°+120°= 360°. In pattern block F, the total number that fit around one vertex is 12 and the interior angle measurement is 360° ÷ 12 which is 30° and the sum of the angles around the vertex is 30°+30°+30°+30°+30°+30°+30°+30°+30°+30°+30°+30°= 360°. Question 2. Find the measurements of the angles indicated by the arcs. Refer to the below table. Explanation: In pattern block A, the total number angle measurement which 150° and the sum of the angles around the vertex is 60°+90° which is 150°. In pattern block B, the total number angle measurement which 180° and the sum of the angles around the vertex is 60°+120° which is 180°. In pattern block C, the total number angle measurement which 210°, and the sum of the angles around the vertex is 120°+90° which is 210°. Question 3. Use two or more pattern blocks to figure out the measurements of the angles indicated by the arcs. Refer to the below table. Explanation: In pattern block A, the total number angle measurement which 60°, and the sum of the angles around the vertex is 30°+30° which is 60°. In pattern block B, the total number angle measurement which 210°, and the sum of the angles around the vertex is 120°+90° which is 210°. In pattern block C, the total number angle measurement which 120°, and the sum of the angles around the vertex is 90°+30° which is 120°. ### Eureka Math Grade 4 Module 4 Lesson 9 Exit Ticket Answer Key Question 1. Describe and sketch two combinations of the blue rhombus pattern block that create a straight angle. Refer to the below image. Explanation: Here, we have constructed two combinations of the blue rhombus pattern block that create a straight angle. Question 2. Describe and sketch two combinations of the green triangle and yellow hexagon pattern block that create a straight angle. Refer to the below image. Explanation: Here, we have constructed two combinations of the green triangle and yellow hexagon pattern block that create a straight angle. ### Eureka Math Grade 4 Module 4 Lesson 9 Homework Answer Key Sketch two different ways to compose the given angles using two or more pattern blocks. Write an addition sentence to show how you composed the given angle. Question 1. point A, B, and C form a straight line. Refer to the below image. Explanation: Here, we have constructed three points A, B, C, and formed a straight line, and formed two different ways to compose the given angles using two or more pattern blocks. In the first image, we have divided 180° angle as 90°+30°+60°, so the addition sentence is 90°+30°+60°= 180°. And in the second image, we have constructed three points A, B, C, and formed a straight line we have divided 180° angle as 120°+60°, so the addition sentence is 120°+60°= 180°. Question 2. ∠DEF = 90° The addition sentence is 30°+60°= 90°. Explanation: Here, we have constructed <DEF a and formed two different ways to compose the given angles using two or more pattern blocks. In the first image, we have divided 90° angle as 30°+60°, so the addition sentence is 30°+60°= 90°. And in the second image, we have constructed <DEF and we have divided 90° angle as 30°+60°, so the addition sentence is 30°+60°= 90°. Question 3. ∠GHI = 120° The addition sentence for the <GHI is 30°+90°= 120°. The addition sentence for the <GHI is 30°+90°= 120° Explanation: Here, we have constructed <GHI a and formed two different ways to compose the given angles using two or more pattern blocks. In the first image, we have divided 120° angle as 30°+90°, so the addition sentence is 30°+90°= 120°. And in the second image, we have constructed <GHI and we have divided 90° angle as 60°+60°, so the addition sentence is 60°+60°= 120°. Question 4. x° = 270° The addition sentence for the first image is 60°+90°+120°= 270°. The addition sentence for the second image is 60°+60°+30°+60°+60°= 270°. Explanation: Here, we have constructed <LKJ a and formed two different ways to compose the given angles using two or more pattern blocks. In the first image, we have divided 270° angle as 60°+90°+120°, so the addition sentence is 60°+90°+120°= 270°. And in the second image, we have constructed <LKJ and we have divided 270° angle as 60°+60°+30°+60°+60°, so the addition sentence is 60°+60°+30°+60°+60°= 270°. Question 5. Micah built the following shape with his pattern blocks. Write an addition sentence for each angle indicated by an arc and solve. The first one is done for you. a. y° = 120° + 90° y° = 210° b. z° = _____________ z° = _____________ z° = 60°+30° z° = 90° Explanation: Here, the value of z is 60°+30° which is 90° c. x° = _____________ x° = _____________<\|endoftext\|>	4.59375
411	In this activity, students will watch a video about Antoine Lavoisier, who many consider to be the father of modern chemistry. They will answer questions as they learn about oxygen, hydrogen, and the first proposal of the Law of the Conservation of Mass. Middle, and High School By the end of this activity, students should be able to - Explain the Law of Conservation of Mass. - Understand how Lavoisier is widely considered to be the father of modern chemistry. This activity supports students’ understanding of - Conservation of mass - History of Chemistry - SI units Teacher Preparation: minimal Lesson: 10 minutes - No specific safety precautions need to be observed for this activity. - The Antoine Lavoisier video was developed as a part of the AACT original video series, Founders of Chemistry. The entire series can be found here. - The running time of this video is six minutes. - This video is intended for students to watch, and for teachers to integrate into their curriculum. - The student questions/answers are presented in sequential order in the video. - An answer key has also been provided for teacher reference. - Videos can be shown with the use of a classroom projector, or teachers can generate a Student Video Pass through their AACT membership to allow students to independently access the video. For the Student While watching the Founders of Chemistry Video about Antoine Lavoisier, answer the following questions: - What elements did Lavoisier discover and name? - How did Lavoisier impact the naming of chemical compounds? - What is the Law of Conservation of Mass? - Why does this Law make sense logically? - What does this Law build the foundation for? - How did Lavoisier disprove the phlogiston theory? - How did Lavoisier’s wife, Marie-Anne Paulze Lavoisier, help her husband’s scientific experiments and discoveries?<\|endoftext\|>	3.921875
1,705	# Graphing THE CENTER OF CURVATURE AND THE EVOLUTE. Ellipse. Definition of the Ellipse To read how the ellipse got its name, and what it means, see Parabola. That page also contains some background information on conic sections and other topics that also applies to ellipses, that won't be repeated here. Finding the arc length of an ellipse, which introduces elliptic integrals, and Jacobian elliptic functions, are treated in their own articles. An oval is generally regarded as any ovum (egg)-shaped smooth, convex closed curve. Convex means that any chord connecting two points of the curve lies completely within the curve, and smooth means that the curvature does not change rapidly at any point. Curvature. Arc Length Curvature Arc Length For a parametrically defined curve we had the definition of arc length. Since vector valued functions are parametrically defined curves in disguise, we have the same definition. Osculating Circles for Plane Curves. Untitled. So, you want to more? Here are all the gruesome details for calculating the tangent, normal, curvature and osculating circle of a curve! Think of your favourite smooth curve in a plane that can be described by a function, . Parametric Equation of an Ellipse. Scarpelli Assignment 10. Notice from the graphs above when a = b, the paramaterization of x and y is a circle centered at the origin (0,0) with radius equal to a. When a = b = 1, we have a graph of the unit circle. Since the domain of t is between 0 and 2π (approximately 6.28) or 360 degrees, the graph starts at t = 0, the point (a,0) or in this case (1,0), and moves in the counterclockwise direction around the origin to create a full circle that ends at t =2π, which is also the point (a,0) or in this case (1,0). The same holds for when a = b = 2 except the circle begins at t = 0, the point (2,0), and ends at t = 2π, also the point (2,0). To further investigate, hypothesize what will happen if your t - values range from 0 to π instead of 0 to 2π. How will your graph be different than the graph above? T hat, N hat, B hat and kappa. This figure shows T(0), N(0), and B(0). Click on the figure to see the movie of T(t), N(t) and B(t). This figure shows the osculating ("kissing") circle of the helix at t = 0. (There is no animation for this figure.) The ellipse x2/4 + y2/9 = 1 is shown in these figures. Combo. GraphSketch - Parametric. GraphSketch.com - Parametric Click here to download this graph. Permanent link to this graph page. Mode: Functions Parametric Enter Graph Equations: Plot y^2+z^2-1. X Image InputX Upload an image from your computer: Enter a URL for an image: Terms \| Privacy All images »Recent images Loading... Sample images Data InputX Give numeric, tabular, or other data. Antiderivatives - Application Center. Antiderivatives Copyright Maplesoft, a division of Waterloo Maple Inc., 2007 Introduction This application is one of a collection of examples teaching Calculus with Maple. These applications use Clickable Calculus methods to solve problems interactively. Steps are given at every stage of the solution, and many are illustrated using short video clips. Buttons to watch the videos. Problem Statement. Math Forum - Ask Dr. Math. Calculate the integral $\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt$, by deformation theorem. Ellipse. Matematicas Visuales. From Wolfram MathWorld. An ellipse is a curve that is the locus of all points in the plane the sum of whose distances and from two fixed points (the foci) separated by a distance of is a given positive constant (Hilbert and Cohn-Vossen 1999, p. 2). Where. Archimedes and the area of an ellipse: an intuitive approach. Algebra Conic Ellipse Equation Derivation. Study the following diagram for an ellipse: Recall the definition of an ellipse requires that d₁ + d₂ = 2a. Let (x,y) be the coordinates for the point P. In other words, P can be any point on the ellipse as shown below. Here we wave two right triangles, the blue and the green. We will write the distances d₁ and d₂ in terms of the legs of these right triangles using Pythagorean's Theorem and simplify the result. D₁ + d₂ = 2a The general equation involves translation of coordinates from one coordinate system to another, that is moving from one origin to another. Conic Sections: Ellipse with Foci. Using Greens Theorem to find Area of an Ellipse. k12math - Algebra Conic Ellipse Equation Derivation. Using Green's theorem to find area - Math Insight. Typically we use Green's theorem as an alternative way to calculate a line integral \dlint. If, for example, we are in two dimension, \dlc is a simpleclosed curve, and \dlvf(x,y) is defined everywhere inside \dlc, we can use Green's theorem to convert the line integral into to double integral. Instead of calculating line integral \dlint directly, we calculate the double integral \begin{align} \iint_\dlr \left(\pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}\right) dA \end{align} Can we use Green's theorem to go the other direction? If we are given a double integral, can we use Green's theorem to convert the double integral into a line integral and calculate the line integral? If we are given the double integral \begin{align} \iint_\dlr f(x,y) dA, \end{align} we can use Green's theorem only if there happens to be a vector field \dlvf(x,y) so that \begin{align*} f(x,y) = \pdiff{\dlvfc_2}{x} - \pdiff{\dlvfc_1}{y}. Where \dlvf(x,y) = (-y/2,x/2). Area enclosed by an ellipse. Area enclosed by an ellipse From Latin: area - "level ground, an open space," The number of square units it takes to exactly fill the interior of an ellipse. Try this Drag the orange dots to move and resize the ellipse. As the shape and size of the ellipse changes, the area is recalculated. An ellipse is actually a line, one that connects back to itself making a loop. Desmos Graphing Calculator. Desmos Graphing Calculator. Area between curves with multiple boundaries. Area between Curves. Area under a curve. How to calculate areas under a function graph. Theory: Area Under A Curve. Shading. Functions Domain Calculator - Symbolab. Domain of √(x + 2) / (x - 1) X Image InputX. Find Domain of a Function - Calculator. Composition of Functions - Domains. Visual Calculus - Drill - Domains of Functions. Calculus and algebra vs topology - Intelligent Perception. We revisit the subject of Topology via Calculus here but put it on a more solid, algebraic foundation. Once again, we start with dimension 1 and no forms. Suppose R is the union of m disjoint open intervals, I_1,... ,I_m in {\bf R}. Set-Builder Notation. How to describe a set by saying what properties its members have. Mathematics Calculus. What Is Domain and Range in a Function? - Video & Lesson Transcript. ActiveMath Applet: Arithmetic operations with real functions. Highschool Content of LeAM_calculus. Finding Domain and Range. Finding Domain and Range.<\|endoftext\|>	4.5
525	# How many number of vertices does a rectangular prism have? ## How many number of vertices does a rectangular prism have? 8 Cuboid/Number of vertices ## How do you find the edges of a rectangular prism? If your base has four sides to find the faces you will add two. If you know the base has four sides you will multiple by two to get the number of vertices. If you know your base has four sides you multiple by three to get the edges. Does a rectangular prism have more than 6 vertices? A rectangular prism has 6 faces, 8 vertices (or corners) and 12 edges. How many faces are on a rectangular prism? 6 Cuboid/Number of faces ### What is the formula to a rectangular prism? The volume of a rectangular prism follows the simple method, multiply all three dimensions – length, height, and width. Thus, the volume of the rectangular prism is given by the formula V= l × w × h where”V”, “l” “w”, and “h” are the volume, length, width, and height of the rectangular prism respectively. ### Does a rectangular prism have 6 faces 8 edges and 10 vertices? Properties of Rectangular Prism A rectangular prism has 6 faces, 8 vertices, and 12 edges. Its base and top are always rectangles. The side faces are rectangles for a right rectangular prism whereas the side faces of an oblique rectangular prism are parallelograms. It has 3 dimensions, length, width, and height. Does a triangular prism have more than 6 vertices? A triangular prism is a polyhedron, (three-dimensional shape) made up of two triangular bases and three rectangular sides. It has 5 faces, 6 vertices and 9 edges in total. What is the example of rectangular prism? i.e., every two opposite faces are identical in a rectangular prism. It has three dimensions, length, width, and height. Some examples of a rectangular prism in real life are rectangular tissue boxes, school notebooks, laptops, fish tanks, large structures such as cargo containers, rooms, storage sheds, etc. #### What is the height of rectangular prism? Re-work the equation to solve for h: h = 3V / b. So the height is three times the volume, divided by the base’s area. #### What is the prism formula? The Prism Formula is as follows, The surface area of a prism = (2×BaseArea) +Lateral Surface Area. The volume of a prism =Base Area× Height.<\|endoftext\|>	4.4375
1,625	# Give the combinatorial proof of the identity $\sum_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$ Given the identity $$\sum_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}$$ Need to give a combinatorial proof a) in terms of subsets b) by interpreting the parts in terms of compositions of integers I should not use induction or any other ways... - HINTS: 1. Consider a $k$-element subset of $[n+k]=\{1,\dots,n+k\}$; it has a maximum element, which can be anything from $k$ through $n+k$. How many such subsets are there with maximum element $k+i$ for $i=0,\dots,n$? 2. There are $\binom{k-1+i}{k-1}$ compositions of $k+i$ with $k$ terms. There are $\binom{n+k}k$ compositions of $n+k+1$ with $k+1$ terms. - 2)Let's split n+k into two parts: 1 and n+k-1. including 1, we have $\binom{n+k-1}{k-1}$, otherwise $\binom{n+k-1}{k}$. Hence $\binom{n+k}{k}$= $\binom{n+k-1}{k-1}$+$\binom{n+k-1}{k}$ Using $\binom{n+k-1}{k}$,repeat the procedure till $\binom{k-1}{k-1}$. 1)I'm still confused here. I understand that the answer to your question should be the identity, however I don't quite understand how this works. – John Lennon Mar 5 '13 at 22:09 (1) Oops: the maximum should have been $k+i$, not $k-1+i$. If $k+i$ is the maximum element of a $k$-element subset of $[n+k]$, the other $k-1$ elements can be any $k-1$ elements of the set $[k-1+i]$, and there are $\binom{k-1+i}{k-1}$ of them. Thus, the summation is just counting the $k$-element subsets of $n+k$ in groups corresponding to their maximum elements. (2) You’re suggesting an informal proof by induction, not a combinatorial argument. You do not want an argument by induction; you want to show that the two sides are counting the same set of compositions in two different ways. – Brian M. Scott Mar 5 '13 at 22:15 "There are $\binom{k-1+i}{k-1}$ compositions of $k+i$ with $k$ terms. There are $\binom{n+k}k$ compositions of $n+k+1$ with $k+1$ terms." Is this already an answer? If so could you explain it to me? – John Lennon Mar 19 '13 at 23:20 @vercammen: Suppose that you have a composition of $n+k+1$ with $k+1$ terms; when you throw away the last term, what’s left is a composition with $k$ terms of some number between $k$ and $k+n$ inclusive. – Brian M. Scott Mar 19 '13 at 23:30 finally clear. thank you! – John Lennon Mar 19 '13 at 23:42 Use $\binom{k}{r} = \binom{k-1}{r} + \binom {k-1}{r-1}$ repeatedly on the expansion of sum. - $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum_{j = 0}^{n}{k - 1 + j \choose k - 1}} & = \sum_{j = 0}^{n}{k - 1 + j \choose j} = \sum_{j = 0}^{n}{-k + 1 - j + j - 1 \choose j}\pars{-1}^{j} = \sum_{j = 0}^{n}{-k \choose j}\pars{-1}^{j} \\[3mm] & = \sum_{j = -\infty}^{n}{-k \choose j}\pars{-1}^{j} = \sum_{j = -n}^{\infty}{-k \choose -j}\pars{-1}^{-j} = \sum_{j = 0}^{\infty}{-k \choose n - j}\pars{-1}^{j + n} = \\[3mm] & = \pars{-1}^{n}\sum_{j = 0}^{\infty}\pars{-1}^{j}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{-k} \over z^{n - j + 1}}\,{\dd z \over 2\pi\ic} \\[3mm] & = \pars{-1}^{n}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{-k} \over z^{n + 1}}\sum_{j = 0}^{\infty}\pars{-z}^{j} \,{\dd z \over 2\pi\ic} = \pars{-1}^{n}\oint_{\verts{z} = 1^{-}} {\pars{1 + z}^{-k - 1} \over z^{n + 1}}\,{\dd z \over 2\pi\ic} \\[3mm] & = \pars{-1}^{n}{-k - 1 \choose n} = \pars{-1}^{n}{k + 1 + n - 1 \choose n}\pars{-1}^{n} = {n + k \choose n} = \color{#f00}{n + k \choose k} \end{align} -<\|endoftext\|>	4.4375
1,056	Methicillin-Resistant Staph Infection (MRSA; Methicillin-Resistant Staphylococcus aureus Infection; Infection, Methicillin-Resistant; Methicillin-Resistant Staphylococcus aureus Community-Acquired MRSA; CA-MRSA; Methicillin-Resistant Staphylococcus aureus Nosocomial MRSA; Healthcare-Associated MRSA; HA-MRSA) by Krisha McCoy, MS Methicillin-resistant staphylococcus aureus (MRSA) is a potentially serious infection that resists this antibiotic used to treat infections. It may be resistant to many antibiotics, especially if the infection was contracted in a healthcare setting. MRSA can affect the skin, blood, bones, or lungs. A person can either be infected or colonized with MRSA. There are two types of MRSA infections: MRSA can spread several ways: MRSA is caused by bacteria that adapt to repeated exposure to antibiotics, building up a resistance to them. Factors that may increase your chance of developing MRSA include: MRSA may not cause any symptoms in people who are colonized, but not infected, with the bacteria. In those that have symptoms, MRSA may cause: Your doctor will ask about your symptoms and medical history. A physical exam will be done. Your bodily fluids and tissues may be tested. This can be done with: Talk with your doctor about the best treatment plan for you. Treatment options include: Antibiotics may given to kill the bacteria based on the location and severity of the infection, and any underlying illness. Only a few antibiotics are available that can treat MRSA. One that will treat MRSA may be chosen based on what usually works and your culture results. Incision and Drainage of an Abscess Your doctor may open the abscess and allow the fluid to drain. Do not attempt to do this on your own. In many cases this will be the only treatment needed. Cleansing of the Skin Do the following to treat the infection and to keep it from spreading: Decolonization is a process to help rid your body of the bacteria so you do not reinfect yourself. This process may involve using nasal ointments, washing with special soap, and taking medications, including antibiotics. Decolonization is only recommended in certain cases. To help reduce your chances of MRSA: Centers for Disease Control and Prevention National Institute of Allergy and Infectious Diseases Canadian Dermatology Association Public Health Agency of Canada Barton M, Hawkes M, Moore D, et al. Guidelines for the prevention and management of community-associated methicillin-resistant Staphylococcus aureus: A perspective for Canadian health care practitioners. Can J Infect Dis Med Microbiol. 2006;17(Suppl C):4C. Methicillin-resistant Staphylococcus aureus (MRSA) infection. EBSCO DynaMed Plus website. Available at: http://www.dynamed... . Updated January 6, 2017. Accessed August 18, 2017. MRSA decolonization. Aurora BayCare Medical Center website. Available at: https://ahc.aurorahealthcare.org/fywb/baycare/x34012bc.pdf. Accessed August 18, 2017. Methicillin-resistant Staphylococcus aureus (MRSA). Centers for Disease Control and Prevention website. Available at: https://www.cdc.gov/mrsa. Updated May 16, 2016. Accessed August 18, 2017. MRSA. Kids Health—Nemours Foundation website. Available at: ...(Click grey area to select URL) Updated June 2014. Accessed August 18, 2017. 6/4/2018 DynaMed Plus Systematic Literature Surveillance http://www.dynamed... : Gualandi N, Mu Y, Bamberg WM, et al. Racial disparities in invasive methicillin-resistant staphylococcus aureus infections, 2005-2014. Clin Infect Dis. 2018 Apr 5 [Epub ahead of print]. Last reviewed September 2018 by EBSCO Medical Review Board Michael Woods, MD, FAAP Last Updated: 6/4/2018 EBSCO Information Services is fully accredited by URAC. URAC is an independent, nonprofit health care accrediting organization dedicated to promoting health care quality through accreditation, certification and commendation. This content is reviewed regularly and is updated when new and relevant evidence is made available. This information is neither intended nor implied to be a substitute for professional medical advice. Always seek the advice of your physician or other qualified health provider prior to starting any new treatment or with questions regarding a medical condition. To send comments or feedback to our Editorial Team regarding the content please email us at [email protected]. Our Health Library Support team will respond to your email request within 2 business days.<\|endoftext\|>	3.796875
590	Setting standards has gradually generated agreement about the meaning of two key concepts--academic content standards and performance standards. The following definitions, illustrated below with examples from Delaware, are consistent with those suggested by the National Education Goals Panel (1993) and enacted into law by the Goals 2000: Educate America Act: Academic content standards describe what every student should know and be able to do in the core academic content areas (e.g., mathematics, science, geography). Content standards should apply equally to students of all races and ethnicities, from all linguistic and cultural backgrounds, both with and without special learning needs. Performance standards answer the question, "How good is good enough?" They define how students demonstrate their proficiency in the skills and knowledge framed by states' content standards. An Example of One of DELAWARE's Content and Performance Standards in ReadingThe content standard in reading requires students to: construct, examine, and extend the meaning of literacy informative, and technical texts through listening, reading, and viewing. For example, to demonstrate their knowledge of this standard, fifth graders must read a full-length passage from a text and answer questions requiring both brief and detailed responses. Based on how students' answers demonstrate their understanding of the passage, the performance standard indicates they "meet or exceed" the standard if their answer: Academic content standards are meant to apply to all children. In some states, however, although the standards apply to all students, the strategies for achieving them may differ. For example, New York State has established two ninth grade earth science courses: a two semester course and a three semester course. The course expectations are the same, students take the same Regents exam, and they meet the same high standard, only the length of time to learn the material differs. In other states, like Kentucky, some percentage of students who have severe disabilities are often excluded from meeting the state's standards of performance (See box below on applying standards to students with disabilities). Standards development is supported by state legislatures, local communities, private foundations, and the federal government. States and communities establish their own standards without direction from outside sources, but they often adapt their standards to those set by teachers and professional organizations such as the National Council of Teachers of Mathematics (NCTM), the National Council of Teachers of English (NCTE), the National Science Teachers' Association (NSTA), the American Association for the Advancement of Science (AAAS), the National Academy of Sciences (NAS), or New Standards, a collaborative of 19 states, 6 urban districts, and the Learning Research and Development Center of Pittsburgh and the National Center on Education and the Economy. In 1989, NCTM was the first professional organization to issue its discipline's standards, Curriculum and Evaluation Standards for School Mathematics, and other discipline-based organizations have been developing their standards since 1992. Applying State Standards to Students with Disabilities<\|endoftext\|>	3.875
1,277	Problems top top coin toss probability are described here with different examples. You are watching: What is the probability of getting zero heads in five tosses When us flip a coin over there is always a probability to get a head or a tail is 50 percent.Suppose a coin tossed climate we obtain two feasible outcomes one of two people a ‘head’ (H) or a ‘tail’ (T), and it is impossible to predict whether the result of a toss will be a ‘head’ or ‘tail’. The probability for equally likely outcomes in an event is: Number that favourable outcomes ÷ Total number of possible outcomes Total variety of possible outcomes = 2 (i) If the favourable result is head (H). Number that favourable outcomes = 1. number of favorable outcomes = P(H) = total number of possible outcomes = 1/2. (ii) If the favourable result is tail (T). Number the favourable outcomes = 1. Therefore, P(getting a tail) variety of favorable outcomes = P(T) = total variety of possible outcomes = 1/2. Word difficulties on Coin Toss Probability: 1. A coin is tossed double at random. What is the probability the getting (i) at the very least one head (ii) the very same face? Solution: The feasible outcomes space HH, HT, TH, TT. So, total variety of outcomes = 4. (i) number of favourable outcomes for event E = number of outcomes having actually at least one head = 3 (as HH, HT, TH are having actually at least one head). So, by definition, P(F) = $$\frac34$$. (ii) variety of favourable outcomes for event E = variety of outcomes having actually the very same face = 2 (as HH, TT are have the very same face). So, by definition, P(F) = $$\frac24$$ = $$\frac12$$. 2. See more: What Are The Plastic Thing On The End Of A Shoelace S Called? If three fair coins are tossed randomly 175 times and also it is discovered that three heads appeared 21 times, two heads appeared 56 times, one head appeared 63 times and also zero head showed up 35 times. What is the probability the getting Solution: Total variety of trials = 175. Number the times 3 heads showed up = 21. Number the times 2 heads showed up = 56. Number of times one head appeared = 63. Number of times zero head showed up = 35. Let E1, E2, E3 and E4 it is in the events of gaining three heads, 2 heads, one head and also zero head respectively. variety of times three heads appeared = P(E1) = total variety of trials = 21/175 = 0.12 variety of times 2 heads showed up = P(E2) = total number of trials = 56/175 =0.32 number of times one head appeared = P(E3) = total number of trials =63/175 = 0.36 number of times zero head showed up = P(E4) = total number of trials = 35/175 = 0.20 Note: Remember when 3 coinsare tossed randomly, the only feasible outcomes space E2, E3, E4 andP(E1) + P(E2) + P(E3) + P(E4) = (0.12 + 0.32 + 0.36 + 0.20) = 1 3. 2 coins space tossed randomly 120 times and also it is uncovered that 2 tailsappeared 60 times, one tailappeared 48 times and no tail appeared 12 times. If two coins room tossed in ~ random, what is theprobability of obtaining (i) 2 tails, (ii) 1 tail, (iii) 0 tail Solution: Total number oftrials = 120 Number of times 2 tails appear= 60 Number of times 1 tail appears= 48 Number of time 0 tail appears= 12 Let E1, E2 and E3 it is in the occasions of gaining 2 tails, 1 tail and 0 tail respectively. (i) P(getting2 tails) number of times 2 tails appear = P(E1) = total variety of trials = 60/120 = 0.50 (ii) P(getting 1 tail) number of times 1 tail appear = P(E2) = total variety of trials = 48/120 = 0.40 (iii) P(getting0 tail) number of times no tail appear = P(E3) = total number of trials = 12/120 = 0.10 Note: Remember while tossing 2 coins simultaneously, the only feasible outcomes are E1, E2, E3 and, P(E1) + P(E2) + P(E3) = (0.50 + 0.40 + 0.10) = 1 4. Suppose a fair coin is randomlytossed for 75 times and also it is uncovered that head turns up 45times and tail 30 times. What is the probability of acquiring (i)a head and also (ii) a tail? Solution: Total number of trials = 75. Number of times head transforms up = 45 Number of time tail turns up = 30 (i) let X it is in the occasion ofgetting a head. number of times head turns up = P(X) = total variety of trials = 45/75 = 0.60 (ii) let Y bethe event of getting a tail. P(getting a tail) number of times tail transforms up = P(Y) = total variety of trials = 30/75 = 0.40 Note: Remember as soon as afair coin is tossed and also then X and Y arethe only feasible outcomes, and also<\|endoftext\|>	4.625
2,437	6-8 > Mathematics Grade level: 6-8 Subject: Mathematics Duration: Two class periods Objectives \| Materials \| Procedures \| Adaptations \| Discussion Questions \| Evaluation \| Extensions \| Suggested Readings \| Vocabulary \| Academic Standards \| Credit Students will: 1 understand that ratios are used to create scale models of buildings and structures; 2 understand the principles of ratio and apply these principles in the solution of problems; and 3 understand how to calculate scale using ratio. The class will need the following: • 0.25-inch graph paper • map(s) of the United States • pencils • ruler (metric or inches) • tape measure • Take-Home Activity Sheet: Home Measurements 1. Begin by introducing the concept of scale. Write the wordscaleon the board and brainstorm examples of where scales are found and what they measure. For example, we use scales to measure the weight of an object, the temperature of air, the length of an object, and so on. 2. Show students a map of the United States and point out the scale in the map key. Remind them that this map is a smaller, scaled-down representation of the United States, not an actual representation. Explain that sometimes we shrink objects or make them larger so they are easier to work with. The map is a scale model of an object that is too large to represent on paper. Other scale models represent objects that are too small, such as a diagram of an atom or a magnified view of a computer chip. Review the scale on the map. For example, the scale may say that 1 inch is equal to 50 miles. Explain that a scale is a ratio used to determine the size of a model of a real object. In this case, the map of the United States is the model. 3. A ratio is a relationship between two objects in quantity, size, or amount. For example, four quarters are in a dollar, so the ratio of quarters to dollar is 4 to 1. In other words, a quarter is one-fourth the value of a dollar. Have students think of other examples of how money can be turned into a scale, such as dimes to dollars (10:1 or 1:10) or pennies to dollars (100:1 or 1:100). 4. Illustrate how to draw an object to scale. Use a ruler to draw a square on the board with sides that equal 10 inches in length. Ask students how they might use this square to draw another that is half its size. Explain that an object is not simply cut in half when it is scaled down. The whole object is shrunkproportionally, meaning that it doesn’t change shape but is reduced to a smaller size. For example, if you could scale a carrot to half its size, you wouldn’t simply cut the carrot in half. All parts of the carrot need to shrink equally in size. 5. Now measure and draw a second square with 5-inch sides. Explain that when an object is scaled down, the length of its sides must be reduced by the same amount. Compare the corresponding sides of the two squares. The ratio of the small square to the larger is 5:10. Explain that a ratio can be expressed in three ways: 5:10, 5 to 10, or 5/10, which is a fraction that reduces to 1/2. 6. Remind students that the perimeter of an object is the sum of the length of its sides. So if an object has been scaled down proportionally, the perimeter of the object will scale down by the same ratio. For example, the perimeter of the smaller square is 20, or 5 × 4, which is half the perimeter of the larger square, which is 40, or 10 × 4. 7. Explain that students will use ratio to make a scale drawing of the classroom floor plan. First invite students to brainstorm a list of the kinds of people who might use scale drawings. (Examples include architects, construction workers, and cartographers.) 8. Divide students into teams of four. Explain that each team will measure the surface areas of objects in the classroom—the desks, tables, closets, and so on. The class may choose to use either metric or English measurements. Explain to students that their floor plan will show objects in the classroom as seen from above. Each group should have access to a tape measure, pencils, and paper to record their measurements. 9. Construct a class data table on the board with three columns labeled “object,” “measurement,” and “scaled measurement.” Students should copy this table in their notebooks and fill in the answers as they measure the objects. 10. Once teams have recorded all their data, they will decide on the scale of their floor plan. Distribute graph paper. With the class, discuss the proportions that would allow students to draw the entire room on one sheet of 8.5" × 11" graph paper. (For example, if the longest wall in the classroom is 16 feet long, then a scale of 1" = 1’ will not work. But 0.5" = 1’ will work perfectly.) 11. Use the agreed-upon ratio to create the proportion for your classroom. Then have groups convert their measurements into scaled equivalents. For example, if a desktop measures 2 feet in width and the scale is 0.5" = 1’, use the following equation to figure out how large the scaled drawing of the desktop should be. 0.5 inches divided by 1 foot = the scaled down length of the object divided by 2 feet Or, written as an equation of two ratios: 0.5 inches = yinches 1 foot 2 feet y= 1 inch 12. Students can determine their scaled equivalents by cross-multiplying. Students should recall that when both sides of an equation are multiplied by the same amount, the equation remains balanced. In cross-multiplication, both sides of an equation are multiplied by the denominators (the bottom numbers in the fractions). The result is the same as multiplying across the “equals” sign diagonally (i.e., the “bottom left” number times “top right” number equal to the “top left” number times the “bottom right” number). Have students consider the following example: 13. Have students use their scaled measurement, rulers, and graph paper to draw the floor plan their team measured. Remind them to include a title, labels, and a scale. 14. As students complete their drawings, encourage them to calculate the perimeter of their classrooms. What is the relationship between the perimeter of the drawing and the perimeter of the actual classroom? 15. For homework, ask students to complete the sheet, asking them to make a floor plan of a room in their home. Have students determine the relationship between the area of the drawing and the area of the actual classroom. They should notice that the ratio of these areas is the square of the scale they chose. For example, if a scale of 0.5 inch = 1 foot was used, the ratio of areas of the drawing to the actual room will be (0.5 inches)2 = (1 foot)2 or 0.25 square inches = 1 square foot. Students can also conduct experiments to determine how volume changes with scale. 1 Using what you have learned about ratios, proportions, and scale models, create four word problems for other students in your class to solve. For example: A square carpet measures 8 feet × 4 feet. Suppose the scale of a drawing containing the carpet is 1 foot to 1/4 inch. What are the dimensions of the carpet in the drawing?The answer: 2 inches × 1 inch. 2 Is it possible to draw scale models that are completely accurate? Why is accuracy important in the creation of maps, blueprints, and other scale models? 3 Compare your classroom floor plan to that of another student. How are they similar and different? Which would be more useful to a construction worker trying to build a classroom in a new school? Why? 4 List other instances in which you use ratio to compare objects in your daily life. Why is it important to maintain the same scale for each measurement you record when making your model? 5 Debate the merits of using the metric system and the English system to measure lengths. Explain how to convert between the two systems. 6 Compare your classroom to a nearby classroom using scale models of each. Explain how you could use estimation to create a scale model. Would the model be more or less accurate? You can evaluate students’ work using the following three-point rubric: Three points:records and converts all of the measurements accurately; uses measurements to draw a classroom floor plan to scale in precise detail. Two points:records and converts most of the measurements correctly; uses measurements to draw a classroom floor plan that is not entirely accurate. One point:records and converts some or few of the measurements accurately; is unable to create a classroom floor plan that is accurate. Distance Brenda Walpole. Gareth Stevens, 1995.Learn how standardized measurements developed, as early civilizations used parts of the body for measurements like cubits and fathoms, which gradually became inches and feet. Ways of estimating distances and heights are included along with lots of easy measuring experiments you can do with just a few simple objects. A timeline of important measurement “events” shows the progress of standardization to the present. The Story of Weights and Measures Anita Ganeri. Oxford University Press, 1996.An excellent introduction to the concepts of weight and measurement is encompassed in this slim book. Learn about the history of the development of instruments for accurate weighing and measuring. A short timeline and glossary are included, as well as illustrations and short descriptions equivalent Definition:Being the same or effectively the same; equal. Context:The length of the front wall isequivalentto the length of the back wall in our rectangular classroom. perimeter Definition:The boundary, or border, of a closed, two-dimensional figure or area. Context:We built a fence around theperimeterof our yard to keep the dog from running away. ratio Definition:The relation of one part to another or to a whole. Context:We have twice as many girls as boys in our class. Therefore theratioof girls to boys is 2 to 1, or 2:1. scale Definition:The ratio of the size of a model or other representation, such as a map, to the actual size of the object represented. Context:By looking at thescale, we could tell that 1 inch represented 1 mile on our map of New York. symmetry Definition:A state in which parts on opposite sides of a plane, line, or point display the same size, form, or arrangement. Context:The butterfly’s wings were exactly alike, displaying perfectsymmetry. This lesson plan may be used to address the academic standards listed below. These standards are drawn from Content Knowledge: A Compendium of Standards and Benchmarks for K-12 Education: 2nd Edition and have been provided courtesy of theMid-continent Research for Education and Learningin Aurora, Colorado. Grade level:6-8 Subject area:Mathematics Standard: Understands and applies basic and advanced properties of the concepts of measurement. Benchmarks: Solves problems involving units of measurement and converts answers to a larger or smaller unit within the same system (i.e., standard or metric). Grade level:6-8 Subject area:Mathematics Standard: Understands and applies basic and advanced properties of the concepts of measurement. Benchmarks: Understands formulas for finding measures (e.g., area, volume, surface area) Jessi Hempel, communications team member, Bay Area School Reform Collaborative, San Francisco, California; former fourth-grade teacher in New York City Public School 92.<\|endoftext\|>	4.65625
269	An optical profiler is an interference microscope, and it measures height variations on surfaces with great precision using the wavelength of light as the ruler. Optical Profiling uses the wave properties of light to compare the optical path difference between a test surface and a reference surface. Inside an optical interference profiler, a light beam is split, reflecting half the beam from a test material which is passed through the focal plane of a microscope objective, and the other half of the split beam is reflected from the reference mirror. When the distance from the beam splitter to the reference mirror is the same distance as the beam splitter is from the test surface, the recombined two beams are in phase, resulting in constructive interference and hence bright band. Dark band occurs due to destructive interference when the beam paths have the half wavelength difference. These light and dark bands known as interference fringes. In the interference image (an “interferogram”), each transition from light to dark represents one-half a wavelength of difference between the reference path and the test path. The reference mirror is as close to perfect flatness as possible; hence the optical path differences are due to height variances in the test surface. If the wavelength is known, it is possible to calculate height differences across a surface. A 3D surface map is obtained.<\|endoftext\|>	3.796875
271	Researchers at Ohio State University have demonstrated the first device that utilizes the spin of electrons to read and write data. An alternative to traditional microelectronics, so-called "spintronics" could store more data in less space, process data faster, and consume less power. Arthur J. Epstein and colleagues described the material as a hybrid of a semiconductor that is made from organic materials and a special magnetic polymer semiconductor. Researchers have long known that electrons can be polarized to orient in particular directions, like a bar magnet. They refer to this orientation as spin-either "spin up" or "spin down"-and have been working on a way to store data using spin. The resulting electronics, dubbed spintronics, would effectively let computers store and transfer twice as much data per electron. Flipping the spin of an electron requires less energy and produces minimal heat - so they can run on smaller batteries - which means lesser power consumption. "We would love to take portable electronics to a spin platform," said Epstein. "Any place that makes computer chips could do this. Plus, in this case, we made the device at room temperature, and the process is very eco-friendly," said researcher Jung-Woo Yoo. The study is published in the August 2010 issue of the journal Nature Materials.<\|endoftext\|>	4.21875
1,036	# Area Of Sector Of A Circle Before knowing about a sector of a circle, let’s know how the area of a circle is calculated. When it comes to the area, it is always related to two-dimensions. Anything which is two dimensional can form a plane. So, any two-dimensional figure will have area. What about a circle? Definition 1: A circle is the collection of all the points in a plane which are at a fixed distance from a fixed point. The fixed point is known as the center of the circle and the fixed distance is known as the radius of the circle. Definition 2: If all the points which lie inside and on the circle are taken together, the plane constructed is known as a disk. A disk is basically the region bounded by a circle. So, the area of a circle will always be that of the disk. The area, A of the circle with radius r is given by $$A$$ = $$π~r^2$$ Definition 3: The portion of the circle enclosed by two radii and the corresponding arc is known as the sector of a circle. Basically, a sector is the portion of a circle. It would hence be right to say that a semi-circle or a quarter-circle is a sector of the given circle. In fig.1, OPAQ is called the minor sector and OPBQ is called the major sector because of lesser and greater areas. The angles subtended by the arcs PAQ and PBQ are equal to the angle of the sectors OPAQ and OPBQ respectively. When the angle of the sector is equal to 180°, there is no minor or major sector. ## Area of sector In a circle with radius r and center at O, let ∠POQ = θ (in degrees) be the angle of the sector. Then, the area of a sector of circle formula is calculated using the unitary method. When angle of the sector is 360°, area of the sector i.e. the whole circle = $$πr^2$$ When the angle is 1°, area of sector = $$\frac{πr^2}{360°}$$ So, when the angle is θ, area of sector, OPAQ, $$A$$ = $$\frac{θ}{360°}~×~ πr^2$$ Similarly, length of the arc (PQ) of the sector with angle θ, l = $$\frac{θ}{360°}×2πr$$ If the length of the arc of the sector is given instead of the angle of the sector, there is a different way to calculate the area of the sector. Let the length of the arc be l. For the radius of a circle equal to r units, an arc of length r units will subtend 1 radian at the center. It can be hence concluded that an arc of length l will subtend $$\frac{l}{r}$$ angle at the center. So, if l is the length of the arc, r is the radius of circle and θ is the angle subtended at center, $$θ$$ = $$\frac{l}{r}$$, where θ is in radians When angle of the sector is 2π, area of the sector i.e. the whole circle = $$πr^2$$ When the angle is 1, area of the sector = $$\frac{πr^2}{2π}$$ = $$\frac{r^2}{2}$$ So, when the angle is θ, area of the sector = $$θ~×~\frac{r^2}{2}$$ = $$\frac{l}{r}~×~\frac{r^2}{2}$$ = $$\frac{lr}{2}$$ Some examples for better understanding are discussed from here on. Example 1: If the angle of the sector with radius 4 units is 45°, area = $$\frac{θ}{360°}~×~ πr^2$$ = $$\frac{45°}{360°}~×~\frac{22}{7}~×~4~×~4$$ = $$\frac{44}{7}$$ square units The length of the same sector = $$\frac{θ}{360°}~×~ 2πr$$ = $$\frac{45°}{360°}~×~2~×~\frac{22}{7}~×~4$$ = $$\frac{22}{7}$$ units Example 2: If the length of the arc of a circle with radius 16 units is 5 units, the area of the sector corresponding to that arc = $$\frac{lr}{2}$$ = $$\frac{5~×~16}{2}$$ = $$40$$ square units To practice more on are of sector of a circle, download BYJU’S – The Learning App from the Google Play Store.<\|endoftext\|>	4.78125
704	Digital cameras are very useful for taking holiday snaps, but they are also useful tools for imaging inside and outside the body. For example, dentists use digital imaging to improve diagnosis and treatment. Cameras on the end of probes that can move around inside your mouth are linked to screens, so the dentist can see more clearly inside the mouth and show the patient exactly what is going on. Digital camera technology includes endoscopy, capsule endoscopy and digital image-based elasto-tomography (DIET). With endoscopy (Greek for “look inside”), a very small camera on the end of a flexible tube is inserted into the body through an opening, and manoeuvred around. The camera transmits pictures and allows doctors to see what is happening inside. This method is commonly used to check out a patient’s large intestine, to look for problems like tumours or inflammation. It can also be used to view the oesophagus and upper gastrointestinal tract. At the beginning of last century, doctors tried to use lighted telescopes to look inside the human body. Most of these systems had rigid tubes, but a system with a flexible tube was invented by Dr Rudolph Schindler in the 1930s. The results from these devices were often frustrating, prompting young doctor Basil Hirschowitz to experiment with the use of fibre optics in the early 1950s. At first, there was too much light lost in the process, so Hirschowitz, along with physics Professor C.W. Peters and student Larry Curtiss, decided to refine the technology. They discovered that, if they coated the fibres in another glass, loss of light was completely eliminated because the coating created a condition of total internal reflection where the light being transmitted was bounced off the glass cladding. This also meant the signal would still be transmitted if the fibre was curved. Hirschowitz and his co-workers created a working prototype in 1957 and he used it to view the inside of his own stomach. In 1960, his invention – now in commercial production – was used to examine the stomach of a patient. His fellow doctors were impressed by the clarity of the images and adopted the endoscope to view other parts of the human body such as the lungs, colon and abdomen. Today, the technology can be used for more than just viewing the body – the end of the probe can include instruments, allowing doctors to treat the conditions that they find. Endoscopes can also be hooked up to video screens where medical staff can view a real time tour of your insides – a sort of medical YouTube. With capsule endoscopy, a camera, a wireless radio transmitter, four LED lights and a battery are fitted into a pill-sized unit that is swallowed by the patient. The pill moves through the digestive tract naturally and is expelled at the end in the usual manner. Images are constantly transmitted to antenna pads on the body, then to a recording unit worn around the patient’s waist, and the pictures are downloaded to a computer and strung together into a movie. Digital image-based elasto-tomography (DIET) Digital image-based elasto-tomography (DIET) uses digital cameras to image motion on the outside of the body to work out what is going on in the tissue underneath. Dr Eli van Houten (University of Canterbury) uses this technique in his research into better breast cancer screening (detection).<\|endoftext\|>	3.921875
781	# Know Your Numbers: What's Half Way? (1) In this worksheet, students find the number that is exactly half way between the two given numbers. Key stage: KS 2 Curriculum topic: Maths and Numerical Reasoning Curriculum subtopic: Place Value Difficulty level: ### QUESTION 1 of 10 There are several ways to find the number exactly halfway between two given numbers. We shall use two easy numbers, such as 60 and 70 to illustrate three different methods. Method 1 Find the difference between the two numbers, which here is 70 - 60 = 10 Halve this difference, which here is 10 ÷ 2 = 5 Add this to the smaller number, which here is 60 + 5 = 65 Method 2 Find the difference between the two numbers, which here is 70 - 60 = 10 Halve this difference, which here is 10 ÷ 2 = 5 Subtract this from the larger number, which here is 70 - 5 = 65 Method 3 Find the mean of the two numbers by adding them together and halving the result. The total is 60 + 70 = 130 Halve the result to get 130 ÷ 2 = 65 Write down the number which is halfway between: 50 and 60 Write down the number which is halfway between: 54 and 64 Write down the number which is halfway between: 540 and 640 Write down the number which is halfway between: 54 and 74 Write down the number which is halfway between: 5400 and 7400 Write down the number which is halfway between: 5000 and 7400 Write down the number which is halfway between: 5000 and 9400 Write down the number which is halfway between: 5400 and 9400 Write down the number which is halfway between: 5400 and 8400 Write down the number which is halfway between: 5400 and 10400 • Question 1 Write down the number which is halfway between: 50 and 60 55 • Question 2 Write down the number which is halfway between: 54 and 64 59 • Question 3 Write down the number which is halfway between: 540 and 640 590 • Question 4 Write down the number which is halfway between: 54 and 74 64 • Question 5 Write down the number which is halfway between: 5400 and 7400 6400 • Question 6 Write down the number which is halfway between: 5000 and 7400 6200 • Question 7 Write down the number which is halfway between: 5000 and 9400 7200 • Question 8 Write down the number which is halfway between: 5400 and 9400 7400 • Question 9 Write down the number which is halfway between: 5400 and 8400 6900 • Question 10 Write down the number which is halfway between: 5400 and 10400 7900 ---- OR ---- Sign up for a £1 trial so you can track and measure your child's progress on this activity. ### What is EdPlace? We're your National Curriculum aligned online education content provider helping each child succeed in English, maths and science from year 1 to GCSE. With an EdPlace account you’ll be able to track and measure progress, helping each child achieve their best. We build confidence and attainment by personalising each child’s learning at a level that suits them. Get started<\|endoftext\|>	4.625
755	Most people with an interest in conservation have heard by now about the ordeal vultures are going through worldwide. The decrease in their populations has been dramatically visible in India, where almost all vulture species suffered big losses as a consequence of the administration of the anti-inflammatory drug Diclofenac to treat cattle, which is fatal for these birds. As a result, 99.7% of the population of White-rumped Vulture (Gyps bengalensis) has disappeared in less than 10 years, making it the fastest collapse in any avian population ever recorded in history. Overall, it’s estimated that over 95-99% of Indian vultures have disappeared in the last 2 decades. The joint efforts of the scientific community, NGOs and government have managed to stop this decline, and the actions undertaken to revert the crisis have now started to show some encouraging results. Even so, some experts believe vulture populations might never fully recover from this catastrophe. The disappearance of a species from an ecosystem creates an “ecological vacuum”, an unstable state in which other species will step in to take over the extinct species’ role in the environment and proliferate until the system is balanced again. This situation alters the whole ecosystem at every level, and the changes are even more extreme following the loss of a key species such as vultures. Unfortunately, it possible that by the time vulture populations are on the road to recovery, the ecosystem would have undergone an ecological re-arrangement and no longer be suitable to sustain the same numbers as it did before. In India, the outcome of the “vulture crisis” is clearly visible: The decrease in vulture populations caused an explosion in the numbers of small scavengers such as rats and feral dogs, which are not as efficient at disposing of carcasses as vultures, and unlike these birds can carry diseases from corpses such as rabies or anthrax, making them an important vector for the outbreak of infections in humans. As of 2015, the cost of the decrease in Indian vulture populations has been estimated at around 25 billion USD. Another consequence of this vulture decline impacted one of the world´s oldest cultures: One of the ancient Zoroastrian culture’s best known customs is their traditional “sky burial”, where the bodies of deceased Parsis are brought to the tower of death, where vultures consume the dead and leave the bones untouched, which are then retrieved and buried. After the disappearance of vultures from the Indian skies, bodies were not able to be disposed of this way. This has forced the Parsi community to start considering alternatives to the sky burial; a century-old and sustainable tradition that might now be lost. More information here: http://www.newindianexpress.com/nation/With-vultures-gone-Mumbai-Parsis-turn-to-cremations/2016/06/07/article3470549.ece Despite all the information out there, sadly most people have never heard how Diclofenac affects these birds, or even know that vultures are on the brink of extinction all across their distribution range. Moreover, vultures still have one of the worst – yet undeserved – reputations in the animal kingdom. Ugly, disgusting, malevolent, foul…these are but a few of the adjectives most commonly used to describe them. One of the main goals of organizations such as the VCF is to help turn this negative attitude towards these incredible birds, and raise awareness on their invaluable role as “nature’s waste managers”.<\|endoftext\|>	3.765625
598	# How Do I Figure the Interest Rate on a Loan? by Ellis Davidson Most loans are advertised with a set interest rate, which can be fixed or variable over the lifetime of a loan. However, some loans are structured such that a fixed dollar amount is loaned at the beginning of the loan period, and another fixed dollar amount is due to service the loan at the end of that period. These loans can be compared with interest-based structures by calculating the interest rate that results in the amount due. 1. Divide the amount of the additional payment by the amount loaned to determine the simple interest rate. For example, consider a loan of \$1,000, which must be repaid in one year with the amount of \$1,300. This is \$300/\$1,000, or 30 percent per year, which is a hefty interest rate. However, if the loan was \$10,000, and required a repayment of \$10,300, that is an interest rate of \$300/\$10,000, or 3 percent per year, which is extremely low. 2. Calculate the compound interest rate, in which you are paying interest on both the amount of the loan and the interest accrued, by using exponentials. As the interest rate is charged against the total balance each period, the formula is (1 + Interest Rate)^Periods. For example, we have determined that \$300 on a \$1,000 loan is 30 percent in simple interest. To compare this with a credit card, we must determine the compound interest rate. First, determine the number of compounding periods over the lifetime of the loan; in the example, this is 365 (days in a year). This yields the following algebraic formula, where X is the daily interest rate: 1,000 * (1 + X)^365 = 1,300 (1 + X)^365 = 1.3 (1 + X) = 1.000719 X = .000719 This can be compared with the daily interest rate on a credit card statement, or multiplied by 365 to compare it to the credit card's Annual Percentage Rate (APR). This shows that the APR of this loan is 26.2 percent, but after compounding, it results in total paid interest of 30 percent. 3. Determine the cost of a short-term loan by expanding it over the course of a year. For example, a payday loan store may charge \$50 for a \$1,000 loan, due in two weeks' time. This is a rate of 5 percent for the two weeks, but once those two weeks are multiplied by 26 (the number of two-week periods in a year), the simple interest is shown to be 130 percent--a very expensive loan. If additional fees and interest are levied after the initial two week period, this interest rate can skyrocket even higher.<\|endoftext\|>	4.5
591	The original purpose of the British Canal Network was for commerce and travel. At the time roads were just being constructed and were very muddy. Being so muddy made the roads hard to travel so for any type of mass transit the canals were the safer and faster route to take. The first canals were actually constructed in the Roman times. These were mainly used as land drainage canals or for irrigation purposes. However, there were short canals that connected the rivers that could be navigated. In the Middle Ages the canals were improved so that building materials for various projects could be transported in safety. During this period castles and monasteries were built in abundance which meant that having the canals were more important than in previous years. With the improvement of the canals the government began regulating the goods transported on the waterways. During the 16th century is when the major improvements of the canal navigation system began. In fact 29 major navigation improvements for the riverways took place between the 16th and 17th centuries. This set the stage for the Industrial Revolution that began in the late 18th century. The Industrial Revolution brought the modern canal system into being, mainly due to the demand for a economical way to safely transport goods in large quantities. Locks were installed in abundance to help travelers and goods avoid the dangerous rivers or rivers that were simply difficult to navigate. It was during this time that the idea of a “pure” canal came into being. Instead of letting where a river was located determine how goods were transported the “pure” canals could be constructed so that the goods could be shipped where they needed to go in the least amount of time. There are two canals that are debated about as having claim to being the first “pure” canal. One is the Sankey Brook Navigation and the other is the Bridgewater Canal. Transportation along the canals has changed over time as well. One of the first ways that boats were moved up and down the river was by having horses pull the boats along. This method of moving the boats was highly successful. One horse could pull a boat that was carrying about thirty tons of weight. With such a high rate of success this method was actually used until the 1950s but with the introduction of steam-powered boats, it was soon obsolete. In the beginning there were no families that lived and worked on the canal boats. Instead the crews were all male and their families occupied homes along the banks. This continued until after World War One. During the late 19th century wives and children began coming aboard to provide extra labor and reduce the costs of extra hires and rent for their homes. From their meager beginnings the canals have evolved to provide not only recreation but jobs for many people. Even though some are artificially made they still have served a purpose and without them who knows how different the transportation of goods would have been over the centuries.<\|endoftext\|>	3.921875
588	You must remember that “Digestion” has a specific meaning in Biology. It is the term used for the process that involves the chemical breakdown of large, insoluble food molecules into smaller, simpler molecules that can be absorbed into the blood. Many of the molecules in food are polymers – that is macromolecules made from long chains of repeating subunits. Examples of dietary macromolecules include proteins, polysaccharides and fats. These molecules are too large to be able to pass into the blood in the villi of the small intestine and so the body has evolved to chemically break them down into their constituent monomers or building blocks. Digestion is the process in the alimentary canal that achieves this. Digestion reactions are also known as hydrolysis reactions because a molecule of water is required in the reaction to break the covalent bond holding the monomers together. These reactions are all catalysed (sped up) by specific molecules called digestive enzymes. Why do different food types need different digestive enzymes to speed up their breakdown in the digestive system? (If you are unsure, you need to revise the way enzymes work to catalyse reactions by a “lock and key” theory?) Digestion of Carbohydrates Many simple carbohydrates (e.g. glucose) do not need digesting. This is because they are already small enough to be absorbed into the blood directly in the ileum (small intestine). But larger disaccharide sugars (e.g. maltose and sucrose) do need to be broken down, as do all polysaccharides (e.g starch). The family of enzymes that break down carbohydrates are called carbohydrases. Starch is a large polysaccharide made up of many hundreds of glucose residues linked together. It is way too big to be able to cross the epithelial lining of the small intestine and so needs to be digested. This happens in a two-stage process. Firstly there is an enzyme amylase that can catalyse the following reaction: starch + water ——-> maltose Amylase is made in the salivary glands and so works in the mouth. But the main region for the digestion of starch is in the duodenum. This is because amylase is also made in the pancreas. Maltose is a disaccharide molecule made of two glucose residues joined together. Maltose itself requires digesting to its constituent glucose molecules in order to be absorbed. So the second stage in the digestion of starch involves a second enzyme, maltase that is found embedded into the epithelial lining of the ileum. Maltase catalyses the breakdown of a molecule of maltose into two molecules of glucose which can be absorbed into the blood. maltose + water ——> glucose<\|endoftext\|>	4.125
1,715	### The Study of Vibrations - Examples Example 1.1. A spring is stretched 100 mm by a 10 kg block. If the block is displaced 50 mm downward from its equilibrium position and given a downward velocity of 3 m/s determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t=4 s. ## Figure 1.1 – a) Model of a system, b) Free body diagrams for external and effective forces Solution: From previous figure we can derive the differential equation. \begin{align} & +\downarrow \sum{{{F}_{y}}}=m{{a}_{y}} \\ & mg-k\left( y+{{y}_{st}} \right)=m\ddot{y} \\ \end{align} where $$k{{y}_{st}}=mg$$ Now we need to substitute the previous equation into the differential equation. With this operation we get: \begin{align} & mg-ky-mg=m\ddot{y}, \\ & \ddot{y}+\frac{k}{m}y=0, \\ \end{align} Before we can determine the natural frequency of the system we need to determine the stiffness of the spring. We know that the force the stretch the spring is proportion to the stiffness of the spring and the elongation of the spring. In this case we don’t have a force which acts on spring but we have a block which is attached to the sprig. So we will determine the stiffness of the spring by following expression \begin{align} & F=ky,F=G=mg \\ & k=\frac{G}{y}=\frac{mg}{y}=\frac{10\cdot 9.81}{0.1}=981\text{ N/m}{{\text{m}}^{2}} \\ \end{align} The next step is to determine the natural frequency and we will do that by following expression: $$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{981}{10}}=9.9045\text{ }{{\text{s}}^{-1}}$$ When we have determined the natural frequency we can insert it into differential equation. Whit this operation we get: \begin{align} & \ddot{y}+{{(9.9045)}^{2}}y=0, \\ & \ddot{y}+98.1y=0, \\ \end{align} All that is left is to apply the boundary conditions. At $$t=0,y=0,05$$ By applying this boundary conditions we get: \begin{align} & 0.05=A\sin 0+b\cos 0 \\ & b=0.05 \\ \end{align} At $$t=0,{{v}_{0}}=3m/s$$ \begin{align} & {{v}_{0}}=A\omega \cos 0-0=A \\ & A=\frac{{{v}_{0}}}{\omega }=\frac{3}{9.9045}=0.30289 \\ \end{align} Example 1.2. A 3 kg block is suspended from a spring having a stiffness of k=200N/m. If the block is pushed 50 mm upward from its equilibrium position and then released from rest, determine the equation that describes the motion. What are the amplitude and the frequency of the vibration? Assume that the positive displacement is downward. From the given data we can determine the frequency of the vibrations. \begin{align} & \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{3}}=8.165 \\ & f=\frac{\omega }{2\pi }=\frac{8.165}{2\pi }=1.299=1.3Hz \\ & x=A\sin \omega t+B\cos \omega t, \\ \end{align} Boundary conditions are: \begin{align} & t=0,x=-0.05, \\ & t=0,v=0 \\ \end{align} By applying the boundary condition to the solution of the differential equation we get: \begin{align} & 0.05=A\sin 0+B\cos 0, \\ & b=-0.05. \\ & v=A\omega \cos \omega t-B\omega \sin \omega t, \\ & 0=A\omega \cos 0-B\omega \sin 0 \\ & A=0 \\ \end{align} In order to determine the amplitude of the system we will apply the following equation. $$C=\sqrt{{{A}^{2}}+{{B}^{2}}}=\sqrt{2.5\cdot {{10}^{-3}}}=0.05m=50mm$$ Example 1.3 A spring has a stiffness of 800 N/m. If a 2 kg block is attached to the spring, pushed 50 mm above its equilibrium position, and released from rest, determine the equation that describes the block’s motion. Assume that positive displacement is downward. \begin{align} & \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{800}{2}}=20{{s}^{-1}}, \\ & y=A\sin \omega t+B\cos \omega t, \\ \end{align} For t=0, y=-0.05 m \begin{align} & -0.05=A\sin 0+B\cos 0, \\ & B=-0.05 \\ \end{align} \begin{align} & y=A\sin \omega t+B\cos \omega t, \\ & v=A\omega \cos \omega t-B\omega \sin \omega t, \\ \end{align} For t=0, v=0 \begin{align} & v=A\omega \cos \omega t-B\omega \sin \omega t, \\ & 0=A\omega \cos 0-B\omega \sin 0, \\ & A=0 \\ \end{align} Thus we get: $$y=-0.05\cos \left( 20t \right)$$ Example 1.4. A spring is stretched 200 mm by a 15 kg block. If the bloc is displaced 100 mm downward from its equilibrium position and given downward velocity of 0.75 m/s determine the equation which describes the motion. What is a phase angle? Assume that positive displacement is downward \begin{align} & F=ky\Rightarrow k=\frac{F}{y}=\frac{15\cdot 9.81}{0.2}=735.75N/m \\ & \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{735.75}{15}}=7{{s}^{-1}} \\ & y=A\sin \omega t+B\cos \omega t \\ \end{align} When t=0 then y=0.1 m and When t=0 then v=0.75 m/s \begin{align} & y=A\sin \omega t+B\cos \omega t \\ & 0.1=A\sin 0+B\cos 0 \\ & B=0.1m \\ \end{align} \begin{align} & v=A\omega \cos \omega t-B\omega \sin \omega t \\ & 0.75=A\omega \cos 0-B\omega \sin 0 \\ & 0.75=A\omega \\ & A=\frac{0.75}{7}=0.107 \\ \end{align} Now we can insert the values of A and B into the solution. \begin{align} & y=0.107\sin (7t)+0.100\cos (7t) \\ & \phi =arctan\left( \frac{B}{A} \right)=\arctan \left( \frac{0.100}{0.107} \right)={{43.0}^{\circ }} \\ \end{align}<\|endoftext\|>	4.65625
774	# How to Define a Zero and Negative Exponent / / درس 6 ### توضیح مختصر The zero and negative exponent properties are two you will use quite a lot in mathematics. The negative exponent property can be confusing, but when you remember a couple fun ideas, you will get it right every time! • زمان مطالعه 3 دقیقه • سطح متوسط ### دانلود اپلیکیشن «زوم» این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید ## Formulas for negative exponents In math, we like to write exponents with a positive number. So what happens if I get a negative exponent? What about a zero exponent? Before we get started, I need to tell you something important here: x ^- a does not mean - x ^ a . The negative exponent has nothing to do with positive or negative numbers. Let’s look at the two formulas. x ^- a = 1/( x ^ a ) 1/( x ^- a ) = x ^ a I want to make sure that you understand that x ^- a does not mean - x ^ a . Once again, the negative exponent has nothing to do with positive or negative numbers. If the exponential is negative in the numerator, or the top, it tells us the exponential is actually positive in the denominator. If the exponential is negative in the denominator, or the bottom, it tells us the exponential is actually positive in the numerator. If you see a negative exponent, flip it to a positive. That is, if the exponent is negative in the numerator, flip it positive to the denominator. If the exponent is negative in the denominator, flip it positive to the numerator. Let’s say I have x ^-4. Now remember, x ^-4 can be written as a fraction (x^-4)/1. Remember, if it’s negative in the numerator, you flip it positive to the denominator. So that’s the same thing as saying 1/(x^4). In this next example, x ^-7 is in the denominator. We’re going to flip it positive to the numerator. x ^7 is the same thing as 1/( x ^-7). ## The formula for zero exponents I think the zero exponent is really fun! Basically, the formula says that anything - any number, any letter - raised to the zero exponent is always one. So when I have x ^0, it’s 1. Let’s say I had something funny like 999,999^0. It’s still 1! The Quotient of Powers Rule says that when we divide exponentials with the same base, we subtract their exponents. So, I have ( x ^3)/( x ^3). Remember, we subtract their exponents. So that would be x ^(3-3), which is x^0. But where does the 1 come from? We have ( x ^3)/( x ^3). That is, ( x x x ) on the top, or numerator, and ( x x x ) on the bottom, or denominator. x / x is 1, x / x is 1, and x / x is 1. When we multiply those together (1 1 1), we get 1! So that pretty much proves it: x ^0 = 1! ## Lesson Objectives By the end of this lesson you’ll understand the zero and negative exponent properties. ### مشارکت کنندگان در این صفحه تا کنون فردی در بازسازی این صفحه مشارکت نداشته است. 🖊 شما نیز می‌توانید برای مشارکت در ترجمه‌ی این صفحه یا اصلاح متن انگلیسی، به این لینک مراجعه بفرمایید.<\|endoftext\|>	4.53125
853	# How do I solve for m and n? mx-y=23 nx+y=12 hala718 \| High School Teacher \| (Level 1) Educator Emeritus Posted on mx - y = 23.............(1) nx + y = 12............(2) Let us add (1) and (2): ==> nx + mx = 35 ==> (n+m) x = 35 ==> n + m = 35/ x ==> n = 35/x - m But : mx - y = 23 ==> m = (y+ 23)/x ==> n = 35/x - (y+23)/ x => m = 35/x - n But : nx + y = 12 ==> n = 12-y)/ x ==> m = (35/x) - (12-y)/x justaguide \| College Teacher \| (Level 2) Distinguished Educator Posted on We have to find m and n using the equations : mx - y = 23...(1) nx + y = 12...(2) Now we see that there are 4 terms m, n, x and y in the two equations given. So we can only write 2 of them in terms of the others. Using (1), we can write, m in terms of x and y as mx - y = 23 => m = (23 + y) / x Using (2), we can write n in terms of x and y as nx + y = 12 => n = (12 - y) / x Therefore m and n in terms of x and y are m=(23 + y)/x and n=(12 - y)/x givingiswinning \| Student, Grade 10 \| (Level 1) Valedictorian Posted on mx - y = 23 mx = y + 23 `m = (y + 23)/x` The second one : nx + y= 12 nx = 12 - y ` n = (12 - y)/x ` atyourservice \| Student, Grade 11 \| (Level 3) Valedictorian Posted on mx-y=23 the first step is to add y mx = 23 + y divide by x because were are trying to solve for m `x = 23/x + y/x ` this can also be written as `x = (23 + y) / x` nx+y=12 subtract y nx = 12 - y divide by x `n = 12/x - y/x` which can also be written as `n=(12-y)/x` Wiggin42 \| Student, Undergraduate \| (Level 2) Valedictorian Posted on mx-y=23 nx+y=12 To solve for a variable in terms of another variable means to isolate the variable you want from the rest. Lets deal with the first one : mx - y = 23 mx = y + 23 m = (y + 23)/x Now the second one : nx + y= 12 nx = 12 - y n = (12 - y)/x neela \| High School Teacher \| (Level 3) Valedictorian Posted on To solve for m and n in equations: mx-y=23...(1). nx+y=12...(2). We can treat both the graphs as independent, as there are no relationship given between the graphs at (1) and (2). Both equations represent different straight lines. From the first equation, mx-y = 23, mx = 23+y m = (23+y)/x is the solution for m. From the 2nd equation nx+y = 12. we get: nx = 12-y n = (12-y)/x. Therefore the solution for n and m are m = (23+y)/x and n = (12-y)/x.<\|endoftext\|>	4.53125
542	# 2001 AMC 8 Problems/Problem 18 ## Problem Two dice are thrown. What is the probability that the product of the two numbers is a multiple of 5? $\text{(A)}\ \dfrac{1}{36} \qquad \text{(B)}\ \dfrac{1}{18} \qquad \text{(C)}\ \dfrac{1}{6} \qquad \text{(D)}\ \dfrac{11}{36} \qquad \text{(E)}\ \dfrac{1}{3}$ ## Solution 1 This is equivalent to asking for the probability that at least one of the numbers is a multiple of $5$, since if one of the numbers is a multiple of $5$, then the product with it and another integer is also a multiple of $5$, and if a number is a multiple of $5$, then since $5$ is prime, one of the factors must also have a factor of $5$, and $5$ is the only multiple of $5$ on a die, so one of the numbers rolled must be a $5$. To find the probability of rolling at least one $5$, we can find the probability of not rolling a $5$ and subtract that from $1$, since you either roll a $5$ or not roll a $5$. The probability of not rolling a $5$ on either dice is $\left(\frac{5}{6} \right) \left(\frac{5}{6} \right)=\frac{25}{36}$. Therefore, the probability of rolling at least one five, and thus rolling two numbers whose product is a multiple of $5$, is $1-\frac{25}{36}=\frac{11}{36}, \boxed{\text{D}}$ ## Solution 2 The only way to get a multiple of 5 is to have at least one 5. If the first dice rolls a 5, there are 6 ways to get a multiple of 5. If the second dice rolls a 5, there are also 6 ways. However, there is one case that is repeated: both dice roll a 5. Therefore, there are 6 + 6 - 1 = 11, and there is a total of 6 x 6 ways, so the probability is $\text{(D)}\ \dfrac{11}{36}$ Solution by ILoveMath31415926535 ## Video Solution https://youtu.be/4aX9-DZHgNw Soo, DRMS, NM<\|endoftext\|>	4.78125

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