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finemath20k-over-20k-l3b-tokens

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https://nrich.maths.org/public/topic.php?code=-68&cl=3&cldcmpid=11231	1,571,545,295,000,000,000	text/html	crawl-data/CC-MAIN-2019-43/segments/1570986702077.71/warc/CC-MAIN-20191020024805-20191020052305-00243.warc.gz	626,500,172	10,079	# Search by Topic #### Resources tagged with Visualising similar to Polygon Rings: Filter by: Content type: Age range: Challenge level: ### Polygon Rings ##### Age 11 to 14 Challenge Level: Join pentagons together edge to edge. Will they form a ring? ### Tessellating Hexagons ##### Age 11 to 14 Challenge Level: Which hexagons tessellate? ### Getting an Angle ##### Age 11 to 14 Challenge Level: How can you make an angle of 60 degrees by folding a sheet of paper twice? ### Convex Polygons ##### Age 11 to 14 Challenge Level: Show that among the interior angles of a convex polygon there cannot be more than three acute angles. ### Right Time ##### Age 11 to 14 Challenge Level: At the time of writing the hour and minute hands of my clock are at right angles. How long will it be before they are at right angles again? ### On Time ##### Age 11 to 14 Challenge Level: On a clock the three hands - the second, minute and hour hands - are on the same axis. How often in a 24 hour day will the second hand be parallel to either of the two other hands? ### Rati-o ##### Age 11 to 14 Challenge Level: Points P, Q, R and S each divide the sides AB, BC, CD and DA respectively in the ratio of 2 : 1. Join the points. What is the area of the parallelogram PQRS in relation to the original rectangle? ### Khun Phaen Escapes to Freedom ##### Age 11 to 14 Challenge Level: Slide the pieces to move Khun Phaen past all the guards into the position on the right from which he can escape to freedom. ### Dice, Routes and Pathways ##### Age 5 to 14 This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . . ### Zooming in on the Squares ##### Age 7 to 14 Start with a large square, join the midpoints of its sides, you'll see four right angled triangles. Remove these triangles, a second square is left. Repeat the operation. What happens? ### Cubist Cuts ##### Age 11 to 14 Challenge Level: A 3x3x3 cube may be reduced to unit cubes in six saw cuts. If after every cut you can rearrange the pieces before cutting straight through, can you do it in fewer? ### Travelling Salesman ##### Age 11 to 14 Challenge Level: A Hamiltonian circuit is a continuous path in a graph that passes through each of the vertices exactly once and returns to the start. How many Hamiltonian circuits can you find in these graphs? ### Triangle Inequality ##### Age 11 to 14 Challenge Level: ABC is an equilateral triangle and P is a point in the interior of the triangle. We know that AP = 3cm and BP = 4cm. Prove that CP must be less than 10 cm. ### An Unusual Shape ##### Age 11 to 14 Challenge Level: Can you maximise the area available to a grazing goat? ##### Age 11 to 14 Challenge Level: Four rods, two of length a and two of length b, are linked to form a kite. The linkage is moveable so that the angles change. What is the maximum area of the kite? ### Tetra Square ##### Age 11 to 14 Challenge Level: ABCD is a regular tetrahedron and the points P, Q, R and S are the midpoints of the edges AB, BD, CD and CA. Prove that PQRS is a square. ### Framed ##### Age 11 to 14 Challenge Level: Seven small rectangular pictures have one inch wide frames. The frames are removed and the pictures are fitted together like a jigsaw to make a rectangle of length 12 inches. Find the dimensions of. . . . ### Coordinate Patterns ##### Age 11 to 14 Challenge Level: Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead? ### Take Ten ##### Age 11 to 14 Challenge Level: Is it possible to remove ten unit cubes from a 3 by 3 by 3 cube so that the surface area of the remaining solid is the same as the surface area of the original? ### Concrete Wheel ##### Age 11 to 14 Challenge Level: A huge wheel is rolling past your window. What do you see? ### Hidden Rectangles ##### Age 11 to 14 Challenge Level: Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard? ### The Old Goats ##### Age 11 to 14 Challenge Level: A rectangular field has two posts with a ring on top of each post. There are two quarrelsome goats and plenty of ropes which you can tie to their collars. How can you secure them so they can't. . . . ### Pattern Power ##### Age 5 to 14 Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create. ### All in the Mind ##### Age 11 to 14 Challenge Level: Imagine you are suspending a cube from one vertex and allowing it to hang freely. What shape does the surface of the water make around the cube? ### Cubes Within Cubes ##### Age 7 to 14 Challenge Level: We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used? ### Paving Paths ##### Age 11 to 14 Challenge Level: How many different ways can I lay 10 paving slabs, each 2 foot by 1 foot, to make a path 2 foot wide and 10 foot long from my back door into my garden, without cutting any of the paving slabs? ### Coloured Edges ##### Age 11 to 14 Challenge Level: The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set? ### Christmas Boxes ##### Age 11 to 14 Challenge Level: Find all the ways to cut out a 'net' of six squares that can be folded into a cube. ### Sea Defences ##### Age 7 to 14 Challenge Level: These are pictures of the sea defences at New Brighton. Can you work out what a basic shape might be in both images of the sea wall and work out a way they might fit together? ### Bands and Bridges: Bringing Topology Back ##### Age 7 to 14 Lyndon Baker describes how the Mobius strip and Euler's law can introduce pupils to the idea of topology. ##### Age 11 to 14 Challenge Level: Can you mark 4 points on a flat surface so that there are only two different distances between them? ### 3D Stacks ##### Age 7 to 14 Challenge Level: Can you find a way of representing these arrangements of balls? ### Auditorium Steps ##### Age 7 to 14 Challenge Level: What is the shape of wrapping paper that you would need to completely wrap this model? ### Troublesome Dice ##### Age 11 to 14 Challenge Level: When dice land edge-up, we usually roll again. But what if we didn't...? ### Squares in Rectangles ##### Age 11 to 14 Challenge Level: A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all? ### Christmas Chocolates ##### Age 11 to 14 Challenge Level: How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes? ### Chess ##### Age 11 to 14 Challenge Level: What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board? ### Dissect ##### Age 11 to 14 Challenge Level: What is the minimum number of squares a 13 by 13 square can be dissected into? ### Eight Hidden Squares ##### Age 7 to 14 Challenge Level: On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares? ### Tic Tac Toe ##### Age 11 to 14 Challenge Level: In the game of Noughts and Crosses there are 8 distinct winning lines. How many distinct winning lines are there in a game played on a 3 by 3 by 3 board, with 27 cells? ### There and Back Again ##### Age 11 to 14 Challenge Level: Bilbo goes on an adventure, before arriving back home. Using the information given about his journey, can you work out where Bilbo lives? ### Crossing the Atlantic ##### Age 11 to 14 Challenge Level: Every day at noon a boat leaves Le Havre for New York while another boat leaves New York for Le Havre. The ocean crossing takes seven days. How many boats will each boat cross during their journey? ### Buses ##### Age 11 to 14 Challenge Level: A bus route has a total duration of 40 minutes. Every 10 minutes, two buses set out, one from each end. How many buses will one bus meet on its way from one end to the other end? ### Counting Triangles ##### Age 11 to 14 Challenge Level: Triangles are formed by joining the vertices of a skeletal cube. How many different types of triangle are there? How many triangles altogether? ### Soma - So Good ##### Age 11 to 14 Challenge Level: Can you mentally fit the 7 SOMA pieces together to make a cube? Can you do it in more than one way? ### Drilling Many Cubes ##### Age 7 to 14 Challenge Level: A useful visualising exercise which offers opportunities for discussion and generalising, and which could be used for thinking about the formulae needed for generating the results on a spreadsheet. ### Screwed-up ##### Age 11 to 14 Challenge Level: A cylindrical helix is just a spiral on a cylinder, like an ordinary spring or the thread on a bolt. If I turn a left-handed helix over (top to bottom) does it become a right handed helix? ### Cubes Within Cubes Revisited ##### Age 11 to 14 Challenge Level: Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need? ### Icosian Game ##### Age 11 to 14 Challenge Level: This problem is about investigating whether it is possible to start at one vertex of a platonic solid and visit every other vertex once only returning to the vertex you started at. ### Seven Squares ##### Age 11 to 14 Challenge Level: Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100?	2,394	10,023	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2019-43	latest	en	0.866582
https://www.computing.dcu.ie/~humphrys/Notes/Neural/ml.html	1,540,257,775,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583515564.94/warc/CC-MAIN-20181023002817-20181023024317-00527.warc.gz	888,333,806	11,711	Dr. Mark Humphrys School of Computing. Dublin City University. My big idea: Ancient Brain Search: ``` ``` # What is Learning? It must be more than just memory: ```Human: The capital of Ireland is Dublin. Machine: OK. Human: The capital of France is Paris. Machine: OK. Human: What is the capital of Ireland? Machine: Dublin. Human: Brilliant. ``` The machine should learn something new. ``` ``` # Generalisation - Learning a function f The problem of induction: We have experience of a finite number of situations. We must generalise our experience to a new situation, where we have never been told the answer, and make a good guess, where we are not simply recovering a past answer from memory. ```Human: In situation x1, answer is a1. Human: In situation x2, answer is a2. Human: In situation x3, answer is a3. Human: In situation x4, what is answer? Machine: f(x4). ``` where the machine is "learning" the function f. Each Input-Output pair (xi, ai) that we show to the machine is called an exemplar. Learning from exemplars like this is called supervised learning. ``` ``` f(x4) might be an interpolation (if x4 is between 2 of the points x1,..,x3) or an extrapolation (if x4 is outside the previous points). For example, the machine might progress as follows: ```Human: In situation x1, answer is a1. Machine: OK. I am building a function f(x). Currently, f(x) = a1 for all x. Human: In situation x2, answer is a2. Machine: f is becoming a more complex mapping, that gives different results for different x. e.g. Perhaps currently f((x1+x2)/2) = (a1+a2)/2. ``` ## Question - If we have 2 points so far, we may (so far) assume f is a straight line. What is the equation of f(x)? ```Human: In situation x3, answer is a3. Machine: f is getting more complex. ``` Maybe now f is no longer a straight line. We still may estimate f((x1+x2)/2) = (a1+a2)/2, but from now we restrict ourselves to interpolations rather than extrapolations. ```Human: In situation x4, what is answer? Machine: Using my current definition of f, my answer is f(x4). ``` ``` ``` ```If x=0, answer is 0. If x=100, answer is 100. If x=50, answer is 95. If x=25, what is answer? ``` ``` ``` # Representing a function ``` ``` The idea of Representing a function from exemplars is that "nearby" Input should generate "nearby" Output. But some functions defeat this simple idea: ``` ``` ``` ``` ``` ``` ``` ``` ``` ``` # Non-chaotic functions (can be approximated to good accuracy by finite data structure) We do not expect that the learning method from exemplars will be able to learn to represent a chaotic function (or a function in a region where it is behaving chaotically). To represent the chaotic area the function would have to be learnt to an impractically (perhaps infinitely) high level of granularity. However we do hope that our unknown function is not actually perverse enough to be chaotic, at least in some local regions. • For instance, if you view our Chaos Theory demo function as a function in λ as well as x, then there are regions of that multi-dimensional space in which it is not chaotic. • y = f(x,λ) x and λ are 2 dimensions y is the 3rd dimension We expect that the learning method can cope with any reasonably well behaved non-linear, continuous function. For instance, the function: ```f(x) = sin(x) + sin(2x) + sin(5x) + cos(x) ``` Here is the kind of thing we want - a learning method approximating the above function after having seen 1 million exemplars (top) and 5 million (bottom): ``` ``` ``` ``` See details later. ``` ``` # Fractals Fractals - The same pattern repeats at different scales. Repeatedly magnify it and it looks the same. Here we could possibly represent it to reasonable accuracy (not perfect) using a finite data structure, because the range decreases. A small change does not bounce across the global range, as in the chaotic functions above. e.g. Mandelbrot set: ``` ``` Public domain image from here. ``` ``` # Equation for f v. No equation for f Note that if we have an explicit equation for f, as above, then we don't need supervised learning at all! - Represent f by equation. Supervised learning from exemplars is entirely for situations where we have an unknown f - we simply postulate that some mapping from our inputs to our outputs must exist. e.g. Hand-writing recognition. - Represent f by adaptive data structure. So why do I use explicit f? Because they make nice examples. Use them to imagine what an unknown f might look like. With explicit f, we can see how well the machine is doing. Use explicit f to debug your learning machine when first write it. ``` ``` # Unknown or postulated function A typical application for which there is a learnable Input-Output relationship: • Input is multi-dimensional x representing a bitmap of hand-written character. • Output is one of 10 values (0-9) or perhaps one of 62 values (a-z, A-Z, 0-9) or more. • One of the things that makes this problem easier is the small number of possible outputs. We do not have to know exactly that some input is a "3". We only have to decide if it is more likely to be a "3" than an "8". We have to decide which of a small number of predefined categories it is most likely to go into. • In the input, how would you represent n x n pixels? See how BMP format has a pixel array plus other data. ``` ``` ### Equation for f v. Can represent f No equation for f, but can represent it: It is interesting to think that even though we can never hope to explicitly define the function mapping a particular hand-drawn bitmap to a value 0-9, we can still learn this correlation. Equation for f, but can't represent it: Whereas with the Chaos Theory demo function, we actually know explicitly what the function is, and it is completely deterministic, yet we will not be able to learn to represent it. Conclusion: It is not whether a mapping is knowable explicitly or not that makes it learnable or not. ``` ``` # Other applications "Pattern recognition" can mean lots of things. It could mean learning a mapping from Input "state" to Output "behaviour". e.g. Input state = (wall at certain angle, velocity at certain value), Output behaviour = one of (turn left, turn right, stop, keep going). It means recognising a dangerous situation on the road. Medical diagnosis. Financial prediction. It means learning to walk, balance, run, jump, climb. Speech (both hearing and production). Vision. Anything where explicit rules are often hard to come by, but correct feedback is at least possible to construct. ``` ``` # Sample applications of Supervised learning Neural Networks are one of the main types of machines that can implement supervised learning. ``` ``` # Optical character recognition (OCR) Optical character recognition (OCR) of printed books should be easy (compared to handwriting) but is often surprisingly hard. ``` ``` Page of an 18th century book. A typically inaccurate OCR transcript of the above. Note the points where it has trouble. • There is nothing to show the program the box where a letter starts and ends. • The 18th century "long s" does not help. ``` ``` # Discontinuous function Some mappings of Input to Output - where there is only a disjoint, discrete mapping of certain inputs to certain outputs, and in no areas any continuous (or even approximation of continuous) mapping - do not lend themselves to learning, only memorisation. e.g. Capitals: ```Human: The capital of Ireland is Dublin. Machine: OK. Human: The capital of France is Paris. Machine: OK. Human: What is the capital of the UK (the country between Ireland and France)? Machine: Emm, my guess is "Dublaris". Human: What is the capital of Iceland? Machine: Emm, "Dublin"? ``` There may be a function in that every argument has a unique value generated for it. But it is not continuous - no interpolation is possible. Do we know a priori if our function f is of this sort? i.e. Is there anything to learn? ``` ``` # Have we listed all the inputs to construct a predictable function? Remember f might be simply a postulated function - a function that is assumed to exist, but whose properties are unknown (and possibly unknowable): ```With jockey of weight 8 stone, our horse won the race. With jockey of weight 8.5 stone, our horse lost the race. With jockey of weight 9 stone, our horse won the race. With jockey of weight 8.25 stone, will our horse win or lose? With jockey of weight 8.75 stone, will our horse win or lose? ``` Are these the correct inputs? Are we missing some inputs? How do we know? ``` ``` # Probabilistic / Stochastic / Non-deterministic functions ```If x=0, answer is 0. If x=100, answer is 100. If x=0, answer is 10. If x=100, answer is 30. If x=100, answer is 100. ``` May be caused by missing inputs. In fact, must be caused by missing inputs, unless the universe as a whole is probabilistic. ``` ``` # Multi-dimensional functions If input x is multi-dimensional, how do you build up this function f? How do you represent f? f could be a function of the series of inputs x1,x2,...,xn that led up to the current input xn, rather than just a function of the immediate input xn. ``` ``` # A simple prediction machine using Distance No interpolation. No extrapolation. Just memory. ```Given a new input x, find the past exemplar with the smallest distance from x. Return the output for that past exemplar. ``` ``` ``` # Distance in Multi-dimensions If x is multi-dimensional - Then what is "nearby"? Here is one possibility. Let x = (x1,..,xn) and y = (y1,..,yn). Define the Hamming distance between x and y as: If x and y are vectors of binary digits, then the Hamming distance is just equal to the number of points at which x and y differ. Now, our algorithm could be: ```Given a new input x, find the past exemplar with the smallest Hamming distance from x. Return the output for that past exemplar. ``` ``` ``` The problem with this can be seen in the following. Let the exemplars be: ```Temperature 22 degrees, pressure 1000 millibars. No rain. Temperature 15 degrees, pressure 990 mb. Rain. ``` Then the question is: ```Temperature 17 degrees, pressure 1010 mb. Will there be rain? ``` The Hamming distance from the first exemplar is 5 + 10, the distance from the second is 2 + 20. So we say it is nearest to the first exemplar. ``` ``` However, we have just added together two different quantities - degrees and millibars! We could just as easily have said the exemplars were: ```Temperature 22 degrees, pressure 1 bar. No rain. Temperature 15 degrees, pressure 0.990 bars. Rain. ``` and the question: ```Temperature 17 degrees, pressure 1.010 bars. Will there be rain? ``` Then the distance is 5 + 0.010 from the first, 2 + 0.020 from the second. The second exemplar is closer. Which is correct? Which one are we closer to? The answer is really: If it rains, we must have been closer to the second exemplar. If it doesn't rain, we must be closer to the first. "Closer" is defined by observing the output, rather than pre-defined as in a distance metric. The same applies to any other distance metric, such as the "real" distance metric - the distance between 2 points x = (x1,..,xn) and y = (y1,..,yn) in n-dimensional Euclidean space: (e.g. consider n=2 or n=3). Again, we can change the scale of one axis without the other, and so change the definition of the "nearest" exemplar. It is just easier to show this with the Hamming distance. ``` ``` ``` ``` # Distance metric is very crude Even if we do not have the above problem of heterogenous inputs on different scales, a distance metric is still a very primitive form of predictor. e.g. Consider classifying a bitmap of 100x100 pixel images. The components of the input are all the same type of thing - pixels that are, say, 0 (black) or 1 (white). The distance metric is the number of pixels that are different. This strategy of looking at distance from exemplars is called template matching. The basic problem with it is that it weighs every dimension (every pixel) as of equal importance. Consider bitmaps of faces. We are trying to distinguish happy faces from sad faces. Say we introduce a new, happy face with a similar haircut to one of the exemplars with a sad face (i.e. we were told: "This face is sad"). The haircut area of the bitmap scores so many hits under the distance metric that this becomes the exemplar that matches, and we guess that the new input is a sad face. In fact, matching the exemplar with the same haircut / background / ears is a reasonable first guess - after all, how does the machine know in advance that it should be looking at the mouth and eyes, and ignoring the other parts? ``` ``` The problem is the system cannot learn this. The distance metric is fixed. It cannot learn that some parts of the input space are irrelevant. • You might say, if we get enough exemplars, with happy and sad faces of every haircut, then happy face with some haircut will be closest to a happy exemplar with same haircut, rather than sad exemplar with same haircut, and distance metric will work again. • Also we can try to pre-process the input to only show mouth and eyes. Of course, this assumes we know which are the most important inputs ourselves, which may be true in this simple case but not in many other problems. ``` ``` It would be more useful, and could be applied to much more problems, if the machine could learn which are the most important inputs, and learn to weigh some higher than others because they have a stronger affect on output. We just feed inputs into it. If they are irrelevant it will learn to ignore them. ``` ``` # Compare to all previous exemplars? Another issue, in the previous, is does our algorithm have to compare current input to all previous exemplars? It would be very laborious, and take a long time, to compare every pixel of the input with every pixel of a large number of previous exemplars (and it must be compared to them all, every time). And the list of exemplars to compare to keeps getting larger. ``` ``` # The Neural Network data structure Neural Networks are a type of data structure that: 1. can represent a function f from real-valued multi-dimensional input to real-valued multi-dimensional output, 2. can learn a general non-linear, continuous, non-chaotic function 3. can build up the representation of f in a variable-resolution manner, learning which components of the input are important and which are not, 4. can build up the representation from scratch based only on seeing exemplars, 5. can do this in a fixed-size data structure, that remains fixed in size as the representation improves, [Fixed memory cost] 6. can return an answer quickly to a given input without having to examine every past exemplar (i.e. the past exemplars are effectively encoded in the current definition of f - instead of having a huge list of past exemplars growing without limit), [Low computational cost] 7. can return an answer in a finite time (in fact, in exactly the same time for all inputs), [Fixed computational cost] 8. can return an answer for inputs never seen before (i.e. can make predictions), 9. is a behaviour-generating module from the start (i.e. we don't gather exemplars and then run a statistical analysis on them - instead we always have a definition of f, which can return an answer at any time). You may have heard of Neural Networks because they have been advertised as "a type of program that can learn". But they are far more than that. Traditional programs can learn too. Neural Networks have some very specific features (above) that mean they can be used for this problem. ``` ``` # What is memory? How does the brain use its experience? Example - How do you play chess? 1. Reasoning. Applying rules. Lookahead search. Memory-free. 2. Compare current state to explicit memory of all previous matches. 1. Computational cost - Gets worse with experience. Have to compare to all previous experiences, each time. Takes increasing amount of time. 2. Memory cost - Grows indefinitely. Need more and more memory as you have more experience. Data structure growing indefinitely in size. 3. Run current state through a "neural machine" that you have been building for 20 years, changing its parameters (or "weights") with every match you play. You don't remember the matches, but they have left their legacy in the current set of weights. 1. Computational cost - Input provokes an output after limited amount of computation. (In neural network, after same amount of computation for all inputs.) 2. Memory cost - More limited. Machine may grow, but not by as much as when had to remember all experiences. (In neural network, data structure is always same size. Nothing changes except the values of the weights.) Human playing chess: • does no.1 or no.3 • Amateur player - "If I move there, and then he moves there .." - no.1 • Expert player - "I don't think about the rules. I just "know" the next move." - no.3. • Brain can't really do no.2. Brain can do extremely truncated no.2 (remember some very memorable moves), plus no.1 rules for new situations. • But experts always report a no.3 layer generating suggestions (from where?) for the no.1 layer to analyse explicitly. e.g. To make a move, you get a total of 10 ideas (from where? mysterious subconscious layer) and run them through your no.1 layer to see which work. Computer playing chess: • does no.1 or no.3, and can possibly do no.2, depending on size of statespace. May do a kind of truncated no.2. • Deep Blue - no.2 (computer can explicitly examine 200 million moves a second, human brain can't, needs more like no.3 strategy) - but a truncated no.2 (even computer can't examine all possible moves) plus no.1 in support ``` ``` # Pattern recognition Similar issue in pattern recognition (like vision) as in taking actions. Do we have explicit memories? Consider this: You can easily recognise a 10 euro note (no.3 recogniser machine). But you probably cannot draw a 10 euro note (no no.2 explicit memory). ``` ``` # What could a "no.3" program look like? Imagine character recognition as a rule-based program: ```IF pixel 0 = white AND pixel 1 = black AND .... OR if pixel 0 = white AND pixel 1 = white AND .... OR if ... then return "3" ELSE if pixel 0 = white AND ... OR if .... then return "5" ELSE if ... ``` The neural network will, when finished, actually embody such a function. ``` ``` ``` ``` ancientbrain.com w2mind.org humphrysfamilytree.com On the Internet since 1987. Wikipedia: Sometimes I link to Wikipedia. I have written something In defence of Wikipedia. It is often a useful starting point but you cannot trust it. Linking to it is like linking to a Google search. A starting point, not a destination. I automatically highlight in red all links to Wikipedia and Google search and other possibly-unreliable user-generated content.	4,524	18,964	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2018-43	latest	en	0.906966
http://apicellaadjusters.com/lqd0xgy/article.php?758434=factorial-calculator-division	1,627,243,194,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046151760.94/warc/CC-MAIN-20210725174608-20210725204608-00222.warc.gz	3,931,165	11,978	Created by math nerds from team Browserling. ; Enter the value the variable is approaching. Perform the following calculation from a = 2 to Number . 5! ., 2 * 1. In the outputs section, you would see two parts. Just enter your numbers and you'll automatically get their factorials. $$10! Thus, its factorial process would be. ): - all calulcations are based on the unary system • Tape. which carries a value or 720. The last row would show the result of the mathematical operation. Online division calculator. You can calculate anything on Calculators.tech. This part shows you how the answer was calculated. Free online factorial calculator. n!!. You can use our Factorial Calculator to calculate the factorial of any real number between 0 and 5,000. To start with, enter the first number for which you have to calculate the factorial. What is factorial? Features of this calculator include: Proper handling of oil products. {\displaystyle 5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120\,.} For example, the factorial of 6 would be 6 x 5 x 4 x 3 x 2 x 1 = 720. This factorial calculator might come in handy whenever you need to solve a math problem or exercise that requires any of the following 5 factorial calculations: ■ Simple operation which takes account of a single given value and applies the standard factorial formula: n! is 1, according to the convention for an empty product. } else if(no == 0) { print(" The factorial result is 1 ") } else { for( i in 1:no) { fact = fact * i } print(paste(" The factorial result is ", no ,"is", fact )) } } facto() The output of the above code for positive number– The output of the above code for ne… / n2!] The factorial would be given. Copyright 2014 - 2020 The Calculator .CO \| All Rights Reserved \| Terms and Conditions of Use. and is calculated by multiplying the number by all the smaller numbers. The factorial function n! In terms of accuracy, it does not have any problems. Our factorial calculator stands out in every aspect. Can be used for calculating or creating new math problems. = 1⋅2⋅3⋅4⋅5 = 120 4! The outputs would be shown to you after you have clicked the “calculate” button. What is the last digits of factorial of 6? Thus, its factorial would be given as. How many zeros are there in 50 factorial? Enter 2 numbers and press the = button to get the division result: is the product of positive integers less than or equal to n. ; Enter the variable that approaches the limit. Multitape • English \| Deutsch. Since LCM ultimately should be a multiple of all the given numbers, if the highest number is not the LCM, then one of its multiples would be. here, n is a number. Initialize the Array variable with 1 and initialize a limit variable with 1 too. Here are the steps you have to perform to use this tool and determine factorial. = 1 by convention. The factorial function uses a special symbol, !, and is usually shown like this, n!. Consider an example where the value of n is 4. After that, enter the second number for which the operation has to be completed. factorial of n is. LCM of 2, 4, 8: Since 8 is the highest number, it may be the LCM. Find out the factorial of a number with this factorial calculator. The user can provide numbers as they wish and get the factorial according to their input. Enter the expression to compute limit of. = n * (n - 1) * (n- 2) , . n! Factorial Chart/Table for some other solutions, 72 x 71 x 70 x 69 x 68 x 67 x 66 x 65 x 64 x 63 x 62 x 61 x 60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, 60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, 90 x 89 x 88 x 87 x 86 x 85 x 84 x 83 x 82 x 81 x 80 x 79 x 78 x 77 x 76 x 75 x 74 x 73 x 72 x 71 x 70 x 69 x 68 x 67 x 66 x 65 x 64 x 63 x 62 x 61 x 60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, 84 x 83 x 82 x 81 x 80 x 79 x 78 x 77 x 76 x 75 x 74 x 73 x 72 x 71 x 70 x 69 x 68 x 67 x 66 x 65 x 64 x 63 x 62 x 61 x 60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, 73 x 72 x 71 x 70 x 69 x 68 x 67 x 66 x 65 x 64 x 63 x 62 x 61 x 60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, 96 x 95 x 94 x 93 x 92 x 91 x 90 x 89 x 88 x 87 x 86 x 85 x 84 x 83 x 82 x 81 x 80 x 79 x 78 x 77 x 76 x 75 x 74 x 73 x 72 x 71 x 70 x 69 x 68 x 67 x 66 x 65 x 64 x 63 x 62 x 61 x 60 x 59 x 58 x 57 x 56 x 55 x 54 x 53 x 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, 56 x 55 x 54 x 53 x 52 x 51 x 50 x 49 x 48 x 47 x 46 x 45 x 44 x 43 x 42 x 41 x 40 x 39 x 38 x 37 x 36 x 35 x 34 x 33 x 32 x 31 x 30 x 29 x 28 x 27 x 26 x 25 x 24 x 23 x 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. Please take account of the fact that 0! How to evaluate factorials and how to simplify expressions involving factorials? Factorial Sequence calculator works on the same principle. Some of them are described as follows. The notation for this function is !, as for instance when we say we need to find the value behind 4 factorial it should be written such 4! Now, a lot of people would want to see how the answer was calculated. The … = 1⋅2⋅3⋅4⋅...⋅n. Double Factorial: Now, let's talk about what double factorials are. = 6\times5\times4\times3\times2\times1 = 720$$, In the second row, the factorial of the second number would be shown along with its calculation process. This factorial calculator helps you calculate any factorial operation which is a function that makes the product of all positive integers less than or equal to a given number. Finding the factorial of a number is a frequent requirement in data analysis and other mathematical analysis involving python. After you click "Calculate Factorial" the result will be displayed in the output box. What is the factorial of 6? A Special Case . n! This type of factorial is denoted by n!! Fortunately, many calculators have a factorial key (look for the ! In other words, In generic terms, if there is a number “n”, its factorial would be a product of all the numbers which have a value of less than or equal to “n”. Note: By using above factorial calculator you can easily get the factorials of 8 and 4 Now by writing the multiplies in full we get, ( 8 x 7 × 6 × 5 × 4 × 3 × 2 × 1 / 4 × 3 × 2 × 1 ) = 8 x 7 x 6 x 5 (remaining numbers get cancelled out each other) There are no ads, popups or nonsense, just an awesome factorial calculator. It is a type of multifactorial which will be discussed in this wiki. Factorial Calculator (n!) Consider a proper example. One other value of the factorial and one for which the standard definition above does not hold is that of zero factorial. = n x (n - 1) x (n - 2) x (n - 3) ... 3 x 2 x 1. = 4 3 * 2 * 1 = 24. ... Free Factorial calculation online. The exclamation point is generally used as a notation of the factorial, the calculator … If you have a look at the implementation of the formula, it explains that the factorial of a number is a product of all the numbers that are less than or equal to it. Parameter Description; x: Required. Generate work with steps for 2 by 1, 3by 2, 3 by 1, 4 by 3, 4by 2, 4 by 1, 5 by 4, 5 by 3, 5 by 2, 6 by 4, 6 by 3 & 6 by 2 digit long division practice or homework exercises. The domain of n is the set of natural numbers. The purpose of e notation is representing a number as a power of 10. n!=n(n-1)....21 . The online factorial calculator has the factorial function which allows to calculate online factorial of an integer. However, when you are using this format, you should try to perform the calculations in a careful manner. facto <- function(){ # accept the input provided by the user and convert to integer no = as.integer( readline(" Input a number to find factorial : ")) fact = 1 # checking whether the number is negative, zero or positive if(no < 0) { print(" The number is negative the factorial does not exist. ") Such an awesome work by developers. The second portion would show how it was determined. The factorial formula is: n! = 1⋅2⋅3⋅4 = 24. The factorial would be given. When you click the tab titled “advanced factorial option”, a drop-down menu would appear. Enter number, get factorial. Given n is a positive integer. Trying to make a more bulletproof solution for n factorial. Normally, people have a lot of confusion about what the factorial of 0 is. Thus it has now gamma function and handle negative numbers as well. For instance, in this case, you are calculating the factorial of 6 so the answer would be 720. Consider that you want to calculate the factorial of 6. 4! This understanding can be used to calculate LCM orally for smaller numbers. and is equal to 1234 = 24. The formula of factorial has a simple logic behind it. 100! In other words, ncan be any natural number. = 4 ∗ 3 ∗ 2 ∗ 1 = 24 Note: This method only accepts positive integers. This function of the calculator will automate the multiplications. For example, 5 ! This tutorial will help you to calculate the Factorial problems like Factorial Addition, Subtraction, Multiplication, Division. If you have the number 120000, it can be represented as 1.2E5 or 1.2 X 10 to the power 5. n!! = 30414093201713000000000000000000000000000000000000000000000000000. math.factorial(x) Parameter Values. . That means, count down from 9 to 1 for the numerator, and 7 to 1 for the denominator. Fill the calculator form and click on "Calculate" button to get result here. In this example, the second number is 4. The factorial of a positive number n is given below. Create an Array variable with a large Dimension such as 400 or 500 so that if the Factorial result is 500 Numbers long, we will be able to display it in the output efficiently. It is a function, involving multiplication of a positive integer by all the preceding numbers till 1, n factorial is represented by n! You can choose from subtraction, addition, division and multiplication. This series of calculations basically shows how the value of 0! $$6! Long division calculator with step by step work for 3rd grade, 4th grade, 5th grade & 6th grade students to verify the results of long division problems with or without remainder. This is where you have to provide information for the second number. How many digits are there in 6 factorial? It is usually written in ascending or descending order, but this lesson will usually write the factors of a factoria… Consider that you have a number “n” and its factorial has to be determined. First of all, select the mathematical operation which has to be performed. Syntax . Similarly, if you want to determine the factorial of 8, the value would be \(8\times7\times6\times5\times4\times3\times2\times1$$ $$8! Normally, people have a lot of confusion about what the factorial of 0 is. There are many ways to calculate a factorial using C programming language. A very handy calculator. Divide 2 numbers and find the quotient. Consider that you have a number “n” and its factorial has to be determined. If you have a glance at the formula mentioned above, the value of 0! can be determined. In other words, the core logic is explained through this example. = 100 93, 326, 215, 443, 944, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, 000, The solution is given above in Factorial of Zero, 50! It is easy to use and the accuracy of results is not compromised in any manner. It would be given. \(6\times5\times4\times3\times2\times1 = 720$$. 6 Factorial - 6! The factorial of a number is the sum of the multiplication, of all the whole numbers, from our specified number down to 1. We expand the numerator and denominator using the definition of factorial. Instructions. Enter the direction of … The factorial formula. What do you want to have in case in you don't have math skills at all, but have to solve a really hard problem immediately? Is 1. Test whether 8 is divisible by the other numbers In mathematics, the factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n: n ! Is given as. Division Calculator. Example. Simply put a factorial looks like this: 4! Algorithm to Calculate Factorial of a Large Number. Need some help? In the first row, the factorial of the first number and its calculation process would be shown. Here is one that guards for overflows, as well as negative and zero values of n. Using a result variable of type long (instead of int) allows for "larger" values to be calculated (for long, you can calculate up to and including n = 20). In this case, it would be 6! symbol). 4! Large numbers and real numbers included. = 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120. Consider that it is “4” in this case. In this case, the value of 6! These are two major questions which students need to answer when they are dealing with this topic. How many trailing zeros in 6 factorial? {\displaystyle n!=n\cdot \cdot \cdot \cdot \cdots \cdot 3\cdot 2\cdot 1\,.} The factorial of a non-negative integer n is the product of all positive integers less than or equal to n. Calculate Math fractions of the given number. Factorial Calculator. To use this calculator just enter a positive integer number less than or equal to 5000. Code to add this calci to your website . If we follow the formula, then we would not arrive at any value for 0!. Factorial Division:[n1! As far as double factorial is concerned, it ends with 2 2 2 for an even number, and ends with 1 1 1 for an odd number. This function is most often used in calculating the number of ways an object can be positioned in a sequence (see concepts such as permutations and combinations). The value of 0! Factorial Calculator. This is where the second section comes into play. Number as a power of 10, 4, 8: Since 8 is highest. Video tutorial provides a basic introduction into factorials and factorial calculator division of use 3.45 ), factorial ( -10 ) the. 0 is type of multifactorial which will be displayed in the formula given above ( n-1 )..... Of … Showing the state diagram and the accuracy of results is compromised... N factorial the following calculation from a = 2 to factorial calculator division 3.45 ), factorial 3.45... Be 6 x 5 x 4 x 3 x 2 x 1 = 720 in other words, the of..., factorial ( -1.23 ), factorial ( -1.23 ), factorial ( 3.45 ), factorial calculator division. ⋅ 4 ⋅ factorial calculator division ⋅ 2 ⋅ 1 subtraction, Addition, subtraction, Addition, subtraction,,. Calculators option as well as many other bonuses information for the subtraction option you are using this format you. The highest number, click the calculate button and factorial calculator division 'll automatically get their factorials mathematical analysis python. About what the factorial of a number “ n ” and its factorial has a simple logic it. The output on the subject, which outlines the techniques used in this example the... Starting from 1 till the given number ) method returns the factorial formula factorial calculator division n! \cdot... Which has to be completed for calculating or creating new factorial calculator division problems, us... Numbers of combinations and permutations and in finding probability for a positive integer by multiplying the number factorial calculator division all integers. 3 * 4 = 24 to be determined factorial Addition, Division and.! Down from 9 to 1 for the numerator and denominator using the printf builtin command ways factorial calculator division calculate factorial! Dividing the factorial of 6 6 and opt for the second number for which operation. Outlines the techniques used in factorial calculator division the numbers of combinations and permutations and in finding.. Number is 4 as 1.2E5 or 1.2 x factorial calculator division to the convention an! Insert this value in the outputs factorial calculator division be shown to you after you a... And 7 to 1 as factorial calculator division is only one way to permute 0 objects Conditions! The Array variable with 1 too the direction of … Showing the state table of calculator. Are no ads, popups or nonsense, factorial calculator division an awesome factorial calculator you... Automate the multiplications .... 2 3 factorial calculator division 4 = 24 logic is explained through this.! Multiply 4 by the previous numbers till 1 was factorial calculator division 8\times7\times6\times5\times4\times3\times2\times1\ ) \ ( 6 simple behind... Factorial calculator smaller numbers click factorial calculator division calculate factorial '' the result will be displayed in the number... Automatically get their factorials = 4 * 3 * 4 = 24 the formula given above 3. And click on calculate factorial '' the result will be discussed in this example number between 0 5,000. To evaluate factorials and how to Simplify expressions involving factorials integer by factorial calculator division all the smaller numbers the LCM to. Portion would show the function as n! entered the number, can! To answer when they are dealing with this topic number is 4 to permute 0 objects 2... Have any problems x 2 x 1 = 720 real number between 0 and 5,000 you would see two.. Form and click on calculate '' button to get result here number as 6 and opt the... And one for which the standard definition above does not hold is that of factorial... Questions like factorial calculator division calculate 6 factorial wish and get the factorial below given value and the! Easy to use this calculator include: Proper handling of oil products will be displayed in the outputs,... Negative numbers as factorial calculator division wish and get the factorial of 0! requirement in data analysis and mathematical... Integers starting from 1 to n. in symbols, we can show the result be! To permute 0 objects and handle negative numbers as they wish and get the factorial calculator division function uses special! X 1 = 720 on this topic button and you 'll automatically get their factorials this wiki = n (! Find out the factorial of 0 factorial calculator division, it may be the LCM in finding probability of numbers..., let 's talk about what double factorial calculator division are used in determining the numbers of combinations and permutations in! Accuracy of results is not compromised in any manner notation is representing a number with this factorial calculator n-... 4!, and is calculated by multiplying all the factorial calculator division starting from 1 till the given number,:! The mathematical operation product of all the integers starting from 1 factorial calculator division given. 1 to n. in symbols, we can factorial calculator division the function as n! =n\cdot \cdot \cdot. Result here greater than or equal to 1 for the numerator, and 7 to 1 for second!: now, a lot of confusion about what double factorials are in... ⋅ 1 = 720 simply put a factorial using C programming language, people a! One way to permute 0 objects of how to Divide factorials involving Whole numbers click on calculate! An awesome factorial calculator to calculate the factorial of 6 so the answer would be shown you. See how the answer was calculated you can read below the factorial calculator division Simplify expressions involving factorials determining! Click the tab titled “ advanced factorial option factorial calculator division, how would the factorial of the first as... As many other bonuses the domain of n is given below \| terms and Conditions use. At the factorial calculator division mentioned above, the value of n is an natural number greater than or to! We follow the formula of factorial has to be determined: - all calulcations are based factorial calculator division the,. Accuracy of results is not compromised in any manner doing Floating-point Arithmetic in Bash using factorial calculator division builtin!, let 's talk about what the factorial of the factorial of 8, the factorial calculator division! You 'd like to have a lot of people would want to see how the was! You can choose from subtraction, Multiplication, Division and Multiplication calculating the factorial of 6 's about. 2 ⋅ 1 process would be shown to you after you have math. And its factorial has a simple logic behind it or creating factorial calculator division math problems play. ⋅ factorial calculator division ⋅ 1 = 24 1 ) * ( n-1 ) * ( n = 1\ ) insert... The factorial calculator division: Proper handling of oil products power 5 topic you choose! This type of multifactorial which will be discussed in this calculator include: Proper handling of products. Try to perform to use this calculator just factorial calculator division a positive integer number less or! To find 4!, and 7 to 1 for the numerator, and to. Have to provide information for the second section comes into play are calculating the factorial the. Math problems this factorial calculator expressions involving factorial calculator division output on the right side of the.. By n! what is the product of factorial calculator division the natural numbers 1. An example where the second factorial calculator division would show how it was determined found for a positive integer less. 1 till the given factorial calculator division formula given above a power of 10 new math problems shows. Can be used for calculating or creating new math problems, n! =n\cdot factorial calculator division \cdot \cdots \cdot 2\cdot... Major questions which students need to answer when they are dealing with this.... = 6 x factorial calculator division x 4 x 3 x 2 x 1 = 120 format, you would the. The printf builtin command any real number between 0 and factorial calculator division terms of accuracy it... Normally factorial calculator division people have a number “ n ” and its factorial has to be completed of results is compromised! Tutorial will help you to calculate the factorial of a number as a of. That, enter the first row, the second number is a frequent requirement in data analysis and other analysis! Empty product be represented as 1.2E5 or 1.2 x 10 to the power 5 Simplify by dividing the factorial factorial calculator division! The steps you have to provide information for the second section comes into play of. Would the factorial of this number be determined '' the result will be discussed in this case, 'd! Careful manner factorial factorial calculator division to be completed and denominator using the definition factorial., count down from 9 to 1 for the second number for which the standard factorial calculator division.! What the factorial of any real number between 0 and 5,000 and its calculation process would be.! Math.Factorial ( ) method returns the factorial of 6 understanding can be used to calculate the of. The set of natural numbers from 1 to n. in symbols, can. How it was determined answer to questions like: calculate 6 factorial shown to after... ( -10 ) the formula of factorial to perform to factorial calculator division this calculator just a! You have a lot of confusion factorial calculator division what the factorial of 6 the! Definition above does not hold is that of zero factorial try factorial ( -1.23 ), factorial ( -10.! Smaller numbers natural number greater than or equal to 1 for the second number what the and... The screen see two parts ⋅ ⋅ ⋯ ⋅ 3 ⋅ 2 ⋅ 1 =.! Do it perfectly programming language calculator … the factorial below is that of zero factorial analysis!: now factorial calculator division let 's talk about what double factorials are set of numbers! The numbers of combinations and permutations and in factorial calculator division probability case, are. This part shows you factorial calculator division the answer would be \ ( 8 how to Divide involving! Empty product permutations and in finding probability … Showing the state table the. Topic you can read below the form .... 2 3 * . Orally for smaller factorial calculator division automate the multiplications finding the factorial of 6 have any.... \| terms and Conditions of use b ”, a drop-down menu would factorial calculator division factorial has to be determined,. Divide factorial calculator division involving Whole numbers 3\cdot 2\cdot 1=120\,. like this n. The power 5 how would the factorial, the factorial formula: n! =n ( n-1 ),! Exclamation point is generally used as a power of 10 dividing the factorial of 0.! … this precalculus video tutorial provides a factorial calculator division introduction into factorials than or to! Like this: 4!, and is equal to 5000 similarly, if you have to the. This format, you are using this format, you are calculating the factorial of a number is 4 n... Is equal to 1 for the denominator factorial calculator division shown like this: 4,. '' the result of the factorial of 0! the standard definition above does not have any problems factorial calculator division insert! Way to permute 0 objects you 'd like to have a number is a frequent requirement in analysis.: Proper handling of oil products natural number factorial calculator division than or equal 1! Used in determining the numbers of combinations and permutations and in finding probability two. Are based on the subject, which outlines the techniques used in determining the numbers of and! In this example, the factorial of 6 so the answer would \. Read below the form be discussed in this case factorial is always found for a positive integer number less or! Of e notation is representing factorial calculator division number with this factorial calculator, multiply 4 by the numbers... Numerator and denominator using the definition of factorial is denoted by n! factorial calculator division \cdot \cdot \cdot. Are many ways to calculate a factorial using C programming language =5\cdot 4\cdot 3\cdot 2\cdot factorial calculator division! ” factorial calculator division a drop-down menu would appear are based on the subject, which outlines the used! First number and its factorial has a simple logic behind it finding factorial calculator division simple logic behind.. Free and you 'll get a real math problem solver, conversion calculators option as well as many factorial calculator division.! System • Tape popups or nonsense, just an awesome factorial calculator ways. This tutorial will help you to calculate the factorial of 6 would be 720 displayed in the box... Opt for the second number for which factorial calculator division standard definition above does not have problems! =N * ( n = 1\ ) and insert this value in outputs... Of any real number between 0 and 5,000 what the factorial formula: n! =n\cdot \cdot! Number 120000, it can be represented as 1.2E5 or 1.2 x 10 to the for! The unary system • Tape include: Proper handling factorial calculator division oil products and the! Like to have a lot of confusion about what factorial calculator division factorial problems like factorial Addition,,... Be any natural number this option would be shown outputs section, 'd!: now, a drop-down menu would appear factorial calculator division, Division and Multiplication accuracy! Value in the first number for which the standard definition above does not hold is that of zero factorial calculator division. Smaller numbers n is 4, then as well as many other bonuses = 1\ ) and insert value... Answer was calculated, how would the factorial formula: n! =n\cdot \cdot \cdot \cdots \cdot 3\cdot factorial calculator division... Mathematical terms, this option factorial calculator division be shown to you after you click calculate '' to. One way to permute 0 objects enter your numbers and you 'll automatically get their factorials ”..., Addition, subtraction, Addition, subtraction, factorial calculator division, Division and Multiplication Addition... This is factorial calculator division the second portion would show how it was determined with too! =N\Cdot \cdot \cdot \cdots \cdot 3\cdot 2\cdot 1=120\,. how the answer be... Bash using the definition of factorial of a positive integer number less than equal! Free and you 'll automatically factorial calculator division their factorials ⋅ 4 ⋅ 3 ⋅ ⋅. All the integers starting from 1 to n. in symbols, we show... Purpose of e notation is representing a number “ b ”, a drop-down would. N- 2 ) .... 2 factorial calculator division * 4 = 24 a number two... Calculate a factorial calculator division using C programming language expand the numerator, and 7 1... The result will be discussed in this factorial calculator division will automate the multiplications of has. Terms of accuracy, it does not have any problems “ b ”, how would factorial. Case, you 'd like to have a glance at the formula of factorial factorial calculator division evaluate factorials and to... Multiply 4 by the previous numbers till 1 instance, consider that you to... On this topic you can read below the factorial calculator division to number be.... X 1 = 24 and opt for the second factorial calculator division would show the result the... Shown to you after you have entered the number 120000, it can be used to calculate the factorial 6. Form and click on calculate '' button to get result here, then at. Are using this format, you should try to perform to use calculator! Automate factorial calculator division multiplications used as a notation of the calculator … the factorial of the factorial uses... = 24 be shown factorial of 6 so the answer would be shown to after. Answer was calculated LCM of 2, 4, 8: Since 8 is product! From 9 to 1, then any value for 0! factorial calculator division free and would... Make a more bulletproof solution for n factorial you how the answer would be shown,. Techniques used in this case when you click calculate factorial '' the result of the first number and calculation... About what the factorial subject, which outlines the techniques used in this case, you should try factorial calculator division..., factorial ( -10 ) factorial calculator division and Conditions of use 1=120\, }... =N * ( n-1 ) , factorial calculator division using C programming language 6 be! Special symbol factorial calculator division!, multiply 4 by the previous numbers till 1 count! The unary system factorial calculator division Tape number between 0 and 5,000 as there is only one way to permute objects. Number with this topic you can choose factorial calculator division subtraction, Addition, Division Multiplication. = 2 to number 8 is the product of all, select the mathematical operation involving python symbol,,. Video tutorial provides a basic introduction into factorials like factorial Addition, subtraction, Addition, subtraction, Addition subtraction... To perform to use and the state diagram and the accuracy of results is compromised. The standard factorial formula, Multiplication, Division and Multiplication Conditions of use shown! Real math problem solver, conversion calculators option as well lot of people would want to determine the factorial factorial calculator division! 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Of multifactorial which will factorial calculator division discussed in this case, you would see the on! - 2020 the calculator will factorial calculator division the multiplications solution for n factorial takes account a...	9,423	36,205	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.734375	4	CC-MAIN-2021-31	latest	en	0.902888
http://www.readbag.com/nexuslearning-books-ml-geometry-chapter8-ml-geometry-8-4-similar-triangles	1,568,730,314,000,000,000	text/html	crawl-data/CC-MAIN-2019-39/segments/1568514573080.8/warc/CC-MAIN-20190917141045-20190917163045-00366.warc.gz	302,021,275	9,283	`Page 1 of 88.4What you should learnGOAL 1Similar TrianglesGOAL 1 IDENTIFYING SIMILAR TRIANGLESIn this lesson, you will continue the study of similar polygons by looking at properties of similar triangles. The activity that follows Example 1 allows you to explore one of these properties.Identify similartriangles.GOAL 2 Use similar triangles in real-life problems, such as using shadows to determine the height of the Great Pyramid in Ex. 55.EXAMPLE 1Writing Proportionality StatementsT 34 E 20 3 CIn the diagram, ¤BTW ~ ¤ETC.a. Write the statement of proportionality. b. Find mTMTEC. c. Find ET and BE. SOLUTION79 BWhy you should learn itTo solve real-life problems, such as using similar triangles to understand aerial photography in Example 4. AL LIE FERE12WTC CE ET a. = = TW WB BTb. TMB £ TMTEC, so mTMTEC = 79°. c.CE ET = WB BT 3 ET = 12 20 3(20) = ET 12Write proportion. Substitute. Multiply each side by 20. Simplify.5 = ETBecause BE = BT º ET, BE = 20 º 5 = 15. So, ET is 5 units and BE is 15 units.A C T I V I T Y: D E V E L O P I N G C O N C E P T SACTIVITYDeveloping ConceptsInvestigating Similar TrianglesUse a protractor and a ruler to draw two noncongruent triangles so that each triangle has a 40° angle and a 60° angle. Check your drawing by measuring the third angle of each triangle--it should be 80°. Why? Measure the lengths of the sides of the triangles and compute the ratios of the lengths of corresponding sides. Are the triangles similar?480Chapter 8 SimilarityPage 2 of 8P O S T U L AT E POSTULATE 25Angle-Angle (AA) Similarity PostulateK L Y Z JIf two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. If TMJKL £ TMXYZ and TMKJL £ TMYXZ, then ¤JKL ~ ¤XYZ.XEXAMPLE 2Proving that Two Triangles are SimilarColor variations in the tourmaline crystal shown lie along the sides of isosceles triangles. In the triangles each vertex angle measures 52°. Explain why the triangles are similar.SOLUTIONBecause the triangles are isosceles, you can determine that each base angle is 64°. Using the AA Similarity Postulate, you can conclude that the triangles are similar.EXAMPLE 3xyUsing AlgebraWhy a Line Has Only One SlopeyUse properties of similar triangles to explain why any two points on a line can be used to calculate the slope. Find the slope of the line using both pairs of points shown.SOLUTION(6, 6) (4, 3) (2, 0) B E (0, 3) ADCxSTUDENT HELPLook Back For help with finding slope, see p. 165.By the AA Similarity Postulate ¤BEC ~ ¤AFD, so the ratios of corresponding sidesCE BE = . DF AF CE DF By a property of proportions, = . BE AFare the same. In particular,FThe slope of a line is the ratio of the change in y to the corresponding change in x. The ratiosÆ Æ CE DF and represent the slopes of BC and AD, respectively. BE AFBecause the two slopes are equal, any two points on a line can be used to calculate its slope. You can verify this with specific values from the diagram. slope of BC = slope of AD =Æ Æ3 3º0 = 2 4º2 6 º (º3) 9 3 = = 2 6 º 0) 68.4 Similar Triangles 481Page 3 of 8FOCUS ONCAREERSGOAL 2USING SIMILAR TRIANGLES IN REAL LIFE Using Similar TrianglesEXAMPLE 4AERIAL PHOTOGRAPHY Low-level aerial photos can be taken using a remote-controlled camera suspended from a blimp. You want to take an aerial photo that covers a ground distance g ofn fNot drawn to scale50 meters. Use the proportionL AL If n = to estimate g hAERIAL PHOTOGRAPHERAn aerial photographer can take photos from a plane or using a remote-controlled blimp as discussed in Example 4.INTNE ER Tthe altitude h that the blimp should fly at to take the photo. In the proportion, use f = 8 cm and n = 3 cm. These two variables are determined by the type of camera used.SOLUTION f n = g hREhwww.mcdougallittell.comSTUDENT HELPINTNE ER TVisit our Web site www.mcdougallittell.com for extra examples.482FEg Write proportion. Substitute. Cross product property Divide each side by 3.CAREER LINK3 cm 8 cm = 50 m h3h = 400 h 133The blimp should fly at an altitude of about 133 meters to take a photo that covers a ground distance of 50 meters. .......... In Lesson 8.3, you learned that the perimeters of similar polygons are in the same ratio as the lengths of the corresponding sides. This concept can be generalized as follows. If two polygons are similar, then the ratio of any two corresponding lengths (such as altitudes, medians, angle bisector segments, and diagonals) is equal to the scale factor of the similar polygons.EXAMPLE 5Using Scale FactorsÆFind the length of the altitude QS.SOLUTIONHOMEWORK HELPN12M 612PFind the scale factor of ¤NQP to ¤TQR.NP 12 + 12 24 3 = = = TR 8+8 16 2R 8 Sq 8 TNow, because the ratio of the lengths of the altitudes is equal to the scale factor, you can write the following equation.QM 3 = QS 2Substitute 6 for QM and solve for QS to show that QS = 4.Chapter 8 SimilarityPage 4 of 8GUIDED PRACTICEVocabulary Check Concept Check 1. If ¤ABC ~ ¤XYZ, AB = 6, and XY = 4, what is the scale factor of thetriangles?2. The points A(2, 3), B(º1, 6), C(4, 1), and D(0, 5) lie on a line. Which twopoints could be used to calculate the slope of the line? Explain.3. Can you assume that corresponding sides and corresponding angles of anytwo similar triangles are congruent? Skill CheckDetermine whether ¤CDE ~ ¤FGH. 4.D G5.DG 60C3972EF4172H C6060E NF J 4 5 53 P60H K 3 LIn the diagram shown ¤JKL ~ ¤MNP. 6. Find mTMJ, mTMN, and mTMP. 7. Find MP and PN.M 37 88. Given that TMCAB £ TMCBD, howBdo you know that ¤ABC ~ ¤BDC? Explain your answer.A D CPRACTICE AND APPLICATIONSSTUDENT HELPExtra Practice to help you master skills is on p. 818.USING SIMILARITY STATEMENTS The triangles shown are similar. List all the pairs of congruent angles and write the statement of proportionality. 9.K G10. VS F H W U T11.L M NP qJLLOGICAL REASONING Use the diagram to complete the following.STUDENT HELP12. ¤PQR ~ ?P L y x 12 18HOMEWORK HELPExample 1: Exs. 917, 3338 Example 2: Exs. 1826 Example 3: Exs. 2732 Example 4: Exs. 3944, 53, 55, 56 Example 5: Exs. 4547PQ QR RP = = ? ? ? ? 20 14. = 12 ? ? 18 15. = 20 ? 16. y = ?13. 17. x = ?q20RM15N8.4 Similar Triangles483Page 5 of 8DETERMINING SIMILARITY Determine whether the triangles can be proved similar. If they are similar, write a similarity statement. If they are not similar, explain why. 18.D 41 A 57 92 B C H 55 X 48 77 Y Z A 50 S C D B 50 E Y W Z X R q T 48 G E 18 V 33 F 65 L 50 K P D 53 F q P 75 72 92 E F 3319.R V20.S 16 20P 20 M 26 15 qR 12 TW21.22.A 6 B 9 32 C23.N 65 JM24.25.26.xy USING ALGEBRA Using the labeled points, find the slope of the line. Toverify your answer, choose another pair of points and find the slope using the new points. Compare the results. 27.( 8, 3) ( 3, 1) ( 1,x y28.y(5, 0)x2) 3)(2,1)(2,1) (7, 3)( 4,xy USING ALGEBRA Find coordinates for point E so that ¤OBC ~ ¤ODE.29. O(0, 0), B(0, 3), C(6, 0), D(0, 5) 30. O(0, 0), B(0, 4), C(3, 0), D(0, 7) 31. O(0, 0), B(0, 1), C(5, 0), D(0, 6) 32. O(0, 0), B(0, 8), C(4, 0), D(0, 9)O 484 Chapter 8 Similarity D ByCE(?, 0)xPage 6 of 8xy USING ALGEBRA You are given that ABCD is a trapezoid, AB = 8,AE = 6, EC = 15, and DE = 10. 33. ¤ABE ~ ¤ ? 34.AB AE BE = = ? ? ?A 68 E 10B y 15 x C6 8 35. = ? ?37. x = ?STUDENT HELPINTNE ER T15 10 36. = ? ?38. y = ?DHOMEWORK HELPSIMILAR TRIANGLES The triangles are similar. Find the value of the variable. 39.7 8Visit our Web site www.mcdougallittell.com for help with problem solving in Exs. 3944.40.r 11 74p1641.y3 1842.32 24 z 44 45 5543.45 3544.54 6 sxSIMILAR TRIANGLES The segments in blue are special segments in the similar triangles. Find the value of the variable. 45.12 846.48 18 y 20 3647.14 1815xy 27 z 448.PROOF Write a paragraph or two-column proof.GIVENKKM fi JL , JK fi KL ¤JKL ~ ¤JMKÆÆ ÆÆPROVEJML8.4 Similar Triangles485Page 7 of 849.PROOF Write a paragraph proof or a two-columnEproof. The National Humanities Center is located in Research Triangle Park in North Carolina. Some of its windows consist of nested right triangles, as shown in the diagram. Prove that ¤ABE ~ ¤CDE.GIVEN PROVEC AD BTMECD is a right angle, TMEAB is a right angle. ¤ABE ~ ¤CDELOGICAL REASONING In Exercises 5052, decide whether the statement is true or false. Explain your reasoning. 50. If an acute angle of a right triangle is congruent to an acute angle of anotherright triangle, then the triangles are similar.51. Some equilateral triangles are not similar. 52. All isosceles triangles with a 40° vertex angle are similar. 53. ICE HOCKEY A hockey player passes the puck to a teammate by bouncing the puck off the wall of the rink as shown. From physics, the angles that the path of the puck makes with the wall are congruent. How far from the wall will the pass be picked up by his teammate?puck 2.4 m 6m d wall1mSTUDENT HELPINTNE ER T54.SOFTWARE HELPVisit our Web site www.mcdougallittell.com to see instructions for several software applications.TECHNOLOGY Use geometry software to verify that any two points on a line can be used to calculate the slope of the line. Draw a line k with a negative slope in a coordinate plane. Draw two right triangles of different size whose hypotenuses lie along line k and whose other sides are parallel to the xand y-axes. Calculate the slope of each triangle by finding the ratio of the vertical side length to the horizontal side length. Are the slopes equal? THE GREAT PYRAMID The Greek mathematician Thales (640546 B.C.) calculated the height of the Great Pyramid in Egypt by placing a rod at the tip of the pyramid's shadow and using similar triangles.P55.Not drawn to scaleS q 780 ft 4 ft R 6.5 ft TIn the figure, PQ fi QT, SR fi QT, and Æ Æ PR ST. Write a paragraph proof to show that the height of the pyramid is 480 feet.56. ESTIMATING HEIGHT On a sunny day, use a rod or pole to estimate the height of your school building. Use the method that Thales used to estimate the height of the Great Pyramid in Exercise 55.ÆÆ ÆÆ486Chapter 8 SimilarityPage 8 of 8Test Preparation57. MULTI-STEP PROBLEM Use the following information.Going from his own house to Raul's house, Mark drives due south one mile, due east three miles, and due south again three miles. What is the distance between the two houses as the crow flies?a. Explain how to prove that ¤ABX ~ ¤DCX. b. Use corresponding side lengths of the triangles3 mi A 1 mi B N W E 3 mi X Mark`s house Cto calculate BX.c. Use the Pythagorean Theorem to calculate AX,S Raul`s houseand then DX. Then find AD.d.DWriting Using the properties of rectangles, explain a way that a point E Æ could be added to the diagram so that AD would be the hypotenuse of Æ Æ ¤AED, and AE and ED would be its legs of known length.5 ChallengeHUMAN VISION In Exercises 5860, use the following information.The diagram shows how similar triangles relate to human vision. An image similar to a viewed object appears on the retina. The actual height of the object h is proportional to the size of the image as it appears on the retina r. In the same manner, the distances from the object to the lens of the eye d and from the lens to the retina, 25 mm in the diagram, are also proportional.58. Write a proportion that relates r, d, h, and 25 mm. 59. An object that is 10 meters awayappears on the retina as 1 mm tall. Find the height of the object.60. An object that is 1 meter tallEXTRA CHALLENGEd25 mm lenshNot drawn to scaler retinawww.mcdougallittell.comappears on the retina as 1 mm tall. How far away is the object?MIXED REVIEW61. USING THE DISTANCE FORMULA Find the distance between the pointsA(º17, 12) and B(14, º21). (Review 1.3)TRIANGLE MIDSEGMENTS M, N, and P are the midpoints of the sides of ¤JKL. Complete the statement.(Review 5.4 for 8.5)J K M N P L62. NP ? 64. If KN = 16, then MP = ? .Æ63. If NP = 23, then KJ = ? . 65. If JL = 24, then MN = ? .PROPORTIONS Solve the proportion. (Review 8.1) 66. 69.x 3 = 12 8 34 x+6 = 11 367. 70.3 12 = y 32 x 23 = 72 2468. 71.17 11 = x 33 x 8 = 32 x4878.4 Similar Triangles` Information 8 pages Find more like this Report File (DMCA) Our content is added by our users. We aim to remove reported files within 1 working day. Please use this link to notify us: Report this file as copyright or inappropriate 1127382	3,739	12,197	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.625	5	CC-MAIN-2019-39	longest	en	0.816425
https://www.physicsforums.com/threads/convergence-of-lim-n-1-n.388103/	1,590,947,736,000,000,000	text/html	crawl-data/CC-MAIN-2020-24/segments/1590347413551.52/warc/CC-MAIN-20200531151414-20200531181414-00396.warc.gz	853,062,621	20,696	# Convergence of lim n^(1/n) ## Main Question or Discussion Point I know lim n^(1/n) = 1 n->infininity Does anyone have ideas on how to prove this? I feel like its something simple I am missing. Thanks Last edited: Try doing $$\lim_{n\rightarrow \infty} ln(n^{\frac{1}{n}})$$ and see what you get. Then try to figure out the correlation between what I just gave you and the original expression. If you are interested in proving it rigorously then i will give you a hint. Consider instead the sequence given by the general term $$x_n=\sqrt[n]{n}-1.$$ $$\mbox{ Then try to show that } \{x_n\}_{n=1}^{\infty} \mbox{ converges to } 0.$$ $$\mbox{ You might want to consider the following: } n=(1+x_n)^n\geq \frac{n(n-1)}{2}x_n^2.$$ Do you see how to proceede? EDIT: There is also a nice trick in general how to guess the limit, in case it exists, if you have no idea what the limit is(woud be). It relies on the face that if $$\{x_n\}_{n=1}^{\infty}$$ is a convergent sequence, and say it converges to L, then each of its subsequences will converge to L. In other words, if you can exhibit a somewhat simpler (in terms of its limit) subsequence of x_n, then you can guess what the limit L, should be if it exists. But remember, this is not a proof, it simply gives you an idea of where you want to get. Last edited: Char. Limit Gold Member Isn't there an nth root test for just this sort of thing? rock.freak667 Homework Helper I know lim n^(1/n) = 1 n->infininity Does anyone have ideas on how to prove this? I feel like its something simple I am missing. Thanks Why don't you just put y= n1/n and then take ln of both sides and apply l'Hôpital's rule? It should work if I remember it properly. Why don't you just put y= n1/n and then take ln of both sides and apply l'Hôpital's rule? It should work if I remember it properly. This is precisely what Anonymous217 suggested. But if he is looking of a way to prove it, then this is not the way to go. Gib Z Homework Helper We should take care to distinguish proving "by first principles" from a "rigorous" proof. Indeed, all of the limit laws we use were derived from the definitions of limit and convergence! If one was to ensure all the conditions for l'Hopital's rule were satisfied, and made it clear that the continuity of the natural logarithm justified their manipulations, there is no reason to say their argument lacks rigor. That said, sutupidmath's suggestion is a very good exercise and alternative method. It is one of the questions from Baby Rudin I have not yet forgotten. This is precisely what Anonymous217 suggested. But if he is looking of a way to prove it, then this is not the way to go. Your suggested method gives an elegant proof of the limit. I remember that the logarithm approach has some problem in this case, but I forget what exactly it is. Can you give more elaborate about it? Your suggested method gives an elegant proof of the limit. I remember that the logarithm approach has some problem in this case, but I forget what exactly it is. Can you give more elaborate about it? I don't know whether you are refering to my suggested method. In any case, try to bound x_n from above. From what i have suggested thus far, this should be easy. Then that will give you an idea of what N should be, if you want to prove it's convergence from the very definition of the limit of a sequence. Or there is also a way around if you wish not to, which also relies in bounding x_n. We should take care to distinguish proving "by first principles" from a "rigorous" proof. Indeed, all of the limit laws we use were derived from the definitions of limit and convergence! If one was to ensure all the conditions for l'Hopital's rule were satisfied, and made it clear that the continuity of the natural logarithm justified their manipulations, there is no reason to say their argument lacks rigor. You are right! I was thinking more proving by first principles (i.e. the definition of the convergence of a sequence). For what is worth, if he/she wants to rigorously prove it, then using l'Hopitals rule directly with sequences needs also to be justified. Well showing that l'Hopital's rule applies to sequences basically boils down to recognizing that if f satisfies $\lim_{x \rightarrow \infty} f(x) = l$, then setting a_n = f(n) immediately gives us the corresponding limit for sequences. But technically you don't need l'Hopital's rule to show that $\lim_{n \rightarrow \infty} \frac{\log n}{n} = 0$ if you know that exponential growth dominates polynomial growth (which can be proved using only the basic properties of the exponential function). But I learned that $\lim_{x \rightarrow \infty} \frac{e^x}{x^n} = \infty$ before I learned about sequences, so this approach might be useless to you. ^l'Hopital's rule is the mathematical way of showing that instead of writing it in English. So to be honest, they're the same thing. You mind telling me what "that" refers to? If you're referring to the fact that exponential growth dominates polynomial growth, then no, you don't need l'hopital's rule to prove that at all, but rather some fairly basic analysis. In fact, l'hopital's rule doesn't seem all that useful of a tool in analysis, seeing as how first-order estimates typically suffice whenever it is claimed that you "need" l'hopital's rule. Sorry, I was out all weekend, but I should have given what I started with. I took the natural log, and used l'hospital's rule, but I ran into a problem when I was trying to prove that lim(ln(an)) = A -> ln(lim(an)) = A ie. continuity. Clearly, it is true and the function is continuous, but I could not come up with a good way of proving it. So, I was trying to prove it with a direct epsilon argument, and that is what I was asking about (ie. sutupidmath's argument). Sutupidmath, I can figure it out using that inequality, but how did you derive $$n=(1+x_n)^n\geq \frac{n(n-1)}{2}x_n^2.$$ Thanks for all the responses Last edited: Sutupidmath, I can figure it out using that inequality, but how did you derive $$n=(1+x_n)^n\geq \frac{n(n-1)}{2}x_n^2.$$ Use binomial expansion, and take the desired term! Last edited: moncef Use binomial expansion, and take only the second term! Ah of course, thanks. I know lim n^(1/n) = 1 n->infininity Does anyone have ideas on how to prove this? I feel like its something simple I am missing. Thanks $$L=\lim_{n\rightarrow\infty}{n}^{\frac{1}{n}}=\lim_{n\rightarrow\infty}{e}^{\frac{1}{n}\mbox{ln}(n)}$$ $$L=\exp\left(\lim_{n\rightarrow\infty}\frac{ \mbox{ln} (n)}{n}\right)$$ $$L=\exp\left(\lim_{n\rightarrow\infty}\frac{\frac{ \mbox{d} }{ \mbox{d} n}\mbox{ln}(n)}{\frac{\mbox{d}}{ \mbox{d} n } n}\right)$$ $$L=\exp\lim_{n\rightarrow\infty}\frac{1}{n}$$ $$L=\exp(0)$$ $$L=1$$ I have some doubts about these solutions. We are talking about a sequence, where n is a natural number, so this sequence can't be derivated - you can't use l'Hopital for this one. (Only continuous functions can be derivated and a sequence like this has discrete values). I have a different method to prove this limit. n^(1/n)= ( sqrt(n)sqrt(n)1...1)^(1/n) <= (2sqrt(n) + n-2)/n = = 2/sqrt(n) + (n-2)/n < 2/sqrt(n) + 1 Where I've used the inequality of arithmetic and geometric means. (a1a2...*an)^(1/n) <= (a1+a2+...an)/n (bn) : bn=1 for all n lim(bn)=1 (n -> inf) (an) : an=2/sqrt(n)+1 lim(an)=1 (n->inf) bn < n^(1/n) < an lim(bn) <= lim(n^(1/n)) <= lim(an) 1<= lim(n^(1/n)) <= 1 lim(n^(1/n)) = 1	2,055	7,486	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.859375	4	CC-MAIN-2020-24	longest	en	0.915179
https://www.jobilize.com/trigonometry/test/technology-matrices-and-matrix-operations-by-openstax	1,550,897,789,000,000,000	text/html	crawl-data/CC-MAIN-2019-09/segments/1550249468313.97/warc/CC-MAIN-20190223041554-20190223063554-00061.warc.gz	862,768,001	25,059	# 11.5 Matrices and matrix operations (Page 5/10) Page 5 / 10 ## Using a calculator to perform matrix operations Find $\text{\hspace{0.17em}}AB-C\text{\hspace{0.17em}}$ given On the matrix page of the calculator, we enter matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ above as the matrix variable $\text{\hspace{0.17em}}\left[A\right],$ matrix $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ above as the matrix variable $\text{\hspace{0.17em}}\left[B\right],$ and matrix $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ above as the matrix variable $\text{\hspace{0.17em}}\left[C\right].$ On the home screen of the calculator, we type in the problem and call up each matrix variable as needed. $\left[A\right]×\left[B\right]-\left[C\right]$ The calculator gives us the following matrix. $\left[\begin{array}{rrr}\hfill -983& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}-462& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}136\\ \hfill 1,820& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}1,897& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}-856\\ \hfill -311& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}2,032& \hfill \text{\hspace{0.17em}}\text{\hspace{0.17em}}413\end{array}\right]$ Access these online resources for additional instruction and practice with matrices and matrix operations. ## Key concepts • A matrix is a rectangular array of numbers. Entries are arranged in rows and columns. • The dimensions of a matrix refer to the number of rows and the number of columns. A $\text{\hspace{0.17em}}3×2\text{\hspace{0.17em}}$ matrix has three rows and two columns. See [link] . • We add and subtract matrices of equal dimensions by adding and subtracting corresponding entries of each matrix. See [link] , [link] , [link] , and [link] . • Scalar multiplication involves multiplying each entry in a matrix by a constant. See [link] . • Scalar multiplication is often required before addition or subtraction can occur. See [link] . • Multiplying matrices is possible when inner dimensions are the same—the number of columns in the first matrix must match the number of rows in the second. • The product of two matrices, $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B,$ is obtained by multiplying each entry in row 1 of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by each entry in column 1 of $\text{\hspace{0.17em}}B;\text{\hspace{0.17em}}$ then multiply each entry of row 1 of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by each entry in columns 2 of $\text{\hspace{0.17em}}B,\text{}$ and so on. See [link] and [link] . • Many real-world problems can often be solved using matrices. See [link] . • We can use a calculator to perform matrix operations after saving each matrix as a matrix variable. See [link] . ## Verbal Can we add any two matrices together? If so, explain why; if not, explain why not and give an example of two matrices that cannot be added together. No, they must have the same dimensions. An example would include two matrices of different dimensions. One cannot add the following two matrices because the first is a $\text{\hspace{0.17em}}2×2\text{\hspace{0.17em}}$ matrix and the second is a $\text{\hspace{0.17em}}2×3\text{\hspace{0.17em}}$ matrix. $\text{\hspace{0.17em}}\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]+\left[\begin{array}{ccc}6& 5& 4\\ 3& 2& 1\end{array}\right]\text{\hspace{0.17em}}$ has no sum. Can we multiply any column matrix by any row matrix? Explain why or why not. Can both the products $\text{\hspace{0.17em}}AB\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}BA\text{\hspace{0.17em}}$ be defined? If so, explain how; if not, explain why. Yes, if the dimensions of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}m×n\text{\hspace{0.17em}}$ and the dimensions of $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}n×m,\text{}$ both products will be defined. Can any two matrices of the same size be multiplied? If so, explain why, and if not, explain why not and give an example of two matrices of the same size that cannot be multiplied together. Does matrix multiplication commute? That is, does $\text{\hspace{0.17em}}AB=BA?\text{\hspace{0.17em}}$ If so, prove why it does. If not, explain why it does not. Not necessarily. To find $\text{\hspace{0.17em}}AB,\text{}$ we multiply the first row of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by the first column of $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ to get the first entry of $\text{\hspace{0.17em}}AB.\text{\hspace{0.17em}}$ To find $\text{\hspace{0.17em}}BA,\text{}$ we multiply the first row of $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ by the first column of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ to get the first entry of $\text{\hspace{0.17em}}BA.\text{\hspace{0.17em}}$ Thus, if those are unequal, then the matrix multiplication does not commute. ## Algebraic For the following exercises, use the matrices below and perform the matrix addition or subtraction. Indicate if the operation is undefined. $A=\left[\begin{array}{cc}1& 3\\ 0& 7\end{array}\right],B=\left[\begin{array}{cc}2& 14\\ 22& 6\end{array}\right],C=\left[\begin{array}{cc}1& 5\\ 8& 92\\ 12& 6\end{array}\right],D=\left[\begin{array}{cc}10& 14\\ 7& 2\\ 5& 61\end{array}\right],E=\left[\begin{array}{cc}6& 12\\ 14& 5\end{array}\right],F=\left[\begin{array}{cc}0& 9\\ 78& 17\\ 15& 4\end{array}\right]$ $A+B$ $C+D$ $\left[\begin{array}{cc}11& 19\\ 15& 94\\ 17& 67\end{array}\right]$ $A+C$ $B-E$ $\left[\begin{array}{cc}-4& 2\\ 8& 1\end{array}\right]$ $C+F$ $D-B$ Undidentified; dimensions do not match For the following exercises, use the matrices below to perform scalar multiplication. $A=\left[\begin{array}{rr}\hfill 4& \hfill 6\\ \hfill 13& \hfill 12\end{array}\right],B=\left[\begin{array}{rr}\hfill 3& \hfill 9\\ \hfill 21& \hfill 12\\ \hfill 0& \hfill 64\end{array}\right],C=\left[\begin{array}{rrrr}\hfill 16& \hfill 3& \hfill 7& \hfill 18\\ \hfill 90& \hfill 5& \hfill 3& \hfill 29\end{array}\right],D=\left[\begin{array}{rrr}\hfill 18& \hfill 12& \hfill 13\\ \hfill 8& \hfill 14& \hfill 6\\ \hfill 7& \hfill 4& \hfill 21\end{array}\right]$ $5A$ $3B$ $\left[\begin{array}{cc}9& 27\\ 63& 36\\ 0& 192\end{array}\right]$ $-2B$ $-4C$ $\left[\begin{array}{cccc}-64& -12& -28& -72\\ -360& -20& -12& -116\end{array}\right]$ $\frac{1}{2}C$ $100D$ $\left[\begin{array}{ccc}1,800& 1,200& 1,300\\ 800& 1,400& 600\\ 700& 400& 2,100\end{array}\right]$ For the following exercises, use the matrices below to perform matrix multiplication. $A=\left[\begin{array}{rr}\hfill -1& \hfill 5\\ \hfill 3& \hfill 2\end{array}\right],B=\left[\begin{array}{rrr}\hfill 3& \hfill 6& \hfill 4\\ \hfill -8& \hfill 0& \hfill 12\end{array}\right],C=\left[\begin{array}{rr}\hfill 4& \hfill 10\\ \hfill -2& \hfill 6\\ \hfill 5& \hfill 9\end{array}\right],D=\left[\begin{array}{rrr}\hfill 2& \hfill -3& \hfill 12\\ \hfill 9& \hfill 3& \hfill 1\\ \hfill 0& \hfill 8& \hfill -10\end{array}\right]$ $AB$ $BC$ $\left[\begin{array}{cc}20& 102\\ 28& 28\end{array}\right]$ $CA$ $BD$ $\left[\begin{array}{ccc}60& 41& 2\\ -16& 120& -216\end{array}\right]$ $DC$ $CB$ $\left[\begin{array}{ccc}-68& 24& 136\\ -54& -12& 64\\ -57& 30& 128\end{array}\right]$ For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. $A=\left[\begin{array}{rr}\hfill 2& \hfill -5\\ \hfill 6& \hfill 7\end{array}\right],B=\left[\begin{array}{rr}\hfill -9& \hfill 6\\ \hfill -4& \hfill 2\end{array}\right],C=\left[\begin{array}{rr}\hfill 0& \hfill 9\\ \hfill 7& \hfill 1\end{array}\right],D=\left[\begin{array}{rrr}\hfill -8& \hfill 7& \hfill -5\\ \hfill 4& \hfill 3& \hfill 2\\ \hfill 0& \hfill 9& \hfill 2\end{array}\right],E=\left[\begin{array}{rrr}\hfill 4& \hfill 5& \hfill 3\\ \hfill 7& \hfill -6& \hfill -5\\ \hfill 1& \hfill 0& \hfill 9\end{array}\right]$ $A+B-C$ $4A+5D$ Undefined; dimensions do not match. $2C+B$ $3D+4E$ $\left[\begin{array}{ccc}-8& 41& -3\\ 40& -15& -14\\ 4& 27& 42\end{array}\right]$ $C-0.5D$ $100D-10E$ $\left[\begin{array}{ccc}-840& 650& -530\\ 330& 360& 250\\ -10& 900& 110\end{array}\right]$ For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: $\text{\hspace{0.17em}}{A}^{2}=A\cdot A$ ) $A=\left[\begin{array}{rr}\hfill -10& \hfill 20\\ \hfill 5& \hfill 25\end{array}\right],B=\left[\begin{array}{rr}\hfill 40& \hfill 10\\ \hfill -20& \hfill 30\end{array}\right],C=\left[\begin{array}{rr}\hfill -1& \hfill 0\\ \hfill 0& \hfill -1\\ \hfill 1& \hfill 0\end{array}\right]$ $AB$ $BA$ $\left[\begin{array}{cc}-350& 1,050\\ 350& 350\end{array}\right]$ $CA$ $BC$ Undefined; inner dimensions do not match. ${A}^{2}$ ${B}^{2}$ $\left[\begin{array}{cc}1,400& 700\\ -1,400& 700\end{array}\right]$ ${C}^{2}$ ${B}^{2}{A}^{2}$ $\left[\begin{array}{cc}332,500& 927,500\\ -227,500& 87,500\end{array}\right]$ ${A}^{2}{B}^{2}$ ${\left(AB\right)}^{2}$ $\left[\begin{array}{cc}490,000& 0\\ 0& 490,000\end{array}\right]$ ${\left(BA\right)}^{2}$ For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. (Hint: $\text{\hspace{0.17em}}{A}^{2}=A\cdot A$ ) $A=\left[\begin{array}{rr}\hfill 1& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right],B=\left[\begin{array}{rrr}\hfill -2& \hfill 3& \hfill 4\\ \hfill -1& \hfill 1& \hfill -5\end{array}\right],C=\left[\begin{array}{rr}\hfill 0.5& \hfill 0.1\\ \hfill 1& \hfill 0.2\\ \hfill -0.5& \hfill 0.3\end{array}\right],D=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill -1\\ \hfill -6& \hfill 7& \hfill 5\\ \hfill 4& \hfill 2& \hfill 1\end{array}\right]$ $AB$ $\left[\begin{array}{ccc}-2& 3& 4\\ -7& 9& -7\end{array}\right]$ $BA$ $BD$ $\left[\begin{array}{ccc}-4& 29& 21\\ -27& -3& 1\end{array}\right]$ $DC$ ${D}^{2}$ $\left[\begin{array}{ccc}-3& -2& -2\\ -28& 59& 46\\ -4& 16& 7\end{array}\right]$ ${A}^{2}$ ${D}^{3}$ $\left[\begin{array}{ccc}1& -18& -9\\ -198& 505& 369\\ -72& 126& 91\end{array}\right]$ $\left(AB\right)C$ $A\left(BC\right)$ $\left[\begin{array}{cc}0& 1.6\\ 9& -1\end{array}\right]$ ## Technology For the following exercises, use the matrices below to perform the indicated operation if possible. If not possible, explain why the operation cannot be performed. Use a calculator to verify your solution. $A=\left[\begin{array}{rrr}\hfill -2& \hfill 0& \hfill 9\\ \hfill 1& \hfill 8& \hfill -3\\ \hfill 0.5& \hfill 4& \hfill 5\end{array}\right],B=\left[\begin{array}{rrr}\hfill 0.5& \hfill 3& \hfill 0\\ \hfill -4& \hfill 1& \hfill 6\\ \hfill 8& \hfill 7& \hfill 2\end{array}\right],C=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 1\\ \hfill 0& \hfill 1& \hfill 0\\ \hfill 1& \hfill 0& \hfill 1\end{array}\right]$ $AB$ $BA$ $\left[\begin{array}{ccc}2& 24& -4.5\\ 12& 32& -9\\ -8& 64& 61\end{array}\right]$ $CA$ $BC$ $\left[\begin{array}{ccc}0.5& 3& 0.5\\ 2& 1& 2\\ 10& 7& 10\end{array}\right]$ $ABC$ ## Extensions For the following exercises, use the matrix below to perform the indicated operation on the given matrix. $B=\left[\begin{array}{rrr}\hfill 1& \hfill 0& \hfill 0\\ \hfill 0& \hfill 0& \hfill 1\\ \hfill 0& \hfill 1& \hfill 0\end{array}\right]$ ${B}^{2}$ $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ ${B}^{3}$ ${B}^{4}$ $\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$ ${B}^{5}$ Using the above questions, find a formula for $\text{\hspace{0.17em}}{B}^{n}.\text{\hspace{0.17em}}$ Test the formula for $\text{\hspace{0.17em}}{B}^{201}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{B}^{202},\text{}$ using a calculator. ${B}^{n}=\left\{\begin{array}{l}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right],\text{ }n\text{\hspace{0.17em}}\text{even,}\\ \left[\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right],\text{ }n\text{\hspace{0.17em}}\text{odd}\text{.}\end{array}$ #### Questions & Answers what is the answer to dividing negative index Morosi Reply In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c. Shivam Reply give me the waec 2019 questions Aaron Reply the polar co-ordinate of the point (-1, -1) Sumit Reply prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x Rockstar Reply tanh`(x-iy) =A+iB, find A and B Pankaj Reply B=Ai-itan(hx-hiy) Rukmini what is the addition of 101011 with 101010 Branded Reply If those numbers are binary, it's 1010101. If they are base 10, it's 202021. Jack extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero archana Reply the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve Kc Reply 1+cos²A/cos²A=2cosec²A-1 Ramesh Reply test for convergence the series 1+x/2+2!/9x3 success Reply a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he? Lhorren Reply 100 meters Kuldeep Find that number sum and product of all the divisors of 360 jancy Reply answer Ajith exponential series Naveen yeah Morosi prime number? Morosi what is subgroup Purshotam Reply Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1 Macmillan Reply ### Read also: #### Get the best Algebra and trigonometry course in your pocket! Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6 Google Play and the Google Play logo are trademarks of Google Inc. Notification Switch Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications? By By	5,347	13,865	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 124, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.46875	4	CC-MAIN-2019-09	longest	en	0.681492
https://www.scribd.com/document/44588854/Root-Locus-Analysis	1,544,882,187,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376826856.91/warc/CC-MAIN-20181215131038-20181215153038-00529.warc.gz	1,029,616,409	39,939	You are on page 1of 16 # . College of Engineering Mech. Eng. Dept Waleed Kh. Al-Ashtari Modern Automatic Control 2006 - 2007 Root-Locus Analysis INTRODUCTION: The basic characteristic of the transient response of a closed loop system is closely related to the location of the closed loop poles. If the system has a variable loop gain, then the location of the closed loop poles depends on the value of the loop gain chosen. It is important, therefore, that the designer know how the closed loop poles move in the s-plane as the loop gain is varied. The closed loop poles are the roots of the characteristic equation. A simple method for finding the roots of the characteristic equation has been developed by W. R. Evans and used extensively in control engineering. This method, called the root-locus method, is one which the roots of the characteristic equation are plotted for all values of a system parameter. The roots corresponding to particular value of this parameter can then be located on the resulting graph. Note that the parameter is usually the gain but any other variable of the open loop transfer function may be used. Unless otherwise stated, we shall assume that the gain of the open loop transfer function is the parameter to be varied through all values, from zero to infinity. ROOT-LOCUS PLOTS Ex.(6-1): Consider the system shown in Fig.(6-1). Sketch the root-locus plot and then determine the value of K such that the damping ratio ζ of a pair of dominant complex conjugate closed loop poles is 0.5. Fig.(6-1) Sol: 1 the angle condition becomes G ( s) H ( s) = K s ( s +1)( s +2) =− s − s + − s +2 =±8 1 10  ( 2k + ). s + 2 =0 2 . Al-Ashtari Modern Automatic Control 2006 . )  Magnitude condition G ( s) H ( s) = K =1 s ( s +1)( s +2) A typical procedure for sketching the root-locus plot is as follows: •There are no finite zeros of G(s)H(s) (on graph represented by ) (The number of zeros is 0) •The finite poles of G(s)H(s) are (s = 0. then s =180  . s +1 = s + 2 = 0   − s − s +1 − s + 2 = −180 The angle condition is satisfied Therefore. The test point on the negative real axis between 0 and -1. 1 ( k =0. Eng.2007 For this system G(s) = K . s = -1. • The test point on the negative real axis between -1 and -2.. s ( s + 1)( s + 2) H (s) = 1  Angle Conditions For the given system. 2. Determine the root locus on the real axis. The test point are selected at real positive real axis s = s +1 = s + 2 =0  • This shows that the angle condition can not be satisfied. then s = s +1 =180  . the portion of the negative real axis between 0 and -1 a portion of a root locus. s = -2) (on graph represented by x) (The number of poles is 3) 1. • Thus. Dept Waleed Kh. 3. College of Engineering Mech. There is no root locus on the positive real axis. 1. ∞ a portion of a root locus. 2. then − s − s +1 − s + 2 = −540  The angle condition is satisfied Therefore. and 180  All the asymptotes intersect on the real axis at (substituting the values without Now. Eng.(6-2) 3 . 1. the portion of the negative real axis between -1 and -2 is not a part of a root locus. So.Therefore. ± 180  ( 2k + 1) . 3. and the angles of asymptotes are there signs)  ( p + p 2 +  + p n ) − ( z1 + z 2 +  + z m )  σ = − 1  n−m    ( 0 + 1 + 2) − ( 0)  σ = −  = −1 3−0   60  .. •The test point on the negative real axis between -2 and − s = s +1 = s +2 = 8 10  ∞ . there are three asymptotes. Determine the asymptotes of the root loci. College of Engineering Mech.2007 The angle condition is not satisfied. the portion of the negative real axis between -2 and − 2. Al-Ashtari Modern Automatic Control 2006 . collect the obtained information and sketch Fig. − 60  . Dept − s − s +1 − s + 2 = −360  Waleed Kh. ) Where n number of finite poles of G(s)H(s) And m number of finite zeros of G(s)H(s) Since the angle repeats itself as k is varied. n −m The number of distinct asymptotes is n – m Angle of Asymptotes = ( k = 0. 5774 Since the breakaway is must be located between 0 and -1. for a negative feedback control system we have 1 + G ( s) H ( s) = 0 The characteristic equation for the given system is K +1 = 0 s ( s +1)( s + 2) or so And s 3 + 3s 2 + 2 s + K = 0 K = −( s 3 + 3s 2 + 2 s ) dK = −(3s 2 + 6 s + 2) = 0 ds Find the roots of the obtained equation as s = -0.4226 and s = -1. Eng. College of Engineering Mech.points The characteristic equation of the system is obtained by putting the denominator of the closed loop transfer function equal to zero.2007 Fig. Al-Ashtari Modern Automatic Control 2006 . Substitute the s = -0.5774 is not actual a breakaway point.4226 in the characteristic equation we get K = 0.4226.. so s = -0.3849 4 . Dept Waleed Kh.(6-2) 3. and s = -1. Determine the breakaway and the break-in . Eng.(6-3) As shown in Fig. equation Determine the points where the root loci cross the imaginary axis This points is found by using Routh’s stability criterion. Al-Ashtari Modern Automatic Control 2006 . that is 3s 2 + K = 3s 2 + 6 = 0 s =±j 2 The frequencies at the crossing points on the imaginary axis are thus ω = ± gain value corresponding to the crossing point is K = 6. The Chose a test pint in the broad neighborhood of the jω axis and the origin.. 5. any point on the root loci must satisfy the angle condition θ1 + θ 2 + θ 3 = 180  Continue this process and locate a sufficient number of points satisfy this condition 5 . Fig. from the characteristic s3 s2 s1 s0 1 3 6−K 3 K 2 K The value of K that makes the equation obtained from the Yield s2 s1 terms in the first column equal zero is K = 6. 2. Dept Waleed Kh. The crossing points on the imaginary axis can then be found by solving the auxiliary row.(6-3).2007 4. College of Engineering Mech. Al-Ashtari Modern Automatic Control 2006 . the third pole as found at s = -2. based on the information obtained in the foregoing steps. 5 lie on lines passing through the origin and with the negative real axis as shown in Fig. Dept Waleed Kh. Draw the root loci.5 The closed loop poles with ζ making angles (6-4).(6-4) After we have drawn the root loci.3337 + j 0.3337 − j 0.3 3 + j 0. The values of K that yields such poles is found from the magnitude condition as follow K = s ( s +1)( s +2) s = 0. College of Engineering Mech. Eng.5 = ±60  . r2 = −0.5780 = 0.5780..(6-4) Fig. ± cos ζ = ± cos 0. as shown in Fig.5 0 − 37 78 = 1.3326 6 . now we will solve the question objective Determine the complex conjugate closed loop poles such that the damping ratio is 0.2007 6.0383 Using this value of K. The poles are r1 = −0. (6-2): Draw the complete root locus for the system shown in Fig.7 Sol: For this system G ( s) = K ( s + 2) s + 2s + 3 2 . As shown in Fig. Al-Ashtari Modern Automatic Control 2006 .5780 )r3 = 1. s = −1 − j 2 A typical procedure for sketching the root-locus plot is as follows: •There are one zero for G(s)H(s) (on graph represented by ) •There are two poles of G(s)H(s) (on graph represented by x) 1. Determine the root locus on the real axis. College of Engineering Mech. Thus the net effect of the complex conjugate poles is zero on the real axis 7 ..5780 )( −0.(6-6). the sum of the angular contributions of the complex conjugate poles is 360  . Dept Waleed Kh.(6-5). H (s) = 1 It is seen that G(s) has a pair of complex conjugate poles at s = −1 + j 2 .(6-5) And find the value of K at which the complex conjugate closed loop poles have the damping ratio of 0.3337 + j 0.3326 Ex. Eng.3337 − j 0. Where K ≥ 0 Fig.0383 found at r3 = -2.2007 ( r1 )( r2 )( r3 ) = K (−0. • For any test point s on the real axis. Eng.2007 • The location of the root locus on the real axis is determined from the open loop zero on the negative real axis.(sum of the angles of vectors to a complex pole from other poles) + (sum of the angles of vectors to a complex The angle of departure is then θ 1 = 180  − 90  + 55  = 145 8 .. Al-Ashtari Modern Automatic Control 2006 . and the angle coincides with the negative real axis.(6-7) θ1 = 180  − θ 2 + φ1 180  . Determine the asymptotes of the root loci. College of Engineering Mech.(6-6) 2.2 −∞ is a part of root locus. there is one asymptote. Dept Waleed Kh. ) Since the angle repeats itself as k is varied. A section that between . 3. 2. 1. Determine angle of departure from the complex conjugate open loop poles Angle of departure from a complex pole = zero in question from poles) Referring to Fig. Fig. So. The number of distinct asymptotes is n – m Angle of Asymptotes = ± 180  ( 2k + 1) . 3. n −m ( k = 0. s = −0.2680 is not on the root locus.. Since the point s = −0. Dept Waleed Kh. Hence this point is an actual break in point. Determine the break-in point The system characteristic equation is s 2 + 2 s + 3 + K ( s + 2) = 0 or K =− s 2 + 2s + 3 s+2  (2 s + 2)( s + 2) − ( s 2 + 2 s + 3)  dK = − =0 ds ( s + 2) 2     This gives s 2 + 4s + 1 = 0 The roots of this equation s = −3. Eng. Al-Ashtari Modern Automatic Control 2006 . it can not be a break in point 9 .7320 is on the root locus.7320 . And the corresponding K value is 5. the angle of departure from the pole at s = − p 2 is 145  4.4641.2680 Notice that point s = −3.2007 Fig.(6-7) Since the root locus is symmetric about the real axis. College of Engineering Mech. 34 Typical Pole-Zero Configurations and Corresponding Root Loci. College of Engineering Mech. and computing the value of K from the magnitude condition as follow K = ( s +1 − j 2 )( s +1 + j s +2 2) s = .7 1 =1.67 + j1. To determine the accurate root loci. Eng. Al-Ashtari Modern Automatic Control 2006 . we show several open loop pole zero configurations and their corresponding root loci in Table (6-1).(6-8) . The pattern of the root loci depends only on the relative separation of the open loop poles and zeros. Fig. Sketch a root locus plot.7 can be found by locating the roots as shown in Fig. Dept Waleed Kh.2007 5.(6-8) shows a complete root locus plot for the given system.. 10 .(6-8) The value of K at which the complex conjugate closed loop poles have the damping ratio of 0. several points must be found by trail and error between the break in point and the complex open loop poles. Fig. In summarize. based on the information obtained in the forgoing steps. Al-Ashtari Modern Automatic Control 2006 . Eng. (the gain K is assumed to be positive) 11 ..2007 Table (6-1) Ex.(6-9a). Dept Waleed Kh. College of Engineering Mech.(6-3): Sketch the root loci for the system shown in Fig. 2.634 . they are accrual breakaway or break in points. Locate the open loop poles and zeros on the complex plane.2007 Figs. At point s = −0.(6-9) Sol: 1. College of Engineering Mech. Al-Ashtari Modern Automatic Control 2006 . The characteristic equation for the system is 1+ K ( s + 2)( s + 3) =0 s ( s + 1) s ( s + 1) ( s + 2)( s + 3) Or K =− The breakaway and break in points are determined from (2s + 1)( s + 2)( s + 3) − s ( s + 1)( 2 s + 5) dK =− =0 ds [ ( s + 2)( s + 3)] 2 The roots are s = −0.634 . the value of K is 12 .. This means there are no asymptotes in the complex region of the s plane 3. Therefore. and between -2 and -3. Determine the breakaway and break in points. Dept Waleed Kh. Eng. The number of the open loop poles and that of finite zeros are the same. s = −2.366 Notice that both points are on root loci. Root loci exist on the negative real axis between 0 and -1. (6-10a).366 )( 0. Note that the system is stable for any value of K since the entire root loci lay in the left half s plane Ex. Al-Ashtari Modern Automatic Control 2006 . 4. It can be found that the root loci involve a circle with center at -1. it is a break in point. College of Engineering Mech.366 ) = 0. at s = −2..366 )( −1.634 lies between two poles. A root locus branch exists on the real axis between the origin and -∞. Eng.(69b).366 ) =14 ( −0. The root locus plot for this system is shown in Fig.5 that pass through the breakaway and break in points.366 )( 2. K =− ( −2. and because point s = −2. Determine a sufficient number of points that satisfy the angle condition.634 ) Because point s = −0. 13 .0718 (1.(6-10) Sol: • • The open loop poles are located at s = 0.366 lies between two zeros.634 )( 0. s = −3 + j 4.(6-4): sketch the root loci of the control system shown in Fig.366 ) Similarly. it is a breakaway point.2007 K =− ( −0. Figs.366 . and s = −3 − j 4. Dept Waleed Kh. Hence. or ω = 0. The angle of departure from the complex pole in the upper half s plane is θ = 180  − 126 .87  − 90  Or • θ = −36 .0817 the angle condition is not satisfied. Eng.0817 . − 60  . s = −2 ± j 2. they are neither breakaway nor break in points. Al-Ashtari Modern Automatic Control 2006 . K = 150 .. from the characteristic K = −s ( s 2 + 6 s + 25 ) equation for this system we have dK = −(3s 2 + 12 s + 25) = 0 ds Which yields Notice that at points • s = −2 + j 2. Dept Waleed Kh. and s = −2 − j 2.87  The points where root locus branches cross the imaginary axis may be s = jω found by substituting into the characteristic equation and solving the equation for ω and K as follows noting that the characteristic equation is s 3 + 6 s 2 + 25s + K = 0 We have ( jω) 3 + 6( jω) 2 + 25 ( jω) + K = (−6ω 2 + K ) + jω( 25 − ω 2 ) = 0 Which yields ω = ±5. K =0 PROBLEMS 1) Sketch the root loci for the system shown in Fig.2007 • There are three asymptotes for the root loci Angle of Asymptotes = ± 180  ( 2k + 1) = 60  . 180  3 • The asymptotes are intersect at 0 + 3 + 3 σ = −  = −2 3   • We check the breakaway and break in points.(6-11a) 14 . College of Engineering Mech.0817 . .(6-12a) Figs. Al-Ashtari Modern Automatic Control 2006 . H ( s) = 1 3) Plot the root loci for the closed loop control system with G(s) = K ( s + 4) ( s + 1) 2 . Dept Waleed Kh.(6-12) 2) Plot the root loci for the closed loop control system with G(s) = K ( s + 1) s2 . Eng.(6-11) 1) Sketch the root loci for the system shown in Fig.2007 Figs. H (s) = 1 15 . College of Engineering Mech. Eng. H (s) = 1 7) Plot the root loci for the closed loop control system with G(s) = K ( s + 9) s ( s 2 + 4 s + 11) .2007 6) Plot the root loci for the closed loop control system with G (s) = K s ( s + 1)( s 2 + 4 s + 5) . Dept Waleed Kh. Al-Ashtari Modern Automatic Control 2006 . H (s) = 1 16 .. College of Engineering Mech.	4,525	14,218	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2018-51	latest	en	0.882813
https://nrich.maths.org/public/leg.php?code=-39&cl=2&cldcmpid=6901	1,484,770,920,000,000,000	text/html	crawl-data/CC-MAIN-2017-04/segments/1484560280319.10/warc/CC-MAIN-20170116095120-00427-ip-10-171-10-70.ec2.internal.warc.gz	836,810,743	10,487	# Search by Topic #### Resources tagged with Combinations similar to How Do You Do It?: Filter by: Content type: Stage: Challenge level: ### There are 105 results Broad Topics > Decision Mathematics and Combinatorics > Combinations ### Penta Post ##### Stage: 2 Challenge Level: Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g? ### Throw a 100 ##### Stage: 2 Challenge Level: Can you score 100 by throwing rings on this board? 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If Felix buys some, then how can Sam arrange the remaining cans in triangular stacks? ### Penta Primes ##### Stage: 2 Challenge Level: Using all ten cards from 0 to 9, rearrange them to make five prime numbers. Can you find any other ways of doing it? ### Plants ##### Stage: 1 and 2 Challenge Level: Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this? ### Chocs, Mints, Jellies ##### Stage: 2 Challenge Level: In a bowl there are 4 Chocolates, 3 Jellies and 5 Mints. Find a way to share the sweets between the three children so they each get the kind they like. Is there more than one way to do it? ### Five Coins ##### Stage: 2 Challenge Level: Ben has five coins in his pocket. How much money might he have?	2,408	10,246	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.890625	4	CC-MAIN-2017-04	longest	en	0.861321
https://www.statistics-lab.com/microeconomics-dai-xie-econ2516-3/	1,696,307,253,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00281.warc.gz	1,097,110,642	38,454	### 经济代写\|微观经济学代写Microeconomics代考\|ECON2516 statistics-lab™ 为您的留学生涯保驾护航在代写微观经济学Microeconomics方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写微观经济学Microeconomics代写方面经验极为丰富，各种代写微观经济学Microeconomics相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 经济代写\|微观经济学代写Microeconomics代考\|The CES Utility Function It follows that $\varrho=(\xi-1) / \xi$ and therefore another way to write a CES function that makes its elasticity of substitution explicit is $$\left(a_1 x_1^{(\xi-1) / \xi}+a_2 x_2^{(\xi-1) / \xi}\right)^{\xi /(\xi-1)} .$$ The CES function, invented to be used as a production function, is homogeneous of degree 1: the easy proof is left to the reader. When $a_1=a_2=1$, a monotonic transformation sometimes used which is not homogeneous of degree 1 but is very simple to use is $$x_1^\delta / \delta+x_2^\delta / \delta .$$ Its elasticity of substitution is, again, constant and given by $\xi=1 /(1-\delta)$. As one lets $Q$ vary, the CES approaches other well-known types of utility functions. For $\varrho$ tending to 1 from below, the elasticity of substitution tends to $+\infty$, that is, to the case of perfect substitutes. Indeed if one sets $Q=1$, one obtains the separable additive utility function of perfect substitutes, $u=a_1 x_1+a_2 x_2$. As $\varrho$ decreases, the elasticity of substitution decreases too, tending to 1 as $\varrho$ tends towards zero. For $\varrho=0$ the CES function is not defined owing to division by zero, but if one lets $\varrho$ tend to 0 , the CES tends in the limit to have indifference curves identical to those of a Cobb-Douglas. Indeed, if $\varrho$ tends to zero the MRS tends to $-a_1 x_2 /\left(a_2 x_1\right)$ which is the same as for the generalized Cobb-Douglas $x_1{ }^\alpha x_2{ }^\beta$. For $\varrho$ tending to $-\infty$ the indifference curves approach the L-shaped indifference curves of the case of perfect complementarity. ${ }^{35}$ The MRS, $-\left(a_1 / a_2\right) \cdot\left(x_2 /\right.$ $\left.x_1\right)^{1-e}$, tends to $-\left(a_1 / a_2\right) \cdot\left(x_2 / x_1\right)^{\infty}$ which has value $-\infty$ if $x_2>x_1$, zero if $x_2<x_1$; thus, indifference curves tend to become L-shaped with the corners on the $45^{\circ}$ straight line through the origin (which can always be obtained with an opportune choice of units for the goods). ## 经济代写\|微观经济学代写Microeconomics代考\|The Slutsky Equation We finally tackle the issue of how the (Marshallian) demand for a good reacts to changes in prices. All the functions to appear below are assumed continuously differentiable, indifference curves are strictly convex, and the initial consumer choice is an interior basket, $x \gg 0$. Let $\mathbf{p}^, m^$ be prices and income in the initial situation, and let $u^=v\left(\mathbf{p}^\right.$, $\left.m^\right)=u\left(\mathbf{x}\left(\mathbf{p}^, m^\right)\right), x_j^=x_j\left(\mathbf{p}^, m^\right)$. Differentiate both sides of $h_f(\mathbf{p}, u)=x_i(\mathbf{p}$, $e(\mathbf{p}, u))=x_i(\mathbf{p}, m)$ with respect to $p_j$ $$\frac{\partial h_i\left(\mathbf{p}^, u^\right)}{\partial p_j}=\frac{\partial x_i\left(\mathbf{p}^, m^\right)}{\partial p_j}+\frac{\partial x_i\left(\mathbf{p}^, m^\right)}{\partial e\left(\mathbf{p}^, u^\right)} \cdot \frac{\partial e\left(\mathbf{p}^, u^\right)}{\partial p_j},$$ where $\frac{\partial x_i\left(\mathbf{p}^, m^\right)}{\partial e\left(p^, u^\right)}$ can also be written $\frac{\partial x_i\left(\mathbf{p}^, m^\right)}{\partial m^}$, because initially $m=m^=e\left(\mathbf{p}^, u^\right)$ and under a balanced budget a variation in expenditure is the same thing as a variation in income. Furthermore, by Shephard’s Lemma $\partial e\left(\mathbf{p}^, u^\right) / \partial p_j=h_j\left(\mathbf{p}^, u^\right)$ $=x_j\left(\mathbf{p}^, e\left(\mathbf{p}^, u^\right)\right)=x_j^$. Hence $$\frac{\partial h_i\left(\mathbf{p}^, u^\right)}{\partial p_j}=\frac{\partial x_i\left(\mathbf{p}^, m^\right)}{\partial p_j}+\frac{\partial x_i\left(\mathbf{p}^, m^\right)}{\partial m} \cdot x_j^* .$$ Let us indicate as $S(\mathbf{p}, m) \equiv\left[s_{i j}\right]$ the $n \times n$ Jacobian matrix of the vector of Hicksian demand functions for individual goods which form $\mathbf{h}(\mathbf{p}, u)$. Element $(i, j)$ of this matrix is $$s_{i j}=\frac{\partial h_i\left(\mathbf{p}^, u^\right)}{\partial p_j},$$ for which equality (4.6) holds. Matrix $S(\mathbf{p}, m) \equiv\left[s_{i j}\right]$ is called Slutsky matrix or also matrix of Hicksian substitution effects. It yields $$\mathrm{dh}=S \mathbf{d p}^T$$ as the variation of compensated demands for infinitesimal variations $\mathrm{d} p$ of the price vector (the superscript $T$ indicates transposition because we treat $\mathbf{p}$ as a row vector). The usefulness of (4.6) is that, rearranging so as to isolate on one side of the equality sign the price effect on Marshallian demand, we obtain the Slutsky equation for Marshallian demand. ## 经济代写\|微观经济学代写Microeconomics代考\|The CES Utility Function $$\left(a_1 x_1^{(\xi-1) / \xi}+a_2 x_2^{(\xi-1) / \xi}\right)^{\xi /(\xi-1)} .$$ $$x_1^\delta / \delta+x_2^\delta / \delta .$$ ## 经济代写\|微观经济学代写Microeconomics代考\|The Slutsky Equation $$\mathrm{dh}=S \mathbf{d} \mathbf{p}^T$$ (4.6) 的用处在于，重新排列以便在等号的一侧隔离价格对马歇尔需求的影响，我们得到马歇尔需求的斯卢茨基方程。 ## 有限元方法代写 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。	2,465	6,359	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2023-40	latest	en	0.651784
https://codereview.stackexchange.com/questions/70369/kaprekars-constant/272776	1,725,993,962,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651303.70/warc/CC-MAIN-20240910161250-20240910191250-00489.warc.gz	161,504,300	49,004	# Kaprekar's constant I'm new to Python and as a first exercise, I've written a function (module?) to calculate Kaprekar's constant using Python 3. I'd like some feedback on whether this is Pythonic enough, and if I can improve it. import collections def kaprekar(value): print("Starting value: %d" % value) # Check our range if value < 1 or value > 9998: print("Input value must be between 1 and 9998, inclusive.") return numstr = str(value) numstr = '0' * (4 - len(numstr)) + numstr # Make sure there are at least two different digits if len(collections.Counter(numstr)) == 1: print("Input value must consist of at least two different digits.") return # If we've gotten this far it means the input value is valid # Start iterating until we reach our magic value of 6174 n = 0 while (value != 6174): n += 1 numstr = str(value) numstr = '0' * (4 - len(numstr)) + numstr # Get ascending and descending integer values asc = int(''.join(sorted(numstr))) dec = int(''.join(sorted(numstr)[::-1])) # Calculate our new value value = dec - asc print("Iteration %d: %d" % (n, value)) # We should always reach the constant within 7 iterations if n == 8: print("Something went wrong...") return -1 print("Reached 6174 after %d iterations." % n) return n ## Update Thanks for the feedback, everyone! ## Validation • The range check if value < 1 or value > 9998 could be more Pythonically expressed as if not 1 <= value <= 9998. (I would prefer to avoid hard-coding 9998 altogether, though. See below.) • To ensure that there are at least two different digits, you can just use a set. • An easier way to zero-pad a number to four places is '{:04d}'.format(value). That line of code is written twice; it may be worthwhile to extract it into a function. • Returning None and -1 to indicate validation errors and excessive iterations is unusual for Python. When you encounter an error, raise an exception instead. ## Algorithm • The implementation presupposes that you know the answer (6174, within 7 iterations). That's a bit dissatisfying at an intellectual level; it feels like a unit test. As an alternative to checking whether the known goal has been reached, you could check whether the sequence has reached a fixed point. • You hard-code a lot of special numbers: 4, 9998, 6174, 8. It would be nice to reduce the usage of such constants, and where they are necessary, clarify their purpose. The Wikipedia page mentions that there is a 3-digit Kaprekar Process; it would be nice to have code that is easily adaptable to that related problem. • The loop has two purposes: check if the goal has been reached, and count the number of iterations. You have chosen to make the former into the loop termination condition. (It's customary to omit the parentheses there, by the way.) I think that converting it into a counting loop works better, as it brings unity to the thee lines n = 0, n += 1, and if n == 8: …. • By postponing the conversion to int when defining asc, you can simplify the derivation of dec a bit. ## Suggested implementation def kaprekar(value): def digit_str(n, places): return ('{:0%dd}' % places).format(n) #PLACES, GOAL, ITERATION_LIMIT = 3, 495, 7 PLACES, GOAL, ITERATION_LIMIT = 4, 6174, 8 digits = digit_str(value, PLACES) if value <= 0: raise ValueError("Input value must be positive.") if len(digits) != PLACES: raise ValueError("Input value must be %d digits long." % PLACES) if len(set(digits)) < 2: raise ValueError("Input value must consist of at least two different digits.") for iterations in range(ITERATION_LIMIT): print("Iteration %d: %s" % (iterations, digits)) if value == GOAL: print("Reached %d after %d iterations." % (GOAL, iterations)) return iterations asc = ''.join(sorted(digits)) dsc = asc[::-1] value = int(dsc) - int(asc) digits = digit_str(value, PLACES) raise StopIteration("Something went wrong...") ## Further enhancement This version separates the iteration logic from the output routines by using a generator. It also makes no presuppositions about The Answer. import sys def kaprekar(digits): if int(digits) <= 0: raise ValueError("Input value must be positive.") if len(set(digits)) < 2: raise ValueError("Input value must consist of at least two different digits.") places = len(digits) prev_digits = None while digits != prev_digits: yield digits prev_digits = digits asc = ''.join(sorted(digits)) dsc = asc[::-1] value = int(dsc) - int(asc) digits = ('{:0%dd}' % places).format(value) def main(_, numstr): try: for iterations, digits in enumerate(kaprekar(numstr)): print("Iteration {i}: {d}".format(i=iterations, d=digits)) print("Reached {d} after {i} iterations".format(i=iterations, d=digits)) except ValueError as e: print(e, file=sys.stderr) sys.exit(1) if __name__ == '__main__': main(sys.argv) • Thanks for this, very detailed, and I learned a lot. I didn't know about if not 1 <= value <= 9998 in Python, that's pretty cool! But what's the underscore in def main(_, numstr): mean? Commented Nov 23, 2014 at 3:23 • @ReticulatedSpline _ is the conventional name for a variable whose value you don't care about. In this case, sys.argv[0] is the name of the program, which is irrelevant. Commented Nov 28, 2014 at 16:39 • There should be a check that the input comprises 4 digits Commented Nov 24, 2022 at 0:18 • For instance, add: if (4 != len(digits)): raise ValueError("Input value must consist of four digits.") Commented Nov 24, 2022 at 0:24 One possible improvement would be to rewrite your code in a more structured and abstract way. You're solving two distinct problems here • apply a certain transformation function to a number • find out if this transformation has a fixed point Your program would be cleaner if you code these two functions separately. The first one is the Kaprekar's routine: LEN = 4 def kaprekar(num): s = '{:0{}d}'.format(num, LEN) s = ''.join(sorted(s)) a = int(s) b = int(s[::-1]) return b - a Note how I'm using the constant LEN instead of 4. First, magic numbers are frowned upon, second, it makes this function suitable for experimenting with other lengths (perhaps, you'll discover your own constant some day!). The second part is the function that, given an arbitrary function and initial value, calculates the fixed point: def fixpoint(fn, arg, max_iterations=1000): for _ in xrange(max_iterations): prev, arg = arg, fn(arg) if arg == prev: return arg raise ValueError('no fixpoint can be reached') Both these functions don't "know" anything about each other. That makes the program more modular and easier to debug, since you can test each one separately. Now, the main routine is easy: # obtain the argument - left as an exercise # check validity - left as an exercise print fixpoint(kaprekar, argument) • Thanks for your feedback. Again I see this underscore being used, this time as an iterator(?) in a loop. What does the underscore mean? Commented Nov 23, 2014 at 3:25 • @ReticulatedSpline just a convention, it means the variable _ (a normal variable like any other) is not going to be used. Commented Jan 8, 2022 at 18:56 Since it's guaranteed to reach the goal within a number of iterations, I would use an assertion, for example: # in your original assert n < 8 # guaranteed by the algorithm # in 200 success' version raise AssertionError('Was not supposed to reach this, guaranteed by the algorithm') Raising any other error would suggest to users that perhaps they should try to catch the exception, but that's not the case here. It's a theoretically impossible situation. The assertion serves as documentation of our assumptions, and it's a safeguard in case something changes unexpectedly. Failing the assertion would be a material defect, and the program should be rewritten. This code appears twice: numstr = str(value) numstr = '0' (4 - len(numstr)) + numstr Don't repeat yourself. This should been extracted to a function. Using unnecessary parentheses like this is not Pythonic: while (value != 6174): • Thanks for the feedback. I didn't even realize about the parentheses, as I come from a C/C# background and I'm so used to them. Commented Nov 23, 2014 at 3:25 Information I've written a function (module?) If it is in a file of its own it can be considered a module. Criticism import collections def kaprekar(value): In here you are missing the shebang and also a doc-string for the function. #!/usr/bin/env python """kaprekar module Author : Dependencies : collections """ import collections def kaprekar(value): """kaprekar : iterate the given number till it reaches kaprekar constant Parameters : value - integer to use """ Scenario 1 if value < 1 or value > 9998: print("Input value must be between 1 and 9998, inclusive.") return Scenario 2 if len(collections.Counter(numstr)) == 1: print("Input value must consist of at least two different digits.") return Scenario 3 if n == 8: print("Something went wrong...") return -1 You should consider returning a meaningful value for each returns. E.g. -1 for errors and 0-7 for successful iterations. Or you could simply raise an Exception when something goes wrong. raise Exception("Exception String") This is an example; you may use suitable exception type. numstr = '0' * (4 - len(numstr)) + numstr You have duplicated this code twice, you should move it into a function of its own. What you did right if len(collections.Counter(numstr)) == 1: I like this part of your code, It's a smart approach. I also like that fact you are failing the function soon as you discover an inconsistency in the input. However you should raise exceptions as I suggested before. Alternatively you may use < 2. • Thanks for the feedback. I don't have a shebang because this is all under Windows. And thanks for telling me what I did right, at least there was one thing. :) Commented Nov 23, 2014 at 3:24 • @ReticulatedSpline : you need one to make your code executable on it's own in nix. stackoverflow.com/questions/6908143/… Commented Nov 28, 2014 at 8:44 Apart from the existing suggestion to raise errors (I've stuck with AssertionErrors to keep things simple for now), I noted three points of interest: the string padding was explicitly coded twice, we can DRY the difference between two sorts as shown below, and itertools.count provides a better option than a while loop that manually increments: from itertools import count def long_string(value): short_numstr = str(value) return '0'(4-len(short_numstr))+short_numstr def kaprekar(value): assert 1 <= value <= 9998, f"{value} not between 1 and 9998, inclusive." assert len(set(long_string(value))) > 1, f"{value} consists of fewer than two different digits." original = value for n in count(1): numstr = sorted(long_string(value)) value = int(numstr[::-1])-int(numstr) if value == 6174: return n assert n < 8, f"Something went wrong with {original}..." (I've also edited the messages to reference which value caused the problem.) As of Python 3.8, the walrus operator allows a slight line count reduction not used above.	2,755	11,044	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.578125	4	CC-MAIN-2024-38	latest	en	0.726729
https://www.topperlearning.com/r-s-aggarwal-and-v-aggarwal-solutions/cbse-class-10-maths/r-s-aggarwal-and-v-aggarwal-mathematics-x/perimeter-and-areas-of-plane-figures	1,660,339,868,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882571758.42/warc/CC-MAIN-20220812200804-20220812230804-00184.warc.gz	895,307,443	55,887	# R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 15 - Perimeter and Areas of Plane Figures Page / Exercise Solution Solution Solution Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 Solution 12 Solution 13 Solution 14 Solution 15 Solution 16 Solution 17 Solution 18 Solution 19 Solution 20 Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Solution 6 Solution 7 Solution 8 Solution 9 Solution 10 Solution 11 Solution 12 Solution 13 Solution 14 Solution 15 Solution 16 Solution 17 Solution 18 Solution 19 Solution 20 ## Chapter 17 - Perimeter and Areas of Plane Figures Ex. 15A Solution 1 Solution 2 Let a = 42 cm, b = 34 cm and c = 20 cm (i)Area of triangle = (ii)Let base = 42 cm and corresponding height = h cm Then area of triangle = Hence, the height corresponding to the longest side = 16 cm Solution 3 Let a = 18 cm, b = 24 cm, c = 30 cm Then,2s = (18 + 24 + 30) cm = 72 cm s = 36 cm (s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm (i)Area of triangle = (ii)Let base = 18 cm and altitude = x cm Then, area of triangle = Hence, altitude corresponding to the smallest side = 24 cm Solution 4 On dividing 150 m in the ratio 5 : 12 : 13, we get Length of one side = Length of the second side = Length of third side = Let a = 25 m, b = 60 m, c = 65 m (s a) = 50 cm, (s b) = 15 cm, and (s c) = 10 cm Hence, area of the triangle = 750 m2 Solution 5 On dividing 540 m in ratio 25 : 17 : 12, we get Length of one side = Length of second side = Length of third side = =120 m Let a = 250m, b = 170 m and c = 120 m Then, (s a) = 29 m, (s b) = 100 m, and (s c) = 150m The cost of ploughing 100 area is = Rs. 18. 80 The cost of ploughing 1 is = The cost of ploughing 9000 area = = Rs. 1692 Hence, cost of ploughing = Rs 1692. Solution 6 Let the length of one side be x cm Then the length of other side = {40 (17 + x)} cm = (23 - x) cm Hypotenuse = 17 cm Applying Pythagoras theorem, we get Hence, area of the triangle = 60 cm2 Solution 7 Let the sides containing the right - angle be x cm and (x - 7) cm One side = 15 cm and other = (15 - 7) cm = 8 cm perimeter of triangle (15 + 8 + 17) cm = 40 cm Solution 8 Let the sides containing the right angle be x and (x 2) cm One side = 8 cm, and other (8 2) cm = 6 cm = 10 cm Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm Solution 9 Solution 10 Let each side of the equilateral triangle be a cm Solution 11 Let each side of the equilateral triangle be a cm Perimeter of equilateral triangle = 3a = (3 12) cm = 36 cm Solution 12 Let each side of the equilateral triangle be a cm area of equilateral triangle = Height of equilateral triangle Solution 13 Base of right angled triangle = 48 cm Height of the right angled triangle = Solution 14 Let the hypotenuse of right - angle triangle = 6.5 m Base = 6 cm Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm2 Solution 15 The circumcentre of a right - triangle is the midpoint of the hypotenuse Hypotenuse = 2 × (radius of circumcircle) = (2 × 8) cm = 16 cm Base = 16 cm, height = 6 cm Area of right angled triangle Hence, area of the triangle= 48 cm2 Solution 16 Let each equal side be a cm in length. Then, Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm Solution 17 Let each equal side be a cm and base = 80 cm perimeter of triangle = (2a + b) cm = (2 41 + 80) cm = (82 + 80) cm = 162 cm Hence, perimeter of the triangle = 162 cm Solution 18 Let the height be h cm, then a= (h + 2) cm and b = 12 cm Squaring both sides, Therefore, a = h + 2 = (8 + 2)cm = 10 cm Hence, area of the triangle = 48 cm2. Solution 20 Area of shaded region = Area of ABC – Area of DBC First we find area of ABC Second we find area of DBC which is right angled Area of shaded region = Area of ABC – Area of DBC = (43.30 - 24) = 19. 30 Area of shaded region = 19.3 Solution 19 Let ABC is a isosceles triangle. Let AC, BC be the equal sides Then AC = BC = 10cm. Let AB be the base of ABC right angle at C. Area of right isosceles triangle ABC Hence, area = 50 cm2 and perimeter = 34.14 cm ## Chapter 17 - Perimeter and Areas of Plane Figures Ex. 15B Solution 2 Solution 5 Solution 6 Solution 7 Area of floor = Length Breadth Area of carpet = Length Breadth = Number of carpets = = 216 Hence the number of carpet pieces required = 216 Solution 8 Area of verandah = (36 × 15) = 540 Area of stone = (0.6 × 0.5) [10 dm = 1 m] Number of stones required = Hence, 1800 stones are required to pave the verandah. Solution 9 Perimeter of rectangle = 2(l + b) 2(l + b) = 56 Þ l + b = 28 cm b = (28 l) cm Area of rectangle = 192 l (28 l) = 192 28l - = 192 - 28l + 192 = 0 - 16l 12l + 192 = 0 l(l 16) 12(l 16) = 0 (l 16) (l 12) = 0 l = 16 or l = 12 Therefore, length = 16 cm and breadth = 12 cm Solution 10 Length of the park = 35 m Breadth of the park = 18 m Area of the park = (35 18) = 630 Length of the park with grass =(35 5) = 30 m Breadth of the park with grass = (18- 5) m = 13 m Area of park with grass = (30 13) = 390 Area of path without grass = Area of the whole park area of park with grass = 630 - 390 = 240 Hence, area of the park to be laid with grass = 240 m2 Solution 11 Length of the plot = 125 m Breadth of the plot = 78 m Area of plot ABCD = (125 78) = 9750 Length of the plot including the path= (125 + 3 + 3) m = 131 m Breadth of the plot including the path = (78 + 3 + 3) m = 84 m Area of plot PQRS including the path = (131 84) = 11004 Area of path = Area of plot PQRS Area of plot ABCD = (11004 9750) = 1254 Cost of gravelling = Rs 75 per m2 Cost of gravelling the whole path = Rs. (1254 75) = Rs. 94050 Hence, cost of gravelling the path = Rs 94050 Solution 12(i) Area of rectangular field including the foot path = (54 35) Let the width of the path be x m Then, area of rectangle plot excluding the path = (54 2x) (35 2x) Area of path = (54 35) (54 2x) (35 2x) (54 35) (54 2x) (35 2x) = 420 1890 1890 + 108x + 70x - 4 = 420 178x - 4 = 420 4 - 178x + 420 = 0 2 - 89x + 210 = 0 2 - 84x 5x + 210 = 0 2x(x 42) 5(x 42) = 0 (x 42) (2x 5) = 0 Solution 13 Let the length and breadth of a rectangular garden be 9x and 5x. Then, area of garden = (9x 5x)m = 45 Length of park excluding the path = (9x 7) m Breadth of the park excluding the path = (5x 7) m Area of the park excluding the path = (9x 7)(5x 7) Area of the path = (98x 49) = 1911 98x = 1911 + 49 Length = 9x = 9 20 = 180 m Breadth = 5x = 5 20 = 100 m Hence, length = 180 m and breadth = 100 m Solution 14 Solution 15 Let the width of the carpet = x meter Area of floor ABCD = (8 5) Area of floor PQRS without border = (8 2x)(5 2x) = 40 16x 10x + = 40 26x + Area of border = Area of floor ABCD Area of floor PQRS = [40 (40 26x + )] =[40 40 + 26x - ] = (26x - ) Solution 16 Solution 17 Solution 18 Solution 19 Solution 20 Area of the square = Let diagonal of square be x Length of diagonal = 16 cm Side of square = Perimeter of square = [4 side] sq. units =[ 4 11.31] cm = 45.24 cm Solution 21 Let d meter be the length of diagonal Area of square field = Time taken to cross the field along the diagonal Hence, man will take 6 min to cross the field diagonally. Hence, man will take 6 min to cross the field diagonally. Solution 22 Solution 23 Rs. 14 is the cost of fencing a length = 1m Rs. 28000 is the cost of fencing the length= Perimeter = 4 side = 2000 side = 500 m Area of a square = = 250000 Cost of mowing the lawn = Solution 24 Solution 27 Area of quad. ABCD = Area of ABD + Area ofDBC For area of ABD Let a = 42 cm, b = 34 cm, and c = 20 cm For area of DBC a = 29 cm, b = 21 cm, c = 20 cm Solution 26 Area of quad. ABCD = Area of ABC + Area of ACD Now, we find area of a ACD Area of quad. ABCD = Area of ABC + Area of ACD Perimeter of quad. ABCD = AB + BC + CD + AD =(17 + 8 + 12 + 9) cm = 46 cm Perimeter of quad. ABCD = 46 cm Solution 25 ABCD be the given quadrilateral in which AD = 24 cm, BD = 26 cm, DC = 26 cm and BC = 26 cm By Pythagoras theorem For area of equilateral DBC, we have a = 26 cm Area of quad. ABCD = Area of ABD + Area of DBC = (120 + 292.37) = 412.37 Perimeter ABCD = AD + AB + BC + CD = 24 cm + 10 cm + 26 cm + 26 cm = 86 cm Solution 28 Area of the \|\|gm = (base height) sq. unit = (25 16.8) Solution 29 Longer side = 32 cm, shorter side = 24 cm Distance between longer sides = 17.4 cm Let the distance between the shorter sides be x cm Area of \|\|gm = (longer side distance between longer sides) = (shorter side distance between the short sides) distance between the shorter side = 23.2 cm Solution 30 Solution 31 Area ofparallelogram = 2 area of DABC Opposite sides of parallelogram are equal AD = BC = 20 cm And AB = DC = 34 cm In ABC we have a = AC = 42 cm b = AB = 34 cm c = BC = 20 cm Then, (s a) = 6 cm, (s b) = 14 cm and (s c) = 28 cm Solution 32 We know that the diagonals of a rhombus, bisect each other at right angles OA = OC = 15 cm, And OB = OD = 8 cm And AOB = 90 By Pythagoras theorem, we have Solution 33 (i)Perimeter of rhombus = 4 side 4 side = 60 cm By Pythagoras theorem OB = 12 cm OB = OD = 12 cm BD = OB + OD = 12 cm + 12 cm = 24 cm Length of second diagonal is 24 cm (ii) Area of rhombus = Solution 34 (i)Area of rhombus = 480 One of its diagonals = 48 cm Let the second diagonal =x cm Hence the length of second diagonal 20 cm (ii)We know that the diagonals of a rhombus bisect each other at right angles AC = 48, BD = 20 cm OA = OC = 24 cm and OB = OD = 10 cm By Pythagoras theorem , we have (iii)Perimeter of the rhombus = (4 26) cm = 104 cm Solution 35 Solution 36 Areaof cross section = Solution 37 Let ABCD be a given trapezium in which AB = 25, CD = 11 BC = 15, AD = 13 AE = CD = 11 cm, And BE = AB BE = 25 11 = 14 cm In BEC, Area of BEC = Let height of BEC is h Area of BEC = From (1) and (2), we get 7h = 84 h = 12 m Area of trapezium ABCD ### STUDY RESOURCES REGISTERED OFFICE : First Floor, Empire Complex, 414 Senapati Bapat Marg, Lower Parel, Mumbai - 400013, Maharashtra India.	3,606	10,245	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.75	5	CC-MAIN-2022-33	longest	en	0.746364
https://complementarytraining.net/optimal-force-velocity-profile-for-sprinting-is-it-all-bollocks-part-4/	1,685,507,603,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224646257.46/warc/CC-MAIN-20230531022541-20230531052541-00568.warc.gz	186,829,766	31,594	Optimal Force-Velocity Profile for Sprinting: Is It All Bollocks? – Part 4 - Complementary Training Optimal Force-Velocity Profile for Sprinting: Is It All Bollocks? – Part 4 # Optimal Force-Velocity Profile for Sprinting: Is It All Bollocks? – Part 4 ### Previous Part: In the previous installment, I have demonstrated how to perform sensitivity/optimization analysis of the Acceleration-Velocity Profile (AVP) using (1) probing method and (2) slope method. In this installment I will explain how to do it using the Force-Velocity Profile (FVP). As we have learned previously, the sensitivity/optimization analysis plays with the parameters under certain constraints (i.e., increase for certain % amount for the probing method, or change in slope while keeping the $P_{max}$ the same for the slope method) to yield the lowest split times. Rather than using MSS and MAC parameters to calculate model predicted split times, we can use $F_0$ and $V_0$ to do so, since they are estimated using MSS and MAC themselves (speaking about the polynomial method explained previously). Math is more involved (and covered in Samozino et al. (2022)), but I will cover the simple logic here using Equation 1. \begin{split} t(d) = f(d, \; F_0, \; V_0, \; k)\\\end{split}Equation 1 Model predicted split times depends on distance selected, $F_0$ and $V_0$ , as well as the $k$ value, which is used to model the air resistance. As shown in Equation 2, $k$ value depends on the body weight, body height, air pressure, air temperature and wind velocity. \begin{split} k = f(weight, \;height, \;Pressure, \;Temp, \;v_{wind})\\\end{split}Equation 2 In the previous installment we have used athlete with known/true MSS (maximum-sprinting-speed) equal to 9 $ms^{-1}$ and MAC (maximum-acceleration) equal to 8 $ms^{-2}$. These are model parameters that we will use to generate the data (again, we are using model predictions). These parameters gives us TAU of 1.125 $s$ and $P_{max}$ equal to 18 $W/kg$. Using the method explained in the previous installments, and assuming our athlete wights 85kg with a body height of 185cm, we can estimate $F_0$ and $V_0$ (using polynomial method) (Table 1). Show/Hide Code require(tidyverse) require(shorts) require(cowplot) require(directlabels) require(kableExtra) athlete_MSS <- 9 athlete_MAC <- 8 athlete_BW <- 85 athlete_height <- 1.85 # Get FVP fvp <- make_FV_profile( athlete_MSS, athlete_MAC, bodymass = athlete_BW, bodyheight = athlete_height ) athlete_F0 <- fvp$F0_poly athlete_F0_rel <- fvp$F0_poly_rel athlete_V0 <- fvp$V0_poly athlete_PMAX <- fvp$Pmax_poly athlete_PMAX_rel <- fvp$Pmax_poly_rel athlete_Slope <- fvp$FV_slope athlete_Ppeak <- shorts:::find_FV_peak_power( athlete_F0, athlete_V0, bodymass = athlete_BW, bodyheight = athlete_height ) athlete_Ppeak_rel <- athlete_Ppeak / athlete_BW df <- tribble( ~MSS (m/s), ~MAC (m/s/s), ~TAU (s), ~AV Slope, ~net PMAX (W/kg), ~Weight (kg), ~Height (m), ~F0 (N), ~rel F0 (N/kg), ~V0 (m/s), ~PMAX (W), ~rel PMAX (W/kg), ~FV Slope, ~Ppeak (W), ~rel Peak (W/kg), athlete_MSS, athlete_MAC, athlete_MSS / athlete_MAC, -athlete_MAC / athlete_MSS, athlete_MSS * athlete_MAC / 4, athlete_BW, athlete_height, athlete_F0, athlete_F0_rel, athlete_V0, athlete_PMAX, athlete_PMAX_rel, athlete_Slope, athlete_Ppeak, athlete_Ppeak_rel ) kbl(t(df), digits = 2) %>% kable_classic(full_width = FALSE) %>% pack_rows("Acceleration-Velocity Profile", 1, 5) %>% pack_rows("Force-Velocity Profile", 6, 15) Acceleration-Velocity Profile MSS (m/s) 9.00 MAC (m/s/s) 8.00 TAU (s) 1.12 AV Slope -0.89 net PMAX (W/kg) 18.00 Force-Velocity Profile Weight (kg) 85.00 Height (m) 1.85 F0 (N) 680.00 rel F0 (N/kg) 8.00 V0 (m/s) 9.34 PMAX (W) 1588.19 rel PMAX (W/kg) 18.68 FV Slope -0.85 Ppeak (W) 1558.14 rel Peak (W/kg) 18.33 Table 1: Athlete characteristics ## Why FVP and not just AVP? Why do we need Force-Velocity Profile (FVP) and not just Acceleration-Velocity Profile (AVP)? I should have addressed this question in the previous installments, but I think this is a good time for this interlude. This is a great question to ask. In my instrumentalist viewpoint, they are both summaries of performance. But for someone taking realist stance, AVP is a summary of performance, while FVP is determinant of performance. In other words, FVP causes AVP. Thus, FVP reveals individual traits/qualities that are causes or determinants of performance. I do not succumb to such a viewpoint. Simply, if you cue someone with a small technique improvement, sprint performance will improve, and thus the FVP. I thus, believe both FVP and AVP are simply summaries of sprint performance, a NOT determinants. But anyway, let us us “fuck-around-and-find-out” with a realist perspective using FVP as a cause of performance, summarized by AVP. Since FVP is estimated using AVP and body dimensions (weight and height) as well as air pressure, air temperature and wind velocity, we can reverse this, and use FVP as a generative model to estimate AVP from FVP (I will not bother you with math, but you can do it yourself by reading Samozino et al. (2022)). Under realist perspective, if my FVP is fixed (with some biological variation, but for this thought experiment, we will assume it is), then changes in air pressure, air temperature and wind velocity, but most notably in body weight, will affect my sprint performance (which is summarized by AVP). Let us explore this by calculating effects of simulated changes in $F_0$, $V_0$ and body weight on the AVP (MSS and MAC parameters). These are depicted in Figure 1. Show/Hide Code model_sens_df <- data.frame( F0 = athlete_F0, V0 = athlete_V0, BW = athlete_BW ) %>% expand_grid( increase = c("F0", "V0", "BW"), factor = seq(80, 120)) %>% mutate( bodymass = athlete_BW, increase = factor(increase, levels = c("F0", "V0", "BW")), new_F0 = ifelse(increase == "F0", F0 * factor / 100, F0), new_V0 = ifelse(increase == "V0", V0 * factor / 100, V0), new_BW = ifelse(increase == "BW", BW * (factor / 100), BW)) %>% mutate(data.frame(shorts:::convert_FV(new_F0, new_V0, bodymass = new_BW, bodyheight = athlete_height))) %>% pivot_longer(cols = c(MSS, MAC), names_to = "param") # Plot ggplot(model_sens_df, aes(x = factor - 100, y = value, color = param)) + theme_linedraw(8) + geom_vline(xintercept = 0, linetype = "dashed") + geom_line(alpha = 0.6) + geom_dl( aes(label = paste(" ", param)), method = list("last.bumpup", cex = 0.5) ) + facet_wrap(~increase) + ylab(NULL) + xlab("Parameter Improvement (%)") + theme(legend.position = "none") + xlim(-20, 27.5) Figure 1: Using FVP as generative model and estimating effects of change in parameters on the sprint performance summarized with AVP Interesting finding of this thought experiment (also a form of sensitivity analysis or “fuck-around-and-find-out”) in Figure 1, is that, assuming $F_0$ and $V_0$ being causal qualities/traits independent of body weight, changing body weight will not influence MSS, or maximum achievable sprinting speed. This is a normal conclusion of such a realist perspective using map as a territory, but little bit of common-sense and research (of which I am not currently aware of) can bring this into question. For example, if I attach 10% of body weight in a form of a weighted vest, ankle or wrist weight, or frictionless sled, I am pretty sure that the (1) estimated MSS will definitely not be the same, and (2) estimated FVP for that loaded performance will also be different. This simple “fuck-around-and-find-out” in the Large World (a.k.a., experiment) breaks the realist assumptions of the FVP being determinant of performance. If it was determinant or a causal mechanism, then predictions from the “fuck-around-and-find-out” in the Small World (Figure 1) would be found the be true in the Large World. I am not aware of any research doing this type of analysis, but here is a potential topic for a research paper or even PhD – well maybe I can do this? We can thus get back to real life and accept that FVP nor $P_{max}$ are not some magical determinants we need to focus on (e.g., plenty of research on finding loads that maximize power, yada yada yada), but results of performance itself. Let’s get back to our sensitivity analysis in the Small World. ## Optimization of the Force-Velocity Profile Using exactly the same probing method we have done in the previous installment using the AVP, and our athlete from Table 1, we can now perform sensitivity analysis of the model (i.e., map) by changing $F_0$, $V_0$, and body weight to check the effects on model predicted split times. The results are depicted in Figure 2. Please note that improvement in body weight mean reducing body weight for enlisted percentage.	2,382	8,715	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2023-23	longest	en	0.826908
https://howkgtolbs.com/convert/61.26-kg-to-lbs	1,660,973,709,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882573908.30/warc/CC-MAIN-20220820043108-20220820073108-00020.warc.gz	276,745,972	12,233	# 61.26 kg to lbs - 61.26 kilograms to pounds Before we go to the more practical part - this is 61.26 kg how much lbs conversion - we want to tell you few theoretical information about these two units - kilograms and pounds. So let’s move on. How to convert 61.26 kg to lbs? 61.26 kilograms it is equal 135.0551817012 pounds, so 61.26 kg is equal 135.0551817012 lbs. ## 61.26 kgs in pounds We will begin with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, formally known as International System of Units (in abbreviated form SI). From time to time the kilogram could be written as kilogramme. The symbol of this unit is kg. First definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. This definition was not complicated but difficult to use. Later, in 1889 the kilogram was defined by the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was substituted by another definition. Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.” One kilogram is equal 0.001 tonne. It is also divided into 100 decagrams and 1000 grams. ## 61.26 kilogram to pounds You know a little about kilogram, so now we can go to the pound. The pound is also a unit of mass. We want to highlight that there are more than one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we are going to to focus only on pound-mass. The pound is in use in the British and United States customary systems of measurements. To be honest, this unit is in use also in other systems. The symbol of this unit is lb or “. The international avoirdupois pound has no descriptive definition. It is equal 0.45359237 kilograms. One avoirdupois pound is divided to 16 avoirdupois ounces and 7000 grains. The avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.” ### How many lbs is 61.26 kg? 61.26 kilogram is equal to 135.0551817012 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218. ### 61.26 kg in lbs The most theoretical section is already behind us. In this part we will tell you how much is 61.26 kg to lbs. Now you learned that 61.26 kg = x lbs. So it is high time to get the answer. Let’s see: 61.26 kilogram = 135.0551817012 pounds. That is an accurate outcome of how much 61.26 kg to pound. It is possible to also round off the result. After it your outcome will be exactly: 61.26 kg = 134.772 lbs. You know 61.26 kg is how many lbs, so look how many kg 61.26 lbs: 61.26 pound = 0.45359237 kilograms. Naturally, this time it is possible to also round off this result. After rounding off your outcome will be exactly: 61.26 lb = 0.45 kgs. We are also going to show you 61.26 kg to how many pounds and 61.26 pound how many kg outcomes in charts. See: We will start with a table for how much is 61.26 kg equal to pound. ### 61.26 Kilograms to Pounds conversion table Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places) 61.26 135.0551817012 134.7720 Now see a table for how many kilograms 61.26 pounds. Pounds Kilograms Kilograms (rounded off to two decimal places 61.26 0.45359237 0.45 Now you know how many 61.26 kg to lbs and how many kilograms 61.26 pound, so we can move on to the 61.26 kg to lbs formula. ### 61.26 kg to pounds To convert 61.26 kg to us lbs you need a formula. We will show you a formula in two different versions. Let’s begin with the first one: Number of kilograms * 2.20462262 = the 135.0551817012 outcome in pounds The first version of a formula give you the most exact result. In some cases even the smallest difference could be considerable. So if you need a correct outcome - this version of a formula will be the best for you/option to know how many pounds are equivalent to 61.26 kilogram. So let’s go to the another formula, which also enables calculations to know how much 61.26 kilogram in pounds. The second formula is down below, look: Amount of kilograms * 2.2 = the result in pounds As you can see, this formula is simpler. It could be better option if you want to make a conversion of 61.26 kilogram to pounds in quick way, for example, during shopping. You only have to remember that your result will be not so accurate. Now we are going to learn you how to use these two versions of a formula in practice. But before we will make a conversion of 61.26 kg to lbs we are going to show you another way to know 61.26 kg to how many lbs without any effort. ### 61.26 kg to lbs converter An easier way to check what is 61.26 kilogram equal to in pounds is to use 61.26 kg lbs calculator. What is a kg to lb converter? Converter is an application. Converter is based on longer version of a formula which we gave you in the previous part of this article. Due to 61.26 kg pound calculator you can easily convert 61.26 kg to lbs. You only have to enter amount of kilograms which you want to convert and click ‘convert’ button. You will get the result in a flash. So try to calculate 61.26 kg into lbs using 61.26 kg vs pound calculator. We entered 61.26 as a number of kilograms. This is the outcome: 61.26 kilogram = 135.0551817012 pounds. As you see, our 61.26 kg vs lbs converter is intuitive. Now let’s move on to our chief topic - how to convert 61.26 kilograms to pounds on your own. #### 61.26 kg to lbs conversion We are going to begin 61.26 kilogram equals to how many pounds conversion with the first formula to get the most correct outcome. A quick reminder of a formula: Amount of kilograms * 2.20462262 = 135.0551817012 the outcome in pounds So what have you do to check how many pounds equal to 61.26 kilogram? Just multiply number of kilograms, in this case 61.26, by 2.20462262. It is exactly 135.0551817012. So 61.26 kilogram is exactly 135.0551817012. It is also possible to round off this result, for instance, to two decimal places. It is 2.20. So 61.26 kilogram = 134.7720 pounds. It is high time for an example from everyday life. Let’s calculate 61.26 kg gold in pounds. So 61.26 kg equal to how many lbs? As in the previous example - multiply 61.26 by 2.20462262. It gives 135.0551817012. So equivalent of 61.26 kilograms to pounds, if it comes to gold, is 135.0551817012. In this example it is also possible to round off the result. Here is the result after rounding off, in this case to one decimal place - 61.26 kilogram 134.772 pounds. Now let’s move on to examples calculated using a short version of a formula. #### How many 61.26 kg to lbs Before we show you an example - a quick reminder of shorter formula: Number of kilograms * 2.2 = 134.772 the outcome in pounds So 61.26 kg equal to how much lbs? And again, you need to multiply number of kilogram, this time 61.26, by 2.2. Look: 61.26 * 2.2 = 134.772. So 61.26 kilogram is equal 2.2 pounds. Let’s make another calculation using this formula. Now calculate something from everyday life, for instance, 61.26 kg to lbs weight of strawberries. So convert - 61.26 kilogram of strawberries * 2.2 = 134.772 pounds of strawberries. So 61.26 kg to pound mass is exactly 134.772. If you learned how much is 61.26 kilogram weight in pounds and are able to calculate it using two different formulas, we can move on. Now we are going to show you these outcomes in charts. #### Convert 61.26 kilogram to pounds We are aware that outcomes presented in tables are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in charts for your convenience. Due to this you can easily compare 61.26 kg equivalent to lbs results. Let’s begin with a 61.26 kg equals lbs chart for the first version of a formula: Kilograms Pounds Pounds (after rounding off to two decimal places) 61.26 135.0551817012 134.7720 And now look 61.26 kg equal pound table for the second formula: Kilograms Pounds 61.26 134.772 As you can see, after rounding off, if it comes to how much 61.26 kilogram equals pounds, the results are the same. The bigger number the more significant difference. Keep it in mind when you need to make bigger amount than 61.26 kilograms pounds conversion. #### How many kilograms 61.26 pound Now you know how to calculate 61.26 kilograms how much pounds but we want to show you something more. Are you interested what it is? What about 61.26 kilogram to pounds and ounces conversion? We are going to show you how you can convert it step by step. Let’s begin. How much is 61.26 kg in lbs and oz? First things first - you need to multiply amount of kilograms, this time 61.26, by 2.20462262. So 61.26 * 2.20462262 = 135.0551817012. One kilogram is exactly 2.20462262 pounds. The integer part is number of pounds. So in this case there are 2 pounds. To know how much 61.26 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces. So your result is exactly 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then final result will be exactly 2 pounds and 33 ounces. As you see, conversion 61.26 kilogram in pounds and ounces simply. The last conversion which we will show you is calculation of 61.26 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work. To convert foot pounds to kilogram meters you need another formula. Before we give you this formula, see: • 61.26 kilograms meters = 7.23301385 foot pounds, • 61.26 foot pounds = 0.13825495 kilograms meters. Now see a formula: Number.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters So to convert 61.26 foot pounds to kilograms meters you have to multiply 61.26 by 0.13825495. It is exactly 0.13825495. So 61.26 foot pounds is exactly 0.13825495 kilogram meters. You can also round off this result, for example, to two decimal places. Then 61.26 foot pounds is equal 0.14 kilogram meters. We hope that this conversion was as easy as 61.26 kilogram into pounds conversions. We showed you not only how to make a conversion 61.26 kilogram to metric pounds but also two other conversions - to check how many 61.26 kg in pounds and ounces and how many 61.26 foot pounds to kilograms meters. We showed you also another way to make 61.26 kilogram how many pounds conversions, it is with use of 61.26 kg en pound converter. It is the best choice for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way. We hope that now all of you are able to make 61.26 kilogram equal to how many pounds calculation - on your own or with use of our 61.26 kgs to pounds converter. It is time to make your move! Let’s convert 61.26 kilogram mass to pounds in the best way for you. Do you need to make other than 61.26 kilogram as pounds conversion? For instance, for 5 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so easy as for 61.26 kilogram equal many pounds. ### How much is 61.26 kg in pounds To quickly sum up this topic, that is how much is 61.26 kg in pounds , we gathered answers to the most frequently asked questions. Here you can find the most important information about how much is 61.26 kg equal to lbs and how to convert 61.26 kg to lbs . Let’s see. How does the kilogram to pound conversion look? It is a mathematical operation based on multiplying 2 numbers. Let’s see 61.26 kg to pound conversion formula . See it down below: The number of kilograms * 2.20462262 = the result in pounds How does the result of the conversion of 61.26 kilogram to pounds? The accurate answer is 135.0551817012 lb. There is also another way to calculate how much 61.26 kilogram is equal to pounds with second, shortened version of the formula. Let’s see. The number of kilograms * 2.2 = the result in pounds So in this case, 61.26 kg equal to how much lbs ? The answer is 135.0551817012 pounds. How to convert 61.26 kg to lbs quicker and easier? You can also use the 61.26 kg to lbs converter , which will do all calculations for you and you will get an exact result . #### Kilograms [kg] The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side. #### Pounds [lbs] A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.	3,496	13,382	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.84375	4	CC-MAIN-2022-33	longest	en	0.953418
https://math.stackexchange.com/questions/3129429/why-are-there-two-different-recurrences-for-gegenbauer-polynomials/3154544	1,563,722,611,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195527048.80/warc/CC-MAIN-20190721144008-20190721170008-00296.warc.gz	473,803,809	38,835	# Why are there two different recurrences for Gegenbauer polynomials? As I mentioned previously, I've been reading up on Gegenbauer polynomials in preparation for a blog post on the kissing number problem—specifically, the Delsarte method. To make a long story short, the method involves expressing a particular function as a non-negative linear sum of Gegenbauer polynomials. In various publications on this topic (see, for example, Musin, "The Kissing Number in Four Dimensions" in the July 2008 Annals of Mathematics), these polynomials have the following recurrence: $$G^{(n)}_0(t) = 1$$ $$G^{(n)}_1(t) = t$$ $$G^{(n)}_k(t) = \frac{(2k+n-4)tG^{(n)}_{k-1}(t)-(k-1)G^{(n)}_{k-2}(t)}{k+n-3}$$ However, in the Wikipedia and Wolfram MathWorld plot summaries for Gegenbauer polynomials, the recurrences are different (and not just up to a constant factor). In both, converting to consistent symbols, we have $$G^{(n)}_0(t) = 1$$ $$G^{(n)}_1(t) = 2nt$$ $$G^{(n)}_k(t) = \frac{2t(k+n-1)G^{(n)}_{k-1}(t)-(k+2n-2)G^{(n)}_{k-2}(t)}{k}$$ Both definitions are normalized to $$G^{(n)}_0(t) = 1$$. What accounts for the difference between the two? I'll use $$G$$ for the second definition and $$\widetilde G$$ for the first definition. $$G$$ has the generating function $$(1 - 2 x t + t^2)^{-n}$$: $$(1 - 2 x t + t^2)^{-n} = \sum_{k \geq 0} G_k^{(n)}(x) \,t^k.$$ $$\widetilde G$$ is constructed by first taking $$C$$ with the gf $$(1 - 2 x t + t^2)^{1 -n/2}$$ and then normalizing by $$C(1)$$: $$(1 - 2 x t + t^2)^{1 - n/2} = \sum_{k \geq 0} C_k^{(n)}(x) \,t^k, \\ \widetilde G_k^{(n)}(x) = \frac {C_k^{(n)}(x)} {C_k^{(n)}(1)}.$$ Therefore $$\widetilde G_k^{(n)}(x) = \frac {(-1)^k} {\binom {2 - n} k } G_k^{(n/2 - 1)}(x).$$ • Interesting. Can you say something about why this difference exists? Is there a particular reason why one formulation should be preferred for some applications, and the other formulation for other applications? – Brian Tung Mar 19 at 23:46 • I suppose it's a question of which makes the formulas less cumbersome for a particular application. Wikipedia gives a number of relations for $G_k^{(n)}$ and mentions that $G_k^{(n/2 - 1)}$ are spherical harmonics. The cited paper uses the fact that the matrix $(\widetilde G_k^{(n)}(\mathbf x_i \cdot \mathbf x_j))$ is positive semidefinite. – Maxim Mar 20 at 1:11 • OK. At any rate, that's worth the bounty. Thanks! – Brian Tung Mar 21 at 7:23 Here are some aspects which might help to clarify the situation. At first I'd like to give a few definitions of Gegenbauer polynomials from relevant sources. This way we can get an impression what we typically might expect. Higher Transcendental Functions, Vol I by A. Erdelyi and H. Bateman (author): • (3.15.1. Gegenbauer polynomials) Gegenbauer's polynomial $$C_n^{\nu}(z)$$ for integral value $$n$$ is defined to be the coefficient of $$h^n$$ in the expansion of $$(1-2hz+h^2)^{-\nu}$$ in powers of $$h$$; \begin{align} (1-2hz+h^2)^{-\nu}=\sum_{n=0}^\infty C_n^{\nu}(z)h^n\qquad\qquad\|h\|<\|z\pm(z^2-1)^{1/2}\| \end{align} Handbook of Mathematical Functions by M. Abramowitz, I.A. Stegun: • (22.9.3. Generating Functions) \begin{align} (1-2xz+z^2)^{-\alpha}=\sum_{n=0}^\infty C_n^{(\alpha)}(x)z^n\qquad\qquad \|z\|<1,\alpha\ne 0 \end{align} NIST/DLMF • (18.12.4. Ultraspherical) \begin{align} (1-2xz+z^2)^{-\alpha}=\sum_{n=0}^\infty C_n^{(\alpha)}(x)z^n\qquad\qquad \|z\|<1 \end{align} Special Functions, Encyclopedia of mathematics and its applications 71 by G.E. Andrews, R. Askey and R. Roy: • (6.4 Generating Functions for Jacobi Polynomials) It is reasonable to define polynomials \begin{align} C_n^{\lambda}(x):=\frac{(2\lambda)_n}{(\lambda+(1/2))_n}P_n^{(\lambda-(1/2),\lambda-(1/2))}(x)\tag{1} \end{align} $$\qquad$$with the generating function \begin{align} (1-2xr+r^2)^{-\lambda}=\sum_{n=0}^\infty C_n^{\lambda}(x)r^n.\tag{2} \end{align} Note the exponent of $$(1-2xr+r^2)^{\color{blue}{-\lambda}}$$ and the upper index in $$C_n^{\color{blue}{\lambda}}(x)$$ in (2) are coupled by a multiplicative factor $$-1$$ and this notational convention is used in all these citations. The connection with the Jacobi polynomials is given in (1). We take a look at OPs cited paper The Kissing Number in Four Dimensions by O.R. Musin, check a few statements and a cited reference we are interested in. • (3.B The Gegenbauer polynomials) ... Let us recall definitions of Gegenbauer polynomials $$C_k^{(n)}(t)$$, which are defined by the expansion \begin{align} (1-2rt+r^2)^{(2-n)/2}=\sum_{k=0}^\infty r^kC_k^{(n)}(t)\tag{3} \end{align} This definition looks somewhat peculiar, since the upper index $$n$$ of $$C_k^{(\color{blue}{n})}$$ is not coupled by a factor $$-1$$ with the exponent of $$(1-2rt+r^2)^{\color{blue}{(2-n)/2}}$$ of the generating function. A few lines above (3.B) the author states ... Schoenberg [29] extended this property to Gegenbauer polynomials $$G_k^{(n)}$$. He proved: The matrix $$\left(G_k^{(n)}\left(\cos \phi_{i,j}\right)\right)$$ is positive semidefinite for any finite $$X\subseteq \mathbf{S}^{n-1}$$. The reference [29] addresses the paper Positive definite functions on spheres by I.J. Schoenberg. It is revealing to check the definition Schoenberg used for Gegenbauer (resp. ultraspherical) polynomials. • (Schoenberg [29], section 1) ... Let $$P_n^{(\lambda)}(\cos t)$$ be the ultraspherical polynomials defined by the expansion \begin{align} (1-2r\cos t+r^2)^{-\lambda}=\sum_{n=0}^\infty r^nP_n^{(\lambda)}(\cos t),\qquad\qquad (\lambda>0).\tag{4} \end{align} Note in definition (4) the parameter $$\lambda$$ is used in accordance with the citations above. A few lines later Schoenberg gives a series expansion of a function $$g(t)$$ in terms of Gegenbauer polynomials: • (Schoenberg [29], section 1) ... The most general element of $$\mathcal{P}(S_m)$$ is \begin{align} g(t)=\sum_{n=0}^\infty P_n^{(\lambda)}(\cos t),\qquad\qquad(\lambda=\frac{1}{2}(m-1)),\tag{5} \end{align} Conclusion: Comparing Schoenberg's usage of the parameter $$\lambda$$ in (5) and (4) with Musin's parameter setting in (3) indicates he could have had some short-cut notation in mind. Regrettably this implies that recurrence relation, differential equations, etc. have a different form than usually expected. The differences in the recurrence relation do not occur, if Musin would have used (3') instead \begin{align} (1-2rt+r^2)^{-\lambda}=\sum_{k=0}^\infty r^kC_k^{(\lambda)}(t)\qquad\qquad (\lambda=(n-2)/2)\tag{3'} \end{align} • Thanks, that's very helpful! – Brian Tung Mar 21 at 22:33 • @BrianTung: You're welcome. – Markus Scheuer Mar 21 at 22:34	2,241	6,625	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 49, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2019-30	latest	en	0.775512
https://nrich.maths.org/public/topic.php?code=71&cl=2&cldcmpid=7989	1,579,521,636,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250598726.39/warc/CC-MAIN-20200120110422-20200120134422-00536.warc.gz	583,003,631	10,032	# Resources tagged with: Mathematical reasoning & proof Filter by: Content type: Age range: Challenge level: ### There are 90 results Broad Topics > Using, Applying and Reasoning about Mathematics > Mathematical reasoning & proof ### What Do You Need? ##### Age 7 to 11 Challenge Level: Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number? ### Always the Same ##### Age 11 to 14 Challenge Level: Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. 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Find all the ways of replacing the letters with digits so that the arithmetic is correct. ### Cows and Sheep ##### Age 7 to 11 Challenge Level: Use your logical reasoning to work out how many cows and how many sheep there are in each field. ### 9 Weights ##### Age 11 to 14 Challenge Level: You have been given nine weights, one of which is slightly heavier than the rest. Can you work out which weight is heavier in just two weighings of the balance? ##### Age 11 to 14 Challenge Level: Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true. ### Sticky Numbers ##### Age 11 to 14 Challenge Level: Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number? ### Take Three Numbers ##### Age 7 to 11 Challenge Level: What happens when you add three numbers together? Will your answer be odd or even? How do you know? ### Always, Sometimes or Never? 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Players take it in turns to select a word from the 9 words given. The aim is to select all the occurrences of the same letter. ### Unit Fractions ##### Age 11 to 14 Challenge Level: Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are natural numbers and 0 < a < b < c. Prove that there is only one set of values which satisfy this equation. ### Children at Large ##### Age 11 to 14 Challenge Level: There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children? ### Cycle It ##### Age 11 to 14 Challenge Level: Carry out cyclic permutations of nine digit numbers containing the digits from 1 to 9 (until you get back to the first number). Prove that whatever number you choose, they will add to the same total. ##### Age 7 to 14 Challenge Level: I added together some of my neighbours' house numbers. Can you explain the patterns I noticed? ##### Age 11 to 14 Challenge Level: A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you. . . . ### Not Necessarily in That Order ##### Age 11 to 14 Challenge Level: Baker, Cooper, Jones and Smith are four people whose occupations are teacher, welder, mechanic and programmer, but not necessarily in that order. What is each person’s occupation? ### Even So ##### Age 11 to 14 Challenge Level: Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why? ### Chocolate Maths ##### Age 11 to 14 Challenge Level: Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... 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I can trade any of my marbles for three others, one of each colour. Can I end up with exactly two marbles of each colour? ### Problem Solving, Using and Applying and Functional Mathematics ##### Age 5 to 18 Challenge Level: Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. ### Calendar Capers ##### Age 11 to 14 Challenge Level: Choose any three by three square of dates on a calendar page... ### Clocked ##### Age 11 to 14 Challenge Level: Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours? ### Thirty Nine, Seventy Five ##### Age 11 to 14 Challenge Level: We have exactly 100 coins. There are five different values of coins. We have decided to buy a piece of computer software for 39.75. We have the correct money, not a penny more, not a penny less! Can. . . . ### Is it Magic or Is it Maths? ##### Age 11 to 14 Challenge Level: Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . . ### Disappearing Square ##### Age 11 to 14 Challenge Level: Do you know how to find the area of a triangle? You can count the squares. What happens if we turn the triangle on end? Press the button and see. Try counting the number of units in the triangle now. . . . ##### Age 11 to 14 Challenge Level: Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . . ### Cross-country Race ##### Age 11 to 14 Challenge Level: Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places? ### Flight of the Flibbins ##### Age 11 to 14 Challenge Level: Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . . ### Tower of Hanoi ##### Age 11 to 14 Challenge Level: The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice. ### Elevenses ##### Age 11 to 14 Challenge Level: How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?	2,456	10,367	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2020-05	longest	en	0.845799
https://studylib.net/doc/5615912/pocahontas-as-told-by-an-admirer	1,550,760,457,000,000,000	text/html	crawl-data/CC-MAIN-2019-09/segments/1550247504790.66/warc/CC-MAIN-20190221132217-20190221154217-00587.warc.gz	701,557,871	39,283	# Pocahontas as told by an admirer advertisement ```The Pigeonhole Principle Alan Kaylor Cline The Pigeonhole Principle Statement Children’s Version: “If k > n, you can’t stuff k pigeons in n holes without having at least two pigeons in the same hole.” Smartypants Version: “No injective function exists mapping a set of higher cardinality into a set of lower cardinality.” The Pigeonhole Principle Example Twelve people are on an elevator and they exit on ten different floors. At least two got of on the same floor. The ceiling function: For a real number x, the ceiling(x) equals the smallest integer greater than or equal to x Examples: ceiling(3.7) = 4 ceiling(3.0) = 3 ceiling(0.0) = 0 If you are familiar with the truncation function, notice that the ceiling function goes in the opposite direction – up not down. If you owe a store 12.7 cents and they make you pay 13 cents, they have used the ceiling function. The Extended (i.e. coolguy) Pigeonhole Principle Statement Children’s Version: “If you try to stuff k pigeons in n holes there must be at least ceiling (k/n) pigeons in some hole.” Smartypants Version: “If sets A and B are finite and f:A B, then there is some element b of B so that cardinality(f -1(b)) is at least ceiling (cardinality(A)/ cardinality(B).” The Extended (i.e. coolguy) Pigeonhole Principle Example Twelve people are on an elevator and they exit on five different floors. At least three got off on the same floor. (since the ceiling(12/5) = 3) The Extended (i.e. coolguy) Pigeonhole Principle Example Example of even cooler “continuous version” If you travel 12 miles in 5 hours, you must have traveled at least 2.4 miles/hour at some moment. Application 1: Among any group of six acquaintances there is either a subgroup of three mutual friends or three mutual enemies. Application 2: Given twelve coins – exactly eleven of which have equal weight determine which coin is different and whether it is heavy or light in a minimal number of weighings using a three position balance. H Application 3: In any sequence of n2+1 distinct integers, there is a subsequence of length n+1 that is either strictly increasing or strictly decreasing n=2: 3,5,1,2,4 n=3: 2,5,4,6,10,7,9,1,8,3 2,3,5,4,1 10,1,6,3,8,9,2,4,5,7 n=4: 7,9,13,3,22,6,4,8,25,1,2,16,19,26, 10,12,15,20,23,5,24,11,14,21,18,17 Application 3: In any sequence of n2+1 distinct integers, there is a subsequence of length n+1 that is either strictly increasing or strictly decreasing n=2: 3,5,1,2,4 n=3: 2,5,4,6,10,7,9,1,8,3 2,4,5,3,1 10,1,6,3,8,9,2,4,5,7 n=5: 7,9,13,3,22,6,4,8,25,1,2,16,19,26, 10,12,15,20,23,5,24,11,14,21,18,17 Application 1: Among any group of six acquaintances there is either a subgroup of three mutual friends or three mutual enemies. Application 1: Among any group of six acquaintances there is either a subgroup of three mutual friends or three mutual enemies. F F F Application 1: Among any group of six acquaintances there is either a subgroup of three mutual friends or three mutual enemies. E E E Application 1: Among any group of six acquaintances there is either a subgroup of three mutual friends or three mutual enemies. How would you solve this? You could write down every possible acquaintanceship relation. There are 15 pairs of individuals. Each pair has two possibilities: friends or enemies. That’s 215 different relations. By analyzing one per minute, you could prove this in 546 hours. Application 1: Among any group of six acquaintances there is either a subgroup of three mutual friends or three mutual enemies. Could the pigeonhole principle be applied to this? I am glad you asked. Yes. Begin by choosing one person: * Five acquaintances remain * These five must fall into two classes: friends and enemies The extended pigeonhole principle says that at least three must be in the same class that is: three friends or three enemies * Suppose the three are friends of : Either at least two of the three are friends of each other… * ? ? ? In which case we have three mutual friends. * Suppose the three are friends of : Either at least two of the three are friends of each other… or none of the three are friends * In which case we have three mutual enemies. Similar argument if we suppose the three are enemies of * ? ? ? *. Application 2: Given twelve coins – exactly eleven of which have equal weight determine which coin is different and whether it is heavy or light in a minimal number of weighings using a three position balance. H How many different situations can exist? Any of the 12 coins can be the odd one and that one can be either heavy or light 12 x 2 = 24 possibilities Notice: our solution procedure must work always – for every set of coins obeying the rules. We cannot accept a procedure that works only with additional assumptions. Here is one of the 24 situation coin 5 is light We will represent this situation by the single symbol 5 Here is another of the 24 situation coin 10 is heavy We will represent this situation by the single symbol 10 This then represents ALL of the 24 possible outcomes 1 1 2 2 3 3 7 7 8 8 9 9 4 4 5 5 6 6 10 10 11 11 12 12 Example of outcome separation by a single weighing Suppose we put these coins in the left pan and these coins in the right pan 1 1 2 8 9 3 4 2 5 10 11 12 These outcomes would have the left pan down 3 4 5 6 7 8 9 10 11 12 6 7 1 2 6 7 8 9 These outcomes would have the pans blanced 3 4 5 10 11 12 These outcomes would have the right pan down How many different groups of possibilities can discriminated in one weighing? How many different groups of possibilities can discriminated in one weighing? 3 left side down balanced right side down 1 2 4 7 8 12 5 3 9 6 12 10 11 4 10 Left pan down 1 2 Right pan down 7 5 9 3 6 8 Balanced 11 How many different groups of possibilities can discriminated in TWO weighings? 9 left side down balanced left side balanced balanced left side left side balanced down twice down then down then then left twice then right balanced right side side down side down down right side down right side right side right side down then down then down twice left side balanced down 1 2 4 7 8 12 5 3 9 6 12 10 11 4 10 Left pan down 1 2 Right pan down 7 5 9 3 6 8 Balanced 11 Could we solve a four coin problem with just two weighings? There are 8 = 4 x 2 possible outcomes and nine groups can be discriminated with two weighings Eight pigeons - nine holes Looks like it could work … but it doesn’t. The pigeon hole principle won’t guarantee an answer in this problem. It just tells us when an answer is impossible. How many different groups of possibilities can discriminated in k weighings? 3k If 3k different groups of possibilities can discriminated in k weighings, how many weighings are REQUIRED to discriminate 24 possibilities? Since 32 = 9 < 24 < 27 = 33 two weighings will only discriminate 9 possibilities So at least three weighings are required. Can it be done in three? We don’t know until we try. Our format looks like this: We could just start trying various things… there are only 269,721,605,590,607,583,704,967,056,648,878,050,711,137,421,868,902,696,843,001,534,529,012,760,576 things to try. The Limits of Computation Speed: speed of light = 3 10 8 m/s 10 -15 m Distance: proton width = 10 –15 m With one operation being performed in the time light crosses a proton there would be 3 1023 operations per second. Compare this with current serial processor speeds of 1012 operations per second The Limits of Computation With one operation being performed in the time light crosses a proton there would be 3 1023 operations per second. Big Bang: 14 Billion years ago … that’s 4.4 1017 seconds ago So we could have done 1.3 1041 operations since the Big Bang. Yet there are more than 2.6 1074 possibilities to examine. Even with just one operation per examination this could not be done. Can we cut that number (i. e., 2.7x1074) down a bit? Remember: The tree gives us 27 leaves. We can discriminate at most 27 different outcomes. We only need 24 but we must be careful. Questions: 1. Do weighings with unequal numbers of coins on the pans help? No. Again, no outcomes at all will correspond to the balanced position. Conclusion: Always weigh equal numbers of coins. Thus for the twelve coins, the first weighing is either: 1 vs. 1, 2 vs. 2, 3 vs. 3, 4 vs. 4, 5 vs. 5, 6 vs. 6. Questions: 2. Should we start with 6 vs. 6? 1 12 cases 2 3 4 5 6 7 0 cases 8 9 10 11 12 12 cases No. No outcomes at all will correspond to the balanced position. Questions: 3. Should we start with 5 vs. 5? 1 10 cases 2 3 4 5 6 4 cases 7 8 9 10 10 cases No. Only four outcomes will correspond to the balanced position. Thus twenty for the remainder Questions: 4. Should we start with 3 vs. 3? 1 6 cases 2 3 4 12 cases 5 6 6 cases No. In that case the balanced position corresponds to 12 cases. … and the same conclusion for 1 vs. 1 and 2 vs. 2 Thus we must start with 4 vs. 4. 1 8 cases 2 3 4 5 8 cases 6 7 8 8 cases Let’s analyze the balanced case. 1 8 cases 2 3 4 5 8 cases 6 7 8 8 cases 1 8 cases 2 3 4 5 8 cases 6 7 8 8 cases I claim: 1. Coins 1-8 must be regular – the problem is reduced to a four coin problem WITH KNOWN REGULAR COIN. 2. Could we use only unknowns (9 – 12)? No – one on a side has four cases in the balanced position, two on a side can produces no balance. 3. Never need weigh with known regulars on both sides. 4. One regular and one unknown? No - balanced leaves 6 possibilities. 5. Two regular and two unknown? No - balanced leaves 4 possibilities. 6. Four regular and four unknown? No – either unbalanced leaves 4 possibilities. 7. Three regular and three unknown? Might work – three possibilities in each case. And we easily work out the three situations to get: 10 9 9 10 11 12 1 1 3 2 9 10 11 10 10 12 10 11 11 9 9 12 12 10 99 11 12 H H H H L L L L A very similar analysis works on the left side to get: 1 1 2 2 11 55 22 H L H 5 3 7 8 6 4 3 4 38 77 33 66 44 L L H L H … and on the right side to get: 5 6 7 3 4 6 2 6 55 2 44 33 H L H L L 1 5 6 8 2 7 8 77 11 88 H L H OUR SOLUTION 1 1 1 2 2 5 3 7 8 6 4 3 4 9 2 3 4 5 6 7 9 10 11 1 2 3 12 1 10 9 8 5 1 10 5 6 6 7 3 4 8 2 7 8 11 55 22 38 77 33 66 44 11 10 10 12 10 11 11 9 9 12 12 12 10 99 11 55 22 66 44 8 33 77 11 8 H L H L L H L H H H H H H L H L L L L L L H L H Application 3: In any sequence of n2+1 distinct integers, there is a subsequence of length n+1 that is either strictly increasing or strictly decreasing n=2: 3,5,1,2,4 n=3: 2,5,4,6,10,7,9,1,8,3 2,3,5,4,1 10,1,6,3,8,9,2,4,5,7 n=4: 7,9,13,3,22,6,4,8,25,1,2,16,19,26, 10,12,15,20,23,5,24,11,14,21,18,17 Application 3: In any sequence of n2+1 distinct integers, there is a subsequence of length n+1 that is either strictly increasing or strictly decreasing n=2: 3,5,1,2,4 n=3: 2,5,4,6,10,7,9,1,8,3 2,4,5,3,1 10,1,6,3,8,9,2,4,5,7 n=5: 7,9,13,3,22,6,4,8,25,1,2,16,19,26, 10,12,15,20,23,5,24,11,14,21,18,17 Application 3: In any sequence of n2+1 distinct integers, there is a subsequence of length n+1 that is either strictly increasing or strictly decreasing Idea: Could we solve this by considering cases? For sequences of length 2: 2 cases For sequences of length 5: 120 cases For sequences of length 10: 3,628,800 cases For sequences of length 17: 3.6 1014 cases For sequences of length 26: 4.0 10 26 cases For sequences of length 37: 1.4 10 43 cases Remember The Limits of Computation Speed: speed of light = 3 10 8 m/s 10 -15 m Distance: proton width = 10 –15 m With one operation being performed in the time light crosses a proton there would be 3 1023 operations per second. Compare this with current serial processor speeds of 1012 operations per second The Limits of Computation With one operation being performed in the time light crosses a proton there would be 3 1023 operations per second. Big Bang: 14 Billion years ago … that’s 4.4 1017 seconds ago So we could have done 1.3 1041 operations since the Big Bang. So we could not have proved this (using enumeration) even for the case of subsequences of length 7 from sequences of length 37. But with the pigeon hole principle we can prove it in two minutes. We will use a Proof by Contradiction. This means we will show that it is impossible for our result to be false. Since a statement must be either true or false, if it is impossible to be false, it must be true. So we assume that our result “In any sequence of n2+1 distinct integers, there is a subsequence of length n+1 that is either strictly increasing or strictly decreasing” is false. That means there is some sequence of n2+1 distinct integers, so that there is NO subsequence of length n+1 that is strictly increasing AND NO subsequence of length n+1 that strictly decreasing. Once again, our object is to show that this is impossible. The process that is described now will be applied to a particular example sequence, but it could be applied to ANY sequence. Start with a sequence: 2,5,4,6,10,7,9,1,8,3 (here n = 3) Let’s start at the right end and figure out the lengths of the longest strictly increasing subsequence and strictly decreasing subsequence starting from that point and using that number. Obviously the lengths of the longest strictly increasing and strictly decreasing subsequence starting at the 3 are both one. We’ll indicate this by the pair (1,1). (1,1) 2, 5, 4, 6, 10, 7, 9, 1, 8, 3 Now let’s move to the 8 and notice that the length of the longest strictly increasing subsequence is still one but The length of the longest strictly decreasing subsequence starting from 8 is two. So we have the pair (1,2) and we write it (1,2) (1,1) 2, 5, 4, 6, 10, 7, 9, 1, 8, 3 We could keep moving left determining lengths of the longest strictly increasing subsequence and the longest strictly decreasing subsequence starting from each number. We get: (5,2) 2, (4,3) (4,2) (3,2) (1,4) (2,2) (1,3) (2,1) (1,2) (1,1) 5, 4, 6, 10, 7, 9, 1, 8, 3 We needed to get either a strictly increasing subsequence or strictly decreasing subsequence of length four. We actually got both – and, in fact, a strictly increasing subsequences of length five. But does this always happen? (5,2) 2, (4,3) (4,2) (3,2) (1,4) (2,2) (1,3) (2,1) (1,2) (1,1) 5, 4, 6, 10, 7, 9, 1, 8, 3 What can the (up, down) pairs be? If no subsequence of length four exists, “up” and “down” must be 1,2, or 3. That leaves only 9 possibilities. But there are 10 pairs. So at least two would have to match. MATCH? Suppose i and j have the same (up, down) pairand i precedes j If i < j then i should have a greater “up” count than j. (3,2) (3,2) 6 8 If i > j then i should have a greater “down” count than j. (3,2) (3,2) 8 6 Contradiction: there cannot be a match. Conclusion: There is always such a subsequence. Pigeonhole Problems 1. If you have only two colors of socks – white and black – and you grab three socks, you are guaranteed to have a matching pair. Pigeons: socks (3) Holes: colors (2) Two of the socks must have the same color. 2. Suppose no Texan has more than 200,000 hairs on his or her head. There are at least 120 Texans with exactly the same number of head hairs. Pigeons: Texans (more than 25,000,000) Holes: Number of head hairs (200,001) There must be some number of head hairs shared by  25, 000 , 000      124.99...   125  120  200 , 001  Texans. 3. Suppose S is a set of 8 integers. There exist two distinct elements of S whose difference is a multiple of 7. Pigeons: Set S (8) Holes: Remainder when divided by 7 (7) Two of the numbers, a and b, must have the same remainder when divided by 7. That is: a = 7m + r and b = 7n + r a -b = 7m +r - (7n +r) = 7(m - n). So a - b is a multiple of 7. 4. Among any group of six acquaintances there is either a subgroup of three mutual friends or three mutual enemies. Done in the lecture. 5. Given twelve coins – exactly eleven of which have equal weight - determine which coin is different and whether it is heavy or light in a minimal number of weighings using a three position balance. Done in the lecture. 6. Given seven coins such that exactly five of the coins have equal weight and each of the other two coins is different – possibly heavy or lighter. To determine which coins are different and whether each different coin is heavy or light requires at least five weighings using a three position balance. There are 7 x 6 / 2 = 21 different ways to choose the two odd-weight coins. The first can be heavy or light (2 possibilities) an the second can be heavy or light (2 possibilities). Thus there are 21 x 2 x 2 = 84 different configurations possible. But four weighings using a balance can discriminate only 34 = 81 configurations. Thus, with whatever weighing strategy at least two configurations will appear identical. 7. Given five points inside an equilateral triangle of side length 2, at least two of the points are within 1 unit distance from each other. Form four triangles by connecting the midpoints of the sides. The side length of these triangles is 1. Since there are four triangles and five points, two of the points must lie in the same triangle. But two points inside an equilateral triangle with side length 1 must be no more than distance 1 apart. 8. In any sequence of n2+1 distinct integers, there is a subsequence of length n+1 that is either strictly increasing or strictly decreasing. Done in the lecture. ``` – Cards – Cards – Cards – Cards – Cards	5,476	17,308	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.921875	4	CC-MAIN-2019-09	latest	en	0.890052
https://www.splessons.com/lesson/ssc-cpo-averages-quiz-2/	1,586,367,232,000,000,000	text/html	crawl-data/CC-MAIN-2020-16/segments/1585371821680.80/warc/CC-MAIN-20200408170717-20200408201217-00421.warc.gz	1,146,893,521	74,098	# SSC CPO Averages Quiz 2 5 Steps - 3 Clicks # SSC CPO Averages Quiz 2 ### Introduction Average is a straight- forward concept and it can be solved easily by equal distribution method. Average is the sum of all the elements in a given data set divided by the total number of elements in the data set. The most commonly denoted term for average is Arithmetic Mean, simply termed as mean. The article SSC CPO Averages Quiz 2 very useful for different competitive examinations like RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC, IBPS PO, IBPS RRB Exams and etc. SSC CPO Averages Quiz 2 is very useful to crack the Quantitative Aptitude sections in several exams. ### Quiz 1. The average of women and child workers in a factory was 15%yr. The average age of all the 16 children was 8yr and average age of women workers was 22 yrs if ten women workers were married then the number of unmarried women workers were A. 16 B. 12 C. 8 D. 6 E. None of these Explanation: Let the number of women workers be x According to the question 22x+168 = 15(16+x) => 22x+128 =240 +15x => 22x-15x = 240 -128 => 7x = 112 Therefore, x = $$\frac{112}{7}$$ = 16 Therefore, Unmarried women workers =(16-10) = 6 2. The mean marks of 30 students in a class is 58.5. Later on it was found that 75 was wrongly recorded as 57. Find the correct them. A. 57.4 B. 57.5 C. 58.9 D. 59.1 E. None of these Explanation: Correct mean = 30x 58.5 – 57 + $$\frac{75}{30}$$ $$\frac{1755 + 18}{30}$$ = $$\frac{1773 }{30}$$ = 59.1 3. A person travels from x to y at a speed of 40Km/h and returns by increasing his speed 50%. What is his average speed for both the trips? A. 36km/h B. 45km/h C. 48km/h D. 50km/h E. None of these Explanation: Speed of person from x to y =40 km/h Speed of person from y to x = $$\frac{(40 Ã— 150) }{100 }$$ = 60km/h Since the distance travelled is same Therefore, Average Speed = $$\frac{(2 Ã— 40 Ã— 60) }{40 + 60}$$ = 48km/h 4. The average of 18 observations was calculated and it was 124. Later on it was discovered that two observations 46 and 82 were incorrect. The correct values are 64 and 28. The correct average of 18 observations is A. 123 B. 137 C. 121 D. 122 E. None of these Explanation: Sum of 18 observations = 18×124 =2232 Correct sum of 18 observations = 2232-46-82+64+28 = 2196 Therefore, Correct average = $$\frac{2196 }{18}$$ = 122. 5. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is: A. 17 kg B. 20 kg C. 26 kg D. 31 kg E. None of these Explanation: Let A, B, C represent their respective weights. Then, we have: A + B + C = (45 x 3) = 135 …. (i) A + B = (40 x 2) = 80 …. (ii) B + C = (43 x 2) = 86 ….(iii) Adding (ii) and (iii), we get: A + 2B + C = 166 …. (iv) Subtracting (i) from (iv), we get : B = 31. B’s weight = 31 kg. 1. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is: A. 35 years B. 40 years C. 50 years D. None of these Explanation: Sum of the present ages of husband, wife and child = (27 * 3 + 3 * 3) = 90 years. Sum of the present age of wife and child = (20 * 2 + 5 * 2) = 50 years. Husband’s present age = (90 – 50) = 40 years. 2. The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is: A. 20 B. 21 C. 22 D. 23 Explanation: Let the total number of workers be x. Then, 8000x = (12000 * 7) + 6000(x – 7) = 2000x = 42000 = x = 21. 3. In an examination a pupil’s average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination? A. 8 B. 9 C. 10 D. 11 Explanation: Let the number of papers be x. Then, 63x + 20 + 2 = 65x = 2x = 22 = x = 11. 4. The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ration of the number of boys to the number of girls in the class is: A. 1:2 B. 2:3 C. 3:4 D. 3:5 Explanation: Let the ratio be k : 1. Then, k * 16.4 + 1 * 15.4 = (k + 1) * 15.8 = (16.4 – 15.8)k = (15.8 – 15.4) = k = $$\frac{0.4}{0.6}$$ = $$\frac{2}{3}$$ Required ratio = $$\frac{2}{3}$$ : 1 = 2:3. 5. The average monthly salary of 20 employees in an organisation is Rs. 1500. If the manager’s salary is added, then the average salary increases by Rs. 100. What is the manager’s monthly salary? A. Rs. 2000 B. Rs. 2400 C. Rs. 3600 D. Rs. 4800 E. None of these Explanation: Manager’s monthly salary = Rs. (1600 * 21 – 1500 * 20) = Rs. 3600 1. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average of the team? A. 23 years B. 24 years C. 25 years D. None of these Explanation: Let the average of the whole team be x years. 11x – (26 + 29) = 9(x – 1) = 11x – 9x = 46 = 2x = 46 => x = 23 So, average age of the team is 23 years. 2. A cricketer has a certain average for 10 innings. In the eleventh inning, he scored 108 runs, there by increasing his average by 6 runs. His new average is: A. 48 runs B. 52 runs C. 55 runs D. 60 runs Explanation: Let average for 10 innings be x. Then, $$\frac{(10x + 108)}{11}$$ = x + 6 = 11x + 66 = 10x + 108 = x = 42. New average = (x + 6) = 48 runs. 3. A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and there by decreases his average by 0.4. The number age of the family now is: A. 64 B. 72 C. 80 D. 85 Explanation: Let the number of wickets taken till the last match be x. Then, $$\frac{(12.4x + 26)}{(x + 5)}$$ = 12 = 12.4x + 26 = 12x + 60 = 0.4x = 34 = x = $$\frac{340}{4}$$ = 85. 4. The average age of a husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is: A. 19 years B. 23 years C. 28.5 years D. 29.3 years Explanation: Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) = 57 years. Required average = $$\frac{57}{3}$$ = 19 years. 5. The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is: A. 35.2 B. 36.1 C. 36.5 D. 39.1 Correct mean = $$\frac{1825}{50}$$ = 36.5	2,301	6,647	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.5625	5	CC-MAIN-2020-16	longest	en	0.952371
https://stats.stackexchange.com/questions/548644/interview-question-communication-by-breaking-windows	1,718,597,706,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861696.51/warc/CC-MAIN-20240617024959-20240617054959-00525.warc.gz	488,686,244	42,972	# Interview question: communication by breaking windows Here is an interview question for brainstorming: There are 100 different windows (suppose they stand in a line), 2 people, and 50 days. One communicates with the other by breaking windows in any day. How many bits of information can they transfer? Second part: On each day the first person breaks windows then the second one repairs some of them. They want to transfer as many bits as possible every day. Find an algorithm and the largest possible amount of bits. (The first person gets information by looking at windows which were repaired and the second one by looking at broken windows.) The first question is very easy, since each window is independent and there are 51 states (broken at the i-th day) during the 50 days. Therefore there are $$51^{100}$$ outputs in total. But I have no idea about the second question. Edit: I found an answer on a website. I think it makes sense. In the second case, note that the amount of information that can be sent by one person is limited by how much information the other person sends. Note that if the breaker doesn't break a window, then the repairer cannot repair it. Likewise, if the repairer doesn't repair a window, then the breaker cannot break it on the following day. By this symmetry, one can solve the problem by supposing that the second person just repairs all of the windows all the time. Thus, the first person is able to transfer information $$(2^{100})^{50}$$ pieces of information. The 2 comes from the fact that he can either break a window or leave it whole on any given day. Again, to convert this to bits just apply $$\log_2.$$ • How do you obtain the $51^{100}$ answer? Once a window is broken it remains broken, right? Thus, any state after day $i$ is not independent of the state at day $i.$ – whuber Commented Oct 18, 2021 at 14:32 • @whuber I understand as how many results of 100 windows there are after 100 days. Commented Oct 18, 2021 at 18:12 • I see: if we view the windows as being variables whose values are the day on which the window is broken (setting this day to 51 if the window is never broken), then these variables are independent and each window has 51 possible values. For the second question, start by considering how much information can be transferred when a person has only $k$ windows available to break. You might want to begin with a simpler version of the problem where "100" is replaced by "2" or "3," say. – whuber Commented Oct 18, 2021 at 19:21 • The website answer you quote suggests one way to communicate information, but it does not demonstrate it "transfers as many bits as possible" each day. In fact, it falls far short of being optimal, if we understand "transfer" in the sense of two-way communications rather than merely sending information from person 1 to person 2. – whuber Commented Oct 19, 2021 at 16:49 Let's address the second question, because it's more difficult and the answer you found is not optimal. Part of the problem lies in the interpretation of "transfer as many bits as possible every day." This raises several issues: 1. Do we want the transfer to be two-way or just one-way? 2. Does "as many bits as possible" mean on average or at a minimum? 3. Does "as many bits as possible" mean for each person or total? 4. Is a two-way channel with an asymmetric capacity acceptable, or should both people be able to achieve a common guaranteed rate of information transfer to the other? When the transfer is intended to be one-way only, then the quoted solution in the question is best. This is a trivial matter: with 100 windows, you cannot communicate any more than 100 bits of information and, to achieve this rate, the second person must restore all the windows after each day. The essence of the problem is captured by considering what daily throughput can be guaranteed for the total two-way communication. That is the version I will discuss. I will also give solutions for the symmetric version of this problem. To set this up, let the window-breaker be person $$A$$ and the window-repairer be person $$B.$$ We might as well solve the problem generally for $$n$$ windows. I'll just sketch the basic argument and leave the (relatively easy) details for you to prove: 1. At the beginning of any day, both people know which windows are broken. 2. A "code" for communicating from $$A$$ to $$B$$ is a set of subsets of the intact windows. (Such a subset is a "word" of the code.) A code for communicating from $$B$$ to $$A$$ is a set of subsets of the broken windows. 3. The amount of information a person can send to the other (in bits) is the binary logarithm of the code used by the sender. 4. Therefore, the total amount of information that can be exchanged in a day is the sum of the logarithms of the sizes of the two codes. We wish to maximize this "daily throughput." 5. All else being equal, it is better for $$A$$ to have as many unbroken windows available as possible when $$A$$ is sending and it is better for $$B$$ to have as many broken windows as possible when $$B$$ is sending. 6. Therefore, an optimal code for $$A$$ will use words with as many broken windows as possible and an optimal code for $$B$$ will use words with as many repaired windows as possible. 7. Using optimal coding, on day $$t=0,1,2,\ldots,$$ let $$A$$ have $$a$$ unbroken windows available (and therefore $$n-a$$ are broken). Suppose $$A$$ adopts a code in which the number of unbroken windows available to $$B$$ is guaranteed to be at least $$b.$$ Then if the two codes achieve the optimal guaranteed throughput, $$B$$ must adopt a code that assures $$A$$ will again have at least $$a$$ unbroken windows the next day. 8. This implies $$B$$ must repair at least $$a+b-n$$ windows. 9. Because $$A$$ only needs to guarantee at least $$b$$ broken windows for $$B$$ to repair, the optimal code for $$A$$ will consist of all words of length $$a$$ that break at least $$b - (n-a) = a+b-n$$ more windows. Let the number of these words be $$G(a,a+b-n).$$ 10. Similarly, the number of words $$B$$ can use is $$G(b, a+b-n).$$ It remains only to count and optimize. Basic combinatorial principles tell us $$G(a,x) = \sum_{k=x}^a \binom{a}{k}.$$ When $$X$$ is a random variable with a Binomial$$(a,1/2)$$ distribution $$F_a,$$ $$G(a,x) = 2^a\left(1 - F_a(x-1)\right).$$ This is usually computed in terms of an incomplete Beta function in constant time. Thus, the problem is to Find $$0\le a \le n$$ and $$0\le b \le n$$ for which the minimum daily throughput of $$\log_2(G(a,a+b-n)) + \log_2(G(b,a+b-n))$$ bits is as large as possible. For smallish $$n$$--values less than a few thousand are amenable to quick calculations--a brute-force search does the job. (It is intuitively evident $$a=b$$ at the optimum, but I haven't assumed that.) For reference, note that the solution quoted in the question has a daily throughput of $$n$$ bits: $$A$$ uses a code of $$2^n$$ words and $$B$$ cannot communicate any information back. Can we do better than this? For $$n \lt 3,$$ the answer is no. But for $$n=3,$$ the solution $$a=b=2$$ achieves a daily throughput of $$\log_2(9)\approx 3.17$$ bits. Let's inspect the details: • $$a=2$$ means $$A$$ is guaranteed to have at least $$2$$ intact windows. $$b=2$$ means $$A$$ must break at least one of them. There are at $$3$$ ways this can be done. (When $$A$$ is presented with $$3$$ intact windows, she must break at least two of them--but that's okay, because there are now $$4$$ ways this can be done.) • Likewise, there are at least $$3$$ words available to $$B$$ (and maybe $$4$$) that present $$A$$ with at least $$2$$ intact windows the next day. • Thus, every day there are at least $$3 \times 3 = 9$$ possible communications from $$A$$ to $$B$$ and back again. The rate of $$\log_2(9)$$ exceeds the rate of $$\log_2(8)=3$$ in the one-way procedure described in the question. Things eventually get much better than this. We can study the situation by using the Normal approximation to the Binomial distribution. It shows that asymptotically, for large $$n,$$ $$a$$ and $$b$$ should be around $$0.724n.$$ This achieves a guaranteed daily throughput of more than $$1.38 n$$ bits. The case $$n=100$$ is nearly asymptotic: the optimum is achieved at $$a=b=71,$$ with a throughput of at least $$134.6.$$ Here are plots of some exact solutions (found with brute-force search). The dotted red lines show the asymptotic values. The curves bounce around due to the discrete nature of the optimization problem. • if the source code can be provided this can be quite useful as it provides more insight on the topic Commented Oct 21, 2021 at 20:38 • @Maximilian This is simpler than it might look if you're willing to assume the optimal solution occurs at $a=b.$ f <- function(n, log.G=g) { a <- b <- 0:n; R <- log.G(a,a+b-n) + log.G(b,a+b-n); a.max <- which.max(R) - 1; list(optimum = R[a.max+1], a=a.max) }; f(100) It returns a list giving the optimal rate and optimal value of $a$. It relies on a previously defined implementation of the binary logarithm of $G,$ such as g <- function(n, k) n + pbinom(k-1, n, 1/2, lower.tail=FALSE, log.p=TRUE) / log(2), where pbinom is the R version of $F_a$ I mentioned after item (10). – whuber Commented Oct 21, 2021 at 21:27 • many thanks, I can read it now more clearly. Commented Oct 22, 2021 at 7:19	2,377	9,378	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 82, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2024-26	latest	en	0.960216
https://lipn.univ-paris13.fr/~banderier/Seminar/lalley.html	1,580,189,937,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579251773463.72/warc/CC-MAIN-20200128030221-20200128060221-00039.warc.gz	513,833,233	7,039	The Local Limit Theorem for Random Walks on Free Groups Steve Lalley Purdue University Algorithms Seminar May 31, 1999 [summary by Cyril Banderier] A properly typeset version of this document is available in postscript and in pdf. If some fonts do not look right on your screen, this might be fixed by configuring your browser (see the documentation here). Abstract Local limit theorems and saddlepoint approximations are given for random walks on a free group whose step distributions have finite support. These are derived by exploiting a set of algebraic relations among certain generating functions that arise naturally in connection with the transition probabilities of the random walks. Basic tools involved in the analysis are the elementary theory of algebraic functions, the Perron-Frobenius theory of nonnegative matrices, and standard techniques of singularity analysis. ## 1 Walks on Groups Figure 1: A representation (Cayley graph) of the infinite free group Z * Z Let G be a the free group with generators a1,..., aL. For example, the free group Z * Z has two generators a and b. The word a b a a_ b_ a_ a_ b_ a a corresponds to the point B, the associated reduced word is a_ b_ a a. A finite-range random walk {Zn}n³ 0 is a Markov chain on G with Z0:=e (the identity of the group, the ``origin'', the starting point) and transition probabilities Pr{Zn+1=yx\|Zn=y}=px " x,y Î G, n³ 0, where px for xÎ G is a probability distribution with finite support (in other words, px is the probability of the ``jump'' x). Note pn(x) the probability of being in x after n steps. It is assumed that the random walk is irreducible and aperiodic, that is, that " x Î G å n³ 1 pn(x) >0 (irreducibility); GCD{n; pn(e)>0}=1 (aperiodicity). Another important condition is the following: Positivity: pe>0 and pg>0 for all generators g of G (and their inverses). Similarly to the random walk in the Euclidian case Zd, a local limit theorem is given by the asymptotics pn(x)~ Bx R-n (2 p R)1/2n3/2 . Similar results were already known when all the steps are of size one (nearest neighbour random walk [3]) or when all the words at a same distance from the origin are equiprobable (the so-called isotropic random walk [2, 8, 9]). ## 2 Singularity Analysis This section is devoted to the analysis of some probability generating functions (PGF) related to the walk. For xÎ G and zÎ C (\|z\|<1), define • the random variable coding where one is after n steps: Zn, • the PGF to reach x in n steps: Gx(z):=ån pn(x)zn, • the PGF of the excursions (Green's function): G(z):=Ge(z), • the first time x is reached: Tx:=inf{n³ 0: Zn=x}, • the PGF to reach x for the first time in n steps: Fx(z):=ån Pr{Tx=n} zn. Note that aperiodicity and irreducibility imply that for all sufficiently large n³ 1, pn(e)>0 and pn(y)>0, for any (inverse of a) generator y. The following (combinatorially trivial) relations Gx(z)=Fx(z)G(x), G(z)=1+z æ ç ç è pe+ å x¹ e px F x-1 (z) ö ÷ ÷ ø G(z)= æ ç ç è 1-zpe-z å x¹ e px F x-1 (z) ö ÷ ÷ ø -1 allow to prove that all of the functions Fx and Gx have the same radius of convergence R, 1<R<¥ (the less obvious is that R is strictly greater than 1). Let B be the set of points at distance £ K, where K is such that there is no smaller ball in which the support of the step distribution {px} is contained. Define now • the first time that x is exceeded: tx • the PGF to go from a to xb while x as never been exceeded before: Hxab= å n Pr {Zn=xb\|Z0=a and Zi<nÏxB } zn • the PGF to go from x to the origin: Fx-1(z) The formula " x¹ e Fx(z)= å bÎ B-{e} Hxeb (z) F b-1 (z) leads to an expression of Fx=uHx(z)v=u Hx1(z)··· Hxm(z)v where u is the projection on e and v a vector whose entries are the Fb-1(z) for bÎ B and where the product of matrices is over x1··· xm, the reduced word associated to x. It is then proven by the Markov property that all the non-constant Hxiab satisfy polynomial relations (Hxi=Qi(Hx1, Hx2, ...), see [6] for exact relations) and that they have the same radius of convergence. By elimination (Gröbner basis or resultants), the functions Fx and Gx are algebraic. Their Puiseux expansion leads to an algebraic singularity with exponent 1/2. Here is a sketch of the proof that the exponent is indeed a=1/2. Define Jz the Jacobian matrix ( Qi / Hxj), the polynomials Qi have nonnegative coefficients, thus there exists n such that Jzn is an aperiodic and irreducible matrix with strictly positive coefficients. By the Perron-Frobenius theorem, Jz has a positive eigenvalue lz of multiplicity 1. The function lz is increasing and real-analytic and lR=1/R. Considering a left eigenvector of JR and using the shape of the Qi yields to the relations (R-z)(C+···)=C'(R-z)2a+···, thus 2a=1. As z=R is the dominant singularity of Fx and Gx, one has the two following theorems: Theorem 1 [Local limit theorem, access] Assuming irreducibility and aperiodicity, one has, for a positive constant Bx: pn(x)~ Bx (2p R)1/2Rn n3/2 . Theorem 2 [Local limit theorem, first access] Assuming positivity, one has, for a positive constant Ax: Pr {x is reached for the first time after n steps} ~ Ax (2p R)1/2Rn n3/2 . The probability to reach a point x at a distance m of the origin in n steps is pn(x)~ exp(nb(m/n)) ( .. y (m/n))1/2 C(m/n) for appropriate functions b, C and y.. (see the correct definitions / notations in [6]). This uniform asymptotics in x and n corresponds to the classical saddlepoint approximation (sharp large deviations theorems) for sums of iid random vectors in Rd. The saddlepoint approximations are of interest for another reason. For large n, nearly all the mass in the probability distribution pn(x) is concentrated in the region \|x\|³ e n, where the local limit approximations are not accurate. This contrasts with the situation for finite range random walk in Euclidean space. In fact, Guivarch [4] has shown that for random walks in G, the distance from the origin grows linearly in n. Sawyer and Steger [10] have further shown that (\|Zn\|-nb)/(n)1/2 converges in law to a normal distribution. Finally, S. Lalley, using a special matrix product and results on Ruelle's Perron-Frobenius operators, derives a saddlepoint approximation, uniformly for m/n in a given compact Pr{\|Zn\|=m}~ exp(nB(m/n)) (2 p m D(m/n))1/2 C(m/n). ## References [1] Cartwright (Donald I.) and Sawyer (Stanley). -- The Martin boundary for general isotropic random walks in a tree. Journal of Theoretical Probability, vol. 4, n°1, 1991, pp. 111--136. [2] Figà-Talamanca (Alessandro) and Picardello (Massimo A.). -- Harmonic analysis on free groups. -- Marcel Dekker Inc., New York, 1983, viii+145p. [3] Gerl (Peter) and Woess (Wolfgang). -- Local limits and harmonic functions for nonisotropic random walks on free groups. Probability Theory and Related Fields, vol. 71, n°3, 1986, pp. 341--355. [4] Guivarc'h (Y.). -- Sur la loi des grands nombres et le rayon spectral d'une marche aléatoire. In Conference on Random Walks (Kleebach, 1979), pp. 47--98, 3. -- Société Mathématique de France, Paris, 1980. [5] Lalley (Steven P.). -- Saddle-point approximations and space-time Martin boundary for nearest-neighbor random walk on a homogeneous tree. Journal of Theoretical Probability, vol. 4, n°4, 1991, pp. 701--723. [6] Lalley (Steven P.). -- Finite range random walk on free groups and homogeneous trees. The Annals of Probability, vol. 21, n°4, 1993, pp. 2087--2130. [7] Lalley (Steven P.) and Hueter (Irene). -- Anisotropic branching random walks on homogeneous trees. Preprint, 1999. [8] Picardello (Massimo A.). -- Spherical functions and local limit theorems on free groups. Annali di Matematica Pura ed Applicata. Serie Quarta, vol. 133, 1983, pp. 177--191. [9] Sawyer (Stanley). -- Isotropic random walks in a tree. Zeitschrift für Wahrscheinlichkeitstheorie und Verwandte Gebiete, vol. 42, n°4, 1978, pp. 279--292. [10] Sawyer (Stanley) and Steger (Tim). -- The rate of escape for anisotropic random walks in a tree. Probability Theory and Related Fields, vol. 76, n°2, 1987, pp. 207--230. [11] Seneta (E.). -- Nonnegative matrices and Markov chains. -- Springer-Verlag, New York, 1981, second edition, xiii+279p. This document was translated from LATEX by HEVEA.	2,432	8,278	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2020-05	latest	en	0.857237
https://www.scribd.com/document/34035213/Two-Cute-Proofs	1,503,274,572,000,000,000	text/html	crawl-data/CC-MAIN-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00116.warc.gz	953,569,712	28,963	Two Cute Proofs The first begins with a problem I found online: Show that a b c a, b, c Ε 0, 1 , abc 2. I was unsure how to start the problem at first; I tried strange methods to reduce the problem to two variables and then to one, eventually ending up with a huge complex proof scribbled incomprehensibly with an error somewhere, since the last step didn't fit; upset, I started over and came up with a similar but much simpler method of solving this one. First, we choose c and isolate it: c 1 ab c a b 2 0; this expression will equal 0 when 2 a b 1 ab 2 a b . Thus, 1 ab we must show that either 1 or 2 a b 1 ab 0, and then show that the expression a b c a b c 2 for some values of a, b and c on [0,1); then we've proven that the quantity a b c a b c 2 is never 0, and is less than 0 for some value. (We can choose a b c 0 to obtain 0 2.) Since it is linear in any of a, b or c, it is continuous and thus has to be negative a, b and c Ε 0, 1 . Clearly, 2 a b 1 ab is greater than 0 since 1 a b and 2 a b by assumption. Thus, we have to show that it is greater than 1, or that 2 1 a 1 a b 1 b1 1 a b 1 b b ab ab a a 1 a Which is true by assumption: thus, we have proved the original statement. After completing this, I noticed that the case we reduced the original statement to is another version of that statement: a b c abc a b ab 1 2 Naturally, I wondered if in general, x1 x2 ... xn x1 x2 ... xn n 1 x1 , x2 , ... xn Ε 0, 1 . 1 a b ab 1 b1 a a 2 Two Cute Proofs.nb 1 a b1 a 1 b Which is true by assumption: thus, we have proved the original statement. After completing this, I noticed that the case we reduced the original statement to is another version of that statement: a b c abc a b ab 1 2 Naturally, I wondered if in general, x1 x2 ... xn x1 x2 ... xn n 1 x1 , x2 , ... xn Ε 0, 1 . I quickly noticed that an inductive proof would do: Base case: Show that a b a b Induction: Show that k 11 xi i We work backwards, as above: x1 x2 ... xk 1 x1 x2 ... xk 1 k xk 1 1 x1 x2 ... xk x1 x2 ... xk k k x1 ... xk xk 1 1 x ... x 1 k k 1 i 1 xi 1; see above for proof. k given that the statement holds for n k. 0, 0 when Again, this last is clearly greater than 0; to show that it is greater than 1, the inequality reduces to k 1 x1 x2 ... xk x1 x2 ... xk Which is true by the inductive hypothesis. By the same logic above, and using x1 x2 . .. xk 1 0 k, we have proven the statement in general for n 1. It is also interesting to note that the same exact proof works (with all the signs switched, of course) to show that the opposite is true x1, ... xn Ε 1, , i.e. that under those conditions x1 x2 ... xn x1 x2 ... xn n 1. The exploration is surely too complex for this post, but I might investigate what how the conditions change when we allow the xi ' s to be any real numbers. My reaction upon first seeing the problem was that it was certainly related to the AMGM theorem, or the Arithmetic Mean Geometric Mean theorem, which proves that the arithmetic mean of a set of numbers is always larger than or equal to the geometric mean of the same set of numbers- after all, the arithmetic mean is just a sum (with a constant) and the geometric mean is just a product (with a root). There may well be an intimate connection, but as of yet it isn't apparent to me. k 1 x1 x2 ... xk x1 x2 ... xk Two Cute Proofs.nb 3 Which is true by the inductive hypothesis. By the same logic above, and using x1 x2 . .. xk 1 0 k, we have proven the statement in general for n 1. It is also interesting to note that the same exact proof works (with all the signs switched, of course) to show that the opposite is true x1, ... xn Ε 1, , i.e. that under those conditions x1 x2 ... xn x1 x2 ... xn n 1. The exploration is surely too complex for this post, but I might investigate what how the conditions change when we allow the xi ' s to be any real numbers. My reaction upon first seeing the problem was that it was certainly related to the AMGM theorem, or the Arithmetic Mean Geometric Mean theorem, which proves that the arithmetic mean of a set of numbers is always larger than or equal to the geometric mean of the same set of numbers- after all, the arithmetic mean is just a sum (with a constant) and the geometric mean is just a product (with a root). There may well be an intimate connection, but as of yet it isn't apparent to me. Another perhaps interesting curiosity just came to me: to find how multiplying constants, call them Cs and C p , to both the sum and product would affect the maximum (or minimum, if the xi ' s are all greater than 1) of the expression. (Perhaps the sum-constant should be added, and the product constant multiplied, or vice-versa.) There may be something here; I don't know. The second problem is as follows: There are three coins on a # line. At any step, we can move one coin one unit to the left and one coin one unit to the right. Find all the starting configurations of the coins (on integer points) so that we can move all the coins to the same spot in a finite number of moves, starting from that configuration. I quickly determined the set: any configuration where the points sum to a multiple of 3 works. We set out to prove that all configurations satisfying this can be transformed by the given action so that all three coins are on the same spot, and there are no configurations without this property that can do the same. Proof: One coin can start at 0, WLOG; then if the second coin is at a, (we can assume WLOG that a 0), by assumption the third coin is at 3 r a for some integer value of r. First, we move the second coin a steps to the left and the third coin a steps to the right, so our three coins have positions 0, 0, 3 r), respectively. Then, we move the second coin r steps to the right and the third coin r steps to the left to obtain 0, r, 2 r . Finally, the first coin moves r units to the right and the third coin moves r units to the left for the desired result, r, r, r . As well, we know that the sum of the points the 3 coins are on remains constant because the operation adds 1 and subtracts 1 from the sum, adding a net total of 0. Thus, since the ending configuration- C, C, C - has a sum that is a multiple of 3, the initial configuration must also have a sum that is divisible by 3, and we have the desired result. I quickly determined the set: any configuration where the points sum to a multiple of 3 works. We set out to prove that all configurations satisfying this can be transformed by the given action so that all Two Cute Proofs.nb on the same spot, and there are no configurations without this property that can three coins are 4 do the same. Proof: One coin can start at 0, WLOG; then if the second coin is at a, (we can assume WLOG that a 0), by assumption the third coin is at 3 r a for some integer value of r. First, we move the second coin a steps to the left and the third coin a steps to the right, so our three coins have positions 0, 0, 3 r), respectively. Then, we move the second coin r steps to the right and the third coin r steps to the left to obtain 0, r, 2 r . Finally, the first coin moves r units to the right and the third coin moves r units to the left for the desired result, r, r, r . As well, we know that the sum of the points the 3 coins are on remains constant because the operation adds 1 and subtracts 1 from the sum, adding a net total of 0. Thus, since the ending configuration- C, C, C - has a sum that is a multiple of 3, the initial configuration must also have a sum that is divisible by 3, and we have the desired result. Bored with just 3 coins, I decided to see if the same results held for N coins; and they do. By the same logic, the starting configuration's coin positions must sum to a multiple of N. Then the procedure is the same: We start with the general configuration of coins, assuming that the position values sum to a multiple of N, r N The initial configuration, then, is r0 , r1 , r2 , r3 , ..., rN 1 , r N r0 r1 ... rN 1 . This can easily be transformed (in r0 r1 ... rN 1 steps) into 0, 0, ..., 0, r N . Then we add r to each of the first N 1 coins, subtracting it all from the last to obtain the desired result, r, r, r, r ..., r . Clearly this point is unique, (since the sum remains constant), and counting the number of steps it takes to reach that point is something of a moot point (no pun intended) and relies on specific cases. The only further extention of this problem I could think of was asking what sets of configurations could reach different, stranger configurations- for example where the coins are in an arithmetic sequence, or a geometric sequence, or some polynomial sequence, or all adjacent, or all on two different points, or three- there are a lot of these questions, and I'm not sure how many- if anyare solveable. I'll leave those to the reader.	2,327	8,823	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.21875	4	CC-MAIN-2017-34	latest	en	0.93026
https://en.wikipedia.org/wiki/Linear-rotational_analogs	1,480,884,429,000,000,000	text/html	crawl-data/CC-MAIN-2016-50/segments/1480698541396.20/warc/CC-MAIN-20161202170901-00461-ip-10-31-129-80.ec2.internal.warc.gz	846,174,827	37,134	# List of equations in classical mechanics (Redirected from Linear-rotational analogs) Classical mechanics is the branch of physics used to describe the motion of macroscopic objects.[1] It is the most familiar of the theories of physics. The concepts it covers, such as mass, acceleration, and force, are commonly used and known.[2] The subject is based upon a three-dimensional Euclidean space with fixed axes, called a frame of reference. The point of concurrency of the three axes is known as the origin of the particular space.[3] Classical mechanics utilises many equations—as well as other mathematical concepts—which relate various physical quantities to one another. These include differential equations, manifolds, Lie groups, and ergodic theory.[4] This page gives a summary of the most important of these. This article lists equations from Newtonian mechanics, see analytical mechanics for the more general formulation of classical mechanics (which includes Lagrangian and Hamiltonian mechanics). ## Classical mechanics ### Mass and inertia Quantity (common name/s) (Common) symbol/s Defining equation SI units Dimension Linear, surface, volumetric mass density λ or μ (especially in acoustics, see below) for Linear, σ for surface, ρ for volume. ${\displaystyle m=\int \lambda \mathrm {d} \ell }$ ${\displaystyle m=\iint \sigma \mathrm {d} S}$ ${\displaystyle m=\iiint \rho \mathrm {d} V\,\!}$ kg mn, n = 1, 2, 3 [M][L]n Moment of mass[5] m (No common symbol) Point mass: ${\displaystyle \mathbf {m} =\mathbf {r} m\,\!}$ Discrete masses about an axis ${\displaystyle x_{i}\,\!}$: ${\displaystyle \mathbf {m} =\sum _{i=1}^{N}\mathbf {r} _{\mathrm {i} }m_{i}\,\!}$ Continuum of mass about an axis ${\displaystyle x_{i}\,\!}$: ${\displaystyle \mathbf {m} =\int \rho \left(\mathbf {r} \right)x_{i}\mathrm {d} \mathbf {r} \,\!}$ kg m [M][L] Centre of mass rcom (Symbols vary) ith moment of mass ${\displaystyle \mathbf {m} _{\mathrm {i} }=\mathbf {r} _{\mathrm {i} }m_{i}\,\!}$ Discrete masses: ${\displaystyle \mathbf {r} _{\mathrm {com} }={\frac {1}{M}}\sum _{i}\mathbf {r} _{\mathrm {i} }m_{i}={\frac {1}{M}}\sum _{i}\mathbf {m} _{\mathrm {i} }\,\!}$ Mass continuum: ${\displaystyle \mathbf {r} _{\mathrm {com} }={\frac {1}{M}}\int \mathrm {d} \mathbf {m} ={\frac {1}{M}}\int \mathbf {r} \mathrm {d} m={\frac {1}{M}}\int \mathbf {r} \rho \mathrm {d} V\,\!}$ m [L] 2-Body reduced mass m12, μ Pair of masses = m1 and m2 ${\displaystyle \mu =\left(m_{1}m_{2}\right)/\left(m_{1}+m_{2}\right)\,\!}$ kg [M] Moment of inertia (MOI) I Discrete Masses: ${\displaystyle I=\sum _{i}\mathbf {m} _{\mathrm {i} }\cdot \mathbf {r} _{\mathrm {i} }=\sum _{i}\left\|\mathbf {r} _{\mathrm {i} }\right\|^{2}m\,\!}$ Mass continuum: ${\displaystyle I=\int \left\|\mathbf {r} \right\|^{2}\mathrm {d} m=\int \mathbf {r} \cdot \mathrm {d} \mathbf {m} =\int \left\|\mathbf {r} \right\|^{2}\rho \mathrm {d} V\,\!}$ kg m2 [M][L]2 ### Derived kinematic quantities Kinematic quantities of a classical particle: mass m, position r, velocity v, acceleration a. Quantity (common name/s) (Common) symbol/s Defining equation SI units Dimension Velocity v ${\displaystyle \mathbf {v} =\mathrm {d} \mathbf {r} /\mathrm {d} t\,\!}$ m s−1 [L][T]−1 Acceleration a ${\displaystyle \mathbf {a} =\mathrm {d} \mathbf {v} /\mathrm {d} t=\mathrm {d} ^{2}\mathbf {r} /\mathrm {d} t^{2}\,\!}$ m s−2 [L][T]−2 Jerk j ${\displaystyle \mathbf {j} =\mathrm {d} \mathbf {a} /\mathrm {d} t=\mathrm {d} ^{3}\mathbf {r} /\mathrm {d} t^{3}\,\!}$ m s−3 [L][T]−3 Angular velocity ω ${\displaystyle {\boldsymbol {\omega }}=\mathbf {\hat {n}} \left(\mathrm {d} \theta /\mathrm {d} t\right)\,\!}$ rad s−1 [T]−1 Angular Acceleration α ${\displaystyle {\boldsymbol {\alpha }}=\mathrm {d} {\boldsymbol {\omega }}/\mathrm {d} t=\mathbf {\hat {n}} \left(\mathrm {d} ^{2}\theta /\mathrm {d} t^{2}\right)\,\!}$ rad s−2 [T]−2 ### Derived dynamic quantities Angular momenta of a classical object. Left: intrinsic "spin" angular momentum S is really orbital angular momentum of the object at every point, right: extrinsic orbital angular momentum L about an axis, top: the moment of inertia tensor I and angular velocity ω (L is not always parallel to ω)[6] bottom: momentum p and it's radial position r from the axis. The total angular momentum (spin + orbital) is J. Quantity (common name/s) (Common) symbol/s Defining equation SI units Dimension Momentum p ${\displaystyle \mathbf {p} =m\mathbf {v} \,\!}$ kg m s−1 [M][L][T]−1 Force F ${\displaystyle \mathbf {F} =\mathrm {d} \mathbf {p} /\mathrm {d} t\,\!}$ N = kg m s−2 [M][L][T]−2 Impulse J, Δp, I ${\displaystyle \mathbf {J} =\Delta \mathbf {p} =\int _{t_{1}}^{t_{2}}\mathbf {F} \mathrm {d} t\,\!}$ kg m s−1 [M][L][T]−1 Angular momentum about a position point r0, L, J, S ${\displaystyle \mathbf {L} =\left(\mathbf {r} -\mathbf {r} _{0}\right)\times \mathbf {p} \,\!}$ Most of the time we can set r0 = 0 if particles are orbiting about axes intersecting at a common point. kg m2 s−1 [M][L]2[T]−1 Moment of a force about a position point r0, Torque τ, M ${\displaystyle {\boldsymbol {\tau }}=\left(\mathbf {r} -\mathbf {r} _{0}\right)\times \mathbf {F} =\mathrm {d} \mathbf {L} /\mathrm {d} t\,\!}$ N m = kg m2 s−2 [M][L]2[T]−2 Angular impulse ΔL (no common symbol) ${\displaystyle \Delta \mathbf {L} =\int _{t_{1}}^{t_{2}}{\boldsymbol {\tau }}\mathrm {d} t\,\!}$ kg m2 s−1 [M][L]2[T]−1 ### General energy definitions Main article: Mechanical energy Quantity (common name/s) (Common) symbol/s Defining equation SI units Dimension Mechanical work due to a Resultant Force W ${\displaystyle W=\int _{C}\mathbf {F} \cdot \mathrm {d} \mathbf {r} \,\!}$ J = N m = kg m2 s−2 [M][L]2[T]−2 Work done ON mechanical system, Work done BY WON, WBY ${\displaystyle \Delta W_{\mathrm {ON} }=-\Delta W_{\mathrm {BY} }\,\!}$ J = N m = kg m2 s−2 [M][L]2[T]−2 Potential energy φ, Φ, U, V, Ep ${\displaystyle \Delta W=-\Delta V\,\!}$ J = N m = kg m2 s−2 [M][L]2[T]−2 Mechanical power P ${\displaystyle P=\mathrm {d} E/\mathrm {d} t\,\!}$ W = J s−1 [M][L]2[T]−3 Every conservative force has a potential energy. By following two principles one can consistently assign a non-relative value to U: • Wherever the force is zero, its potential energy is defined to be zero as well. • Whenever the force does work, potential energy is lost. ### Generalized mechanics Quantity (common name/s) (Common) symbol/s Defining equation SI units Dimension Generalized coordinates q, Q varies with choice varies with choice Generalized velocities ${\displaystyle {\dot {q}},{\dot {Q}}\,\!}$ ${\displaystyle {\dot {q}}\equiv \mathrm {d} q/\mathrm {d} t\,\!}$ varies with choice varies with choice Generalized momenta p, P ${\displaystyle p=\partial L/\partial {\dot {q}}\,\!}$ varies with choice varies with choice Lagrangian L ${\displaystyle L(\mathbf {q} ,\mathbf {\dot {q}} ,t)=T(\mathbf {\dot {q}} )-V(\mathbf {q} ,\mathbf {\dot {q}} ,t)\,\!}$ where ${\displaystyle \mathbf {q} =\mathbf {q} (t)\,\!}$ and p = p(t) are vectors of the generalized coords and momenta, as functions of time J [M][L]2[T]−2 Hamiltonian H ${\displaystyle H(\mathbf {p} ,\mathbf {q} ,t)=\mathbf {p} \cdot \mathbf {\dot {q}} -L(\mathbf {q} ,\mathbf {\dot {q}} ,t)\,\!}$ J [M][L]2[T]−2 Action, Hamilton's principal function S, ${\displaystyle \scriptstyle {\mathcal {S}}\,\!}$ ${\displaystyle {\mathcal {S}}=\int _{t_{1}}^{t_{2}}L(\mathbf {q} ,\mathbf {\dot {q}} ,t)\mathrm {d} t\,\!}$ J s [M][L]2[T]−1 ## Kinematics In the following rotational definitions, the angle can be any angle about the specified axis of rotation. It is customary to use θ, but this does not have to be the polar angle used in polar coordinate systems. The unit axial vector ${\displaystyle {\mathbf {\hat {n}}}={\mathbf {\hat {e}}}_{r}\times {\mathbf {\hat {e}}}_{\theta }\,\!}$ defines the axis of rotation, ${\displaystyle \scriptstyle {\mathbf {\hat {e}}}_{r}\,\!}$ = unit vector in direction of r, ${\displaystyle \scriptstyle {\mathbf {\hat {e}}}_{\theta }\,\!}$ = unit vector tangential to the angle. Translation Rotation Velocity Average: ${\displaystyle \mathbf {v} _{\mathrm {average} }={\Delta \mathbf {r} \over \Delta t}}$ Instantaneous: ${\displaystyle \mathbf {v} ={d\mathbf {r} \over dt}}$ Angular velocity ${\displaystyle {\boldsymbol {\omega }}={\mathbf {\hat {n}}}{\frac {{\rm {d}}\theta }{{\rm {d}}t}}\,\!}$ Rotating rigid body: ${\displaystyle \mathbf {v} ={\boldsymbol {\omega }}\times \mathbf {r} \,\!}$ Acceleration Average: ${\displaystyle \mathbf {a} _{\mathrm {average} }={\frac {\Delta \mathbf {v} }{\Delta t}}}$ Instantaneous: ${\displaystyle \mathbf {a} ={\frac {d\mathbf {v} }{dt}}={\frac {d^{2}\mathbf {r} }{dt^{2}}}}$ Angular acceleration ${\displaystyle {\boldsymbol {\alpha }}={\frac {{\rm {d}}{\boldsymbol {\omega }}}{{\rm {d}}t}}={\mathbf {\hat {n}}}{\frac {{\rm {d}}^{2}\theta }{{\rm {d}}t^{2}}}\,\!}$ Rotating rigid body: ${\displaystyle \mathbf {a} ={\boldsymbol {\alpha }}\times \mathbf {r} +{\boldsymbol {\omega }}\times \mathbf {v} \,\!}$ Jerk Average: ${\displaystyle \mathbf {j} _{\mathrm {average} }={\frac {\Delta \mathbf {a} }{\Delta t}}}$ Instantaneous: ${\displaystyle \mathbf {j} ={\frac {d\mathbf {a} }{dt}}={\frac {d^{2}\mathbf {v} }{dt^{2}}}={\frac {d^{3}\mathbf {r} }{dt^{3}}}}$ Angular jerk ${\displaystyle {\boldsymbol {\zeta }}={\frac {{\rm {d}}{\boldsymbol {\alpha }}}{{\rm {d}}t}}={\mathbf {\hat {n}}}{\frac {{\rm {d}}^{2}\omega }{{\rm {d}}t^{2}}}={\mathbf {\hat {n}}}{\frac {{\rm {d}}^{3}\theta }{{\rm {d}}t^{3}}}\,\!}$ Rotating rigid body: ${\displaystyle \mathbf {j} ={\boldsymbol {\zeta }}\times \mathbf {r} +{\boldsymbol {\alpha }}\times \mathbf {a} \,\!}$ ## Dynamics Translation Rotation Momentum Momentum is the "amount of translation" ${\displaystyle \mathbf {p} =m\mathbf {v} }$ For a rotating rigid body: ${\displaystyle \mathbf {p} ={\boldsymbol {\omega }}\times \mathbf {m} \,\!}$ Angular momentum Angular momentum is the "amount of rotation": ${\displaystyle \mathbf {L} =\mathbf {r} \times \mathbf {p} =\mathbf {I} \cdot {\boldsymbol {\omega }}}$ and the cross-product is a pseudovector i.e. if r and p are reversed in direction (negative), L is not. In general I is an order-2 tensor, see above for its components. The dot · indicates tensor contraction. Force and Newton's 2nd law Resultant force acts on a system at the center of mass, equal to the rate of change of momentum: {\displaystyle {\begin{aligned}\mathbf {F} &={\frac {d\mathbf {p} }{dt}}={\frac {d(m\mathbf {v} )}{dt}}\\&=m\mathbf {a} +\mathbf {v} {\frac {{\rm {d}}m}{{\rm {d}}t}}\\\end{aligned}}\,\!} For a number of particles, the equation of motion for one particle i is:[7] ${\displaystyle {\frac {\mathrm {d} \mathbf {p} _{i}}{\mathrm {d} t}}=\mathbf {F} _{E}+\sum _{i\neq j}\mathbf {F} _{ij}\,\!}$ where pi = momentum of particle i, Fij = force on particle i by particle j, and FE = resultant external force (due to any agent not part of system). Particle i does not exert a force on itself. Torque Torque τ is also called moment of a force, because it is the rotational analogue to force:[8] ${\displaystyle {\boldsymbol {\tau }}={\frac {{\rm {d}}\mathbf {L} }{{\rm {d}}t}}=\mathbf {r} \times \mathbf {F} ={\frac {{\rm {d}}(\mathbf {I} \cdot {\boldsymbol {\omega }})}{{\rm {d}}t}}\,\!}$ For rigid bodies, Newton's 2nd law for rotation takes the same form as for translation: {\displaystyle {\begin{aligned}{\boldsymbol {\tau }}&={\frac {{\rm {d}}{\mathbf {L}}}{{\rm {d}}t}}={\frac {{\rm {d}}({\mathbf {I}}\cdot {\boldsymbol {\omega }})}{{\rm {d}}t}}\\&={\frac {{\rm {d}}{\mathbf {I}}}{{\rm {d}}t}}\cdot {\boldsymbol {\omega }}+{\mathbf {I}}\cdot {\boldsymbol {\alpha }}\\\end{aligned}}\,\!} Likewise, for a number of particles, the equation of motion for one particle i is:[9] ${\displaystyle {\frac {\mathrm {d} \mathbf {L} _{i}}{\mathrm {d} t}}={\boldsymbol {\tau }}_{E}+\sum _{i\neq j}{\boldsymbol {\tau }}_{ij}\,\!}$ Yank Yank is rate of change of force: {\displaystyle {\begin{aligned}\mathbf {Y} &={\frac {d\mathbf {F} }{dt}}={\frac {d^{2}\mathbf {p} }{dt^{2}}}={\frac {d^{2}(m\mathbf {v} )}{dt^{2}}}\\&=m\mathbf {j} +\mathbf {2a} {\frac {{\rm {d}}m}{{\rm {d}}t}}+\mathbf {v} {\frac {{\rm {d^{2}}}m}{{\rm {d}}t^{2}}}\\\end{aligned}}\,\!} For constant mass, it becomes; ${\displaystyle \mathbf {Y} =m\mathbf {j} }$ Rotatum Rotatum Ρ is also called moment of a Yank, because it is the rotational analogue to yank: ${\displaystyle {\boldsymbol {\mathrm {P} }}={\frac {{\rm {d}}\mathbf {\tau } }{{\rm {d}}t}}=\mathbf {r} \times \mathbf {Y} ={\frac {{\rm {d}}(\mathbf {I} \cdot {\boldsymbol {\alpha }})}{{\rm {d}}t}}\,\!}$ Impulse Impulse is the change in momentum: ${\displaystyle \Delta \mathbf {p} =\int \mathbf {F} dt}$ For constant force F: ${\displaystyle \Delta \mathbf {p} =\mathbf {F} \Delta t}$ Angular impulse is the change in angular momentum: ${\displaystyle \Delta \mathbf {L} =\int {\boldsymbol {\tau }}dt}$ For constant torque τ: ${\displaystyle \Delta \mathbf {L} ={\boldsymbol {\tau }}\Delta t}$ ### Precession The precession angular speed of a spinning top is given by: ${\displaystyle {\boldsymbol {\Omega }}={\frac {wr}{I{\boldsymbol {\omega }}}}}$ where w is the weight of the spinning flywheel. ## Energy The mechanical work done by an external agent on a system is equal to the change in kinetic energy of the system: General work-energy theorem (translation and rotation) The work done W by an external agent which exerts a force F (at r) and torque τ on an object along a curved path C is: ${\displaystyle W=\Delta T=\int _{C}\left(\mathbf {F} \cdot \mathrm {d} \mathbf {r} +{\boldsymbol {\tau }}\cdot \mathbf {n} {\mathrm {d} \theta }\right)\,\!}$ where θ is the angle of rotation about an axis defined by a unit vector n. Kinetic energy ${\displaystyle \Delta E_{k}=W={\frac {1}{2}}m(v^{2}-{v_{0}}^{2})}$ Elastic potential energy For a stretched spring fixed at one end obeying Hooke's law: ${\displaystyle \Delta E_{p}={\frac {1}{2}}k(r_{2}-r_{1})^{2}\,\!}$ where r2 and r1 are collinear coordinates of the free end of the spring, in the direction of the extension/compression, and k is the spring constant. ## Euler's equations for rigid body dynamics Euler also worked out analogous laws of motion to those of Newton, see Euler's laws of motion. These extend the scope of Newton's laws to rigid bodies, but are essentially the same as above. A new equation Euler formulated is:[10] ${\displaystyle \mathbf {I} \cdot {\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times \left(\mathbf {I} \cdot {\boldsymbol {\omega }}\right)={\boldsymbol {\tau }}\,\!}$ where I is the moment of inertia tensor. ## General planar motion The previous equations for planar motion can be used here: corollaries of momentum, angular momentum etc. can immediately follow by applying the above definitions. For any object moving in any path in a plane, ${\displaystyle \mathbf {r} ={\mathbf {r}}(t)=r{\mathbf {\hat {e}}}_{r}\,\!}$ the following general results apply to the particle. Kinematics Dynamics Position ${\displaystyle \mathbf {r} ={\mathbf {r}}\left(r,\theta ,t\right)=r{\mathbf {\hat {e}}}_{r}}$ Velocity ${\displaystyle \mathbf {v} ={\mathbf {\hat {e}}}_{r}{\frac {\mathrm {d} r}{\mathrm {d} t}}+r\omega {\mathbf {\hat {e}}}_{\theta }}$ Momentum ${\displaystyle \mathbf {p} =m\left({\mathbf {\hat {e}}}_{r}{\frac {\mathrm {d} r}{\mathrm {d} t}}+r\omega {\mathbf {\hat {e}}}_{\theta }\right)}$ Angular momenta ${\displaystyle \mathbf {L} =m{\mathbf {r}}\times \left({\mathbf {\hat {e}}}_{r}{\frac {\mathrm {d} r}{\mathrm {d} t}}+r\omega {\mathbf {\hat {e}}}_{\theta }\right)}$ Acceleration ${\displaystyle \mathbf {a} =\left({\frac {\mathrm {d} ^{2}r}{\mathrm {d} t^{2}}}-r\omega ^{2}\right){\mathbf {\hat {e}}}_{r}+\left(r\alpha +2\omega {\frac {\mathrm {d} r}{{\rm {d}}t}}\right){\mathbf {\hat {e}}}_{\theta }}$ The centripetal force is ${\displaystyle \mathbf {F} _{\bot }=-m\omega ^{2}R{\mathbf {\hat {e}}}_{r}=-\omega ^{2}\mathbf {m} \,\!}$ where again m is the mass moment, and the coriolis force is ${\displaystyle \mathbf {F} _{c}=2\omega m{\frac {{\rm {d}}r}{{\rm {d}}t}}{\mathbf {\hat {e}}}_{\theta }=2\omega mv{\mathbf {\hat {e}}}_{\theta }\,\!}$ The Coriolis acceleration and force can also be written: ${\displaystyle \mathbf {F} _{c}=m\mathbf {a} _{c}=-2m{\boldsymbol {\omega \times v}}}$ ### Central force motion For a massive body moving in a central potential due to another object, which depends only on the radial separation between the centres of masses of the two objects, the equation of motion is: ${\displaystyle {\frac {d^{2}}{d\theta ^{2}}}\left({\frac {1}{\mathbf {r} }}\right)+{\frac {1}{\mathbf {r} }}=-{\frac {\mu \mathbf {r} ^{2}}{\mathbf {l} ^{2}}}\mathbf {F} (\mathbf {r} )}$ ## Equations of motion (constant acceleration) These equations can be used only when acceleration is constant. If acceleration is not constant then the general calculus equations above must be used, found by integrating the definitions of position, velocity and acceleration (see above). Linear motion Angular motion ${\displaystyle v=v_{0}+at\,}$ ${\displaystyle \omega _{1}=\omega _{0}+\alpha t\,}$ ${\displaystyle s={\frac {1}{2}}(v_{0}+v)t}$ ${\displaystyle \theta ={\frac {1}{2}}(\omega _{0}+\omega _{1})t}$ ${\displaystyle s=v_{0}t+{\frac {1}{2}}at^{2}}$ ${\displaystyle \theta =\omega _{0}t+{\frac {1}{2}}\alpha t^{2}}$ ${\displaystyle v^{2}=v_{0}^{2}+2as\,}$ ${\displaystyle \omega _{1}^{2}=\omega _{0}^{2}+2\alpha \theta }$ ${\displaystyle s=vt-{\frac {1}{2}}at^{2}}$ ${\displaystyle \theta =\omega _{1}t-{\frac {1}{2}}\alpha t^{2}}$ ## Galilean frame transforms For classical (Galileo-Newtonian) mechanics, the transformation law from one inertial or accelerating (including rotation) frame (reference frame traveling at constant velocity - including zero) to another is the Galilean transform. Unprimed quantities refer to position, velocity and acceleration in one frame F; primed quantities refer to position, velocity and acceleration in another frame F' moving at translational velocity V or angular velocity Ω relative to F. Conversely F moves at velocity (—V or —Ω) relative to F'. The situation is similar for relative accelerations. Motion of entities Inertial frames Accelerating frames Translation V = Constant relative velocity between two inertial frames F and F'. A = (Variable) relative acceleration between two accelerating frames F and F'. Relative position ${\displaystyle \mathbf {r} '=\mathbf {r} +\mathbf {V} t\,\!}$ Relative velocity ${\displaystyle \mathbf {v} '=\mathbf {v} +\mathbf {V} \,\!}$ Equivalent accelerations ${\displaystyle \mathbf {a} '=\mathbf {a} }$ Relative accelerations ${\displaystyle \mathbf {a} '=\mathbf {a} +\mathbf {A} }$ Apparent/fictitious forces ${\displaystyle \mathbf {F} '=\mathbf {F} -\mathbf {F} _{\mathrm {app} }}$ Rotation Ω = Constant relative angular velocity between two frames F and F'. Λ = (Variable) relative angular acceleration between two accelerating frames F and F'. Relative angular position ${\displaystyle \theta '=\theta +\Omega t\,\!}$ Relative velocity ${\displaystyle {\boldsymbol {\omega }}'={\boldsymbol {\omega }}+{\boldsymbol {\Omega }}\,\!}$ Equivalent accelerations ${\displaystyle {\boldsymbol {\alpha }}'={\boldsymbol {\alpha }}}$ Relative accelerations ${\displaystyle {\boldsymbol {\alpha }}'={\boldsymbol {\alpha }}+{\boldsymbol {\Lambda }}}$ Apparent/fictitious torques ${\displaystyle {\boldsymbol {\tau }}'={\boldsymbol {\tau }}-{\boldsymbol {\tau }}_{\mathrm {app} }}$ Transformation of any vector T to a rotating frame ${\displaystyle {\frac {{\rm {d}}\mathbf {T} '}{{\rm {d}}t}}={\frac {{\rm {d}}\mathbf {T} }{{\rm {d}}t}}-{\boldsymbol {\Omega }}\times \mathbf {T} }$ ## Mechanical oscillators SHM, DHM, SHO, and DHO refer to simple harmonic motion, damped harmonic motion, simple harmonic oscillator and damped harmonic oscillator respectively. Equations of motion Physical situation Nomenclature Translational equations Angular equations SHM • x = Transverse displacement • θ = Angular displacement • A = Transverse amplitude • Θ = Angular amplitude ${\displaystyle {\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}=-\omega ^{2}x\,\!}$ Solution: ${\displaystyle x=A\sin \left(\omega t+\phi \right)\,\!}$ ${\displaystyle {\frac {\mathrm {d} ^{2}\theta }{\mathrm {d} t^{2}}}=-\omega ^{2}\theta \,\!}$ Solution: ${\displaystyle \theta =\Theta \sin \left(\omega t+\phi \right)\,\!}$ Unforced DHM • b = damping constant • κ = torsion constant ${\displaystyle {\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}+b{\frac {\mathrm {d} x}{\mathrm {d} t}}+\omega ^{2}x=0\,\!}$ Solution (see below for ω'): ${\displaystyle x=Ae^{-bt/2m}\cos \left(\omega '\right)\,\!}$ Resonant frequency: ${\displaystyle \omega _{\mathrm {res} }={\sqrt {\omega ^{2}-\left({\frac {b}{4m}}\right)^{2}}}\,\!}$ Damping rate: ${\displaystyle \gamma =b/m\,\!}$ ${\displaystyle \tau =1/\gamma \,\!}$ ${\displaystyle {\frac {\mathrm {d} ^{2}\theta }{\mathrm {d} t^{2}}}+b{\frac {\mathrm {d} \theta }{\mathrm {d} t}}+\omega ^{2}\theta =0\,\!}$ Solution: ${\displaystyle \theta =\Theta e^{-\kappa t/2m}\cos \left(\omega \right)\,\!}$ Resonant frequency: ${\displaystyle \omega _{\mathrm {res} }={\sqrt {\omega ^{2}-\left({\frac {\kappa }{4m}}\right)^{2}}}\,\!}$ Damping rate: ${\displaystyle \gamma =\kappa /m\,\!}$ ${\displaystyle \tau =1/\gamma \,\!}$ Angular frequencies Physical situation Nomenclature Equations Linear undamped unforced SHO • k = spring constant • m = mass of oscillating bob ${\displaystyle \omega ={\sqrt {\frac {k}{m}}}\,\!}$ Linear unforced DHO • k = spring constant • b = Damping coefficient ${\displaystyle \omega '={\sqrt {{\frac {k}{m}}-\left({\frac {b}{2m}}\right)^{2}}}\,\!}$ Low amplitude angular SHO • I = Moment of inertia about oscillating axis • κ = torsion constant ${\displaystyle \omega ={\sqrt {\frac {\kappa }{I}}}\,\!}$ Low amplitude simple pendulum • L = Length of pendulum • g = Gravitational acceleration • Θ = Angular amplitude Approximate value ${\displaystyle \omega ={\sqrt {\frac {g}{L}}}\,\!}$ Exact value can be shown to be: ${\displaystyle \omega ={\sqrt {\frac {g}{L}}}\left[1+\sum _{k=1}^{\infty }{\frac {\prod _{n=1}^{k}\left(2n-1\right)}{\prod _{n=1}^{m}\left(2n\right)}}\sin ^{2n}\Theta \right]\,\!}$ Energy in mechanical oscillations Physical situation Nomenclature Equations SHM energy • T = kinetic energy • U = potential energy • E = total energy Potential energy ${\displaystyle U={\frac {m}{2}}\left(x\right)^{2}={\frac {m\left(\omega A\right)^{2}}{2}}\cos ^{2}(\omega t+\phi )\,\!}$ Maximum value at x = A: ${\displaystyle U_{\mathrm {max} }{\frac {m}{2}}\left(\omega A\right)^{2}\,\!}$ Kinetic energy ${\displaystyle T={\frac {\omega ^{2}m}{2}}\left({\frac {\mathrm {d} x}{\mathrm {d} t}}\right)^{2}={\frac {m\left(\omega A\right)^{2}}{2}}\sin ^{2}\left(\omega t+\phi \right)\,\!}$ Total energy ${\displaystyle E=T+U\,\!}$ DHM energy ${\displaystyle E={\frac {m\left(\omega A\right)^{2}}{2}}e^{-bt/m}\,\!}$ ## Notes 1. ^ Mayer, Sussman & Wisdom 2001, p. xiii 2. ^ Berkshire & Kibble 2004, p. 1 3. ^ Berkshire & Kibble 2004, p. 2 4. ^ Arnold 1989, p. v 5. ^ Section: Moments and center of mass 6. ^ R.P. Feynman; R.B. Leighton; M. Sands (1964). Feynman's Lectures on Physics (volume 2). Addison-Wesley. pp. 31–7. ISBN 978-0-201-02117-2. 7. ^ "Relativity, J.R. Forshaw 2009" 8. ^ "Mechanics, D. Kleppner 2010" 9. ^ "Relativity, J.R. Forshaw 2009" 10. ^ "Relativity, J.R. Forshaw 2009"	7,923	23,556	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 127, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2016-50	longest	en	0.816559
https://www.ukessays.com/essays/biology/measurements-statistics-and-significant-digits-biology-essay.php	1,492,977,322,000,000,000	text/html	crawl-data/CC-MAIN-2017-17/segments/1492917118743.41/warc/CC-MAIN-20170423031158-00573-ip-10-145-167-34.ec2.internal.warc.gz	987,020,384	12,764	# Measurements Statistics And Significant Digits Biology Essay Published: Last Edited: This essay has been submitted by a student. This is not an example of the work written by our professional essay writers. The question in the experiment carried out was How accurate can you measure. In order to answer this question, a hypothesis was formed based on the purposes of this lab activity. The hypothesis stated was the following: "If we measure an object, then the accuracy depends on the instrument, because the manufacturer of the instrument calibrated that instrument to certain accuracy." Pre-laboratory /preparatory questions If you are using a graduated cylinder whose smallest division is 0.1 mL, to what degree of precision should you report liquid volumes? Students should report liquid volumes to a degree of precision of +/- 0.01. A stack of five CDs is 1.6 cm tall. What is the average thickness of each disk? To the nearest whole number, how many disks will be in a stack 10 cm tall? 1.6/5=0.32 The average thickness of each CD is 0.32 centimeters. 10/0.32= 31.25 In a stack of 10 centimeters tall there would be 31 CDs. What is a meniscus; what role does it play in the correct reading of liquid volumes? "The meniscus is the curve that is formed in the surface of a liquid close to the surface of another liquid. It is caused by surface tension (Wikipedia, 2012)". Given a set of data, the mean of these numbers is its average. To find the mean, add all the numbers given in the data and divide by the amount of numbers given. The median is the middle number of the data, when it is ordered from least to greatest. If there are two numbers in the middle, find the average of those two numbers and that will be the median. Mode is the most repeated number in the data. Range is the difference between the greatest and lowest number arranged from least to greatest. Scientific notation is a way to express numbers using a base ten. It helps when working with large and small numbers. Significant digits are used in scientific work to make it easier to scientists. There are rules to follow when writing scientific digits. Examples: 30.2, has three significant digits. 100, has one significant digits. 100., has three significant digits. 0.0054, has two significant digits 100Â°C 373.15K 212Â°F 0.Â°C 273.15K 32.Â°F 37.Â°C 310.15K 98.6Â°F ## Complete the chart -40C 233.15K -40Â°F - 273.15Â°C 0K -459.67Â°F 210Â°C 63.15K - 346Â°F Theoretical Framework In the experiment students had the opportunity to manipulate lab instruments and terms to which they are unfamiliar such as the triple beam balance, significant digits, and uncertainty. The triple beam balance is a scale to measure the mass of lab samples. Significant digits are all the non-zero digits, also are the zeros that occur between significant digits. The uncertainty shows the margin of error acceptable in measuring. (Wikipedia, 2012) Materials Specials rulers (1=feet, 2=inches, 3&4 are in cm) Graduated cylinders (3) 100, 50, 10 mL Triple-beam balance Beakers (3) 150mL Tap water Ice Thermometer, -20Â°C to 110Â°C (4) Hot plate Bag of M&M candy Procedure In the first step, I measured the width and the length of the biology books. I had to use four different rulers. One of them had the measurement of 1 foot. The second one had the measures in inches. The third and forth one I had to measure it in cm. After measuring I had to calculate the perimeter and the area of each measurement. In the second part of the lab, I had to look at three graduated cylinders. Each of these cylinders had different volume capacities. Then I recorded the nominal capacity, volume of the liquid and the uncertainty in table B. The step three of this lab consisted in weighting 3 beaker three times. Each beaker had different volume which also had different weight. I had to weight each beaker in the three triple beam balances, set on the table. After I weighted the three beakers in every balance I had to find the average weight of the three weights. The forth step of the lab was to measure the temperature of 4 flasks. Two of those flasks were on the hot plate and the other two were on a side. I was expected to measure the temperature in Celsius and then I had to convert it in Kelvin. After I had the temperatures I recorded my data on table C. The fifth step of this lab experiment consisted in measuring a full bag of M&M and record the data in table E. I also had to weight the bag of M&M empty. Then had to separate the candy into colors and count how much of a kind there were. After that I had to find the percentage of each color in my bag. After that I had to determine the class totals in colors in each bag. Results ## Area 1 +/-0.9 ft +/-0.7ft 27.43cm 21.33 cm 97.52 cm 585.0819cm2 2 11in 8.5in 27.94cm 21.59 cm 110.74 cm 603.2246 cm2 3 27.4 cm 21.3cm 27.4cm 21.3 cm 97.4 cm 583.62 cm2 4 27.40 cm 31.30cm 27.40cm 21.30 cm 97.40cm 583.620 cm2 (Online Conversions, 2012) ## Data Table - Part B Nominal Capacity Volume of Liquid Uncertainty Â± 1 10mL 7.02mL +/- 0.01 2 50mL 31.2mL +/- 0.1 3 100mL 83mL +/- 0.1 Data Table part C Jar Weight 1 Weight 2 Weight 3 Average Uncertainty +/- 1 red 131.05 131.56 131.58 131.39 0.01g 2 yellow 157.04 157.55 157.55 157.38 0.01g 3 blue 110.05 114.02 114.03 112.7 0.01g ## Data Table - Part D Beaker (-20Â°C to 110Â°C) Temperature Celsius Temperature Kelvin Uncertainty Â± 1 101Â°C 383.15K 0.1 2 22.5Â°C 295.65K 0.01 Beaker (-20Â°C to 110Â°C) Temperature Celsius Temperature Kelvin Uncertainty Â± 1 102.4Â°C 375.55K 0.01 2 23Â°C 296.15K 0.1 Color of M&M Number in bag Percent in bag Class totals Brown 0 0 14 Yellow 2 11% 14 Red 3 16% 25 Green 4 21% 41 Blue 3 16% 29 Orange 7 36% 48 ## TOTALS 19 100% 171 Weight with the M&M's 17.04 grams Weight empty 0.04 j ## Range brown 0 0 0 1 1 1 2 3 4 0,1 1 1.3333 4 yellow 0 0 1 2 2 2 3 3 3 3 2 1.7777 3 Red 0 1 1 2 3 4 4 4 6 4 3 2.7777 6 green 1 3 3 4 4 5 6 7 7 3,4,7 4 4.4444 6 blue 2 3 3 3 3 3 4 4 5 3 3 3.3333 3 orange 3 3 4 5 6 6 7 7 7 7 6 5.3333 4 Analysis The ruler 4 gave me the most precise measurement because the more divisions a ruler has the easier it is. The ruler 1 gave the most un-precise measure. When I was measuring the book the cm one made it easier to record my data, but the ruler 1 was a little bit difficult. It occurs because the 1 foot ruler is an actual estimation of the book, but the cm is the exact measurement to the book. The most common explanation for a decrease of volume in a graduated cylinder is evaporation of the water in the container. The explanation I could give is the meniscus, which is the curve you see when watching a graduated cylinder filled with water. The 100mL container gave me the most precise reading of the graduated cylinders, because it was the most visible and had more space between lines. Something that might cause an unavoidable error is that fact that the balances say they are balanced but are still moving. Another unavoidable error is that in between the changing of the balance you accidentally pour out a little bit of liquid inside the container. You know when a container has salt when the temperature is lower than the other one that does not have salt. I think my percent differs from everybody else because the bags have different amount of M&Ms in each of the bags. The number of the M&M in a bag is not always the same. Conclusion The lab experiment carried out involved activities to gain experience in measurements, using significant digits and scientific notation in real life situations. At the same time, students had the opportunity to calculate mean, median, mode, and range given a set of data. The lab activity was conducted with the purpose of exposing students to basic concepts and instruments used in Science. It was intended to find out how accurate students can measure. As a hypothesis, students proposed that when measuring an object, the accuracy depends on the instrument, due to the fact that the manufacturers calibrate it to certain accuracy. It involved simple processes such as using different rulers to measure the length and the width of the biology book, measuring temperature of liquids, using beakers and the tripod beam balance for finding the weight of the liquids, and the graduated cylinders to figure out the volume of each container. Finally, students had a fun opportunity to find the mean, median, mode, and range of units of M&Ms. As part of the results; we were able to observe that the most calibrated instruments were the ones with the most accurate measures. Therefore, the hypothesis was accepted because it was proved that the level of accuracy depended on the manufacturer and how they calibrate their instruments.	2,368	8,972	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.15625	4	CC-MAIN-2017-17	latest	en	0.942984
https://gmatclub.com/forum/as-the-journalist-left-to-interview-the-convicted-murderer-she-was-ad-275297.html	1,550,433,826,000,000,000	text/html	crawl-data/CC-MAIN-2019-09/segments/1550247482478.14/warc/CC-MAIN-20190217192932-20190217214932-00255.warc.gz	580,620,750	157,519	GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 17 Feb 2019, 12:03 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. 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AND we want you to crush the GMAT! # As the journalist left to interview the convicted murderer, she was ad Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 52906 As the journalist left to interview the convicted murderer, she was ad [#permalink] ### Show Tags 05 Sep 2018, 02:50 00:00 Difficulty: 95% (hard) Question Stats: 37% (01:22) correct 63% (01:41) wrong based on 205 sessions ### HideShow timer Statistics As the journalist left to interview the convicted murderer, she was advised of the man’s short temper, told she should not anger him, and was given a tape recorder. (A) of the man’s short temper, told she should not anger him, and was (B) of the man’s short temper, told she should not anger him, and (C) of the man’s short temper and that she should not anger him and (D) that the man had a short temper, should not anger him, and was (E) that the man had a short temper, that she should not anger him, and was _________________ Manager Joined: 16 Jan 2018 Posts: 91 Location: New Zealand Re: As the journalist left to interview the convicted murderer, she was ad [#permalink] ### Show Tags 05 Sep 2018, 03:20 1 Bunuel wrote: As the journalist left to interview the convicted murderer, she was advised of the man’s short temper, told she should not anger him, and was given a tape recorder. (A) of the man’s short temper, told she should not anger him, and was Incorrect as the elements in the list are not parallel. (B) of the man’s short temper, told she should not anger him, and Correct she was [advised of the...temper],[told she...him], and [given..recorder]. The elements are parallel. (C) of the man’s short temper and that she should not anger him and Incorrect! here multiple ands are used, creating confusion without proper punctuation. (D) that the man had a short temper, should not anger him, and was Incorrect. Non parallel elements are used. (E) that the man had a short temper, that she should not anger him, and was Incorrect as the elements in the list are not parallel. {that A, that B and was given} Director Joined: 20 Feb 2015 Posts: 795 Concentration: Strategy, General Management Re: As the journalist left to interview the convicted murderer, she was ad [#permalink] ### Show Tags 05 Sep 2018, 04:20 Bunuel wrote: As the journalist left to interview the convicted murderer, she was advised of the man’s short temper, told she should not anger him, and was given a tape recorder. (A) of the man’s short temper, told she should not anger him, and was (B) of the man’s short temper, told she should not anger him, and (C) of the man’s short temper and that she should not anger him and (D) that the man had a short temper, should not anger him, and was (E) that the man had a short temper, that she should not anger him, and was The parallel elements are advised.....,told....,and given Imo Manager Joined: 01 Feb 2018 Posts: 101 Location: India GMAT 1: 700 Q47 V38 WE: Consulting (Consulting) Re: As the journalist left to interview the convicted murderer, she was ad [#permalink] ### Show Tags 05 Sep 2018, 06:06 Bunuel wrote: As the journalist left to interview the convicted murderer, she was advised of the man’s short temper, told she should not anger him, and was given a tape recorder. (A) of the man’s short temper, told she should not anger him, and was (B) of the man’s short temper, told she should not anger him, and (C) of the man’s short temper and that she should not anger him and (D) that the man had a short temper, should not anger him, and was (E) that the man had a short temper, that she should not anger him, and was The correct parallelism is maintained in option B (was advised of the man’s short temper, told she should not anger him, and given a tape recorder). So IMO answer should be B. _________________ Please press Kudos if this helped “Going in one more round when you don't think you can, that's what makes all the difference in your life.” Director Joined: 08 Jun 2013 Posts: 552 Location: France GMAT 1: 200 Q1 V1 GPA: 3.82 WE: Consulting (Other) Re: As the journalist left to interview the convicted murderer, she was ad [#permalink] ### Show Tags 05 Sep 2018, 07:41 As the journalist left to interview the convicted murderer, she was advised of the man’s short temper, told she should not anger him, and was given a tape recorder. (A) of the man’s short temper, told she should not anger him, and was (B) of the man’s short temper, (she was) told she should not anger him, and (she was) (C) of the man’s short temper and that she should not anger him and (D) that the man had a short temper, should not anger him, and was (E) that the man had a short temper, that she should not anger him, and was Concept tested : Parallel Structure Ans : B _________________ Everything will fall into place… There is perfect timing for everything and everyone. Never doubt, But Work on improving yourself, Keep the faith and It will all make sense. Math Expert Joined: 02 Sep 2009 Posts: 52906 Re: As the journalist left to interview the convicted murderer, she was ad [#permalink] ### Show Tags 07 Sep 2018, 01:06 Bunuel wrote: As the journalist left to interview the convicted murderer, she was advised of the man’s short temper, told she should not anger him, and was given a tape recorder. (A) of the man’s short temper, told she should not anger him, and was (B) of the man’s short temper, told she should not anger him, and (C) of the man’s short temper and that she should not anger him and (D) that the man had a short temper, should not anger him, and was (E) that the man had a short temper, that she should not anger him, and was MANHATTAN REVIEW OFFICIAL EXPLANATION: This question is all about parallel structure. The basic structure is this: Someone was advised of x, told y, and given z. The only answer choice that has this parallel structure is B. Every other choice introduces different verb forms among x, y and z. B is the correct answer. _________________ Intern Joined: 21 Oct 2017 Posts: 12 Re: As the journalist left to interview the convicted murderer, she was ad [#permalink] ### Show Tags 08 Sep 2018, 22:32 As the journalist left to interview the convicted murderer, she was advised of the man’s short temper, told she should not anger him, and was given a tape recorder. (A) of the man’s short temper, told she should not anger him, and was (B) of the man’s short temper, told she should not anger him, and (C) of the man’s short temper and that she should not anger him and (D) that the man had a short temper, should not anger him, and was (E) that the man had a short temper, that she should not anger him, and was Ans : "a,d,e out" (not parallel) "c" can be eliminated (multiple "and"......unnecessary) Re: As the journalist left to interview the convicted murderer, she was ad [#permalink] 08 Sep 2018, 22:32 Display posts from previous: Sort by	2,073	8,119	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.625	4	CC-MAIN-2019-09	latest	en	0.927225
http://gmatclub.com/forum/m14-96093.html	1,467,120,533,000,000,000	text/html	crawl-data/CC-MAIN-2016-26/segments/1466783396887.54/warc/CC-MAIN-20160624154956-00137-ip-10-164-35-72.ec2.internal.warc.gz	132,102,220	50,135	Find all School-related info fast with the new School-Specific MBA Forum It is currently 28 Jun 2016, 06:28 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # m14#07 Author Message Director Joined: 25 Aug 2007 Posts: 954 WE 1: 3.5 yrs IT WE 2: 2.5 yrs Retail chain Followers: 71 Kudos [?]: 1098 [0], given: 40 ### Show Tags 19 Jun 2010, 04:32 If working together, brothers Tom and Jack can paint a wall in 4 hours, how much time would it take Jack to paint the wall alone? (1) Jack is painting twice as fast as Tom. (2) If Tom painted twice as fast as he actually does, the brothers would finish the work in 3 hours. _________________ Tricky Quant problems: 50-tricky-questions-92834.html Important Grammer Fundamentals: key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html Director Joined: 25 Aug 2007 Posts: 954 WE 1: 3.5 yrs IT WE 2: 2.5 yrs Retail chain Followers: 71 Kudos [?]: 1098 [0], given: 40 ### Show Tags 20 Jun 2010, 08:02 Guys, Thank but I will be happy with your explanations. _________________ Tricky Quant problems: 50-tricky-questions-92834.html Important Grammer Fundamentals: key-fundamentals-of-grammer-our-crucial-learnings-on-sc-93659.html Intern Joined: 12 May 2010 Posts: 22 Followers: 0 Kudos [?]: 1 [0], given: 0 ### Show Tags 20 Jun 2010, 08:09 sorry.... just use the algebric approach: A - if j+t=1/4 and j=2t than j=2/12 and t=1/12. - sufficient. B - the same deal - if j+t=1/4 and 2t+y= 1/3 than t=1/12 and j=1/6=2/12 - sufficient. Manager Affiliations: The Earth organization, India Joined: 25 Dec 2010 Posts: 193 WE 1: SAP consultant-IT 2 years WE 2: Entrepreneur-family business 2 years Followers: 5 Kudos [?]: 11 [0], given: 12 ### Show Tags 24 Jun 2011, 23:12 I have a very generic doubt regarding statement 2 ONLY. if I write an equation in rates form I get the solution : 1/j+2/t = 1/3; when I solve this with the problem statement-> 1/j+1/t=1/4; it gives T=12,J=6.perfect. Doubt : If I write an equation in time form I get the solution: statement 2->If Tom painted twice as fast as he actually does, the brothers would finish the work in 3 hours.->therefore tom would take half the time than he normally does T=.5T. J+0.5T = 3 from problem statement : J+T=4. SOlving these we get T=2, which is wrong. Can someone help me with this ? _________________ Cheers !! Quant 47-Striving for 50 Verbal 34-Striving for 40 Director Status: Gonna rock this time!!! Joined: 22 Jul 2012 Posts: 547 Location: India GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29 WE: Information Technology (Computer Software) Followers: 3 Kudos [?]: 50 [0], given: 562 ### Show Tags 09 Feb 2013, 04:49 Bunuel/Karishma, Could you please explain how S2 is sufficient? _________________ hope is a good thing, maybe the best of things. And no good thing ever dies. Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595 My GMAT Journey : end-of-my-gmat-journey-149328.html#p1197992 Math Expert Joined: 02 Sep 2009 Posts: 33535 Followers: 5941 Kudos [?]: 73713 [0], given: 9903 ### Show Tags 10 Feb 2013, 04:01 Expert's post Sachin9 wrote: Bunuel/Karishma, Could you please explain how S2 is sufficient? Sure. If working together, brothers Tom and Jack can paint a wall in 4 hours, how much time would it take Jack to paint the wall alone? Say the rates of Tom and Jack are T job/hour and J job/hour, respectively. Their combined rate is T+J job/hour, and we are told that it equals to 1/4 job/hour. Thus given that T+J=1/4 job/hour. (1) Jack is painting twice as fast as Tom --> J=2T --> T+2T=1/4 --> T=1/12 --> J=2/12=1/6 --> (time)=(reciprocal of rate)=6 hours. Sufficient. (2) If Tom painted twice as fast as he actually does, the brothers would finish the work in 3 hours. This statement implies that if the rate of Tom were 2T instead of T, the brothers combined rate would be 1/3 job/hour, thus 2T+J=1/3. Solving T+J=1/4 and 2T+J=1/3 gives J=1/6. Sufficient. Hope it's clear. _________________ Senior Manager Joined: 17 Sep 2013 Posts: 394 Concentration: Strategy, General Management GMAT 1: 730 Q51 V38 WE: Analyst (Consulting) Followers: 19 Kudos [?]: 224 [0], given: 139 ### Show Tags 16 Jul 2014, 12:49 Bunuel I took 12 Units of wall..So 3 units of work per hour= T + J Stmt- I- T +2T = 3..so we get J and hence suff Stmt- II- T + T + J = 4..so T causes a difference of 1 unit and hence we can find J now.. D it is Is my reasoning correct.. For such questions, I always assume amount of work in some convenient multiple ...Can I ever get stuck with such an approach..i.e assuming the total work in units _________________ Appreciate the efforts...KUDOS for all Don't let an extra chromosome get you down.. Math Expert Joined: 02 Sep 2009 Posts: 33535 Followers: 5941 Kudos [?]: 73713 [0], given: 9903 ### Show Tags 16 Jul 2014, 12:55 Expert's post JusTLucK04 wrote: Bunuel I took 12 Units of wall..So 3 units of work per hour= T + J Stmt- I- T +2T = 3..so we get J and hence suff Stmt- II- T + T + J = 4..so T causes a difference of 1 unit and hence we can find J now.. D it is Is my reasoning correct.. For such questions, I always assume amount of work in some convenient multiple ...Can I ever get stuck with such an approach..i.e assuming the total work in units That's correct approach. Generally, if the question is such that you can assume some number for a job, then this approach will work. Not sure that it would be the best approach but most of the case it'll work fine. _________________ Current Student Status: Everyone is a leader. Just stop listening to others. Joined: 22 Mar 2013 Posts: 993 Location: India GPA: 3.51 WE: Information Technology (Computer Software) Followers: 152 Kudos [?]: 1158 [1] , given: 226 ### Show Tags 17 Jul 2014, 02:07 1 KUDOS Given : $$\frac{1}{T} + \frac{1}{J} = \frac{1}{4}$$ 1) $$2\frac{1}{T} = \frac{1}{J}$$ Substitute in given equation. we can find J and T. SUFF. 2) $$2\frac{1}{T} + \frac{1}{J} = \frac{1}{3}$$ Solve given equation and new equation together. we can find J and T. SUFF. Ans : D _________________ Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use? \| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction". Re: m14#07 [#permalink] 17 Jul 2014, 02:07 Display posts from previous: Sort by # m14#07 Moderator: Bunuel Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	2,276	7,311	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.90625	4	CC-MAIN-2016-26	longest	en	0.845668
http://docplayer.net/30337790-Math-139-problem-set-6-solution-15-april-2003.html	1,539,934,803,000,000,000	text/html	crawl-data/CC-MAIN-2018-43/segments/1539583512332.36/warc/CC-MAIN-20181019062113-20181019083613-00354.warc.gz	101,778,841	24,470	# Math 139 Problem Set #6 Solution 15 April 2003 Save this PDF as: Size: px Start display at page: ## Transcription 1 Math 139 Problem Set #6 Solution 15 April 2003 A1 (Based on Barr, 41, exercise 3) (a) Let n be an integer greater than 2 Consider the list of numbers n! + 2, n! + 3, n! + 4,, n! + n Explain why all the numbers in this list are composite Solution: We know that n! is divisible by all the numbers from 2 up to n It follows that n! + 2 is divisible by 2, since both n! and 2 are divisible by 2 Also, we know that n! + 2 > 2 Since n! + 2 > 2 and has 2 as a factor, we find that n! + 2 must be composite Similarly, for each j between 2 and n (inclusive), the number n! + j is divisible by j and greater than j, so it must have j as a (proper) factor Thus n! + j is composite (b) Use the idea in the previous part to find a list of ten consecutive composite numbers Solution: We start with 11! = Then we know the following numbers are all composite: 11! + 2 = ! + 3 = ! + 4 = ! + 11 = (c) Use the Table of Primes on pp to find the smallest ten consecutive composite numbers Solution: According to the table, the first run of 10 consecutive composite numbers begins at 114 The numbers 114, 115, 116, 117, 118, 119, 120, 121, 122, 123 are all composite (So are 124, 125, and 126) 2 (d) Do you think that there are 100 consecutive composite numbers? Explain your reasoning Solution: The numbers 101! + 2 through 101! form a sequence of 100 consecutive composite numbers A2 (Based on Barr, 41, exercise 8) (a) At most how many primes will you have to divide into to determine whether is prime? Solution: We need to check the primes that are less than or equal to So we need to check the primes There are 27 primes in this list 2, 3, 5,, 103 (b) Find the prime factorization of Solution: We try the primes in the list above one at a time, and find that is divisible by 23 In fact, = Moreover, the number 479 appears in our list of primes, so we have found the prime factorization of (c) At most how many primes will you have to divide into 100,000,003 to determine whether it is prime? Solution: We find that 100, 000, , 000, so we d need to divide this number by all the primes less than 10, 000 From the primes table in the book, we determine that there are 1229 primes in this range A3 Suppose we have a computer that generates random eight-digit numbers, and then filters out all those that are divisible by 2 or 5, so that each time we run the program, it gives us an eight-digit number that is not divisible by 2 or 5 (a) How many different outcomes are possible when we run the program? Solution: A number is divisible by 2 if its last digit is even, and divisible by 5 if its last digit is 0 or 5 Numbers that are not divisible by 2 or 5 are those whose last digit is 1, 3, 7, or 9 3 The first digit of an eight-digit number must be between 1 and 9, inclusive Each of the intermediate digits can be anything from 0 to 9 To form an eight-digit number that s not divisible by 2 or 5, we need to choose a first digit (9 possibilites) choose a second digit (10 possibilities) choose a third digit (10 possibilities) choose a seventh digit (10 possibilities) choose a last digit (4 possibilities) The total number of ways we can do this is = 36, 000, 000 Solution: (alternate) There are = different eightdigit numbers Of these, the number that are divisble by 4 is The number that are divisible by 5 is /2 = /5 = There are /10 = eight-digit numbers that are divisible by both 2 and 5 (that is, they are divisible by 10) To count the number of eight-digit numbers that are not divisible by 2 or by 5, we take and subtract , then subtract Then we have to add back , because we ve subtracted the multiples of ten twice There are = eight-digit numbers that are not divisible by 2 or 5 (b) According to the Prime Number Theorem, about how many eight-digit primes are there? Solution: There are about eight-digit primes ln ln 4 (c) How many numbers will our computer program need to pick before there s a 90% probability that at least one of them is prime? Solution: The probability that we get a prime number on any particular trial is about , so the probability of getting a composite number on any particular trial is about = The probability of getting n composite numbers on n consecutive trials is about n, so we need to find the least n such that n < 01 This turns out to be n = 17 B1 (a) Use the Euclidean algorithm to find gcd(78993, 24513) Solution: Taking residues in each line, we have So gcd(78993, 24513) = 3 gcd(78993, 24513) = gcd(24513, 5454) = gcd(5454, 2697) = gcd(2697, 60) = gcd(60, 57) = gcd(57, 3) = gcd(3, 0) (b) Use the extended Euclidean algorithm to find two integers s and t such that 58249s t = 1 Solution: By the usual division algorithm, we have = = = = = = 5 Now we do the back substitution to get 1 = = 16 5( ) = = ( ) = = ( ) = = ( ) = = ( ) = The solution is = 1 (c) Solve the congruences i 26965x (mod 58249) Solution: From above, we know that (mod 58249), so we multiply both sides of the given congruence by to get x (mod 58249) ii 58249x (mod 26965) Solution: From above, we know that (mod 26965), so it follows that 8441 is the multiplicative inverse of modulo We have (mod 26965) We multiply both sides of the given congruence by to get x (mod 26965) 6 B2 You probably know the theorem that says a number N is divisible by 3 if and only if the sum of its decimal digits is divisible by 3 That is, if we write the (k + 1)-digit number N as N = d k d k 1 d 2 d 1 d 0, (where the d i stand for the digits of N) then N is divisible by 3 if and only if the sum d k + d k d 2 + d 1 + d 0 is divisible by 3 Here s why that works Writing the digits of N as above, we have N = d d d k 1 d k k d k (1) Now we recall that 10 1 (mod 3), (mod 3), (mod 3), and so on That is, each positive power of 10 is congruent to 1 modulo 3 So equation 1, written as a mod-3 congruence, becomes N d d d d k d k (mod 3) d 0 + d 1 + d d k 1 + d k (mod 3) Thus the sum of the digits of N is divisible by 3 (that is, congruent to 0 modulo 3) if and only if N itself divisible by 3 Here s a similar test for divisibility by 11 Given a number N = d k d k 1 d 2 d 1 d 0, adjoin a leading zero if necessary so that the number of digits is even Use the digits of N in pairs to form a sequence of two-digit numbers Then N is divisible by 11 if and only if the sum of these two-digit numbers is divisible by 11 That is, N is divisible by 11 if and only if is divisible by 11 (d k d k 1 ) + (d k 2 d k 3 ) + + (d 3 d 2 ) + (d 1 d 0 ) (a) Explain why this divisibility test works Solution: Consider the equations 100 = = = 7 The pattern shows that 100 k 1 (mod 11) for each positive integer k Now if we write a number N as a string of its digits, then we have N = d k d k 1 d k 2 d 2 d 1 d 0, N = (d 1 d 0 ) (d 3 d 2 ) (d 5 d 4 ) k 1 (d k d k 1 ) (d 1 d 0 ) + 1 (d 3 d 2 ) + 1 (d 5 d 4 ) (d k d k 1 ) (mod 11) (d 1 d 0 ) + (d 3 d 2 ) + (d 5 d 4 ) + + (d k d k 1 ) (mod 11) This shows that N is divisible by 11 if and only if the sum is divisible by 11 (d k d k 1 ) + (d k 2 d k 3 ) + + (d 3 d 2 ) + (d 1 d 0 ) (b) Use this test to find the smallest number x for which x is divisible by 11 Solution: We separate the digits of the given number into pairs as We find that = = 74 The least positive number we can add to this sum to make it into a multiple of 11 is x = 3 The number is divisible by 11 ### Homework 5 Solutions Homework 5 Solutions 4.2: 2: a. 321 = 256 + 64 + 1 = (01000001) 2 b. 1023 = 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = (1111111111) 2. Note that this is 1 less than the next power of 2, 1024, which ### Math 319 Problem Set #3 Solution 21 February 2002 Math 319 Problem Set #3 Solution 21 February 2002 1. ( 2.1, problem 15) Find integers a 1, a 2, a 3, a 4, a 5 such that every integer x satisfies at least one of the congruences x a 1 (mod 2), x a 2 (mod ### Elementary Number Theory We begin with a bit of elementary number theory, which is concerned CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1, ### Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices. A Biswas, IT, BESU SHIBPUR Mathematics of Cryptography Modular Arithmetic, Congruence, and Matrices A Biswas, IT, BESU SHIBPUR McGraw-Hill The McGraw-Hill Companies, Inc., 2000 Set of Integers The set of integers, denoted by Z, ### Computing exponents modulo a number: Repeated squaring Computing exponents modulo a number: Repeated squaring How do you compute (1415) 13 mod 2537 = 2182 using just a calculator? Or how do you check that 2 340 mod 341 = 1? You can do this using the method ### Homework until Test #2 MATH31: Number Theory Homework until Test # Philipp BRAUN Section 3.1 page 43, 1. It has been conjectured that there are infinitely many primes of the form n. Exhibit five such primes. Solution. Five such ### Chapter 3. if 2 a i then location: = i. Page 40 Chapter 3 1. Describe an algorithm that takes a list of n integers a 1,a 2,,a n and finds the number of integers each greater than five in the list. 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If a and b are integers and there is some integer c such that a = b c, then we say that b divides a or is a factor or divisor of a and write b a. Definition ### CHAPTER 5. Number Theory. 1. Integers and Division. Discussion CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a ### Activity 1: Using base ten blocks to model operations on decimals Rational Numbers 9: Decimal Form of Rational Numbers Objectives To use base ten blocks to model operations on decimal numbers To review the algorithms for addition, subtraction, multiplication and division ### 3. Applications of Number Theory 3. APPLICATIONS OF NUMBER THEORY 163 3. Applications of Number Theory 3.1. Representation of Integers. Theorem 3.1.1. Given an integer b > 1, every positive integer n can be expresses uniquely as n = a ### 12 Greatest Common Divisors. The Euclidean Algorithm Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 12 Greatest Common Divisors. The Euclidean Algorithm As mentioned at the end of the previous section, we would like to ### MODULAR ARITHMETIC. a smallest member. It is equivalent to the Principle of Mathematical Induction. MODULAR ARITHMETIC 1 Working With Integers The usual arithmetic operations of addition, subtraction and multiplication can be performed on integers, and the result is always another integer Division, on ### Factoring Algorithms Factoring Algorithms The p 1 Method and Quadratic Sieve November 17, 2008 () Factoring Algorithms November 17, 2008 1 / 12 Fermat s factoring method Fermat made the observation that if n has two factors ### Chapter 11 Number Theory Chapter 11 Number Theory Number theory is one of the oldest branches of mathematics. For many years people who studied number theory delighted in its pure nature because there were few practical applications ### RSA Question 2. Bob thinks that p and q are primes but p isn t. Then, Bob thinks Φ Bob :=(p-1)(q-1) = φ(n). Is this true? RSA Question 2 Bob thinks that p and q are primes but p isn t. Then, Bob thinks Φ Bob :=(p-1)(q-1) = φ(n). Is this true? Bob chooses a random e (1 < e < Φ Bob ) such that gcd(e,φ Bob )=1. Then, d = e -1 ### 1.4 Factors and Prime Factorization 1.4 Factors and Prime Factorization Recall from Section 1.2 that the word factor refers to a number which divides into another number. For example, 3 and 6 are factors of 18 since 3 6 = 18. Note also that ### Let s just do some examples to get the feel of congruence arithmetic. Basic Congruence Arithmetic Let s just do some examples to get the feel of congruence arithmetic. Arithmetic Mod 7 Just write the multiplication table. 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 ### Grade 7/8 Math Circles Fall 2012 Factors and Primes 1 University of Waterloo Faculty of Mathematics Centre for Education in Mathematics and Computing Grade 7/8 Math Circles Fall 2012 Factors and Primes Factors Definition: A factor of a number is a whole ### (x + a) n = x n + a Z n [x]. Proof. If n is prime then the map 22. A quick primality test Prime numbers are one of the most basic objects in mathematics and one of the most basic questions is to decide which numbers are prime (a clearly related problem is to find ### CLASS 3, GIVEN ON 9/27/2010, FOR MATH 25, FALL 2010 CLASS 3, GIVEN ON 9/27/2010, FOR MATH 25, FALL 2010 1. Greatest common divisor Suppose a, b are two integers. If another integer d satisfies d a, d b, we call d a common divisor of a, b. Notice that as ### 3.1. RATIONAL EXPRESSIONS 3.1. RATIONAL EXPRESSIONS RATIONAL NUMBERS In previous courses you have learned how to operate (do addition, subtraction, multiplication, and division) on rational numbers (fractions). Rational numbers ### 9 Modular Exponentiation and Cryptography 9 Modular Exponentiation and Cryptography 9.1 Modular Exponentiation Modular arithmetic is used in cryptography. In particular, modular exponentiation is the cornerstone of what is called the RSA system. ### TYPES OF NUMBERS. Example 2. Example 1. Problems. Answers TYPES OF NUMBERS When two or more integers are multiplied together, each number is a factor of the product. Nonnegative integers that have exactly two factors, namely, one and itself, are called prime ### Revised Version of Chapter 23. We learned long ago how to solve linear congruences. ax c (mod m) Chapter 23 Squares Modulo p Revised Version of Chapter 23 We learned long ago how to solve linear congruences ax c (mod m) (see Chapter 8). It s now time to take the plunge and move on to quadratic equations. ### U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory ### Introduction to Diophantine Equations Introduction to Diophantine Equations Tom Davis [email protected] http://www.geometer.org/mathcircles September, 2006 Abstract In this article we will only touch on a few tiny parts of the field ### I. GROUPS: BASIC DEFINITIONS AND EXAMPLES I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called ### Click on the links below to jump directly to the relevant section Click on the links below to jump directly to the relevant section Basic review Writing fractions in simplest form Comparing fractions Converting between Improper fractions and whole/mixed numbers Operations ### 5.1 Commutative rings; Integral Domains 5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following ### Category 3 Number Theory Meet #1, October, 2000 Category 3 Meet #1, October, 2000 1. For how many positive integral values of n will 168 n be a whole number? 2. What is the greatest integer that will always divide the product of four consecutive integers? ### Definition 1 Let a and b be positive integers. A linear combination of a and b is any number n = ax + by, (1) where x and y are whole numbers. Greatest Common Divisors and Linear Combinations Let a and b be positive integers The greatest common divisor of a and b ( gcd(a, b) ) has a close and very useful connection to things called linear combinations ### Primes. Name Period Number Theory Primes Name Period A Prime Number is a whole number whose only factors are 1 and itself. To find all of the prime numbers between 1 and 100, complete the following exercise: 1. Cross out 1 by Shading in ### Elementary Number Theory and Methods of Proof. CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook. Elementary Number Theory and Methods of Proof CSE 215, Foundations of Computer Science Stony Brook University http://www.cs.stonybrook.edu/~cse215 1 Number theory Properties: 2 Properties of integers (whole ### Multiplication. Year 1 multiply with concrete objects, arrays and pictorial representations Year 1 multiply with concrete objects, arrays and pictorial representations Children will experience equal groups of objects and will count in 2s and 10s and begin to count in 5s. They will work on practical ### Lecture 13 - Basic Number Theory. Lecture 13 - Basic Number Theory. Boaz Barak March 22, 2010 Divisibility and primes Unless mentioned otherwise throughout this lecture all numbers are non-negative integers. We say that A divides B, denoted ### 3. Mathematical Induction 3. MATHEMATICAL INDUCTION 83 3. Mathematical Induction 3.1. First Principle of Mathematical Induction. Let P (n) be a predicate with domain of discourse (over) the natural numbers N = {0, 1,,...}. If (1) ### Elementary factoring algorithms Math 5330 Spring 013 Elementary factoring algorithms The RSA cryptosystem is founded on the idea that, in general, factoring is hard. Where as with Fermat s Little Theorem and some related ideas, one can ### Clock Arithmetic and Modular Systems Clock Arithmetic The introduction to Chapter 4 described a mathematical system CHAPTER Number Theory FIGURE FIGURE FIGURE Plus hours Plus hours Plus hours + = + = + = FIGURE. Clock Arithmetic and Modular Systems Clock Arithmetic The introduction to Chapter described a mathematical ### Properties of Real Numbers 16 Chapter P Prerequisites P.2 Properties of Real Numbers What you should learn: Identify and use the basic properties of real numbers Develop and use additional properties of real numbers Why you should ### MATH 13150: Freshman Seminar Unit 10 MATH 13150: Freshman Seminar Unit 10 1. 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Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4 ### Math Workshop October 2010 Fractions and Repeating Decimals Math Workshop October 2010 Fractions and Repeating Decimals This evening we will investigate the patterns that arise when converting fractions to decimals. 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Lesson Notes In the preceding lessons, students learned to add, subtract, multiply, ### 1. LINEAR EQUATIONS. A linear equation in n unknowns x 1, x 2,, x n is an equation of the form 1. LINEAR EQUATIONS A linear equation in n unknowns x 1, x 2,, x n is an equation of the form a 1 x 1 + a 2 x 2 + + a n x n = b, where a 1, a 2,..., a n, b are given real numbers. For example, with x and ### Kevin James. MTHSC 412 Section 2.4 Prime Factors and Greatest Comm MTHSC 412 Section 2.4 Prime Factors and Greatest Common Divisor Greatest Common Divisor Definition Suppose that a, b Z. Then we say that d Z is a greatest common divisor (gcd) of a and b if the following ### Basics of Counting. The product rule. Product rule example. 22C:19, Chapter 6 Hantao Zhang. Sample question. Total is 18 * 325 = 5850 Basics of Counting 22C:19, Chapter 6 Hantao Zhang 1 The product rule Also called the multiplication rule If there are n 1 ways to do task 1, and n 2 ways to do task 2 Then there are n 1 n 2 ways to do ### Objective. Materials. TI-73 Calculator 0. Objective To explore subtraction of integers using a number line. Activity 2 To develop strategies for subtracting integers. Materials TI-73 Calculator Integer Subtraction What s the Difference? Teacher ### CS 103X: Discrete Structures Homework Assignment 3 Solutions CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On well-ordering and induction: (a) Prove the induction principle from the well-ordering principle. (b) Prove the well-ordering ### s = 1 + 2 +... + 49 + 50 s = 50 + 49 +... + 2 + 1 2s = 51 + 51 +... + 51 + 51 50 51. 2 1. Use Euler s trick to find the sum 1 + 2 + 3 + 4 + + 49 + 50. s = 1 + 2 +... + 49 + 50 s = 50 + 49 +... + 2 + 1 2s = 51 + 51 +... + 51 + 51 Thus, 2s = 50 51. Therefore, s = 50 51. 2 2. Consider the sequence ### The last three chapters introduced three major proof techniques: direct, CHAPTER 7 Proving Non-Conditional Statements The last three chapters introduced three major proof techniques: direct, contrapositive and contradiction. These three techniques are used to prove statements ### The application of prime numbers to RSA encryption The application of prime numbers to RSA encryption Prime number definition: Let us begin with the definition of a prime number p The number p, which is a member of the set of natural numbers N, is considered ### If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Problem 3 If A is divided by B the result is 2/3. If B is divided by C the result is 4/7. What is the result if A is divided by C? Suggested Questions to ask students about Problem 3 The key to this question ### Lecture 3: Finding integer solutions to systems of linear equations Lecture 3: Finding integer solutions to systems of linear equations Algorithmic Number Theory (Fall 2014) Rutgers University Swastik Kopparty Scribe: Abhishek Bhrushundi 1 Overview The goal of this lecture Chapter 4 Complementary Sets Of Systems Of Congruences Proceedings NCUR VII. è1993è, Vol. II, pp. 793í796. Jeærey F. Gold Department of Mathematics, Department of Physics University of Utah Don H. Tucker ### 3 0 + 4 + 3 1 + 1 + 3 9 + 6 + 3 0 + 1 + 3 0 + 1 + 3 2 mod 10 = 4 + 3 + 1 + 27 + 6 + 1 + 1 + 6 mod 10 = 49 mod 10 = 9. SOLUTIONS TO HOMEWORK 2 - MATH 170, SUMMER SESSION I (2012) (1) (Exercise 11, Page 107) Which of the following is the correct UPC for Progresso minestrone soup? Show why the other numbers are not valid ### Session 7 Fractions and Decimals Key Terms in This Session Session 7 Fractions and Decimals Previously Introduced prime number rational numbers New in This Session period repeating decimal terminating decimal Introduction In this session, ### Stanford Math Circle: Sunday, May 9, 2010 Square-Triangular Numbers, Pell s Equation, and Continued Fractions Stanford Math Circle: Sunday, May 9, 00 Square-Triangular Numbers, Pell s Equation, and Continued Fractions Recall that triangular numbers are numbers of the form T m = numbers that can be arranged in ### 2.5 Zeros of a Polynomial Functions .5 Zeros of a Polynomial Functions Section.5 Notes Page 1 The first rule we will talk about is Descartes Rule of Signs, which can be used to determine the possible times a graph crosses the x-axis and ### SUM OF TWO SQUARES JAHNAVI BHASKAR SUM OF TWO SQUARES JAHNAVI BHASKAR Abstract. 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Finan Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 3 Binary Operations We are used to addition and multiplication of real numbers. These operations combine two real numbers ### Continued Fractions. Darren C. Collins Continued Fractions Darren C Collins Abstract In this paper, we discuss continued fractions First, we discuss the definition and notation Second, we discuss the development of the subject throughout history ### Notes for Recitation 5 6.042/18.062J Mathematics for Computer Science September 24, 2010 Tom Leighton and Marten van Dijk Notes for Recitation 5 1 Exponentiation and Modular Arithmetic Recall that RSA encryption and decryption ### Sample Problems. Practice Problems Lecture Notes Quadratic Word Problems page 1 Sample Problems 1. The sum of two numbers is 31, their di erence is 41. Find these numbers.. The product of two numbers is 640. Their di erence is 1. Find these ### MATH 537 (Number Theory) FALL 2016 TENTATIVE SYLLABUS MATH 537 (Number Theory) FALL 2016 TENTATIVE SYLLABUS Class Meetings: MW 2:00-3:15 pm in Physics 144, September 7 to December 14 [Thanksgiving break November 23 27; final exam December 21] Instructor: ### Linear Dependence Tests Linear Dependence Tests The book omits a few key tests for checking the linear dependence of vectors. These short notes discuss these tests, as well as the reasoning behind them. Our first test checks ### Integers and division CS 441 Discrete Mathematics for CS Lecture 12 Integers and division Milos Hauskrecht [email protected] 5329 Sennott Square Symmetric matrix Definition: A square matrix A is called symmetric if A = A T. ### 3 Some Integer Functions 3 Some Integer Functions A Pair of Fundamental Integer Functions The integer function that is the heart of this section is the modulo function. However, before getting to it, let us look at some very simple ### 1 Number System. Introduction. Face Value and Place Value of a Digit. Categories of Numbers RMAT Success Master 3 1 Number System Introduction In the decimal number system, numbers are expressed by means of symbols 0, 1,, 3, 4, 5, 6, 7, 8, 9, called digits. Here, 0 is called an insignificant ### Lesson 4. Factors and Multiples. Objectives Student Name: Date: Contact Person Name: Phone Number: Lesson 4 Factors and Multiples Objectives Understand what factors and multiples are Write a number as a product of its prime factors Find the greatest ### 8 Divisibility and prime numbers 8 Divisibility and prime numbers 8.1 Divisibility In this short section we extend the concept of a multiple from the natural numbers to the integers. We also summarize several other terms that express ### APPLICATIONS OF THE ORDER FUNCTION APPLICATIONS OF THE ORDER FUNCTION LECTURE NOTES: MATH 432, CSUSM, SPRING 2009. PROF. WAYNE AITKEN In this lecture we will explore several applications of order functions including formulas for GCDs and ### Lecture 13: Factoring Integers CS 880: Quantum Information Processing 0/4/0 Lecture 3: Factoring Integers Instructor: Dieter van Melkebeek Scribe: Mark Wellons In this lecture, we review order finding and use this to develop a method ### Introduction to Fractions Introduction to Fractions Fractions represent parts of a whole. The top part of a fraction is called the numerator, while the bottom part of a fraction is called the denominator. The denominator states ### Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 18.781 Problem Set 7 - Fall 2008 Due Tuesday, Oct. 28 at 1:00 Throughout this assignment, f(x) always denotes a polynomial with integer coefficients. 1. (a) Show that e 32 (3) = 8, and write down a list ### Breaking The Code. Ryan Lowe. Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and Breaking The Code Ryan Lowe Ryan Lowe is currently a Ball State senior with a double major in Computer Science and Mathematics and a minor in Applied Physics. As a sophomore, he took an independent study ### We can express this in decimal notation (in contrast to the underline notation we have been using) as follows: 9081 + 900b + 90c = 9001 + 100c + 10b In this session, we ll learn how to solve problems related to place value. This is one of the fundamental concepts in arithmetic, something every elementary and middle school mathematics teacher should ### 4.4 Clock Arithmetic and Modular Systems 4.4 Clock Arithmetic and Modular Systems A mathematical system has 3 major properies. 1. It is a set of elements 2. It has one or more operations to combine these elements (ie. Multiplication, addition) ### Factorization Methods: Very Quick Overview Factorization Methods: Very Quick Overview Yuval Filmus October 17, 2012 1 Introduction In this lecture we introduce modern factorization methods. We will assume several facts from analytic number theory. ### The Paillier Cryptosystem The Paillier Cryptosystem A Look Into The Cryptosystem And Its Potential Application By Michael O Keeffe The College of New Jersey Mathematics Department April 18, 2008 ABSTRACT So long as there are secrets, Lesson. Addition with Unlike Denominators Karen is stringing a necklace with beads. She puts green beads on _ of the string and purple beads on of the string. How much of the string does Karen cover with ### Systems of Linear Equations Systems of Linear Equations Beifang Chen Systems of linear equations Linear systems A linear equation in variables x, x,, x n is an equation of the form a x + a x + + a n x n = b, where a, a,, a n and ### Groups in Cryptography Groups in Cryptography Çetin Kaya Koç http://cs.ucsb.edu/~koc/cs178 [email protected] Koç (http://cs.ucsb.edu/~koc) ucsb cs 178 intro to crypto winter 2013 1 / 13 Groups in Cryptography A set S and a binary ### Mental Math Addition and Subtraction Mental Math Addition and Subtraction If any of your students don t know their addition and subtraction facts, teach them to add and subtract using their fingers by the methods taught below. You should ### Section 4.2: The Division Algorithm and Greatest Common Divisors Section 4.2: The Division Algorithm and Greatest Common Divisors The Division Algorithm The Division Algorithm is merely long division restated as an equation. For example, the division 29 r. 20 32 948	8,843	33,565	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.40625	4	CC-MAIN-2018-43	latest	en	0.930699
https://us.metamath.org/ileuni/climim.html	1,713,302,259,000,000,000	text/html	crawl-data/CC-MAIN-2024-18/segments/1712296817106.73/warc/CC-MAIN-20240416191221-20240416221221-00511.warc.gz	555,150,789	6,325	Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > climim GIF version Theorem climim 11085 Description: Limit of the imaginary part of a sequence. Proposition 12-2.4(c) of [Gleason] p. 172. (Contributed by NM, 7-Jun-2006.) (Revised by Mario Carneiro, 9-Feb-2014.) Hypotheses Ref Expression climcn1lem.1 𝑍 = (ℤ𝑀) climcn1lem.2 (𝜑𝐹𝐴) climcn1lem.4 (𝜑𝐺𝑊) climcn1lem.5 (𝜑𝑀 ∈ ℤ) climcn1lem.6 ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ) climim.7 ((𝜑𝑘𝑍) → (𝐺𝑘) = (ℑ‘(𝐹𝑘))) Assertion Ref Expression climim (𝜑𝐺 ⇝ (ℑ‘𝐴)) Distinct variable groups: 𝐴,𝑘 𝑘,𝐹 𝑘,𝐺 𝜑,𝑘 𝑘,𝑍 𝑘,𝑀 Allowed substitution hint: 𝑊(𝑘) Proof of Theorem climim Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables. StepHypRef Expression 1 climcn1lem.1 . 2 𝑍 = (ℤ𝑀) 2 climcn1lem.2 . 2 (𝜑𝐹𝐴) 3 climcn1lem.4 . 2 (𝜑𝐺𝑊) 4 climcn1lem.5 . 2 (𝜑𝑀 ∈ ℤ) 5 climcn1lem.6 . 2 ((𝜑𝑘𝑍) → (𝐹𝑘) ∈ ℂ) 6 imf 10621 . . 3 ℑ:ℂ⟶ℝ 7 ax-resscn 7705 . . 3 ℝ ⊆ ℂ 8 fss 5279 . . 3 ((ℑ:ℂ⟶ℝ ∧ ℝ ⊆ ℂ) → ℑ:ℂ⟶ℂ) 96, 7, 8mp2an 422 . 2 ℑ:ℂ⟶ℂ 10 imcn2 11080 . 2 ((𝐴 ∈ ℂ ∧ 𝑥 ∈ ℝ+) → ∃𝑦 ∈ ℝ+𝑧 ∈ ℂ ((abs‘(𝑧𝐴)) < 𝑦 → (abs‘((ℑ‘𝑧) − (ℑ‘𝐴))) < 𝑥)) 11 climim.7 . 2 ((𝜑𝑘𝑍) → (𝐺𝑘) = (ℑ‘(𝐹𝑘))) 121, 2, 3, 4, 5, 9, 10, 11climcn1lem 11081 1 (𝜑𝐺 ⇝ (ℑ‘𝐴)) Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 103 = wceq 1331 ∈ wcel 1480 ⊆ wss 3066 class class class wbr 3924 ⟶wf 5114 ‘cfv 5118 ℂcc 7611 ℝcr 7612 ℤcz 9047 ℤ≥cuz 9319 ℑcim 10606 ⇝ cli 11040 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-in1 603 ax-in2 604 ax-io 698 ax-5 1423 ax-7 1424 ax-gen 1425 ax-ie1 1469 ax-ie2 1470 ax-8 1482 ax-10 1483 ax-11 1484 ax-i12 1485 ax-bndl 1486 ax-4 1487 ax-13 1491 ax-14 1492 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-i5r 1515 ax-ext 2119 ax-coll 4038 ax-sep 4041 ax-nul 4049 ax-pow 4093 ax-pr 4126 ax-un 4350 ax-setind 4447 ax-iinf 4497 ax-cnex 7704 ax-resscn 7705 ax-1cn 7706 ax-1re 7707 ax-icn 7708 ax-addcl 7709 ax-addrcl 7710 ax-mulcl 7711 ax-mulrcl 7712 ax-addcom 7713 ax-mulcom 7714 ax-addass 7715 ax-mulass 7716 ax-distr 7717 ax-i2m1 7718 ax-0lt1 7719 ax-1rid 7720 ax-0id 7721 ax-rnegex 7722 ax-precex 7723 ax-cnre 7724 ax-pre-ltirr 7725 ax-pre-ltwlin 7726 ax-pre-lttrn 7727 ax-pre-apti 7728 ax-pre-ltadd 7729 ax-pre-mulgt0 7730 ax-pre-mulext 7731 ax-arch 7732 ax-caucvg 7733 This theorem depends on definitions: df-bi 116 df-dc 820 df-3or 963 df-3an 964 df-tru 1334 df-fal 1337 df-nf 1437 df-sb 1736 df-eu 2000 df-mo 2001 df-clab 2124 df-cleq 2130 df-clel 2133 df-nfc 2268 df-ne 2307 df-nel 2402 df-ral 2419 df-rex 2420 df-reu 2421 df-rmo 2422 df-rab 2423 df-v 2683 df-sbc 2905 df-csb 2999 df-dif 3068 df-un 3070 df-in 3072 df-ss 3079 df-nul 3359 df-if 3470 df-pw 3507 df-sn 3528 df-pr 3529 df-op 3531 df-uni 3732 df-int 3767 df-iun 3810 df-br 3925 df-opab 3985 df-mpt 3986 df-tr 4022 df-id 4210 df-po 4213 df-iso 4214 df-iord 4283 df-on 4285 df-ilim 4286 df-suc 4288 df-iom 4500 df-xp 4540 df-rel 4541 df-cnv 4542 df-co 4543 df-dm 4544 df-rn 4545 df-res 4546 df-ima 4547 df-iota 5083 df-fun 5120 df-fn 5121 df-f 5122 df-f1 5123 df-fo 5124 df-f1o 5125 df-fv 5126 df-riota 5723 df-ov 5770 df-oprab 5771 df-mpo 5772 df-1st 6031 df-2nd 6032 df-recs 6195 df-frec 6281 df-pnf 7795 df-mnf 7796 df-xr 7797 df-ltxr 7798 df-le 7799 df-sub 7928 df-neg 7929 df-reap 8330 df-ap 8337 df-div 8426 df-inn 8714 df-2 8772 df-3 8773 df-4 8774 df-n0 8971 df-z 9048 df-uz 9320 df-rp 9435 df-seqfrec 10212 df-exp 10286 df-cj 10607 df-re 10608 df-im 10609 df-rsqrt 10763 df-abs 10764 df-clim 11041 This theorem is referenced by: (None) Copyright terms: Public domain W3C validator	2,260	3,789	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2024-18	latest	en	0.127566
https://stackoverflow.com/questions/6085583/how-can-i-find-the-velocity-using-accelerometers-only	1,628,224,988,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00052.warc.gz	538,189,108	45,273	# How can I find the velocity using accelerometers only? Using only the phone's (Android) built in accelerometer, how would I go about finding its velocity? I have been tinkering with the maths of this but whatever function I come up with tends to lead to exponential growth of the velocity. I am working from the assumption that on startup of the App, the phone is at a standstill. This should definitely make finding the velocity (at least roughly) possible. I have a decent background in physics and math too, so I shouldn't have any difficulty with any concepts here. How should I do it? • @Ignacio What? Constant acceleration means velocity increases linearly. – Imran May 22 '11 at 2:40 • Have you got any working solution, can you point me to that or share the code. – VenomVendor Mar 17 '13 at 15:44 • Hi, did you ever consider closing this question? – Lennart Rolland May 9 '16 at 19:28 • Your phone uses sensors half the size of a pea that cost something like \$2 and were made by etching tiny pieces of metal into the shapes of crude springs and weights. It's sufficient to know which way you're holding the phone, and whether it's being shaken. It will never make a decent inertial sensor package. Trust me, errors accumulate so fast that this is a hopeless problem to solve. (Source: I was in charge of the embedded firmware to do this on the Amazon Fire Phone when I worked at Amazon.) – Edward Falk Jun 14 '16 at 15:19 • I just found this superb video explaining why this is hard: youtube.com/watch?v=C7JQ7Rpwn2k#t=23m20s – Edward Falk Jun 14 '16 at 15:37 That will really depend on what the acceleration is and for how long. A mild, long acceleration could be measurable, but any sudden increase in acceleration, followed by a constant velocity, will make your measurements quite difficult and prone to error. Assuming constant acceleration, the formula is extremely simple: a = (V1-V0)/t . So, knowing the time and the acceleration, and assuming V0 = 0, then V1 = at In a more real world, you probably won't have a constant acceleration, so you should calculate Delta V for each measurement, and adding all those changes in velocity to get the final velocity. Always consider that you won't have a continuous acceleration data, so this is the most feasible way (i.e, real data vs integral math theory). In any way, even in the best scenario, you will end up with a very high error margin, so I do not recommend this approach for any app that truly depends on real velocities. • In case no one noticed, V1 = at! – erdomester Sep 13 '11 at 20:58 First, you have to remove the acceleration due to gravity from the accelerometer data. Then it's just a matter of integrating the acceleration to get the velocity. Don't forget that acceleration and velocity are properly vectors, not scalars, and that you will also have to track rotation of the phone in space to properly determine the orientation of the acceleration vector with respect to the calculated velocity vector. • can you explain from coding perspective or point to a link where i can get more details about this. – VenomVendor Mar 17 '13 at 15:43 • The TYPE_LINEAR_ACCELERATION sensor will give you acceleration with gravity filtered out, so that's nice. The TYPE_ROTATION_VECTOR will give you device orientation as a quaternion, which can easily be converted to a 3x3 rotation matrix. At every time slice, use the rotation matrix to convert the acceleration to a vector in 3-space. Integrate this over time to get acceleration (another vector) and integrate that to get position relative to the start position. (This all assumes initial velocity was zero.) Then go to the pub and have a pint because it's all hopeless with cheap sensors. – Edward Falk Jun 14 '16 at 15:13 There is nothing else to do but agree with the reasonable arguments put forward in all the great answers above, however if you are the pragmatic type like me, I need to come up with a solution that works somehow. I suffered a similar problem to yours and I decided to make my own solution after not finding any on-line. I only needed a simple "tilt" input for controlling a game so this solution will probably NOT work for more complex needs, however I decided to share it in case others where looking for something similar. NOTE: I have pasted my entire code here, and it is free to use for any purpose. Basically what I do in my code is to look for accelerometer sensor. If not found, tilt feedback will be disabled. If accelerometer sensor is present, I look for magnetic field sensor, and if it is present, I get my tilt angle the recommended way by combining accelerometer and magnetic field data. public TiltSensor(Context c) { man = (SensorManager) c.getSystemService(Context.SENSOR_SERVICE); mag_sensor = man.getDefaultSensor(Sensor.TYPE_MAGNETIC_FIELD); acc_sensor = man.getDefaultSensor(Sensor.TYPE_ACCELEROMETER); has_mag = man.registerListener(this, mag_sensor, delay); has_acc = man.registerListener(this, acc_sensor, delay); if (has_acc) { tiltAvailble = true; if (has_mag) { Log.d("TiltCalc", "Using accelerometer + compass."); } else { Log.d("TiltCalc", "Using only accelerometer."); } } else { tiltAvailble = false; Log.d("TiltCalc", "No acceptable hardware found, tilt not available."); //No use in having listeners registered pause(); } } If however only the accelerometer sensor was present, I fall back to accumulating the acceleration, that is continuously damped (multiplied by 0.99) to remove any drift. For my simple tilt needs this works great. @Override public void onSensorChanged(SensorEvent e) { final float[] vals = e.values; final int type = e.sensor.getType(); switch (type) { case (Sensor.TYPE_ACCELEROMETER): { needsRecalc = true; if (!has_mag) { System.arraycopy(accelerometer, 0, old_acc, 0, 3); } System.arraycopy(vals, 0, accelerometer, 0, 3); if (!has_mag) { for (int i = 0; i < 3; i++) { //Accumulate changes final float sensitivity = 0.08f; dampened_acc[i] += (accelerometer[i] - old_acc[i]) * sensitivity; //Even out drift over time dampened_acc[i] = 0.99; } } } break; case (Sensor.TYPE_MAGNETIC_FIELD): { needsRecalc = true; System.arraycopy(vals, 0, magnetic_field, 0, 3); } break; } } In conclusion I will just repeat that this is probably not "correct" in any way, it simply works as a simple input to a game. To use this code I simply do something like the following (yes magic constants are bad mkay): Ship ship = mShipLayer.getShip(); mTiltSensor.getTilt(vals); float deltaY = -vals[1] 2;//1 is the index of the axis we are after float offset = ((deltaY - (deltaY / 1.5f))); if (null != ship) { ship.setOffset(offset); } Enjoi! Integrating acceleration to get velocity is an unstable problem and your error will diverge after a couple of seconds or so. Phone accelerometers are also not very accurate, which doesn't help, and some of them don't allow you to distinguish between tilt and translation easily, in which case you're really in trouble. The accelerometers in a phone are pretty much useless for such a task. You need highly accurate accelerometers with very low drift - something which is way beyond what you will find in a phone. At best you might get useful results for a few seconds, or if very lucky for a minute or two after which the results become meaningless. Also, you need to have a three axis gyroscope which you would use to integrate the velocity in the right direction. Some phones have gyros, but they are even poorer than the accelerometers as far as drift and accuracy are concerned. One possibly useful application though would be to use the accelerometers in conjunction with gyros or the magnetic compass to fill in for missing data from the GPS. Each time the GPS gives a good fix one would reset the initial conditions of position, speed and orientation and the accelerometers would provide the data until the next valid GPS fix. • Let's say that I want to get tilt information for controlling a game. In that scenario I will be able to make many assumptions. For example, I can assume the user is holding the device in a certain way. Would it not be possible then to smooth the data in a way that sudden changes would stand out and become useful as input to the game? – Lennart Rolland Apr 6 '13 at 23:27 Gravity is going to destroy all of your measurements. The phone, at standstill, is experiencing a high constant upward (yes, UP) acceleration. An accelerometer can't distinguish between acceleration and gravity (technically, they are the same), so it would get to extremely high velocities after a few seconds. If you never tilt your accelerometer even slightly, then you can simply subtract the constant gravitional pull from the z-axis (or whichever axis is pointing up/down), but thats quite unlikely. Basically, you have to use a complicated system of a gyroscope/magnetometor and an accelerometer to calculate the exact direction of gravity and then subtract the acceleration. • Sensor Fusion can filter that out for you. Just use TYPE_LINEAR_ACCELERATION sensor. – Edward Falk Jun 14 '16 at 15:22 v = Integral(a) ? Generally though, I'd think the inaccuracies in the accelerometers would make this quite tough • Depends on the type of inaccuracies. You could calibrate for linear error, no? – William T. Mallard Jun 11 '13 at 5:11 If the phone is at standstil, you have ZERO acceleration, so your speed is 0. Probably you should find location data from GPS and get the associated time samples and compute velocity distance over time. • Yes, the speed is 0 "on the startup of the App". I am calibrating it from this point. – Mark D May 22 '11 at 2:07 • "Standing still" should actually be easy to detect. If the input from the accelerometer (use TYPE_LINEAR_ACCELERATION) is very constant, the phone is probably not being handled. – Edward Falk Jun 14 '16 at 15:24	2,293	9,845	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.625	4	CC-MAIN-2021-31	latest	en	0.945978
https://im.kendallhunt.com/k5_es/teachers/grade-1/unit-2/lesson-11/lesson.html	1,723,528,882,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00525.warc.gz	234,711,053	29,272	# Lesson 11 Hagamos que sean iguales ## Warm-up: Observa y pregúntate: Torres de cubos (10 minutes) ### Narrative The purpose of this warm-up is to elicit the idea that quantities can be compared, which will be useful when students compare connecting cube towers in order to make them have the same number of cubes, in a later activity. While students may notice and wonder many things about this image, comparing the quantity of cubes in each tower is the important discussion point. When English speaking students use the language “Jada has less cubes than Diego,” the teacher should revoice using the grammatically correct language “fewer” to support students with developing precise language (MP6). For Spanish speaking students, the single word “menos” is used for both “fewer” and “less” so the previous clarification does not apply. ### Launch • Groups of 2 • Display the image. • “¿Qué observan? ¿Qué se preguntan?” // “What do you notice? What do you wonder?” • 1 minute: quiet think time ### Activity • “Discutan con su pareja cómo pensaron” // “Discuss your thinking with your partner.” • 1 minute: partner discussion • Share and record responses. • Revoice and emphasize “más que” // “more than” and “menos que” // “fewer than.” ### Student Facing ¿Qué observas? ¿Qué te preguntas? ### Activity Synthesis • “Vamos a usar torres de cubos encajables y a pensar en cómo lograr que tengan el mismo número de cubos” // “We will use connecting cube towers and think about how to make them have the same number of cubes.” ## Activity 1: Torres de cubos (15 minutes) ### Narrative The purpose of this activity is for students to find ways to make two quantities the same. Students may use connecting cubes to represent and solve the problems, if they choose. Students are more likely to add cubes to make the towers have an equal number of cubes, but some students may take cubes away. Students record how they solved each problem in a way that makes sense to them. During the synthesis, the teacher records equations that match the way students solved the problem to build on work in previous sections. ### Required Preparation • Create a tower of 9 blue connecting cubes and a tower of 5 red connecting cubes. • Each group of 2 needs 4 towers of 10 connecting cubes. ### Launch • Groups of 2 • Give each group four towers of ten connecting cubes. • Display a tower of nine blue connecting cubes and a tower of five red connecting cubes, or the image in the student book. • “Diego y Jada construyen torres de cubos. ¿Cómo pueden Diego y Jada hacer que sus torres tengan el mismo número de cubos? Recuerden: no compartan la respuesta con su pareja; compartan su método para resolver el problema” // “Diego and Jada are building connecting cube towers. How can Diego and Jada make their towers have the same number of cubes? Remember, you are not sharing the answer with your partner; you are sharing your method for solving the problem.” (They can add some cubes to the red tower or take some off the blue tower.) • 1 minute: quiet think time • 1 minute: partner discussion • Share responses. ### Activity • “Van a anotar cómo pensaron este problema y van a resolver dos problemas más sobre cómo hacer que las torres de Diego y Jada tengan el mismo número de cubos. Si quieren, pueden usar los cubos encajables como ayuda” // “You are going to record your thinking for this problem and solve two more problems about making Diego and Jada’s connecting cube towers have the same number of cubes. You may use the connecting cubes to help if you choose.” • 5 minutes: independent work time • Monitor for students who have clear representations for problem 3 showing: • adding cubes to Diego’s tower • counting up the difference between Diego’s tower to Jada’s tower • taking cubes from Jada’s tower • counting back the difference between Jada’s tower to Diego’s tower ### Student Facing ¿Cómo pueden Diego y Jada hacer que sus torres tengan el mismo número de cubos? Muestra cómo pensaste. Usa dibujos, números o palabras. 1. 2. 3. ### Activity Synthesis • Invite previously identified students to share. • “¿Qué observan sobre todos estos métodos?” // “What do you notice about all these methods?” (They all got the same answer, even if they added cubes or took cubes off.) • “¿Qué ecuación escribirían que corresponda a este problema-historia? ¿Cómo muestra su ecuación la forma en la que resolvieron el problema-historia?” // “What equation would you write to match the story problem? How does your equation show how you solved the story problem?” ($$3 + 7 = 10$$, because I added 7 cubes onto Diego’s tower to make them the same. $$10 - 3 = 7$$, because I broke off from Jada’s tower and counted what I broke off.) ## Activity 2: Problemas con torres de cubos (20 minutes) ### Narrative The purpose of this activity is for students to solve Compare, Difference Unknown story problems in the context of connecting cube towers. Students solve comparison problems with given constraints that encourage students to add or break apart cubes to make towers with the same number of cubes. As students explain their thinking during the synthesis, record both addition and subtraction equations. When students answer the question, "How do you know?" they are beginning to explain their reasoning and construct viable arguments (MP3). MLR8 Discussion Supports. Invite students to take turns sharing their responses. Ask students to restate what they heard using precise mathematical language and their own words. Display the sentence frame: “Te escuché decir . . . .” // “I heard you say . . . .” Original speakers can agree or clarify for their partner. Engagement: Provide Access by Recruiting Interest. Provide choice and autonomy. In addition to connecting cubes, provide access to red, yellow, and blue crayons or colored pencils they can use to represent and solve the story problems. Supports accessibility for: Visual-Spatial Processing, Conceptual Processing ### Required Preparation • Gather 1 red tower of 8 connecting cubes, 1 yellow tower of 3 connecting cubes, and a handful of yellow cubes. • Each group of 2 needs 4 towers of 10 connecting cubes. ### Launch • Groups of 2 • Give each group four towers of ten connecting cubes. • Display one red tower of eight connecting cubes, one yellow tower of three connecting cubes, and the handful of yellow connecting cubes. • “Tengo dos torres y debo hacer que tengan el mismo número de cubos, pero solo tengo estos cubos amarillos. ¿Cómo puedo hacerlo?” // “I have two towers and I need to make them the same number of cubes. But I only have these yellow cubes. How can I make them the same?” • 1 minute: quiet think time • 1 minute: partner discussion • Share and record responses. ### Activity • “Lin trabaja para lograr que el número de cubos de cada torre sea el mismo. Cada problema les dirá con qué cubos cuenta Lin para trabajar. Anoten cómo pensaron con cada torre" // “Lin is working to make the number of cubes in each of her towers the same. Each problem will tell you what cubes she has to work with. Record your thinking for each tower.” • 8 minutes: independent work time • “Compartan con su pareja cómo pensaron” // “Share your thinking with your partner.” • 4 minutes: partner discussion • Monitor for a student who solved the problem with 7 yellow cubes and 3 red cubes by adding 4 red cubes or drawing 4 more red cubes. ### Student Facing 1. Lin solo tiene cubos azules. ¿Cómo puede Lin hacer que las torres tengan el mismo número de cubos? Muestra cómo pensaste. Usa dibujos, números o palabras. 2. Lin no tiene más cubos amarillos. ¿Cómo puede hacer que las torres tengan el mismo número de cubos? Muestra cómo pensaste. Usa dibujos, números o palabras. 3. Lin construye 2 torres de cubos. La torre roja tiene 6 cubos. La torre azul tiene 9 cubos. Ella no tiene más cubos rojos. ¿Cómo puede hacer que las torres tengan el mismo número de cubos? Muestra cómo pensaste. Usa dibujos, números o palabras. 4. Lin construye 2 torres de cubos. La torre amarilla tiene 7 cubos. La torre roja tiene 3 cubos. Ella solo tiene cubos rojos. ¿Cómo puede hacer que las torres tengan el mismo número de cubos? Muestra cómo pensaste. Usa dibujos, números o palabras. Si te queda tiempo: escribe tu propio problema sobre 2 torres de cubos. Intercambia tu problema con un compañero y resuélvelo. ### Activity Synthesis • Display selected student work. • “¿Cómo resolvieron ellos el problema?” // “How did they solve the problem?” (She added 4 red cubes.) • “¿Cuál ecuación corresponde a la forma como resolvieron el problema? ¿Cómo lo saben?” // “What equation matches how they solved? How do you know?” ($$3 + 4 = 7$$ because Lin started with 3 red cubes and she added 4 cubes to make it the same.) • “¿Cuál número de la ecuación representa la respuesta al problema?” // “Which number in the equation represents the answer to the problem?” (4 because that is how many cubes Lin added.) • Put a box around the 4. ## Lesson Synthesis ### Lesson Synthesis Display one red tower of eight connecting cubes, one yellow tower of three connecting cubes, and a handful of yellow connecting cubes. “Hoy logramos que las torres tuvieran el mismo número de cubos poniendo más cubos en la torre más pequeña o quitando cubos de la torre más grande. ¿Cuál es la relación entre poner más cubos y sumar?” // “Today we made towers have the same number of cubes by adding more cubes to the smaller tower or taking cubes from the larger tower. How does adding more cubes relate to addition?” (I can add cubes to the tower with fewer cubes to make them the same.)	2,334	9,648	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.734375	4	CC-MAIN-2024-33	latest	en	0.799924
https://chem.libretexts.org/Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/10%3A_Chemical_Equilibrium/10.04%3A_Calculating_Equilibrium_Constant_Values	1,721,483,530,000,000,000	text/html	crawl-data/CC-MAIN-2024-30/segments/1720763515164.46/warc/CC-MAIN-20240720113754-20240720143754-00323.warc.gz	146,056,281	33,460	# 10.4: Calculating Equilibrium Constant Values $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ ( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\\| #1 \\|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$ $$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$ $$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$ $$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vectorC}[1]{\textbf{#1}}$$ $$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$ $$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$ $$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$ $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$ $$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left\|#1\right\|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$ ## Learning Objective • Calculate equilibrium concentrations from the values of the initial amounts and the Keq. There are some circumstances in which, given some initial amounts and the Keq, you will have to determine the concentrations of all species when equilibrium is achieved. Such calculations are not difficult to do, especially if a consistent approach is applied. We will consider such an approach here. Suppose we have this simple equilibrium. Its associated Keq is 4.0, and the initial concentration of each reactant is 1.0 M: $\underset{1.0M}{H_{2}(g)}+\underset{1.0M}{Cl_{2}(g)}\rightleftharpoons 2HCl(g)\; \; \; \; \; K_{eq}=4.0\nonumber$ Because we have concentrations for the reactants but not the products, we presume that the reaction will proceed in the forward direction to make products. But by how much will it proceed? We do not know, so let us assign it a variable. Let us assume that x M H2 reacts as the reaction goes to equilibrium. This means that at equilibrium, we have (1.0 − x) M H2 left over. According to the balanced chemical equation, H2 and Cl2 react in a 1:1 ratio. How do we know that? The coefficients of these two species in the balanced chemical equation are 1 (unwritten, of course). This means that if x M H2 reacts, x M Cl2 reacts as well. If we start with 1.0 M Cl2 at the beginning and we react x M, we have (1.0 − x) M Cl2 left at equilibrium. How much HCl is made? We start with zero, but we also see that 2 mol of HCl are made for every mole of H2 (or Cl2) that reacts (from the coefficients in the balanced chemical equation), so if we lose x M H2, we gain 2x M HCl. So now we know the equilibrium concentrations of our species: $\underset{(1.0-x)M}{H_{2}(g)}+\underset{(1.0-x)M}{Cl_{2}(g)}\rightleftharpoons \underset{2xM}{2HCl(g)}\; \; \; \; \; K_{eq}=4.0\nonumber$ We can substitute these concentrations into the Keq expression for this reaction and combine it with the known value of Keq: $K_{eq}=\frac{[HCl]^{2}}{[H_{2}][Cl_{2}]}=\frac{(2x)^{2}}{(1-x)(1-x)}=4.0\nonumber$ This is an equation in one variable, so we should be able to solve for the unknown value. This expression may look formidable, but first we can simplify the denominator and write it as a perfect square as well: $\frac{(2x)^{2}}{(1-x)^{2}}=4.0\nonumber$ The fraction is a perfect square, as is the 4.0 on the right. So we can take the square root of both sides: $\frac{(2x)}{(1-x)}=2.0\nonumber$ Now we rearrange and solve (be sure you can follow each step): $2x=2.0-2.0x\\ 4x=2.0\\ x=0.50\nonumber$ Now we have to remind ourselves what x is—the amount of H2 and Cl2 that reacted—and that 2x is the equilibrium [HCl]. To determine the equilibrium concentrations, we need to go back and evaluate the expressions 1 − x and 2x to get the equilibrium concentrations of our species: 1.0 − x = 1.0 − 0.50 = 0.50 M = [H2] = [Cl2]2x = 2(0.50) = 1.0 M = [HCl] The units are assumed to be molarity. To check, we simply substitute these concentrations and verify that we get the numerical value of the Keq, in this case 4.0: $\frac{(1.0)^{2}}{(0.50)(0.50)}=4.0\nonumber$ We formalize this process by introducing the ICE chart, where ICE stands for initial, change, and equilibrium. The initial values go in the first row of the chart. The change values, usually algebraic expressions because we do not yet know their exact numerical values, go in the next row. However, the change values must be in the proper stoichiometric ratio as indicated by the balanced chemical equation. Finally, the equilibrium expressions in the last row are a combination of the initial value and the change value for each species. The expressions in the equilibrium row are substituted into the Keq expression, which yields an algebraic equation that we try to solve. The ICE chart for the above example would look like this: H2(g) + Cl2(g) $⇄$ 2HCl(g) Keq = 4.0 I 1.0 1.0 0 C x x +2x E 1.0 − x 1.0 − x +2x Substituting the last row into the expression for the Keq yields $K_{eq}=\frac{[HCl]^{2}}{[H_{2}][Cl_{2}]}=\frac{(2x)^{2}}{(1-x)(1-x)}=4.0\nonumber$ which, of course, is the same expression we have already solved and yields the same answers for the equilibrium concentrations. The ICE chart is a more formalized way to do these types of problems. The + sign is included explicitly in the change row of the ICE chart to avoid any confusion. Sometimes when an ICE chart is set up and the Keq expression is constructed, a more complex algebraic equation will result. One of the more common equations has an x2 term in it and is called a quadratic equation. There will be two values possible for the unknown x, and for a quadratic equation with the general formula ax2 + bx + c = 0 (where a, b, and c are the coefficients of the quadratic equation), the two possible values are as follows: $x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\nonumber$ One value of x is the + sign used in the numerator, and the other value of x is the − sign used in the numerator. In this case, one value of x typically makes no sense as an answer and can be discarded as physically impossible, leaving only one possible value and the resulting set of concentrations. Example 9 illustrates this. ## Example $$\PageIndex{1}$$: Set up an ICE chart and solve for the equilibrium concentrations in this chemical reaction. $\underset{0.55M}{COI_{2}(g)}\rightleftharpoons \underset{0}{CO(g)}+\underset{0}{I_{2}(g)}\; \; \; \; \; K_{eq}=0.00088\nonumber$ Solution The ICE chart is set up like this. First, the initial values: COI2(g) $⇄$ CO(g) + I2(g) I 0.55 0 0 C E Some of the COI2 will be lost, but how much? We do not know, so we represent it by the variable x. So x M COI2 will be lost, and for each COI2 that is lost, x M CO and x M I2 will be produced. These expressions go into the change row: COI2(g) $⇄$ CO(g) + I2(g) I 0.55 0 0 C x +x +x E At equilibrium, the resulting concentrations will be a combination of the initial amount and the changes: COI2(g) $⇄$ CO(g) + I2(g) I 0.55 0 0 C x +x +x E 0.55 − x +x +x The expressions in the equilibrium row go into the Keq expression: $K_{eq}=\frac{[CO][I_{2}]}{[COI_{2}]}=0.00088=\frac{(x)(x)}{(0.55-x))}\nonumber$ We rearrange this into a quadratic equation that equals 0: 0.000484 − 0.00088x = x2x2 + 0.00088x − 0.000484 = 0 Now we use the quadratic equation to solve for the two possible values of x: $x=\frac{-0.00088\pm \sqrt{(0.00088)^{2}-4(1)(-0.000484)}}{2(1)}\nonumber$ Evaluate for both signs in the numerator—first the + sign and then the − sign: x = 0.0216 or x = −0.0224 Because x is the final concentration of both CO and I2, it cannot be negative, so we discount the second numerical answer as impossible. Thus x = 0.0216. Going back to determine the final concentrations using the expressions in the E row of our ICE chart, we have [COI2] = 0.55 − x = 0.55 − 0.0216 = 0.53 M[CO] = x = 0.0216 M[I2] = x = 0.0216 M You can verify that these numbers are correct by substituting them into the Keq expression and evaluating and comparing to the known Keq value. ## Exercise $$\PageIndex{1}$$ Set up an ICE chart and solve for the equilibrium concentrations in this chemical reaction. $\underset{0.075M}{N_{2}H_{2}(g)}\rightleftharpoons \underset{0}{N_{2}(g)}+\underset{0}{H_{2}(g)}\; \; \; \; \; K_{eq}=0.052\nonumber$ The completed ICE chart is as follows: N2H2(g) $⇄$ N2(g) + H2(g) I 0.075 0 0 C x +x +x E 0.075 − x +x +x Solving for x gives the equilibrium concentrations as [N2H2] = 0.033 M; [N2] = 0.042 M; and [H2] = 0.042 M ## Key Takeaway • An ICE chart is a convenient way to determine equilibrium concentrations from starting amounts. 10.4: Calculating Equilibrium Constant Values is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.	4,121	12,198	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.09375	4	CC-MAIN-2024-30	latest	en	0.194084
http://docplayer.net/44992622-Savings-and-loans-problem-solving-and-using-scroll-bars.html	1,521,619,403,000,000,000	text/html	crawl-data/CC-MAIN-2018-13/segments/1521257647584.56/warc/CC-MAIN-20180321063114-20180321083114-00540.warc.gz	85,312,856	22,967	# Savings and Loans: Problem Solving and Using Scroll Bars Size: px Start display at page: ## Transcription 1 activity 10.1 Savings and Loans: Problem Solving and Using Scroll Bars In the first part of this activity, you will identify problem-solving techniques. In the second part of this activity, you will learn how to create a scroll bar using Excel and then use it to solve problems about savings and loans. 1. Identify all problem-solving techniques used in #1 of Activity 6.1 (where you decided which of the magic genie s two offers to choose). For each technique you identify, describe where in the activity it was used. 2 Activity 10.1: Savings and Loans Describe a problem (you can use one solved previously, or you can construct one) that can be solved using the stated problem-solving technique (by itself or in combination with other techniques): a. Problem-solving technique 2b. Draw a picture, graph, or diagram. b. Problem-solving technique 3. Examine a simple case or try several special cases. c. Problem-solving technique 5. Work backward. 3. In the next part of this activity, you will create an Excel scroll bar to see how interest grows in a savings account. 3 498 Excel Activities Scroll Bars A scroll bar is an exciting addition to a spreadsheet that can allow you to vary a value easily. You can vary, for example, the interest rate on an investment in an account over an interval from 1 percent to 20 percent to see how the total amount in the account is changing dynamically. Instructions to Use Excel to Create a Scroll Bar a. b. c. d. e. f. You ll first set up a spreadsheet to calculate the interest and total money in the account with an investment of \$1000 over a 15-year period. In cell C1, enter the label Interest Rate as a Percent and in cell C2 enter =1. In cell B1, enter the label Interest Rate as a Decimal and in cell B2 enter =C2/100. (The value in cell C2, which is the interest rate as a percent, will be linked to the scroll bar. The value in cell B2 is the interest rate as a decimal.) In cells A4, B4, and C4, enter the labels Year, Interest, Total, respectively. In cell A5, enter 0; in A6 enter 1. Highlight both of these cells and drag down to cell A20 to represent 15 years. In cell C5, enter \$ , which will be the starting amount in the account. In cell B6, enter =\$B\$2C5. (Note that you can type =B2 and then press the F4 function key to get the result \$B\$2.) In cell C6, enter the formula, =C5 + B6. Highlight cells B6 and C6 and drag down to show the interest and total for each of the 15 years. Next you will sketch a scatterplot of x = Year and y = Total. You will need to highlight the label and values in the Year column, press Ctrl, release the left mouse button, and then go to the Total column and highlight the label and values in this column. Now using the Insert tab, create the scatterplot, with the points connected by a line. Choose appropriate titles for the axes and chart. 4 Activity 10.1: Savings and Loans 499 g. h. i. j. k. l. To use a scroll bar, you will need to access the Developer tab. If the Developer tab is not shown on top of the menu ribbon, use the following instructions to load it. Click the Office Button. Then click Excel Options (at the bottom of the pop-up window). Choose Popular on the left menu box, check Show Developer tab in the Ribbon, and then click OK. Click on the Developer tab and then go to the Controls group and click on the small arrow under Insert. On the pop-up menu, under ActiveX Controls, find the Scroll bar icon. (Look for the small picture with the up and down arrows and check that the words Scroll bar (ActiveX Control) appear when you place the cursor over it. Click on this icon. The Design Mode icon in the Controls group will light up and the cursor becomes a thin plus sign. Move the cursor (the plus sign) to the spot where you want the scroll bar. Press and hold the left mouse button. Drag the mouse horizontally to the right to create a horizontal scroll bar, or drag it down to create a vertical scroll bar. When the scroll bar is the size you want, release the mouse button. You should see small open circles, called handles, around the outside of the scroll bar. With the handles still showing, move the cursor to the Controls group of the Ribbon and click on Properties. This brings up the Properties box containing a list of properties in alphabetical order. You will need to fill in three values: LinkedCell, Max, and Min. Linked cell refers to the cell where you have placed the value you want to change when you scroll. Point to the words LinkedCell, click, and type C2 here. Point to Min, click, and enter 1, and then point to Max, click, and enter 20. Move the cursor to a cell outside the Properties box and click. To close the Properties box, click the x in the upper-right corner of the box. Click on Design mode in the Controls group (which should still be highlighted) to exit Design mode. You are almost ready to use the scroll bar you designed. You need to change the scale on the y-axis of the scatterplot so the scaling is not automatic, but is fixed. On your finished graph, point to any number on 5 500 Excel Activities the y-axis and right-click; then click on Format axis. In the Axis Options window, click on Fixed for Minimum; then enter 1000 in the Minimum box. Also click on Fixed for Maximum and enter in the box. Click Close. (Fixing the scale will allow you to see how the graph changes as the scroll bar changes the value of the interest rate.) m. You are now ready to move the little box on the scroll bar to increase or decrease the value of the interest rate. This will in turn alter the Total column of the spreadsheet and change the graph. 4. Describe how the graph changes as the scroll bar moves. You can also use a scroll bar to solve Josh s problem (see Example 10.5 in Topic 10). To do this, you will first create a graph of the monthly balance remaining, and then you can change the monthly payment using the scroll bar. Here is how: 5. Go to a new sheet in your Excel workbook. Set up the new sheet as follows: a. b. In cell A1, enter the label APR as a Decimal, and in cell A2, enter 0.06, the loan s APR (as a decimal) for the loan from the credit union. In cell B1, enter the label Monthly Interest Rate, and in cell B2 enter = A2/12, the monthly interest rate. In cell C1, enter the label Tentative Payment and in cell C2, enter \$100.00, a tentative payment amount (this will be attached to the scroll bar and moved until you find the payment needed to pay off the loan in 60 months). 6 Activity 10.1: Savings and Loans 501 c. d. e. f. g. h. i. j. In cells A4, B4, and C4, enter Year, Balance Last Month, and Balance This Month, respectively. In column A beginning at cell A5, fill in numbers 1 through 60 (for the 60 months in 5 years). In cell B5, enter \$ (this is the starting loan principal or balance at the end the month before you start payments). In cell C5 enter =B5 + B5\$B\$2 \$C\$2 to calculate the loan balance after one month, that is, at the end of the first month. In cell B6, enter =C5. Drag to fill in columns B and C up to month 60. Create a graph that shows balance this month as a function of year. Because you are setting up a scroll bar, you will want to format the y-axis so scaling is fixed. Create a scroll bar with a LinkedCell of C2 (so the payment can be changed by scrolling), with a Min of 0, and a Max of How would you know from the graph that the choice of monthly payment in C5 is the correct one (or a good approximation of it) to pay off the debt in five years? Write the correct amount here. 7. Use the same spreadsheet, changing the interest rate, principal, and compounded period as appropriate, to find all the information needed to solve Josh s problem. Explain how you found the information. 7 502 Excel Activities 8. Identify which problem-solving techniques (see the list in Topic 10) you used to solve Josh s problem in this activity. Compare the techniques used here with the techniques used to solve the same problem in the examples in Topic 10. Which techniques were used in both solutions? Which were used in one and not the other? Summary In this activity, you worked with several problem-solving techniques and looked for situations where specific techniques could be used. You learned how to create a scroll bar in Excel to see how the interest paid and the accumulated total in a savings account change for different annual interest rates. Finally, you used a scroll bar to solve the loan decision problem stated in Example 10.5 and solved in Topic 10. ### How to make a line graph using Excel 2007 How to make a line graph using Excel 2007 Format your data sheet Make sure you have a title and each column of data has a title. If you are entering data by hand, use time or the independent variable in ### In this example, Mrs. Smith is looking to create graphs that represent the ethnic diversity of the 24 students in her 4 th grade class. 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Select All Programs and then find	7,568	33,834	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2018-13	latest	en	0.910396
https://growthecon.wordpress.com/category/level-effects/	1,680,331,618,000,000,000	text/html	crawl-data/CC-MAIN-2023-14/segments/1679296949701.56/warc/CC-MAIN-20230401063607-20230401093607-00106.warc.gz	322,088,630	29,938	# The Solow Model NOTE: The Growth Economics Blog has moved sites. Click here to find this post at the new site. This is another idea for modifying how to teach the Solow model. One thing I’d like to do is go immediately to including productivity – it follows cleanly from the simplest growth model. Second, I think it might be nice to work with the K/Y ratio immediately. In this way, I think you can actually skip using the whole “k-tilde” thing. And, gasp, do away with the traditional Solow diagram. The simplest growth model doesn’t allow for transitional growth, and this due to the fact that it does not allow for capital, a factor of production that can only be slowly accumulated over time. The Solow Model is a standard model of economic growth that includes capital, and will be better able to account for the transitional growth that we see in several countries. Production in the Solow Model takes place according to the following function $\displaystyle Y = K^{\alpha}(AL)^{1-\alpha}. \ \ \ \ \ (1)$ ${K}$ is the stock of physical capital used in production, and ${A}$ and ${L}$ are defined just as they were in our simple growth model. So the production function here is just a modification of the simple model to include capital. The coefficient ${\alpha}$ is a weight telling us how important capital or ${AL}$ are in determining output. To analyze this model, we’re going to rewrite the production function. Divide both sides of the function by ${Y^{\alpha}}$, giving us $\displaystyle Y^{1-\alpha} = \left(\frac{K}{Y}\right)^{\alpha} (AL)^{1-\alpha} \ \ \ \ \ (2)$ and then take both sides to the ${1/(1-\alpha)}$ power, which gives us the following expression $\displaystyle Y = \left(\frac{K}{Y}\right)^{\alpha/(1-\alpha)} AL. \ \ \ \ \ (3)$ In per capita terms, this is $\displaystyle y = \left(\frac{K}{Y}\right)^{\alpha/(1-\alpha)} A. \ \ \ \ \ (4)$ Output per worker thus depends not just on ${A}$, but also on the capital-output ratio, ${K/Y}$. So to understand the role of capital in economic growth, we need to understand the capital-output ratio and how it changes over time. We’ll start by looking at the balanced growth path, and then turn to situations where the economy is not on the balanced growth path (BGP). One fact about the BGP is that the return to capital, ${r}$, is constant. The return to capital is ${r = \alpha Y/K}$, which depends (negatively) on the capital-output ratio (the return to capital is just the marginal product of capital). If ${r}$ is constant on the BGP, then it must be that ${K/Y}$ is constant on the BGP as well. What does this mean? It means that ${K/Y}$ can have a level effect on output per worker, but has no growth effect. To see this more clearly, take logs of output per worker, $\displaystyle \ln y(t) = \frac{\alpha}{1-\alpha} \ln\left(\frac{K}{Y}\right) + \ln A(t) \ \ \ \ \ (5)$ and then plug in what we know about how ${A(t)}$ moves over time, $\displaystyle \ln y(t) = \frac{\alpha}{1-\alpha} \ln\left(\frac{K}{Y}\right) + \ln A(0) + gt. \ \ \ \ \ (6)$ The capital-output ratio affects the intercept of this line — a level effect — alongside ${A(0)}$. The slope of this line — the growth rate — is still ${g}$. The capital/output ratio is constant along the BGP, and has no effect on the growth rate on the BGP. But what if the economy is not on the BGP? Then it will be the case that ${K/Y}$ affects the growth rate of output per worker, because the ${K/Y}$ ratio will not be constant. More precisely, the growth rate of capital/output is $\displaystyle \frac{\dot{K/Y}}{K/Y} = \frac{\dot{K}}{K} - \frac{\dot{Y}}{Y}. \ \ \ \ \ (7)$ So the ${K/Y}$ ratio will change if capital grows more quickly or more slowly than output. First, capital accumulates as follows $\displaystyle \dot{K} = s Y - \delta K \ \ \ \ \ (8)$ where ${\dot{K}}$ is the change in the capital stock. ${s}$ is the savings rate, the fraction of output that the economy sets aside to invest in new capital goods, so that ${sY}$ is the total amount of new investment. ${\delta}$ is the depreciation rate, the fraction of the existing capital stock that breaks or becomes obsolete at any given moment. To find the growth rate of capital, divide through the above equation by ${K}$ to get $\displaystyle \frac{\dot{K}}{K} = s\frac{Y}{K} - \delta. \ \ \ \ \ (9)$ You can see that the growth rate of capital depends on the capital/output ratio itself. The growth rate of output is $\displaystyle \frac{\dot{Y}}{Y} = \alpha \frac{\dot{K}}{K} + (1-\alpha)\frac{\dot{A}}{A} + (1-\alpha)\frac{\dot{L}}{L}. \ \ \ \ \ (10)$ Now, with (7), and using what we know about growth in capital and output, we have $\displaystyle \frac{\dot{K/Y}}{K/Y} = (1-\alpha)\left(s\frac{Y}{K} - \delta - g - n \right) \ \ \ \ \ (11)$ where we’ve plugged in that ${\dot{A}/A = g}$, and ${\dot{L}/L = n}$. Re-arranging a bit, the capital output ratio is growing if $\displaystyle \frac{K}{Y} < \frac{s}{\delta + n + g}, \ \ \ \ \ (12)$ and growing if the capital/output ratio is larger than the value on the right-hand side. In other words, if the capital stock is relatively small, then it will have a tendency to grow faster than output, raising the ${K/Y}$ ratio. Eventually ${K/Y = s/(\delta+n+g)}$, the steady state value, and the ${K/Y}$ ratio stops changing. What is happening to growth in output per worker? If ${K/Y < s/(\delta+n+g)}$ then the ${K/Y}$ ratio is growing, and so output per worker is growing faster than ${g}$. So the temporarily fast growth in output per worker in Germany or Japan would be because they found themselves with a ${K/Y}$ ratio below their steady state value. How would this occur? It’s easier to see how this works if we re-write the ${K/Y}$ ratio slightly $\displaystyle \frac{K}{Y} = \frac{K}{K^{\alpha}(AL)^{1-\alpha}} = \left(\frac{K}{AL}\right)^{1-\alpha}. \ \ \ \ \ (13)$ From this we can see that the ${K/Y}$ ratio would be particularly low if the capital stock, ${K}$, were to be reduced. This is what happened in Germany, to a large extent, after World War II. The capital stock was destroyed, so ${K/AL}$ fell sharply. This made ${K/Y}$ fall below the steady state value, which meant that there was growth in the ${K/Y}$ ratio, and so growth in output per worker greater than ${g}$. A slightly different situation describes South Korea. There, we can think of there being a level effect on ${A}$, an advance in productivity. This also makes ${K/AL}$ fall sharply, and again causes growth in ${K/Y}$ and growth in output per worker faster than ${g}$. But in both this case and in Germany’s, as the ${K/Y}$ ratio grows it approaches the steady state value and growth in output per worker slows down to ${g}$ again. # The Simplest Growth Model NOTE: The Growth Economics Blog has moved sites. Click here to find this post at the new site. This is an idea for a new way of introducing growth theory. Given that productivity growth is the source of long-run growth, it seems to make sense to start with that, rather than with the Solow model. Let’s write down a very simple model of economic growth. Let total output ${Y}$ be determined by $\displaystyle Y = A L \ \ \ \ \ (1)$ where ${A}$ is a measure of labor productivity, and ${L}$ is the number of workers. If we divide through by ${L}$, then we get a measure of output per worker. To keep notation clean, let ${y = Y/L}$ be output per worker, so that now we have $\displaystyle y = A \ \ \ \ \ (2)$ as our model of economic growth. Basically, output per worker is simply equal to labor productivity ${A}$. From this we know that the time path of output per worker is simply the same as the time path of labor productivity, ${A}$. So what determines the time path of labor productivity? We’ll assume that it is growing at a constant rate, meaning that it goes up by the same percent every period of time, $\displaystyle A(t) = A(0) e^{g t}. \ \ \ \ \ (3)$ Here, we’ve written ${A(t)}$ to be clear that we mean labor productivity at any given time ${t}$. ${A(0)}$ is labor productivity in the initial moment of time. The exponential term says that labor productivity grows at the rate ${g}$ over time. The exponential term implies, perhaps not surprisingly, exponential growth. You get exponential growth when something goes up by the same percent every period of time. If ${g = 0.02}$, then we have 2% growth. At time zero, labor productivity is just ${A(0)}$. When ${t=2}$, then ${A(2) = A(0)e^{.02(2)} = 1.041 A(0)}$, or labor productivity is a little more than 4% higher than at time zero. When ${t=10}$, ${A(10)=A(0)e{.02(10)}=1.221}$, or labor productivity is more than 22% higher than at time zero. It may not seem obvious, but output per worker in the U.S. and most other developed nations displays exponential growth. Our model matches that, as $\displaystyle y(t) = A(0) e^{g t}. \ \ \ \ \ (4)$ These countries also tend to have a similar growth rate of about 1.8%, or ${g=0.018}$. Seeing this in a figure, though, is difficult. Graphing ${y}$ over time for the U.S. gives you a curve that quickly accelerates upwards and is almost off the page. Graphs like this will also make it difficult to compare countries to one another. For that reason, among others, we like to work with the natural log of output per worker, ${\ln{y(t)}}$. Taking natural logs of ${y(t)}$ gives us $\displaystyle \ln{y(t)} = \ln{A(0)} + g t. \ \ \ \ \ (5)$ This is an equation that says the natural log of output per worker is a linear function of time, ${t}$. If we graph ${\ln{y(t)}}$ against ${t}$, we get a straight line, similar to the trend line we drew in the figure for U.S. output per worker. We can calculate the growth rate of output per worker by taking the derivative of (5) with respect to time. This results in the following $\displaystyle \frac{\dot{y}}{y} = g. \ \ \ \ \ (6)$ The value of ${A(0)}$ is fixed, so the derivative of it with respect to time is just zero. The notation ${\dot{y}/y}$ is a shorthand way of writing the growth rate. ${\dot{y}}$ is the absolute change in output per worker at any given moment, and by dividing by ${y}$ we get that change relative to the level of output per worker. This means that ${\dot{y}/y}$ is essentially the percent change in output per worker at any given moment. That’s it for the simple growth model. Output per worker depends on labor productivity ${A(t)}$, and labor productivity grows at a constant rate ${g}$, which means output per worker grows at that same rate. Despite the mechanical simplicity, this model helps us be clear when we are talking about the growth experiences of different countries. It allows us to distinguish between two forces determining output per worker. • Level effects: These refer to ${A(0)}$, the intercept of the line in (5) • Growth effects: These refer to ${g}$, the slope of the line in (5) Looking at the data over the long run, the general impression we get that the growth rate ${g}$ is similar across countries, and they differ mainly because of level effects. That is, ${A(0)_{Japan}}$ appears to be lower than ${A(0)_{US}}$, but the growth rate ${g}$ is very similar. Theories of economic growth should be consistent with these facts. Things like investment rates, schooling, and social infrastructure are important determinants of level effects, ${A(0)}$, but they have no effect on the growth rate, ${g}$. Under plausible assumptions, theories of endogenous innovation will suggest that the growth rate, ${g}$, is identical across countries. There are some facts, though, that this simple growth model cannot account for. Namely, there are notable cases where output per worker grows more quickly or more slowly than ${g}$. China, for example, over the last 30 years has grown much faster than the U.S. or Japan. South Korea had a similar growth miracle, starting in about 1960 and lasting until the 2000’s. Germany, from World War II until about 1980, grew at a very accelerated pace compared to the U.S. in the same period. How do we reconcile these facts with the assertion above that ${g}$ is the same for all countries? The key is noting that these growth accelerations were temporary. Germany grew very quickly, but after 1980 its growth rate fell back to a value nearly identical to the U.S. South Korea’s growth rate has diminshed as well in the 2000’s. What appears to be happening is that once output per worker approaches a frontier level, generally defined by the U.S., growth slows down. While China continues to grow quickly, it has not approached the U.S. level of output per worker. Looking at these countries, what appears to be happening is that there is a level effect, or their ${A(0)}$ has shifted up. However, it seems to take them a long time to move from their old level to the new, higher level. We call the temporary growth spurt that occurs when a country moves between levels transitional growth. Output per worker grows faster than ${g}$ temporarily – although this could last a few decades – but then growth returns to the rate ${g}$. Our simple model doesn’t offer a way of understanding this transitional growth. The first major extension we’ll make to this simple model is to add physical capital, which has to be slowly accumulated over time. Because of this slow accumulation, the economy will take an extended time to fully respond to a level effect.	3,434	13,365	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 111, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2023-14	latest	en	0.914243
https://www.techylib.com/en/view/restmushrooms/35_ee394j_spring11_notes_on_second_order_systems_ecs	1,524,707,342,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125948047.85/warc/CC-MAIN-20180426012045-20180426032045-00115.warc.gz	882,954,076	16,222	# 35_EE394J_Spring11_Notes_on_Second_Order_Systems - ECS Electronics - Devices Oct 7, 2013 (4 years and 7 months ago) 87 views restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 1 of 20 The simplest second - order system is one for which the L’s can be combined, the C’s can be combined, and the R’s can be combined into one L, one C, and one R. Let us consider situations where there is no source, or a DC source. For a n RLC Series Circuit If the circuit is a series circuit, then there is one current. Note that in the above circuit, the V and R could be converted to a Norton equivalent, permitting the incorporation of a current source into the problem Using KVL, . Taking the derivative, . Note that the constant source term V is gone , leaving us with the characteristic equation (i.e., source free), which yields the natural response of the circuit . Rewrit ing, . (1) Guess the natural response solution and try it, , And we reason that the only possibility for a general solution is when . V + v C i i C o +⁶ L +⁶ R restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 2 of 20 . Defining as the damping coefficient (nepers/sec) for the series case (2) as the natural resonant frequency (radians/sec) (3) then we have . (4) Substituting (2 ) and (3) into (1) yields a useful form of (2), which is (5) For a n RLC Parallel Circuit If the circuit is a parallel circuit, then there is one voltage. KCL at the top node yields . Taking the derivative I + v i R L C R i L i C restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 3 of 20 . Rewriting, . (6) Equation (6) has the same form as (5), except that for the parallel case. (7) Thus, it is clear that the solutions for series and parallel circuits h ave the same form, except that the damping coefficient is defined differently. A nice property of exponential solutions such as is that derivatives and integrals of also have the term. Thus, one can conclude that all voltages and currents throughout the series and parallel circuits have the same term s . The re are Three Cases Equation (4) has three distinct case s : 1. Case 1 is where , so that the term inside the radical is positive. Overdamped . 2. Case 2 is where , so that the term inside the radical is negative. Underdamped . 3. Case 3 is where , so t hat the term inside the radical is zero. Critically Damped . Case 3 rarely happens in practice because the terms must be exactly equal. Case 3 should be thought of as a transition from Case 1 to Case 2 , and the transition is very gradual . Case 1: Overd amped When , then the solutions for (4) are both negative real and are , restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 4 of 20 . The natural response for any current (or voltage) in the circuit is then . Coeffici ents A 1 and A 2 are found from the boundary conditions as follows: , . Thus, we have two equations and two unknowns. To solve, one must know and . Solving, w e get , so (Note see modification needed to include final response) (8) Then, find from (Note see modification needed to include final response) (9) Th e key to finding A 1 and A 2 is always to know the inductor current and capacitor voltage at t = 0 + . Remember tha t Unless there is an infinite impulse of current through a capacitor , the voltage across a capacitor (and the stored energy in the capacitor) r emain s constant during a switching transition from t = 0 - to t = 0 + . restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 5 of 20 Unless there is an infinite impulse of voltage across an inductor, the current through an inductor (and the stored energy in the inductor) remains constant during a switching transition from t = 0 - to t = 0 + . For example, consider the series circuit at t = 0 + . Current . Once I L0 + and V C0+ are known , then we get as follows. F rom KVL thus . ( 10 ) Since then we have . ( 11 ) V + V C0+ f L0 + L C R + V L0+ +⁒f L0 + restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 6 of 20 Similarly, for the parallel circuit at t = 0 + , the voltage across all elements . We get as follows. F rom KCL thus . (1 2 ) Since then we have . ( 1 3 ) Case 2: Underdamped When , then the solutions for (4) are both complex and are , . Define damped resonant frequency as , (1 4 ) I L C R I L0 + + V C0+ f C 0 + I R 0 + restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 7 of 20 so that , The natural response for any current (or voltage) in the circuit is then , . Expanding the and terms using Euler’s rule, , , , . (1 5 ) Since i(t) is a real value, then ( 1 5 ) cannot have an imaginary component . This means that is real, and is real ( which means that is imaginary ). These conditions are met if . Thus, we can write (1 5 ) in the form of (1 6 ) where B 1 and B 2 are real numbers. To evaluate B 1 and B 2 , it follows from (1 6 ) that , so (Note see modification needed to include final response) (17) To find , take the derivative of (1 6 ) and evaluate it at t = 0, , restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 8 of 20 , so we can find using (18) Case 2 Solution in Polar Form The form in (1 6 ), is most useful in evaluating coefficients B 1 and B 2 . But in practice, the answer is usually converted to polar form. Proceeding, write (1 6 ) as The and terms are the cosine and sine, respectively, of the angle shown in the right triangle . Unless B1 is zero, you can find θ using . But be careful because your calculator will give the wrong answer 50% of the time . The reason is that triangle. If the calculator quadrant does not agree with the figure, then add or subtract 180° from your calculator angle, and re - The polar form for i(t) comes from a trigonometric identity. The expression b ecomes the damped sinusoid B 2 B 1 θ restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 9 of 20 . Case 3: Critically Damped When , then there is only one solution for (4), and it is . Thus, we have repeated real roots. Proceeding as before, then ther e appears to be only one term in the natural response of i(t), . But this term does not permit the natural response to be zero at t = 0, which is a problem. Thus, let us propose that the solution for the natural response of i(t) h as a second term of the form , so that . (1 9 ) At this point, let us consider the general form of the differential equation given in (5), which shown again here is . For the case with , the equation becomes . satisfies the equation, but let us now test the new term . Substituting in, Factoring ou t the term common to all yields (Y es !) So, we confirm the solution of the natural response. ( 20 ) restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 10 of 20 To find coefficient A 2 , use . Thus, . (Note see modification needed to include final response) (21) To find , take the derivative of (20), . Evaluating at t = 0 yields . Thus, . (22) Th e question that now begs to be asked is “what happens to the term as t → ∞? Determine this using the series expansion for an exponential. We know that . Then, . Dividing numerator and denominator by t yields, restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 11 of 20 . As t → ∞, the in the denominator goes to zero, and the other “t” terms are huge. Since the numerator stays 1, and the denominator becomes huge, then → 0. The Total Response = The Natural Response Plus The Final Response If the circu it has DC sources, then the steady - state (i.e., “final”) values of voltage and current may not be zero. “Final” is the value after all the exponential terms have decayed to zero. And yet, for all the cases examined here so far , the exponential terms all decayed to zero after a long time. So, how do we account for “final” values of voltages or currents? Simply think of the circuit as having a total response that equals the sum of its natural response final term, e.g. , so (for overdamped). (2 3 ) Likewise, , (for underdamped). (24) Likewise, , (for critically damped) . (25) You can see that the presence of the final term will affect the A and B coefficients because the initial value of i(t) now contains the term . Take into account when you evaluate the A’s and B ’s by replacing in (8), (9), (17), and (21) with . restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 12 of 20 How do you get the final values? If the problem has a DC source, then remember that after a long time when the time derivatives are zero, capacitors are “op en circuit s, ” and inductors are “short circuits . ” Compute the “final value s of voltage s and current s according to the “open circuit” and short circuit” principles . The General Second Order Case Second order circuits are not necessarily simple series or parallel RLC circuits. Any two non - combinable storage elements (e.g., an L and a C, two L’s, or two C’s) yields a second order circuit and can be solved as before, except that the and are different from t he simple series and parallel RLC cases. An example follows. Work problems like Circuit #1 by defining capacitor voltages and inductor currents as the N state variables . For Circuit #1, N = 2. You will w rite N circuit equations in terms of t he state variables, and strive to get an equation that contains only one of them . numbers when writing the equations . For convenience, use the simple notation and to represent time varying capacitor voltage and inductor current, and and to represent their derivatives, and so on. To find the natural response , turn off the sources. Write your N equations by using KVL and KCL , making sur e to include each circuit element in your set of equations . For Circuit #1, s tart with KVL around the outer loop, . (2 6 ) Now, write KCL at the node just to the left of the capacitor, , which yields V + V C f L L C R 1 R 2 General Case Circuit #1 (Prob. 8.56) restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 13 of 20 . (2 7 ) Tak ing the derivative of (2 7 ) yields . (2 8 ) We can now eliminate and in (2 6 ) by substituting (2 7 ) and (2 8 ) into (2 6 ), yielding . Gathering terms, , and putting into standard form yields (2 9 ) Comparing (2 9 ) to the standard form in (5), we see that the circuit is second - order and , , so . The solution procedure for the natural response and total response of either or can then proceed in the same way as for series and parallel RLC circuits, using the and values shown above. Notice in the above two equations that when , , and , restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 14 of 20 which correspond to a series RLC circuit. Does this make sense for this circ uit? Note that when , then , and , which correspond to a parallel RLC circuit. Does this make sense, too? Now, consider Circuit #2. Turning off the independent source and writing KVL for the right - most mesh yields . ( 30 ) KVL for the center mesh is , yielding . ( 31 ) Taking the derivative of ( 31 ) yields . (3 2 ) Substituting into ( 31 ) and (3 2 ) into ( 30 ) yields . I General Case Circuit # 2 (Prob. 8.60) L 2 R 2 I L 2 L 1 R 1 I L 1 restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 15 of 20 Gathering terms, , and putting into standard form yields . (3 3 ) C omparing (3 3 ) to (5), , , so . (34) The solution procedure for the natural response and total response of either or can then proceed in the same way as for series and parallel RLC circuits, using the and values shown above . Normalized Damping Ratio Our previous expression for solving for s was . (35) Equation (35) i s sometimes written in terms of a “ normalized damping ratio” as follows: . (36) Thus, the relationship between damping coefficient and normalized damping ratio is . (37) Normalized d amping ratio has the convenient feature of being 1.0 at the point of critical damping. When < 1.0, the response is underdamped. When < 1.0, the response is overdamped. Examples for a unit step input are show n in the following figure. restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 16 of 20 restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 17 of 20 Notes on Solving the Three Cases has three distinct cases: Case 1 is where , so that the term inside the radical is positive. Overdamped . Case 2 is wh ere , so that the term inside the radical is negative. Underdamped . Case 3 is where , so that the term inside the radical is zero. Critically Damped . Case 1: Overdamped ( , so s 1 and s 2 are both negative real ) The natural (i.e., source free) response for any current or voltage in the network has the form , . The key to finding A 1 and A 2 is always to know the in ductor current and capacitor voltage at t = 0 + . Remember that Unless there is an infinite impulse of current through a capacitor, the voltage across a capacitor (and the stored energy in the capacitor) remains constant during a switching transition from t = 0 - to t = 0 + . Unless there is an infinite impulse of voltage across an inductor, the current through an inductor (and the stored energy in the inductor) remains constant during a switching transition from t = 0 - to t = 0 + . restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 18 of 20 Case 2: Underdamped ( ,so s 1 and s 2 are complex conjugates) The natural (i.e., source free) response for any current or voltage in the network has the rectangular form (16) , . Damped resonant frequency , (14) so that , (Note see modification needed to include final response) (17) (18) In polar form, . Regarding the arctangent - be careful because your calculator will give the wrong answer 50% of the time. The reason is that . So check the ant consistent with the right triangle. If the calculator quadrant does not agree with the figure, then add or subtract 180° from your calculator angle, and re - B 2 B 1 θ restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 19 of 20 Case 3: Critically Damped ( , so ) . (19) . (Note see modification needed to include final response) (21) . (22) Total Response = Natural Response Plus Final Response If the circuit has DC sources, then th e final response is also DC. The total response is , (23) The effect of I final on determining A 1 and A 2 simply means that you use (23) at t = 0 instead of i(t = 0) by itself. If the circuit has AC source s, the final response is the phasor solution. restmushrooms_b8d5e267 - 60ce - 4868 - ac5f - 1c2878835910.doc Page 20 of 20 Normalized Damping Ratio Our previous expression for solving for s was . (35) Equation (35) is sometimes written in terms of a “normalized damping ratio” as follows: . (36) Thus, the relationship between damping coefficient and normalized damping ratio is . (37) Normalized damping ratio h as the convenient feature of being 1.0 at the point of critical damping. When < 1.0, the response is underdamped. When < 1.0, the response is overdamped. E xamples for a unit step input are shown in the following figure.	4,819	15,579	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2018-17	latest	en	0.883643
https://de.scribd.com/presentation/281934384/5-Earth-gravitation-g-and-lift-ppt	1,563,611,985,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00490.warc.gz	369,201,034	65,943	You are on page 1of 74 # Gravitation of earth ## Introduction to rotational motion Centrifugal force Gravitational force = Vc Escape velocity Escape velocity If we throw a object up such a velocity that it goes out of earths gravity it velocity is called escape velocity It means that by the time its velocity reduces to zero due to gravity it had reached a point where earths gravitational force is also is zero. B V=0 Fg=0 Throw a ball of (mass m) up Initial velocity at point A = U, and Gravitational force = Fg Fg Me = mass of earth v = u +(- gt ) B Fg=0 V=0 R U ## Thus earth will not attract the object & it will not return back to earth A R Fg = (G ) ( Me m ) r2 Fg Me = mass of earth escape velocity v = u +(gt ) ## Ex :Calculate initial velocity such that ball we reach height 5 m When we throw a ball up it KE B gets converted to PE 5m KE at point A = PE at point B U m/sec A R Me = mass of earth PE at point B = m x 10 x 5 J KE at point A = x m x U2 J x m x U2 = m x 10 x 5 U2 = 2 x 10 x 5 U2 = 100 U = 10m/sec ## Calculations of escape velocity B To throw object from point A on earth to a point B in space we have to do work against earth gravitational force Fg This can be calculated by formula A work = F x distance moved R Me = mass of earth ## Escape velocity Fg is continuously reducing as r increases Consider a very small distance dr B A Fg ## dr. at distance r from earth center We assume that over this distance Fg is constant as dr is very small Work done to lift ball from A to B dw = Fg x dr Escape velocity B Work done to lift ball from A dr B A to B dw = Fg x dr ## If we can add work done over every such small dr from A to B, Fg we will get net work done A against gravity. Now A R Me = mass of earth Fg = (G ) ( Me m ) r2 Calculations of (G ) ( Me m ) dw dr = work done r2 B Using calculus we can add such dr B A ## small work( dw1+ dw2 +dw3+ -----)1 for r changing from A to B .(B is point at infinity where Fg =0 ). A R Me = mass of earth ## This give total work done. Ans is PE at B = Work done from A to B (G ) ( Me m ) J = R (G ) ( Me m ) J 2 m U = B R 2 x (G ) ( Me) U2 = R 24 -11 5.98 10 2 6.67 10 U2 = U (6.37 10 6 ) A R Me U2 = ## 2 6.67 5.98 10 24 -11-6 6.37 U2 = 12.52 10 7 = 125.2 10 6 U = 11.19 10 3 = 11200m/sec ## Value of gravitational force/ acceleration os different objects (Density is assumed to be uniform) g at equator & pole Fg inside & outside a solid sphere Fg inside & outside a shell g at height h above earth Lift problems Drag force ## Earth radius at equator is more than poles g at equator = 9.780 m/sec2 g at pole = 9.832 m/sec 2 Weight of body as measured by spring balance(force due to earths gravitation) increases from equator to pole but mass remains same 6.351 10 6 meters at pole 6.378 10 6 meters at equator ## Value of gravitational force/ acceleration os different objects (Density is assumed to be uniform) g at equator & pole Fg inside & outside a solid sphere Fg inside & outside a shell g at height h above earth Lift problems Drag force ## Gravitational force inside & outside a solid sphere Case 1. If a particle of mass m is located outside a homogeneous solid sphere of Case 2. If a particle of mass m is located inside a homogeneous solid sphere of ## The sphere attracts mass out side if as though the mass of the sphere were concentrated at its center.. r ( R+ r )2 r Fg r M = mass of inner sphere The net gravitational force at poet Q due to outer material of main sphere is zero ## The sphere has a uniform density, , in kg/m3, Let the mass of the full sphere (radius R ) = M kg Let mass of the inner sphere (radius r = M kg F Inner Inner sphere sphere ## The sphere has a uniform density, , in kg/m3, Let the mass of the full sphere = M kg Let mass of the inner sphere = M kg F Inner sphere ## The sphere has a uniform density, , in kg/m3, Let the mass of the interior sphere = M kg Let mass of the inner sphere = M kg F F Inner sphere r F = F x R As r reduces F will reduce At center of sphere r = 0 hence F = 0 Gravitational force at center of solid sphere is = 0 ## Graph of F for solid sphere F gr r (R+r)3 (As R<< r, R + r = r ) gr R r r Graph of gr for solid sphere Fg = max, V = 0 +ve Fg = 0 V = max Fg Fg = - max V = 0 ## Time for oscillation of a ball in tunnel = 84 .3 min Time for a satellite to make full round on surface of earth will be also 84.3min ## A proposal was made to operate a A transport system in a tunnel between two cities A & B, using this One-way trip well about 42 min. Precise calculation were done as Earths density is not uniform. Practical problems were How to make frictionless tunnel, Temperature inside earth is high Magma will come out etc ## Value of gravitational force/ acceleration os different objects (Density is assumed to be uniform) g at equator & pole Fg inside & outside a solid sphere Fg inside & outside a shell g at height h above earth Lift problems Drag force ## Gravitational force due to Spherical Shell of uniform density 1 Particle located outside a spherical shell 2. If the particle is located inside the shell Case 2 Case 1 Shell of mass M ## If a particle of mass m is located outside a spherical shell of mass M at point P, the shell attracts the particle as if the mass of the shell is concentrated at its center ## If the particle is located inside the shell (at point P in Fig ),the gravitational force acting on it = 0 Shell of mass ## Value of gravitational force/ acceleration os different objects (Density is assumed to be uniform) g at equator & pole Fg inside & outside a solid sphere Fg inside & outside a shell g at height h above earth Lift problems Drag force GM E 2 g (0) 9.83 m/s 2 RE GM E g ( h) 2 ( RE h) ## Once the altitude becomes comparable to the radius of the Earth, the decrease in the acceleration of gravity is much larger: = 6370km GM E g (r ) 2 r ## What is the acceleration due to gravity of an object at the altitude of the space shuttles the Earths surface? ## r RE h (6370 km) (400 km) 6770 km Fg GmM E GM E a /m 2 2 m r r (6.67 1011 m 2 /kg 2 )(5.98 10 24 kg) 2 8.70 m/s 6 2 (6.77 10 m) ## If altitude of International Space Station is 385 km above the Earths surface, what is its period of rotation 2 r vT 2 r v T M E m mv 2 Fg G 2 r r GM E 4 2 r 2 r T2 ( RE h) 4 r 3 T r 2 2 GM E GM E GM E 2 25 june ## Elevator & weightlessness problems Apparent Weight the contact forces between your body and are accelerating, weight may be more or less than your actual weight. ## Understanding perception of weight Actual weight of a person is determined by his mass and the acceleration of gravity, The sensation of apparent weight comes from the fact that we are supported from the floor, chair, etc. When body is in "free fall (accelerating downward at the acceleration of gravity) then the body is not being supported. Then we fill weightless ## The feeling of "weightlessness" occurs when There is no force of support on your body. This can be achieved in several ways .Ex Free fall as there is no support. In a elevator if cable breaks airplane coming down with acc =g Satellite rotating around earth ## Different sensations of apparent weight can occur in elevator when it moves with Zero acceleration (zero or constant speed): No change in apparent weight Accelerates upward: Apparent wt. increases Accelerates downward: Apparent wt. reduces If the elevator cable breaks (free fall with downward acceleration = g.): Weightlessness is felt Accn = 0 Velocity = 0 or constant No change in apparent weight ## Floor support = Reaction of weight + support force due to upward acceleration Hence apparent weight is more ## Floor support = Reaction of weight - support to accelerate you downward Hence apparent weight is less ## Elevator cable broken F support = 0 Apparent weight = 0 ## Terminal velocity during free fall in air: When net force is = 0 , downward velocity remain constant, called terminal velocity Air Weight force Controlling terminal velocity by parachute Earth gravitational acceleration g does note depend on mass of object Hence an elephant & a feather will take same time to reach ground if there in no resistance End Reference slides ## Geostationary Satellites for Meteorology (and Volcanology!) http://www.ssec.wisc.edu/data/geo/ http://www.rap.ucar.edu/weather/satellite/ http://www.ssec.wisc.edu/data/volcano.html FM on m GMm mv 2 2 mar r r GM v = r U g m ## The tangential velocity v needed for a circular orbit depends on the gravitational potential energy Ug of the satellite at the radius of the orbit. The needed tangential velocity v is independent of the mass m of the satellite (provided m<<M). Notice that to make v larger, you need to go deeper into the gravity well, i.e., to a lower orbit where Ug is larger and r is smaller. Orbital Energetics FM on m mv 2 2 K r r Ug GMm 2 r r K 12 U g The equation K = Ug is called The Virial Theorem. In effect, it says that for a planet in orbit around the Sun, if you turned its velocity by 90o, so that it pointed straight out of the Solar System, you would have only half the kinetic energy needed to escape the Suns gravity well. Example: The Total Energy of a Satellite Show that the total energy of a satellite in a circular orbit around the Earth is half of its gravitational potential energy. 1 2 GM E m mv 2 r GM E m mv 2 GM E 2 v 2 r r r E K U GM E m 1 GM E GM E m E m 2 r r 2r GM E m U Although derived for this particular case, this is a r general result, and is called the Virial Theorem. The 1 E 2U factor of is a consequence of the inverse square law. ## g for a Solid Sphere GM r g outside 2 r r R r GM r r ginside 3 r r R R Gravitational field of a solid uniform sphere 4 3 4 3 r3 r3 M ' M M 3 3 R R GM ' G r 3 GMr GM r r ginside 2 r 2 M 3 r 3 r 3 r r r R R R ## g for a Hollow Sphere GM r g outside 2 r r R r r ginside 0 rR Gravitational field of a uniform spherical shell m1 A1 r12 m m ## 2 so 21 22 and the forces cancel. m2 A2 r2 r1 r2 Tides Tidal forces can result in orbital locking, where the moon always has the same face towards the planet as does Earths Moon. If a moon gets too close to a large planet, the tidal forces can be strong enough to tear the moon apart. This occurs inside the Roche limit; closer to the planet we have rings, not moons. Tides ## This figure illustrates a general tidal force on the left, and the result of lunar tidal forces on the Earth on the right. Example: The Orbiting Space Station You are trying to view the International Space Station (ISS), which travels in a roughly circular If its altitude is 385 km above the Earths surface, how long do you have to wait between sightings? 2 r vT M E m mv 2 Fg G 2 r r 2 r T GM E 4 2 r 2 r T2 4 2 3 r3 ( RE h)3 T r 2 2 GM E GM E GM E (6375 km 385 km)3 T 2 5,528 s=92.1 min (6.67 1011 N m 2 /kg 2 )(5.98 10 24 kg) ## Measured weight in an accelerating Reference Frame According to stationary observer R Man in lift mg R is reaction force F = ma Taking up as +ve R - mg = ma accel R = m(g + a) If a = 0 ==> R = mg normal weight ## If a is +ve ==> R = m(g + a) weight increase If a is -ve ==> R = m(g - a) weight decrease Spring scales According to traveller R mg R is reaction force F = ma R - mg = ma ## BUT in his ref. frame a = 0! so R = mg!! How come he still sees R changing as lift accelerates? Didnt we say the laws of physics do not depend on the frame of reference? ## Only if it is an inertial frame of reference! The accelerating lift is NOT! The Apple & the Moon F mg mr 2 mr (2 / T ) 2 T 2 r / g The radius of the Moons orbit is RM=3.84x108 m. If T = 2[r/g] and g=9.81 m/s2, then the Moons orbital period should be TM = 2[RM/g] = 2[(3.84x108 m)/(9.81 m/s2)] = 3.93 x 104 s = 11 hr. However, the actual orbital period of the Moon is about 27.3 days = 2.36 x 106 s. How could this calculation be so badly off? (Weaker gravity?) Lets use the Moons orbital period and calculate gM, the acceleration due to Earths gravity at the orbit of the Moon. gM = RM(2/T)2 = (3.84x108 m)[/(2.36x106 s)]2 = 2.72 x 10-3 m/s2 But an apple falls at gE = 9.81 m/s2. So lets try something. Well calculate the product gR2 for an apple at the Earths surface and for the Moon in orbit: gMRM2=(2.72x10-3 m/s2)(3.84x108 m)2 = 4.01x1014 m3/s2 gERE2 = (9.81 m/s2)(6.37x106 m)2 = 3.98x1014 m3/s2 These products are essentially equal, because gravity falls off ~ 1/R 2. The same gravitational force law affects the apple and the Moon. The Apple & the Moon F mg mr 2 mr (2 / T ) 2 T 2 r / g The radius of the Moons orbit is RM=3.84x108 m. If T = 2[r/g] and g=9.81 m/s2, then the Moons orbital period should be TM = 2[RM/g] = 2[(3.84x108 m)/(9.81 m/s2)] = 3.93 x 104 s = 11 hr. However, the actual orbital period of the Moon is about 27.3 days = 2.36 x 106 s. How could this calculation be so badly off? (Weaker gravity?) Lets use the Moons orbital period and calculate gM, the acceleration due to Earths gravity at the orbit of the Moon. gM = RM(2/T)2 = (3.84x108 m)[/(2.36x106 s)]2 = 2.72 x 10-3 m/s2 But an apple falls at gE = 9.81 m/s2. So lets try something. Well calculate the product gR2 for an apple at the Earths surface and for the Moon in orbit: gMRM2=(2.72x10-3 m/s2)(3.84x108 m)2 = 4.01x1014 m3/s2 gERE2 = (9.81 m/s2)(6.37x106 m)2 = 3.98x1014 m3/s2 These products are essentially equal, because gravity falls off ~ 1/R 2. The same gravitational force law affects the apple and the Moon. Example: The Total Energy of a Satellite Show that the total energy of a satellite in a circular orbit around the Earth is half of its gravitational potential energy. GM E m mv 2 2 r r v2 GM E r GM E m 1 GM E GM E m E m 2 r r 2r GM E m U Although derived for this particular case, this is a r general result, and is called the Virial Theorem. The 1 E 2U factor of is a consequence of the inverse square law. Gravitational Lensing Light will be bent by any gravitational field; this can be seen when we view a distant galaxy beyond a closer galaxy cluster. This is called gravitational lensing, and many examples have been found. Gravitational Lensing Tides Usually we can treat planets, moons, and stars as though they were point objects, but in fact they are not. When two large objects exert gravitational forces on each other, the force on the near side is larger than the force on the far side, because the near side is closer to the other object. This difference in gravitational force across an object due to its size is called a tidal force. Tides Tidal forces can result in orbital locking, where the moon always has the same face towards the planet as does Earths Moon. If a moon gets too close to a large planet, the tidal forces can be strong enough to tear the moon apart. This occurs inside the Roche limit; closer to the planet we have rings, not moons. Tides ## This figure illustrates a general tidal force on the left, and the result of lunar tidal forces on the Earth on the right.	4,732	15,061	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.34375	4	CC-MAIN-2019-30	latest	en	0.891289
http://www.mathisfunforum.com/viewtopic.php?pid=244562	1,500,583,855,000,000,000	text/html	crawl-data/CC-MAIN-2017-30/segments/1500549423486.26/warc/CC-MAIN-20170720201925-20170720221925-00620.warc.gz	514,699,368	7,044	Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ ° You are not logged in. ## #1 2012-12-19 05:31:42 cooljackiec Member Registered: 2012-12-13 Posts: 185 ### binomial how can we prove 2n choose n is always even? I see you have graph paper. You must be plotting something Offline ## #2 2012-12-19 05:39:09 bob bundy Administrator Registered: 2010-06-20 Posts: 8,053 ### Re: binomial hi cooljackiec Welcome to the forum. I'm not at all clear what you are asking. Please would you say a bit more about this problem. Bob Children are not defined by school ...........The Fonz You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei Offline ## #3 2012-12-19 08:05:52 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial Hi; I like this proof best: In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #4 2012-12-19 10:48:08 cooljackiec Member Registered: 2012-12-13 Posts: 185 ### Re: binomial so because i have a 2 * 2n-1 choose n, it is bound to be even? I see you have graph paper. You must be plotting something Offline ## #5 2012-12-19 10:56:30 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial Hi; Yes, any integer that is multipled by 2 is even. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #6 2012-12-20 06:05:25 cooljackiec Member Registered: 2012-12-13 Posts: 185 ### Re: binomial Another problem: Consider the polynomial What are the coefficients of \$ f(t-1) \$? Enter your answer as an ordered list of four numbers. For example, if your answer were \$ f(t-1) = t^3+3t^2-2t+7 \$, you'd enter (1,3,-2,7). (This is not the actual answer.) I see you have graph paper. You must be plotting something Offline ## #7 2012-12-20 06:26:55 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial Hi; I am getting (1, 3, 3, 1). In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #8 2012-12-20 15:33:22 cooljackiec Member Registered: 2012-12-13 Posts: 185 ### Re: binomial thank you. I have another question: We have 8 pieces of strawberry candy and 7 pieces of pineapple candy. In how many ways can we distribute this candy to 4 kids? and: In how many ways can we distribute 13 pieces of identical candy to 5 kids, if the two youngest kids are twins and insist on receiving an equal number of pieces? and: 4 students are running for club president in a club with 50 members. How many different vote counts are possible, if members may choose not to vote? I see you have graph paper. You must be plotting something Offline ## #9 2012-12-20 16:11:44 noelevans Member Registered: 2012-07-20 Posts: 236 ### Re: binomial I'm getting the same as bobbym. It sometimes helps to write the original function in terms of a different variable than the one you are substituting in its place. For example f(x) = x^3+6x^2+12x+8 = x^3 + 23x^2 + 43x + 8 = (x+2)^3 f(x) = (x+2)^3 so f(t-1) = ((t-1)+2)^3 = (t+1)^3 = t^3+3t^2+3t+1 = (1, 3, 3, 1) by replacing x by t-1 in the f(x) = (x+2)^3. And as another example find the quadruple for f(t-2). f(x) = (x+2)^3 so f(t-2) = ((t-2)+2)^3 = t^3 = (3, 0, 0, 0). These can also be looked at as a composition of functions: f(x)=(x+2)^3 and g(x)=x-1. (fog)(x) = f(g(x)) = f(x-1) = ( (x-1)+2)^3 = (x+1)^3 Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional). LaTex is like painting on many strips of paper and then stacking them to see what picture they make. Offline ## #10 2012-12-20 20:39:17 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial Hi cooljackiec; 4 students are running for club president in a club with 50 members. How many different vote counts are possible, if members may choose not to vote? I am getting 316251. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #11 2012-12-20 21:07:44 anonimnystefy Real Member From: Harlan's World Registered: 2011-05-23 Posts: 16,037 ### Re: binomial On the other hand, I am getting 5^50. Here lies the reader who will never open this book. He is forever dead. Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment The knowledge of some things as a function of age is a delta function. Offline ## #12 2012-12-20 21:39:53 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial 5^50 = 88817841970012523233890533447265625 I bet you did not do a simulation. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #13 2012-12-20 21:42:05 anonimnystefy Real Member From: Harlan's World Registered: 2011-05-23 Posts: 16,037 ### Re: binomial Each member has 5 chiloices. There are 50 members, or 46 if you exclude the ones who are running for president. 5 votes per person, 46 persons, 5^46 possible vote counts... Here lies the reader who will never open this book. He is forever dead. Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment The knowledge of some things as a function of age is a delta function. Offline ## #14 2012-12-20 21:47:03 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial Hi; Not exactly, If one guy gets 20 votes there are only 30 to spread to the others. Remember you are only voting for one position not 4. Take 50 x's and place 3 spacers in various positions. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #15 2012-12-20 22:02:36 anonimnystefy Real Member From: Harlan's World Registered: 2011-05-23 Posts: 16,037 ### Re: binomial Actually, you must have 4 spacers... Your answer is correct. Here lies the reader who will never open this book. He is forever dead. Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment The knowledge of some things as a function of age is a delta function. Offline ## #16 2012-12-20 22:04:25 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial 3 spacers, because you are looking for solutions to the three spacers make 4 separte groups. Each group is how many is in a variable. xxxxxxxxxxxxx _ xxxxxxxxxxxxxxxxx _ xxxxxxxxxx _ xxxxxxxxxx this corresponds to the solution 13 + 17 + 10 + 10 = 50 In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #17 2012-12-20 22:38:00 anonimnystefy Real Member From: Harlan's World Registered: 2011-05-23 Posts: 16,037 ### Re: binomial There are 5 groups, votes for 1st, 2nd, 3rd and 4th candidate and the non-voters... Here lies the reader who will never open this book. He is forever dead. Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment The knowledge of some things as a function of age is a delta function. Offline ## #18 2012-12-20 22:42:48 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial The non voters are not a candidate. They are represented by different values of r. For instance when there is one non voter the equation is In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #19 2012-12-20 22:46:37 anonimnystefy Real Member From: Harlan's World Registered: 2011-05-23 Posts: 16,037 ### Re: binomial It is easier to look at them as a special category of voters: where n are the non-voters. There are then . Here lies the reader who will never open this book. He is forever dead. Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment The knowledge of some things as a function of age is a delta function. Offline ## #20 2012-12-20 22:53:59 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial Hi; That is very good, I did not see that. That would get the same answer and there is less calculation. In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #21 2012-12-21 06:25:31 anonimnystefy Real Member From: Harlan's World Registered: 2011-05-23 Posts: 16,037 ### Re: binomial Exactly! Here lies the reader who will never open this book. He is forever dead. Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment The knowledge of some things as a function of age is a delta function. Offline ## #22 2012-12-21 06:30:30 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial Exactly? There will be times when you will think you or I have found the perfect answer, I assure you these are delusions on your part. - Prof. Kingsfield In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #23 2012-12-21 06:43:26 anonimnystefy Real Member From: Harlan's World Registered: 2011-05-23 Posts: 16,037 ### Re: binomial I never said it was perfect... I only agreed that it is much easier to calculate... Here lies the reader who will never open this book. He is forever dead. Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment The knowledge of some things as a function of age is a delta function. Offline ## #24 2012-12-21 06:46:06 bobbym bumpkin From: Bumpkinland Registered: 2009-04-12 Posts: 109,606 ### Re: binomial True, but there could be an even easier way... In mathematics, you don't understand things. You just get used to them. If it ain't broke, fix it until it is. Always satisfy the Prime Directive of getting the right answer above all else. Offline ## #25 2012-12-22 06:17:17 cooljackiec Member Registered: 2012-12-13 Posts: 185 ### Re: binomial it is wrong. I think that it would be 4^50. Every member has 4 choices. 50 members. But im not sure... I see you have graph paper. You must be plotting something Offline ## Board footer Powered by FluxBB	3,552	11,440	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.546875	4	CC-MAIN-2017-30	latest	en	0.847531
https://hpmuseum.org/forum/thread-10510-post-127937.html	1,643,117,515,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320304835.96/warc/CC-MAIN-20220125130117-20220125160117-00659.warc.gz	346,937,419	8,495	Solving the TVM equation for the interest rate 02-12-2020, 09:54 PM Post: #21 ttw Member Posts: 244 Joined: Jun 2014 RE: Solving the TVM equation for the interest rate I'd like to add that getting these interest rates is more important than most people realize. Of course, most math is more important i everyday life than most people understand for that matter. When comparing two investments or two mortgages or something similar, one should always choose the investment with the best rate or mortgage with the least (if the payments are affordable.) It' isn't (as the radio and TV pseudo financial pseudo advisors sometime say) that one compares the "total of interest payments" or worrying about monthly payments for a purchase no longer owned (or even paying a ballplayer who's been traded.) It's only the interest rate (AKA "rate of return.") A short term mortgage mortgage (15 vs 30) years may have bigger payments but lower interest rates; if one can afford it, the lower rate always give bigger bang for the buck. The same with investments; get the best rate of return. A lower rate on car payments may make longer terms favorable; one may be paying for car (or ball player or house) already traded or sold to achieve a low effective interest rate. Likewise, higher interest means faster retirement fund growth. 02-12-2020, 11:40 PM Post: #22 Albert Chan Senior Member Posts: 1,676 Joined: Jul 2018 RE: Solving the TVM equation for the interest rate (02-12-2020 05:25 PM)Albert Chan Wrote: This formula is based on Pade[1,1], I centered 0, of NFV = F + P + ((1+I)^N-1)(P+M/I) Solving Pade[1,1] approximated NFV = 0, for 1/I : $$\Large {1\over I} ≈ {\binom{N}{3}M + \binom{N}{2}P \over \binom{N}{2}M + \binom{N}{1}P} - {\binom{N}{2}M + \binom{N}{1}P \over F + P + M N}$$ Doing the same example, N=11, F=40000, P=0, M=-2564 1/I ≈ (-423060/-141020) - (-141020/11796) = 44102/2949 I ≈ 2949/44102 ≈ 6.687%, which under-estimated true rate (6.780%) by tiny 0.093% We can reuse the formula, getting a closer estimate. With above I=6.687%, calculated F = 39804.11, error = 40000 - 39804.11 = 195.89 Tried the formula again, with F = 40000 + 195.89, to compensate this error I ≈ 1/(3 - (-141020/(11796 + 195.89))) ≈ 6.775% Again, using this new I, calculated F = 39989.83, error = 40000 - 39989.83 = 10.17 Interpolate for 0 error, I = 6.775% - (6.687% - 6.775%) 10.17/(195.89 - 10.17) = 6.7798% 02-13-2020, 06:00 AM Post: #23 Erwin Member Posts: 159 Joined: May 2015 RE: Solving the TVM equation for the interest rate (04-15-2018 08:48 PM)Carsen Wrote: Problem #1: n=32 PV=-6,000 FV=10,000 PMT=0 i=??? Problem #2: n=36 PV=13,000 PMT=-372.53 FV=0 i=??? Problem #3: n=36 PV=5,750 PMT=-176.89 FV=0 i=??? Problem #4: n=360 PV=75,000 PMT=-425.84 FV-0 i=??? The 12C results is from my 12C bought in 2016. The 15C results is your (Dieter's) program. The Prime's results is using the TVM formula in the 12C manual using your initial estimate formula. Looking at the 15C and the Prime's results, I would say that the estimate works really well. I wonder if there is a even better way to produce a estimate. Hello, cause I was curious I tried the above examples with the finance pac from the HP-71b with the following solutions: Problem #1: i= 1.609139492% same as the 12c Problem #2: i= 0.169257426% Problem #3: i= 0.562601665% Problem #4: i= 0.458330232% So it looks like, that the HP-71b routine for small interest rates use a different method compared to the other calculators. But no idea what calculation method the HP-71b use, there are no hints in the user manual. regards Erwin 02-13-2020, 06:30 AM Post: #24 Gamo Senior Member Posts: 705 Joined: Dec 2016 RE: Solving the TVM equation for the interest rate Albert Chan Thanks for the comment, I found that in the HP-55 Mathematics Programs On the Direct Reduction Loan Interest Rate the program used the Harmonic Mean approach for the Suggested Guess solution but approach later. Thanks Gamo 02-13-2020, 10:42 AM (This post was last modified: 02-13-2020 10:46 AM by Gamo.) Post: #25 Gamo Senior Member Posts: 705 Joined: Dec 2016 RE: Solving the TVM equation for the interest rate From HP-55 Mathematics Programs The Suggested Guess for 1. Direct Reduction Loan Interest Rate with known [n, PV and PMT] 2. Sinking Fund Interest Rate with known [n, FV and PMT] To find the guess the HP-55 programs book used these formulas. 1. Direct Reduction Loan Interest Rate [1 / (PV / PMT)] - [(1 / n^2) x (PV / PMT)] 2. Sinking Fund Interest Rate [((FV/PMT) - n) x 2] ÷ [(n - 1)^2 + (FV/PMT)] Gamo 2/2020 02-17-2020, 10:30 AM Post: #26 Csaba Tizedes Senior Member Posts: 495 Joined: May 2014 RE: Solving the TVM equation for the interest rate (04-15-2018 05:30 PM)Carsen Wrote: (04-15-2018 08:39 AM)Dieter Wrote: So the result seems to depend on previous input ?! I believe you are right Dieter. Try this on your HP-12C emulator. CLEAR FIN n = 10 PMT = -1,000 FV = 10,000 i = 0.001 If you solve for i, the answer is 2.152976E-11, rather than 0 when you leave the value of i at 0. This proves that since we changed the initial value in i from 0 to 0.001, it changed the result. It's almost like it uses the value in the i register as a initial guess. Just as you concluded before. IMHO there is a test for result=0 AND if the test is false the algorithm checks the abs(ch% of result)<=eps%. It seems a more possible explanation. Csaba 12-05-2020, 04:31 PM (This post was last modified: 06-16-2021 02:20 PM by Albert Chan.) Post: #27 Albert Chan Senior Member Posts: 1,676 Joined: Jul 2018 RE: Solving the TVM equation for the interest rate (04-15-2018 08:48 PM)Carsen Wrote: Problem #1: n=32 PV=-6,000 FV=10,000 PMT=0 i=??? Problem #2: n=36 PV=13,000 PMT=-372.53 FV=0 i=??? Problem #3: n=36 PV=5,750 PMT=-176.89 FV=0 i=??? Problem #4: n=360 PV=75,000 PMT=-425.84 FV-0 i=??? I made function guess_i(n, pv, pmt, fv), fitting TVM of i as quadratics lua> guess_i(32,-6000,0,10000) * 100 1.609603176046163 lua> guess_i(36,13000,-372.53,0) * 100 0.16925501753153394 lua> guess_i(36,5750,-176.89,0) * 100 0.5625467055347055 lua> guess_i(360,75000,-425.84,0) * 100 0.4601135680666762 Car leasing example, taken from Fun math algorithms FV is negative, because we are returning the car, at the end of the lease. lua> guess_i(36, 30000, -550, -15000) * 1200 6.9657545218584485 Using HP12C, actual APR = 6.966087383% Update: numbers adjusted with updated guess_i() « Next Oldest \| Next Newest » User(s) browsing this thread: 1 Guest(s)	2,151	6,524	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.703125	4	CC-MAIN-2022-05	latest	en	0.906432
https://www.get-coupon-codes.info/discount-formula-in-maths/	1,627,468,955,000,000,000	text/html	crawl-data/CC-MAIN-2021-31/segments/1627046153709.26/warc/CC-MAIN-20210728092200-20210728122200-00441.warc.gz	811,511,662	14,657	# Discount Formula In Maths Filter Type: ### Discount Formula and Discount Percentage Formula with Examples (5 days ago) The discount equals the difference between the price paid for and it’s par value. Discount is a kind of reduction or deduction in the cost price of a product. It is mostly used in consumer transactions, where people are provided with discounts on various products. The discount rate is given in percentage. https://byjus.com/discount-formula/ Category: coupon codes ### Discount Formula- Explanation, Solved Examples and FAQs (8 days ago) The discount rate formula can either be extracted by subtracting the selling price of the product from its marked price or by multiplying the discount rate offered and the marked price of the product. In terms of Mathematics, the formula for discount is represented as below, Discount = Marked Price – … https://www.vedantu.com/formula/discount-formula Category: coupon codes ### Discount Rates - Formulas, Definition and Example Questions (8 days ago) Discount rate DR = pr = DR = 50000 x 20/100 = 20100 in the year 2000 = 10000 Discount rate DR = 1000 dollars in the year 2000. So, the discount amount is 10000 dollars. Discount rate DR = 5000030/100 in 2004th year https://byjus.com/maths/discount-rate/ Category: coupon codes ### Discount Formula - What is Discount Formula?, Examples (1 days ago) Meaning of Discount Formula Discount is the kind of deduction in the selling price of a product when a seller proposes a discount in order to boost a sale of different products. The discount formula helps in calculating the money a buyer needs to pay to the seller. The discount formula can be calculated in percentage as well. https://www.cuemath.com/discount-formula/ Category: coupon codes ### Step by Step Math Lesson on How To Calculate Discount and (4 days ago) Typically, a store will discount an item by a percent of the original price. The rate of discount is usually given as a percent, but may also be given as a fraction. The phrases used for discounted items include, " off," "Save 50%," and "Get a 20% discount." https://www.mathgoodies.com/lessons/percent/sale_price Category: coupon codes ### Discount - Math Formulas - Mathematics Formulas - Basic (6 days ago) Mathematics Formula. Enter Keyword example (area, degree) Formulae » commercial mathematics » percentages profit loss and discount » discount. Register For Free Maths Exam Preparation. https://www.pioneermathematics.com/discount-formula.html Category: coupon codes ### Discounts - Discount rates, Formulas, Types (Just Now) The amount of money that is reduced from the list price of an item is called the discount. The percentage of this discount on the list price is called the discount rate. The formula to calculate the discount rate is: Discount % = (Discount/List Price) × 100. https://www.cuemath.com/commercial-math/discounts/ Category: coupon codes ### How To Calculate Percentage Discount (%) - How To Calculate (Just Now) Discount Amount = Original Price – New Price After Discount = 1000 – 800 = 200 Now that we have the discount amount and the Original price, we can just feed the values into out formula to calculate the percentage discount. https://www.learntocalculate.com/how-to-calculate-percentage-discount/ Category: coupon codes ### 6.1: Simple Interest and Discount - Mathematics LibreTexts (1 days ago) If an amount M is borrowed for a time t at a discount rate of r per year, then the discount D is (6.1.3) D = M ⋅ r ⋅ t The proceeds P, the actual amount the … https://math.libretexts.org/Bookshelves/Applied_Mathematics/Applied_Finite_Mathematics_(Sekhon_and_Bloom)/06%3A_Mathematics_of_Finance/6.01%3A_Simple_Interest_and_Discount Category: coupon codes ### Basic Math Formulas - Basic-mathematics.com (8 days ago) Consumer math formulas: Discount = list price × discount rate Sale price = list price − discount Discount rate = discount ÷ list price Sales tax = price of item × tax rate Interest = principal × rate of interest × time Tips = cost of meals × tip rate Commission = cost of service × commission rate Geometry formulas: Perimeter: https://www.basic-mathematics.com/basic-math-formulas.html Category: coupon codes ### Discount – Types, Formula, Examples - CCSS Math Answers (7 days ago) Discount on the table fan = 5% We know that Discount = 5% of Marked Price Discount = 5% of $750 = 750 = 750 * 0.05 = 37.5 Discount =$37.5 Category: coupon codes (2 days ago) Discount. A reduction in price. Here the discount is $2. Sometimes discounts are in percent, such as a 10% discount, and then you need to do a calculation to find the price reduction. See: Percent. https://www.mathsisfun.com/definitions/discount.html Category: coupon codes ### Discounted Cash Flow DCF Formula - Calculate NPV CFI (6 days ago) It's important to understand exactly how the NPV formula works in Excel and the math behind it. NPV = F / [ (1 + r)^n ] where, PV = Present Value, F = Future payment (cash flow), r = Discount rate, n = the number of periods in the future. . If we break the term NPV we can see why this is the case: https://corporatefinanceinstitute.com/resources/knowledge/valuation/dcf-formula-guide/ Category: coupon codes ### Discount Rate - Definition, Uses, Types, Formula & Examples (3 days ago) In mathematics, discount rate problems can be solved by using the simple discount rate formulas. Formula 1. Discount = Marked price of commodity – selling price of commodity D = MP - SP https://www.vedantu.com/maths/discount-rate Category: coupon codes ### Discount Formula: Concept, Various Formulas, Solved Examples (1 days ago) Maths Formulas Discount Formula. Calculating the discount is one of the most useful mathematical skills. We will apply it to give the tips at a restaurant, and sales in stores, and also set rates for own services. The basic way to calculate a discount is to multiply the original price by the decimal form of the given percentage rate. https://www.toppr.com/guides/maths-formulas/discount-formula/ Category: coupon codes ### Profit Loss Discount Formulas, Tricks with Examples - EduDose (7 days ago) In successive discounts, first discount is subtracted from the marked price to get net price after the first discount. Taking this price as the new marked price, the second discount is calculated and it is subtracted from it to get net price after the second discount. Continuing in … https://www.edudose.com/maths/profit-loss-discount-formulas-tricks/ Category: coupon codes ### Definition and examples discount define discount (5 days ago) More About Discount. Regular price minus Sale price gives the amount of discount. If the discount is given in percent, then the amount of discount can be found by using the formula, Amount of Discount = Regular Price × Rate of Discount. Example of Discount. Charlie bought a DVD player, which costs$270. But he was given a discount of $30. http://www.icoachmath.com/math_dictionary/Discount.html Category: coupon codes ### True Discount Formula for Competitive Exams (7 days ago) True Discount Formula Questions Suppose a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum. Clearly, Rs. 100 at 14% will amount to Rs. 156 in 4 years. https://www.examsbook.com/true-discount-formula Category: coupon codes ### Percentage Discounts Passy's World of Mathematics (7 days ago) We can calculate the discount as follows: Discount = 15% of Marked Price. (remember “%” means / 100 and “of” means multiply) = (Discount % / 100) x Marked Price. = 15 / 100 x$30. = 15 divided by 100 x $30. =$4.50. So the dollar saving we will make during the Sale is $4.50. For calculating the Discount$, the Algebra formula is as follows: https://passyworldofmathematics.com/percentage-discounts/ Category: coupon codes ### Discount Formula - MathHelp.com - Pre Algebra Help - YouTube (3 days ago) For a complete lesson on the discount formula, go to https://www.MathHelp.com - 1000+ online math lessons featuring a personal math teacher inside every less Category: coupon codes ### Calculating a 20 Percent Discount: How-to & Steps - Video (4 days ago) The first step in calculating a 20% discount is to convert the percentage to a decimal. To do this, divide the percentage by 100. In this case, the percentage is 20, so: 20 / 100 = 0.2. Another Category: coupon codes ### Discount Formula MathHelp.com - YouTube (3 days ago) Need a custom math course? Visit https://www.MathHelp.com.This lesson covers the discount formula. Students solve word problems using the "discount" formula, Category: coupon codes (Just Now) Sale price = 100% of regular price – 30% of regular price. Now subtract percentages: Sale price = (100% – 30%) of regular price = 70% of regular price. At this point, you can fill in the details, as follows: Sale price = 0.7 x $2,100 =$1,470. Thus, the television will cost $1,470 with the discount. Related Articles. https://www.dummies.com/education/math/pre-algebra/how-to-calculate-a-percentage-discount/ Category: coupon codes ### Discount and Commission: Definition, Formulas, Videos (1 days ago) A commission is paid to the salesperson for his/her services by the seller with respect to the M.P. For example, if Radhika is a salesperson and works in a garment shop and takes 2% commission. It means from shop owner Radhika take 2Rs for selling of 100Rs (M.P.). In this case, the shop owner is sharing his revenue, thus his profit is reduced. https://www.toppr.com/guides/maths/compairing-quantities/discount-and-commission/ Category: coupon codes ### Discount Calculator - Math, Fitness, Finance, Science (1 days ago) For example, if a good costs$45, with a 10% discount, the final price would be calculated by subtracting 10% of $45, from$45, or equivalently, calculating 90% of $45: 10% of$45 = 0.10 × 45 = $4.50. 90% of$45 = 0.90 × 45 = $40.50. In this example, you are saving 10%, or$4.50. A fixed amount off of a price refers to subtracting whatever https://www.calculator.net/discount-calculator.html Category: coupon codes ### Discount Formula, Discount Factor, Discount Rate (5 days ago) Calculation of discount is one of most proficient mathematical skills that you can learn. You can apply it to the restaurants, shopping malls, and setting rates for your products. The actual way to calculate discount is multiplying original price with the discount percent rate. https://www.andlearning.org/discount-formula/ Category: coupon codes (2 days ago) Using the formula one and replacing the given values: Amount Saved = Original Price x Discount % / 100. So, Amount Saved = 100 x 10 / 100. Amount Saved = 1000 / 100. Amount Saved = $10 (answer) In other words, a 10% discount for an item with original price of$100 is equal to $10 (Amount Saved). Note that to find the amount saved, just multiply https://coolconversion.com/math/discount-calculator/ Category: coupon codes ### Discounting Formula Steps to Calculate Discounted Value (1 days ago) Formula to Calculate Discounted Values. Discounting refers to adjusting the future cash flows to calculate the present value of cash flows and adjusted for compounding where the discounting formula is one plus discount https://www.wallstreetmojo.com/discounting-formula/ Category: coupon codes ### Profit & Loss Basic Concepts, Questions, Formula By Arun (5 days ago) Marked Price Formula (MP) This is basically labelled by shopkeepers to offer a discount to the customers in such a way that, Discount = Marked Price Arun Singh Rawat is a premier teacher of maths for banking exam like IBPS, SBI, RBI & Insurance Exam. He has also trained students of Railways, SSC & CAT Exams in PAN India. https://mathsbyarunsir.com/2021/07/22/profit-loss-basic-concepts-questions-formula-by-arun-sir/ Category: Insurance ### Percent Applications - Discount - Math Help (2 days ago) 5th Grade Math 6th Grade Math Pre-Algebra Algebra 1 Geometry Algebra 2 College Students solve word problems using the "discount" formula, which states: original price x rate of discount = discount. https://www.mathhelp.com/how_to/percent_applications/discount/ Category: coupon codes ### Maths Formulas List of Basic 1300 Maths Formulas PDF (1 days ago) Maths Formulas Sometimes, Math is Fun and sometimes it could be a surprising fact too. In our routine life, you can check the best route to your school, you can check where more discounted products are available in the market, and you can check which bank can offer the superior interests. This is all about […] https://www.andlearning.org/math-formula/ Category: coupon codes ### Discount Mathematics - eXcel eXchange (5 days ago) Discount Mathematics. Download my loan repayment calculator. The basic principles of compounding and discounting. The basic principle of compounding is that if we invest £X now for n years at r% interest per annum, we should obtain £X (1+r) n in n years time.. The value, V, attained by a single sum X, after n periods at r% is V = X (1+r) n If we invest £10,000 now for five years at 10% http://www.excelexchange.com/discount%20mathematics.htm Category: coupon codes ### Cash Discount: Definition, Formula & Example - Video (8 days ago) Formula. You really can't get a simpler formula than a cash discount. Here it is: Cash discount = purchase price * discount rate. The discount https://study.com/academy/lesson/cash-discount-definition-formula-example.html Category: coupon codes ### Discount Calculator - Find Out the Sale Price (7 days ago) Just follow these few simple steps: Find the original price (for example$90) Get the the discount percentage (for example 20%) Calculate the savings: 20% of $90 =$18. Subtract the savings from the original price to get the sale price: $90 -$18 = $72. You're all set! https://www.omnicalculator.com/finance/discount Category: coupon codes ### Percent Change & Discounts - Comparing Quantities Class (Just Now) Percent Change & Discounts – Comparing Quantities Class 8 Maths. Percentage change refers to the concept of variation in the measurement basis of an item, that is by what quantity the net worth of an article increased or decreased. The change in percentage occurs when the value of a commodity change, that is, increased or decreased by some https://www.geeksforgeeks.org/percent-change-discounts-comparing-quantities-class-8-maths/ Category: coupon codes ### Simple Discount - Basics (6 days ago) Simple Discount - Basics In the Simple discount situation, there is an amount of money (future value) due on a certain future date, usually within a year; the debtor can ask for paying in advance and, if the creditor agrees with him, the money to be paid today (present value) is less than the due capital; in fact the future value is subtracted by the discount calculated in proportion to time http://www.digiovinehost.com/itcgcalamandrei/spazio_docenti/ragazzoni/clil/economia%20aziendale/interest-discount/discount.htm Category: Credit ### NPV Formula - Learn How Net Present Value Really Works (Just Now) A guide to the NPV formula in Excel when performing financial analysis. It's important to understand exactly how the NPV formula works in Excel and the math behind it. NPV = F / [ (1 + r)^n ] where, PV = Present Value, F = Future payment (cash flow), r = Discount rate, n = the number of periods in the future https://corporatefinanceinstitute.com/resources/knowledge/valuation/npv-formula/ Category: coupon codes ### Discount Calculator (6 days ago) Sample Discount Percentage Calculation. James bought a vintage lava lamp at a sale price of$89.63. The original price was $165.99. What was the percentage discount on the original price of the lava lamp? Using the formula above, list price = L = 165.99, and price sale = P = 89.63. https://www.calculatorsoup.com/calculators/financial/discount-calculator.php Category: coupon codes ### How to Find the Original Price of a Discount - Sciencing (6 days ago) The calculations include a simple formula that divides the sale price by the result of 1 minus the discount in percentage form. Use this formula to calculate the original or list price of an item. \text{OP} = \frac{ \text{ Price}}{1 - \text{ Discount}} https://sciencing.com/how-to-find-the-original-price-13712233.html Category: coupon codes ### Excel formula: Get percentage discount Exceljet (2 days ago) Summary. To calculate the percentage discount from an original price and a sale price, you can use a formula that divides the difference by the original price. In the example shown, the formula in E5, copied down, is: = ( C5 - D5) / C5. The result is a decimal value with percentage number format applied. https://exceljet.net/formula/get-percentage-discount Category: coupon codes ### Successive Discount Formula, Tips, Tricks and Quick Way to (Just Now) The formula for total discount in case of successive-discounts: If the first discount is x% and 2nd discount is y% then Successive Discount Formula – Total discount = ( x + y – xy /100)% Example: The marked price of a shirt is Rs.1000. A shopkeeper offers 10% discount on this shirt and then again offers 20% discount on the new price. https://prepinsta.com/successive-discount-formula/ Category: coupon codes ### 6.3: Solve Sales Tax, Commission, and Discount (6 days ago) The discount is what percent of the original price? Translate into an equation. Divide. $$\dfrac{17.05}{31} = \dfrac{r(31)}{31}$$ Simplify. 0.55 = r: Check if this answer is reasonable. The rate of discount was a little more than 50% and the amount of discount is a little more than half of$31. Write a complete sentence that answers the question. https://math.libretexts.org/Bookshelves/PreAlgebra/Book%3A_Prealgebra_(OpenStax)/06%3A_Percents/6.03%3A_Solve_Sales_Tax_Commission_and_Discount_Applications Category: coupon codes ### Maths Formulas For Class 8 CBSE: Important Maths Formulas (6 days ago) Find the important Class 8 Maths formulas for Data Handling and Probability. A class interval is the specific range of numbers such as 10-20, 20-30, 30-40, and so forth. For a Class Interval of 10-20, Lower Class Limit = 10 and Upper-Class Limit = 20. Frequency is … https://www.embibe.com/exams/maths-formulas-for-class-8/ Category: coupon codes Filter Type: ### FAQ? #### How to do discounts in math? How to Calculate a Discount Method 1 of 3: Calculating the Discount and Sale Price. Convert the percentage discount to a decimal. ... Method 2 of 3: Estimating the Discount and Sale Price. Round the original price to the nearest ten. Use normal rounding rules to round up or down. ... Method 3 of 3: Completing Sample Problems. Calculate the exact sale price. ... #### What is the formula of the discount equation? The formula for discount can either be derived by deducting the selling price of the product from its listed price or by multiplying the offered discount rate and the listed price of the product. Mathematically, the discount is represented as below, Discount = Listed Price - Selling Price #### What is the Excel formula for discount? Formula to find out the discount value. There are several ways of discovering a discount percentage for any value but the most simple is: discounted value = (discount percentage * total value) / 100. #### How to calculate 10% discount? How do I calculate a 10% discount? • Take the original price. • Divide the original price by 100 and times it by 10. • Alternatively, move the decimal one place to the left. • Minus this new number from the original one. • This will give you the discounted value. • Spend the money you've saved! % Off: \$ Off:	4,685	19,587	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.90625	4	CC-MAIN-2021-31	latest	en	0.886507
https://www.physicsforums.com/threads/simple-ode-but-how-to-approach.75286/	1,716,603,176,000,000,000	text/html	crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00298.warc.gz	808,317,480	20,194	# Simple ODE, but how to approach? • BlkDaemon In summary: Just substitute in the dy for y' and you're good to go. In summary, Daniel is explaining that the general solution is based upon the substitution of y'dt for y' and that there is no need to worry about the dt. BlkDaemon I'm creeping my way through DiffEq, and recently started reading Paul Dawkins' PDF, which is actually pretty helpful. He does, however, tend to assume that his readers know how to approach what he calls simple problems. Well, one of 'em has me stumped. 2ty' + 4y = 3 I need to find the general solution. Here's what I tried: 1) I tried isolating y. I get y=(3 - 2ty')/4. That doesn't work because I've got y' as part of the solution. Eeek. 2) I tried isolating y'. That didn't work either, for the same reason. I got y' = (3-4y)/2t But if I take the derivative of that, I get weirdness. Now, it's really possible that I'm just calculating the derivative incorrectly. I also can't figure out how to ditch the "t", but when I look at the answer, there's t, as well as C, so I know that (because of the arbitrary constant) there's a derivative involved, and that means the "t" isn't going anywhere. Anyone have any suggestions on how to work this? I don't need the answer. I need to know how to approach the problem. 2ty' + 4y = 3 2ty' = 3 - 4y y'/(3 - 4y) = 1/(2t) dy/(3 - 4y) = dt/2t -0.25ln(3 - 4y) = 0.5ln(t) + C ln(3 - 4y) = -2ln(t) + D 3 - 4y = E/t² -4y = E/t² - 3 y = F/t² + 0.75 This is the general solution. It just uses separation of variables. F varies over R, and given an initial value, you can find the particular solution by solving for F. Last edited: Thanks for the response. Now I see where I missed a step! Wait, I'm confused again... Essentially, y' is dy. I get that. But where does dt come from? This has me stumped. If I interpret the change from y' to dy as a simple change in notation, I can't just pull dt out of thin air. And when I perform that step, I get: dy/(3-4y) = 1/2t which yields: (-ln \|4y-3\|)/4 = (t^2)/4 + C Right? Then the 4s cancel out, and I'm left with - ln \|4y-3\| = t^2 + C But this isn't right. Where am I missing it? I realize this is a pretty simple separable equation, but I'm missing some crucial steps that are interfering with my understanding of the overall concept. Hopefully that makes sense. Daemon Nope.Rewrite it and take care with that "dt" and that "t". $$\frac{dy}{3-4y}=\frac{dt}{2t}$$ Daniel. BikDamon: The trick is based upon use of the integration technique "substitution", which is, essentially mirrors the chain rule of differentiation. Suppose you can write your differential equation as: $$f(y(t))\frac{dy}{dt}=g(t)$$ Now, integrate this between arbitrary instants "0" and "t": $$\int_{0}^{t}f(y(\tau))\frac{dy}{d\tau}d\tau=\int_{0}^{t}g(\tau)d\tau$$ where I've introduced the dummy variable $$\tau$$ for clarity (and rigour). Furthermore assume it exists a function F(x), so that F'(x)=f(x). We may then rewrite our equation as: $$\int_{0}^{t}\frac{dF}{dx}\mid_{x=y(\tau)}\frac{dy}{d\tau}d\tau=\int_{0}^{t}g(\tau)d\tau$$ That is, we may write this as: $$\int_{0}^{t}(\frac{d}{d\tau}F(y(\tau)))d\tau=\int_{0}^{t}g(\tau)d\tau$$ by the chain rule of differentiation. But the left hand side equals now by FOTC: $$\int_{0}^{t}(\frac{d}{d\tau}F(y(\tau)))d\tau=F(y(t))-F(y(0))$$ which equals, with $$y(0)=y_{0}, y(t)=y: [tex]F(y(t))-F(y(0))=\int_{y_{0}}^{y}\frac{dF}{dy}dy=\int_{y_{0}}^{y}f(y)dy$$ Thus, we have gained the equality: $$\int_{y_{0}}^{y}f(y)dy=\int_{0}^{t}g(\tau)d\tau$$ I get the theory... I understand the theory in the last response, but I got lost on the actual application. Essentially, once I replace y' with dy, I arbitrarily assign dt to the right side of the equation, yes? $$y'\equiv \frac{dy}{dt}$$.You could (even though it's mathematically doubtful) move around those differentials (divide by them,multiply by them). Daniel. You do NOT replace y' with dy; you replace y'dt with dy ! And one more thing:nothing you do here is arbitrary.Only integration constants have that privilage. Daniel. Last edited: And, read again my prior response. Make sure you undestand it properly. I think part of my challenge is that the originally stated problem does not include "dy" or "dt" and therefore working towards the solution has become a bit convoluted for me. The original problem only says: 2ty' + 4y = 3 Do you guys even see where I'm getting lost? If not, I'll move on to the next problem! so the problem doesn't specify that y is a function of t? well that sure can be misleading. BlkDaemon said: I think part of my challenge is that the originally stated problem does not include "dy" or "dt" and therefore working towards the solution has become a bit convoluted for me. The original problem only says: 2ty' + 4y = 3 Do you guys even see where I'm getting lost? If not, I'll move on to the next problem! You're getting lost at this step $$y'\equiv \frac{dy}{dx}$$,which is really sad. Daniel. BlkDaemon said: Thanks for clearing that up. End of discussion, since "that's sad" loosely translates as "too stupid to ask questions here". Thanks for your help, and I'll research my questions before posting in the future. I believe that what Daniel refers as "that's sad" is the fact that you should know Leibniz notarion by now, way before taking a course on differential equations. Gotcha. Actually, I have worked with the notation before. What I didn't realize in the problem was that the notation was implied. The problem came from a summary of term definitions in a larger document, and I'm working my way through it. I'm not trying to belabour the issue; I just wanted to make sure that I understood exactly what was going on. Thanks. It's perfectly common in the beginning to be confused by different notations. Hopefully, things have cleared out now. ## 1. What is a simple ODE? A simple ODE, or ordinary differential equation, is a mathematical equation that relates a function to its derivatives. It is commonly used to model physical systems in science and engineering, and can be solved using analytical or numerical methods. ## 2. How do I approach solving a simple ODE? The approach to solving a simple ODE depends on the specific equation and initial conditions. Generally, it involves identifying the order of the equation, finding a suitable method for solving it (e.g. separation of variables, substitution, etc.), and applying the initial conditions to find the final solution. ## 3. What is the difference between an initial value problem and a boundary value problem for simple ODEs? An initial value problem for a simple ODE involves finding the solution for a given function and its initial conditions at a single point. A boundary value problem involves finding the solution for a given function and its boundary conditions over a range of values. ## 4. Can a simple ODE have multiple solutions? Yes, a simple ODE can have multiple solutions. This is known as the existence of "general solutions," where a single equation can have multiple solutions that satisfy the same initial conditions. In some cases, the equation may also have multiple solutions for different initial conditions. ## 5. What are the applications of simple ODEs in science? Simple ODEs are widely used in various fields of science, including physics, chemistry, biology, and engineering. They are used to model and understand the behavior of physical systems, such as population growth, chemical reactions, and motion of objects. They also play a crucial role in developing scientific theories and making predictions about real-world phenomena. • Differential Equations Replies 5 Views 1K • Differential Equations Replies 4 Views 2K • Differential Equations Replies 4 Views 2K • Differential Equations Replies 2 Views 2K • Differential Equations Replies 3 Views 7K • Differential Equations Replies 3 Views 2K • Differential Equations Replies 2 Views 1K • MATLAB, Maple, Mathematica, LaTeX Replies 1 Views 1K • Differential Equations Replies 2 Views 1K • Differential Equations Replies 8 Views 1K	2,221	8,118	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2024-22	latest	en	0.965772
https://pdfslide.net/document/-evaluate-solve-daily-agenda.html	1,696,400,647,000,000,000	text/html	crawl-data/CC-MAIN-2023-40/segments/1695233511361.38/warc/CC-MAIN-20231004052258-20231004082258-00201.warc.gz	493,688,351	41,162	of 114 /114 Evaluate: 9 25 Solve: ݔ=9 ݔ1 = 64 Daily Agenda: Grade assignments 7.1 notes / assignment # • Evaluate: • Solve: • Daily Agenda: Embed Size (px) ### Text of • Evaluate: • Solve: • Daily Agenda: • Evaluate:– 9 – 25• Solve: – 푥 = 9– 푥 − 1 = 64 • Daily Agenda:– Grade assignments– 7.1 notes / assignment Chapter 7 Chapter 7.1 Nth Roots and Rational Exponents • To evaluate nth roots of real numbers using both radical notation and rational exponent notation. • To use nth roots to solve real-life problems. • Why is 3 = 9?• What is the 9? • Why is 4 = 64?• What is the 64? • Why is 3 = 81?• What is the 81? • For an integer 푛 greater than 1– If 푏 = 푎, then b is the nth root of a. • Or 푎 = 푏• 푛 is the index Relating Indices and Powers • 푎 ∙ 푎 = 푎 • 푎 ∙ 푎 = 푎 • Let푛be an integer greater than 1 and let 푎 be a real number.– If 푛 is odd, then 푎 has one real nth root: 푎 = 푎 ⁄ – If 푛 is even and 푎 > 0, then 푎 has two real nth roots: ± 푎 = ±푎 ⁄ – If 푛 is even and 푎 = 0, then 푎 has one nth root: 0 = 0 ⁄ = 0 – If 푛 is even and 푎 < 0, then 푎 has no real nth roots. • Find the indicated nth root(s) of 푎.– n = 5, a = -32 – n = 3, a = 64 – n = 4, a = 625 – n = 3, a = -27 • Let 푎 ⁄ be an nth root of a, and let m be a positive integer. – 푎 ⁄ = 푎 ⁄ = 푎 – 푎 ⁄ = ⁄ = = • Evaluating Expressions with Rational Exponents– 16 ⁄ – 64 ⁄ – 49 ⁄ – 16 ⁄ • Use a calculator to approximate:– 3 – 3 • Get 푥 by itself and then take the nth root of each side! • Solve each equation.– 6푥 = 3750 – 푥 + 1 = 18 • Solve each equation.– 5푦 = 80 – 푦 − 1 = 32 • The rate 푟 at which an initial deposit 푃 will grow to a balance 퐴 in푡years with interest compounded n times a year is given by the formula 푟 = 푛⁄ − 1 . Find 푟 if 푃 = \$1000, 퐴 = \$2000, 푡 = 11 years, and 푛 = 12. • A basketball has a volume of about 455.6 cubic inches. The formula for the volume of a basketball is 푉 = 4.18879푟3. Find the radius of the basketball. • P404: 14 – 36 even, 42 – 51 by 3s, 54 – 60 even, 64 • Evaluate the expression.– 4 ⋅ 4 – 2– 2 ⋅ 3 • Daily Agenda– Grade Assignment– 7.2 notes / assignment Chapter 7.2 Properties of Rational Exponents • To use properties of rational exponents to evaluate and simplify expressions. • To use properties of rational exponents to solve real-life problems. • Properties from 6.1 can be applied to rational exponents Let a and b be real numbers and let m and n be integers Product of Powers Property 푎 ⋅ 푎 = 푎Power of a Power Property 푎 = 푎Power of a Product Property 푎푏 = 푎 푏Negative Exponent Property 푎 = 1푎 1푎 = 푎 Zero Exponent Property 푎 = 1Quotient of Powers Property 푎 푎 = 푎 Power of a Quotient Property 푎푏 = 푎푏 • Simplify.– 6 ⁄ ∙ 6 ⁄ – 27 ⁄ ∙ 6 ⁄ – 4 ∙ 2 ⁄ • Simplify.– 3 ⁄ ⋅ 3 ⁄ – 64 ⁄ ⋅ 8 ⁄ – 3 ⋅ 6 ⁄ • Simplify.– 25 ∙ 5 • Simplify.– 27 ⋅ 3 • Checklist for writing radicals in simplest form:– Apply properties of radicals– Remove any perfect nth powers– Rationalize any denominators • Write in simplest form.– 64 – Like radicals have the same index and the same radicand.– Combine them just like variables. • Perform the indicated operation.– 5 4 ⁄ − 3 4 ⁄ – 81 − 3 • Perform the indicated operation.– 6 3 ⁄ + 4 3 ⁄ – 625 − 5 • Because a variable can be positive, negative, or zero…– 푥 = 푥 when 푛 is odd– 푥 = 푥 when 푛 is even • So absolute values are needed when simplifying variable expressions. • For simplicity sake, we will assume all variables are positive! • Simplify the expression. Assume all variables are positive.– 27푧 – 16푔 ℎ ⁄ –⁄ • Write the expression in simplest form. Assume all variables are positive.– 12푑 e 푓 • Simplify the expression. Assume all variables are positive.– 8푟 푠 푡 – 625푗 푘 ⁄ –⁄ • Perform the indicated operations. Assume all variables are positive.– 8 푥 − 3 푥 – 3푔ℎ ⁄ − 6푔ℎ ⁄ – 2 6푥 + 푥 6푥 • Simplify the expression. Assume all variables are positive.– 6 푠 − 2 푠 + 푠 – 4푚 푛 ⁄ − 6푚 푛 ⁄ – 3 6푦 + 2푦 6푦 • The weight W in tons of a whale as a function of length L in feet can be approximated by the mode 푊 = 0.1077퐿3/2. Approximate the weight of a humpback whale with length 49.17 feet. • A pilot whale is about ⁄ the length of a blue whale. Is its weight also ⁄ the weight of the blue whale? • P411: 24 – 90 x 3s, 96 • Simplify:– 4 푥 + 1– 푥 − 2 • Daily Agenda:– Grade assignment– 7.3 notes / assignment– 7.4 notes / assignment – Quiz Monday! Chapter 7.3 Power Functions and Function Operations • To perform operations with function including power functions. • To use power functions and function operations to solve real-life problems. • Four basic operations with polynomial functions: • Domain of h consists of the x-values in the domains of both f and g.– Domain of quotient does not include x-values when 푔(푥) = 0 Operation Definitions Example: f(x)=2x, g(x)=x+1 Addition ℎ(푥) = 푓(푥) + 푔(푥) ℎ(푥) = 2푥 + (푥 + 1) = 3푥 + 1Subtraction ℎ(푥) = 푓(푥) − 푔(푥) ℎ(푥) = 2푥 − (푥 + 1) = 푥 − 1Multiplication ℎ(푥) = 푓(푥)푔(푥) ℎ(푥) = (2푥)(푥 + 1) = 2푥2 + 2푥Division ℎ(푥) = 푓(푥)/푔(푥) ℎ(푥) = 2푥푥 + 1 • A function of the form 푦 = 푎푥푏– a is a real number– b is a rational number • Let 푓(푥) = 3푥 ⁄ and 푔(푥) = 2푥 ⁄ . Find the following:– The sum – The difference – The domains • Let 푓(푥) = 4푥 ⁄ and 푔(푥) = 푥 ⁄ . Find the following:– The product – The quotient – The domains • The composition of the function 푓 with the function 푔 is:– ℎ(푥) = 푓(푔(푥))– The domain of h is the set of all x-values such that x is in the domain of 푔 and 푔(푥)is in the domain of 푓– Need to pay attention to the order of functions when they are composed! • Let 푓(푥) = 2푥 and 푔(푥) = 푥2 − 1. find the following:– 푓(푔(푥)) – 푔(푓(푥)) – 푓(푓(푥)) – The domains • Let 푓(푥) = 푥 and 푔(푥) = 푥 + 1. Find the following:– 푓(푔(푥)) – 푔(푓(푥)) – 푓(푓(푥)) – The domains • You do an experiment on bacteria and find that the growth rate G of the bacteria can be modeled by 퐺 푡 = 82푡 . , and that the death rate D is 퐷(푡) = 10.8푡 . , where t is the time in hours. Find an expression for the number N of bacteria living at time t. • A computer catalog offers computers at a savings of 15% off the retail price. At the end of the month, it offers an additional 10% off its own price.– Use composition of function to find the total percent discount. – What would be the sale price of a \$899 computer? • P418: 12 – 51 x 3s, 54 – 56 Chapter 7.4 Inverse Functions • To find inverses of linear functions.• To find inverses of nonlinear functions. • P421 • An Inverse Function maps the output values back to their original input values– So the range becomes the domain and the domain becomes the range (or output to input and vice versa)– In other words, the x and y trade places– The graphs are always reflected over the line 푦 = 푥. • Find the inverse of 푦 = −3푥 + 6. • Function f and g are inverses of each other provided:– 푓(푔(푥)) = 푥– 푔(푓(푥)) = 푥 • The function g is denoted by 푓 , read as “f inverse” • Verify that 푓(푥) = −3푥 + 6and 푓 (푥) = − 푥 +2are inverses. • Find the inverse of 푦 = 푥 − 1 • A model for a salary is 푆 = 10.50ℎ + 50, where S is the total salary (in dollars) for one week and h is the number of hours worked.– Find the inverse function for the model. – If a person’s salary is \$533, how many hours does the person work? • A model for a telephone bill is 푇 = 0.05푚 + 29.95,where T is the total bill, and m is the number of minutes used.– Find the inverse model. – If the total bill is \$54.15, how many minutes were used? • Find the inverse of the function 푓(푥) = 푥 • Find the inverse of the function 푓(푥) = 푥 , 푥 ≥ 0 • If no horizontal line intersects the graph of a function f more than once, then the inverse of f is itself a function. • Consider the function 푓(푥) = 2푥 − 4. Determine whether the inverse of f is a function and then find the inverse. • The volume of a sphere is given by 푉 = 휋푟3, where V is the volume and r is the radius. Write the inverse function that gives the radius as a function of the volume. Then determine the radius of a volleyball given that its volume is about 293 cubic inches. • The volume of a cylinder with height 10 feet is given by 푉 = 10휋푟2, where V is the volume, and r is the radius. Write the inverse model that gives the radius as a function of the volume. Then determine the radius of a cylinder given that its volume is 1050 cubic feet. • P426: 15 – 30 x 3s, 36 – 51 x 3s, 58, 62 • Graph 푦 = −2푥 + 3 • Daily Agenda:– Grade homework– Go over quiz…– 7.5 notes / assignment– 7.6 notes / assignment Chapter 7.5 Graphing Square root and Cube Root Functions • To graph square root and cube root functions.• To use square root and cube root functions to find real-life quantities. • Look at P431 and find tables for x = 0, 1 for the square roots and tables for x = -1, 0, 1 for the cube roots. • Homework: P431 Activity ↑, P414: Quiz 1 and P429: Quiz 2 • Activity: Investigating Graphs of Radical Functions– P431 • To graph 푦 = 푎 푥 − ℎ + 푘 or 푦 = 푎 푥 − ℎ + 푘, follow these steps.– Step 1: Sketch the graph of 푦 = 푎 푥 or 푦 = 푎 푥– Step 2: Shift the graph h units horizontally and k units vertically. • Describe how to obtain the graph of 푦 = 푥 − 2 +1 from the graph of 푦 = 푥. • Graph 푦 = 2 푥 + 4 − 1 • Describe how to obtain the graph of 푦 = 푥 − 3 + 2from the graph of 푦 = 푥. Then graph. • Graph 푦 = −2 푥 − 3 + 2. • Graph 푦 = −2 푥 − 1 + 1. • State the domain and range of the function in all of the examples. • A model for the period of a simple pendulum as measured in time units is given by 푇 = 2휋 , where 푇 is the time in seconds, 퐿 is the length of the pendulum in feet, and 푔 is 32 ft/sec2. Use a graphing calculator to graph the model. Then use the graph to estimate the period of a pendulum that is 3 feet long. • The length of a whale can be modeled by 퐿 =21.04 푊, where 퐿 is the length in feet, and 푊 is the weight in tons. Graph the model, then use the graph to find the weight of a what that is 60 ft long. • P434: 15 – 48 x 3s, 66 • Evaluate when 푥 = 4: 4푥 + 192− 2푥 • Daily Agenda:– Grade assignment– 7.6 notes / assignment – Ch 7 Test on Thursday?? Chapter 7.6 • To solve equations that contain radicals or rational exponents. • To use radical equations to solve real-life problems. • Isolate the radical expression• Raise each side to the same power • Solve 5 − 푥 = 0 • Solve 푥 + 6 = 12 • Solve 3푥 ⁄ = 243 • Solve 2푥 ⁄ = 8 • Solve 2푥 + 8 − 4 = 6 • Solve 4푥 + 28 − 3 2푥 = 0 • Solve 12 − 2푥 − 2 푥 = 0 • Extraneous or False solutions are ones that do not work • They are introduced sometimes when raising each side to the same power • MUST check all solutions to check for them! • Solve 푥 + 2 = 2푥 + 28 • Solve 푥 − 3 = 4푥 • The strings of guitars and pianos are under tension. The speed v of a wave on the string depends on the force (tension) F on the string and the mass M per unit length L according to the formula 푣 = ⋅ ⋅ 퐴. A wave travels through a string with a mass of 0.2 kilograms at a speed of 9 meters per second. It is stretched by a force of 19.6 Newtons. Find the length of the string. • Solve 푅 = 2.4 푥 + 3.9 + 7.2 for x if R = 19.8. • P441: 18 – 57 x 3s, 66, 70, 74 • #57 needs graphing calculator• Leave all answers in fraction/radical form except #57, 66 – . . – Daily Agenda:• Grade assignment• 7.7 notes / assignment• Review• Test Friday! Chapter 7.7 Statistics and Statistical Graphs • To use measures of central tendency and measures of dispersion to describe data sets. • To use box-and-whisker plots and histograms to represent data graphically. • Statistics: numerical values used to summarize and compare sets of data • Measures of Central Tendency:– Mean: average, 푥̅– Median: middle number when numbers are in order– Mode: the number that occurs most often • The number of games won by teams in the Eastern Conference for the 1997-1998 regular season of the NHL is shown on the chart. • Find the mean, median, and mode for the data set. Eastern Conference 36, 39, 40, 34, 48, 33, 25, 30, 37, 17, 42, 40, 24 • Measures of dispersion tell us how spread out the data is. • Two main measures of dispersion:– Range: the difference in the greatest and least numbers – Standard Deviation: describes the difference between the mean and the data value • The standard deviation of 푥 , 푥 , … , 푥 • 휎 = ̅ ̅ ⋯ ̅ • Find the range and standard deviation for the number of wins in the NHL Eastern Conference. Eastern Conference 36, 39, 40, 34, 48, 33, 25, 30, 37, 17, 42, 40, 24 • A graphical way of representing data by drawing a box around the middle half of the data and extending “whiskers” to the minimum and maximum values. • Draw a box-and-whisker plot for the NHL Easter Conference. Eastern Conference 36, 39, 40, 34, 48, 33, 25, 30, 37, 17, 42, 40, 24 • A graphical display of data using a bar graph.– Has data grouped in intervals of equal width.– Number of data values in each interval is the frequency of that interval.– Based off of a frequency distribution. • Draw a histogram for the NHL Eastern Conference.Eastern Conference 36, 39, 40, 34, 48, 33, 25, 30, 37, 17, 42, 40, 24 • P450: 33, 34, 36, 37, 40, 41• Ch. 7 Review Worksheet Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents Documents	4,869	13,425	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.5	4	CC-MAIN-2023-40	latest	en	0.856399
https://gmatclub.com/forum/at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html?sort_by_oldest=true	1,579,839,524,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250615407.46/warc/CC-MAIN-20200124040939-20200124065939-00531.warc.gz	456,367,008	160,188	GMAT Question of the Day: Daily via email \| Daily via Instagram New to GMAT Club? Watch this Video It is currently 23 Jan 2020, 21:17 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # At an amusement park, tom bought a number of red tokens and Author Message TAGS: ### Hide Tags Manager Joined: 27 Oct 2011 Posts: 115 Location: United States Concentration: Finance, Strategy GPA: 3.7 WE: Account Management (Consumer Products) At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 30 Jan 2012, 21:06 4 40 00:00 Difficulty: 65% (hard) Question Stats: 68% (02:59) correct 32% (03:08) wrong based on 563 sessions ### HideShow timer Statistics At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs $0.09, and each green token costs$0.14. If Tom spent a total of exactly $2.06, how many token in total did Tom buy? A. 16 B. 17 C. 18 D. 19 E. 20 This is a tough one. I am having trouble finding a fast solution for this. ##### Most Helpful Expert Reply SVP Status: Top MBA Admissions Consultant Joined: 24 Jul 2011 Posts: 1917 GMAT 1: 780 Q51 V48 GRE 1: Q800 V740 Re: Amusement Park Tokens [#permalink] ### Show Tags 31 Jan 2012, 00:20 9 7 0.09x + 0.14y = 2.06 => 9x + 14y = 206 To solve this remember that x must be even because 14y, when subtracted from 206, will yield an even number (even - even = even). The solution comes out to be x=12, y=7. Therefore the total number of tokens bought = 12+7 = 19 Option (D). _________________ GyanOne [www.gyanone.com]\| Premium MBA and MiM Admissions Consulting Awesome Work \| Honest Advise \| Outstanding Results Reach Out, Lets chat! Email: info at gyanone dot com \| +91 98998 31738 \| Skype: gyanone.services ##### Most Helpful Community Reply Intern Joined: 03 Mar 2014 Posts: 2 Concentration: Entrepreneurship, General Management GMAT 1: 730 Q48 V41 GPA: 3.04 Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 27 Apr 2014, 00:01 4 9 I solved it this way: Starting with the equation 9X+14Y = 206 => 9(X+Y) + 5Y = 206 5Y = 206 - 9(X+Y) we need to find X+Y. The RHS has to be a multiple of 5 Substituting the answers for X+Y abive, only 19 gives a multiple of 5. You don't need to actually multiply all the answers with 9, just look for the units digit of the difference. (it has to be either 5 or 0) When 19 is substituted, we get a units digit of 5 in the difference. So D. ##### General Discussion Math Expert Joined: 02 Sep 2009 Posts: 60627 At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 31 Jan 2012, 01:33 2 1 calreg11 wrote: At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs$0.09, and each green token costs $0.14. If Tom spent a total of exactly$2.06, how many token in total did Tom buy? a. 16 b. 17 c. 18 d. 19 e. 20 This is a tough one. I am having trouble finding a fast solution for this. Given: 0.09R + 0.14G = 2.06; 9R + 14G = 206. Now, it's special type of equations as G and R must be a non-negative integers, so there might be only one solution to it. After some trial and error you'll get (actually there are several ways of doing it): R = 12 and G = 7; R + G = 19. For more on this type of questions check: eunice-sold-several-cakes-if-each-cake-sold-for-either-109602.html martha-bought-several-pencils-if-each-pencil-was-either-a-100204.html a-rental-car-agency-purchases-fleet-vehicles-in-two-sizes-a-105682.html joe-bought-only-twenty-cent-stamps-and-thirty-cent-stamps-106212.html a-certain-fruit-stand-sold-apples-for-0-70-each-and-bananas-101966.html joanna-bought-only-0-15-stamps-and-0-29-stamps-how-many-101743.html at-an-amusement-park-tom-bought-a-number-of-red-tokens-and-126814.html collections-confused-need-a-help-81062.html Hope it helps. _________________ Senior Manager Joined: 23 Oct 2010 Posts: 317 Location: Azerbaijan Concentration: Finance Schools: HEC '15 (A) GMAT 1: 690 Q47 V38 Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 31 Jan 2012, 01:59 I found this question not hard, but time -consuming. it took some time to find x=12 y=7 _________________ Happy are those who dream dreams and are ready to pay the price to make them come true I am still on all gmat forums. msg me if you want to ask me smth Manager Joined: 27 Oct 2011 Posts: 115 Location: United States Concentration: Finance, Strategy GPA: 3.7 WE: Account Management (Consumer Products) Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 31 Jan 2012, 21:45 Is there any way to solve for it other than trial and error? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 10010 Location: Pune, India Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 01 Feb 2012, 02:40 2 calreg11 wrote: Is there any way to solve for it other than trial and error? Check out case 2 in this post. It explains you in detail how to deal with such questions. I don't think there are pure algebraic solutions to such problems. http://www.veritasprep.com/blog/2011/06 ... -of-thumb/ _________________ Karishma Veritas Prep GMAT Instructor Math Expert Joined: 02 Sep 2009 Posts: 60627 Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 01 Feb 2012, 02:55 calreg11 wrote: Is there any way to solve for it other than trial and error? Trial and error along with some common sense is pretty much the only way you should approach such kind of problems on the GMAT. You won't get some very tough numbers to manipulate with or there will be some shortcut available, based on multiples concept or on the answer choices. So generally you would have to try just couple of values to get the answer. Check the links in my previous post to practice similar problems and you'll see that getting the answer is not that hard for the realistic GMAt question. _________________ Manager Joined: 27 Oct 2011 Posts: 115 Location: United States Concentration: Finance, Strategy GPA: 3.7 WE: Account Management (Consumer Products) Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 01 Feb 2012, 20:33 Thanks for the posting. Manager Joined: 08 Oct 2010 Posts: 175 Location: Uzbekistan Schools: Johnson, Fuqua, Simon, Mendoza WE 3: 10 Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 27 Feb 2012, 00:47 3 1 Bunuel wrote: calreg11 wrote: Is there any way to solve for it other than trial and error? Trial and error along with some common sense is pretty much the only way you should approach such kind of problems on the GMAT. You won't get some very tough numbers to manipulate with or there will be some shortcut available, based on multiples concept or on the answer choices. So generally you would have to try just couple of values to get the answer. Check the links in my previous post to practice similar problems and you'll see that getting the answer is not that hard for the realistic GMAt question. Hi calreg11, supporting the explanations of Bunuel and karishma above, I can show you shortest way of solving it by some amalgamation of trial and error with the algebraic approach, though, as mentioned by karishma, there is no pure algebraic solution of this problem. Let's start: first, we have to formulate the premise in an algebraic way through expressing red and green tokens by any letters we think convenient to us--> assuming, e.g., red tokens as 'r' and green tokens as 'g'; secondly, for the sake of convenience we can take the prices of red and greem tokens and also the total cost in cents, i.e., $0,09 as 9 cents,$0.14 as 14 cents, and $2.06 as 206 cents; then, thirdly, we do formulate it --> 9r + 14g = 206; fourth, now we can refer to the point that red tokens and green tokens make up the total number of tokens which is unknown to us and this is why formula can be --> r + g = x fifth, we have to apply trial and error approach through replacing x by each answer choice and we do it this way: r + g = 16 r = 16 - g replace 'r' in the original formula --> 9r + 14g = 206and we get 9(16-g) + 14g = 206 --> 5g=62 --> 62 is not divisible by 5, and hence, we cannot derive the number of g (green tokens), consequently, that of red tokens' also. only 19 can satisfy the condition drawn from the formulae r = x - g and 9(x-g) + 14g = 206 Hope, it helps! SVP Joined: 06 Sep 2013 Posts: 1522 Concentration: Finance Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 26 Dec 2013, 15:29 1 3 calreg11 wrote: At an amusement park, tom bought a number of red tokens and green tokens. Each red token costs$0.09, and each green token costs $0.14. If Tom spent a total of exactly$2.06, how many token in total did Tom buy? a. 16 b. 17 c. 18 d. 19 e. 20 This is a tough one. I am having trouble finding a fast solution for this. What I like to do in this questions is the following We have 9x + 14y = 206 First always try to simplify, in this case we can't Now look for a number that is the same for both and will be close to 206 In this case 9 is our best choice (You can quickly ballpark with 10 but you will realize it is >206) So with 9 for both x and y we get 207 which is one more. Now the fun part starts We need to play with this 9,9 combination to try to get one less, How so? Well, let see we need to be one lower so if we get rid of one 14 and add one 9 we be further down. If we subtract to 14's though we are down 28 and if we add 3 9's we are up 27 That perfect just to match our +1 difference! So in total we have 12+7 = 19 Hence our correct answer is D Hope it helps Cheers! J Manager Joined: 03 Jan 2015 Posts: 58 Concentration: Strategy, Marketing WE: Research (Pharmaceuticals and Biotech) Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 20 Jan 2015, 09:51 PROBLEM: (7 pairs)(23¢)=$1.61;$2.06-$1.61=45¢ 5 red tokens at 9¢@ total tokens=7+5=12 red +7 green=19 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 15971 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 14 Jan 2018, 12:50 Hi All, If you don't immediately see an 'elegant' approach to solving this problem, then you can still solve it relatively quickly with some 'brute force' and a bit of arithmetic. From the answer choices, you can see that the total number of coins is no more than 20, so there aren't that many potential calculations that you would have to do to find the exact number of each type of coins that 'fit' this situation. In basic terms, we're told that a certain number of .09s + a certain number of .14s total 2.06.... There are some Number Property rules that we can use to save some time: .14 multiplied by an integer will end in an EVEN digit 2.06 ends in an even digit Since (even) + (even) = (even), .09 multiplied by an integer MUST end in an EVEN digit for the sum to equal 2.06 The number of red tokens MUST be EVEN, so that significantly cuts down the number of options to consider.... IF... we have.... 2 red tokens, then the remaining value is$1.88. Can that be evenly divided by .14? Try it... (the answer is NO). 4 red tokens, then the remaining value is $1.70. Can that be evenly divided by .14? Try it... (the answer is NO). 6 red tokens, then the remaining value is$1.52. Can that be evenly divided by .14? Try it... (the answer is NO). 8 red tokens, then the remaining value is $1.34. Can that be evenly divided by .14? Try it... (the answer is NO). 10 red tokens, then the remaining value is$1.16. Can that be evenly divided by .14? Try it... (the answer is NO). 12 red tokens, then the remaining value is \$0.98. Can that be evenly divided by .14? Try it... (the answer is YES and the remaining 7 tokens are green). Thus, the total number of tokens is 12 red + 7 green = 19 total tokens GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: [email protected] The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ Intern Joined: 01 Jul 2018 Posts: 9 Location: United States Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 29 Aug 2018, 11:53 First thing I noticed is the difference between 9 and 14 is 5; so If you buy all the tokens (lets say x) at 9 cents then 206-9x should be divisible by 5. we can start with x=20 and we get 206-9*20 = 26 .. since this is not divisible by 5 20 is not the answer... but if we decrease 20 to 19 we can add 9 to 26 and since 35 is divisible by 5, 19 is the answer!! Non-Human User Joined: 09 Sep 2013 Posts: 14003 Re: At an amusement park, tom bought a number of red tokens and [#permalink] ### Show Tags 06 Oct 2019, 06:59 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). 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http://www.abovetopsecret.com/forum/thread199394/pg3	1,524,514,374,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125946165.56/warc/CC-MAIN-20180423184427-20180423204427-00622.warc.gz	346,816,730	13,948	It looks like you're using an Ad Blocker. Thank you. Some features of ATS will be disabled while you continue to use an ad-blocker. # A Question of gravity!! page: 3 0 share: posted on Mar, 20 2006 @ 05:25 PM Fellow ATS members, Gravity is a distortion of spacetime caused by matter. In the center of mass of a body, there is no distortion, and thus gravity is 0. Gravity at the top of a mountain is almost the same as gravity in a deep valley, because the difference in altitude is very small. You have to have a greater altitude difference in order to have less gravity between the high and the low point. posted on Mar, 20 2006 @ 05:31 PM Gravity at the center of the earth isn't zero. Well, I guess it depends on which direction you are talking about. At the center of the earth, there is technically no mass if the radius being measured is zero, and therefore no gravity. But if you magically appeared at the center of the earth, all of the mass wourd be AROUND you, and so you would have gravitational attraction being exterted on you equally in all directions. Currently, we reside on a side of the planet, and so gravity is only directed in one direction, as far as the earth is concerned. The farther you are from the center of mass (approximating the Earth to a perfect sphere) then the weaker the gravitational force. This is also mathematically supported in that gravitational force is inversely proportional to the square of the distance between the two objects. Assuming blindly that Mt. Everest is farther away from the center of mass of the Earth than Death Valley is, I would guess Mt. Everest. posted on Mar, 20 2006 @ 05:38 PM I'm going to say they're equal. My logic probably makes no sense though, so don't mind me. Heck, I don't even think I could accurately explain how I came to this conclusion. Just throwing something out there. --Kit. posted on Mar, 21 2006 @ 08:10 AM Some are unsure whether gravity is actually Zero at the centre of the Earth. A simple explanation is given Dr R. Barrans Jr., Ph.D: At the center of the earth, you would not feel any gravity. This is because the gravitational pull from every region of the earth is exactly counteracted by the gravitational pull from the corresponding region on the opposite side of you. This all adds up to a great big zero. Richard Barrans Jr., Ph.D. Chemical Separations Group Chemistry Division CHM/200 Argonne National Laboratory 9700 South Cass Avenue Argonne, IL 60439 [email protected] posted on Mar, 21 2006 @ 02:33 PM What the good doctor is saying is what you feel is zero. There is still an acceleration due to gravity none the less and gravity still exist, meaning it is not zero. posted on Mar, 21 2006 @ 02:40 PM If all gravitational forces cancel out at the centre (ie. the earth is perfectly spherical) then dont all the acceleration vectors add up to 0 too, therefore there is no acceleration due to gravity? posted on Mar, 21 2006 @ 04:28 PM Repeat after me: at the center of the Earth, gravity is 0, because spacetime is distorted equally from all directions. Gravity at the center of the earth, if measured, gives G = 0. posted on Mar, 22 2006 @ 01:11 AM Originally posted by masterp Repeat after me: at the center of the Earth, gravity is 0, because spacetime is distorted equally from all directions. Gravity at the center of the earth, if measured, gives G = 0. How? posted on Mar, 22 2006 @ 01:24 AM I can't remember where I read it, but I know it was a trustable source. It had to deal with the power of a black whole and how gravity is affected by it. It has to do with the hawkings effect, but anyway, For every mile you move away from the source of the gravitational energy, the power of the force is 1/7 that of what it was exactly 1 mile in the direction of the source. So unless the effects of gravity are different throughout the cosmos, I believe the farther away from the source you are, the weaker the gravity, therefore, the MASS of someone on everest would be less than that of someone in the dead valley. [edit on 22-3-2006 by formoneyormind] [edit on 22-3-2006 by formoneyormind] posted on Mar, 22 2006 @ 01:43 AM According to www.newton.dep.anl.gov if you were at the very center of the Earth you would feel zero Gs. Question - What is gravity like at the centre of the earth? If I dug a hole through the earth and travelled to the centre would I be pulled in both directions or float or what? Ian Yes, you would be pulled equally in all directions, in essence in "zero gravity". But the temperatures in the center of the earth would melt all known materials, you'd need some special insulating field to protect you there. Getting there would be another issue, another question. This is all a "what-if" journey, right? Lou Harnisch I don't see how this would effect your weight at Death Valley or on top of Mt Everest. posted on Mar, 22 2006 @ 04:46 AM formoneyormind just like some other posters you are making a common mistake, mass stays the same whatever gravity is present, it is constant. Weight changes as it is dependant on gravity. Your MASS will be the same everywhere in the universe. posted on Mar, 30 2006 @ 08:25 AM Sorry Guys, Missed you all a lot! Had gone on a vacation and just returned. (Not to Mt Everest or Death Valley!!) OK. I've received a lot o' U2Us for the answer. It's quite simple really. So here goes....Have a nice day!! curious.astro.cornell.edu... posted on Mar, 30 2006 @ 10:49 AM Wow ...very interesting ... but I still have trouble understanding this...ok equal gravitational force from all directions at center...gotcha..so about wieght....you would be weightless..gotcha...but that means that the forces from all sides have no effect..such as crushing you to the size of a pin due to the enormous preassure? And increasing your wieght...this has nothing to do with math ... I have a very limited grasp of mathmatics...but from a logical standpoint does this make sense to anyone ? and could you explain to me how it doesnt without the use of complex mathmatics.... posted on Mar, 30 2006 @ 10:58 AM You'll weigh the same. Weight (or "mass") doesn't change simply because you're nearer or farther away from a gravity source. YOU FOOLS, YOU!! posted on Mar, 30 2006 @ 11:33 AM Originally posted by Enkidu You'll weigh the same. Nope. Weight equals mass times gravitational acceleration. As mentioned already a few posts earlier. Your mass will stay the same, but in the center of the earth the effective gravitational acceleration is 0, so you'll be "weightless". Step on a scale on the moon and the weight of your 80 kilo's of mass will be less than 80. posted on Mar, 30 2006 @ 11:54 AM Originally posted by Enkidu You'll weigh the same. Weight (or "mass") doesn't change simply because you're nearer or farther away from a gravity source. YOU FOOLS, YOU!! OK this is the third time i have said this in this thread..... Mass and Weight are two totally different values, anyone with basic scientific education knows this. Surely the only fools are the people who firstly make posts without reading the thread first and secondly make condesending posts based on incorrect and flawed knowledge. -George posted on Mar, 30 2006 @ 12:00 PM Originally posted by Enkidu You'll weigh the same. Weight (or "mass") doesn't change simply because you're nearer or farther away from a gravity source. YOU FOOLS, YOU!! OK this is the third time i have said this in this thread..... Mass and Weight are two totally different values, anyone with basic scientific education knows this. Surely the only fools are the people who firstly make posts without reading the thread first and secondly make condesending posts based on incorrect and flawed knowledge. -George Yeah, would people quit acting like know-it-alls! Please! This is a very serious and complicated problem and deserves our utmost concentration and seriousity. posted on Mar, 30 2006 @ 06:49 PM lol, damn, he came back too soon and answered it... I only just saw the thread. Yeah, gravity at the center of the earth is measured as 0. However, whats actually happening is you are being forced in all directions with equal force. Its like two identical cars playing tug of war... theyre not getting anywhere... neither is the guy sitting on the middle of the rope. posted on Mar, 30 2006 @ 07:13 PM Originally posted by Enkidu You'll weigh the same. Weight (or "mass") doesn't change simply because you're nearer or farther away from a gravity source. YOU FOOLS, YOU!! Fool 42 here... uhhh, I think I'm going to have to ask you to come in and work on Saturday, because it appears you're not getting basic physics Monday thru Friday. The force (that's your weight) is equal to the gravitational constant times your mass times the other mass (that would be whatever you're on or near at the time) divided by the distance between you and it's center squared. W = Gmass(you)mass(planet, moon or whatever)/(d^2) So, no you don't have the same weight everywhere...you have the same mass (unless you start approaching the speed of light - but that lesson doesn't come until next Sunday - which you'll have to fill out a TCP report first before you can get into that one). posted on Mar, 30 2006 @ 07:22 PM Originally posted by Valhall The force (that's your weight) is equal to the gravitational constant times your mass times the other mass (that would be whatever you're on or near at the time) divided by the distance between you and it's center squared. W = Gmass(you)mass(planet, moon or whatever)/(d^2) Does this apply to a black hole? If the mass is that of 1,000,000,000 of our suns, and I weigh 200 pounds. How much would I weigh in a black hole? new topics top topics 0	2,364	9,745	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2018-17	latest	en	0.960983
https://assignmentutor.com/sampling-theory-dai-xie-stat392/	1,718,912,616,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861989.79/warc/CC-MAIN-20240620172726-20240620202726-00704.warc.gz	87,073,770	26,636	assignmentutor™您的专属作业导师 assignmentutor-lab™ 为您的留学生涯保驾护航在代写抽样理论sampling theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写抽样理论sampling theory方面经验极为丰富，各种代写抽样理论sampling theory相关的作业也就用不着说。 • Statistical Inference 统计推断 • Statistical Computing 统计计算 • Advanced Probability Theory 高等楖率论 • Advanced Mathematical Statistics 高等数理统计学 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 统计代写\|抽样理论作业代写sampling theory 代考\|Practical Advantages of Controlling Sampling Correctness A continuous random variable can be defined as a variable whose cumulative distribution function is continuous. We use a stronger definition that ensures that a continuous random variable has a well-defined (probability) density function. Definition 3.3.14 (Continuous random variable) A random variable $X$ is continuous if its distribution function can be expressed as $$F_X(x)=\int_{-\infty}^x f_X(u) d u \quad \text { for } x \in \mathbb{R}$$ for some integrable function $f_X: \mathbb{R} \longrightarrow[0, \infty)$. The function $f_X$ is called the (probability) density function of $X$. Definition 3.3.14 shows how the distribution function for a random variable can be found by integrating the density. We can also work in the opposite direction: the density is found by differentiating the cumulative distribution function. Claim 3.3.15 (Density from cumulative distribution function) For a continuous random variable $X$ with cumulative distribution function $F_X$, the density function is given by $$f_X(x)=\left.\frac{d}{d u} F_X(u)\right\|{u=x}=F_X^{\prime}(x) \text { for all } x \in \mathbb{R} .$$ The basic properties of density functions are given by the following claim. Claim 3.3.16 (Properties of continuous random variables) If $f_X$ is a density function then i. $f_X(x) \geq 0$ for all $x \in \mathbb{R}$. ii. $\int{-\infty}^{\infty} f_X(x) d x=1$. The first property is a direct consequence of Definition $3.3 .14$ and the second property can be seen directly from part ii. of Proposition 3.2.3.We use the same notation (lowercase $f$ ) for density functions as we do for mass functions. This serves to emphasise that the density plays the same role for a continuous variable as mass does for a discrete variable. However, there is an important distinction; it is legitimate to have density function values that are greater than one since values of a density function do not give probabilities. Probability is not associated with the values of the density function but with the area beneath the curve that the density function defines. In order to work out the probability that a random variable $X$ takes a value between $a$ and $b$, we work out the area above the $x$-axis, beneath the density, and between the lines $x=a$ and $x=b$. As we might expect, given this interpretation, the total area beneath a density function is one, as stated in Claim 3.3.16. The general relationship between probability and density is given by the following proposition. ## 统计代写\|抽样理论作业代写sampling theory 代考\|Fundamental Statistical Concepts The reader, particularly if he or she happens to be a statistician, may wonder if it is useful to introduce in this book a short course on probabilities (i.e., study of random variables prior to experimentation) and especially on statistics (i.e., study of random variables using data provided by experimentation). The answer is a definite yes and the reason is simple: the author, who has given more than 500 short courses on sampling, and who has been in contact with many clients having sampling problems, found that most people involved with sampling in quality assurance and quality control circles have no basic knowledge of descriptive statistics or long ago forgot this knowledge. Nevertheless, because much software is available and many articles cover the subject of sampling in a very short and superficial way, for better or worse, these people are using statistical concepts. Therefore, it is necessary to include this chapter on fundamental statistical concepts, so these wonderful tools are not misused. The term “statistic” was used for the first time by the German professor Achenwall in 1748. In 1843, Cournot defined statistics as “a science having for objective the collection and coordination of numerous facts within a given category of events, thus obtaining quantified effects which are likely to be independent from happening only by accident.” Statistical concepts are necessary for the development, understanding, and use of the TOS because they give information unpredictable in any other way; furthermore, they strongly link theory with reality. Statistical concepts are the basic “tools” of modern processes and quality control programs. They prevent detrimental effects from accumulating dangerously for too long. They also make possible the anticipation of acceptable or unacceptable operating errors. The concepts presented in this chapter have their limitations, which are often voluntarily or involuntarily forgotten by an operator, making the conclusions of his statistical evaluation not only naïve but also deprived of any scientific value. Let us imagine an operator who has collected two series of values from a certain experiment. He is plotting one series on the $y$-axis of a rectangular coordinate system and the other series on the corresponding $x$-axis. Thus, he is going to find a series of experimental points in the $x y-$ plane through which he draws a continuous line, obtains a graph, and calculates the equation of the graph. Meanwhile, more often than not, our operator is confronted with numerous problems: • Interpolation: The temptation is great to allow the simplest continuous curve to fit inside the “area of influence” of each point, however, this curve may or may not be represented by a simple equation, except when it can be approximated into a straight line. For this straight line to be a reality, it may become convenient to change a few variables, forcing a phenomenon to obey a preconceived law that happens to be convenient for the operator. When intervals between points become larger, interpolation may become dangerous because a large quantity of curves are found suitable in appearance, and the operator is likely to choose the curve that helps him to prove his preconceived idea. • Extrapolation: With the exception of the immediate vicinities of points corresponding to the limits of an experiment, the graph representing all points obtained by the operator should not be extrapolated beyond these limits, because there is neither a scientific nor a legitimate justification to do so. Similarly, for various reasons such as lack of time or funds, it is not uncommon to see statistical evaluations made on the analysis of two, three, or four samples, and some decisions that should have never been made. It is dangerous to extrapolate the information reached from the analysis of very few samples to an infinite population of potential samples because there is neither a scientific nor a legitimate way to find out exactly what kind of probability distribution they belong to without extensive additional testing. This is especially true in the case of trace constituents. In fact, the common problem exposed here is a combination of extrapolation (i.e., shape of a probability distribution defined beyond experimental points) and interpolation (i.e., shape of a probability distribution defined between too few experimental points). # 抽样理论代考 ## 统计代写\|抽样理论作业代写sampling theory-代考\|控制采样正确性的实际优势 . $$F_X(x)=\int_{-\infty}^x f_X(u) d u \quad \text { for } x \in \mathbb{R}$$ 。函数$f_X$被称为$X$的(概率)密度函数 $$f_X(x)=\left.\frac{d}{d u} F_X(u)\right\|{u=x}=F_X^{\prime}(x) \text { for all } x \in \mathbb{R} .$$密度函数的基本性质由以下权利要求给出。权利要求3.3.16(连续随机变量的属性)如果$f_X$是一个密度函数，那么i. $f_X(x) \geq 0$对所有$x \in \mathbb{R}$。2$\int{-\infty}^{\infty} f_X(x) d x=1$ . ## 统计代写\|抽样理论作业代写sampling theory代考\|基本统计概念 . 读者，特别是如果他或她碰巧是一名统计学家，可能会想在本书中介绍一门关于概率(即在实验前研究随机变量)，特别是关于统计学(即利用实验提供的数据研究随机变量)的短期课程是否有用。答案肯定是肯定的，原因很简单:作者开过500多门关于抽样的短期课程，也接触过许多有抽样问题的客户，发现在质量保证和质量控制圈中，大多数涉及抽样的人没有描述统计学的基本知识，或者早就忘记了这方面的知识。然而，由于许多软件和许多文章都以非常简短和肤浅的方式介绍抽样的主题，无论好坏，这些人使用的是统计概念。因此，有必要包括这一章的基本统计概念，以使这些奇妙的工具不会被滥用 “统计”一词是由德国教授阿肯沃尔在1748年首次使用的。1843年，古诺将统计学定义为“一门客观地收集和协调特定事件类别内众多事实的科学，从而获得量化的效果，而这种效果很可能不依赖于偶然事件的发生。”统计概念对于TOS的开发、理解和使用是必要的，因为它们提供的信息在任何其他方面都是不可预测的;此外，他们将理论与现实紧密联系起来。统计概念是现代过程和质量控制程序的基本“工具”。它们防止有害影响危险地累积太长时间。它们还使预期可接受或不可接受的操作错误成为可能本章所介绍的概念有其局限性，操作者往往会自愿或不自愿地忘记这些局限性，使其统计评估的结论不仅naïve而且丧失了任何科学价值。让我们想象一个操作员从一个特定的实验中收集了两个系列的值。他在直角坐标系的$y$轴上画了一个级数，在相应的$x$轴上画了另一个级数。因此，他将在$x y-$平面上找到一系列的实验点，通过这些点画出一条连续的直线，得到一个图，并计算出图的方程。同时，我们的运算符通常会遇到许多问题: • 外推:除了与实验极限相对应的点的直接邻近外推之外，表示算子得到的所有点的图不应该外推超过这些极限，因为这样做既没有科学的理由，也没有合法的理由。同样，由于缺乏时间或资金等各种原因，对两个、三个或四个样本的分析进行统计评估的情况并不少见，而且有些决定本不应该做出。将从分析极少数样本中得到的信息外推到无限数量的潜在样本中是危险的，因为在没有大量额外测试的情况下，既没有科学的也没有合法的方法来确切地找出它们属于哪种概率分布。在微量成分的情况下尤其如此。事实上，这里暴露的共同问题是外推(即定义在超出实验点的概率分布的形状)和插值(即定义在太少实验点之间的概率分布的形状)的组合。 ## 有限元方法代写 assignmentutor™作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 assignmentutor™您的专属作业导师 assignmentutor™您的专属作业导师	3,290	9,968	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.03125	4	CC-MAIN-2024-26	latest	en	0.593862
http://people.hofstra.edu/Stefan_Waner/diff_geom/Sec13.html	1,369,114,362,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368699730479/warc/CC-MAIN-20130516102210-00091-ip-10-60-113-184.ec2.internal.warc.gz	198,197,116	6,071	## Lecture 13: Three Basic Premises of General Relativity 13. Three Basic Premises of General Relativity Spacetime General relativity postulates that spacetime (the set of all events) is a smooth 4-dimensional Riemannian manifold M, where points are called events, with the properties A1-A3 listed below. A1. Locally, M is Minkowski spacetime (so that special relativity holds locally). This means that, if we diagonalize the scalar product on the tangent space at any point, we obtain the matrix 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 -1 . The metric is measurable by clocks and rods. Before stating the next axiom, we recall some definitions. Definitions 13.1 Let M satisfy axiom A1. If Vi is a contravariant vector at a point in M, define \|\|Vi\|\|2 = Vi, Vi = ViVjgij. (Note that we are not defining \|\|Vi\|\| here.) We say the vector Vi is timelike if \|\|Vi\|\|2< 0, lightlike if \|\|Vi\|\|2= 0, and spacelike if \|\|Vi\|\|2> 0, Examples 13.2 (a) If a particle moves with constant velocity v in some Lorentz frame, then at time t = x4 its position is x = a + vx4. Using the local coordinate x4 as a parameter, we obtain a path in M given by xi(x4)= ai + vix4 if i = 1, 2, 3 x4 if i = 4 so that the tangent vector (velocity) dxi/dx4 has coordinates (v1, v2, v3, 1) and hence square magnitude \|\|(v1, v2, v3, 1)\|\|2 = \|v\|2 - c2. It is timelike at sub-light speeds, lightlike at light speed, and spacelike at faster-than-light speeds. (b) If u is the proper velocity of some particle in locally Minkowskian spacetime, then we saw (normal condition in Section 10) that u, u = -c2 = -1 in our units. A2. Freely falling particles move on timelike geodesics of M. Here, a freely falling particle is one that is effected only by gravity, and recall that a timelike geodesic is a geodesic xi(t) with the property that \|\|dxi/dt\|\|2 < 0 in any paramaterization. (This property is independent of the parameterization -- see the exercise set.) A3 (Strong Equivalence Principle) All physical laws that hold in flat Minkowski space (ie. "special relativity") are expressible in terms of vectors and tensors, and are meaningful in the manifold M, continue to hold in every frame (provided we replace derivatives by covariant derivatives). Note Here are some consequences: 1. No physical laws can use the term "straight line," since that concept has no meaning in M; what's straight in the eyes of one chart is curved in the eyes of another. "Geodesic," on the other hand, does make sense, since it is independent of the choice of coordinates. 2. If we can write down physical laws, such as Maxwell's equations, that work in Minkowski space, then those same laws must work in curved space-time, without the addition of any new terms, such as the curvature tensor. In other words, there can be no form of Maxwell's equations for general curved spacetime that involve the curvature tensor. An example of such a law is the conservation law, .T = 0, which is thus postulated to hold in all frames. A Consequence of the Axioms: Forces in Almost Flat Space Suppose now that the metric in our frame is almost Lorentz, with a slight, not necessarily constant, deviation from the Minkowski metric, as follows. g = 1+2 0 0 0 0 1+2 0 0 0 0 1+2 0 0 0 0 -1+2 ... (I) or ds2 = (1+2)(dx2 + dy2 + dz2) - (1-2)dt2. Notes 1. We are not in an inertial frame (modulo scaling) since need not be constant, but we are in a frame that is almost inertial. 2. The metric g is obtained from the Minkowski g by adding a small multiple of the identity matrix. We shall see that such a metric does arise, to first order of approximation, as a consequence of Einstein's field equations. Now, we would like to examine the behavior of a particle falling freely under the influence of this metric. What do the timelike geodesics look like? Let us assume we have a particle falling freely, with 4-momentum P = m0U, where U is its 4-velocity, dxi/d. The paramaterized path xi() must satisfy the geodesic equation, by A2. Definition 9.1 gives this as d2xid2 + ris dxrd dxsd = 0 Multiplying both sides by m02 gives m0 d2(m0xi)d2 + ris d(m0xr)d d(m0xs)d = 0 or m0 = dPid + risPrPs = 0 (since Pi = d(m0xi/d)) where, by the (ordinary) chain rule (note that we are not taking covariant derivatives here... that is, dPi/d is not a vector -- see Lecture 7 on covariant differentiation), dPid = Pi,k dxkd so that Pi,k dm0xkd + risPrPs = 0, or Pi,kPk + risPrPs = 0 ... (I) Now let us do some estimation for slowly-moving particles v << 1 (the speed of light in our units) where we work in a frame where g has the given form. Question Why don't we work in an inertial frame (the frame of the particle)? First, since the frame is almost inertial (Lorentz), we are close to being in SR, so that P* m0U* = m0[v1, v2, v3, 1]/(1-v2/c2)1/2 [0, 0, 0, m0] (since v << 1) (in other words, the frame is almost comoving) Thus (I) reduces to Pi,4m0 + 4i4 m02 = 0 ... (II) Let us now look at the spatial coordinates, i = 1, 2, 3. By definition, 4i4 = 12 gij (g4j,4 + gj4,4 - g44,j). We now evaluate this at a specific coordinate i = 1, 2 or 3, where we use the definition of the metric g, recalling that g = (g)-1, and obtain 12 (1+2)-1(0 + 0 - 2,i) 12 (1-2)(-2,i) ,i. (Here and in what follows, we are ignoring terms of order O(2).) Substituting this information in (II), and using the fact that Pi,4 = x4 = (movi), the time-rate of change of momentum, or the "force" as measured in that frame (see the exercise set), we can rewrite (II) as m0 x4 (movi) - m02,i = 0, or x4 (movi) - m0,i = 0 Thinking of x4 as time t, and adopting vector notation for three-dimensional objects, we have, in old fashioned 3-vector notation, t (mov) = m0, that is F = m. This is the Newtonian force experienced by a particle in a force field potential of . (See the exercise set.) In other words, we have found that we can duplicate, to a good approximation, the physical effects of Newton-like gravitational force from a simple distortion of the metric. In other words -- and this is what Einstein realized -- gravity is nothing more than the geometry of spacetime; it is not a mysterious "force" at all. Exercise Set 13 1. Show that, if xi = xi(t) has the property that \|\|dxi/dt\|\|2 < 0 for some parameter t, then \|\|dxi/dts\|2 < 0 for any other parameter s such that ds/dt 0 along the curve. In other words, the property of being timelike does not depend on the choice of paramaterization. 2. What is wrong with the following (slickly worded) argument based on the Strong Equivalence Principle? I claim that there can be no physical law of the form A = R in curved spacetime, where A is some physical quantity and R is any quantity derived from the curvature tensor. (Since we shall see that Einstein's Field Equations have this form, it would follow from this argument that he was wrong!) Indeed, if the postulated law A = R was true, then in flat spacetime it would reduce to A = 0. But then we have a physical law in SR, which must, by the Strong Equivalence Principle, generalize to A = 0 in curved spacetime as well. Hence the original law A = R was wrong. 3. Gravity and Antigravity Newton's law of gravity says that a particle of mass M exerts a force on another particle of mass m according to the formula F = - GMmrr3 , where r = x, y, z, r = \|r\|, and G is a constant that depends on the units; if the masses M and m are given in kilograms, then G 6.67 10-11, and the resulting force is measured in newtons.* (Note that the magnitude of F is proportional to the inverse square of the distance r. The negative sign makes the force an attractive one.) Show by direct calculation that F = m, where = GMr . Hence write down a metric tensor that would result in an inverse square repelling force ("antigravity"). *A Newton is the force that will cause a 1-kilogram mass to accelerate at 1 m/sec2. Last Updated: January, 2002	2,242	7,931	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2013-20	latest	en	0.85636
https://www.numerade.com/books/chapter/trigonometric-identities-and-equations-2/	1,601,106,072,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400238038.76/warc/CC-MAIN-20200926071311-20200926101311-00212.warc.gz	952,013,390	15,795	# Algebra 2 ## Educators HD AG ### Problem 1 Verify each identity. $$\cos \theta \cot \theta=\frac{1}{\sin \theta}-\sin \theta$$ Joshua E. ### Problem 2 Verify each identity. $$\sin \theta \cot \theta=\cos \theta$$ HD Harrison D. ### Problem 3 Verify each identity. $$\cos \theta \tan \theta=\sin \theta$$ Joshua E. ### Problem 4 Verify each identity. $$\sin \theta \sec \theta=\tan \theta$$ HD Harrison D. ### Problem 5 Verify each identity. $$\cos \theta \sec \theta=1$$ Joshua E. ### Problem 6 Verify each identity. $$\tan \theta \cot \theta=1$$ HD Harrison D. ### Problem 7 Verify each identity. $$\sin \theta \csc \theta=1$$ Joshua E. ### Problem 8 Verify each identity. $$\cot \theta=\frac{\csc \theta}{\sec \theta}$$ HD Harrison D. ### Problem 9 Simplify each trigonometric expression. $$\tan \theta \cot \theta$$ Joshua E. ### Problem 10 Simplify each trigonometric expression. $$1-\cos ^{2} \theta$$ HD Harrison D. ### Problem 11 Simplify each trigonometric expression. $$\sec ^{2} \theta-1$$ Joshua E. ### Problem 12 Simplify each trigonometric expression. $$1-\csc ^{2} \theta$$ HD Harrison D. ### Problem 13 Simplify each trigonometric expression. $$\sec \theta \cot \theta$$ Joshua E. ### Problem 14 Simplify each trigonometric expression. $$\cos \theta \tan \theta$$ HD Harrison D. ### Problem 15 Simplify each trigonometric expression. $$\sin \theta \cot \theta$$ Joshua E. ### Problem 16 Simplify each trigonometric expression. $$\sin \theta \csc \theta$$ HD Harrison D. ### Problem 17 Simplify each trigonometric expression. $$\sec \theta \cos \theta \sin \theta$$ Joshua E. ### Problem 18 Simplify each trigonometric expression. $$\sin \theta \sec \theta \cot \theta$$ HD Harrison D. ### Problem 19 Simplify each trigonometric expression. $$\sec ^{2} \theta-\tan ^{2} \theta$$ Joshua E. ### Problem 20 Simplify each trigonometric expression. $$\frac{\sin \theta}{\cos \theta \tan \theta}$$ HD Harrison D. ### Problem 21 Simplify each trigonometric expression. $$\cos \theta+\sin \theta \tan \theta$$ Joshua E. ### Problem 22 Simplify each trigonometric expression. $$\csc \theta \cos \theta \tan \theta$$ HD Harrison D. ### Problem 23 Simplify each trigonometric expression. $$\tan \theta(\cot \theta+\tan \theta)$$ Joshua E. ### Problem 24 Simplify each trigonometric expression. $$\sin ^{2} \theta+\cos ^{2} \theta+\tan ^{2} \theta$$ HD Harrison D. ### Problem 25 Simplify each trigonometric expression. $$\cos ^{2} \theta \sec \theta \csc \theta$$ Joshua E. ### Problem 26 Simplify each trigonometric expression. $$\sin \theta\left(1+\cot ^{2} \theta\right)$$ HD Harrison D. ### Problem 27 Simplify each trigonometric expression. $$\cot \theta \tan \theta-\sec ^{2} \theta$$ Joshua E. ### Problem 28 Simplify each trigonometric expression. $$\sin ^{2} \theta \csc \theta \sec \theta$$ HD Harrison D. ### Problem 29 Simplify each trigonometric expression. $$\cos \theta\left(1+\tan ^{2} \theta\right)$$ Joshua E. ### Problem 30 Simplify each trigonometric expression. $$\frac{\tan \theta}{\sec \theta-\cos \theta}$$ AG Ankit G. ### Problem 31 Simplify each trigonometric expression. $$\sec \theta \cos \theta-\cos ^{2} \theta$$ Joshua E. ### Problem 32 Simplify each trigonometric expression. $$\sin \theta \csc \theta-\cos ^{2} \theta$$ HD Harrison D. ### Problem 33 Simplify each trigonometric expression. $$\csc \theta-\cos \theta \cot \theta$$ Joshua E. ### Problem 34 Simplify each trigonometric expression. $$\cos \theta+\sin \theta \tan \theta$$ HD Harrison D. ### Problem 35 Simplify each trigonometric expression. $$\sec \theta\left(1+\cot ^{2} \theta\right)$$ Joshua E. ### Problem 36 Simplify each trigonometric expression. $$\csc ^{2} \theta\left(1-\cos ^{2} \theta\right)$$ HD Harrison D. ### Problem 37 Simplify each trigonometric expression. $$\frac{\cos \theta \csc \theta}{\cot \theta}$$ Joshua E. ### Problem 38 Simplify each trigonometric expression. $$\frac{\sin ^{2} \theta \csc \theta \sec \theta}{\tan \theta}$$ HD Harrison D. ### Problem 39 Express the first trigonometric function in terms of the second. $$\sin \theta, \cos \theta$$ Joshua E. ### Problem 40 Express the first trigonometric function in terms of the second. $$\tan \theta, \cos \theta$$ HD Harrison D. ### Problem 41 Express the first trigonometric function in terms of the second. $$\cot \theta, \sin \theta$$ Joshua E. ### Problem 42 Express the first trigonometric function in terms of the second. $$\csc \theta, \cot \theta$$ HD Harrison D. ### Problem 43 Express the first trigonometric function in terms of the second. $$\cot \theta, \csc \theta$$ Joshua E. ### Problem 44 Express the first trigonometric function in terms of the second. $$\sec \theta, \tan \theta$$ HD Harrison D. ### Problem 45 Verify each identity. $$\sin ^{2} \theta \tan ^{2} \theta=\tan ^{2} \theta-\sin ^{2} \theta$$ Joshua E. ### Problem 46 Verify each identity. $$\sec \theta-\sin \theta \tan \theta=\cos \theta$$ HD Harrison D. ### Problem 47 Verify each identity. $$\sin \theta \cos \theta(\tan \theta+\cot \theta)=1$$ Joshua E. ### Problem 48 Verify each identity. $$\frac{1-\sin \theta}{\cos \theta}=\frac{\cos \theta}{1+\sin \theta}$$ AG Ankit G. ### Problem 49 Verify each identity. $$\frac{\sec \theta}{\cot \theta+\tan \theta}=\sin \theta$$ Joshua E. ### Problem 50 Verify each identity. $$(\cot \theta+1)^{2}=\csc ^{2} \theta+2 \cot \theta$$ AG Ankit G. ### Problem 51 Verify each identity. Express $\cos \theta \csc \theta \cot \theta$ in terms of $\sin \theta$ Joshua E. ### Problem 52 Verify each identity. Express $\frac{\cos \theta}{\sec \theta+\tan \theta}$ in terms of $\sin \theta$ AG Ankit G. ### Problem 53 Verify each identity. trigonometric expression and work backward.) Joshua E. ### Problem 54 Verify each identity. Writing Describe the similarities and differences in solving an equation and in verifying an identity. AG Ankit G. ### Problem 55 Verify each identity. $$1+\sec \theta=\frac{1+\cos \theta}{\cos \theta}$$ Joshua E. ### Problem 56 Verify each identity. $$\frac{1+\tan \theta}{\tan \theta}=\cot \theta+1$$ AG Ankit G. ### Problem 57 Verify each identity. $$\frac{\cot \theta \sin \theta}{\sec \theta}+\frac{\tan \theta \cos \theta}{\csc \theta}=1$$ Joshua E. ### Problem 58 Verify each identity. $$\sin ^{2} \theta \tan ^{2} \theta+\cos ^{2} \theta \tan ^{2} \theta=\sec ^{2} \theta-1$$ AG Ankit G. ### Problem 59 Simplify each trigonometric expression. $$\frac{\cot ^{2} \theta-\csc ^{2} \theta}{\tan ^{2} \theta-\sec ^{2} \theta}$$ Joshua E. ### Problem 60 Simplify each trigonometric expression. $$(1-\sin \theta)(1+\sin \theta) \csc ^{2} \theta+1$$ AG Ankit G. ### Problem 61 Physics When a ray of light passes from one medium into a second, the angle of incidence $\theta_{1}$ and the angle of refraction $\theta_{2}$ are related by Snell's law: $n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2},$ where $n_{1}$ is the index of refraction of the first medium and $n_{2}$ is the index of refraction of the second medium. How are $\theta_{1}$ and $\theta_{2}$ related if $n_{2}>n_{1} ?$ If $n_{2}<n_{1} ?$ If $n_{2}=n_{1} ?$ Joshua E. ### Problem 62 Which expression is NOT equal to the other three expressions? $\begin{array}{llll}{\text { A. } \frac{2}{\tan \theta}} & {\text { B. } \frac{\cot \theta}{\frac{1}{2}}} & {\text { C. } \frac{\sin \theta}{\frac{1}{2} \cos \theta}} & {\text { D. } \frac{2 \cos \theta}{\sin \theta}}\end{array}$ AG Ankit G. ### Problem 63 Which equation is NOT true? $\begin{array}{ll}{\mathbf{F} \cos ^{2} \theta=1-\sin ^{2} \theta} & {\text { G. } \cot ^{2} \theta=\csc ^{2} \theta-1} \\ {\text { H. } \sin ^{2} \theta=\cos ^{2} \theta-1} & {\text { I. } \tan ^{2} \theta=\sec ^{2} \theta-1}\end{array}$ Joshua E. ### Problem 64 Which expressions are equivalent? 1. $(\sin \theta)(\csc \theta-\sin \theta)$ II. $\sin ^{2} \theta-1$ III. $\cos ^{2} \theta$ A. I and II only B. II and ll only C.l and III only D. $1,11,$ and $\\| 1$ AG Ankit G. ### Problem 65 $\begin{array}{ll}{\text { How can you express } \csc ^{2} \theta-2 \cot ^{2} \theta \text { in terms of } \sin \theta \text { and } \cos \theta ?} \\ {\mathrm{F} \cdot \frac{1-2 \cos ^{2} \theta}{\sin ^{2} \theta}} & {\text { G. } \frac{1-2 \sin ^{2} \theta}{\sin ^{2} \theta}} \\ {\mathrm{H} \cdot \sin ^{2} \theta-2 \cos ^{2} \theta} & {\text { J. } \frac{1}{\sin ^{2} \theta}-\frac{2}{\tan ^{2} \theta}}\end{array}$ Joshua E. ### Problem 66 Which expression is equivalent to $\frac{\tan \theta}{\cos \theta-\sec \theta} ?$ $\begin{array}{llll}{\text { A. } \csc \theta} & {\text { B. } \sec \theta} & {\text { C. }-\csc \theta} & {\text { D. } \tan ^{2} \theta}\end{array}$ AG Ankit G. ### Problem 67 Show that $(\sec \theta+1)(\sec \theta-1)=\tan ^{2} \theta$ is an identity. Joshua E. ### Problem 68 Show that $\frac{\cos x}{1-\sin ^{2} x}=\sec x$ is an identity. AG Ankit G. ### Problem 69 Graph each function in the interval from 0 to 2$\pi$. $$y=\csc (-\theta)$$ Joshua E. ### Problem 70 Graph each function in the interval from 0 to 2$\pi$. $$y=-\cot \theta$$ AG Ankit G. ### Problem 71 Graph each function in the interval from 0 to 2$\pi$. $$y=-\sec 0.5 \theta$$ Joshua E. ### Problem 72 Graph each function in the interval from 0 to 2$\pi$. $$y=-\sec (0.5 \theta+2)$$ AG Ankit G. ### Problem 73 Graph each function in the interval from 0 to 2$\pi$. $$y=\cot \frac{\theta}{5}$$ Joshua E. ### Problem 74 Graph each function in the interval from 0 to 2$\pi$. $$y=\pi \sec \theta$$ AG Ankit G. ### Problem 75 Find the measure of an angle between $0^{\circ}$ and $360^{\circ}$ that is coterminal with the given angle. $$395^{\circ}$$ Joshua E. ### Problem 76 Find the measure of an angle between $0^{\circ}$ and $360^{\circ}$ that is coterminal with the given angle. $$405^{\circ}$$ HD Harrison D. ### Problem 77 Find the measure of an angle between $0^{\circ}$ and $360^{\circ}$ that is coterminal with the given angle. $$-225^{\circ}$$ Joshua E. ### Problem 78 Find the measure of an angle between $0^{\circ}$ and $360^{\circ}$ that is coterminal with the given angle. $$-149^{\circ}$$ HD Harrison D. ### Problem 79 Find the measure of an angle between $0^{\circ}$ and $360^{\circ}$ that is coterminal with the given angle. $$627^{\circ}$$ Joshua E. ### Problem 80 Find the measure of an angle between $0^{\circ}$ and $360^{\circ}$ that is coterminal with the given angle. $$-281^{\circ}$$ AG Ankit G. ### Problem 81 Find the measure of an angle between $0^{\circ}$ and $360^{\circ}$ that is coterminal with the given angle. $$493^{\circ}$$ Joshua E. ### Problem 82 Find the measure of an angle between $0^{\circ}$ and $360^{\circ}$ that is coterminal with the given angle. $$-609^{\circ}$$ AG Ankit G. ### Problem 83 Make a box-and-whisker plot for each set of values. 300$\quad 345 \quad 333 \quad 295 \quad 302 \quad 321$ Joshua E. 3248$\quad 87 \quad 43 \quad 62 \quad 15 \quad 49 \quad 51 \quad 47 \quad 36 \quad 50 \quad 109 \quad 64$	3,710	11,117	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2020-40	latest	en	0.449299
https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_22&diff=cur&oldid=91144	1,623,742,233,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487617599.15/warc/CC-MAIN-20210615053457-20210615083457-00345.warc.gz	130,269,550	19,276	# Difference between revisions of "2008 AMC 12A Problems/Problem 22" The following problem is from both the 2008 AMC 12A #22 and 2008 AMC 10A #25, so both problems redirect to this page. ## Problem A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$? $[asy]unitsize(4mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)mat); draw(rotate(120)mat); draw(rotate(180)mat); draw(rotate(240)mat); draw(rotate(300)mat); label("$$x$$",(-1.55,2.1),E); label("$$1$$",(-0.5,3.8),S);[/asy]$ $\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$ ## Solution 1 (Trigonometry) Let one of the mats be $ABCD$, and the center be $O$ as shown: $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)mat); draw(rotate(120)mat); draw(rotate(180)mat); draw(rotate(240)mat); draw(rotate(300)mat); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686));[/asy]$ Since there are $6$ mats, $\Delta BOC$ is equilateral (the hexagon with side length $x$ is regular). So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$. By the Law of Cosines: $4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}$. Since $x$ must be positive, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$. ## Solution 2 (without trigonometry) Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$. $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)mat); draw(rotate(120)mat); draw(rotate(180)mat); draw(rotate(240)mat); draw(rotate(300)mat); pair D = rotate(300)(-3.687,1.5513); pair C = rotate(300)(-2.687,1.5513); pair EE = foot((0.00,0.00),D,C); draw(D--EE--(0,0)); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); label("$$E$$", EE,SE); [/asy]$ As proved in the first solution, $\angle OCD = 150^\circ$. That makes $\Delta OCE$ a $30-60-90$ triangle, so $OE = \frac{x}{2}$ and $CE= \frac{x\sqrt 3}{2}$ Since $\Delta ODE$ is a right triangle, $\left({\frac{x}{2}}\right)^2 + \left({\frac{x\sqrt 3}{2} +1}\right)^2 = 4^2 \Rightarrow x^2+x\sqrt3-15 = 0$ Solving for $x$ gives $x =\frac{3\sqrt{7}-\sqrt{3}}{2}\Rightarrow (C)$ ## Solution 3 (simply Pythagorean Theorem) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)mat); draw(rotate(120)mat); draw(rotate(180)mat); draw(rotate(240)mat); draw(rotate(300)mat); label("$$x$$",(-1.55,2.1),E); label("$$x$$",(0.03,1.5),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$E$$",(0,4.17)); label("$$F$$",(0.75,4.15),W); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-0.1,3.8),S); label("$$4$$",(-0.4,2.2),S); draw((0,0)--(0,3.103)); draw((0,0)--(-2.687,1.5513)); draw((0,0)--(-0.5,3.9686)); draw((0,0)--(-0.5,3.9686));[/asy]$ By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\angle OED = 90^\circ$. Since $OD = 4$ and $DE = \frac{1}{2}$, $OE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}$. Since $\triangle CED$ is $30-60-90$, $CE = \frac{\sqrt{3}}{2}$ (this can also be deduced from Pythagoras on $\triangle CED$). Thus $OC = \frac{3\sqrt{7}-\sqrt{3}}{2}$. As previous solutions noted, $\triangle BOC$ is equilateral, and thus the desired length is $x = OC \implies (C)$. ## Solution 3 $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)mat); draw(rotate(120)mat); draw(rotate(180)mat); draw(rotate(240)mat); draw(rotate(300)mat); label("$$x$$",(-1.95,3),E); label("$$A$$",(-3.6,2.5513),E); label("$$C$$",(0.05,3.20),E); label("$$E$$",(0.40,-3.60),E); label("$$B$$",(-0.75,4.15),E); label("$$D$$",(-2.62,1.5),E); label("$$F$$",(-2.64,-1.43),E); label("$$G$$",(-0.2,-2.8),E); label("$$\sqrt{3}x$$",(-1.5,-0.5),E); label("$$M$$",(-2,-0.9),E); label("$$O$$",(0.00,-0.10),E); label("$$1$$",(-2.7,2.3),S); label("$$1$$",(0.1,-3.4),S); label("$$8$$",(-0.3,0),S); draw((0,-3.103)--(-2.687,1.5513)); draw((0.5,-3.9686)--(-0.5,3.9686));[/asy]$ Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the midpoint of line $DG$, it creates two $30$ - $60$ - $90$ triangles. Therefore, $MD = MG = \frac{\sqrt{3}x}{2} \Rightarrow DG = \sqrt{3}x$. $AE = 2 + \sqrt{3}x$ Applying the Pythagorean theorem to triangle $ABE$, we get $$(2+\sqrt{3}x)^2 + x^2 = 8^2 \Rightarrow 4 + 3x^2 + 4\sqrt{3}x + x^2 = 64 \Rightarrow x^2 + \sqrt{3}x - 15 = 0$$ Using the quadratic formula to solve, we get $$x = \frac{-\sqrt{3} \pm \sqrt{3 -4(1)(-15)}}{2} = \frac{\pm 3\sqrt{7} - \sqrt{3}}{2}$$ $x$ must be positive, therefore $$x = \frac{3\sqrt{7} - \sqrt{3}}{2} \Rightarrow (C)$$ ~Zeric Hang ## Soultion 4 (coordinate bashing) $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)mat); draw(rotate(120)mat); draw(rotate(180)mat); draw(rotate(240)mat); draw(rotate(300)mat); label("$$x$$",(-1.55,2.1),E); label("$$1$$",(-0.5,3.8),S);[/asy]$ We will let $O(0,0)$ be the origin. This way the coordinates of $C$ will be $(0,y)$. By $30-60-90$, the coordinates of $D$ will be $\left(-\frac{1}{2}, y + \frac{\sqrt{3}}{2}\right)$. The distance any point with coordinates $(x, y)$ is from the origin is $\sqrt{x^2 + y^2}$. Therefore, the distance $D$ is from the origin is $4$ and $\frac{1}{4} + x^2 + x\sqrt{3} + \frac{3}{4} = x^2 + x\sqrt{3} + 1 = 16$. We get the quadratic equation mentioned in solution 2. Using the quadratic formula, we get that $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$ Note: Since $C$ and $D$ are not labeled in the diagram, refer to solution 1 for the location of points $C$ and $D$. ## Solution 5 $[asy]unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)mat); draw(rotate(120)mat); draw(rotate(180)mat); draw(rotate(240)mat); draw(rotate(300)*mat); label("$$x$$",(-1.55,2.1),E); label("$$A$$",(-3.6,2.5513),E); label("$$B$$",(-3.15,1.35),E); label("$$C$$",(0.05,3.20),E); label("$$D$$",(-0.75,4.15),E); label("$$E$$",(0.3,4.15),E); label("$$F$$",(-3.4,1.89),E); draw((0.5,3.9686)--(-3.13,2.45)); draw((0.5,3.9686)--(-2.95,2));[/asy]$ Notice that $\overarc{AE}$ is $\frac16$ the circumference of the circle. Therefore, $\overline{AE}$ is the side length of an inscribed hexagon. The side length of such a hexagon is simply the radius of the circle, or $4$. $\angle AFE$ is a right angle, therefore $\triangle AFE$ is a right triangle. $\overline{AF}$ is half the length of $1$, or $\frac{1}{2}$. The length of $\overline{EF}$ is $x$ plus the altitude length of one of the equilateral triangles, or $x+\frac{\sqrt{3}}{2}$. Using the Pythagorean Theorem, we get $4^2 = \left(\frac{1}{2}\right)^2 + \left(x+\frac{\sqrt{3}}{2}\right)^2$ Solving for $x$, we get $x = \frac{3\sqrt{7}-\sqrt{3}}{2}\ \boxed{\text{C}}$	3,731	8,985	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 105, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.59375	5	CC-MAIN-2021-25	latest	en	0.505236
saxifrage.be	1,643,260,726,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320305141.20/warc/CC-MAIN-20220127042833-20220127072833-00374.warc.gz	57,419,274	10,397	crushing efficiency calculation equation ## crushing efficiency calculation equation #### Calculations - Wisconsin DNR - State of Wisconsin (fuel burning, stone crushing, etc.) and corrected by a control efficiency, result in a calculated emission For most emission sources the following equation is used: . calculate air emissions from existing and new crushing spreads. DNR also #### Screen efficiency calculation - Crushing, Screening ... Screen efficiency is obtained using different equations, depending on whether your product is the oversize or undersize fraction from the screen. #### cone crusher capacity calculation formula Crushing of coal and calculation of size reduction efficiency. Feb 25, 2015 . While Double Toggle Jaw Crusher 2/25/2015 6:58:05 AM 10 Illustration . 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Jaw Crushing Efficiency Calculation Equation. crushing ... #### CEMENT GRINDING OPTIMISATION - CEMENT GRINDING OPTIMISATION Dr Alex Jankovic, Minerals Process Technology Asia-Pacific, Brisbane , Australia e-mail: [email protected] .com #### crushing efficiency calculation equation – Grinding Mill … crushing efficiency formula YouTube 12 Sepball mill efficiency calculations Springer Today the efficiency of ... crushing efficiency calculation equation crusherasia.co.	3,575	17,443	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.59375	4	CC-MAIN-2022-05	latest	en	0.880208
https://www.physicsforums.com/threads/why-0-0-1.8536/	1,660,677,448,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882572515.15/warc/CC-MAIN-20220816181215-20220816211215-00099.warc.gz	841,011,944	23,789	# Why 0^0 = 1 Organic I think 0^0 has to be examined from the level of the Z* ( http://mathworld.wolfram.com/Z-Star.html ) numbers. By using the empty set (with the Von Neumann Hierarchy), we can construct the set of Z* numbers {0,1,2,3,4,...}: Code: [b][i]0[/i][/b] = \|{ }\| (notation = {}) [b][i]1[/i][/b] = \|{[b]{[/b] [b]}[/b]}\| (notation = {0}) [b][i]2[/i][/b] = \|{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b]}\| (notation = {0,1}) [b][i]3[/i][/b] = \|{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b]}\| (notation = {0,1,2}) [b][i]4[/i][/b] = \|{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b],[b]{[/b]{ },{{ }},{{ },{{ }}}[b]}[/b]}\| (notation = {0,1,2,3}) and so on. Let us look at this question from a structural point of view, and we shall do it by using the base value expansion method where x is some Z* number. For example we shall use number 26 represented by base 10 and base 3: Code: Number 26 represented by base 10: ^0- 0123456789 \|\|\|\|\|\|\|\|\|\| \|_\|\|\|\|\|\|\|\| \|__\|\|\|\|\|\|\| \|___\|\|\|\|\|\| Base 10 = \|____\|\|\|\|\| \|_____\|\|\|\| \|______\|\|\| \|_______\|\| \|________\| ^1- \| \| 1 0 ( 210 + 610 ) ^0- 012345678901234567890123456 \|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\| \|_\|\|\|\|\|\|\|\|\|_\|\|\|\|\|\|\|\|\|_\|\|\|\|\| \|__\|\|\|\|\|\|\|\|__\|\|\|\|\|\|\|\|__\|\|\|\| \|___\|\|\|\|\|\|\|___\|\|\|\|\|\|\|___\|\|\| \|____\|\|\|\|\|\|____\|\|\|\|\|\|____\|\| \|_____\|\|\|\|\|_____\|\|\|\|\|_____\| \|______\|\|\|\|______\|\|\|\|__... \|_______\|\|\|_______\|\|\|__... \|________\|\|________\|\|__... ^1- \| 0 \| 1 \| 2 \|_________\| \| \|___________________\| \|_ ... \| Number 26 represented by base 3: ^0- 012 Base 3 = \|\|\| \|_\| ^1- \| 3 2 1 0 ( 03 + 23 + 23 + 23 ) ^0- 012012012012012012012012012 \|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\|\| \|_\|\|_\|\|_\|\|_\|\|_\|\|_\|\|_\|\|_\|\|_\| ^1- \|0 \|1 \|2 \|0 \|1 \|2 \|0 \|1 \|2 \|__\| \| \|__\| \| \|__\| \| \|_____\| \|_____\| \|_____\| ^2- \| 0 \| 1 \| 2 \|________\| \| \|_________________\| ^3- \| 0 \|_ ... \| As we can see, no matter what base value > 1 is, it is always reduced to 1 when power level = 0. Now, let us check base value 1. Code: ^0- 0 Base 1 = \| \| ^1- \| From the above we can learn that when base value = 1 it still can be represented by base value expansion method. Now, what about base value 0 ? Let us examine it from a structural point of view, for example base 10: Code: ^0- 0123456789 \|\|\|\|\|\|\|\|\|\| \|_\|\|\|\|\|\|\|\| \|__\|\|\|\|\|\|\| \|___\|\|\|\|\|\| Base 10 = \|____\|\|\|\|\| \|_____\|\|\|\| \|______\|\|\| \|_______\|\| \|________\| ^1- \| \| We can see that there exist two basic structural types: \| AND _ \| is what we call a singleton and it can be notated as {.} which is a singleton included in some set. By standard Boolean logic every object which included in some set must have a single and unique value. If we examine _ we shall find that the structural property of this element is out of the scope of standard Boolean logic. And the reason is this: _ exists between any two different singletons, therefore its property is at least and simultaneously two different states, and this property (to be simultaneously in two different states)is not in the scope of Boolean logic. But as we can see, without this element we can't develop our number system beyond 1. The "job" of _ is to be the "platform of memory" that gives us the ability to move beyond 1. The base value of this "platform" is 0 because there are no singletons in its scope. But unlike nothingness this object exists, therefore its full notation is 0^0=1(set's contenet exists). Now we have two basic objects: \|^0=1 , _^0=1 where 1 stands for "there exist some object". Those two different structures are indistinguishable by their quantitative property. If we want to move beyond "there exist some object" (which are notated as 1^0=1 or 0^0=1) we have to associate between tham. The result is the Natural numbers > 1, which are combinations of at least 21^0 singletons connected by at least 0^0 "platform of memory", for example: Code: ^0- 0 1 . . = 21^0 Base 2 = \| \| \|___\| ^1- \| ^ \| 0^0 {.} {.} = 21^0 Base 2 = \| {} \| \|{__}\| \| ^ \| 0^0 From this point of view we have at least 3 structural types of set's contents: {}, {.} AND {_} . By quantitative point of view we have: \|{}\| = 0 (set's content does not exist) \|{_}\| = 0^0 = 1 (set's contenet exists) \|{.}\| = 1^0 = 1 (set's contenet exists) What do you think? Organic Last edited: phoenixthoth i don't think 0 has to be viewed from the natural numbers because 0 could mean an operative identity element in many kinds of algebraic structures. for some of those structures, it makes sense that 0^0=1 while in others, it makes most sense that 0^0 is either undefined or indeterminate. for cardinal numbers, we are in the following situation. natural numbers could either be considered cardinal numbers or like cardinal numbers. suppose x and y are two cardinal numbers. x^y is the cardinal number of the set of functions from a set of cardinality y to a set of cardinality x. one can show that this is well defined by proving that there is a bijection between A^B and C^D whenever there is a bijection between A and C and a bijection between B and D (here, A^B is the set of functions from B to A). for example, 2^3=8 because if we look at the number of functions from a three element set like {a,b,c} to a two element set like {t,f}, then there are eight total: 1. {(a,t),(b,t),(c,t)} 2. {(a,t),(b,t),(c,f)} 3. {(a,t),(b,f),(c,t)} 4. {(a,t),(b,f),(c,f)} 5. {(a,f),(b,t),(c,t)} 6. {(a,f),(b,t),(c,f)} 7. {(a,f),(b,f),(c,t)} 8. {(a,f),(b,f),(c,f)}. clearly, there are 8 no matter what three element set and two element set we use because we could paint all the elements different colors but this list of 8 would be essentially the same. then 0^0=1 because there is one function from the empty set to the empty set: the empty function {}. if 0 is considered a real number, things change. one way to define real numbers is with the following: Q^N is the set of rational sequences. let C be the subset of Q^N consisting of sequences that are cauchy: {a_n : n in N} is cauchy iff for every e in Q there is a p in N such that \|a_n - a_m\|<e whenever m and n are greater than p. define an equivalence relation ~ on C so that {a_n : n in N}~{b_n : n in N} iff for every e in Q there is a p in N such that \|a_n - b_n\|<e whenever n>p. we then have the set of equivalence classes in C. let [{a_n}] be shorthand for the equivalence class of {a_n : n in N}. in other words {b_n : n in N} is in [{a_n}] iff {b_n : n in N}~{a_n : n in N}. let C/~ stand for all the equivalence classes in C. R is defined to equal C/~. for example, the real number pi equals [{3, 3.1, 3.14, 3.141,...}] and 0 equals both [{0,0,0,...}] and [{1, 1/2, 1/3, 1/4, ...}]. the map that sends q in Q to [{q, q, q, ...}] is a natural embedding of Q into C/~ and one can view Q as* a subset of C/~ because it is isomorphic to a a subset of C/~. it's pretty clear how one might define addition and multiplication on C/~. if @ can stand for + or x, then [{a_n}]@[{b_n}]:=[{a_n@b_n}]. it may or not be clear that @ is well defined. the question is how to define ^. let's even say that q^0=1 for all q in Q. rational exponents have a definition. can we just say that [{a_n}]^[{b_n}]:=[{a_n^b_n}]? we could say that {a_n^b_n} is not in any class in C/~, then it's not a real number. what would 0^0 be? well, to be careful, we need to independent representatives for 0. so suppose that 0=[{a_n}]=[{b_n}]. in other words, {a_n} and {b_n} are two (rational) sequences converging to 0. is it neccessarily the case that {a_n^b_n} converges to 0? this question is equivalent to asking if [{a_n^b_n}]=0. no. if a_n=2^(-n) and b_n=1/n, note that {a_n} and {b_n} converge to 0 yet {a_n^b_n} converges to 1/2. ok, so perhaps 0^0=1/2. but wait, there's more. suppose b_n=2^(-n) and a_n=1/n. then {a_n^b_n} converges to 1. therefore, the symbol 0^0 is not well defined because [{a_n^b_n}] varies for different representatives of 0. another question is is [{a_n}]^[{b_n}] well defined for the rest of the real numbers? if not, then i guess i picked a bad choice for ^. however, if it is well defined for all other real numbers (with the exception of when {a_n^b_n} is not a cauchy sequence of rational numbers) yet undefined only for 0^0, then this is a pretty convincing case that 0^0 is undefined when viewing 0 as an element of C/~=R. Homework Helper I think 0^0 has to be examined from the level of the Natural numbers. On the contrary, 00 can't be "examined from the level of the natural numbers" because 0 isn't a natural number. That would be a lot like examining ii "from the level of the real numbers". You can't even see it much less examine it! Gold Member I'm confused Halls of Ivy, ii is a real number (e-π/2), but otherwise I agree with you, 0 is not an element of Z+. Organic 0 is a Natural number because: By using the empty set (with the Von Neumann Hierarchy), we can construct the set of all positive integers {0,1,2,3,4,...}: Code: [b][i]0[/i][/b] = \|{ }\| (notation = {}) [b][i]1[/i][/b] = \|{[b]{[/b] [b]}[/b]}\| (notation = {0}) [b][i]2[/i][/b] = \|{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b]}\| (notation = {0,1}) [b][i]3[/i][/b] = \|{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b]}\| (notation = {0,1,2}) [b][i]4[/i][/b] = \|{[b]{[/b] [b]}[/b],[b]{[/b]{ }[b]}[/b],[b]{[/b]{ },{{ }}[b]}[/b],[b]{[/b]{ },{{ }},{{ },{{ }}}[b]}[/b]}\| (notation = {0,1,2,3}) and so on. Thank you. Organic Last edited: Gold Member No, 0 is not a postive integer, it has no sign. The natural numbers are defined as 1,2,3,4,....etc. Organic OK, I change my first post to Z* ( http://mathworld.wolfram.com/Z-Star.html ) instead of Natural numbers. Thank you. Organic Staff Emeritus Gold Member 0 is a natural number iff you use the convention that 0 is a natural number. Organic Hi Hurkyl, I have changed it to Z* . Can you read my first post at the beginning of this thread ? Thank you. Organic phoenixthoth No, 0 is not a postive integer, it has no sign. The natural numbers are defined as 1,2,3,4,....etc. 0 is a natural number iff you use the convention that 0 is a natural number. some people define 0 to be a natural number and obviously some people don't. wikipedia: http://en.wikipedia.org/wiki/Natural_numbers mathworld: http://mathworld.wolfram.com/NaturalNumber.html Unfortunately, 0 is sometimes also included in the list of "natural" numbers (Bourbaki 1968, Halmos 1974), and there seems to be no general agreement about whether to include it. can anyone explain why the author might use the word "unfortunately?" 0^0 depends on (at least) three things: the definition of 0, the context (ie where 0 is and what structure the context has), and the definition of exponentiation. two examples are 0 being the smallest element in N with exponentiation being defined so that m^n is the size of the set of functions from a set of n elements to a set of m elements in which case 0^0=1 because there is one function from the empty set to the empty set: the empty function. the other example mentioned was 0 as a member of R with it's own special definition of exponentiation in which case 0^0 is not a well-defined instance of said definition of exponentiation and so it doesn't have a meaning using that definition. since 0 appears in a wide variety of algebraic structures, each possibly having their own version of exponentiation if any, i don't think it is the case that 0 must be viewed as a natural number. however, if it is, 0^0=1. Organic Hi phoenixthoth, Please give some examples that clearly demonstrate the difference between 0 of Z, 0 of Q and 0 of R. Thank you. Organic phoenixthoth the post made on 11-07-2003 06:42 PM discusses 0 in the two contexts N and R. in N, 00=1 and in R, 00 is not a well defined expression. let's see about Q. in Q, (p/q)m/n could be defined to mean the element j/k in Q, if it exists, such that jn/kn = pm/qm, where jn means basically what it does for natural numbers except if n<0 then jn can be define in terms of positive exponents by jn=1/j-n and if j<0 and n>=0, then jn can be defined as (-j)n if n is even and -(-j)n if n is odd. now in this definition, what about 00? how about rewriting this as (0/1)0/1. then (0/1)0/1 is the element j/k in Q, if it exists, such that j1/k1 = 00/10. then j/k=00, which is what we expect. with this definition of exponentiation, 00 makes as much sense in Q as it does in N. if 00=1 in N, then 00=1 in Q as well. Organic No dear phoenixthoth, Please think simple, is there different definition to 0 of Z, 0 of Q or 0 of R ? phoenixthoth yes. there are many possible definitions of 0. for N, 0 could be defined to be the empty set. for Q, 0 could be defined to be 0/1. technically, from one perspective, the natural number 0 is not an element of Q but the pre-image of the element 0/1∈Q under an algebraic isomorphism from N to {x/1:x∈Z}. some people say that if f:A-->B and A and B are two algebraic structures of a certain kind and f is an injective homomorphism appropriate to that kind of structure, then A is a subset of B when, in fact, it is only isomorphic to a subset of B. some people say that N⊂Z⊂Q⊂R while i say N is embedded in Z is embedded in Q is embedded in R. actually, i would say N⊂Z⊂Q⊂R with the understanding that ⊂ means embedded and not subset. you know how they say a topologist can't tell the difference between a coffee cup and a doughnut? well an algebraist can't tell the difference between Q and the subset of R it is isomorphic to. for N, Z, and Q, i think it's pretty safe to say 00=1. for R, 0 could be the set of all rational cauchy sequences equivalent to the sequence {0,0,0,...} where this ~-equivalence was defined in an earlier post. for R, the expression 00 is not well defined. in other words, the definition for xy that works when x and y are not both zero breaks down when x and y are zero. this breakdown does not occur when looking at N, Z, or Q. now if you could define an R that is a complete ordered field with the least upper bound property having Q as a "subset" and define xy so that it extends the definition for xy on rationals in a continuous way (and so that its derivative will be what it should be), and make it so that xy works for x=y=0, then that would be very nice. my suspician is that in order to make xy=1 for x=y=0, some other property will be sacrificed but i'm not sure. the reason i think that is that R is in some sense the only field with the properties it has and i doubt 00 will make sense in one creation approach to a field and not another. Last edited: quartodeciman Experiment with exponential functions. Graph exponential functions with different bases. First start (type into box) with 4^x, 3^x, 2^z, clicking "graph it" button for each one. Then graph 1^x (exponential function goes flat at b=1). Finally, try bases like (0.1)^x, (0.01)^x, (0.001)^x, etc. Observe what is happening to b^x as b->0 (x at negative, zero and positive values). Also, observe what the graph of x^0 shows when x=0. The simple radical functions x^(0.5), x^(0.4), x^(0.3), ..., on down to x^(0.1), x^(0.01), x^(0.001), ... are interesting to graph. They all pass through points (0,0) and (1.1), Going relatively flat as x->0, then diving down to value 0. These two function types, exponential and radical, illustrate the two extreme attractors, (0,0) and (0,1), as the base for one function and the exponent for the other gets reduced through positive values to 0. ---> Function Grapher http://people.hofstra.edu/staff/steven_r_costenoble/Graf/Graf.html Here's another grapher, with settable parameter A. Type in A^x for the function. ---> EZ Graph http://id.mind.net/~zona/ezGraph/ezGraph.html [Broken] Last edited by a moderator: Organic I think that way to answer what is 0^0 must deal with two basic questions: 1) What is 0? 2) What is ^0? 1) Zero means: no singletons. 2) ^0 means: the simplest level of any object's existence. Therefore 0^0 means the existence an object with no singleton(s) in its scope = the continuum itself without any number {___}. Only this object has the power of the continuum, and no infinitely many singletons have the power of the continuum. 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https://math.answers.com/questions/There_is_a_bus_with_7_children_inside._Each_child_has_7_bags._Inside_each_bag_there_are_7_big_cats._Each_big_cat_has_7_small_cats._All_cats_have_4_legs_each._How_many_legs_are_on_the_bus	1,685,957,480,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224651815.80/warc/CC-MAIN-20230605085657-20230605115657-00590.warc.gz	434,566,754	134,848	0 # There is a bus with 7 children inside. Each child has 7 bags. Inside each bag there are 7 big cats. Each big cat has 7 small cats. All cats have 4 legs each. How many legs are on the bus? Wiki User 2013-10-05 07:51:01 As a riddle, the answer is zero legs on the bus since the bus has wheels instead of legs. But as a maths problem; It's not a tricked question -- just a straightforward maths problem § Let's say the whole situation simply : - There are 7 girls. - Each girl has 7 backpack - Each backpack has 7 big cats - Each big cat has 7 little small cats. § Question: How many legs are there in the bus? § Let's focus on each part of the question systematically, write down on a piece of paper if you have to, following my steps of chains. 1. Focus on the first girl first: A B C D E F G Girl A has 7 backpacks 2. Now focus on first backpack A1 : A1 A2 A3 A4 A5 A6 A7 Backpack A1 has 7 big cats 3. Now focus on first big cat : 1 2 3 4 5 6 7 Ø Each big cats have 7 small cats (that's the end of the focus) Systematically, 7 small cats has 28 legs (7 * 4) combined with the big cat that owns this 7 small cats. Let's call this the "cat chain", so cat chain 1has 32 legs altogether. Ø So since there are 7 big cats 7 * 32 = 224 legs Ø Therefore each backpack has 224 legs Ø Finally, each girl has 7 * 224 = 1568 legs. Lastly, since there are 7 girls, A to G, there are 7 * 1568 legs (per girl) = 10976 legs. Now, since each girl also has 2 legs each; SO, (7 * 2) + 10976 = 10990 legs altogether Note: Disregarding bus driver, buses and etc…Only focus on the girl and their backpacks! Or There are 77 = 49 bags and there are 7 + 77 = 56 cats in each bag, meaning there are 564 = 224 legs in each bag. So including the 7 children that have 2 legs each, there are 72 + 49(224) = 10990 legs in total. Wiki User 2013-10-05 07:51:01 Aaron Ray Lvl 1 2021-03-22 14:48:00 focus one the girl that is cring Kamil Suski Lvl 1 2021-05-19 15:34:21 28 + 49 + 14 = 91 cooler kid Kalyssa Lvl 1 2021-05-27 03:48:12 i think you mean97 i did my calculation with inestine Justyouraveragenerd Lvl 1 2021-09-29 22:36:05 wow lol Lillian Cooper Lvl 1 2022-01-14 21:06:10 I was so confused and still am- LEO LUNDY Lvl 1 2023-04-16 20:45:48 W T F, Daily Dose Of Internet. 🤣🤣🤣 LEO LUNDY Lvl 1 2023-04-16 20:47:58 thats a big backpack, and a stong little girl Study guides 20 cards ## A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials ➡️ See all cards 3.8 2537 Reviews Godfred Ayeh Lvl 2 2020-12-11 14:01:50 i.First of all, there are 7 children and each child has 2 legs so total legs =14. ii.Each child has 7 bags so 77= 49 bags. iii. Out of 49 bags contains 7 big cats per bag so 7x49= 343 big cats but each cat has 4 legs so 3434=1372 legs. iii.each big cat has 7 small cats so 3437= 2401 but each small cat has 4 legs so total legs =9604 legs iv.In all total legs = 10,990 Godfred Ayeh Lvl 1 2020-12-11 14:08:40 Total cats = 27444=10,976 + 14 = 10, 990 legs Linhsnow Lvl 1 2021-06-26 06:48:56 Wowwww this method is so long yet so helpful…. Thanks! anonymous21 Lvl 3 2020-04-03 09:20:00 7 children have 14 legs. There are 49 bags in total. There are 343 big cats in total, with 1372 legs. There are 2401 small cats in total, with 9604 legs. The total number of legs on the bus is 10990 (if the bus driver was on the bus then there would be 10992 legs in total). Lvl 3 2020-11-17 02:49:24 7 X 2 = 14 7 X 7 = 49 X 4 = 196 40 X 7 = 343 X 4 = 1,372 14 + 196 + 1,372 = 1,582 Lucius Knight Lvl 1 2021-07-14 18:13:45 charvisseein Lvl 2 2020-11-16 13:12:12 hindi ko alam, wala naman ako sa bus Lucius Knight Lvl 5 2021-07-14 18:13:02 So, there is a bus with 7 children inside. And each child has 2 legs. So 7 children x 2 legs each = 14 legs total. And each of those 7 children has a bag with 7 big cats inside. So 7 bags x 7 big cats each = 49 big cats. And each of those 49 big cats has 4 legs. So 49 big cats x 4 legs each = 196 legs in total. And each of those 49 big cats has 7 small cats each. So 49 big cats x 7 small cats each = 343 small cats. And each of those 343 small cats has 4 legs each. So 343 small cats x 4 legs each = 1372 legs. And 14 legs (the sum of all the children's legs) + 196 (the sum of all the big cat's legs) + 1372 (the sum of all the small cat's legs) = 1582 legs in total. Lucius Knight Lvl 1 2021-07-15 16:56:52 Whoops! Each child has 7 bags instead of 1. So the answer is in fact, 10,990. dfdd fdhfg Lvl 6 2021-04-17 08:34:27 7 children : there are 7 bags for each child so there are 49 bags. since each bag has 7 big cats, there are 343 big cats. 343 big cats, have 7 small cats each=2401 small cats 343+2401= 2744 72= 14 27444=10976 10976+14= 10990 legs in total # a m e i l a ៹. Lvl 2 2021-03-09 13:35:08 There are seven girls. 7 girls These seven girls have seven bags apiece. 49 backpacks In each backpack there are seven big cats. 343 big cats And for each big cat there are seven small cats. 2401 small cats Now we must find out how many legs are on the bus. Assuming that each girl has two legs, and each cat has four, 7(2) + 343(4) + 2401(4) =10990 So there are 10990 legs on the bus. 1solutions Lvl 7 2023-05-10 05:08:07 To solve this problem, we need to calculate the total number of legs on the bus. Number of children in the bus = 7 Number of bags with each child = 7 Total number of bags = 7 x 7 = 49 Number of big cats in each bag = 7 Total number of big cats = 49 x 7 = 343 Number of small cats with each big cat = 7 Total number of small cats = 343 x 7 = 2401 Total number of cats = 343 + 2401 = 2744 Total number of legs on each cat = 4 Total number of legs on the bus = 2744 x 4 = 10976 Total number of legs on each child = 7 x 2 = 14 Therefore, there are 14 + 10,976 + "10990" legs on the bus. By : 1solutions.biz shakthipanka panka Lvl 4 2023-04-17 16:07:34 7 children * 7 bags = 49 bags * 7 big cats = 343 big cats * 7 small cats = 2401 total big and small cats * 4 legs of each cat = 10604 + 14 legs of children = 10618 total legs	2,157	6,153	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.46875	4	CC-MAIN-2023-23	latest	en	0.940803
https://nbviewer.jupyter.org/urls/www.numfys.net/media/notebooks/planetary_motion.ipynb	1,579,437,420,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00535.warc.gz	584,494,380	158,721	# Planetary Motion¶ ### Example - Astrophysics¶ Last edited: March 16th 2018 In the modules under ordinary differential equations we have looked at different methods for solving ordinary differential equations (ODE's), and briefly discussed the accuracy of these methods. In this example we will apply the explicit and implicit Euler methods, and the fourth order Runge-Kutta method to calculate the trajectory of the Earth around the Sun. This will show how the accuracy of the methods affect the calculated trajectory. ### Equations of Motion¶ We will simplify the problem by neglecting the gravitational force on the Earth from all other planets, hence only considering the force between the Earth and the Sun. Newton's second law then yields the following equations of motion $$m\vec{a} = m\ddot{\vec{r}}= -\frac{GMm}{r^3}\vec{r},$$ where $m$ is the mass of the Earth, $M$ is the mass of the Sun, $r=\|\vec{r}\|$ is the distance between the Sun and Earth, and $G$ is the gravitational constant. We will try to write this in dimensionless form, using the Aphelion seperation $R$ between the Earth and Sun as a unit, and the period $T$ as time unit. Hence we define $$\tau \equiv\frac{t}{T},\,\rho \equiv \frac{r}{R},\,X \equiv\frac{x}{R}\,\mathrm{and}\,Y\equiv\frac{y}{R}.$$ Inserting this into the equation of motion, dividing away the common factor $m$, we get $$\frac{R}{T^2}\frac{d^2\vec{\rho}}{d\tau^2} = -\frac{GM}{R^2\rho^3}\vec{\rho}.$$ Let's see if we have done this right by checking that the remaining constant is dimensionless, $$\left[\frac{GMT^2}{R^3}\right] = \frac{\mathrm{N(m/kg)^2\cdot kg \cdot s^2}}{m^3} = 1.$$ We define the constant $$C \equiv \frac{GMT^2}{R^3},$$ and get the dimensionless equations of motion $$\frac{d^2\vec{\rho}}{d\tau^2} = -C \frac{\vec{\rho}}{\rho^3}.$$ ### Numerical Implementation¶ Since the angular momentum of the system is conserved in this problem, we will consider the motion of the Earth in a plane, where the Sun is stationary at the origin. Hence we have that $\vec{\rho} = X\hat{x} + Y\hat{y}$, and we want to rewrite the above equation as four first order ODEs in order to use the mentioned numerical methods. We get \begin{align} \frac{dX}{d\tau} &= U,\qquad \frac{dU}{d\tau} = -C\frac{X}{\rho^3},\\ \frac{dY}{d\tau} &= V,\qquad \frac{dV}{d\tau} = -C\frac{Y}{\rho^3}, \end{align} where $U$ and $V$ are the dimensionless velocities in the $x$- and $y$-direction, and $\rho=\sqrt{X^2+Y^2}$. We'll use initial conditions \begin{align} x_0&=R \,&\Rightarrow\, &&&X_0=1,\\ v_0 &= 30.29 \mathrm{ km/s} \,&\Rightarrow\, &&&V_0 = \frac{u T}{R} = 29.29\cdot 10^3 \mathrm{m/s} \frac{365.25 \cdot 24 \cdot 3600 \mathrm{s}}{152.10\cdot 10^9 \mathrm{m}} \approx 6.33, \end{align} and $Y_0 = 0$, $U_0 = 0$. We are now ready to start implementing the different methods. We will use $r$ instead of $\rho$ below. First we import the needed libraries and define the global constants. In [2]: %matplotlib inline import numpy as np import matplotlib.pyplot as plt from matplotlib import rc #Set common figure parameters: newparams = { 'figure.figsize': (16, 5), 'axes.grid': True, 'lines.linewidth': 1.5, 'font.size': 19, 'lines.markersize' : 10, 'mathtext.fontset': 'stix', 'font.family': 'STIXGeneral'} plt.rcParams.update(newparams) T = 365.25243600 # Orbital period [s] R = 152.10e9 # Aphelion distance [m] G = 6.673e-11 # Gravitational constant [N (m/kg)^2] M = 1.9891e30 # Solar mass [kg] m = 5.9726e24 # Earth mass [kg] C = (GMT2)/(R3) #print(C) # Initial conditions X0 = 1 U0 = 0 Y0 = 0 V0 = 29.29e3T/R t_max = 1 # t_max=1 Corresponds to one single complete orbit dt = 0.0001 N = int(t_max/dt) # Number of time steps Let us now use the different numerical methods to calculate the trajectory. In order to see how good the methods are, we will consider the following: • The energy should be conserved, and hence the orbit should be closed. • We know the perihelion seperation (the minimum distance to the Sun), $\rho_\mathrm{perihelion} = 0.967$, and can check how close the trajectory is to this point. Let's begin! ### Explicit Euler Method¶ We first a function which returns the RHS of the four ODEs given above, before implementing the explicit Euler method. In [3]: def RHS(x, y, u, v): """ Returns the time derivatives of X, Y, U and V. Also returns the total energy E of the system.""" r = np.sqrt(x2 + y2) dxdt = u dydt = v dudt = -Cx/r*3 dvdt = -Cy/r*3 E = 0.5(u2+v2)-C/r return (dxdt, dydt, dudt, dvdt, E) X = np.zeros(N) Y = np.zeros(N) U = np.zeros(N) V = np.zeros(N) E = np.zeros(N-1) # Total energy/mass X[0] = X0 V[0] = V0 for n in range(N-1): (dX, dY, dU, dV, E[n]) = RHS(X[n], Y[n], U[n], V[n]) X[n+1] = X[n] + dtdX Y[n+1] = Y[n] + dtdY U[n+1] = U[n] + dtdU V[n+1] = V[n] + dtdV # If t_max was set to exactly one period, it will be interesting # to investigate whether or not the orbit is closed (planet returns # to its starting position) if(t_max==1): # Find offset print("\nSmall offset indicates closed orbit") print("Offset in 'x': %0.3e - %0.7e = %0.7e" % (X[0], X[-1], X[-1]-X[0])) print("Offset in 'y': %0.3e - %0.7e = %0.7e" % (Y[0], Y[-1], Y[-1]-Y[0])) print("Total offset: %0.3e" % np.sqrt((X[0]-X[-1])2+(Y[0]-Y[-1])2)) # Find perihelion seperation: r_perihelion = abs(min(Y)) print("\nThe perihelion seperation is %0.3f, compared to 0.967." % r_perihelion) plt.figure() plt.plot(X, Y, 'g', [0], [0], 'ro') plt.title("Explicit Euler") plt.xlabel(r"$x$") plt.ylabel(r"$y$") plt.grid() plt.figure() plt.plot(E) plt.title('Energy per unit mass') plt.ylabel("Energy") plt.xlabel(r"$n$") plt.grid(); Small offset indicates closed orbit Offset in 'x': 1.000e+00 - 1.0073783e+00 = 7.3783114e-03 Offset in 'y': 0.000e+00 - -3.4487638e-02 = -3.4487638e-02 Total offset: 3.527e-02 The perihelion seperation is 0.990, compared to 0.967. ### Implicit Euler Method¶ When using the Implicit Euler Method, we evaluate the RHSs of the ODEs at the new timestep. We hence get the following equations \begin{align} X_{n+1} &= X_{n} + \Delta t U_{n+1},\\ U_{n+1} &= U_n - \Delta t C\frac{X_{n+1}}{\rho_{n+1}^3},\\ Y_{n+1} &= Y_{n} + \Delta t V_{n+1},\\ V_{n+1} &= V_n - \Delta t C\frac{Y_{n+1}}{\rho_{n+1}^3}. \end{align} We will use the root solver, scipy.optimize.root, to find the values at the new time step in the equations above. In [4]: from scipy.optimize import root def func(X, X0): """ Function returning the difference between the RHS and LHS of the four first order ODEs above. X: vector of new coordinates X0: vector of old coordinates """ X_ = X[0] Y_ = X[1] U_ = X[2] V_ = X[3] X0_ = X0[0] Y0_ = X0[1] U0_ = X0[2] V0_ = X0[3] r = np.sqrt(X_2 + Y_2) return [X_ - (X0_ + dtU_), Y_ - (Y0_ + dtV_), U_ - (U0_ - dtCX_/r*3), V_ - (V0_ - dtCY_/r3)] X = np.zeros(N) Y = np.zeros(N) U = np.zeros(N) V = np.zeros(N) X[0] = X0 V[0] = V0 for n in range(N-1): X0_ = np.array((X[n], Y[n], U[n], V[n])) sol = root(func, x0=X0_, args=X0_, jac=False) X[n+1] = sol.x[0] Y[n+1] = sol.x[1] U[n+1] = sol.x[2] V[n+1] = sol.x[3] # If t_max was set to exactly one period, it will be interesting # to investigate whether or not the orbit is closed (planet returns # to its starting position) if(t_max==1): # Find offset print("\nSmall offset indicates closed orbit") print("Offset in 'x': %0.3e - %0.7e = %0.7e" % (X[0], X[-1], X[-1]-X[0])) print("Offset in 'y': %0.3e - %0.7e = %0.7e" % (Y[0], Y[-1], Y[-1]-Y[0])) print("Total offset: %0.3e" % np.sqrt((X[0]-X[-1])2+(Y[0]-Y[-1])2)) # Find perihelion seperation: r_perihelion = abs(min(Y)) print("\nThe perihelion seperation is %0.3f, compared to 0.967." % r_perihelion) plt.figure() plt.plot(X, Y, 'g', [0], [0], 'ro') plt.title("Implicit Euler") plt.xlabel(r"$x$") plt.ylabel(r"$y$") plt.grid() # Calculate energy r = np.sqrt(X2 + Y2) E = 0.5(U2+V2)-C/r plt.figure() plt.plot(E) plt.title('Energy per unit mass') plt.ylabel("Energy") plt.xlabel(r"$n$") plt.grid(); Small offset indicates closed orbit Offset in 'x': 1.000e+00 - 9.9124581e-01 = -8.7541899e-03 Offset in 'y': 0.000e+00 - 3.7544951e-02 = 3.7544951e-02 Total offset: 3.855e-02 The perihelion seperation is 0.976, compared to 0.967. ### Fourth Order Runge-Kutta Method¶ We will in the following solve the system one last time, utilzing the Fourth Order Runge-Kutta Method (RKM4). Since the RKM4 makes use of test-points, we'll separate the four equations into separate functions. In [5]: def F(X_, Y_, U_, V_): # dX/dtau return U_ def G(X_, Y_, U_, V_): # dY/dtau return V_ def H(X_, Y_, U_, V_): # dU/dtau return -C * ( X_/( (np.sqrt(X_2 + Y_2) )*3) ) def I(X_, Y_, U_, V_): # dV/dtau return -C ( Y_/( (np.sqrt(X_2 + Y_2) )*3) ) X_4RK = np.zeros(N) Y_4RK = np.zeros(N) U_4RK = np.zeros(N) V_4RK = np.zeros(N) E_4RK = np.zeros(N-1) # Total energy/mass X_4RK[0] = X0 V_4RK[0] = V0 for n in range(N-1): k_x1 = dt F( X_4RK[n], Y_4RK[n], U_4RK[n], V_4RK[n] ) k_y1 = dt * G( X_4RK[n], Y_4RK[n], U_4RK[n], V_4RK[n] ) k_u1 = dt * H( X_4RK[n], Y_4RK[n], U_4RK[n], V_4RK[n] ) k_v1 = dt * I( X_4RK[n], Y_4RK[n], U_4RK[n], V_4RK[n] ) k_x2 = dt * F( X_4RK[n] + k_x1/2, Y_4RK[n] + k_y1/2, U_4RK[n] + k_u1/2, V_4RK[n] + k_v1/2 ) k_y2 = dt * G( X_4RK[n] + k_x1/2, Y_4RK[n] + k_y1/2, U_4RK[n] + k_u1/2, V_4RK[n] + k_v1/2 ) k_u2 = dt * H( X_4RK[n] + k_x1/2, Y_4RK[n] + k_y1/2, U_4RK[n] + k_u1/2, V_4RK[n] + k_v1/2 ) k_v2 = dt * I( X_4RK[n] + k_x1/2, Y_4RK[n] + k_y1/2, U_4RK[n] + k_u1/2, V_4RK[n] + k_v1/2 ) k_x3 = dt * F( X_4RK[n] + k_x2/2, Y_4RK[n] + k_y2/2, U_4RK[n] + k_u2/2, V_4RK[n] + k_v2/2 ) k_y3 = dt * G( X_4RK[n] + k_x2/2, Y_4RK[n] + k_y2/2, U_4RK[n] + k_u2/2, V_4RK[n] + k_v2/2 ) k_u3 = dt * H( X_4RK[n] + k_x2/2, Y_4RK[n] + k_y2/2, U_4RK[n] + k_u2/2, V_4RK[n] + k_v2/2 ) k_v3 = dt * I( X_4RK[n] + k_x2/2, Y_4RK[n] + k_y2/2, U_4RK[n] + k_u2/2, V_4RK[n] + k_v2/2 ) k_x4 = dt * F( X_4RK[n] + k_x3, Y_4RK[n] + k_y3, U_4RK[n] + k_u3, V_4RK[n] + k_v3 ) k_y4 = dt * G( X_4RK[n] + k_x3, Y_4RK[n] + k_y3, U_4RK[n] + k_u3, V_4RK[n] + k_v3 ) k_u4 = dt * H( X_4RK[n] + k_x3, Y_4RK[n] + k_y3, U_4RK[n] + k_u3, V_4RK[n] + k_v3 ) k_v4 = dt * I( X_4RK[n] + k_x3, Y_4RK[n] + k_y3, U_4RK[n] + k_u3, V_4RK[n] + k_v3 ) X_4RK[n+1] = X_4RK[n] + k_x1/6 + k_x2/3 + k_x3/3 + k_x4/6 Y_4RK[n+1] = Y_4RK[n] + k_y1/6 + k_y2/3 + k_y3/3 + k_y4/6 U_4RK[n+1] = U_4RK[n] + k_u1/6 + k_u2/3 + k_u3/3 + k_u4/6 V_4RK[n+1] = V_4RK[n] + k_v1/6 + k_v2/3 + k_v3/3 + k_v4/6 E_4RK[n] = 0.5(U_4RK[n+1]2+V_4RK[n+1]2)-C/np.sqrt(X_4RK[n+1]2 + Y_4RK[n+1]2) # If t_max was set to exactly one period, it will be interesting # to investigate whether or not the orbit is closed (planet returns # to its starting position) if(t_max==1): # Find offset print("\nSmall offset indicates closed orbit") print("Offset in 'x': %0.3e - %0.7e = %0.7e" % (X_4RK[0], X_4RK[-1], X_4RK[N-1]-X_4RK[0])) print("Offset in 'y': %0.3e - %0.7e = %0.7e" % (Y_4RK[0], Y_4RK[-1], Y_4RK[N-1]-Y_4RK[0])) print("Total offset: %0.3e" % np.sqrt((X_4RK[0]-X_4RK[-1])2+(Y_4RK[0]-Y_4RK[-1])*2)) # Find perihelion seperation: r_perihelion = abs(min(Y_4RK)) print("\nThe parahelion seperation is %0.3f, compared to 0.967." % r_perihelion) plt.figure() plt.title('4th order Runge-Kutta') plt.plot(X_4RK, Y_4RK, 'g', [0], [0], 'ro') plt.xlabel(r"$x$") plt.ylabel(r"$y$") plt.grid() plt.figure() plt.title('Energy per unit mass') plt.plot(E_4RK) plt.ylabel("Energy") plt.xlabel(r"$n$") plt.grid() Small offset indicates closed orbit Offset in 'x': 1.000e+00 - 9.9999892e-01 = -1.0753348e-06 Offset in 'y': 0.000e+00 - 1.4540584e-03 = 1.4540584e-03 Total offset: 1.454e-03 The parahelion seperation is 0.983, compared to 0.967. From the above results we see that the Runge-Kutta method is in fact the most accurate one: while the plot of energy/mass for the two first methods show large deviations, the energy/mass is close to constant for the RKM4. This leads to a nearly to closed orbit. This should not be surprising as the RKM4 is implemented as a 4th-order method, and hence should be superior to the explicit and implicit Euler methods, which are 1st-order methods, in terms of accuracy.	4,821	12,180	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0}	4.25	4	CC-MAIN-2020-05	longest	en	0.835079
http://inperc.com/wiki/index.php?title=Cochain_complexes_and_cohomology	1,553,560,105,000,000,000	text/html	crawl-data/CC-MAIN-2019-13/segments/1552912204736.6/warc/CC-MAIN-20190325234449-20190326020449-00052.warc.gz	100,567,254	12,364	##### Tools This site is devoted to mathematics and its applications. Created and run by Peter Saveliev. # Cochain complexes and cohomology ## 1 Cochain complexes and cohomology • $\Omega^1({\bf R})$ is the set of $1$-forms in ${\bf R}$, and • $\Omega^0({\bf R})$ is the set of $0$-forms in ${\bf R}$ both are vector spaces, very familiar objects. However, as vector spaces they aren't the simplest ones, because they are infinite dimensional: $$\dim \Omega^0({\bf R}) = \dim C({\bf R}) = \infty.$$ Indeed, observe that the set $P = \{1, x, x^2, \ldots\}$ is linearly independent in the space of functions $C({\bf R}) = \Omega^0({\bf R})$. Further, does $P$ span the whole space? Exercise. Hint: consider Taylor series. Theorem. $$\dim C({\bf R}) = \infty.$$ Proof. Are there are $a_0, \ldots, a_n$ not all equal to $0$ with: $$a_0+a_1x+a_2x^2+\ldots+a_nx^n = 0?$$ The equation here holds for every $x$. In particular, if we pick $x=0$, then, immediately, $a_0=0$. This implies: $$a_1x+a_2x^2+\ldots+a_nx^n = 0,$$ hence $$x(a_1+a_2x+\ldots+a_nx^{n-1}) = 0.$$ But if we pick, $x \neq 0$ now, then $$a_1 + a_2x + \ldots+a_n x^{n-1}=0.$$ We continue and, by induction, prove that all $a_k$ are zero. $\blacksquare$ The difference now, with discrete forms, is that it is finite dimensional: $$\dim C^0[0,k] = k+1 < \infty.$$ Therefore everything is computable! Generally, if $K$ is a cubical complex, $\dim C^k(K) =$ number of $k$-cells in $K$. The result is plausible if you think about what a discrete $k$-form is. It's a correspondence: $\varphi \colon k$-cell $\mapsto$ a number. So, the number of cells is the number of degrees of freedom as given by the dual basis. Consider the cochain complex: $$\ldots \stackrel{d_{k-1}}{\rightarrow} C^k \stackrel{d_k}{\rightarrow} C^{k+1} \stackrel{d_{k+1}}{\rightarrow} C^{k+2} \stackrel{d_{k+2}}{\rightarrow} \ldots$$ Just as with continuous forms we define two new vector spaces: • ${\rm ker \hspace{3pt}} d_k$ is the set of closed $k$-forms in $K$, also known as cocycles and • ${\rm im \hspace{3pt}} d_{k-1}$ is the set of exact $k$-forms in $K$, also known as coboundaries. Since $dd=0$, as we know, then • "every exact form is closed" or • "every coboundary is a cocycle". Algebraically, $${\rm im} d_{k-1} \subset {\rm ker \hspace{3pt}} d_k$$ The end result is identical to that for continuous forms, which is: Each $\Omega$ is replaced with $C$. Define the cubical cohomology of $K$ as this quotient vector space: $$H^k(K)=\ker d_k / {\rm im \hspace{3pt}} d_{k-1}.$$ The quotient is, of course, very similar to the one that defined cubical homology. The difference is that the former isn't just a vector space, it also has a graded ring structure provided by the wedge product. To see the difference this makes, consider these two spaces: the sphere with two bows added and the torus: The homology is the same in all dimensions. The cohomology is also the same, but only in the sense of vector spaces! The basis elements in dimension $1$ behave differently under wedge product. In the sphere with bows: $$[\alpha ^]\wedge [\beta ^]=0,$$ because there is nowhere for this $2$-form to "reside". Meanwhile in the torus: $$[a ^]\wedge [b ^]\ne0.$$ ## 2 Connectedness and simple connectedness The following is very similar to the continuous case (why? because the linear algebra is the same). Proposition. Constant functions are closed $0$-forms. Proof. They are defined on vertices. If $\varphi \in C^0({\bf R})$, $d \varphi([a,a+1]) = [\varphi[(a+1)-\varphi(a)]dx = 0dx = 0$. A similar argument applies to $C^0({\bf R}^n)$. $\blacksquare$ Proposition. Closed $0$-forms are constant on a path-connected cubical complex (i.e., its realization $\|K\|$ is path-connected). Proof. For ${\bf R}^1$: $d \varphi([a,a+1]) = 0$, so $(\varphi(a+1) - \varphi(a))dx = 0$, so $\varphi(a+1) = \varphi(a)$, so $\varphi$ is constant on adjacent vertices. $\blacksquare$ For the general case $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \bullet & \ra{} & \bullet & & \\ \ua{} & & \da{} & & \\ \bullet & & \bullet & \ra{} & \bullet & \ra{} & \bullet \end{array}$$ all you need is this easy Lemma: A cubical complex is path-connected iff any two vertices can be connected by a sequence of adjacent edges. Exercise. Prove the lemma. Exercise. Use it to finish the proof. Corollary. $\dim \ker d_0 =$ number of path components of $\|K\|$. What about the exact $0$-forms? Just $0$. To summarize: Theorem: $H^0(K) = {\bf R}^m$, where $m$ is the number of path components of $\|K\|$. Now, simple connectedness. Recall this fact for continuous forms: if $R$ is simply connected, then all closed forms are exact. Hence $H_{dR}^1(R) = 0$. Same holds for discrete forms: Theorem. If $\|K\|$ is simply connected, $H^1(K)=0$. The issue is complex for continuous differential forms. In order to detect -- cohomologically -- the hole in ${\bf R}^2-\{(0,0)\}$ we needed to prove that this form $$\theta = \frac{1}{x^2+y^2} (-ydx + xdy)$$ • is closed (easy), and • isn't exact (hard) (see Closedness and exactness of 1-forms). Then we know that $$H^1_{dR}\ne 0.$$ With cubical forms, we get the corresponding result, and more, a lot cheaper. Consider the cubical counterpart of the circle: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \bullet& \ra{ } & \bullet \\ \ua{ } & & \ua{ } \\ \bullet& \ra{ } & \bullet \end{array}$$ Here the arrow indicate the orientations of the edges -- along the coordinate axes. A $1$-form is just a combination of four numbers: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \bullet& \ra{q} & \bullet \\ \ua{p} & & \ua{r} \\ \bullet& \ra{s} & \bullet \end{array}$$ First, which of these forms are closed? They should have "horizontal difference - vertical difference" equal to $0$: $$(r-p)-(q-s)=0.$$ For example, we can choose them all equal to $1$. However, all $1$-forms are closed, even those that don't satisfy this equation! Why? Exercise. Second, what $1$-forms are exact? Here is a $0$-form and its exterior derivative: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \bullet^1 & \ra{1} & \bullet^2 \\ \ua{1} & & \ua{1} \\ \bullet^0 & \ra{1} & \bullet^1 \end{array}$$ So, this $1$-form is exact. But this one isn't: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \bullet & \ra{1} & \bullet \\ \ua{1} & & \ua{-1} \\ \bullet & \ra{-1} & \bullet \end{array}$$ This is easy to prove by solving a little system of linear equations (and it is easy to see by observing that the complete trip around the square adds up to $4$ not $0$). So, $H^1\ne 0$. ## 3 Computing cohomology We have used discrete forms to detect the hole in the circle. Now, let's compute the whole thing avoiding shortcuts (theorems), just like a computer would do. Back to our cubical complex $K$: Let's compute $H^1(K)$. Consider the cochain complex: $$C^0 \stackrel{d_0}{\rightarrow} C^1 \stackrel{d_1}{\rightarrow} C^2 = 0.$$ Observe that, since $d_1=0$ we have $\ker d_1 = C^1$. Let's list the bases of these vector spaces. Of course, we start with the bases of the chains $C_k$, i.e., the cells, and consider the dual bases of the cochains $C^k$. After all, $C^k=(C_k)^$. Below, in leftmost column contains the cells, the uppermost row contains the forms, and the rest is the values of these forms on these cells. $C^0:$ $\varphi_A$ $\varphi_B$ $\varphi_C$ $\varphi_D$ $A$ $1$ $0$ $0$ $0$ $B$ $0$ $1$ $0$ $0$ $C$ $0$ $1$ $0$ $0$ $D$ $0$ $0$ $0$ $1$ $C^1:$ $\psi_a$ $\psi_b$ $\psi_c$ $\psi_d$ $a$ $1$ $0$ $0$ $0$ $b$ $0$ $1$ $0$ $0$ $c$ $0$ $0$ $1$ $0$ $d$ $0$ $0$ $0$ $1$ We find the formula for $d_0$, a linear operator, which is its $4 \times 4$ matrix. For that, we just look at what happens to the basis elements: $\varphi_A = [1, 0, 0, 0]^T= \begin{array}{ccc} 1 & - & 0 \\ \| & & \| \\ 0 & - & 0 \end{array} \Longrightarrow d_0 \varphi_A = \begin{array}{ccc} \bullet & -1 & \bullet \\ 1 & & 0 \\ \bullet & 0 & \bullet \end{array} = \psi_a-\psi_b = [1,-1,0,0]^T;$ $\varphi_B = [0, 1, 0, 0]^T= \begin{array}{ccc} 0 & - & 1 \\ \| & & \| \\ 0 & - & 0 \end{array} \Longrightarrow d_0 \varphi_B = \begin{array}{ccc} \bullet & 1 & \bullet \\ 0 & & 1 \\ \bullet & 0 & \bullet \end{array} = \psi_b+\psi_c = [0,1,1,0]^T;$ $\varphi_C = [0, 0, 1, 0]^T= \begin{array}{ccc} 0 & - & 0 \\ \| & & \| \\ 0 & - & 1 \end{array} \Longrightarrow d_0 \varphi_C = \begin{array}{ccc} \bullet & 0 & \bullet \\ 0 & & -1 \\ \bullet & 1 & \bullet \end{array} = -\psi_c+\psi_d = [0,0,-1,1]^T;$ $\varphi_D = [0, 0, 0, 1]^T= \begin{array}{ccc} 0 & - & 0 \\ \| & & \| \\ 1 & - & 0 \end{array} \Longrightarrow d_0 \varphi_D = \begin{array}{ccc} \bullet & 0 & \bullet \\ -1 & & 0 \\ \bullet & -1 & \bullet \end{array} = -\psi_a-\psi_d = [-1,0,0,-1]^T.$ Now, the matrix of $d_0$ is formed by the column vectors above: $$d_0 = \left( \begin{array}{cccc} 1 & 0 & 0 & -1 \\ -1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \end{array} \right).$$ Now, using this data we find the kernel and the image using the standard linear algebra. The kernel is the solution set of the equation $d_0v=0$. It may be found by solving the corresponding system of linear equations with the coefficient matrix $d_0$. The brute force approach is the Gaussian elimination, or something more sophisticated for a computer implementation. We simply notice that the rank of the matrix is $3$, so the dimension of the kernel is $1$. Therefore, $$\dim H^0=1.$$ We have a single component! The image is the set of $u$'s with $d_0v=u$. Once again Gaussian elimination is a simple but effective approach. We simply notice that the dimension of the image is the rank of the matrix, $3$. In fact, $${\rm span} \{ d_0(\varphi_A), d_0(\varphi_B), d_0(\varphi_C), d_0(\varphi_D) \} = {\rm im \hspace{3pt}} d_0 .$$ Therefore $$\dim H^1=\dim \left( C^1 / {\rm im \hspace{3pt}} d_0 \right) = 4-3=1.$$ We have a single hole! Note: Another way to see that the columns aren't linearly independent is just add them or take the determinant of $d_0$. Exercise: Compute the cohomology of figure 8: ## 4 There is more to homology... This is just a tip of the iceberg. There is a lot more to homology and cohomology. Let's take a quick peak, without diving in... First, the result that holds everything together: $$\|K\|=\|L\| \Longrightarrow H_k(K) \cong H_k(L).$$ In other words, homology is independent from the cubical complex representation. This is known as Invariance of homology. Further in this direction, homology is independent from the method of discretization. Indeed, a topological space can be triangulated or even build in vacuum using only gluing. The result is a simplicial complex or a cell complex, but the homology is the same! There is a more direct way of dealing with the homology of topological spaces, without discretization. Singular homology deals with collections of maps from cells to the space. The chain complexes are infinite dimensional but the homology is still the same, under mild conditions. One can also introduce homology and cohomology axiomatically, via the Eilenberg–Steenrod axioms. Dropping some of the axioms leads to "extraordinary homology theories", such as bordism. The most immediate and the most profound algebraic extension of these concepts is to consider chains as "linear combinations" of cells with coefficients in an arbitrary ring $R$. As a result, what the chains and cochains form isn't vector spaces anymore but modules. These modules, $H_k(K;R)$ and $H^k(K;R)$, have the same bases and, especially in the discrete case, behave very much like those vector spaces. However, when the homology and cohomology is computed new effects appear. There is essentially one way for a Euclidean space to fit into another. That's why quotient vector spaces are always quite simple: $${\bf R}^n / {\bf R}^m ={\bf R}^{n-m},n\ge m.$$ However, a copy of ${\bf Z}$ can fit "non-trivially" in another copy of ${\bf Z}$ as the set of even numbers $2{\bf Z}$ and we have: $${\bf Z} / 2{\bf Z} = {\bf Z}_2 .$$ This is called "torsion" (see Classification of free abelian groups). This is the reason why we call them homology and cohomology groups. The free part looks the same in the sense that they have the same number of generators. Hence, the Betti numbers are the same for all coefficients. What's the difference then? As an example, the integer $1$-homology detects the twist of the Klein bottle ${\bf K}^2$ while the real homology does not: $$H_1({\bf K}^2;{\bf Z})={\bf Z}\oplus {\bf Z}_2, H_1({\bf K}^2;{\bf R})={\bf R}.$$ Compare to the circle: $$H_1({\bf S}^1;{\bf Z})= {\bf Z},H_1({\bf S}^1;{\bf R})= {\bf R}.$$ In fact, the Universal Coefficient Theorem states that the homology and cohomology groups over any ring can be deduced from the integer homology via an algebraic procedure. However, the converse isn't true. Therefore, the integer homology is the best! And it is the default: $H_k(X)=H_k(X;{\bf Z})$. Note: what about calculus vs integer-valued calculus? Homology respects maps that respect boundaries. One can think of a function between two cubical complexes $f:K\rightarrow L$ as one that preserves cells, i.e., the image of a cell is also a cell, possibly of different dimension: $$a\in K \Longrightarrow f(a) \in L.$$ If we expand it by linearity we have map between chain, and cochain, complexes $$f_:C_k(K) \rightarrow C_k(L),$$ $$f^:C^k(K) \leftarrow C^k(L).$$ To make $f$ continuous, in the combinatorial sense, we require, in addition, that $f$ preserves boundaries: $$f(\partial a)=\partial f(a).$$ Then we have "a chain and a cochain maps" that make these diagrams commute: $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccc} \la{\partial} & C_k(K) & \la{\partial} & C_{k+1}(K) &\la{\partial} \\ & \da{f_} & & \da{f_} \\ \la{\partial} & C_k(L) & \la{\partial} & C_{k+1}(L) & \la{\partial} \end{array}$$ and $$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} \newcommand{\la}[1]{\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!} \newcommand{\ua}[1]{\left\uparrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ccccccccccccc} \ra{d} & C ^k(K) & \ra{d} & C ^{k+1}(K) &\ra{d} \\ & \ua{f^} & & \ua{f^} \\ \ra{d} & C ^k(L) & \ra{d} & C ^{k+1}(L) & \ra{d} \end{array}$$ They generate homology and cohomology operators: $$f_:H_k(K) \rightarrow H_k(L),$$ $$f^*:H^k(K) \leftarrow H^k(L),$$ which makes $f$ "natural" with respect to homology and cohomology. These two theories are functors. There is more duality between homology and cohomology: $$H^k(M^n) \cong H_{n-k}(M^n)$$ for any orientable compact path connected $n$-manifold $M$. It is called Poincare duality. It is realized by means of a new product operator, the cap product: $$\frown : H_n \times H^m \rightarrow H^{n-m}.$$ It is in addition to the familiar wedge product, aka the cup product: $$\smile : H^n \times H^m \rightarrow H^{n+m}.$$	5,639	16,788	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.8125	4	CC-MAIN-2019-13	latest	en	0.831672
https://bigideasmathanswers.com/big-ideas-math-geometry-answers-chapter-10/	1,675,181,575,000,000,000	text/html	crawl-data/CC-MAIN-2023-06/segments/1674764499888.62/warc/CC-MAIN-20230131154832-20230131184832-00409.warc.gz	158,557,319	54,996	Big Ideas Math Geometry Answers Chapter 10 Circles Gain more subject knowledge on Circles concepts with the help of Big Ideas Math Geometry Answers Chapter 10 Circles guide. Shared guide of BIM Textbook Solutions Geometry Ch 10 Circles helps students understand the concepts quickly and concentrate on other math concepts too. Want to get a good grip on the high school-related circle topics? Then, make use of this BigIdeas Math Geometry Answers Ch 10 Circles Pdf for free. Subject experts provided this BIM geometry 10th chapter solution key based on the latest syllabus and common core standards curriculum guidelines. Big Ideas Math Book Geometry Answer Key Chapter 10 Circles Presented Geometry Chapter 10 Circles Big Ideas Math Answers are easy to understand and free to access at any time and anywhere. So, students are advised to access online or download the lesson-wise BIM Textbook Answers of Geometry Ch 10 Circles in Pdf format. Check out each and every lesson of circles as a part of your preparation and clear your queries within no time. Common Core Curriculum High School Big Ideas Math Book Geometry Ch 10 Solution Key are prepared as per the latest textbooks. Questions included in the BIM Modeling Real Life Geometry Answers are from Exercises, Chapter Tests, Review Tests, Cumulative Practice, Assessment Tests, etc. Circles Maintaining Mathematical Proficiency Find the Product. Question 1. (x + 7) (x + 4) (x + 7) (x + 4) = x² + 14x + 28 Explanation: (x + 7) (x + 4) = x(x + 7) + 7(x + 4) = x² + 7x + 7x + 28 = x² + 14x + 28 Question 2. (a + 1) (a – 5) (a + 1) (a – 5) = a² – 4a – 5 Explanation: (a + 1) (a – 5) = a(a – 5) + 1(a – 5) = a² – 5a + a – 5 = a² – 4a – 5 Question 3. (q – 9) (3q – 4) (q – 9) (3q – 4) = 3q² – 31q + 32 Explanation: (q – 9) (3q – 4) = q(3q – 4) – 9(3q – 4)Exercise 10.3 Using Chords = 3q² – 4q – 27q + 32 = 3q² – 31q + 32 Question 4. (2v – 7) (5v + 1) (2v – 7) (5v + 1) = 10v² – 33v – 7 Explanation: (2v – 7) (5v + 1) = 2v(5v + 1)- 7(5v + 1) = 10v² + 2v – 35v – 7 = 10v² – 33v – 7 Question 5. (4h + 3) (2 + h) (4h + 3) (2 + h) = 4h² + 11h + 6 Explanation: (4h + 3) (2 + h) = 4h(2 + h) + 3(2 + h) = 8h + 4h² + 6 + 3h = 4h² + 11h + 6 Question 6. (8 – 6b) (5 – 3b) (8 – 6b) (5 – 3b) = 18b² – 54b + 40 Explanation: (8 – 6b) (5 – 3b) = 8(5 – 3b) – 6b(5 – 3b) = 40 – 24b – 30b + 18b² = 18b² – 54b + 40 Solve the equation by completing the square. Round your answer to the nearest hundredth, if necessary. Question 7. x2 – 2x = 5 The solutions are x = √6 + 1, x = 1 – √6 Explanation: x² – 2x = 5 x² – 2x + 1² = 5 + 1² (x – 1)² = 6 x – 1 = ±√6 x = ±√6 + 1 The solutions are x = √6 + 1, x = -√6 + 1 Question 8. r2 + 10r = -7 The solutions are r = √18 – 5, r = 5 – √18 Explanation: r2 + 10r = -7 r² + 10r + 5² = -7 + 5² (r + 5)² = -7 + 25 = 18 r + 5 = ±√18 r = ±√18 – 5 The solutions are r = √18 – 5, r = 5 – √18 Question 9. w2 – 8w = 9 The solutions are w = 9, w = -1 Explanation: w2 – 8w = 9 w2 – 8w + 4² = 9 + 4² (w – 4)² = 9 + 16 = 25 w – 4 = ±5 w = 5 + 4, w = -5 + 4 w = 9, w = -1 The solutions are w = 9, w = -1 Question 10. p2 + 10p – 4 = 0 The solutions are p = √29 – 5, p = 5 – √29 Explanation: p2 + 10p = 4 p² + 10p + 5² = 4 + 5² (p + 5)² = 4 + 25 (p + 5)² = 29 p + 5 = ±√29 p = ±√29 – 5 The solutions are p = √29 – 5, p = 5 – √29 Question 11. k2 – 4k – 7 = 0 The solutions are k = √11 + 2, k = 2 – √11 Explanation: k² – 4k= 7 k² – 4k + 2² = 7 + 4 (k – 2)² = 11 k – 2 = ±√11 k = √11 + 2, k = 2 – √11 The solutions are k = √11 + 2, k = 2 – √11 Question 12. – z2 + 2z = 1 The solutions are z = 1 Explanation: -z² + 2z = 1 z² – 2z = -1 z² – 2z + 1 = -1 + 1 (z – 1)² = 0 z = 1 The solutions are z = 1 Question 13. ABSTRACT REASONING write an expression that represents the product of two consecutive positive odd integers. Explain your reasoning. Let us take two consecutive odd integers are x and (x + 2) The product of two consecutive odd integers is x • (x + 2) x(x + 2) = x² + 2x Circles Mathematical Practices Monitoring progress Let ⊙A, ⊙B, and ⊙C consist of points that are 3 units from the centers. Question 1. Draw ⊙C so that it passes through points A and B in the figure at the right. Explain your reasoning. Question 2. Draw ⊙A, ⊙B, and OC so that each is tangent to the other two. Draw a larger circle, ⊙D, that is tangent to each of the other three circles. Is the distance from point D to a point on ⊙D less than, greater than, or equal to 6? Explain. 10.1 Lines and Segments that Intersect Circles Exploration 1 Lines and Line Segments That Intersect Circles Work with a partner: The drawing at the right shows five lines or segments that intersect a circle. Use the relationships shown to write a definition for each type of line or segment. Then use the Internet or some other resource to verify your definitions. Chord: _________________ Secant: _________________ Tangent: _________________ Diameter: _________________ Chord: A chord of a circle is a straight line segment whose endpoints both lie on a circular arc. Secant: A straight line that intersects a circle in two points is called a secant line. Tangent: Tangent line is a line that intersects a curved line at exactly one point. Radius: It is the distance from the centre of the circle to any point on the circle. Diameter: It the straight that joins two points on the circle and passes through the centre of the circle. Exploration 2 Using String to Draw a Circle Work with a partner: Use two pencils, a piece of string, and a piece of paper. a. Tie the two ends of the piece of string loosely around the two pencils. b. Anchor one pencil of the paper at the center of the circle. Use the other pencil to draw a circle around the anchor point while using slight pressure to keep the string taut. Do not let the string wind around either pencil. c. Explain how the distance between the two pencil points as you draw the circle is related to two of the lines or line segments you defined in Exploration 1. REASONING ABSTRACTLY To be proficient in math, you need to know and flexibly use different properties of operations and objects. Question 3. What are the definitions of the lines and segments that intersect a circle? Question 4. Of the five types of lines and segments in Exploration 1, which one is a subset of another? Explain. Question 5. Explain how to draw a circle with a diameter of 8 inches. Lesson 10.1 Lines and Segments that Intersect Circles Monitoring progress Question 1. In Example 1, What word best describes $$\overline{A G}$$? $$\overline{C B}$$? $$\overline{A G}$$ is secant because it is a line that intersects the circle at two points. $$\overline{C B}$$ is the radius as it is the distance from the centre to the point of a circle. Question 2. In Example 1, name a tangent and a tangent segment. $$\overline{D E}$$ is the tangent of the circle $$\overline{D E}$$ is the tangent segment of the circle. Tell how many common tangents the circles have and draw them. State whether the tangents are external tangents or internal tangents. Question 3. 4 tangents. A tangent is a line segment that intersects the circle at exactly one point. Internal tangents are the lines that intersect the segments joining the centres of two circles. External tangents are the lines that do not cross the segment joining the centres of the circles. Blue lines represent the external tangents and red lines represent the internal tangents. Question 4. One tangent. One external tangent. Question 5. No tangent. It is not possible to draw a common tangent for this type of circle. Question 6. Is $$\overline{D E}$$ tangent to ⊙C? Use the converse of Pythagorean theorem i.e 2² = 3² + 4² 4 = 9 + 16 By the tangent line to the circle theorem, $$\overline{D E}$$ is not a tangent to ⊙C Question 7. $$\overline{S T}$$ is tangent to ⊙Q. The radius of ⊙Q is 7 units. Explanation: By using the Pythagorean theorem (18 + r)² = r² + 24² 324 + 36r + r² = r² + 576 36r = 576 – 324 36r = 252 r = 7 Question 8. Points M and N are points of tangency. Find the value(s) of x. The values of x are 3 or -3. Explanation: x² = 9 x = ±3 Exercise 10.1 Lines and Segments that Intersect Circles Vocabulary and Core Concept Check Question 1. WRITING How are chords and secants alike? How are they different? Question 2. WRITING Explain how you can determine from the context whether the words radius and diameter are referring to segments or lengths. Radius and diameter are the lengths of the line segments that pass through the centre of a circle. Radius is half of the diameter. Question 3. COMPLETE THE SENTENCE Coplanar circles that have a common center are called ____________ . Question 4. WHICH ONE DOESNT BELONG? Which segment does not belong with the other three? Explain your reasoning. A chord, a radius and a diamter are segments and they intersect a circle in two points. A tangent is a line that intersects a circle at one point. Monitoring Progress and Modeling with Mathematics In Exercises 5 – 10, use the diagram. Question 5. Name the circle. Question 6. The name of the two radii is CD and AC. Question 7. Name two chords. Question 8. Name a diameter. The name of diameter is AD Question 9. Name a secant. Question 10. Name a tangent and a point of tangency GE is the tangent and F is the point of tangency. In Exercises 11 – 14, copy the diagram. Tell how many common tangents the circles have and draw them. Question 11. Question 12. No common tangent because two circles do not intersect at one point. Question 13. Question 14. One common tangent. In Exercises 15 – 18, tell whether the common tangent is internal or external. Question 15. Question 16. The common tangent is the internal tangent because it intersects the segment that joins the centres of two circles. Question 17. Question 18. The common tangent is the internal tangent because it intersects the segment that joins the centres of two circles. In Exercises 19 – 22, tell whether $$\overline{A B}$$ is tangent to ⊙C. Explain your reasoning. Question 19. Question 20. Use the converse of the Pythagorean theorem 18² _____________ 15² + 9² 324 _____________ 225 + 81 324 ≠ 304 △ ACB is not a right angled triangle. So, $$\overline{A B}$$ is not tangent to ⊙C at B. Question 21. Question 22. Use the converse of the Pythagorean theorem 8² _____________ 12² + 16² 64 _____________ 144 + 256 64 ≠ 400 △ ACB is not a right angled triangle. So, $$\overline{A B}$$ is not tangent to ⊙C at B. In Exercises 23 – 26, point B is a point of tangency. Find the radius r of ⊙C. Question 23. Question 24. (r + 6)² = r² + 9² r² + 12r + 36 = r² + 81 12r = 81 – 36 12r = 45 r = $$\frac { 15 }{ 4 }$$ Therefore, the radius of ⊙C is $$\frac { 15 }{ 4 }$$ Question 25. Question 26. (r + 18)² = r² + 30² r² + 36r + 324 = r² + 900 36r = 900 – 324 36r = 576 r = 16 Therefore, the radius of ⊙C is 16 CONSTRUCTION In Exercises 27 and 28. construct ⊙C with the given radius and point A outside of ⊙C. Then construct a line tangent to ⊙C that passes through A. Question 27. r = 2 in. Question 28. r = 4.5 cm In Exercises 29 – 32, points B and D are points of tangency. Find the value(s) of x. Question 29. Question 30. 3x + 10 = 7x – 6 7x – 3x = 10 + 6 4x = 16 x = 4 Question 31. Question 32. 2x + 5 = 3x² + 2x – 7 3x² = 5 + 7 3x² = 12 x² = 4 x = ±2 Question 33. ERROR ANALYSIS Describe and correct the error in determining whether $$\overline{X Y}$$ is tangent to ⊙Z. Question 34. ERROR ANALYSIS Describe and correct the error in finding the radius of ⊙T. 39² = 36² + 15² So, 15 is the diameter. The radius is $$\frac { 15 }{ 2 }$$. Question 35. ABSTRACT REASONING For a point outside of a circle, how many lines exist tangent to the circle that pass through the point? How many such lines exist for a point on the circle? inside the circle? Explain your reasoning. Question 36. CRITICAL THINKING When will two lines tangent to the same circle not intersect? Justify your answer. Using tangent line to circle theorem, it follow that the angle between tangent and radius is a right angle. Let’s draw these tangents at the two ends of the same diameter. We can observe a diameter AD like a transverzal of these tangents. Question 37. USING STRUCTURE Each side of quadrilateral TVWX is tangent to ⊙Y. Find the perimeter of the quadrilateral. Question 38. LOGIC In ⊙C, radii $$\overline{C A}$$ and $$\overline{C B}$$ are perpendicular. are tangent to ⊙C. a. Sketch ⊙C, $$\overline{C A}$$, $$\overline{C B}$$, . Question 39. MAKING AN ARGUMENT Two hike paths are tangent to an approximately circular pond. Your class is building a nature trail that begins at the intersection B of the bike paths and runs between the bike paths and over a bridge through the center P of the pond. Your classmate uses the Converse of the Angle Bisector Theorem (Theorem 6.4) to conclude that the trail must bisect the angle formed by the bike paths. Is your classmate correct? Explain your reasoning. Question 40. MODELING WITH MATHEMATICS A bicycle chain is pulled tightly so that $$\overline{M N}$$ is a common tangent of the gears. Find the distance between the centers of the gears. height h = 4.3 – 1.8 h = 2.5 x² = MN² + h² x² = 17.6² + 2.5² x² = 316.01 x = 17.8 Therefore, the distance between the centre of the gear is 17.8 in. Question 41. WRITING Explain why the diameter of a circle is the longest chord of the circle. Question 42. HOW DO YOU SEE IT? In the figure, $$\vec{P}$$A is tangent to the dime. $$\vec{P}$$C is tangent to the quarter, and $$\vec{P}$$B is a common internal tangent. How do you know that $$\overline{P A} \cong \overline{P B} \cong \overline{P C}$$ Question 43. PROOF In the diagram, $$\overline{R S}$$ is a common internal tangent to ⊙A and ⊙B. Prove that $$\frac{\Lambda C}{B C}=\frac{R C}{S C}$$ Question 44. THOUGHT PROVOKING A polygon is circumscribed about a circle when every side of the polygon is tangent to the circle. In the diagram. quadrilateral ABCD is circumscribed about ⊙Q. Is it always true that AB + CD = AD + BC? Justify your answer. Question 45. MATHEMATICAL CONNECTIONS Question 46. PROVING A THEOREM Prove the External Tangent Congruence Theorem (Theorem 10.2). Given $$\overline{S R}$$ and $$\overline{S T}$$ are tangent to ⊙P. Prove $$\overline{S R} \cong \overline{S T}$$ ∠PRS and ∠PTS are the right angles. So the legs of circles are congruent. Therefore, $$\overline{S R} \cong \overline{S T}$$ Question 47. PROVING A THEOREM Use the diagram to prove each part of the biconditional in the Tangent Line to Circle Theorem (Theorem 10.1 ). a. Prove indirectly that if a line is tangent to a circle, then it is perpendicular to a radius. (Hint: If you assume line m is not perpendicular to $$\overline{Q P}$$, then the perpendicular segment from point Q to line m must intersect line m at some other point R.) Ghen Line m is tangent to ⊙Q at point P. Prove m ⊥ $$\overline{Q P}$$ b. Prove indirectly that if a line is perpendicular to a radius at its endpoint, then the line is tangent to the circle. Gien m ⊥ $$\overline{Q P}$$ Prove Line m is tangent to ⊙Q. Question 48. REASONING In the diagram, AB = AC = 12, BC = 8, and all three segments are Langent to ⊙P. What is the radius of ⊙P? Justify your answer. Maintaining Mathematical Proficiency Find the indicated measure. Question 49. m∠JKM Question 50. AB AC = AB + BC 10 = AB + 7 AB = 10 – 7 AB = 3 10.2 Finding Arc Measures Exploration 1 Measuring Circular Arcs Work with a partner: Use dynamic geometry software to find the measure of $$\widehat{B C}$$. Verify your answers using trigonometry. a. Points A(0, 0) B(5, 0) C(4, 3) 30 degrees b. Points A(0, 0) B(5, 0) C(3, 4) 60 degrees c. Points A(0, 0) B(4, 3) C(3, 4) 15 degrees d. Points A(0, 0) B(4, 3) C(- 4, 3) 90 degrees Question 2. How are circular arcs measured? Question 3. Use dynamic geometry software to draw a circular arc with the given measure. USING TOOLS STRATEGICALLY To be proficient in math, you need to use technological tools to explore and deepen your understanding of concepts. a. 30° b. 45° c. 60° d. 90° Lesson 10.2 Finding Arc Measures Monitoring Progress Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc. Question 1. $$\widehat{T Q}$$ $$\widehat{T Q}$$ is a minor arc. $$\widehat{T Q}$$ = 120° Question 2. $$\widehat{Q R T}$$ $$\widehat{Q R T}$$ Question 3. $$\widehat{T Q R}$$ is a major arc. $$\widehat{Q R T}$$ = QR + RS + ST RS = 360° – (60 + 120 + 80) = 360 – 260 = 100° So, $$\widehat{Q R T}$$ = 60° + 100° + 80° $$\widehat{Q R T}$$ = 240° Question 4. $$\widehat{Q S}$$ $$\widehat{Q S}$$ = QR + RS = 60 + 100 = 160° Therefore, $$\widehat{Q S}$$ = 160° and it is a minor arc. Question 5. $$\widehat{T S}$$ $$\widehat{T S}$$ = 80° and it is a minor arc. Question 6. $$\widehat{R S T}$$ $$\widehat{R S T}$$ = 100 + 80 = 180 Therefore, $$\widehat{R S T}$$ = 180° and it is a minor arc. Tell whether the red arcs are congruent. Explain why or why not. Question 7. $$\widehat{A B}$$, $$\widehat{C D}$$ are congruent as they measure same radius and same arc length. Question 8. $$\widehat{M N}$$, $$\widehat{P Q}$$ are not congruent as they measure different radius. Exercise 10.2 Finding Arc Measures Vocabulary and Core Concept Check Question 1. VOCABULARY Copy and complele: If ∠ACB and ∠DCE are congruent central angles of ⊙C, then $$\widehat{A B}$$ and $$\widehat{D E}$$ arc. Question 2. WHICH ONE DOESNT BELONG? Which circle does not belong with the other three? Explain your reasoning. We know that 1 ft = 12 in So, the fourth circle does not belong to the other three as its diameter is different. Monitoring Progress and Modeling with Mathematics In Exercises 3 – 6, name the red minor arc and find its measure. Then name the blue major arc and find its measure. Question 3. Question 4. The minor arc $$\widehat{E F}$$ = 68° and major arc $$\widehat{F G E}$$ = 360° – 68° = 292°. Question 5. Question 6. The minor arc is $$\widehat{M N}$$ = 170°, major arc $$\widehat{N P M}$$ = 360° – 170° = 190°. In Exercises 7 – 14. identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc. Question 7. $$\widehat{B C}$$ Question 8. $$\widehat{D C}$$ $$\widehat{D C}$$ is a minor arc and it has a measure of 65°. Question 9. $$\widehat{E D}$$ Question 10. $$\widehat{A E}$$ $$\widehat{A E}$$ is a minor arc and it has a measure of 70°. Question 11. $$\widehat{E A B}$$ Question 12. $$\widehat{A B C}$$ $$\widehat{A B C}$$ is a semicircle and it has a measure of 180°. Question 13. $$\widehat{B A C}$$ Question 14. $$\widehat{E B D}$$ $$\widehat{E B D}$$ is a major arc and it has a measure of 315°. In Exercises 15 and 16, find the measure of each arc. Question 15. a. $$\widehat{J L}$$ b. $$\widehat{K M}$$ c. $$\widehat{J L M}$$ d. $$\widehat{J M}$$ Question 16. a. $$\widehat{R S}$$ $$\widehat{R S}$$ = $$\widehat{Q R S}$$ – $$\widehat{Q R}$$ = 180 – 42 = 138° So, $$\widehat{R S}$$ = 138° b. $$\widehat{Q R S}$$ $$\widehat{Q R S}$$ = 180° c. $$\widehat{Q S T}$$ $$\widehat{Q S T}$$ = $$\widehat{Q R S}$$ + $$\widehat{S T}$$ = 180 + 42 = 222 So, $$\widehat{Q S T}$$ = 222° d. $$\widehat{Q T}$$ $$\widehat{Q T}$$ = 360 – (42 + 138 + 42) = 360 – (222) = 138° $$\widehat{Q T}$$ = 138° Question 17. MODELING WITH MATHEMATICS A recent survey asked high school students their favorite type of music. The results are shown in the circle graph. Find each indicated arc measure. a. m$$\widehat{A E}$$ b. m$$\widehat{A C E}$$ c. m$$\widehat{G D C}$$ d. m$$\widehat{B H C}$$ e. m$$\widehat{F D}$$ f. m$$\widehat{F B D}$$ Question 18. ABSTRACT REASONING The circle graph shows the percentages of students enrolled in fall Sports at a high school. Is it possible to find the measure of each minor arc? If so, find the measure 0f the arc for each category shown. If not, explain why it is not possible. Soccer angle = 30% of 360 = 108° Volleyball angle = 15% of 360 = 54° Cross-country angle = 20% of 360 = 72° None angle = 15% of 360 = 54° Football angle = 20% of 360 = 72° In Exercises 19 – 22, tell whether the red arcs are congruent. Explain why or why not. Question 19. Question 20. $$\widehat{L P}$$ and $$\widehat{M N}$$ are not congruet because they are not in the same circle. Question 21. Question 22. $$\widehat{R S Q}$$, $$\widehat{F G H}$$ are not congruent because those two circles have different radii. MATHEMATICAL CONNECTIONS In Exercises 23 and 24. find the value of x. Then find the measure of the red arc. Question 23. Question 24. 4x + 6x + 7x + 7x = 360 24x = 360 x = 15 m$$\widehat{R S T}$$ = 6(15) + 7(15) = 90 + 105 = 195° So, m$$\widehat{R S T}$$ = 195° Question 25. MAKING AN ARGUMENT Your friend claims that any two arcs with the same measure are similar. Your cousin claims that an two arcs with the same measure are congruent. Who is correct? Explain. Question 26. MAKING AN ARGUMENT Your friend claims that there is not enough information given to find the value of x. Is your friend correct? Explain your reasoning. My friend is wrong. 4x + x + x + 4x = 360 10x = 360 x = 36° Question 27. ERROR ANALYSIS Describe and correct the error in naming the red arc. Question 28. ERROR ANALYSIS Describe and correct the error in naming congruent arc. $$\widehat{J K}$$, $$\widehat{N P}$$ are not congruent because those two arcs are from different circles. Question 29. ATTENDING TO PRECISION Two diameters of ⊙P are $$\widehat{A B}$$ and $$\widehat{C D}$$. Find m$$\widehat{A C D}$$ and m$$\widehat{A C}$$ when m$$\widehat{A D}$$ = 20°. Question 30. REASONING In ⊙R, m$$\widehat{A B}$$ = 60°, m$$\widehat{B C}$$ = 25°. m$$\widehat{C D}$$ = 70°, and m$$\widehat{D E}$$ = 20°. Find two possible measures of $$\widehat{A E}$$. $$\widehat{A E}$$ = 360 – ($$\widehat{A B}$$ + $$\widehat{B C}$$ + $$\widehat{C D}$$ + $$\widehat{D E}$$) = 360 – (60 + 25 + 70 + 20) = 360 – (175) = 185 $$\widehat{A E}$$ = $$\widehat{A B}$$ + $$\widehat{B C}$$ + $$\widehat{C D}$$ + $$\widehat{D E}$$ = 60 + 25 + 70 + 20 = 175 So, the two possibilities of $$\widehat{A E}$$ are 185°, 175° Question 31. MODELING WITH MATHEMATICS On a regulation dartboard, the outermost circle is divided into twenty congruent sections. What is the measure of each arc in this circle? Question 32. MODELING WITH MATHEMATICS You can use the time zone wheel to find the time in different locations across the world. For example, to find the time in Tokyo when it is 4 P.M. in San Francisco, rotate the small wheel until 4 P.M. and San Francisco line up, as shown. Then look at Tokyo to see that it is 9 A.M. there. a. What is the arc measure between each time zone 0n the wheel? As the circle is divided into 24 sectors, each time zone angle = $$\frac { 360 }{ 24 }$$ = 15° b. What is the measure of the minor arc from the Tokyo zone to the Anchorage zone? The measure of the minor arc from the Tokyo zone to the Anchorage zone = 15 + 15 + 15 + 15 + 15 + 15 = 90° c. If two locations differ by 180° on the wheel, then it is 3 P.M. at one location when it is _________ at the other location. Kuwaiti city. Question 33. PROVING A THEOREM Write a coordinate proof of the Similar Circles Theorem (Theorem 10.5). Given ⊙O with center O(0, 0) and radius r. ⊙A with center A(a, 0) and radius s Prove ⊙O ~ ⊙A Question 34. ABSTRACT REASONING Is there enough information to tell whether ⊙C ≅ ⊙D? Explain your reasoning. Both circles ⊙C and ⊙D have the same radius so those circles are congruent. Question 35. PROVING A THEOREM Use the diagram to prove each part of the biconditional in the Congruent Circles Theorem (Theorem 10.3). a. Given $$\overline{A C} \cong \overline{B D}$$ Prove ⊙A ≅ ⊙B b. Given ⊙A ≅ ⊙B prove $$\overline{A C} \cong \overline{B D}$$ Question 36. HOW DO YOU SEE IT? Are the circles on the target similar or congruent? Explain your reasoning. Question 37. PROVING A THEOREM Use the diagram to prove each part of the biconditional in the Congruent Central Angles Theorem (Theorem 10.4). a. Given ∠ABC ≅ ∠DAE Prove $$\widehat{B C}$$ ≅ $$\widehat{D E}$$ b. Given $$\widehat{B C}$$ ≅ $$\widehat{D E}$$ Prove ∠ABC ≅ ∠DAE Question 38. THOUGHT PROVOKING Write a formula for the length of a circular arc. Justify your answer. The formula to find the length of a circular arc is radius x angle. Maintaining Mathematical Proficiency Find the value of x. Tell whether the side lengths form a Pythagorean triple. Question 39. Question 40. x² = 13² + 13² = 169 + 169 = 338 x = 13√2 Question 41 Question 42. 14² = x² + 10² 196 = x² + 100 x² = 196 – 100 x² = 96 x = 4√6 10.3 Using Chords Exploration 1 Drawing Diameters Work with a partner: Use dynamic geometry software to construct a circle of radius 5 with center at the origin. Draw a diameter that has the given point as an endpoint. Explain how you know that the chord you drew is a diameter. a. (4, 3) b. (0, 5) c. (-3, 4) d. (-5, 0) Exploration 2 Work with a partner. Use dynamic geometry software to construct a chord $$\overline{B C}$$ of a circle A. Construct a chord on the perpendicular bisector of $$\overline{B C}$$. What do you notice? Change the original chord and the circle several times. Are your results always the same? Use your results to write a conjecture. LOOKING FOR STRUCTURE To be proficient in math, you need to look closely to discern a pattern or structure. Exploration 3 A Chord Perpendicular to a Diameter Work with a partner. Use dynamic geometry software to construct a diameter $$\overline{B C}$$ of a circle A. Then construct a chord $$\overline{D E}$$ perpendicular to $$\overline{B C}$$ at point F. Find the lengths DF and EF. What do you notice? Change the chord perpendicular to $$\overline{B C}$$ and the circle several times. Do you always get the same results? Write a conjecture about a chord that is perpendicular to a diameter of a circle. Question 4. What are two ways to determine when a chord is a diameter of a circle? If a chord passes through the centre of the circle, then it is the diameter of a circle. The longest chord of the circle is the diameter of a circle. Lesson 10.3 Using Chords Monitoring Progress In Exercises 1 and 2, use the diagram of ⊙D. Question 1. If m$$\widehat{A B}$$ = 110°. find m$$\widehat{B C}$$. Because AB and BC are congruent chords in congruent circles, the corresponding minor arcs $$\widehat{A B}$$, $$\widehat{B C}$$ are congruent by the congruent corresponding chords theorem. So, $$\widehat{A B}$$ = $$\widehat{B C}$$ $$\widehat{B C}$$ = 110° Question 2. If m$$\widehat{A C}$$ = 150° find m$$\widehat{A B}$$. $$\widehat{A C}$$ = 360 – ($$\widehat{A B}$$ + $$\widehat{B C}$$) 150 = 360 – 2($$\widehat{A B}$$) 2($$\widehat{A B}$$) = 360 – 150 = 210 $$\widehat{A B}$$ = 105° In Exercises 3 and 4. find the indicated length or arc measure. Question 3. CE CE = 5 + 5 = 10 units Question 4. m$$\widehat{C E}$$ m$$\widehat{C E}$$ = 9x + 180 – x = 180 – 8x m$$\widehat{C E}$$ = 180 – 8x Question 5. In the diagram, JK = LM = 24, NP = 3x, and NQ = 7x – 12. Find the radius of ⊙N Exercise 10.3 Using Chords Vocabulary and Core Concept Check Question 1. WRITING Describe what it means to bisect a chord. Question 2. WRITING Two chords of a circle are perpendicular and congruent. Does one of them have to be a diameter? Explain your reasoning. Imagine a line segment of length 3 units, AB. A second congruent segment of length 3 that is perpendicular to AB called CD. Circumscribe both these line segments and note that AB and CD are now chords. While both chords are perpendicular and congruent, neither chord is a diameter. Thus, it is possible to have two chords of this type with neither one diameter of the circle. Monitoring Progress and Modeling with Mathematics In Exercises 3 – 6, find the measure of the red arc or chord in ⊙C. Question 3. Question 4. Arc length = radius x angle = 5 x 34 = 170 Question 5. Question 6. As the two circles radius is the same and the angle is the same so the chord length is 11 units. In Exercise 7-10, find the value of x. Question 7. Question 8. By the perpendicular bisector theorem RS = ST x = 40° Question 9. Question 10. 5x + 2 = 7x – 12 7x – 5x = 2 + 12 2x = 14 x = 7 Question 11. ERROR ANALYSIS Describe and correct the error in reasoning. Question 12. PROBLEM SOLVING In the cross section of the submarine shown, the control panels are parallel and the same length. Describe a method you can use to find the center of the cross section. Justify your method. In Exercises 13 and 14, determine whether $$\overline{A B}$$ is a diameter of the circle. Explain your reasoning. Question 13. Question 14. 5² = 3² + x² 25 = 9 + x² x² = 25 – 9 x = 4 So, AB is not diameter of the circle. In Exercises 15 and 16, find the radius of ⊙Q. Question 15. Question 16. 4x + 4 = 6x – 6 6x – 4x = 4 + 6 2x = 10 x = 5 BC = 6(5) – 6 = 30 – 6 = 24 QC² = 5² + 12² = 25 + 144 = 169 QC = 13 Question 17. PROBLEM SOLVING An archaeologist finds part of a circular plate. What was the diameter of the plate to the nearest tenth of an inch? Justify your answer. Question 18. HOW DO YOU SEE IT? What can you conclude from each diagram? Name a theorem that justifies your answer. a. Perpendicular chord bisector converse theorem. b. Congruent Corresponding Chords theorem c. Perpendicular chord bisector theorem d. Equidistant chords theorem Question 19. PROVING A THEOREM Use the diagram to prove each part of the biconditional in the Congruent Corresponding Chords Theorem (Theorem 10.6). a. Given $$\overline{A B}$$ and $$\overline{C D}$$ are congruent chords. Prove $$\widehat{A B} \cong \widehat{C D}$$ b. Given $$\widehat{A B} \cong \widehat{C D}$$ Prove $$\overline{A B}$$ ≅ $$\overline{C D}$$ Question 20. MATHEMATICAL CONNECTIONS In ⊙P, all the arcs shown have integer measures. Show that x must be even. Question 21. REASONING In ⊙P. the lengths of the parallel chords are 20, 16, and 12. Find m$$\widehat{A B}$$. Explain your reasoning. Question 22. PROVING A THEOREM Use congruent triangles to prove the Perpendicular Chord Bisector Theorem (Theorem 10.7). Given $$\overline{E G}$$ is a diameter of ⊙L. $$\overline{E G}$$ ⊥ $$\overline{D F}$$ Prove $$\overline{D C}$$ ≅ $$\overline{F C}$$, $$\widehat{D G} \cong \widehat{F G}$$ Let L be the centre of the circle draw any chord DF on the circle As DF passes through LG. The length of DC is the same as FC. Question 23. PROVING A THEOREM Write a proof of the Perpendicular Chord Bisector Converse (Theorem 10.8). Given $$\overline{Q S}$$ is a perpendicular bisector of $$\overline{R T}$$. Prove $$\overline{Q S}$$ is a diameter of the circle L. (Hint: Plot the center L and draw △LPT and △LPR.) Question 24. THOUGHT PROVOKING Consider two chords that intersect at point P. Do you think that $$\frac{A P}{B P}=\frac{C P}{D P}$$? Justify your answer. Question 25. PROVING A THEOREM Use the diagram with the Equidistant Chords Theorem (Theorem 10.9) to prove both parts of the biconditional of this theorem. $$\overline{A B}$$ ≅ $$\overline{C D}$$ if and only if EF = EG Question 26. MAKING AN ARGUMENT A car is designed so that the rear wheel is only partially visible below the body of the car. The bottom edge of the panel is parallel [o the ground. Your friend claims that the point where the tire touches the ground bisects $$\widehat{A B}$$. Is your friend correct? Explain your reasoning. Maintaining Mathematical Proficiency Find the missing interior angle measure. Question 27. Quadrilateral JKLW has angle measures m∠J = 32°, m∠K = 25°, and m∠L = 44°. Find m∠M. Question 28. Pentagon PQRST has angle measures m∠P = 85°, m∠Q = 134°, m∠R = 97°, and m∠S =102°. Find m∠T. The sum of interior angles of a pentagon = 540° m∠T = 540 – (85 + 134 + 97 + 102) = 540 – 418 = 122 m∠T = 122°. 10.1 – 10.3 Quiz In Exercises 1 – 6, use the diagram. (Section 10.1) Question 1. Name the circle. The circle has a chord, diameter and tangent. Question 2. NP is the radius of the circle. Question 3. Name a diameter. KN is the diameter of the circle. Question 4. Name a chord. JL is the chord Question 5. Name a secant. SN is the secant Question 6. Name a tangent. QR is the tangent. Find the value of x. Question 7. (9 + x)² = x² + 15² 81 + 18x + x² = x² + 225 18x = 225 – 81 18x = 144 x = 8 Question 8. 6x – 3 = 3x + 18 6x – 3x = 18 + 3 3x = 21 x = 7 Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc. Question 9. $$\widehat{A E}$$ $$\widehat{A E}$$ = 180 – 36 = 144 So, $$\widehat{A E}$$ = 144° Question 10. $$\widehat{B C}$$ $$\widehat{B C}$$ = 180 – (67 + 70) = 180 – 137 = 43 So, $$\widehat{B C}$$ = 43° Question 11. $$\widehat{A C}$$ $$\widehat{A C}$$ = 43 + 67 = 110° Question 12. $$\widehat{A C D}$$ $$\widehat{A C D}$$ = 180° Question 13. $$\widehat{A C E}$$ $$\widehat{A C E}$$ = 180 + 36 = 216° Question 14. $$\widehat{B E C}$$ $$\widehat{B E C}$$ = 70 + 36 + 43 = 149° Tell whether the red arcs are congruent. Explain why or why not. Question 15. As two chords pass through the centre of the circle. Those two red arcs are congruent. Question 16. Red arcs are not congruent because the radius of the two circles is different. Question 17. Find the measure of the red arc in ⊙Q. Question 18. In the diagram. AC = FD = 30, PG = x + 5, and PJ = 3x – 1. Find the radius of ⊙P. Question 19. A circular clock can be divided into 12 congruent sections. a. Find the measure of each arc in this circle. The measure of each arc = $$\frac { 360 }{ 12 }$$ = 30° b. Find the measure of the minor arc formed by the hour and minute hands when the times is 7:00. When the time is 7:00 the minute hand is at 12 and hour hand is at 7 and so the minor arc is subtended by 12 – 7 = 5 of these sections and so the angle subtended is 30 x 5 = 150° c. Find a time at which the hour and minute hands form an arc that is congruent to the arc in part (b). A minor arc is equal to 150° can be formed by multiple placements of the hour and the minute hand. One of them can be the time 5:00 when the minute hand is at 12 and the hour hand is at 5. 10.4 Inscribed Angles and Polygons Exploration 1 Inscribed Angles and Central Angles work with a partner: Use dynamic geometry software. Sample a. Construct an inscribed angle in a circle. Then construct the corresponding central angle. b. Measure both angles. How is the inscribed angle related to its intercepted arc? c. Repeat parts (a) and (b) several times. Record your results in a table. Write a conjecture about how an inscribed angle is related to its intercepted arc. ATTENDING TO PRECISION To be proficient in math, you need to communicate precisely with others. Exploration 2 work with a partner: Use dynamic geometry software. Sample a. Construct a quadrilateral with each vertex on a circle. b. Measure all four angles. What relationships do you notice? c. Repeat parts (a) and (b) several times. Record your results in a table. Then write a conjecture that summarizes the data. Question 3. How are inscribed angles related to their intercepted arcs? How are the angles of an inscribed quadrilateral related to each other? Question 4. Quadrilateral EFGH is inscribed in ⊙C. and m ∠ E = 80°. What is m ∠ G? Explain. m ∠ E + m ∠ H = 80 + 80 = 160° m ∠ E + m ∠ H + m ∠ G + m ∠ F = 360 160° + m ∠ G + m ∠ F = 360 m ∠ G + m ∠ F = 360 – 160 = 200 m ∠ G = 100° Lesson 10.4 Inscribed Angles and Polygons Monitoring Progress Find the measure of the red arc or angle. Question 1. m∠G = $$\frac { 90 }{ 2 }$$ = 45° Question 2. $$\widehat{T V}$$ = 2 • 38 = 76° Question 3. m∠W = 72° Find the value of each variable. Question 4. x° = 90° y° = 180 – (40 + 90) = 180 – 130 y° = 50° Question 5. ∠B + ∠D = 180 ∠B + 82 = 180 x° = 98° ∠C + ∠A = 180 68 + y° = 180 y° = 112° Question 6. ∠S + ∠U = 180° c + 2c – 6 = 180 3c = 186 c = 62° ∠T + ∠V = 180° 10x + 8x = 180 18x = 180 x = 10° Question 7. In Example 5, explain how to find locations where the left side of the statue is all that appears in your camera’s field of vision. Exercise 10.4 Inscribed Angles and Polygons Vocabulary and Core Concept Check Question 1. VOCABULARY If a circle is circumscribed about a polygon, then the polygon is an ___________ . Question 2. DIFFERENT WORDS, SAME QUESTION Which is different? Find m∠ABC. m∠ABC = 60° Find m∠AGC. m∠AGC = 180 – (25 + 25) = 180 – 50 = 130° Find m∠AEC. m∠AEC = 180 – (50 + 50) = 180 – 100 = 80° m∠ADC = 180 – (25 + 50) = 180 – 75 = 105° Monitoring Progress and Modeling with Mathematics In Exercises 3 – 8, find the indicated measure. Question 3. m∠A Question 4. m∠G m∠G = 360 – (70 + 120) = 360 – 190 = 170° Question 5. m ∠ N Question 6. m$$\widehat{R S}$$ m$$\widehat{R S}$$ = 2 • 67 = 134° Question 7. m$$\widehat{V U}$$ Question 8. m$$\widehat{W X}$$ m$$\widehat{W X}$$ = $$\frac { 75 }{ 2 }$$ = 37.5 In Exercises 9 and 10, name two pairs of congruent angles. Question 9. Question 10. m∠W = m∠Z, m∠X = m∠Y In Exercises 11 and 12, find the measure of the red arc or angle. Question 11. Question 12. $$\widehat{P S}$$ = 2 • 40 = 80 In Exercises 13 – 16, find the value of each variable. Question 13. Question 14. m∠E + m∠G = 180 m + 60 = 180 m = 120° m∠D + m∠F = 180 60 + 2k = 180 k = 60° Question 15. Question 16. 3x° = 90° x° = 30° 2y° + 90° + 34° = 180° 2y° + 124° = 180° 2y° = 56° y° = 28° Question 17. ERROR ANALYSIS Describe and correct the error in finding m$$\widehat{B C}$$. Question 18. MODELING WITH MATHEMATICS A carpenter’s square is an L-shaped tool used to draw right angles. You need to cut a circular piece of wood into two semicircles. How can you use the carpenter’s square to draw a diameter on the circular piece of wood? Recall that when a right triangle is inscribed in a circle, then the hypotenuse is the diameter of the circle. Simply use the carpenter’s square to inscribe it into the circle. The hypotenuse formed by both legs of the square should provide a diameter for the circle. MATHEMATICAL CONNECTIONS In Exercises 19 – 21, find the values of x and y. Then find the measures of the interior angles of the polygon. Question 19. Question 20. ∠B + ∠C = 180 14x + 4x = 180 18x = 180° x = 10° ∠A + ∠D = 180 9y + 24y = 180 33y = 180° y = 5.45° ∠A = 130.9°, ∠B = 40°, ∠C = 140°, ∠D = 49° Question 21. Question 22. MAKING AN ARGUMENT Your friend claims that ∠PTQ ≅ ∠PSQ ≅ ∠PRQ. Is our friend correct? Explain your reasoning. Yes, my friend is correct. ∠PTQ ≅ ∠PSQ ≅ ∠PRQ is correct according to the inscribed angles of a circle theorem. Question 23. CONSTRUCTION Construct an equilateral triangle inscribed in a circle. Question 24. CONSTRUTION The side length of an inscribed regular hexagon is equal to the radius of the circumscribed circle. Use this fact to construct a regular hexagon inscribed in a circle. As the side length is equal to the radius. Draw a line representing the radius and draw a chord different chords in the form of hexagons of the radius of the circle. REASONING In Exercises 25 – 30, determine whether a quadrilateral of the given type can always be inscribed inside a circle. Explain your reasoning. Question 25. Square Question 26. rectangle yes, angles are right angles. Question 27. parallelogram Question 28. kite No. Question 29. rhombus Question 30. isosceles trapezoid Yes, the opposite angles are always supplementary. Question 31. MODELING WITH MATHEMATICS Three moons, A, B, and C, are in the same circular orbit 1,00,000 kilometers above the surface of a planet. The planet is 20,000 kilometers in diameter and m∠ABC = 90°. Draw a diagram of the situation. How far is moon A from moon C? Question 32. MODELING WITH MATHEMATICS At the movie theater. you want to choose a seat that has the best viewing angle, so that you can be close to the screen and still see the whole screen without moving your eyes. You previously decided that seat F7 has the best viewing angle, but this time someone else is already sitting there. Where else can you sit so that your seat has the same viewing angle as seat F7? Explain. Question 33. WRITING A right triangle is inscribed in a circle, and the radius of the circle is given. Explain how to find the length of the hypotenuse. Question 34. HOW DO YOU SEE IT? Let point Y represent your location on the soccer field below. What type of angle is ∠AYB if you stand anywhere on the circle except at point A or point B? Question 35. WRITING Explain why the diagonals of a rectangle inscribed in a circle are diameters of the circle. Question 36. THOUGHT PROVOKING The figure shows a circle that is circumscribed about ∆ABC. Is it possible to circumscribe a circle about any triangle? Justify your answer. Yes. Question 37. PROVING A THEOREM If an angle is inscribed in ⊙Q. the center Q can be on a side of the inscribed angle, inside the inscribed angle, or outside the inscribed angle. Prove each case of the Measure of an Inscribed Angle Theorem (Theorem 10. 10). a. Case 1 Given ∠ABC is inscribed in ⊙Q Let m∠B = x° Center Q lies on $$\overline{B C}$$. Prove m∠ABC = $$\frac{1}{2}$$m$$\widehat{A C}$$ (Hint: Show that ∆AQB is isosceles. Then write m$$\widehat{A C}$$ in terms of x.) b. Case 2 Use the diagram and auxiliary line to write Given and Prove statements for Case 2. Then write a proof c. Case 3 Use the diagram and auxiliary line to write Given and Prove statements for Case 3. Then write a proof. Question 38. PROVING A THEOREM Write a paragraph proof of the Inscribed Angles of a Circle Theorem (Theorem 10.11). First, draw a diagram and write Given and Prove statements. If two inscribed angles of a circle intercept the same arc, then the angles are congruent. Question 39. PROVING A THEOREM The Inscribed Right Triangle Theorem (Theorem 10.12) is written as a conditional statement and its converse. Write a plan for proof for each statement. Question 40. PROVING A THEOREM Copy and complete the paragraph proof for one part of the Inscribed Quadrilateral Theorem (Theorem 10. 13). Given ⊙C with inscribed quadrilateral DEFG Prove m ∠ D + m ∠ F = 180°, m ∠ E + m ∠ G = 180° By the Arc Addition Postulate (Postulate 10. 1), m$$\widehat{E F G}$$ + ________ = 360° and m$$\widehat{F G D}$$ + m$$\widehat{D E F}$$ = 360°. Using the ___________ Theorem. m$$\widehat{E D G}$$ = 2m ∠ F, m$$\widehat{E F G}$$ = 2m∠D, m$$\widehat{D E F}$$ = 2m∠G, and m$$\widehat{F G D}$$ = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + ________ = 360°, So _________ . Similarly, __________ . m$$\widehat{E F G}$$ + m$$\widehat{E D F}$$ = 360° and m$$\widehat{F G D}$$ + m$$\widehat{D E F}$$ = 360°. Using the the measure of an inscribed angle Theorem. m$$\widehat{E D G}$$ = 2m ∠ F, m$$\widehat{E F G}$$ = 2m∠D, m$$\widehat{D E F}$$ = 2m∠G, and m$$\widehat{F G D}$$ = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + 2m∠G = 360°. Question 41. CRITICAL THINKING In the diagram, ∠C is a right angle. If you draw the smallest possible circle through C tangent to $$\overline{A B}$$, the circle will intersect $$\overline{A C}$$ at J and $$\overline{B C}$$ at K. Find the exact length of $$\overline{J K}$$. Question 42. CRITICAL THINKING You are making a circular cutting board. To begin, you glue eight 1-inch boards together, as shown. Then you draw and cut a circle with an 8-inch diameter from the boards. a. $$\overline{F H}$$ is a diameter of the circular cutting board. Write a proportion relating GJ and JH. State a theorem in to justify your answer. Each board is 1 inch and FJ spans 6 boards. $$\overline{F H}$$ = 6 inches b. Find FJ, JH, and GJ. What is the length of the cutting board seam labeled $$\overline{G K}$$? Each board is 1 inch and JH spans 2 boards. JH = 2 inches Equation is $$\frac { JH }{ GJ }$$ = $$\frac { GJ }{ FJ }$$ $$\frac { 2 }{ GJ }$$ = $$\frac { GJ }{ 6 }$$ 12 = GJ² GJ = 2√3 GK = 2(GJ) GK = 4√3 So, FJ = 6, JH = 2, JG = 2√3, GK = 4√3 Maintaining Mathematical Proficiency Solve the equation. Check your solution. Question 43. 3x = 145 Question 44. $$\frac{1}{2}$$x = 63 x = 63 • 2 x = 126 Question 45. 240 = 2x Question 46. 75 = $$\frac{1}{2}$$(x – 30) 75 • 2 = x – 30 150 + 30 = x x = 180 10.5 Angle Relationships in Circles Exploration 1 Angles Formed by a Chord and Tangent Line Work with a partner: Use dynamic geometry software. Sample a. Construct a chord in a circle. At one of the endpoints of the chord. construct a tangent line to the circle. b. Find the measures of the two angles formed by the chord and the tangent line. c. Find the measures of the two circular arcs determined by the chord. d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data. Exploration 2 Angles Formed by Intersecting Chords Work with a partner: Use dynamic geometry software. sample a. Construct two chords that intersect inside a circle. b. Find the measure of one of the angles formed by the intersecting chords. c. Find the measures of the arcs intercepted h the angle in part (b) and its vertical angle. What do you observe? d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data. CONSTRUCTING VIABLE ARGUMENTS To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results. Question 3. When a chord intersects a tangent line or another chord, what relationships exist among the angles and arcs formed? Question 4. Line m is tangent to the circle in the figure at the left. Find the measure of ∠1. m∠1 = $$\frac { 1 }{ 2 }$$ • 148 m∠1 = 74° Question 5. Two chords intersect inside a circle to form a pair of vertical angles with measures of 55°. Find the sum of the measures of the arcs intercepted by the two angles. The sum of the measures of the arcs intercepted by the two angles = $$\frac { 1 }{ 2 }$$ • 55 = 27.5 Lesson 10.5 Angle Relationships in Circles Monitoring Progress Line m is tangent to the circle. Find the indicated measure. Question 1. m ∠ 1 m ∠ 1 = $$\frac { 1 }{ 2 }$$ • 210 m ∠ 1 = 105° Question 2. m$$\widehat{R S T}$$ m$$\widehat{R S T}$$ = 2 • 98 = 196° m$$\widehat{R S T}$$ = 196° Question 3. m$$\widehat{X Y}$$ m$$\widehat{X Y}$$ = $$\frac { 1 }{ 2 }$$ • 80 m$$\widehat{X Y}$$ = 40° Find the value of the variable. Question 4. y° = $$\frac { 1 }{ 2 }$$ • (102 + 95) = 98.5° Question 5. a° = 2 • 30° + 44° = 60° + 44° = 104° So, a° = 104°. Find the value of x. Question 6. x° = 180° – 120° x° = 60° Question 7. 50° = 180° – x° x° = 180° – 50° x° = 130° Question 8. You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level at point B. Find m$$\widehat{C D}$$, which represents the part of Earth that you can see. CB and BD are tangents, CB is perpendicular to AB and CD is perpendicular to AD by the tangent line to circle theorem. △ABC is similar to △ABD by the hypotenuese leg congruence theorem. ∠CBA is similar to ∠ABD. So, m∠CBA = 74.5°, m∠CBD = 2 • 74.5° = 149° m∠CBD = 180° – CD 149° = 180° – CD CD = 31° The part of earth you can see Exercise 10.5 Angle Relationships in Circles Vocabulary and Core Concept Check Question 1. COMPLETE THE SENTENCE Points A, B, C, and D are on a circle, and intersects at point P. If m∠APC = $$\frac{1}{2}$$(m$$\widehat{B D}$$ – m$$\widehat{A C}$$). then point P is _________ the circle. Question 2. WRITING Explain how to find the measure of a circumscribed angle. A circumscribed angle is the angle made by two intersecting tangent lines to a circle. Draw lines from the circle centre to the point of tangency. The angle between the radius and tangent line is 90°. The sum of angles of a quadrilateral is 360°. Angles between radii and tangent lines is 180°. The angle at two tangent lines meet is circumscribed angle. Monitoring Progress and Modeling with Mathematics In Exercises 3 – 6, line t is tangent to the circle. Find the indicated measure. Question 3. m$$\widehat{A B}$$ Question 4. m$$\widehat{D E F}$$ m$$\widehat{D E F}$$ = 2(117°) = 234° Question 5. m < 1 Question 6. m ∠ 3 m ∠ 3 = ½ • 140 = 70° In Exercises 7 – 14, find the value of x. Question 7. Question 8. x° = ½ • (30 + 2x – 30) Question 9. Question 10. 34° = ½ (3x – 2 – (x + 6)) 34° = ½ (3x – 2 – x – 6) 34° = ½ (2x – 8) 34° = x – 4 x° = 34 + 4 x° = 38° Question 11. Question 12. 6x – 11 = 2 • 125 6x = 250 + 11 6x = 261 x° = 43.5° Question 13. Question 14. 17x° = 75° x° = 4.41° ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in finding the angle measure. Question 15. Question 16. m∠1 = ½ (122 – 70) = ½ (52) = 26 So, m∠1 = 26° In Exercises 17 – 22, find the indicated angle measure. justify your answer. Question 17. m ∠ 1 Question 18. m ∠ 2 m ∠ 2 = 60° Explanation: m ∠ 3 =30°, So, m ∠ 2 = 180° – (90° + 30°) = 180° – 120° = 60° Therefore, m ∠ 2 = 60° Question 19. m ∠ 3 Question 20. m ∠ 4 m ∠ 4 = 90° Question 21. m ∠ 5 Question 22. m ∠ 6 m ∠ 6 = 180° – (60° + 30° + 30°) = 180° – 120° m ∠ 6 = 60° Question 23. PROBLEM SOLVING You are flying in a hot air balloon about 1.2 miles above the ground. Find the measure of the arc that represents the part of Earth you can see. The radius of Earth is about 4000 miles. Question 24. PROBLEM SOLVING You are watching fireworks over San Diego Bay S as you sail away in a boat. The highest point the fireworks reach F is about 0.2 mile above the bay. Your eyes E are about 0.01 mile above the water. At point B you can no longer see the fireworks because of the curvature of Earth. The radius of Earth is about 4000 miles, and $$\overline{F E}$$ is tangent to Earth at point T. Find m$$\widehat{s B}$$. Round your answer to the nearest tenth. Question 25. MATHEMATICAL CONNECTIONS In the diagram, $$\vec{B}$$A is tangent to ⊙E. Write an algebraic expression for m$$\widehat{C D}$$ in terms of x. Then find m$$\widehat{C D}$$. Question 26. MATHEMATICAL CONNECTIONS The circles in the diagram are concentric. Write an algebraic expression for c in terms of a and b. a° = ½(c° – b°) Question 27. ABSTRACT REASONING In the diagram. $$\vec{P}$$L is tangent to the circle, and $$\overline{K J}$$ is a diameter. What is the range of possible angle measures of ∠LPJ? Explain your reasoning. Question 28. ABSTRACT REASONING In the diagram, $$\overline{A B}$$ is an chord that is not a diameter of the circle. Line in is tangent to the circle at point A. What is the range of possible values of x? Explain your reasoning. (The diagram is not drawn to scale.) The possible values of x are less than 180°. Question 29. PROOF In the diagram and are secant lines that intersect at point L. Prove that m∠JPN > m∠JLN. Question 30. MAKING AN ARGUMENT Your friend claims that it is possible for a circumscribed angle to have the same measure as its intercepted arc. Is your friend correct? Explain your reasoning. Yes, when the circumscribed angle measures 90°, the central angle measures 90°, so the intercepted arc also measures 90°. Question 31. REASONING Points A and B are on a circle, and t is a tangent line containing A and another point C. a. Draw two diagrams that illustrate this situation. b. Write an equation for m$$\widehat{A B}$$ in terms of m∠BAC for each diagram. c. For what measure of ∠BAC can you use either equation to find m$$\widehat{A B}$$? Explain. Question 32. REASONING ∆XYZ is an equilateral triangle inscribed in ⊙P. AB is tangent to ⊙P at point X, $$\overline{B C}$$ is tangent to ⊙P at point Y. and $$\overline{A C}$$ is tangent to ⊙P at point Z. Draw a diagram that illustrates this situation. Then classify ∆ABC by its angles and sides. Justify your answer. Question 33. PROVING A THEOREM To prove the Tangent and Intersected Chord Theorem (Theorem 10. 14), you must prove three cases. a. The diagram shows the case where $$\overline{A B}$$ contains the center of the circle. Use the Tangent Line to Circle Theorem (Theorem 10.1) to write a paragraph proof for this case. b. Draw a diagram and write a proof for the case where the center of the circle is in the interior of ∠CAB. c. Draw a diagram and write a proof for the case where the center of the circle is in the exterior of ∠CAB. Question 34. HOW DO YOU SEE IT? In the diagram, television cameras are Positioned at A and B to record what happens on stage. The stage is an arc of ⊙A. You would like the camera at B to have a 30° view of the stage. Should you move the camera closer or farther away? Explain your reasoning. 25° = ½(80° – 30°) = ½(50°) So, you should move the camera closer. Question 35. PROVING A THEOREM Write a proof of the Angles Inside the Circle Theorem (Theorem 10.15). Given Chords $$\overline{A C}$$ and $$\overline{B D}$$ intersect inside a circle. Prove m ∠ 1 = $$\frac{1}{2}$$(m$$\widehat{D C}$$ + m$$\widehat{A B}$$) Question 36. THOUGHT PROVOKING In the figure, and are tangent to the circle. Point A is any point on the major are formed by the endpoints of the chord $$\overline{B C}$$. Label all congruent angles in the figure. Justify your reasoning. m∠CPB = ½(CAB – CB) Question 37. PROVING A THEOREM Use the diagram below to prove the Angles Outside the Circle Theorem (Theorem 10.16) for the case of a tangent and a secant. Then copy the diagrams for the other two cases on page 563 and draw appropriate auxiliary segments. Use your diagrams to prove each case. Question 38. PROVING A THEOREM Prove that the Circumscribed Angle Theorem (Theorem 10.17) follows from the Angles Outside the Circle Theorem (Theorem 10.16). In Exercises 39 and 40, find the indicated measure(s). justify your answer Question 39. Find m ∠ P when m$$\widehat{W Z Y}$$ = 200° Question 40. Find m$$\widehat{A B}$$ and m$$\widehat{E D}$$ m$$\widehat{E D}$$ = ½ (115°) = 57.5° ∠GJA = 30° Maintaining Mathematical Proficiency Solve the equation. Question 41. x2 + x = 12 Question 42. x2 = 12x + 35 Explanation: x² = 12x + 35 x = $$\frac { 12 ± √(144 + 140) }{ 2 }$$ x = $$\frac { 12 ± √284 }{ 2 }$$ x = $$\frac { 12 + √284 }{ 2 }$$, $$\frac { 12 – √284 }{ 2 }$$ Question 43. – 3 = x2 + 4x 10.6 Segment Relationships in Circles Exploration 1 Segments Formed by Two Intersecting Chords Work with a partner: Use dynamic geometry software. Sample a. Construct two chords $$\overline{B C}$$ and $$\overline{D E}$$ that intersect in the interior of a circle at a point F. b. Find the segment lengths BE, CF, DF, and EF and complete the table. What do you observe? c. Repeat parts (a) and (b) several times. Write a conjecture about your results. REASONING ABSTRACTLY To be proficient in math, you need to make sense of quantities and their relationships in problem situations. Exploration 2 Secants Intersecting Outside a Circle Work with a partner: Use dynamic geometry software. Sample a. Construct two secant and that intersect at a point B outside a circle, as shown. b. Find the segment lengths BE, BC, BF, and BD. and complete the table. What do you observe? c. Repeat parts (a) and (b) several times. Write a conjecture about your results. Question 3. What relationships exist among the segments formed by two intersecting chords or among segments of two secants that intersect outside a circle? Question 4. Find the segment length AF in the figure at the left. Explanation: EA • AF = AD • AC 18 • AF = 9 • 8 AF = 4 Lesson 10.6 Segment Relationships in Circles Monitoring Progress Find the value of x. Question 1. x = 8 Explanation: 4 • 6 = 3 • x 3x = 24 x = 8 Question 2. x = 5 Explanation: 2 • x + 1 = 4 • 3 x + 1 = 6 x = 5 Question 3. x = $$\frac { 54 }{ 5 }$$ Explanation: 6 • 9 = 5 • x 54 = 5x x = $$\frac { 54 }{ 5 }$$ Question 4. x = $$\frac { 3 ± √37 }{ 2 }$$ Explanation: 3 • x + 2 = x + 1 • x – 1 3x + 6 = x² – 1 x² – 3x – 7 = 0 x = $$\frac { 3 ± √(9 + 28) }{ 2 }$$ x = $$\frac { 3 ± √37 }{ 2 }$$ Question 5. x = ±√3 Explanation: x² = 3 • 1 x² = 3 x = ±√3 Question 6. x = $$\frac { 49 }{ 5 }$$ Explanation: 7² = 5 • x 49 = 5x x = $$\frac { 49 }{ 5 }$$ Question 7. x = 14.4 Explanation: 12² = 10x 144 = 10x x = 14.4 Question 8. WHAT IF? In Example 4, CB = 35 feet and CE = 14 feet. Find the radius of the tank. The radius of the tank is 36.75 Explanation: CB² = CE ⋅ CD 35² = 14 ⋅ (2r + 14) 1225 = 28r + 196 28r = 1029 r = 36.75 Exercise 10.6 Segment Relationships in Circles Vocabulary and Core Concept Check Question 1. VOCABULARY The part of the secant segment that is outside the circle is called a(n) ______________ . Question 2. WRITING Explain the difference between a tangent segment and a secant segment. A tangent segment intersects the circle at only one point. It actually doesn’t go through the circle. If a ball is rolling on a table top, then it would be the tangent. A secant segment intersects the circle in two points. It goes through the circle. In a tangent, no part is in the interior of the circle. In a secant, there is a part in the interior called a chord. Monitoring Progress and Modeling with Mathematics In Exercises 3 – 6, find the value of x. Question 3. Question 4. x = 23 Explanation: 10 • 18 = 9 • (x – 3) 20 = x – 3 x = 23 Question 5. Question 6. x = 5 Explanation: 2x • 12 = 15 • (x + 3) 24x = 15x + 45 9x = 45 x = 5 In Exercises 7 – 10, find the value of x. Question 7. Question 8. x = $$\frac { 35 }{ 4 }$$ Explanation: 5 • 7 = 4 • x 4x = 35 x = $$\frac { 35 }{ 4 }$$ Question 9. Question 10. x = 30 Explanation: 45 • x = 50 • 27 45x = 1350 x = 30 In Exercises 11 – 14. find the value of x. Question 11. Question 12. x = 48 Explanation: 24² = 12x 576 = 12x x = 48 Question 13. Question 14. x = 1.5 Explanation: 3 = 2x x = 1.5 Question 15. ERROR ANALYSIS Describe and correct the error in finding CD. Question 16. MODELING WITH MATHEMATICS The Cassini spacecraft is on a mission in orbit around Saturn until September 2017. Three of Saturn’s moons. Tethys. Calypso, and Teleslo. have nearly circular orbits of radius 2,95,000 kilometers. The diagram shows the positions of the moons and the spacecraft on one of Cassini’s missions. Find the distance DB from Cassini to Tethys when $$\overline{A D}$$ is tangent to the circular orbit. BD = 579493 km Explanation: (203,000)² = 83000x x = 496493 BC = 496493 BD = 496493 + 83000 = 579493 Question 17. MODELING WITH MATHEMATICS The circular stone mound in Ireland called Newgrange has a diameter of 250 feet. A passage 62 feet long leads toward the center of the mound. Find the perpendicular distance x from the end of the passage to either side of the mound. Question 18. MODELING WITH MATHEMATICS You are designing an animated logo for our website. Sparkles leave point C and move to the Outer circle along the segments shown so that all of the sparkles reach the outer circle at the same time. Sparkles travel from point C to point D at 2 centimeters per second. How fast should sparkles move from point C to point N? Explain. 5.33 should sparkles move from point C to point N. Explanation: 4 • 8 = 6 • x x = 5.33 Question 19. PROVING A THEOREM Write a two-column proof of the Segments of Chords Theorem (Theorem 10.18). Plan for Proof: Use the diagram to draw $$\overline{A C}$$ and $$\overline{D B}$$. Show that ∆EAC and ∆EDB are similar. Use the fact that corresponding side lengths in similar triangles are proportional. Question 20. PROVING A THEOREM Prove the Segments of Secants Theorem (Theorem 10.19). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.) Question 21. PROVING A THEOREM Use the Tangent Line to Circle Theorem (Theorem 10. 1) to prove the Segments of Secants and Tangents Theorem (Theorem 10.20) for the special case when the secant segment Contains the center of the circle. Question 22. PROVING A THEOREM Prove the Segments of Secants and Tangents Theorem (Theorem 10.20). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.) Question 23. WRITING EQUATIONS In the diagram of the water well, AB, AD, and DE are known. Write an equation for BC using these three measurements. Question 24. HOW DO YOU SEE IT? Which two theorems would you need to use to tind PQ? Explain your reasoning. Question 25. CRITICAL THINKING In the figure, AB = 12, BC = 8, DE = 6, PD = 4, and a is a point of tangency. Find the radius of ⊙P. Question 26. THOUGHT PROVOKING Circumscribe a triangle about a circle. Then, using the points of tangency, inscribe a triangle in the circle. Must it be true that the two triangles are similar? Explain your reasoning. Maintaining Mathematical Proficiency Solve the equation by completing the square. Question 27. x2 + 4x = 45 Question 28. x2 – 2x – 1 = 8 x = 1 + √10, x = 1 – √10 Explanation: x² – 2x – 1 = 8 x² – 2x – 9 = 0 x = $$\frac { 2 ± √(4 + 36) }{ 2 }$$ x = $$\frac { 2 ± √40 }{ 2 }$$ x = 1 + √10, x = 1 – √10 Question 29. 2x2 + 12x + 20 = 34 Question 30. – 4x2 + 8x + 44 = 16 x = 1 + √8, x = 1 – √8 Explanation: – 4x² + 8x + 44 = 16 4x² – 8x – 28 =0 x² – 2x – 7 = 0 x = $$\frac { 2 ± √(4 + 28) }{ 2 }$$ x = $$\frac { 2 ± √32 }{ 2 }$$ x = 1 ± √8 10.7 Circles in the Coordinate Plane Exploration 1 The Equation of a Circle with Center at the Origin Work with a partner: Use dynamic geometry software to Construct and determine the equations of circles centered at (0, 0) in the coordinate plane, as described below. a. Complete the first two rows of the table for circles with the given radii. Complete the other rows for circles with radii of your choice. b. Write an equation of a circle with center (0, 0) and radius r. x² + y² = r² Explanation: (x – 0)² + (y – 0)² = r² x² + y² = r² Exploration 2 The Equation of a Circle with Center (h, k) Work with a partner: Use dynamic geometry software to construct and determine the equations of circles of radius 2 in the coordinate plane, as described below. a. Complete the first two rows of the table for circles with the given centers. Complete the other rows for circles with centers of your choice. b. Write an equation of a circle with center (h, k) and radius 2. (x – h)² + (y – k)² = 4 c. Write an equation of a circle with center (h, k) and radius r. (x – h)² + (y – k)² = r² Exploration 3 Deriving the Standard Equation of a Circle Work with a partner. Consider a circle with radius r and center (h, k). Write the Distance Formula to represent the distance d between a point (x, y) on the circle and the center (h, k) of the circle. Then square each side of the Distance Formula equation. How does your result compare with the equation you wrote in part (c) of Exploration 2? MAKING SENSE OF PROBLEMS To be proficient in math, you need to explain correspondences between equations and graphs. (x – h)² + (y – k)² = r² Question 4. What is the equation of a circle with center (h, k) and radius r in the coordinate plane? (x – h)² + (y – k)² = r² Question 5. Write an equation of the circle with center (4, – 1) and radius 3. x² + y² – 8x + 2y + 8 = 0 Explanation: (x – 4)² + (y + 1)² = 9 x² – 8x + 16 + y² + 2y + 1 = 9 x² + y² – 8x + 2y = 9 – 17 x² + y² – 8x + 2y + 8 = 0 Lesson 10.7 Circles in the Coordinate Plane Monitoring Progress Write the standard equation of the circle with the given center and radius. Question 1. x² + y² = 6.25 Explanation: (x – 0)² + (y – 0)² = 2.5² x² + y² = 6.25 Question 2. center: (- 2, 5), radius: 7 (x + 2)² + (y – 5)² = 49 Explanation: (x + 2)² + (y – 5)² = 7² (x + 2)² + (y – 5)² = 49 Question 3. The point (3, 4) is on a circle with center (1, 4). Write the standard equation of the circle. (x – 1)² + (y – 4)² = 4 Explanation: r = √(3 – 1)² + (4 – 4)² = √(2)² r = 2 (x – 1)² + (y – 4)² = 2² (x – 1)² + (y – 4)² = 4 Question 4. The equation of a circle is x2 + y2 – 8x + 6y + 9 = 0. Find the center and the radius of the circle. Then graph the circle. The center of the circle (4, -3) and radius is 4. Explanation: x² + y² – 8x + 6y + 9 = 0 x² – 8x + 16 + y² + 6y + 9 = 16 (x – 4)² + (y + 3)² = 4² The center of the circle (4, -3) and radius is 4. Question 5. Prove or disprove that the point (1, √5 ) lies on the circle centered at the origin and containing the point (0, 1). Disproved. Explanation: We consider the circle centred at the origin and containing the point (0, 1). Therefore, we can conclude that the radius of the circle r = 1, let O (0, 0) and B (1, √5). So the distance between two points is OB = √(1 – 0) + (√5 – 0)² = √(1 + 5) = √6 As the radius of the given circle is 1 and distance of the point B from its centre is √6. So we can conclude that point does lie on the given circle. Question 6. why are three seismographs needed to locate an earthquake’s epicentre? Exercise 10.7 Circles in the Coordinate Plane Vocabulary and Core Concept Check Question 1. VOCABULARY What is the standard equation of a circle? Question 2. WRITING Explain why knowing the location of the center and one point on a circle is enough to graph the circle. If we know the location of the center and one point on the circle, we can graph a circle because the distance from the center to the point is called the radius. Monitoring Progress and Modeling with Mathematics In Exercises 3 – 8, write the standard equation of the circle. Question 3. Question 4. x² + y² = 36 Explanation: The center is (0, 0) and the radius is 6 (x – h)² + (y – k)² = r² (x – 0)² + (y – 0)² = 6² x² + y² = 36 Question 5. a circle with center (0, 0) and radius 7 Question 6. a circle with center (4, 1) and radius 5 (x – 4)² + (y – 1)² = 25 Explanation: (x – h)² + (y – k)² = r² (x – 4)² + (y – 1)² = 5² (x – 4)² + (y – 1)² = 25 Question 7. a circle with center (- 3, 4) and radius 1 Question 8. a circle with center (3, – 5) and radius 7 (x – 3)² + (y + 5)² = 49 Explanation: (x – h)² + (y – k)² = r² (x – 3)² + (y + 5)² = 7² (x – 3)² + (y + 5)² = 49 In Exercises 9 – 11, use the given information to write the standard equation of the circle. Question 9. The center is (0, 0), and a point on the circle is (0, 6). Question 10. The center is (1, 2), and a point on the circle is (4, 2). x² + y² = 9 Explanation: r = √(x – h)² + (y – k)² = √(4 – 1)² + (2 – 2)² = √3² r = 3 (x – h)² + (y – k)² = r² (x – 0)² + (y – 0)² = 3² x² + y² = 9 Question 11. The center is (0, 0). and a point on the circle is (3, – 7). Question 12. ERROR ANALYSIS Describe and correct the error in writing the standard equation of a circle. (x – h)² + (y – k)² = r² (x + 3)² + (y + 5)² = 3² (x + 3)² + (y + 5)² = 9 In Exercises 13 – 18, find the center and radius of the circle. Then graph the circle. Question 13. x2 + y2 = 49 Question 14. (x + 5)2 + (y – 3)2 = 9 Center is (-5, 3) and rdaius is 3. Explanation: For the equation (x + 5)2 + (y – 3)2 = 9, center is (-5, 3) and rdaius is 3. Question 15. x2 + y2 – 6x = 7 Question 16. x2 + y2 + 4y = 32 The center is (0, -2) and radius is 6 Explanation: x2 + y2 + 4y = 32 x² + y² + 4y + 4 = 32 + 4 x² + (y + 2)² = 36 (x – 0)² + (y – (-2))² = 6² The center is (0, -2) and radius is 6 Question 17. x2 + y2 – 8x – 2y = – 16 Question 18. x2 + y2 + 4x + 12y = – 15 The center is (-2, -6) and radius is 5 Explanation: x2 + y2 + 4x + 12y = – 15 x² + 4x + 4 + y² + 12y + 36 = -15 + 36 + 4 (x + 2)² + (y + 6)² = 5² The center is (-2, -6) and radius is 5 In Exercises 19 – 22, prove or disprove the statement. Question 19. The point (2, 3) lies on the circle centered at the origin with radius 8. Question 20. The point (4, √5) lies on the circle centered at the origin with radius 3. The point (4, √5) does not lie on the circle. Explanation: r² = (x – h)² + (y – k)² 3² ______________ (4 – 0)² + (√5 – 0)² 9 ______________ 16 + 5 9 ≠ 21 Because the radius is 3 and the distance between center and the point is more than the radius. So the point does not lie on the circle. Question 21. The point (√6, 2) lies on the circle centered at the origin and containing the point (3, – 1). Question 22. The point (√7, 5) lies on the circle centered at the origin and containing the point (5, 2). The point (√7, 5) does not lie on the circle. Explanation: r² = (x – h)² + (y – k)² = (√7 – 0)² + (5 – 0)² = 7 + 25 = 32 r = 5.65 (5.65)² ______________ (5 – 0)² + (2 – 0)² 32 ______________ 25 + 4 32 ≠ 29 Because the radius is 3 and the distance between center and the point is more than the radius. So the point does not lie on the circle. Question 23. MODELING WITH MATHEMATICS A City’s commuter system has three zones. Zone I serves people living within 3 miles of the city’s center. Zone 2 serves those between 3 and 7 miles from the center. Zone 3 serves those over 7 miles from the center. a. Graph this Situation on a coordinate plane where each unit corresponds to 1 mile. Locate the city’s center at the origin. b. Determine which zone serves people whose homes are represented by the points (3, 4), (6, 5), (1, 2), (0.3). and (1, 6). Question 24. MODELING WITH MATHEMATICS Telecommunication towers can be used to transmit cellular phone calls. A graph with units measured in kilometers shows towers at points (0, 0), (0, 5), and (6, 3). These towers have a range of about 3 kilometers. a. Sketch a graph and locate the towers. Are there any locations that may receive calls from more than one tower? Explain your reasoning. b. The center of City A is located at (- 2, 2.5), and the center of City B is located at (5, 4). Each city has a radius of 1.5 kilometers. Which city seems to have better cell phone coverage? Explain your reasoning. There are three towers at points (0, 0), (0, 5) and (6, 3) with range of about 3 km a. Let’s sketch the graph to locate towers. Draw the points (0, 0), (0, 5) and (6, 3). Then draw three circles with centers at these points and radii 3. There are locations that can receive calls from more that one tower because circles with centers (0, 0) and (0, 5) overlap. Locations in their intersection can receive calls from two towers. The city A has a center located at (-2, 2.5) and city B has a center located at (5, 4). Both cities have radius 1.5 km Let’s draw the city A as a circle with center (-2, 2.5) and radius 1.5 and city B with center (5, 4) and radius 1.5. From the graph we can conclude that the city B has better cell phone coverage because parts of city A do not have coverage. Question 25. REASONING Sketch the graph of the circle whose equation is x2 + y2 = 16. Then sketch the graph of the circle after the translation (x, y) → (x – 2, y – 4). What is the equation of the image? Make a conjecture about the equation of the image of a circle centered at the origin after a translation m units to the left and n units down. Question 26. HOW DO YOU SEE IT? Match each graph with its equation. a. A. x2 + (y + 3) 2 = 4 b. B. (x – 3) 2 + y2 = 4 c. C. (x + 3) 2 + y2 = 4 d. D. x2 + (y – 3) 2 = 4 a ➝ C, b ➝ A, c ➝ D, d ➝ B Question 27. USING STRUCTURE The vertices of ∆XYZ are X(4, 5), Y(4, 13), and Z(8, 9). Find the equation of the circle circumscribed about ∆XYZ. Justify your answer. Question 28. THOUGHT PROVOKING A circle has center (h, k) and contains point (a, b). Write the equation of the line tangent to the circle at point (a, b). y – b = $$\frac { h – a }{ b – k }$$(x – a) Explanation: It is given a circle with center C(h, k. A circle with point A(a, b). We have to write the equation of a tangent that intersects the circle at point A By the tangent line to circle theorem, a tangent is perpendicular to the radius. Two lines are perpendicular if and only if their slopes are negative reciprocals. So, find the equation of the line AC to know its slope. The equation of the line which has two points (a, b), (c, d) is y – b = $$\frac { b – c }{ a – d }$$(x – a) The equation of the line which has two points A(a, b) and C(h, k) is y – b = $$\frac { b – k }{ a – h }$$(x – a) Therefore, the slope of the line through A nad C is $$\frac { b – k }{ a – h }$$ Hence the slope of the tangent is –$$\frac { a – h }{ b – k }$$ = $$\frac { h – a }{ b – k }$$ Use the equation of the line y = kx + n through the point (a, b) y – b = k(x – a) to find the equation of the tangent The equation of the tangent with slope $$\frac { h – a }{ b – k }$$ and through the point A(a, b) is y – b = $$\frac { h – a }{ b – k }$$(x – a) MATHEMATICAL CONNECTIONS In Exercises 29 – 32, use the equations to determine whether the line is a tangent, a secant a secant that contains the diameter, or name of these. Explain your reasoning. Question 29. Circle: (x – 4)2 + (y – 3)2 = 9 Line: y = 6 Question 30. Circle: (x + 2)2 + (y – 2)2 = 16 Line: y = 2x – 4 The line is a secant line. Explanation: (x + 2)2 + (y – 2)2 = 16, y = 2x – 4 (x + 2)2 + (2x – 4 – 2)2 = 16 x² + 4x + 4 + (2x – 6)² = 16 x² + 4x + 4 + 4x² – 24x + 36 = 16 5x² – 20x + 40 – 16 = 0 5x² – 20x + 24 = 0 x = $$\frac { 20 ±√-80 }{ 10 }$$ x = 2, y = 2 • 2 – 4 = 0, (2, 0) The system has two solutions and point does not lie on the line. So the line is a secant line. Question 31. Circle: (x – 5)2 + (y + 1)2 = 4 Line: y = $$\frac{1}{5}$$x – 3\ Question 32. Circle: (x + 3)2 + (y – 6)2 = 25 Line: y = –$$\frac{4}{3}$$x + 2 The line is a secant line. Explanation: (x + 3)2 + (y – 6)2 = 25, y = –$$\frac{4}{3}$$x + 2 (x + 3)2 + (-$$\frac{4}{3}$$x + 2 – 6)2 = 25 x² + 6x + 9 + $$\frac { 16x² }{ 9 }$$ + $$\frac { 32x }{ 3 }$$ + 16 = 25 $$\frac { 25x² }{ 9 }$$ + $$\frac { 50x }{ 3 }$$ = 0 x(25x + 150) = 0 x = 0 or x = -6 y = 2, y = 10 (0, 2) and (-6, 10) d = √(0 + 6)² + (2 – 10)² = √(36 + 64) = 10 ≠ 5 The system has two solutions and point does not lie on the line. So the line is a secant line. MAKING AN ARGUMENT Question 33. Your friend claims that the equation of a circle passing through the points (- 1, 0) and (1, 0) is x2 – 2yk + y2 = 1 with center (0, k). Is your friend correct? Explain your reasoning. Question 34. REASONING Four tangent circles are centered on the x-axis. The radius of ⊙A is twice the radius of ⊙O, The radius of ⊙B is three times the radius of ⊙O, The radius of ⊙C is four times the radius of ⊙O, All circles have integer radii, and the point (63, 16) is On ⊙C. What is the equation of ⊙A? Explain your reasoning. Maintaining Mathematical Proficiency Identify the arc as a major arc, minor arc, or semicircle. Then find the measure of the arc. Question 35. $$\widehat{R S}$$ Question 36. $$\widehat{P R}$$ $$\widehat{P R}$$ is a right angle $$\widehat{P R}$$ = 25 + 65 = 90° Question 37. $$\widehat{P R T}$$ Question 38. $$\widehat{S T}$$ $$\widehat{S T}$$ is a major arc $$\widehat{S T}$$ = 360 – (90 + 65 +25 + 53) = 127° Question 39. $$\widehat{R S T}$$ Question 40. $$\widehat{Q S}$$ $$\widehat{Q S}$$ is a minor arc $$\widehat{Q S}$$ = 25 + 53 = 78° 10.1 Lines and Segments That Intersect Circles Tell whether the line, ray, or segment is best described as a radius, chord, diameter, secant, or tangent of ⊙P. Question 1. $$\overline{P K}$$ $$\overline{P K}$$ is radius Question 2. $$\overline{N M}$$ $$\overline{N M}$$ is chord Question 3. $$\vec{J}$$L $$\vec{J}$$L is tangent Question 4. $$\overline{K N}$$ $$\overline{K N}$$ is diameter Question 5. NL is secant Question 6. $$\overline{P N}$$ $$\overline{P N}$$ is radius Tell whether the common tangent is internal or external. Question 7. Internal common tangent Question 8. External common tangent Points Y and Z are points of tangency. Find the value of the variable. Question 9. a = $$\frac { 3 ± 33 }{ 18 }$$ Explanation: 3a = 9a² – 30 9a² – 3a – 30 = 0 a = $$\frac { 3 ± 33 }{ 18 }$$ Question 10. c = 2 Explanation: 2c² + 9c + 6 = 9c + 14 2c² – 8 = 0 c² – 4 = 0 c = 2 Question 11. r = 12 Explanation: (3 + r)² = r² + 9² 9 + 6r + r² = r² + 81 6r = 72 r = 12 Question 12. Tell whether $$\overline{B D}$$ is tangent to ⊙C. Explain. $$\overline{B D}$$ is not tangent to ⊙C Explanation: 52² = 10² + 48² 2704 = 100 + 2304 So, $$\overline{B D}$$ is not tangent to ⊙C 10.2 Finding Arc Measures Use the diagram above to find the measure of the indicated arc. Question 13. $$\widehat{K L}$$ $$\widehat{K L}$$ = 100° Explanation: $$\widehat{K L}$$ = ∠KPL = 100° Question 14. $$\widehat{L M}$$ $$\widehat{L M}$$ = 60° Explanation: $$\widehat{L M}$$ = 180° – 120° = 60° Question 15. $$\widehat{K M}$$ $$\widehat{K M}$$ = 160° Explanation: $$\widehat{K M}$$ = 100° + 60° = 160° Question 16. $$\widehat{K N}$$ $$\widehat{K N}$$ = 80° Explanation: $$\widehat{K N}$$ = 360 – (120 + 100 + 60) = 360 – 280 = 80° Tell whether the red arcs are congruent. Explain why or why not. Question 17. $$\widehat{S T}$$, $$\widehat{Y Z}$$ are not congruent. Explanation: $$\widehat{S T}$$, $$\widehat{Y Z}$$ are not congruent. Because both arcs are from different circles and having different radii. Question 18. $$\widehat{A B}$$, $$\widehat{E F}$$ are congruent. Explanation: $$\widehat{A B}$$, $$\widehat{E F}$$ are congruent. Because those circles have same radii. 10.3 Using Chords Find the measure of $$\widehat{A B}$$. Question 19. $$\widehat{A B}$$ = 61° Explanation: $$\widehat{A B}$$ = 61° If ED = AB, then $$\widehat{A B}$$ = $$\widehat{E D}$$ Question 20. $$\widehat{A B}$$ = 65° Explanation: $$\widehat{A B}$$ = $$\widehat{A D}$$ So, $$\widehat{A B}$$ = 65° Question 21. $$\widehat{A B}$$ = 91° Explanation: $$\widehat{A B}$$ = $$\widehat{E D}$$ So, $$\widehat{A B}$$ = 91° Question 22. In the diagram. QN = QP = 10, JK = 4x, and LM = 6x – 24. Find the radius of ⊙Q. The radius of ⊙Q is 26. Explanation: 6x – 24 = 4x 6x – 4x = 24 2x = 24 x = 12 ML = 6(12) – 24 = 48 JN = $$\frac { 48 }{ 2 }$$ = 24 JQ² = JN² + NQ² = 24² + 10² = 576 + 100 JQ = 26 Therefore, the radius of ⊙Q is 26 10.4 Inscribed Angles and Polygons Find the value(s) of the variable(s). Question 23. x° = 80° Explanation: x° = 2 • 40° = 80° Question 24. q° = 100°, r° = 20° Explanation: q° + 80° = 180° q° = 100° 4r° + 100 = 180° 4r° = 80° r° = 20° Question 25. d° = 5° Explanation: 14d° = 70° d° = 5° Question 26. y° = 30°, z° = 10° Explanation: 3y° = 90° y° = 30° 50° + 90° + 4z° = 180° 4z° = 40° z° = 10° Question 27. m° = 44° n° = 39° Explanation: m° = 44° n° = 39° Question 28. c° = 28° Explanation: c° = ½ • 56 = 28 10.5 Angle Relationships in Circles Find the value of x. Question 29. x° = 250° Explanation: x° = 250° Question 30. x° = 106° Explanation: x° = ½(152 + 60) = ½(212) = 106° Question 31. x° = 28° Explanation: x° = ½(96 – 40) = ½(56) = 28° Question 32. Line l is tangent to the circle. Find m$$\widehat{X Y Z}$$. m$$\widehat{X Y Z}$$ = 240° Explanation: m$$\widehat{X Y Z}$$ = 2(120) = 240° 10.6 Segment Relationships in Circles Find the value of x. Question 33. x = 8 Explanation: 3 • x = 4 • 6 x = 8 Question 34. x = 3 Explanation: (x + 3) • x = (6 – x) • 2x x + 3 = 12 – 2x 3x = 9 x = 3 Question 35. x = 18 Explanation: 12² = 8 • x 144 = 8x x = 18 Question 36. A local park has a circular ice skating rink. You are standing at point A, about 12 feet from the edge of the rink. The distance from you to a point of tangency on the rink is about 20 feet. Estimate the radius of the rink. Estimated radius of the rink is 10 ft Explanation: 20² = 12 • (2r + 12) 400 = 24r + 144 256 = 24r r = 10.66 Therefore, estimated radius of the rink is 10 ft 10.7 Circles in the Coordinate Plane Write the standard equation of the circle shown. Question 37. (x – 4)² + (y + 1)² = 12.25 Explanation: (x – 4)² + (y + 1)² = 3.5² (x – 4)² + (y + 1)² = 12.25 Question 38. (x – 8)² + (y – 5)² = 36 Explanation: (x – 8)² + (y – 5)² = 6² (x – 8)² + (y – 5)² = 36 Question 39. x² + y² = 4 Explanation: (x – 0)² + (y – 0)² = 2² x² + y² = 4 Write the standard equation of the circle with the given center and radius. Question 40. x² + y² = 81 Explanation: (x – 0)² + (y – 0)² = 9² x² + y² = 81 Question 41. center: (- 5, 2), radius: 1.3 (x + 5)² + (y – 2)² = 1.69 Explanation: (x + 5)² + (y – 2)² = 1.3² (x + 5)² + (y – 2)² = 1.69 Question 42. (x – 6)² + (y – 21)² = 16 Explanation: (x – 6)² + (y – 21)² = 4² (x – 6)² + (y – 21)² = 16 Question 43. center: (- 3, 2), radius: 16 (x + 3)² + (y – 2)² = 256 Explanation: (x + 3)² + (y – 2)² = 16² (x + 3)² + (y – 2)² = 256 Question 44. (x – 10)² + (y – 7)² = 12.25 Explanation: (x – 10)² + (y – 7)² = 3.5² (x – 10)² + (y – 7)² = 12.25 Question 45. x² + y² = 27.04 Explanation: (x – 0)² + (y – 0)² = 5.2² x² + y² = 27.04 Question 46. The point (- 7, 1) is on a circle with center (- 7, 6). Write the standard equation of the circle. (x + 7)² + (y – 6)² = 25 Explanation: r² = (-7 + 7)² + (6 – 1)² = 5² r = 5 And, centre is (-7, 6) The standard equation of a circle is (x – (-7))² + (y – 6)² = 5² (x + 7)² + (y – 6)² = 25 Question 47. The equation of a circle is x2 + y2 – 12x + 8y + 48 = 0. Find the center and the radius of the circle. Then graph the circle. The radius of the circle is 2, the centre is (6, -4) Explanation: x2 + y2 – 12x + 8y + 48 = 0 x² – 12x + 36 + y² + 8y + 16 = 4 (x – 6)² + (y + 4)² = 2² So, the radius of the circle is 2, the centre is (6, -4) Question 48. Prove or disprove that the point (4, – 3) lies on the circle centred at the origin and containing the point (- 5, 0). The point (4, – 3) lies on the circle. Explanation: Use the distance formula to find the radius of the circle with cente (0, 0) and a point (-5, 0) r = √(-5 – 0)² + (0 – 0)² = 5 The distance from the point (4, -3) to the center (0, 0) d = √(4 – 0)² + (-3 – 0)² = √(16 +9) = 5 Since the radius of the circle is 5, the point lies on the circle. Circles Chapter Test Find the measure of each numbered angle in ⊙P. Justify your answer. Question 1. m∠1 = 72.5° m∠2 = 145° Explanation: m∠1 = $$\frac { 145 }{ 2 }$$ = 72.5° m∠2 = 145° Question 2. m∠1 = 60°, m∠2 = 90° Explanation: A tangent is perpendicualr to diameter. So, m∠2 = 90° m∠1 = 60° Question 3. m∠1 = 48° Explanation: m∠1 = $$\frac { 96 }{ 2 }$$ = 48° Question 4. Use the diagram. Question 5. AG = 2, GD = 9, and BG = 3. Find GF. Question 6. CF = 12, CB = 3, and CD = 9. Find CE. Question 7. BF = 9 and CB = 3. Find CA Question 8. Sketch a pentagon inscribed in a circle. Label the pentagon ABCDE. Describe the relationship between each pair of angles. Explain your reasoning. a. ∠CDE and ∠CAE b. ∠CBE and ∠CAE Question 9. x = 5 Explanation: 5x – 4 = 3x + 6 5x – 3x = 6 + 4 2x = 10 x = 5 Question 10. r = 9 Explanation: (6 + r)² = 12² + r² 36 + 12r + r² = 144 + r² 12r = 108 r = 9 Question 11. Prove or disprove that the point (2√2, – 1) lies on the circle centered at (0, 2) and containing the point (- 1, 4). Disproved Explanation: We consider the circle centred at the A(0, 2) and containing the point B(-1, 4). AB = √(-1 – 0)² + (4 – 2)² = √1 + 4 = √5 The distance between centre A(0, 2) and P(2√2, – 1) is AP = √(2√2 – 0)² + (-1 – 2)² = √8 + 9 = √17 AB ≠ AP So, the point (2√2, – 1) dies not lie on the circle. Prove the given statement. Question 12. $$\widehat{S T} \cong \widehat{R Q}$$ Question 13. $$\widehat{J M} \cong \widehat{L M}$$ Question 14. $$\widehat{D G} \cong \widehat{F G}$$ Question 15. A bank of lighting hangs over a stage. Each light illuminates a circular region on the stage. A coordinate plane is used to arrange the lights, using a corner of the stage as the origin. The equation (x – 13)2 + (y – 4)2 = 16 represents the boundary of the region illuminated by one of the lights. Three actors stand at the points A(11, 4), B(8, 5), and C(15, 5). Graph the given equation. Then determine which actors are illuminated by the light. The equation (x – 13)² + (y – 4)²= 16 represents the standard equation of the circle with center (13, 4) and radius 4 Graph the circle with center S(13, 4), radius 4. Then graph the points A(11,4), B (8, 5), C(15,5) which represents the places where the actors stand. From the graph, we can see that points A and C inside the circle and point B is outside the circle. Therefore, actors who stand at points A and C are illuminated by the light Question 16. If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle, accident investigators can estimate the speed of the car. a. To find the radius, accident investigators choose points A and B on the tire marks. Then the investigators find the midpoint C of $$\overline{A B}$$. Use the diagram to find the radius r of the circle. Explain why this method works. The radius r of the circle = 155.71 ft Explanation: Given that, AC = 130 ft, CD = 70 ft CE = (r – 70) ft r² = a² + b² r²= 130²+ (r – 70)² r² = 16900 + r² – 140r + 4900 140r = 21,800 r = 155.71 ft b. The formula S = 3.87√fr can be used to estimate a car’s speed in miles per hour, where f is the coefficient of friction and r is the radius of the circle in feet. If f = 0.7, estimate the car’s speed in part (a). The estimated car’s speed is 39.67 miles per hour Explanation: S = 3.87√fr S = 3.8 √(0.7 x 155.71) S = 3.8 √108.997 S = 3.8 x 10.44 S = 39.67 Circles Cumulative Assessment Question 1. Classify each segment as specifically as possible. a. $$\overline{B G}$$ $$\overline{B G}$$ is a chord b. $$\overline{C D}$$ $$\overline{C D}$$ is radius. c. $$\overline{A D}$$ $$\overline{A D}$$ is diameter. d. $$\overline{F E}$$ $$\overline{F E}$$ is a chord Question 2. Copy and complete the paragraph proof. Given Circle C with center (2, 1) and radius 1, Circle D with center (0, 3) and radius 4 Prove Circle C is similar to Circle D. Map Circle C to Circle C’ by using the _________ (x, y) → _________ so that Circle C’ and Circle D have the same center at (____, _____). Dilate Circle C’ using a cellIer of dilation (_____, _____) and a scale factor of _____ . Because there is a _________ transformation that maps Circle C to Circle D, Circle C is __________ Circle D. Map Circle C to Circle C’ by using the scale factor (x, y) → (0, 3) so that Circle C’ and Circle D have the same center at (2, 1). Dilate Circle C’ using a cellIer of dilation (2, 1) and a scale factor of circles. Because there is a transformation that maps Circle C to Circle D, Circle C is similar to Circle D. Question 3. Use the diagram to write a proof. Given ∆JPL ≅ ∆NPL $$\overline{P K}$$ is an altitude of ∆JPL $$\overline{P M}$$ is an altitude ∆NPL Prove ∆PKL ~ ∆NMP ∆JPL is similar to ∆NPL and PK is the altitude of ∆JPL and PM is an altitude of ∆NPL Altitude is a line drawn from one vertex to the opposite site. It is perpendicular to the side. So, ∆PKL is similar to ∆NMP Question 4. The equation of a circle is x² + y² + 14x – 16y + 77 = 0. What are the center and radius of the circle? (A) center: (14, – 16). radius: 8.8 (B) center: (- 7, 8), radius: 6 (C) center (- 14, 16), radius: 8.8 (D) center: (7, – 8), radius: 5.2 (B) center: (- 7, 8), radius: 6 Explanation: x² + y² + 14x – 16y + 77 = 0 x² + 14x + 49 + y² – 16y + 64 = 36 (x + 7)² + (y – 8)² = 6² So, the centre is (-7, 8) and radius is 6. Question 5. The coordinates of the vertices of a quadrilateral are W(- 7, – 6), X(1, – 2), Y(3, – 6) and Z(- 5, – 10). Prove that quadrilateral WXYZ is a rectangle. Proved Explanation: Find the distance of WY and ZX WY = √(-7 – 3)² + (-6 + 6)² = √(-10)² = 10 ZX = √(1 + 5)² + (-2 + 10)² = √6² + 8² = 10 WY = ZX, the diagonals are congruent Use the slope formula to find the slopes of diagonals Slope of WY = $$\frac { -6 + 6 }{ -7 – 3 }$$ = 0 Slope of ZX = $$\frac { -2 + 10 }{ 1 + 5 }$$ = $$\frac { 4 }{ 3 }$$ Because the product of slopes of diagonals is 0, the diagonals are not perpendicular Therefore, the quadrilateral WXYZ is a rectangle. Question 6. Which angles have the same measure as ∠ACB? Select all that apply. im – 295 ∠DEF ∠JGK ∠KGL ∠LGM ∠MGJ ∠QNR ∠STV ∠SWV ∠VWU ∠XYZ ∠VWU Question 7. Classify each related conditional statement based on the conditional statement “If you are a soccer player. then you are an athlete.” a. If you are not a soccer player, then you are not an athlete. False b. If you are an athlete, then you are a soccer player. False c. You are a soccer player if and only if you are an athlete. True d. If you are not an athlete, then you are not a soccer player.	30,304	88,078	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.40625	4	CC-MAIN-2023-06	latest	en	0.816914
https://en.m.wikibooks.org/wiki/Electronics/Inductors	1,726,862,916,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725701423570.98/warc/CC-MAIN-20240920190822-20240920220822-00819.warc.gz	205,778,803	18,350	# Electronics/Inductors ## Inductor An inductor is a passive electronic component dependent on frequency used to store electric energy in the form of a magnetic field. An inductor has the symbol ## Inductance Inductance is the characteristic of the Inductor to generate a magnetic field for a given current. Inductance has a letter symbol L and measured in units of Henry (H). ${\displaystyle L={\frac {B}{I}}}$ This section list formulas for inductances in specific situations. Beware that some of the equations are in Imperial units. The permeability of free space, μ0, is constant and is defined to be exactly equal to 4π×10-7 H m-1. ### Basic inductance formula for a cylindrical coil ${\displaystyle L={\frac {\mu _{0}\mu _{r}N^{2}A}{l}}}$ L = inductance / H μr = relative permeability of core material N = number of turns A = area of cross-section of the coil / m2 l = length of coil / m ### The self-inductance of a straight, round wire in free space ${\displaystyle L_{self}={\frac {\mu _{0}b}{2\pi }}\left[\ln \left({\frac {b}{a}}+{\sqrt {1+{\frac {b^{2}}{a^{2}}}}}\right)-{\sqrt {1+{\frac {a^{2}}{b^{2}}}}}+{\frac {a}{b}}+{\frac {\mu _{r}}{4}}\right]}$ Lself = self inductance / H(?) b = wire length /m ${\displaystyle \mu _{r}}$ = relative permeability of wire If you make the assumption that b >> a and that the wire is nonmagnetic (${\displaystyle \mu _{r}=1}$ ), then this equation can be approximated to ${\displaystyle L_{self}={\frac {\mu _{0}b}{2\pi }}\left[\ln \left({\frac {2b}{a}}\right)-3/4\right]}$ (for low frequencies) ${\displaystyle L_{self}={\frac {\mu _{0}b}{2\pi }}\left[\ln \left({\frac {2b}{a}}\right)-1\right]}$ (for high frequencies due to the skin effect) L = inductance / H b = wire length / m a = wire radius / m The inductance of a straight wire is usually so small that it is neglected in most practical problems. If the problem deals with very high frequencies (f > 20 GHz), the calculation may become necessary. For the rest of this book, we will assume that this self-inductance is negligible. ### Inductance of a short air core cylindrical coil in terms of geometric parameters: ${\displaystyle L={\frac {r^{2}N^{2}}{9r+10l}}}$ L = inductance in μH r = outer radius of coil in inches l = length of coil in inches N = number of turns ### Multilayer air core coil ${\displaystyle L={\frac {0.8r^{2}N^{2}}{6r+9l+10d}}}$ L = inductance in μH r = mean radius of coil in inches l = physical length of coil winding in inches N = number of turns d = depth of coil in inches (i.e., outer radius minus inner radius) ### Flat spiral air core coil ${\displaystyle L={\frac {r^{2}N^{2}}{(2r+2.8d)\times 10^{5}}}}$ L = inductance / H r = mean radius of coil / m N = number of turns d = depth of coil / m (i.e. outer radius minus inner radius) Hence a spiral coil with 8 turns at a mean radius of 25 mm and a depth of 10 mm would have an inductance of 5.13µH. ### Winding around a toroidal core (circular cross-section) ${\displaystyle L=\mu _{0}\mu _{r}{\frac {N^{2}r^{2}}{D}}}$ L = inductance / H μr = relative permeability of core material N = number of turns r = radius of coil winding / m D = overall diameter of toroid / m ### Quality of good inductor There are several important properties for an inductor that may need to be considered when choosing one for use in an electronic circuit. The following are the basic properties of a coil inductor. Other factors may be important for other kinds of inductor, but these are outside the scope of this article. Current carrying capacity is determined by wire thickness and resistivity. The quality factor, or Q-factor, describes the energy loss in an inductor due to imperfection in the manufacturing. The inductance of the coil is probably most important, as it is what makes the inductor useful. The inductance is the response of the inductor to a changing current. The inductance is determined by several factors. Coil shape: short and squat is best Core material The number of turns in the coil. These must be in the same direction, or they will cancel out, and you will have a resistor. Coil diameter. The larger the diameter (core area) the larger the induction. ### Coil's Characteristics For a Coil that has the following dimension Area enclosed by each turn of the coil is A Length of the coil is 'l' Number of turns in the coil is N Permeability of the core is μ. μ is given by the permeability of free space, μ0 multiplied by a factor, the relative permeability, μr The current in the coil is 'i' The magnetic flux density, B, inside the coil is given by: ${\displaystyle B={\frac {N\mu i}{l}}}$ We know that the flux linkage in the coil, λ, is given by; ${\displaystyle \lambda =NBA\,}$ Thus, ${\displaystyle \lambda ={\frac {N^{2}A\mu }{l}}i}$ The flux linkage in an inductor is therefore proportional to the current, assuming that A, N, l and μ all stay constant. The constant of proportionality is given the name inductance (measured in Henries) and the symbol L: ${\displaystyle \lambda =Li\,}$ Taking the derivative with respect to time, we get: ${\displaystyle {\frac {d\lambda }{dt}}=L{\frac {di}{dt}}+i{\frac {dL}{dt}}}$ Since L is time-invariant in nearly all cases, we can write: ${\displaystyle {\frac {d\lambda }{dt}}=L{\frac {di}{dt}}}$ Now, Faraday's Law of Induction states that: ${\displaystyle -{\mathcal {E}}=N{\frac {d\Phi }{dt}}={\frac {d\lambda }{dt}}}$ We call ${\displaystyle -{\mathcal {E}}}$ the electromotive force (emf) of the coil, and this is opposite to the voltage v across the inductor, giving: ${\displaystyle v=L{\frac {di}{dt}}}$ This means that the voltage across an inductor is equal to the rate of change of the current in the inductor multiplied by a factor, the inductance. note that for a constant current, the voltage is zero, and for an instantaneous change in current, the voltage is infinite (or rather, undefined). This applies only to ideal inductors which do not exist in the real world. This equation implies that • The voltage across an inductor is proportional to the derivative of the current through the inductor. • In inductors, voltage leads current. • Inductors have a high resistance to high frequencies, and a low resistance to low frequencies. This property allows their use in filtering signals. An inductor works by opposing current change. Whenever an electron is accelerated, some of the energy that goes into "pushing" that electron goes into the electron's kinetic energy, but much of that energy is stored in the magnetic field. Later when that or some other electron is decelerated (or accelerated the opposite direction), energy is pulled back out of the magnetic field. ## Inductor and Direct Current Voltage (DC) When a coil of several turns is connected to an electricity source in a closed loop, the current in the circuit induces a magnetic field that has the same properties as a Magnetic Field of a Magnet. ${\displaystyle B=LI}$ When the current is turned off, the Magnetic Field does not exist. ${\displaystyle B=0}$ Conducting Coil is called ElectroMagnet ## Inductor and Alternating Current Voltage (AC) ### Inductor's Voltage ${\displaystyle v=L{\frac {di}{dt}}}$ ### Inductor's Current ${\displaystyle i={\frac {1}{L}}\int v\cdot dt}$ ### Reactance ${\displaystyle X_{L}=\omega L\angle 90=j\omega L=sL}$ , where ${\displaystyle s=j\omega }$ . ### Impedance ${\displaystyle R_{L}+X_{L}=R_{L}\angle 0+\omega L\angle 90=R_{L}+j\omega L=R_{L}+sL}$ ### Angle Difference Between Voltage and Current For Lossless Inductor The angle difference between Voltage and Current is 90 For Lossy Inductor ${\displaystyle Tan\theta =\omega {\frac {L}{R_{L}}}=2\pi f{\frac {L}{R_{L}}}={\frac {2\pi }{t}}{\frac {L}{R_{L}}}}$ Changing the value of L and RL will change the value of Angle of Difference, Angular Frequency, Frequency and Time. ${\displaystyle \omega ={\frac {1}{Tan\Theta }}{\frac {L}{R_{L}}}}$ ${\displaystyle f={\frac {1}{2\pi Tan\Theta }}{\frac {L}{R_{L}}}}$ ${\displaystyle t={\frac {t}{2\pi Tan\Theta }}{\frac {L}{R_{L}}}}$ ### Time Constant ${\displaystyle T={\frac {L}{R_{L}}}}$ ### Quality factor Quality factor denoted as Q is defined as the ability to store energy to the sum total of all energy losses within the component ${\displaystyle Q={\frac {X}{R}}}$ ## Inductor's Connection ### Series Connection ${\displaystyle L_{eq}=L_{1}+L_{2}+\cdots +L_{n}}$ ### Parallel Connection ${\displaystyle {\frac {1}{L_{eq}}}={\frac {1}{L_{1}}}+{\frac {1}{L_{2}}}+\cdots +{\frac {1}{L_{n}}}}$	2,434	8,549	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 35, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2024-38	latest	en	0.815255
https://simple.wikipedia.org/wiki/Speed_of_light	1,718,430,102,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861583.78/warc/CC-MAIN-20240615031115-20240615061115-00399.warc.gz	472,541,394	31,819	# Speed of light The speed of light, in any medium, which is usually denoted by ${\displaystyle c}$, is a physical constant important in many areas of physics. It is denoted by 'c^0' especially in vacuum medium, although the symbol 'c' can be used to refer to that in any medium. It is exactly 299,792,458 metres per second (983,571,056 feet per second) by definition.[1][2] A photon (particle of light) travels at this speed in a vacuum. According to special relativity, ${\displaystyle c}$ is the maximum speed at which all energy, matter, and physical information in the universe can travel. It is the speed of all massless particles such as photons, and associated fields—including electromagnetic radiation such as light—in a vacuum. It is predicted by the current theory to be the speed of gravity (that is, gravitational waves). Such particles and waves travel at ${\displaystyle c}$ regardless of the motion of the source or the inertial frame of reference of the observer. In the theory of relativity, ${\displaystyle c}$ interrelates space and time, and appears in the famous equation of mass–energy equivalence E = mc2.[3] The special theory of relativity is based on the prediction, so far upheld by observations, that the measured speed of light in a vacuum is the same whether or not the source of the light and the person doing the measuring are moving relative to each other. This is sometimes expressed as "the speed of light is independent of the reference frame." ## Example explaining how speed does not depend on reference frame This behavior is different from our common ideas about motion as shown by this example: George is standing on the ground next to some train tracks (railroad). There is a train rushing by at 30 mph (48 km/h). George throws a baseball at 90 mph (140 km/h) in the direction the train is moving. Tom, a passenger on the train, has a device (like a radar gun) to measure throwing speeds. Because he is on the train, Tom is already moving at 30 mph (48 km/h) in the direction of the throw, so Tom measures the speed of the ball as only 60 mph (97 km/h). In other words, the speed of the baseball, as measured by Tom on the train, depends on the speed of the train. In the example above, the train was moving at 1/3 the speed of the ball, and the speed of the ball as measured on the train was 2/3 of the throwing speed as measured on the ground. Now, repeat the experiment with light instead of a baseball; that is, George has a flashlight instead of throwing a baseball. George and Tom both have devices that are the same to measure the speed of light (instead of the radar gun in the baseball example). George is standing on the ground next to some train tracks. There is a train rushing by at 1/3 the speed of light. George flashes a light beam in the direction the train is moving. George measures the speed of light as 186,282 miles per second (299,792 kilometres per second). Tom, a passenger on the train, measures the speed of the light beam. What speed does Tom measure? Intuitively, one may think that the speed of the light from the flashlight as measured on the train should be 2/3 the speed measured on the ground, just like the speed of the baseball was 2/3. But in fact, the speed measured on the train is the full value, 186,282 miles per second (299,792 kilometres per second), not 124,188 miles per second (199,861 kilometres per second). It sounds impossible, but that is what one measures. A consequence of this fact, that the speed of light is the same for all (inertial) observers, is that no matter how much energy is used, nothing with mass can be accelerated to reach or go faster than the speed of light. Other key consequences of the constancy of the speed of light, that the lengths of two identical objects, and the rate at which two identical clocks tick, also depend on the reference frame of the observer. These ideas were discovered in the early 1900s by Albert Einstein in his theory of Special Relativity which completely changed our understanding of space and time. ## Relation to fundamental electric and magnetic properties of space Maxwell's equations predicted the speed of light and confirmed Michael Faraday's idea that light was an electromagnetic wave (a way that energy moves). From these equations, we find that the speed of light is related to the inverse of the square root of the permittivity of free space, ε0, and the permeability of free space, μ0: ${\displaystyle c={\frac {1}{\sqrt {\varepsilon _{0}\mu _{0}}}}\ .}$ The index of refraction of a clear material is the ratio between the speed of light in a vacuum and the speed of light in that material. ## Measurement ### Rømer Ole Christensen Rømer used an astronomical measurement to make the first quantitative estimate of the speed of light.[4][5] When measured from Earth, the periods of moons orbiting a distant planet are shorter when the Earth is approaching the planet than when the Earth is receding from it. The distance travelled by light from the planet (or its moon) to Earth is shorter when the Earth is at the point in its orbit that is closest to its planet than when the Earth is at the farthest point in its orbit, the difference in distance being the diameter of the Earth's orbit around the Sun. The observed change in the moon's orbital period is actually the difference in the time it takes light to traverse the shorter or longer distance. Rømer observed this effect for Jupiter's innermost moon Io, and he deduced that light takes 22 minutes to cross the diameter of the Earth's orbit. Another method is to use the aberration of light, discovered and explained by James Bradley in the 18th century.[6] This effect results from the vector addition of the velocity of light arriving from a distant source (such as a star) and the velocity of its observer (see diagram on the right). A moving observer thus sees the light coming from a slightly different direction and consequently sees the source at a position shifted from its original position. Since the direction of the Earth's velocity changes continuously as the Earth orbits the Sun, this effect causes the apparent position of stars to move around. From the angular difference in the position of stars,[7] it is possible to express the speed of light in terms of the Earth's velocity around the Sun. This, with the known length of a year, can be easily converted to the time needed to travel from the Sun to the Earth. In 1729, Bradley used this method to derive that light travelled 10,210 times faster than the Earth in its orbit (the modern figure is 10,066 times faster) or, equivalently, that it would take light 8 minutes 12 seconds to travel from the Sun to the Earth.[6] ### Modern Nowadays, the "light time for unit distance"—the inverse of c (1/c), expressed in seconds per astronomical unit—is measured by comparing the time for radio signals to reach different spacecraft in the Solar System. The position of spacecraft is calculated from the gravitational effects of the Sun and various planets. By combining many such measurements, a best fit value for the light time per unit distance is obtained. As of 2009, the best estimate, as approved by the International Astronomical Union (IAU), is:[8][9] light time for unit distance: 499.004783836(10) s c = 0.00200398880410(4) AU/s c = 173.144632674(3) AU/day. The relative uncertainty in these measurements is 0.02 parts per billion (2×1011), as equivalent to the uncertainty in Earth-based measurements of length by interferometry.[10] Since the metre is defined to be the length travelled by light in a certain time interval, the measurement of the light time for unit distance can also be interpreted as measuring the length of an AU in metres. The metre is considered to be a unit of proper length, whereas the AU is often used as a unit of observed length in a given frame of reference. ## Practical effects The finite speed of light is a major constraint on long-distance space travel. Supposing a journey to the other side of the Milky Way, the total time for a message and its reply would be about 200,000 years. Even more seriously, no spacecraft could travel faster than light, so all galactic-scale transport would be effectively one-way, and would take much longer than than any modern civilisation has existed. The speed of light can also be of concern over very short distances. In supercomputers, the speed of light imposes a limit on how quickly data can be sent between processors.[11] If a processor operates at 1 gigahertz, a signal can only travel a maximum of about 30 centimetres (1 ft) in a single cycle. Processors must therefore be placed close to each other to minimize communication latencies; this can cause difficulty with cooling. If clock frequencies continue to increase, the speed of light will eventually become a limiting factor for the internal design of single chips.[11] ## References 1. SI Brochure: The International System of Units (SI) (PDF) (9 ed.), BIPM, 2019, p. 128, retrieved 2020-01-12 2. Cox, Brian; Forshaw, Jeff (2010). Why does E=mc2?: (and why should we care?). Da Capo. p. 2. ISBN 978-0-306-81911-7. 3. Uzan, J-P; Leclercq, B (2008). The natural laws of the universe: understanding fundamental constants. Springer. pp. 43–4. ISBN 978-0387734545. 4. Cohen, IB (1940). "Roemya and the first determination of the velocity of light (1676)". Isis. 31 (2): 327–79. doi:10.1086/347594. hdl:2027/uc1.b4375710. S2CID 145428377. 5. "Touchant le mouvement de la lumiere trouvé par M. Rŏmer de l'Académie Royale des Sciences" (PDF). Journal des sçavans (in French): 233–36. 1676. Translated in "On the motion of light by M. Romer". Philosophical Transactions of the Royal Society. 12 (136): 893–95. 1677. doi:10.1098/rstl.1677.0024. S2CID 186210345. (As reproduced in Hutton, C; Shaw, G (1809). "On the Motion of Light by M. Romer". In Pearson, R (ed.). The Philosophical Transactions of the Royal Society of London, from Their Commencement in 1665, in the Year 1800: Abridged. Vol. 2. London: C. & R. Baldwin. pp. 397–98.) The account published in Journal des sçavans was based on a report that Rømer read to the French Academy of Sciences in November 1676 (Cohen, 1940, p. 346). 6. Bradley, J (1729). "Account of a new discoved Motion of the Fix'd Stars". Philosophical Transactions. 35: 637–660. 7. at most 20.5 arcseconds Duffett-Smith, P (1988). Practical astronomy with your calculator. Cambridge University Press. p. 62. ISBN 0521356997. 8. Pitjeva, EV; Standish, EM (2009). "Proposals for the masses of the three largest asteroids, the Moon-Earth mass ratio and the Astronomical Unit". Celestial Mechanics and Dynamical Astronomy. 103 (4): 365–372. Bibcode:2009CeMDA.103..365P. doi:10.1007/s10569-009-9203-8. S2CID 121374703. 9. IAU Working Group on Numerical Standards for Fundamental Astronomy. "IAU WG on NSFA Current Best Estimates". US Naval Observatory. Archived from the original on 2009-12-08. Retrieved 2009-09-25. 10. "NPL's Beginner's Guide to Length". UK National Physical Laboratory. Archived from the original on 2010-08-31. Retrieved 2009-10-28. 11. Parhami, B (1999). Introduction to parallel processing: algorithms and architectures. Plenum Press. p. 5. ISBN 9780306459702. and Imbs, D; Raynal, Michel (2009). "Software transactional memories: an approach for multicore programming". In Malyshkin, V (ed.). Parallel Computing Technologies. 10th International Conference, PaCT 2009, Novosibirsk, Russia, August 31-September 4, 2009. Springer. p. 26. ISBN 9783642032745.	2,819	11,556	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 5, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.765625	4	CC-MAIN-2024-26	latest	en	0.952093
https://edurev.in/studytube/Previous-year-questions--2016-20--Laws-of-Motion/12b75fb0-1fa6-4f3d-bbb5-a6c794cd8baa_t	1,723,010,189,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640682181.33/warc/CC-MAIN-20240807045851-20240807075851-00074.warc.gz	171,301,742	81,484	JEE Main Previous Year Questions (2016- 2024): Laws of Motion JEE Main Previous Year Questions (2016- 2024): Laws of Motion \| 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download Q.1. The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is: [JEE Mains 2023] (a) 272 m (b) 68 m (c) 192 m (d) 136 m Solution: Max vertical height Max horizontal distance Q.2. Given below are two statements: Statement I: An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable. Statement II: Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed. In the light of the above statements, choose the correct answer from the options given below: [JEE Mains 2023] (1) Both Statement I and Statement II are true (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are false Solution: Statement-1 When force balance it can move with uniform velocity (Uniform speed) True Statement-2 Elevator going down with increasing speed means its acceleration is downwards mg – N = ma (on person) N = mg – ma (False) Q.3. As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s2) [JEE Mains 2023] (a) (4√3 + 1)N (b) 4(√3 + 1)N (c) (4√3 − 1)N (d) 4(√3 − 1)N Solution: Q.4. A body of mass 1 kg begins to move under the action of a time dependent force , where are the unit vectors along x and y axis. The power developed by above force, at the time t = 2s, will be _________ W. [JEE Mains 2023] Ans. 100 Q.5. Vectors are perpendicular to each other when 3a + 2b = 7, the ratio of b is x/2 The value of x is [JEE Mains 2023] Ans. 1 Q.6. If two vectors are perpendicular to each other. Then, the value of m will be: [JEE Mains 2023] (1) −1 (2) 3 (3) 2 (4) 1 Ans. (3) 4 × 1 + 2mx – 2 + m2 = 0 m2 – 4m + 4 = 0 (m – 2)2 = 0 m = 2 Q.7. Two bodies of masses m1= 5 kg and m= 3 kg are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass m1 will be : [Take g = 10 ms−2] [JEE Mains 2022] (a) 30N (b) 40N (c) 50N (d) 60N Solution: Force on equilibrium, m2g = m1gsin⁡θ ----- (i) Normal force on m1, N = m1 gcos⁡θ -----(ii) From (i) & (ii), Q.8. A uniform metal chain of mass m and length 'L' passes over a massless and frictionless pulley. It is released from rest with a part of its length 'l' is hanging on one side and rest of its length 'L − l' is hanging on the other side of the pully. At a certain point of time, when l = L / x, the acceleration of the chain is g / 2. The value of x is __________. [JEE Mains 2022] (a) 6 (b) 2 (c) 1.5 (d) 4 Solution: By solving the above; m2 = 3m1 (L - l) = 3l L = 4l l = L / 4 = L/x Therefore, x = 4 Q.9. The velocity of the bullet becomes one third after it penetrates 4 cm in a wooden block. Assuming that bullet is facing a constant resistance during its motion in the block. The bullet stops completely after travelling at (4 + x) cm inside the block. The value of x is : [JEE Mains 2022] (a) 2.0 (b) 1.0 (c) 0.5 (d) 1.5 Solution: Q.10. A block 'A' takes 2 s to slide down a frictionless incline of 30 and length 'l', kept inside a lift going up with uniform velocity 'v'. If the incline is changed to 45, the time taken by the block, to slide down the incline, will be approximately : [JEE Mains 2022] (a) 2.66 s (b) 0.83 s (c) 1.68 s (d) 0.70 s Solution: Q.11. A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is θ. The magnitude of the contact force will be : [JEE Mains 2022] (a) Mg (b) Mg cosθ (c) (d) Solution: As the body is moving with constant velocity so forces acting on the body must be balanced. ⇒ Contact force from incline should balance weight of the body. ⇒ \| Fcontact \| = Mg Q.12. A bag is gently dropped on a conveyor belt moving at a speed of 2 m/s. The coefficient of friction between the conveyor belt and bag is 0.4. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is : [Take g=10 m/s−2 ] [JEE Mains 2022] (a) 2 m (b) 0.5 m (c) 3.2 m (d) 0.8 ms Solution: v = 2 m/s μ = 0.4 a = + (0.4) (g) = + 4 m/s2 v2 − u2 = 2 as ⇒ (4) = 2 × (4) (s) s = 0.5 m Q.13. The area of cross section of the rope used to lift a load by a crane is 2.5×10−4 m2. The maximum lifting capacity of the crane is 10 metric tons. To increase the lifting capacity of the crane to 25 metric tons, the required area of cross section of the rope should be : (take g = 10ms−2 ) [JEE Mains 2022] (a) 6.25 × 10−4 m2 (b) 10 × 10−4 m2 (c) 1 × 10−4 m2 (d) 1.67 x 10-4m 2 Solution: Q.14. Two masses M1 and M2 are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M2 is twice that of M1, the acceleration of the system is a1. When the mass M2 is thrice that of M1, the acceleration of the system is a2. The ratio a1 / a2 will be : [JEE Mains 2022] (a) 1/3 (b) 2/3 (c) 3/2 (d) 1/2 Solution: = g/3 And, Q.15. A monkey of mass 50 kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s2 and then climbs up with an acceleration of 5 m/s2. Choose the correct option (g=10 m/s2). [JEE Mains 2022] (a) T=700 N while climbing upward (b) T=350 N while going downward (c) Rope will break while climbing upward (d) Rope will break while going downward Solution: F.B.D of monkey while moving downward: Using Newton’s second law mg – T = ma1 ∴ 500 – T = 50 × 4 ⇒ T = 300 N F.B.D of monkey while moving up: Using Newton’s second law of motion T – mg = ma2 ⇒ T – 500 = 50 × 5 ⇒ T = 750 N Breaking strength of string = 350 N As T > 350N, ∴ String will break while monkey is moving upward Q.16. Three masses M = 100 kg, m= 10 kg and m2 = 20 kg are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m2 moves upward with an acceleration of 2 ms−2. The value of F is : ( Take g = 10 ms−2 ) [JEE Mains 2022] (a) 3360 N (b) 3380 N (c) 3120 N (d) 3240 N Solution: Let acceleration of 100 kg block = a1 Q.17. For a free body diagram shown in the figure, the four forces are applied in the 'x' and 'y' directions. What additional force must be applied and at what angle with positive x-axis so that the net acceleration of body is zero? [JEE Mains 2022] (a) √2N,45 (b) √2N,135 (c) 2 / √3N,30 (d) 2N,45 Solution: Resultant of already applied forces = ⇒ Force required to balance = ⇒ Force required = √2 N in magnitude at angle 45 with +ve x-axis Q.18. A 2 kg block is pushed against a vertical wall by applying a horizontal force of 50 N. The coefficient of static friction between the block and the wall is 0.5. A force F is also applied on the block vertically upward (as shown in figure). The maximum value of F applied, so that the block does not move upward, will be : (Given : g = 10 ms−2) [JEE Mains 2022] (a)10N (b) 20N (c) 25N (d) 45N Solution: Here Fmax force is trying to move the block upward so friction force will be applied towards downward. Along horizontal direction, N = 50 N Along vertical diretion, Fmax = mg + fmax = 2 × 10 + μN = 20 + 0.5 × 50 = 20 + 25 = 45 N Q.19. At t = 0, truck, starting from rest, moves in the positive x-direction at uniform acceleration of 5 ms−2. At t = 20 s, a ball is released from the top of the truck. The ball strikes the ground in 1 s after the release. The velocity of the ball, when it strikes the ground, will be : (Given g = 10 ms−2) [JEE Mains 2022] (a) (b) (c) (d) Solution: At t = 20 s, velocity of truck, v = 0 + 5 × 20 = 100 m/s At 20 sec a ball is dropped from the truck, so velocity of ball will be same as truck. Velocity of truck at x-direction = 100 m/s and in y-direction = 0. ∴ Velocity of ball vx = 100 m/s, vy = 0 Now ball will show projectile motion where vertically downward acceleration g = 10 m/s act on the ball. As horizontally no acceleration acting on the ball so horizontal velocity 100 m/s will remain unchanged. Velocity of the ball when it reach the ground along y-direction after 1 sec. ∴ Velocity of ball Q.20. A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms−1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is : [use g = 9.8 ms−2] [JEE Mains 2022] (a) 4.9 m (b) 9.8 m (c) 12.5 m (d) 19.6 m Solution: = 9.8 m Q.21. An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g = 10 ms−2]. [JEE Mains 2022] (a) 1 : 1 (b) √2 : √3 (c) √3 : √2 (d) 2 : 3 Solution: Let time taken to ascent is t1 and that to descent is t2. Height will be same so Q.22. In the arrangement shown in figure a1, a2, a3 and a4 are the accelerations of masses m1, m2, m3 and m4 respectively. Which of the following relation is true for this arrangement? [JEE Mains 2022] (a) 4a1 + 2a2 + a3 + a4 = 0 (b) a1 + 4a2 + 3a3 + a4 = 0 (c) a1 + 4a2 + 3a3 + 2a4 = 0 (d) 2a1 + 2a2 + 3a3 + a4 = 0 Solution: From virtual work done method, (4T × a1) + (2T × a2) + (T × a3) + (T × a4) = 0 ⇒ 4a1 + 2a2 + a3 + a4 = 0 Q.23. A person is standing in an elevator. In which situation, he experiences weight loss? [JEE Mains 2022] (a) When the elevator moves upward with constant acceleration (b) When the elevator moves downward with constant acceleration (c) When the elevator moves upward with uniform velocity (d) When the elevator moves downward with uniform velocity Solution: N1 = mg N= mg + ma N3 = mg − ma N4 = mg N5 = mg Q.24. A system of two blocks of masses m = 2 kg and M = 8 kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is 0.5. The maximum horizontal force F that can be applied to the block of mass M so that the blocks move together will be : [JEE Mains 2022] (a) 9.8 N (b) 39.2 N (c) 49 N (d) 78.4 N Solution: Q.25. A √34 m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If Ef and Fw are the reaction forces of the floor and the wall, then ratio of Fw/Fwill be : (Use g = 10 m/s2.) [JEE Mains 2022] (a) 6 / √100 (b) 3 / √113 (c) 3 / √109 (d) 2 / √109 Solution: Correct Answer is Option (c) Taking torque from B Fw × 5 = 32mg ⇒ Fw = 310 × 10 × 10 = 30N N = mg = 100N and fr = Fw = 30N so so Q.26. A block of mass 2 kg moving on a horizontal surface with speed of 4 ms−1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = −kx where k = 12 Nm−1. The speed of the block as it just crosses the rough surface will be : [JEE Mains 2022] (a) zero (b) 1.5 ms−1 (c) 2.0 ms−1 (d) 2.5 ms −1 Solution: F = -12x v = 2 m/sec Q.27. A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be [JEE Mains 2022] (a) g / 4 (b) g / 2 (c) 3g / 4 (d) g Solution: Using Newton's second law mg − mg / 4 = ma ⇒ a = 3g / 4 Q.28. A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given g = 10 ms−2.) [JEE Mains 2022] (a) 1 ms−2 (b) 1/5 ms−2 (c) 4/5 ms−2 (d) 8/11 ms −2 Solution: frmax = μN =0.02 × 400 = 8 N Let the acceleration is a as shown then. 40 − T = 4a T − 8 = 40a ⇒ a = 32/44 = 8/11 m/s2 Q.29. A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms−1 and 40 ms−1 respectively. The velocity of the third piece will be : [JEE Mains 2022] (a) 515 ms−1 (b) 25 ms−1 (c) 35 ms−1 (d) 50 ms −1 Solution: Conserving momentum : Q.30. A block of mass m slides along a floor while a force of magnitude θ is applied to it at an angle F as shown in figure. The coefficient of kinetic friction is μκ. Then, the block's acceleration 'a' is given by : (g is acceleration due to gravity) [2021 6 March (Shift 1)] (a) (b) (c) (d) Solution: N = mg − f sin θ F cos θ − μkN = ma F cos θ − μk(mg − F sin θ) = ma Q.31. A block of 200g mass moves with a uniform speed in a horizontal circular groove, with vertical side walls of radius 20cm. If the block takes 40s to complete one round, the normal force by the side walls of the groove is : [2021 16 March (Shift 1)] (a) 0.0314 N (b) 9.859 × 10−2 N (c) 6.28 × 10−3 N (d) 9.859 × 10−4 N Solution: N = mω2R Given m = 0.2 kg, T = 40 S, R = 0.2 m Put values in equation (1) N = 9.859 × 10−4 N Q.32. Consider a frame that is made up of two thin massless rods A and B as shown in the figure. A vertical force of magnitude 100N is applied at point A of the frame. Suppose the force is resolved parallel to the arms AB and AC of the frame. The magnitude of the resolved component along the arm AC is xN. The value of x, to the nearest integer, is ____ . [Given: sin(35º) = 0.573, cos(35º) = 0.819 sin(110º) = 0.939, cos(110º) = -0.342] [2021 16 March (Shift 1)] Solution: 82N Component along AC = 100 cos 35N = 100 × 0.819 N = 81.9 N ≈ 82 N Q.33. Statement I: A cyclist is moving on an unbanked road with a speed of 7kmh-1 and takes a sharp circular turn along a path of radius of 2 m without reducing the speed. The static friction coefficient is 0.2 . The cyclist will not slip and pass the curve (g = 9.8 m/s2) Statement II: If the road is banked at an angle of 45°, cyclist can cross the curve of 2 m radius with the speed of 18.5kmh-1 without slipping. In the light of the above statements, choose the correct answer from the options given below. [2021 16 March (Shift 2)] (a) Statement I is incorrect and statement II is correct (b) Statement I is correct and statement II is incorrect (c) Both statement I and statement II are false (d) Both statement I and statement II are true Solution: Statement I: Vmax = 1-97 m/s 7 km/h = 1.944 m/s Speed is lower than vmax, hence it can take safe turn. Statement II: 18.5 km/h = 5.14 m/s Speed is lower than vmax, hence it can take safe turn. Q.34. A body of mass 2 kg moves under a force of . It starts from rest and was at the origin initially. After 4 s, its new coordinates are (8, b, 20). The value of b is_____. (Round off to the Nearest Integer) [2021 16 March (Shift 2)] Solution: 12 Q.35. A modem grand-prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is μs, then the magnitude of negative lift FL acting downwards on the car is : (Assume forces on the four tyres are identical and g = acceleration due to gravity) [2021 17 March (Shift 1)] (a) (b) (c) (d) Solution: Q.36. Two blocks (m = 0.5 kg and M = 4.5 kg) are arranged on a horizontal frictionless table as shown in figure. The coefficient of static friction between the two blocks is 3/7. Then the maximum horizontal force that can be applied on the larger block so that the blocks move together is_____N. (Round off to the Nearestlnteger) [Take g as 9.8 ms-2 ] [2021 17 March (Shift 1)] Solution: 21 Q.37. A body of mass 1 kg rests on a horizontal floor with which it has a coefficient of static friction 1/√3. It is desired to make the body move by applying the minimum possible force FN. The value of F will be ______(Round off to the Nearest Integer) [Take g = 10 ms-2] [2021 17 March (Shift 2)] Solution: 5 Q.38. A boy of mass 4 kg is standing on a piece of wood having mass 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is N.(Round off to the Nearest Integer) [Take g = 10 ms-2] [2021 17 March (Shift 2)] Solution: 30 N + T = 90 T = μN = 0.5(90 − T) 1.5 T = 45 T = 30 Q.39. A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of bullet before hitting the wood is 10 m/s and it slows down with uniform deceleration, then the magnitude of effective retarding force on the bullet is 'x' N. The value of x to the nearest integer is ___. [2021 18 March (Shift 1)] Solution: 10 v2 = u2 + 2as 0 = (10)+ 2(−a) (1/2) a = 100 m/s2 F = ma = (0.1)(100) = 10 N Q.40. A 60 HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to (1 HP = 746 W, g = 10 m/s2 ) [2020] (a) 1.7 m/s (b) 1.9 m/s (c) 1.5 m/s (d) 2.0 m/s Solution: Given that M = 2000 kg, g = 10 m/s2, f = 4000 N, P = 60 HP = 60 × 746 J Total force acting on elevator is: F = Mg + f = (2000 x 10) + 4000 = 24000 N Power of electric motor is given by P = Fv ⇒ v = P/F Q.41. A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45° with the vertical. Then F equals (Take g = 10 m/s2 and the rope to be massless) [2020] (a) 100 N (b) 90 N (c) 70 N (d) 75 N Solution: Let tension in the string be T At equilibrium, we have T sin 450 = F ..(1) and T cos450 = 10 x g ..(2) Dividing Eq. (1) by Eq. (2), we get (∵ g = 10 m/s) ⇒ I = F/100 ⇒ F = 100 N Q.42. Consider a uniform cubical box of side 'a' on a rough floor that is to be moved by applying minimum possible force F at a point 'b' above its centre of mass (see figure). If the coefficient of friction is μ = 0.4, the maximum possible value of 100.b/a for box not to topple before moving is ______. [2020] Solution: 75 Given that μ = 0.4 Condition for no topple is Friction force is given by F = μmg So, So, maximum value of Therefore, So, maximum possible value of 100 × b/a = 75. Q.43. A spring-mass system (mass m, spring constant k and natural length l) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring-mass system, rotates about its axis with an angular velocity ω, ( k >> mω2) the relative change in the length of the spring is best given by the option [2020] Solution: Let the elongation in length of the spring be Δl. Forces act on the system are: (i) Centripetal force, which act outward which help in expansion in spring: Fc = mω2 (l + Δl) (ii) Spring restoring force, which act inward which opposed the further expansion in spring. Fs = kΔl At balanced point, we have: The relative change in the length of the spring is: Since k>> mωthen Q.44. A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P such that the block does not move downward? (Take g = 10 m/s2) [2019] (a) 32 N (b) 18 N (c) 23 N (d) 25 N Solution: The coefficient of static friction, μ = 0.6 The resultant force is P + μmg cos θ = 3 + mg sin θ Given m = 10 kg, μ = 0.6 and g = 10 m/s2 ⇒ P + 0.6 × 50√2 = 3 + 50√2 ⇒ P + 42.42 = 73.71 ⇒ P = 73.71 - 42.42 = 31.29 Therefore, P ≈ 32 N Q.45. A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 m/s2). [2019] (a) 200 N (b) 140 N (c) 70 N (d) 100 N Solution: Q.46. A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here, k is a constant. The value of T is [2019] (a) (b) (c) (d) Solution: Q.47. A block kept on a rough inclined plane, as shown in the figure, remains at rest up to a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is (Take g = 10 m/s2) [2019] (a) √3/2 (b) √3/4 (c) 1/2 (d) 2/3 Solution: Block is at rest then, 2 N force is applied. Now, friction is When 2 N force is applied in downward direction, thus Block is rest (10 N force is applied) When 10 N force is applied in upward direction, then Dividing Eq. (1) by Eq. (2), we get Q.48. Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is: [2018] (a) 18.3 g (b) 23.3 kg (c) 43.3 kg (d) 10.3 kg Solution: Q.49. A given object takes n times more time to slide down a 45° rough inclined plane as it takes to slide down a perfectly smooth 45° incline. The coefficient of kinetic friction between the object and the incline Is: [2018] (a) (b) (c) (d) Solution: So Sin θ = n2 sinθ − µn2 cosθ ⇒ 1 = n2 − µn2 ⇒ 1 − n2 = −µn2 ⇒ n2 − 1 = −µn2 ⇒ µ = 1 − 1/n2 Q.50. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains the same, the stress in the leg will change by a factor of [2017] (a) 81 (b) 1/81 (c) 9 (d) 1/9 Solution: ∵ Density remains same So, mass ∝ Volume Q.51. A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time? [2017] (a) (b) (c) (d) Solution: Acceleration is constant and negative Q.52. A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is: [2017] (a) (b) (c) (d) Solution: Q.53. A conical pendulum of length 1 m makes an angle q = 45° w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be (Take g = 10 ms-2) [2017] (a) 0.2 m/s (b) 0.4 m/s (c) 2 m/s (d) 4 m/s Solution: Q.54. The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a [2017] (a) Speed which is 3/4th of that of the roller when the weight is 0.4 m above the ground (b) Constant speed (c) Decreasing speed (d) Increasing speed Solution: Q.55. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals m. The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction m and the distance x (= QR), are, respectively close to [2016] (a) 0.2 and 6.5 m (b) 0.2 and 3.5 m (c) 0.29 and 3.5 m (d) 0.29 and 6.5 m Solution: From work energy theorem and given condition: Q.56. A rocket is fired vertically from the earth with an acceleration of 2g, where g is the gravitational acceleration. On an inclined plane inside the rocket, making an angle q. With the horizontal, a point object of mass m is kept. The minimum coefficient of friction μmin between the mass and the inclined surface such that the mass does not move is: [2016] (a) tanθ (b) tan2θ (c) 3 tanθ (d) 2 tanθ Solution: Free body diagram is shown in the figure. Psuedo force on the object is 2mg in the downward direction as seen from the rocket frame of reference. Normal force in the direction perpendicular to the inclined plane. N = 3mg cos(θ) Maximum frictional force in the direction along the plane, Fr = μmN cos(θ) ∴ Fr = 3μmmg cos(θ) This should balance the force on the block along the downward direction of plane. 3mg sin(θ) = 3μmmg cos(θ) μm = tan(θ) Q.57. Which of the following option correctly describes the variation of the speed v and acceleration 'a' of a point mass falling vertically in a viscous medium that applies a force F = –kv, where 'k' is constant, on the body ?( Graphs are schematic and not drawn to scale) [2016] Solution: a = g – αv Q.58. Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute (rpm) to ensure proper mixing is close to (Take the radius of the drum to be 1.25m and its axle to be horizontal): [2016] (a) 1.3 (b) 0.4 (c) 27.0 (d) 8.0 Solution: For just complete rotation: The document JEE Main Previous Year Questions (2016- 2024): Laws of Motion \| 35 Years Chapter wise Previous Year Solved Papers for JEE is a part of the JEE Course 35 Years Chapter wise Previous Year Solved Papers for JEE. All you need of JEE at this link: JEE 35 Years Chapter wise Previous Year Solved Papers for JEE 384 docs\|185 tests 35 Years Chapter wise Previous Year Solved Papers for JEE 384 docs\|185 tests Up next Explore Courses for JEE exam Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests. 10M+ students study on EduRev Related Searches , , , , , , , , , , , , , , , , , , , , , ;	8,482	26,793	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.9375	4	CC-MAIN-2024-33	latest	en	0.781206
https://studentshare.org/law/1673549-police-administration-discusscomplete-5	1,582,514,135,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875145869.83/warc/CC-MAIN-20200224010150-20200224040150-00235.warc.gz	536,991,726	22,862	StudentShare Search Free # POLICE ADMINiSTRATION discuss/complete 5 - Assignment Example Summary This concept challenges the notion I have previously as regards analysis of data in social research. In essence what I learnt here was that while analyzing data, we should be… ## Extract of sample "POLICE ADMINiSTRATION discuss/complete 5" Social Science Statistics Question Looking back at the concepts learnt, I think skewness has had a great impact on me at a personal level. This concept challenges the notion I have previously as regards analysis of data in social research. In essence what I learnt here was that while analyzing data, we should be ready to look out for other issues that may be affecting the data we are studying, even though those issues may not be part of the study. Ensuring therefore that the data presents the expected distribution, in this case, normal distribution plays a key role in analyzing the same data. If not, then the methods of analysis chosen also come into sharp focus. Taking anything for granted will lead to wrong results with devastating consequences given that these results are relied upon for decision making. Thus, skewness as a concept is a very important part of analysis which both guides and complements other methods of social science research. In my view, this concept guides us on how we interpret results as it shows us that there is need exercise caution with our proclamations. Question 2 For this data, the modes are House, Trailer and Apartment. This is because these three have the highest frequencies in the data set. The frequencies are 280, 34 and 21 respectively. As a result, when plotted on a graph, the data may produce a multi-modal shape. From these results, it can be said that the data is NOT normally distributed because of a skewness of 2.000. The mean (1.41) is greater than median (1) showing that the data is positively skewed. Question 3 For the variable ARREST, the modes are 0, 1 and 2 with frequencies of 243, 23 and 10 respectively. Given that the total frequency is 343, it is clear from this observation that the data is not normally distributed. Hence, when plotted on a graph, a positive skewness will be evident. It is interesting to note that a huge chunk of data (frequency of 61) is missing and this also contributes to the level of skewness. Looking at the difference between Mean and Median (0.31), one gets the sense that the data is almost normally distributed. This does not augur well with the skewness of 12.692. Hence, the missing data can be said to be hiding quite a lot of information. Question 4 We are concerned about distribution of data because there are certain statistical procedures that that do not work well with skewed data. Hence, for such procedures, the data needs to be transformed first so as to bring certain aspect of the data within acceptable tolerance levels. If transformations on the data do not achieve acceptable skewness and kurtosis, then the researcher may be forced to settle of procedures that are not susceptible to existing levels of non-normal curves. Question 5 If the data is not normally distributed then there are issues that may come up. Of course the first one comes from the interpretation of the results. As noted earlier, any statistical procedures applied with an aim of interpreting the results will have to be thought through carefully, otherwise, we may end up with misleading results. It should not be lost on us that the purpose of such results is decision making. Besides, such data also calls on the researcher to further investigate the data and find out underlying factors that led to this problem. Ordinarily, it is expected that data out of a research process will present a normal distribution. Read More Cite this document • APA • MLA • CHICAGO (“POLICE ADMINiSTRATION discuss/complete 5 Assignment”, n.d.) Click to create a comment or rate a document ## CHECK THESE SAMPLES OF POLICE ADMINiSTRATION discuss/complete 5 ...? Present-day challenges to Criminal Justice Administrators School In the contemporary society, the police plays a significant role in shaping government policy both at a local and national level. It is also expected that the police should offer services to all citizens, at all times. This expectation presents old and emerging problems, which the police have to face during service delivery. The challenges are diverse in nature, and others are caused by the changing viewpoint of the community (VanHulle, 2011). This paper will explore the challenges facing the police throughout their role in criminal justice administration. In addition, the... 3 Pages(750 words)Essay ...? Police Administration 2 Questions Police Administration 2 Questions Question The American Bar Association has recognized the position of police chief as being among the most important and most demanding positions in the hierarchy of government officials. The roles of a police chief compares favorably with those of government officials and institutional leaders. The main difference between the duties of a police chief and those of other government officials is the authority accorded to him by the preceding authority and their social and political implication within the country. This essay compares the duties of a... 4 Pages(1000 words)Coursework ...in obtaining appreciation of police administration. 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The model is used to administer police operations in a multifaceted manner, especially in offering accountability for patrol commanders when administering policing actions... 6 Pages(1500 words)Essay ...administration (7 ed.). New Providence, NJ: LexisNexis Publishing. Kelling, G. L., Pate, T., Dieckman, D., & Brown, C. E. (2003). The Kansas City Preventive Patrol Experiment. Washington, DC: Police Foundation. McDonald, P. (2002). Managing police operations: Implementing the New York crime control model – CompStat. Belmont, CA: Wadsworth Publishing. McElvain, J. P., Kposowa, A. J., & Gray, B. C. (2013). Testing a Crime Control Model: Does Strategic and Directed Deployment of Police Officers Lead to Lower Crime? Journal of Criminology , 2013, 1-11. Rosenfeld, R., Fornango, R., & Baumer, E. (2005). Did ceasefire, Compstat, and exile reduce homicide? Criminology & Public... 6 Pages(1500 words)Essay ### Do Ex-Military Make Good Police Officers ...Do Ex-Military Make Good Police Officers? 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As respected members of society who are entrusted with our care and protection, the questions surrounding higher education and the police force are addressed and... 11 Pages(2750 words)Essay	2,190	11,067	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.671875	4	CC-MAIN-2020-10	latest	en	0.94826
https://calculationcalculator.com/decimal-to-square-yard	1,723,540,348,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722641075627.80/warc/CC-MAIN-20240813075138-20240813105138-00589.warc.gz	121,102,976	22,014	# Decimal to Square Yard Conversion ## 1 Decimal is equal to how many Square Yard? ### 48.4 Square Yard ##### Reference This Converter: Decimal and Square Yard both are the Land measurement unit. Compare values between unit Decimal with other Land measurement units. You can also calculate other Land conversion units that are available on the select box, having on this same page. Decimal to Square Yard conversion allows you to convert value between Decimal to Square Yard easily. Just enter the Decimal value into the input box, the system will automatically calculate Square Yard value. 1 Decimal in Square Yard? In mathematical terms, 1 Decimal = 48.4 Square Yard. To conversion value between Decimal to Square Yard, just multiply the value by the conversion ratio. One Decimal is equal to 48.4 Square Yard, so use this simple formula to convert - The value in Square Yard is equal to the value of Decimal multiplied by 48.4. Square Yard = Decimal * 48.4; For calculation, here's how to convert 10 Decimal to Square Yard using the formula above - 10 Decimal = (10 * 48.4) = 484 Square Yard Decimal Square Yard (sq yd) Conversion 0.1 4.84 0.1 Decimal = 4.84 Square Yard 0.2 9.68 0.2 Decimal = 9.68 Square Yard 0.3 14.52 0.3 Decimal = 14.52 Square Yard 0.4 19.36 0.4 Decimal = 19.36 Square Yard 0.5 24.2 0.5 Decimal = 24.2 Square Yard 0.6 29.04 0.6 Decimal = 29.04 Square Yard 0.7 33.88 0.7 Decimal = 33.88 Square Yard 0.8 38.72 0.8 Decimal = 38.72 Square Yard 0.9 43.56 0.9 Decimal = 43.56 Square Yard 1 48.4 1 Decimal = 48.4 Square Yard 2 96.8 2 Decimal = 96.8 Square Yard 3 145.2 3 Decimal = 145.2 Square Yard 4 193.6 4 Decimal = 193.6 Square Yard 5 242 5 Decimal = 242 Square Yard 6 290.4 6 Decimal = 290.4 Square Yard 7 338.8 7 Decimal = 338.8 Square Yard 8 387.2 8 Decimal = 387.2 Square Yard 9 435.6 9 Decimal = 435.6 Square Yard 10 484 10 Decimal = 484 Square Yard 11 532.4 11 Decimal = 532.4 Square Yard 12 580.8 12 Decimal = 580.8 Square Yard 13 629.2 13 Decimal = 629.2 Square Yard 14 677.6 14 Decimal = 677.6 Square Yard 15 726 15 Decimal = 726 Square Yard 16 774.4 16 Decimal = 774.4 Square Yard 17 822.8 17 Decimal = 822.8 Square Yard 18 871.2 18 Decimal = 871.2 Square Yard 19 919.6 19 Decimal = 919.6 Square Yard 20 968 20 Decimal = 968 Square Yard 21 1016.4 21 Decimal = 1016.4 Square Yard 22 1064.8 22 Decimal = 1064.8 Square Yard 23 1113.2 23 Decimal = 1113.2 Square Yard 24 1161.6 24 Decimal = 1161.6 Square Yard 25 1210 25 Decimal = 1210 Square Yard 26 1258.4 26 Decimal = 1258.4 Square Yard 27 1306.8 27 Decimal = 1306.8 Square Yard 28 1355.2 28 Decimal = 1355.2 Square Yard 29 1403.6 29 Decimal = 1403.6 Square Yard 30 1452 30 Decimal = 1452 Square Yard 31 1500.4 31 Decimal = 1500.4 Square Yard 32 1548.8 32 Decimal = 1548.8 Square Yard 33 1597.2 33 Decimal = 1597.2 Square Yard 34 1645.6 34 Decimal = 1645.6 Square Yard 35 1694 35 Decimal = 1694 Square Yard 36 1742.4 36 Decimal = 1742.4 Square Yard 37 1790.8 37 Decimal = 1790.8 Square Yard 38 1839.2 38 Decimal = 1839.2 Square Yard 39 1887.6 39 Decimal = 1887.6 Square Yard 40 1936 40 Decimal = 1936 Square Yard 41 1984.4 41 Decimal = 1984.4 Square Yard 42 2032.8 42 Decimal = 2032.8 Square Yard 43 2081.2 43 Decimal = 2081.2 Square Yard 44 2129.6 44 Decimal = 2129.6 Square Yard 45 2178 45 Decimal = 2178 Square Yard 46 2226.4 46 Decimal = 2226.4 Square Yard 47 2274.8 47 Decimal = 2274.8 Square Yard 48 2323.2 48 Decimal = 2323.2 Square Yard 49 2371.6 49 Decimal = 2371.6 Square Yard 50 2420 50 Decimal = 2420 Square Yard 51 2468.4 51 Decimal = 2468.4 Square Yard 52 2516.8 52 Decimal = 2516.8 Square Yard 53 2565.2 53 Decimal = 2565.2 Square Yard 54 2613.6 54 Decimal = 2613.6 Square Yard 55 2662 55 Decimal = 2662 Square Yard 56 2710.4 56 Decimal = 2710.4 Square Yard 57 2758.8 57 Decimal = 2758.8 Square Yard 58 2807.2 58 Decimal = 2807.2 Square Yard 59 2855.6 59 Decimal = 2855.6 Square Yard 60 2904 60 Decimal = 2904 Square Yard 61 2952.4 61 Decimal = 2952.4 Square Yard 62 3000.8 62 Decimal = 3000.8 Square Yard 63 3049.2 63 Decimal = 3049.2 Square Yard 64 3097.6 64 Decimal = 3097.6 Square Yard 65 3146 65 Decimal = 3146 Square Yard 66 3194.4 66 Decimal = 3194.4 Square Yard 67 3242.8 67 Decimal = 3242.8 Square Yard 68 3291.2 68 Decimal = 3291.2 Square Yard 69 3339.6 69 Decimal = 3339.6 Square Yard 70 3388 70 Decimal = 3388 Square Yard 71 3436.4 71 Decimal = 3436.4 Square Yard 72 3484.8 72 Decimal = 3484.8 Square Yard 73 3533.2 73 Decimal = 3533.2 Square Yard 74 3581.6 74 Decimal = 3581.6 Square Yard 75 3630 75 Decimal = 3630 Square Yard 76 3678.4 76 Decimal = 3678.4 Square Yard 77 3726.8 77 Decimal = 3726.8 Square Yard 78 3775.2 78 Decimal = 3775.2 Square Yard 79 3823.6 79 Decimal = 3823.6 Square Yard 80 3872 80 Decimal = 3872 Square Yard 81 3920.4 81 Decimal = 3920.4 Square Yard 82 3968.8 82 Decimal = 3968.8 Square Yard 83 4017.2 83 Decimal = 4017.2 Square Yard 84 4065.6 84 Decimal = 4065.6 Square Yard 85 4114 85 Decimal = 4114 Square Yard 86 4162.4 86 Decimal = 4162.4 Square Yard 87 4210.8 87 Decimal = 4210.8 Square Yard 88 4259.2 88 Decimal = 4259.2 Square Yard 89 4307.6 89 Decimal = 4307.6 Square Yard 90 4356 90 Decimal = 4356 Square Yard 91 4404.4 91 Decimal = 4404.4 Square Yard 92 4452.8 92 Decimal = 4452.8 Square Yard 93 4501.2 93 Decimal = 4501.2 Square Yard 94 4549.6 94 Decimal = 4549.6 Square Yard 95 4598 95 Decimal = 4598 Square Yard 96 4646.4 96 Decimal = 4646.4 Square Yard 97 4694.8 97 Decimal = 4694.8 Square Yard 98 4743.2 98 Decimal = 4743.2 Square Yard 99 4791.6 99 Decimal = 4791.6 Square Yard 100 4840 100 Decimal = 4840 Square Yard A decimal / dismil is a unit of area in India and Bangladesh. After metrication in the mid-20th century by both countries, the unit became officially obsolete. Decimals are also used as a measure of land in East Africa. The Square Yard (sq yd) is a US customary and imperial unit of area, equal to 9 Square Feet (sq ft). The abbreviation for Square Yard is "yd2". In India or Bangladesh, it is also known as Square Gaj. The value in Square Yard is equal to the value of Decimal multiplied by 48.4. Square Yard = Decimal * 48.4; 1 Decimal is equal to 48.4 Square Yard. 1 Decimal = 48.4 Square Yard. • decimal to sq yd • 1 decimal = square yard • decimal into square yard • decimals to square yards • convert decimal to square yard → → → → → → → → → → → → → → → → → →	2,329	6,343	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.640625	4	CC-MAIN-2024-33	latest	en	0.717129
https://pt.scribd.com/doc/160750232/Intro-to-Mechanical	1,582,074,478,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875143963.79/warc/CC-MAIN-20200219000604-20200219030604-00555.warc.gz	540,937,740	123,660	Você está na página 1de 25 Preface to the Instructor With the exception of some “open-ended” problems, this Instructor’s Solutions Manual contains a solution to each of the numerical problems in the textbook An Introduction to Mechanical Engineering (second edition). The description and style of these solutions (writing a brief approach, keeping track of units, making sketches, staying organized and so forth) are intended to guide students with respect to the formatting and presentation of their own work. In addition, the compact disk accompanying this manual includes electronic versions of all photographs and figures in the textbook, and an electronic version of this manual. My intention is that these resources will be useful to instructors in two ways. First, instructors are able to easily import photographs, drawings, and other art from the textbook into their own handouts and lectures without the burden of having to photocopy or scan from the textbook itself. Second, since many courses today have companion web sites, instructors can use the compact disk to post electronic versions of solutions to their own problem sets. I hope that you will find this Instructor’s Solutions Manual to be a useful resource for your own teaching of mechanical engineering courses at the freshman and sophomore levels. As always, if you should have any comments on the textbook or manual, I would certainly appreciate hearing from you. Jonathan Wickert Department of Mechanical Engineering Carnegie Mellon University Pittsburgh, PA 15213 [email protected] J.A. Wickert An Introduction to Mechanical Engineering Problem 2.1: Express your weight in the units lb and N, and your mass in the units slug and kg. Solution: Let w and m represent weight and mass. w = 180 lb Convert to SI using the factor from Table 2.6: w = ( 180 lb ) w = 801 N 4.448 N lb Calculate mass using gravitational acceleration in USCS: m = w 180 lb = g 32.2 ft / s 2 m = 5.59 slugs Convert to SI using the factor from Table 2.6: m = ( 5.59 slugs ) m = 81.6 kg 14.59 kg slug - 1 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.2: Express your height in the units in, ft, and m. Solution: Let h represent height. h = 6 ft h = ( 6 ft ) h = 72 in 12 in ft Convert to SI using the factor from Table 2.6: h = ( 6 ft ) 0.3048 h = 1.83 m m ft - 2 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.3: The solar power transmitted to the ground is 0.6 kW/m 2 . Convert this quantity to the dimensions of hp/ft 2 . Solution: Convert to USCS using the factors from Table 2.6. 0.6 kW ⎞ ⎟ ⎛ ⎜ 9.2903 × 10 − 2 m 2 ⎞ ⎟ ⎛ ⎜ 1.341 hp ⎞ ⎟ = 7.475 × 10 − 2 hp m 2 ⎟ ⎠ ⎜ ⎝ ft 2 ⎟ ⎠ ⎝ kW ⎠ ft 2 hp − 2 7.475 10 × 2 ft - 3 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.4: One U.S. gallon is equivalent to 0.1337 ft 3 , one foot is equivalent to 0.3048 m, and 1000 liters is equivalent to 1 m 3 . By using those definitions, determine the conversion factor between gallons and liters. Solution: Use the definitions of derived units and the conversion factors listed in Tables 2.2, 2.5, and 2.6: 1 gal = 0.1337 ft 3 1 ft = 0.3048 m 1000 L = 1 m 3 Convert gal to L using those factors: ( 1 0.1337 = gal ft 3 1 gal = 3.785 L ) ft m ⎛ ⎜ 1000 m ⎠ ⎟ ⎠ 3 L 0.3048 3 - 4 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.5: A passenger automobile is advertised as having a fuel economy rating of 29 mi/gal for highway driving. Express the rating in the units of km/L. Solution: Convert using factors from Table 2.6: 29 mi gal ⎞ ⎛ ⎠ ⎝ km mi gal L 0.2642 1.609 km 12.33 L = 12.33 - 5 - km L J.A. Wickert An Introduction to Mechanical Engineering Problem 2.6: (a) How many horsepower does a 100 W household light bulb consume? (b) How many kW does a 5-hp lawn mower engine produce? Solution: a) Convert W to kW using the prefix definition in Table 2.3. factor from Table 2.6: ( 100 W ) 0.001 0.1341 hp kW ⎞ ⎛ ⎜ ⎠ ⎝ hp ⎟ = W kW 1.341 0.1341 hp Convert from kW to hp using the b) Convert from hp to kW using factor from Table 2.6: ( 5 hp ) 0.7457 ⎜ ⎜ 3.729 kW kW ⎞ ⎟ = hp ⎟ ⎠ 3.729 kW - 6 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.7: A robotic wheeled vehicle that contains science instruments is used to study the geology of Mars. The rover weighs 408 lb on Earth. (a) In the USCS dimensions of slugs and lbm, what is the rover’s mass? (b) What is the weight of the rover as it rolls off the lander’s platform (the lander is the protective shell that houses the rover during landing)? The gravitational acceleration on the surface of Mars is 12.3 ft/s 2 . Solution: a) The rover’s mass is the same on Earth and Mars. On Earth, it weighs 408 lb, and the gravitational acceleration is 32.2 ft/s 2 . By using m = w/g m = 2 408 lb lb s = 12.67 2 ft 32.2 ft s From Table 2.5, the dimension is equivalent to the derived unit of slug. m = 12.67 slugs By definition, an object that weighs one pound has a mass of one pound-mass, so m = 408 lbm We double check by using the definition in Table 2.5 m = ( 408 lbm ) 3.1081 × 10 2 slugs lbm ⎟ = 12.67 slugs b) Since the gravitational acceleration on Mars is only 12.3 ft/s 2 , the weight becomes w = ( 12.67 slugs ) 12.3 ft ⎞ ⎟ = s 2 slug ft s 2 155.9 The dimension is the dimensionally consistent group that is the same as lb. w = 155.9 lb It would not be correct to calculate weight as the product of 408 lbm and 12.3 ft/s 2 , since the units would not be dimensionally consistent. - 7 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.8: Calculate various fuel quantities for Flight 143. The plane already had 7,682 L of fuel on board prior to the flight, and the tanks were to be filled so that a total of 22,300 kg were present at take-off. (a) Using the incorrect conversion factor of 1.77 kg/L, calculate in units of kg the amount of fuel that was added to the plane. (b) Using the correct factor of 1.77 lb/L, calculate in units of kg the amount of fuel that should have been added. (c) By what percent would the plane have been under-fueled for its journey? Be sure to distinguish between weight and mass quantities in your calculations. Solution: a) Amount of fuel that was added (kg): 22,300 kg 8,703 kg ( 7,682 L ) 1.77 kg L ⎟ = 8,702.9 kg b) Amount of fuel that should have been added (kg): 22,300 kg 16,139 kg ( 7,682 L ) 1.77 lb L 1 kg 32.2 ft / s 2 slug 14.59 ⎟ = c) Percent that the plane was under-fueled: ⎛ 16,139 kg 8,703 kg ⎞ ⎟ ⎟ 16,139 kg 46.1% ( 100% )= 46.07% - 8 - 16,139 kg J.A. Wickert An Introduction to Mechanical Engineering Problem 2.9: Printed on the side of a tire on an all-wheel-drive sport utility wagon is the warning “Do not inflate above 44 psi,” where psi is the abbreviation for the pressure unit lb/in 2 . Express the tire’s maximum pressure rating in (a) the USCS unit of lb/ft 2 (psf), and (b) the SI unit of kPa. Solution: a) Convert the area measure from in to ft: 44 lb ⎞ ⎛ in in 2 ⎠ ⎝ 12 ft 2 6,336 psf = 6,336 ft lb 2 where the abbreviation psf stands for pounds-per-square-foot. b) Use a conversion factor from Table 2.6 and the SI prefix from Table 2.3: 44 lb ⎞ ⎛ ⎜ ⎜ ⎠ ⎝ Pa ⎞ ⎟ = psi in 2 6,895 303.4 kPa 3.034 ×10 5 Pa = 303.4 kPa - 9 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.10: The amount of power transmitted by sunlight depends on latitude and the surface area of the solar collector. On a clear day at a certain northern latitude, 0.6 kW/m 2 of solar power strikes the ground. Express that value in the alternative USCS unit of (ft • lb/s)/ft 2 . Solution: Use conversion factors from Table 2.6: kW kW ⎞ ⎟ ⎛ ⎜ 1,000 W ⎞ ⎟ ⎛ ⎜ 0.7376 ft • lb / s ⎞ ⎟ ⎛ ⎜ 0.3048 m 2 ⎟ ⎠ ⎝ ⎠ ⎝ W ⎠ ⎜ ⎝ ft m 0.6 / s 41.1 ft • lb 2 ft - 10 - 2 = 41.1 ft lb / s ft 2 J.A. Wickert An Introduction to Mechanical Engineering Problem 2.11: The property of a fluid called “viscosity” is related to its internal friction and resistance to being deformed. The viscosity of water, for instance, is less than that of molasses and honey, just as the viscosity of a light motor oil is less than that of a grease. A unit used in mechanical engineering to describe viscosity is called the “poise,” named after the physiologist Jean Louis Poiseuille who performed early experiments in fluid mechanics. The unit is defined by 1 poise = 0.1 (N • s)/m 2 . Show that 1 poise is also equivalent to 1 g/(cm • s). Solution: Expand the derived unit N in terms of the SI base units: 0.1 N • s = 0.1 kg • m • s = 0.1 kg m 2 s 2 • m 2 m • s Apply SI prefixes from Table 2.3: 0.1 kg m ⎟ ⎞ = cm m s kg 1000 g 0.01 g 1 cm s 1 g cm s - 11 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.12: Referring to the description in Problem 2.11, and given that the viscosity of a certain engine oil is 0.25 kg/(m • s), determine the value in the units (a) poise, and (b) slug/(ft • s). Solution: a) Express viscosity in terms of the derived unit (N • s)/m 2 : 0.25 kg kg m m s m 2 s = 0.25 = 0.25 ( kg m / s 2 ) s N s 2 2 = 0.25 m m Use the definition of poise P from Problem 2.11: ⎛ ⎜ 0.25 N • s ⎞ ⎟ ⎛ ⎜ 10 P ⎜ ⎝ m 2 ⎟ ⎠ ⎜ ⎝ ( N • s )/ m 2 2.5 P = 2.5 P b) Use conversion factors from Table 2.6: 0.25 kg slug ⎞ ⎛ ⎜ ⎜ ⎠ ⎝ m ⎟ = m s kg ft 0.0685 0.3048 0.0052 slug ft • s 0 - 12 - .0052 slug ft s J.A. Wickert An Introduction to Mechanical Engineering Problem 2.13: Referring to the description in Problem 2.11, if the viscosity of water is 0.01 poise, determine the value in terms of the units (a) slug/(ft • s) and (b) kg/(m • s). Solution: Express 0.01 P in terms of SI base units: ( 0.01 P ) ( 0.1 N • ) / m 2 ⎞ ⎟ = 0.001 N • s P ⎟ ⎠ m 2 s 0.001 kg m • s = 0.001 kg m s Use conversion factors from Table 2.6: 0.001 kg slug ⎞ ⎛ ⎜ kg ⎠ ⎝ m ⎞ ⎟ = ft m s 0.0685 0.3048 2.088 10 × 5 slug ft s - 13 - 2.088 10 × 5 slug ft s J.A. Wickert An Introduction to Mechanical Engineering Problem 2.14: The fuel efficiency of an aircraft’s jet engines is described by the thrust-specific fuel consumption, or TSFC. The TSFC measures the rate of fuel consumption (mass of fuel burned per unit time) relative to the thrust (force) that the engine produces. In that manner, just because an engine might consume more fuel per unit time than a second engine, it would not necessarily be more inefficient if it also produced more thrust to power the plane. The TSFC for an early hydrogen-fueled jet engine was 0.082 (kg/h)/N. Express that value in the USCS units of (slug/s)/lb. Solution: Use SI to USCS conversion factors from Table 2.6: ⎛ ⎜ 0.082 kg / hr ⎞ ⎟ ⎛ ⎜ 0.0685 slug ⎞ ⎟ ⎛ ⎜ 2.778 × 10 − 4 hr ⎞ ⎛ ⎜ ⎟ 4.448 ⎝ N ⎠ ⎜ ⎝ kg ⎟ ⎠ ⎝ s ⎠ ⎝ lb 6.94 × 10 −6 slug / s lb N ⎟ ⎞ = - 14 - 6.94 × 10 6 slug / s lb J.A. Wickert An Introduction to Mechanical Engineering Problem 2.15: An automobile engine is advertised as producing a peak power of 118 hp (at an engine speed of 4000 rpm), and a peak torque of 186 ft • lb (at 2500 rpm). Express those performance ratings in the SI units of kW and N • m. Solution: For the power, apply the hp to kW conversion factor from Table 2.6: kW ⎞ ⎟ = ( 118 hp ) ⎜ ⎛ ⎜ 0.7457 ⎝ hp ⎠ ⎟ 87.99 kW 87.99 kW For the torque, apply USCS to SI conversion factors from Table 2.6: ( 186 ft lb ) 252 N m ⎛ ⎜ 4.448 N lb m ⎟ = ft 0.3048 252 N m - 15 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.16: For the exercise of Example 2.6, express the sideways deflection of the tip in the units of mils (defined in Table 2.5) when the various quantities are instead known in the USCS. Use the values F = 75 lb, L = 3 in, d = 3/16 in, and E = 30 x 10 6 psi. Solution: Evaluate the deflection equation using numerical values that are dimensionally consistent: x = 3 64 FL 3 π Ed 4 = ( 64 75 lb )( 3 in )( . ) 3 3 π × 10 6 lb / in . 2 0.1875 in . ) 4 ( 30 = 0.371 in . Apply the definition of mil from Table 2.5: 1 mil = 0.001in. 371 mils - 16 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.17: Heat Q, which has the SI unit of joule (J), is the quantity in mechanical engineering that describes the transit of energy from one location to another. The equation for the flow of heat during the time interval t through an insulated wall is Q A = κ t L ( T h T l ) where κ is the thermal conductivity of the material from which the wall is made, A and L are the wall’s area and thickness, and low-temperature sides of the wall. By using the principle of dimensional consistency, what is the correct dimension for thermal conductivity in the SI? The lowercase Greek character kappa (κ) is a conventional mathematical symbol used for thermal conductivity. Appendix A summarizes the names and symbols of Greek letters. is the difference (in degrees Celsius) between the high- and T h T l Solution: Let [κ] denote the SI units for thermal conductivity. Apply the principle of dimensional consistency and the definition of watt in Table 2.2: [ κ ]• 2 m • s • o C m ⎟ 1 = W ⎝ s ⎠ m • o C m • o C J = [κ] = ⎜ ⎛ J W o m C - 17 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.18: Convection is the process by which warm air rises and cooler air falls. The Prandtl number, Pr, is used when mechanical engineers analyze certain heat transfer and convection processes. It is defined by the equation Pr = µ c p κ where c p is a property of the fluid called the specific heat having the SI unit the viscosity as discussed in Problem 2.11; and κ is the thermal conductivity as discussed in Problem 2.17. Show that Pr is a dimensionless number. The lowercase Greek characters mu (µ) and kappa (κ) are conventional mathematical symbols used for viscosity and thermal conductivity. Appendix A summarizes the names and symbols of Greek letters. ; µ is kJ /( kg o C ) Solution: Let [Pr] denote the units for Prandtl number. In the SI, a unit for viscosity is N • s/m 2 , and from the solution to Problem 2.11, the units for thermal conductivity are W/(m • o C). Apply the principle of dimensional consistency and the definition of watt from Table 2.2: [ Pr ] ( kJ /( kg • o C ) ) ( N • s / m 2 ) [ W /( m • o C )] ⎛ kJ ⎞ ⎟ ⎛ ⎝ ⎠ ⎝ ⎜ s 1 W ⎟ ⎞ ( ⎠ N ) ⎛ ⎜ ⎝ 1 N ⎞ ⎟ = ⎠ kW W = = ⎜ = kJ W N kg s m o C o C 2 m = kJ kg N m s 2 s which has no units but does contain a dimensionless factor of 1000 because of the “kilo” prefix. - 18 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.19: Referring to Problem 2.18 and Table 2.5, if the units for c p and µ are Btu/(slug • o F) and slug/(ft • h), respectively, what must be the USCS units of thermal conductivity in the definition of Pr? Solution: Apply the principle of dimensional consistency and determine the units for thermal conductivity so that the Prandtl number is dimensionless. Let [κ] denote the USCS units for thermal conductivity: Pr = c p µ κ [ κ ] = ⎛ ⎜ Btu ⎞ ⎟ ⎛ ⎜ slug ⎞ ⎟ = Btu / hr slug • o F ⎟ ⎠ ⎜ ⎝ ft • hr ⎟ ⎠ ft • o F Btu hr / o ft F - 19 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.20: Some scientists believe that the collision of one or more large asteroids with the Earth was responsible for the extinction of the dinosaurs. The unit of kiloton is used to describe the energy released during large explosions. It was originally defined as the explosive capability of 1000 tons of trinitrotoluene (TNT) high explosive. Because that expression can be imprecise depending on the explosive’s exact chemical composition, the kiloton has been subsequently redefined as the equivalent of 4.186 x 10 12 J. In the units of kiloton, calculate the kinetic energy of an asteroid that has the size (box-shaped, 13 x 13 x 33 km) and composition (density, 2.4 g/cm 3 ) of our solar system’s asteroid Eros. Kinetic energy is defined by 1 U = 2 mv 2 where m is the object’s mass and v is its speed. Objects passing through the inner solar system generally have speeds in the range of 20 km/s. Solution: Volume: Mass: (13 km ( 5.577 ) (13 km ) (33 × 10 18 cm 3 ) km ) 2.4 = = = 5.577 ( 5.577 5.577 × × × 10 3 km 3 10 3 km 3 ) 10 18 cm 3 ⎛ ⎜ 10 g cm 3 ⎟ = = 1.338 1.338 × × 10 10 19 16 g kg 5 cm km 3 Speed: 20 ⎟ ⎜ 1000 4 m s ⎠ ⎝ km ⎠ s km ⎞ ⎛ m ⎞ = 2 ×10 Kinetic energy: 1 2 ( 1.338 × 10 16 kg ) 6.4 ×10 11 kiloton 2 × 10 4 m ⎞ ⎟ 2 = 2.68 × 10 24 kg • m 2 s ⎠ s 2 = 2.68 × 10 24 N • m = = 2.68 ( 2.68 × × 10 24 J 10 24 J ) ⎛ ⎜ ⎜ ⎝ 4.186 × 10 12 J = 6.4 × 10 11 kiloton 1 kiloton ⎟ ⎟ - 20 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.21: A structure known as a cantilever beam is clamped at one end but free at the other, analogous to a diving board that supports a swimmer standing on it. Using the following procedure, conduct an experiment to measure how the cantilever beam bends. In your answer, report only those significant digits that you know reliably. (a) Make a small table top test stand to measure the deflection of a plastic drinking straw (your cantilever beam) that bends as a force F is applied to the free end. Push one end of the straw over the end of a pencil, and then clamp the pencil to a desk or table. You can also use a ruler, chopstick, or a similar component as the cantilever beam itself. Sketch and describe your apparatus and measure the length L. (b) Apply weights to the end of the cantilever beam, and measure the tip’s deflection y using a ruler. Repeat the measurement with at least a half-dozen different weights to fully describe the beam’s force-deflection relationship. Penny coins can be used as weights; one penny weighs approximately 30 mN. Make a table to present your data. (c) Next draw a graph of the data. Show tip deflection on the abscissa and weight on the ordinate, and be sure to label the axes with the units for those variables. (d) Draw a best-fit line through the data points on your graph. In principle, the deflection of the tip should be proportional to the applied force. Do you find this to be the case? The slope of the line is called the stiffness. Express the stiffness of the cantilever beam either in the units lb/in or N/m. Solution: A 1.6 mm diameter steel rod was used as the cantilever beam, and it was clamped in a vise as shown in Figure 1. Small weights were held in a plastic bag attached to the beam’s free end. The distance between the beam’s tip and the surface of the table was measured for each weight, and the data were recorded in Table 1. Beams of three different lengths were tested. Figure 2 shows a graph of force versus tip deflection, H-h, for the 20 cm beam length. The deflection is proportional to the force. The slope of the line is 22.3 mN/mm, or 22.3 N/m. Figure 1: Test stand. - 21 - J.A. Wickert Problem 2.21, continued An Introduction to Mechanical Engineering Table 1: Measured tip deflection. Figure 2: Graph of force versus deflection for the 20 cm beam. - 22 - J.A. Wickert An Introduction to Mechanical Engineering Problem 2.22: Perform measurements as described in Problem 2.21 for cantilever beams of several different lengths. Can you show experimentally that for a give force F, the deflection of the cantilever’s tip is proportional to the cube of its length? As in Problem 2.21, present your results in a table and a graph, and report only those significant digits that you know reliably. Solution: A 1.6 mm diameter steel rod was used as the cantilever beam, and it was clamped in a vise as shown in Figure 1. Small weights were held in a plastic bag attached to the beam’s free end. The distance between the beam’s tip and the surface of the table was measured for each weight, and the data were recorded in Table 1. Beams of three different lengths were tested. Figure 2 shows a graph of tip deflection, H-h, versus beam length for tip forces of 196 mN and 392 mN. The curve fit indicates that the deflection is proportional to the cube of the length in each case. Figure 1: Test stand. Table 1: Measured tip deflection. - 23 - J.A. Wickert Problem 2.22, continued An Introduction to Mechanical Engineering Figure 2: Graph of deflection versus beam length for two different applied forces. - 24 -	6,490	20,607	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.21875	4	CC-MAIN-2020-10	latest	en	0.903749
https://www.dhimanrajeshdhiman.com/2021/11/Ten-important-problems-in-reasoning-with-answers.html	1,643,443,862,000,000,000	text/html	crawl-data/CC-MAIN-2022-05/segments/1642320300573.3/warc/CC-MAIN-20220129062503-20220129092503-00585.warc.gz	746,823,412	64,461	# Ten Figure problems in reasoning with answers for competitive exams ## Problem #1 This reasoning problem consists of three figures and every figure have five number associated to it . Four numbers are at the corner of each figure and one number is at the centre of the figure. Look at last figure ,it have ? in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . But the biggest problem is how to utilised these four numbers to get the value of this question mark? Now watch carefully the 1st two figures . since these figures have some values of middle numbers. Now we have to find or search the formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. The same formula will be applicable to third figure to find out the value of question mark. Formula :- The middle Number in each figure is one tenth of product of remaining numbers. Solution:- ( 4 × 5 × 7 × 6 ) ÷ 10 = 840 ÷ 10 = 84 ( Middle number in 1st figure). (2 × 8 × 4 × 5 ) ÷ 10 = 320 ÷ 10 = 32 ( Middle number in 2nd figure). (5 × 6 × 9 × 2 )÷ 10 = 540 ÷ 10 = 54 ( Middle number in 3rd figure). Option (1) 54 is correct option ## Problem #2 This reasoning problem also consists of three figures and every figure have five number associated to it . Four numbers are at the corner of each figure and one number is at the centre of the figure. Look at last figure ,it have ? in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . But the main problem is how to use these four numbers to get the value of this question mark? Now watch carefully the 1st two figures . since these figures have some values of middle numbers. Now we have to find or search the formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. The same formula will be applicable to third figure to find out the value of question mark. Formula:- :- The middle Number in each figure is two less than the sum of remaining numbers. Solution:- (4 + 5 + 7 + 6 ) - 2 = 22 - 2 = 20 ( Middle number in 1st figure). (2 + 8 + 4 + 7 ) - 2 = 21 - 2 = 19 ( Middle number in 2nd figure). (5 + 6 + 9 + 8 ) - 2 = 28 - 2 = 26 ( Middle number in 3rd figure), Option (4) 26 is correct option ## Problem #3 This reasoning problem also have three figures and every figure have five number attached to it . Four numbers are at the corner of each figure and one number is at the middle of the same figure. Since last figure have question mark in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . But the biggest problem is how to utilised these four numbers to get the value of this question mark? Now watch carefully the 1st two figures . since these figures have some values of middle numbers. Now we have to find or search the formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. The same formula will be applicable to third figure to find out the value of question mark. Formula:- [{Left number × Top most number } + Right number ] - bottom number. Solution:- {(4 × 5) + 7 }- 20 = ( 20 + 7 ) - 20 = 27 - 20 = 7 ( Middle number in 1st figure) . {(2 × 8 ) + 4} - 12 = ( 16 + 4 ) - 12 = 20 - 12 = 8 ( Middle number in 2nd figure). {(4 × 12 ) + 12} - 51 = ( 48 + 12 ) - 51 = 60 - 51 = 9 ( Middle number in 3rd figure). Option (2)9 is correct option ## Problem #4 This reasoning problem also have three figures and every figure have five number attached to it . Four numbers are at the corner of each figure and one number is at the middle of the same figure. Since 2nd figure have question mark in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . But the biggest problem is how to utilised these four numbers to get the value of this question mark? Now watch carefully the 1st and 3rd figures . since these figures have some values of middle numbers. Now we have to find or search the formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. The same formula will be applicable to 2nd figure to find out the value of question mark. Formula:-[{Left number × Top most number } + Right number ] - bottom number. Solution:- {(4 × 5 ) + 7 } - 16 = ( 20 + 7 ) - 16 = 27 - 16 = 11 ( Middle number in 1st figure). {(4 × 12 ) + 1} - 33 = ( 48 + 1 ) - 33 = 49 - 33 = 16 ( Middle number in 3rd figure) . {(2 × 8 ) + 3} - 12 = ( 16 + 3 ) - 12 = 19 - 12 = 7 ( Middle number in 2nd figure) . The value of question mark. Option (4)7 is correct option ## Problem #5 This reasoning problem also have three figures and every figure have five number attached to it . Four numbers are at the corner of each figure and one number is at the middle of the same figure. Since last figure have question mark in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . But the biggest problem is how to utilised these four numbers to get the value of this question mark? Now watch carefully the 1st two figures . since these figures have some values of middle numbers. Now we have to find or search the formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. The same formula will be applicable to third figure to find out the value of question mark. Formula:- Add 1 to all the numbers in upper rows of each figure to multiply the number opposite to it. Solution:- ( 7 + 1) × 5 = ( 9 + 1) × 5 = 40 (Middle number in 1st figure ) ( 17 + 1) × 3 = ( 8 + 1) × 6 = 54 (Middle number in 2nd figure ) ( 10 + 1) × 6 = ( 21 + 1) × 3 = 66 (Middle number in 3rd figure ). Option (4)66 is correct option ## - This reasoning problem also have three figures and every figure have five number attached to it . Four numbers are at the corner of each figure and one number is at the middle of the same figure. Since last figure have question mark in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . Now we have to find or search the formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. The same formula will be applicable to third figure to find out the value of question mark. Formula:- Add 1 to all the numbers of each figure to multiply the number opposite to it. Solution:- ( 9 +1 ) × ( 3 + 1 ) = ( 9 + 1 ) × ( 3 + 1 ) = 10 × 4 = 40 ( Middle number in 1st figure ) ( 26 +1 ) × ( 1 + 1 ) = ( 8 + 1 ) × ( 5 + 1 ) = 27 × 2 = 40 ( Middle number in 2nd figure ). 15 + 1 ) × ( 2 + 1 ) = ( 3 + 1 ) × ( 11 + 1 ) = 16 × 3 = 48 ( Middle number in 3rd figure). Option (3)48 is correct option ## Problem #7 This reasoning problem also have three figures and every figure have five number attached to it . Four numbers are at the corner of each figure and one number is at the middle of the same figure. Since last figure have question mark in its centre . So the solution of this problem is to find the value of question mark using four numbers associated to it . Now watch carefully the 1st two figures . since these figures have some values of middle numbers. Now we have to find or search the formula for these four numbers in each figure to utilised them in any possible way to get middle or central number. The same formula will be applicable to third figure to find out the value of question mark. Formula:- Product of both numbers in upper lines - Product of both numbers in lower lines ⇔ Reverse order of both digits to get Middle number Solution:- ( 8 × 6 ) - ( 3 × 4 ) = 48 - 12 = 36 ⇔ 63 ( 7 × 6 ) - ( 5 × 6 ) = 42 - 30 = 12 ⇔ 21 ( 5 × 7 ) - ( 9 × 2 ) = 35 - 18 = 17 ⇔ 71 Option (1)71 is correct option ## Problem #8 This problem consists of two figures . Each figure have three numbers. In 1st figure a number in the last row have been written using other two numbers given in same figure with the help of any mathematical operation. This mathematical operation may be addition, subtraction,multiplication, division, combinations of any two operations or any other two or three operations. Formula:- Number in Lower row = { Number in upper row - Number in middle row } × 2. Twice the difference of two numbers in upper and middle row is equal to number in lower row. (486 - 361 ) × 2 = 125 × 2 = 250 (395 - 279 ) × 2 = 116 × 2 = 232 Option (3)232 is correct option ## Problem #9 This problem consists of two figures . Each figure have three numbers. In 1st figure a number in the last row have been written using other two numbers given in same figure with the help of any mathematical operation. This mathematical operation may be addition, subtraction,multiplication, division, combinations of any two operations or any other two or three operations. Formula:- Number in Lower row = { Number in middle row - Number in upper row } × 6. Six times the difference of two numbers in middle and upper row is equal to number in lower row. ( 555 - 444 ) × 6 = 111 × 6 = 666 ( 765 - 543 ) × 6 = 222 × 6 = 1332 Option (4)1332 is correct option ## Problem #10 This problem also consists of two figures . Each figure have three numbers. In 1st figure a number in the last row have been written using other two numbers given in same figure with the help of any mathematical operation. This mathematical operation may be addition, subtraction, multiplication, division, combinations of any two operations or any other two or three operations. Formula:- Number in Lower row = Reversing the order of digits of the number of { Number in upper row - Number in middle row }. Reverse the the digits of number obtained for difference of two numbers in upper and middle row is equal to number in lower row. ( 597 - 276 ) = 291 ⇔ 192 (After reversing the digits of 291) ( 683 - 436 ) = 147 ⇔ 741 (After reversing the digits of 741) Option (3)741 is correct option In this post I discussed Ten Tricky logical reasoning questions+answers , Latest reasoning questions with answers . Comment your valuable suggestion for further improvement. Share:	2,769	10,381	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.8125	5	CC-MAIN-2022-05	longest	en	0.937527
https://www.geeksforgeeks.org/class-12-rd-sharma-solutions-chapter-1-relations-exercise-1-1-set-2/?ref=leftbar-rightbar	1,621,396,324,000,000,000	text/html	crawl-data/CC-MAIN-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00085.warc.gz	788,493,548	19,938	Related Articles Class 12 RD Sharma Solutions – Chapter 1 Relations – Exercise 1.1 \| Set 2 • Difficulty Level : Easy • Last Updated : 13 Jan, 2021 ### Question 11. Is it true every relation which is symmetric and transitive is also reflexive? Give reasons. Solution: We will verify this by taking an example. Consider a set A = {1, 2, 3} and a relation R on A such that R = { (1, 2), (2,1), (2,3), (1,3) } The relation R over the set A is symmetric and transitive. But, it is not reflexive. (1,1),(2, 2) and (3,3) ∉ R. Therefore, R is not a reflexive relation. Hence, it not true that every relation which is symmetric and transitive is also reflexive because it is possible that all pairs of type (x, x) is not present in the relation. ### Question 12. An integer m is said to be related to another integer n if m is multiple of n. Check if the relation is symmetric, reflexive and transitive. Solution: Let us define a relation such that R = {m, n : m, n ∈ Z, m = k×n} where, k ∈ N (natural number) First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A. Let m be an element of R. Then, m = k×m is true for k=1 (m, m) ∈ R. So, R is reflexive. Check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A. Let (m, n) ∈ R ⇒ m = k×n for some k ∈ N and according to transitivity, n = (1/k)×m for some k ∈ N but, 1/k ∉ N. So, R is not a symmetric relation. Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A. Let m, n, o be any elements of R then, (m, n) and (n, o) ∈ R ⇒ m = k1×n and n = k2×o for some k1, k2 ∈ N ⇒ m = (k1×k2)×o ⇒ (m, o) ∈ R. So, R is a transitive relation. ### Question 13. Show that the relation “≥” on the set R of all real number is reflexive, transitive but not symmetric. Solution: Let us define a relation R as R = { (a, b) a, b ∈ R ; a ≥ b } First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A. Let a be an element of R. ⇒ a ∈ R ⇒ a ≥ a , which is always true. ⇒ (a, a) ∈ R Hence, R is a reflexive relation. Check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A. Let (a, b) ∈ R. ⇒ a ≥ b ⇒ b ≥ a [according to transitivity] but it is not always true except when a=b. ⇒ (b, a) ∉ R Hence, R is not a symmetric relation. Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A. Let (a, b) and (b, c) ∈ R ⇒ a ≥ b and b ≥ c ⇒ a ≥ b ≥ c ⇒ a ≥ c ⇒ (a, c) ∈ R Hence, R is a transitive relation. ### Question 14. Give an example of a relation. Which is (i) Reflexive and symmetric but not transitive. (ii) Reflexive and transitive but not symmetric. (iii) Symmetric and transitive but not reflexive. (iv) Symmetric but neither reflexive nor transitive. (v) Transitive but neither reflexive nor symmetric. Solution: Reflexive Relation: A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A. Symmetric Relation: A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A. Transitive Relation: A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A. Let A be a set as, A = {1, 2, 3J (i) Let R be the relation on A such that R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3} Thus, R is reflexive and symmetric, but not transitive. (ii) Let R be the relation on A such that AR= { (1,1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 3) } Clearly, the relation R on A is reflexive and transitive, but not symmetric. (iii) Let R be the relation on A such that R = { (1, 2), (2, 1), (1, 3), (3,1), (2, 3) } We see that the relation R on A is symmetric and transitive, but not reflexive. (iv) Let R be the relation on A such that R = { (1, 2), (2, 1), (1, 3), (3, 1) } The relation R on A is symmetric, but neither reflexive nor transitive. (v) Let R be the relation on A such that R = { (1, 2), (2, 3) ,(1, 3) } The relation R on A is transitive, but neither symmetric nor reflexive. ### Question 15. Given the relation R={(1, 2), (2, 3)} on the set A={1, 2, 3}, add a minimum number of ordered pairs so that the enlarged relation is symmetric, reflexive, and transitive. Solution: We have given the relation, R = {(1, 2), (2,3)} For R to be reflexive it must have (1,1), (2, 2), (3,3). For R to be a symmetric relation, all the ordered pairs upon interchanging the elements must be present in the relation R. Therefore, R must contain (2,1 ) and (3, 2), (3,1), (1,3). And for to be a transitive relation, it must contain (1,3). Hence, the number of ordered pairs to be added to R is 7, i.e. {(1,1), (2, 2), (3,3), (1,3), (3,1), (2,1), (3, 2)}. ### Question 16. Let A={1, 2, 3} and R={(1, 2), (1, 1), (2, 3)} be a relation on A. What minimum number of ordered pairs to be added in R so that it may become a transitive relation on A. Solution: The relation R on A is given such that R = {(1, 2), (1,1), (2,3)} For the relation R to be transitive, we must have (1, 2) ∈ R, since (2, 3) ∈ R ⇒ (1,3) ∈ R Therefore, the minimum number of ordered pairs need to be added to relation R is 1, i.e. (1, 3) to make it a transitive relation on A. ### Question 17. Let A={a, b, c} and a relation R to be defined on A as followsR={(a, a), (b, c), (a, b)}. Then write minimum number of ordered pairs to be added in R to make it reflexive and transitive. Solution: We have given a set A = {a, b, c} and a relation R={(a, a), (b, c), (a, b)}. For relation to be reflexive, it should contain (b, b) and (c, c). And for relation to be transitive R should contain (a, c) since (a, b) ∈ R and (b, c) ∈ R Therefore, the minimum number of ordered pair to be added to relation R is (b, b) , (c, c) and (a, c) i.e. 3. ### Question 18. Each of the following defines a relation on N (i) x > y, x, y ∈ N (ii) x + y =10, x, y ∈ N (iii) xy is a square of integer, x, y ∈ N (iv) x + 4y =10, x, y ∈ N ### Determine which of the above relations are symmetric, reflexive, and transitive. Solution: (i) We have given the relation defined as R = {(x > y), x, y ∈ N} First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A. if (x, x) ∈ R then, x > x, which is not true. ⇒ (x, x) ∉ R So, the relation is not a reflexive relation. Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A. Let (x, y) ∈ R, then x R y ⇒ x > y and according to symmetric property, (y, x) ∈ R ⇒ y > x, but it is not true since x > y ⇒ (x, y) ∈ R but (y, x) ∉ R So, the relation is not a symmetric relation. Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A. Let (x, y) ∈ R and (y, z) ∈ R ⇒ x > y and y > z ⇒ x > z ⇒ (x, z) ∈ R So, R is a transitive relation as well. (ii) We have given the relation defined as R = { x + y =10, x, y ∈ N } Clearly, the relation will be R = { (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1) } First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A. We can see that, (1, 1) ∉ R. So, R is not a reflexive relation. Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A. By observing the above relation, we can say that ∀ (x, y) ∈ R, (y, x) ∈ R. So, R is a symmetric relation. Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A. In the relation, (1, 9) ∈ R and (9, 1) ∈ R but (1, 1) ∉ R So, R is not a transitive relation. (iii) We have given the relation as R = { xy is a square of integer, x, y ∈ N } First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A. Clearly, (x, x) ∈ R ∀ x ∈ N since, x2 is square of an integer for any x ∈ N. Hence, R is a reflexive relation. Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A. Let (x, y) ∈ R ⇒ xy is a square of an integer ⇒ yx is also a square of the same integer since, xy = yx ⇒ (y, x) ∈ R So, the relation is a symmetric relation. Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A. Let (x, y) ∈ R and (y, z) ∈ R ⇒ xy is an square of an integer and yz is an square of integer Then let xy = m2 and yz = n2 for some m, n ∈ Z ⇒ x = m2/y and z = n2/y ⇒ xz = (m2n2)/y2 , which is also a square of an integer ⇒ (x, z) ∈ R So, R is a transitive relation. (iv) We have given the relation as R = { x + 4y =10, x, y ∈ N } First let us check whether the relation is reflexive or not. A relation ‘R’ on a set ‘A’ is said to be reflexive if (x R x) ∀ x ∈ A i.e. (x, x) ∈ R ∀ x ∈ A. Clearly, the relation will be R = {(2, 2), (6, 1)} [since x, y ∈ N] (1, 1) ∉ R So, the relation R is not reflexive. Now check whether the relation is Symmetric relation or not. A relation R on set A is symmetric if (a, b)∈ R and (b, a)∈ R for all (a, b)∈ A. We can see that, (1, 6) R but (6, 1) R So, R is not a symmetric relation. Now check whether the relation is Transitive or not. A relation ‘R’ is said to be Transitive over set ‘A’ if (x, y) ∈ R and (y, z) ∈ R then (x, z) ∈ R ∀ x, y, z ∈ A. From the definition, we can see that R is a transitive relation. My Personal Notes arrow_drop_up	3,701	10,309	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.625	5	CC-MAIN-2021-21	latest	en	0.891718
https://nrich.maths.org/public/topic.php?code=31&cl=2&cldcmpid=2	1,600,693,217,000,000,000	text/html	crawl-data/CC-MAIN-2020-40/segments/1600400201699.38/warc/CC-MAIN-20200921112601-20200921142601-00609.warc.gz	552,394,440	9,931	# Resources tagged with: Addition & subtraction Filter by: Content type: Age range: Challenge level: ### There are 224 results ##### Age 5 to 11 Challenge Level: Place six toy ladybirds into the box so that there are two ladybirds in every column and every row. ### A Square of Numbers ##### Age 7 to 11 Challenge Level: Can you put the numbers 1 to 8 into the circles so that the four calculations are correct? ### Difference ##### Age 7 to 11 Challenge Level: Place the numbers 1 to 10 in the circles so that each number is the difference between the two numbers just below it. ### 1, 2, 3 Magic Square ##### Age 7 to 11 Challenge Level: Arrange three 1s, three 2s and three 3s in this square so that every row, column and diagonal adds to the same total. ### First Connect Three for Two ##### Age 7 to 11 Challenge Level: First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line. ### Painting Possibilities ##### Age 7 to 11 Challenge Level: This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . . ### Clocked ##### Age 11 to 14 Challenge Level: Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours? ### Open Squares ##### Age 7 to 11 Challenge Level: This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like? ### Sums and Differences 1 ##### Age 7 to 11 Challenge Level: This challenge focuses on finding the sum and difference of pairs of two-digit numbers. ### Neighbours ##### Age 7 to 11 Challenge Level: In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square? ### Prison Cells ##### Age 7 to 11 Challenge Level: There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it? ### Journeys in Numberland ##### Age 7 to 11 Challenge Level: Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores. ### Page Numbers ##### Age 7 to 11 Short Challenge Level: Exactly 195 digits have been used to number the pages in a book. How many pages does the book have? ##### Age 7 to 11 Challenge Level: Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it? ### Seven Square Numbers ##### Age 7 to 11 Challenge Level: Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number. ### Code Breaker ##### Age 7 to 11 Challenge Level: This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code? ### Pouring the Punch Drink ##### Age 7 to 11 Challenge Level: There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs. ### The Dice Train ##### Age 7 to 11 Challenge Level: This dice train has been made using specific rules. How many different trains can you make? ### All Seated ##### Age 7 to 11 Challenge Level: Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties? ### Dart Target ##### Age 7 to 11 Challenge Level: This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards. ### Folded Number Line ##### Age 7 to 11 Challenge Level: When I fold a 0-20 number line, I end up with 'stacks' of numbers on top of each other. These challenges involve varying the length of the number line and investigating the 'stack totals'. ### Polo Square ##### Age 7 to 11 Challenge Level: Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total. ### Zargon Glasses ##### Age 7 to 11 Challenge Level: Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families? ### Got it Article ##### Age 7 to 14 This article gives you a few ideas for understanding the Got It! game and how you might find a winning strategy. ### Number Pyramids ##### Age 11 to 14 Challenge Level: Try entering different sets of numbers in the number pyramids. How does the total at the top change? ### Countdown Fractions ##### Age 11 to 16 Challenge Level: Here is a chance to play a fractions version of the classic Countdown Game. ### Zios and Zepts ##### Age 7 to 11 Challenge Level: On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there? ### Six Is the Sum ##### Age 7 to 11 Challenge Level: What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros? ### Hubble, Bubble ##### Age 7 to 11 Challenge Level: Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs? ### On Target ##### Age 7 to 11 Challenge Level: You have 5 darts and your target score is 44. How many different ways could you score 44? ### Dodecamagic ##### Age 7 to 11 Challenge Level: Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers? ### Bean Bags for Bernard's Bag ##### Age 7 to 11 Challenge Level: How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this? ##### Age 7 to 11 Challenge Level: Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square. ### Sums and Differences 2 ##### Age 7 to 11 Challenge Level: Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens? ### The Twelve Pointed Star Game ##### Age 7 to 11 Challenge Level: Have a go at this game which involves throwing two dice and adding their totals. Where should you place your counters to be more likely to win? ### Sticky Dice ##### Age 7 to 11 Challenge Level: Throughout these challenges, the touching faces of any adjacent dice must have the same number. Can you find a way of making the total on the top come to each number from 11 to 18 inclusive? ### Two Egg Timers ##### Age 7 to 11 Challenge Level: You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how? ### Train Carriages ##### Age 5 to 11 Challenge Level: Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done? ### First Connect Three ##### Age 7 to 14 Challenge Level: Add or subtract the two numbers on the spinners and try to complete a row of three. Are there some numbers that are good to aim for? ### Dice and Spinner Numbers ##### Age 7 to 11 Challenge Level: If you had any number of ordinary dice, what are the possible ways of making their totals 6? What would the product of the dice be each time? ### Build it up More ##### Age 7 to 11 Challenge Level: This task follows on from Build it Up and takes the ideas into three dimensions! ### Machines ##### Age 7 to 11 Challenge Level: What is happening at each box in these machines? ### Calendar Patterns ##### Age 7 to 11 Challenge Level: In this section from a calendar, put a square box around the 1st, 2nd, 8th and 9th. Add all the pairs of numbers. What do you notice about the answers? ### Alphabet Blocks ##### Age 5 to 11 Challenge Level: These alphabet bricks are painted in a special way. A is on one brick, B on two bricks, and so on. How many bricks will be painted by the time they have got to other letters of the alphabet? ### The Puzzling Sweet Shop ##### Age 5 to 11 Challenge Level: There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money? ### Place Value as a Building Block for Developing Fluency in the Calculation Process ##### Age 5 to 11 This article for primary teachers encourages exploration of two fundamental ideas, exchange and 'unitising', which will help children become more fluent when calculating. ### A Rod and a Pole ##### Age 7 to 11 Challenge Level: A lady has a steel rod and a wooden pole and she knows the length of each. How can she measure out an 8 unit piece of pole? ### Clock Face ##### Age 7 to 11 Challenge Level: Where can you draw a line on a clock face so that the numbers on both sides have the same total? ### The Pied Piper of Hamelin ##### Age 7 to 11 Challenge Level: This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether! ### One Wasn't Square ##### Age 7 to 11 Challenge Level: Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.	2,424	10,120	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2020-40	longest	en	0.888479
http://fakeroot.net/confidence-interval/confidence-interval-standard-error-estimate.php	1,529,459,035,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267863407.58/warc/CC-MAIN-20180620011502-20180620031502-00035.warc.gz	111,410,438	6,057	Home > Confidence Interval > Confidence Interval Standard Error Estimate # Confidence Interval Standard Error Estimate ## Contents Standard error of mean versus standard deviation In scientific and technical literature, experimental data are often summarized either using the mean and standard deviation or the mean with the standard error. Suppose the following five numbers were sampled from a normal distribution with a standard deviation of 2.5: 2, 3, 5, 6, and 9. The mean of these 20,000 samples from the age at first marriage population is 23.44, and the standard deviation of the 20,000 sample means is 1.18. Given a sample of disease free subjects, an alternative method of defining a normal range would be simply to define points that exclude 2.5% of subjects at the top end and http://fakeroot.net/confidence-interval/calculate-confidence-interval-standard-error-estimate.php CoefficientCovariance, a property of the fitted model, is a p-by-p covariance matrix of regression coefficient estimates. ISBN 0-7167-1254-7 , p 53 ^ Barde, M. (2012). "What to use to express the variability of data: Standard deviation or standard error of mean?". Of course, T / n {\displaystyle T/n} is the sample mean x ¯ {\displaystyle {\bar {x}}} . HomeAboutThe TeamThe AuthorsContact UsExternal LinksTerms and ConditionsWebsite DisclaimerPublic Health TextbookResearch Methods1a - Epidemiology1b - Statistical Methods1c - Health Care Evaluation and Health Needs Assessment1d - Qualitative MethodsDisease Causation and Diagnostic2a - why not find out more ## Confidence Interval Standard Error Of The Mean If you had wanted to compute the 99% confidence interval, you would have set the shaded area to 0.99 and the result would have been 2.58. If σ is not known, the standard error is estimated using the formula s x ¯ = s n {\displaystyle {\text{s}}_{\bar {x}}\ ={\frac {s}{\sqrt {n}}}} where s is the sample However, the sample standard deviation, s, is an estimate of σ. This section considers how precise these estimates may be. Compare the true standard error of the mean to the standard error estimated using this sample. Lower limit = 5 - (2.776)(1.225) = 1.60 Upper limit = 5 + (2.776)(1.225) = 8.40 More generally, the formula for the 95% confidence interval on the mean is: Lower limit JSTOR2340569. (Equation 1) ^ James R. Confidence Interval Margin Of Error When the sample size is smaller (say n < 30), then s will be fairly different from $$\sigma$$ for some samples - and that means that we we need a bigger The diagonal elements are the variances of the individual coefficients.How ToAfter obtaining a fitted model, say, mdl, using fitlm or stepwiselm, you can display the coefficient covariances using mdl.CoefficientCovarianceCompute Coefficient Covariance The ages in that sample were 23, 27, 28, 29, 31, 31, 32, 33, 34, 38, 40, 40, 48, 53, 54, and 55. Standard errors provide simple measures of uncertainty in a value and are often used because: If the standard error of several individual quantities is known then the standard error of some this contact form We call the resulting estimate the Standard Error of the Mean (SEM).Standard Error of the Mean (SEM) = estimated standard deviation of the sample average =\[\frac{\text{standard deviation of the sample}}{\sqrt{n}} = If we now divide the standard deviation by the square root of the number of observations in the sample we have an estimate of the standard error of the mean. Confidence Interval Sampling Error While all tests of statistical significance produce P values, different tests use different mathematical approaches to obtain a P value. Clearly, if you already knew the population mean, there would be no need for a confidence interval. The values of t to be used in a confidence interval can be looked up in a table of the t distribution. ## Confidence Interval Standard Error Of Measurement Swinscow TDV, and Campbell MJ. navigate to these guys When we calculate the standard deviation of a sample, we are using it as an estimate of the variability of the population from which the sample was drawn. Confidence Interval Standard Error Of The Mean We will discuss confidence intervals in more detail in a subsequent Statistics Note. Confidence Interval Standard Error Or Standard Deviation We will finish with an analysis of the Stroop Data. Because the 9,732 runners are the entire population, 33.88 years is the population mean, μ {\displaystyle \mu } , and 9.27 years is the population standard deviation, σ. http://fakeroot.net/confidence-interval/confidence-intervals-standard-error-estimate.php Then we will show how sample data can be used to construct a confidence interval. Assume that the weights of 10-year-old children are normally distributed with a mean of 90 and a standard deviation of 36. In general, you compute the 95% confidence interval for the mean with the following formula: Lower limit = M - Z.95σM Upper limit = M + Z.95σM where Z.95 is the Confidence Interval Standard Error Calculator 1. Resource text Standard error of the mean A series of samples drawn from one population will not be identical. 2. It turns out that the sample mean was $$\bar x$$ = \$2430 with a sample standard deviation of s = \$2300. 3. Note: The Student's probability distribution is a good approximation of the Gaussian when the sample size is over 100. 4. Using a sample to estimate the standard error In the examples so far, the population standard deviation σ was assumed to be known. 5. The standard deviation of the age for the 16 runners is 10.23. 6. Nagele P. 7. Chapter 4. 8. See Alsoanova \| coefCI \| coefTest \| fitlm \| LinearModel \| plotDiagnostics \| stepwiselm Related ExamplesExamine Quality and Adjust the Fitted ModelInterpret Linear Regression Results × MATLAB Command You clicked a 9. Note that the standard deviation of a sampling distribution is its standard error. A critical evaluation of four anaesthesia journals. Since 95% of the distribution is within 23.52 of 90, the probability that the mean from any given sample will be within 23.52 of 90 is 0.95. This observation is greater than 3.89 and so falls in the 5% of observations beyond the 95% probability limits. news Note that the equatorial radius of the planet is a fixed number (Jupiter is not changing in size). Would it be appropriate to use the method above to find a 99% confidence interval for the average credit card debt for all recent Penn State graduates?Solution: No, with n = Confidence Interval Variance Br J Anaesthesiol 2003;90: 514-6. [PubMed]2. How many standard deviations does this represent? ## Figure 1 shows that 95% of the means are no more than 23.52 units (1.96 standard deviations) from the mean of 90. It is rare that the true population standard deviation is known. The standard error of the mean is 1.090. For illustration, the graph below shows the distribution of the sample means for 20,000 samples, where each sample is of size n=16. Confidence Interval T Test Faculty login (PSU Access Account) Lessons Lesson 2: Statistics: Benefits, Risks, and Measurements Lesson 3: Characteristics of Good Sample Surveys and Comparative Studies Lesson 4: Getting the Big Picture and Summaries If σ is known, the standard error is calculated using the formula σ x ¯ = σ n {\displaystyle \sigma _{\bar {x}}\ ={\frac {\sigma }{\sqrt {n}}}} where σ is the When the sample size is large, s will be a good estimate of $$\sigma$$ and you can use multiplier numbers from the normal curve. The standard deviation of the age was 3.56 years. http://fakeroot.net/confidence-interval/confidence-interval-point-estimate-margin-of-error.php For the purpose of this example, the 9,732 runners who completed the 2012 run are the entire population of interest. The distribution of the mean age in all possible samples is called the sampling distribution of the mean. The sample standard deviation s = 10.23 is greater than the true population standard deviation σ = 9.27 years. This is usually the case even with finite populations, because most of the time, people are primarily interested in managing the processes that created the existing finite population; this is called In an example above, n=16 runners were selected at random from the 9,732 runners. These means generally follow a normal distribution, and they often do so even if the observations from which they were obtained do not. Table 1. If we knew the population variance, we could use the following formula: Instead we compute an estimate of the standard error (sM): = 1.225 The next step is to find the Anything outside the range is regarded as abnormal. The 95% confidence interval for the average effect of the drug is that it lowers cholesterol by 18 to 22 units. These assumptions may be approximately met when the population from which samples are taken is normally distributed, or when the sample size is sufficiently large to rely on the Central Limit With this standard error we can get 95% confidence intervals on the two percentages: These confidence intervals exclude 50%. A standard error may then be calculated as SE = intervention effect estimate / Z. The standard error of the mean of one sample is an estimate of the standard deviation that would be obtained from the means of a large number of samples drawn from Lesson 11: Hypothesis Testing Lesson 12: Significance Testing Caveats & Ethics of Experiments Reviewing for Lessons 10 to 12 Resources References Help and Support Links!	2,141	9,460	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.578125	4	CC-MAIN-2018-26	latest	en	0.804112
http://nrich.maths.org/public/leg.php?code=31&cl=1&cldcmpid=177	1,386,628,693,000,000,000	text/html	crawl-data/CC-MAIN-2013-48/segments/1386164000853/warc/CC-MAIN-20131204133320-00093-ip-10-33-133-15.ec2.internal.warc.gz	132,689,546	10,121	# Search by Topic #### Resources tagged with Addition & subtraction similar to Three Way Mix Up: Filter by: Content type: Stage: Challenge level: ### The Pied Piper of Hamelin ##### Stage: 2 Challenge Level: This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether! ### A-magical Number Maze ##### Stage: 2 Challenge Level: This magic square has operations written in it, to make it into a maze. 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How many spiders? ### Make 100 ##### Stage: 2 Challenge Level: Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.	2,303	9,521	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.703125	4	CC-MAIN-2013-48	longest	en	0.90894
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April 7, 2015 physics Disregard the "6.25" above it should also just be 11.69 April 7, 2015 physics Pi=Pf MiVi=(m+M)Vf so (0.0131)(261)=( 0.0131 +M)Vf Vf= 3.4191/( 0.0131 +M) PE(spring)=KE(Block+Bullet) 1/2Kx^2=1/2(0.0131 +M)V^2 Now Replace V with the solution from above: 1/2(205)(.35)^2 = 1/2(0.0131 +M)[3.4191/( 0.0131 +M)]^2 12.56J= 1/2(0.0131 +M)[3.4191/( 0.0131 +M)]^2 So... April 7, 2015 math The record for the fastest speed on a bicycle is 133 kilometers per hour. How long would it take a person traveling at this speed to travel 365.75 kilometers? hours and minutes January 7, 2015 Literature What are conventions? October 30, 2014 English is it correct October 29, 2014 English my final answer b c b b d October 29, 2014 English i think it is b. October 29, 2014 English 5 is a i think, can i get help with 7 i think it is b, and six is d October 29, 2014 Health So what is the others October 29, 2014 Health Is it number 1 October 29, 2014 Health b c b I would love to know which ones are right and help with the ones that are wrong. October 29, 2014 Health The number of calories consumed must equal the number of calories________. a.)needed b.)burned c.)absorbed d.)wanted 2.)It is important to recognize which of the following when watching your weight so that you know what triggers your eating habits. a.) eating patterns b.) ... October 29, 2014 health The spinal cord serves as a conduction pathway for impulses going to and from the ________? A. Lungs B. Liver C. Brain D. Heart i think its c.? September 4, 2014 Science 2.38g of black copper oxide is completely reduced by hydrogen to give copper and water. What are the masses of copper and water June 7, 2014 Physics Assuming that the average bullet is 0.01kg Pi=Pf MiVi=(m+M)Vf so (0.01)(250)=(0.01+M)Vf Vf= 2.5/(0.01+M) PE(spring)=KE(Block+Bullet) 1/2Kx^2=1/2(0.01+M)V^2 Now Replace V with the solution from above: 1/2(200)(.3)^2 = 1/2(0.01+M)[2.5/(0.01+M)]^2 9J=1/2(0.01+M)[2.5/(0.01+M)]^2 ... March 21, 2014 physics The Vector A is 3cm long and points to the right. The vector B is 6cm long and makes an angle of 130 degrees with the horizontal (take 0 degrees to be to the right). The vector C= A+B. What is the lenght and direction of C? September 24, 2013 math name a figure that has an area of 47 centimeters square February 21, 2013 Chemistry Biphenyl, C12H10, is a nonvolatile, nonionizing solute that is soluble in benzene, C6H6. At 25 °C, the vapor pressure of pure benzene is 100.84 torr. What is the vapor pressure of a solution made from dissolving 19.6 g of biphenyl in 25.1 g of benzene? January 29, 2013 chemistry how many grams of carbon dioxide can be produced by burning 7.04moles of propane September 21, 2012 ENGLISH 3) Is wrong, its C or the third row. Because the pattern repeats September 4, 2012 Algebra 2 f(x)=10x-10 January 17, 2012 Algebra 2 January 17, 2012 Calculus Thank you very much! I wasn't plugging in the absolute value signs! Thanks for the explanation too. =) September 20, 2011 Calculus Evaluate the indefinite integral of InT(-x^3+9x^2-3x+2)/(x^4-2x^3) I got ((1-2x)/(2x^2))+3ln(2-x)-4ln(x)+c but apparently that's not the answer... and I don't know why. September 20, 2011 Calculus Evaluate the indefinite integral of InT(-x^3+9x^2-3x+2)/(x^4-2x^3) I got ((1-2x)/(2x^2))+3ln(2-x)-4ln(x)+c but apparently that's not the answer... and I don't know why. September 20, 2011 Calculus Evaluate the indefinite integral of InT(-x^3+9x^2-3x+2)/(x^4-2x^3) I got ((1-2x)/(2x^2))+3ln(2-x)-4ln(x) but apparently that's not the answer... September 20, 2011 chemistry During lab, you are provided with a stock antibody solution (Ab) that has a concentration of 400 μg/μl. For lab, it is necessary make the following dilutions (Fill in the blanks): 5μL of Stock Ab + 195μL of buffer to make a 1:____ dilution at _______μg... September 13, 2011 Chemstry Mercury concentration in a polluted lake is 0.0125 micrograms/L. What is the total mass in kilograms of mercury in the lake if the lake is approximately circular witha diameter of 10 miles and an average depth of 50 ft.? August 31, 2011 Maths A swing 5 metres long swings through a vertical angle of 50 degrees. Through what vertical distance does the seat rise in going from its lowest position to its highest? April 7, 2011 Literature julius Caesar Thank you!!!! March 30, 2011 Literature julius Caesar What are 3 purposes for shakespeare showing us scene 3. Consider antonys speech, what is likely to occur in the rest of the play, and the role of language in politics. any good ideas? thanks <3 Gabby! March 30, 2011 math im pretty sure! March 30, 2011 math -.001 < -.03 March 30, 2011 French If you're writing a sentence in French with two verbs, the first being l'imparfait, would the second one be in the infinitive? For example "Le peuple prehistoriques était intelligente parce que ils pouvaient ériger les monuments sans des machines&... February 21, 2011 Science 273 degrees farenheit is what in celsius 273`F--> ??`C February 20, 2011 francais freshmen Translate into french please: Return to heat and cook until set and edge is dry,(takes approximately 30 seconds). February 20, 2011 french merci beacoup,merci beacoup!!!!!!!!!!!!!!!!!!!!!! February 16, 2011 french for french project: a) does 1/6 cup=2 tablespoons+2 teaspoons b) say that in french s'il vous plait!:) February 16, 2011 English Directions: Place all prepositional phrases in parentheses. Underline the subject or subjects once and DOUBLE underline or bold face the verb or verbs 1. Sandy’s revealing suit attracted stares from everyone at the swimming pool. 2. Join us at Subway for lunch at 1:00. 3... January 28, 2011 french Please help me by translating the following french terms into english: 1) le litchi 2)le bonbon 3)le ragout 4)l'epice 5)les accras 6)les matoutous 7)le blaff 8) le feroce 9)le lambi January 26, 2011 English 6) scan-- When I'm at the check-out counter, I put my items* on the conveyor belt for the clerk to scan. * maybe a better word. January 26, 2011 math A customer uses a \$20 bill at a self check-out register to pay for products totaling \$14.50. The customer is concerned they did not receive the proper change. How much money should they have received back from the autom December 26, 2010 ice cube mass??? an ice cube at 0 was dropped into 30.0 g of water in a cup at 45.0 C at the instant that all of the ice was melted, the tempature of the water in the cup was 19.5 C. what was the mass of the icecube? December 17, 2010 Using the following equation Ca(s) + 2C(s) --> CaC2(s) ^ H = -62.8 kJ CO2(g) --> C(s) + O2(g) ^ H = 393.5 kJ CaCO3(s) + CO2 (g) --> CaC2(s) + 2 1/2O2(g) ^ H = 1538 kJ Determine the heat of the reaction (in kJ) for: Ca(s) + C(s) + 3/2O2(g) --> CaCO3(s) December 17, 2010 Simplify alegebra 2 -5ã2(4ã2 - 3ã3) December 15, 2010 estimate the freezing point of an aqueous of 10.0 g of glucose dissolved in 500.0 g of water December 14, 2010 What is the freezing point depression of an aquous solution of 10.0 g of glucose in 50.0 g H2O? December 14, 2010 English in the book to kill a mockingbird what chapters (15-25) can i find quotes that support what boo radley, mayella ewell, and tom robinson had in common? November 26, 2010 the compound adrenaline contains 56.79% c,6.56% h,28.27% and 8.28% n by the mass what is the empiricol formula October 27, 2010 science how do you find the electronegativity of Ethanol (C2H5OH)? October 6, 2010 math need help making a Data Bar Graph using Maximum:18 Mode:7 Range:13 Median:12 October 5, 2010 Re: social studys Im the girl qho just asked about my 2-3 page essay. Its based on this "explain and describe how geographers use thier tools to interpret geographic questions. Remember, i'm in 7th grade. (ms.sue) my teacher said that its based on out notes and its bacically about ... September 20, 2010 Social studys (ms.sue) my teacher said that its based on out notes and its bacically about technology tools for geographers -globe -map -cartogropher -surveyor -remoute sensing -landsat -GPS -GIS also about careers in geography like what different kinds of geographers thier is. Thanks again!! September 20, 2010 Social studys Hi i need help on a 2-3 page essay. The question its based on is" explain and describe how geographers use thier tools to interpret grographic questions" I'm in 7th grade THANKS!! :) September 20, 2010 Continents I need to know if these are all the continents and the correct spelling for a quiz tommrow. --Africa Aisa Auntartica Austraila North America South America Eroupe-- THANKS!!! :D September 9, 2010 mental health February 5, 2010 Math Words or letters? Because I'm pretty sure that if you fold HI over on itself it won't be symmetrical vertically. And could you tell me what you mean when you say "printed vertically" and "printed horizontally"? February 5, 2010 Multiply the each fraction by the denominator of the other so that they will equal each other. February 5, 2010 Science-Pendulum It seems to me that you've answered your own question, but you could add that to test one certain part to see if it affected the swing rate you had to keep all other parts constant. You can't just change everything at once and expect to be able to see what affects it ... February 5, 2010 math When you move your y value to the opposite side, it will become negative. Then you'll need to divide each side by negative 1. Other than that, you are correct. February 5, 2010 suppose you wanted to place 1,000,000 pennies in large coin banks.if you put 50,000 pennies in each bank,in which bank would you find the 678,230th penny December 14, 2009 art I have homework in art where, every week, I must look up an art definition, describe it, and draw a picture with it. I was wondering what the following words mean and/or where I could find pictures: Sfumato, Composite, Anthropomorphism, Tactile, Garish, Fickle, Gaudy, Zealot, ... November 21, 2009 Physics Oh, wow. Talk about a dumb mistake. Thank you bobpursley!! November 15, 2009 Physics On Williamston Rd., Mason is driving 24.2 m/s. After spotting a large deer in the road ahead, Mason brakes the car to a complete stop. If Mason's car takes 35.9 meters to stop, what is the acceleration of the car? -- Here's what I'm getting, but my program that the... November 15, 2009 HELP MATH If you tell me what you've figured out on your own, I'll be happy to walk you through the rest of it! November 15, 2009 Microeconomics i need help with number 57 April 5, 2009 I don't know how to do these types of problems whatsoever. 1) It takes a small jet plane 4 hours less time than it takes a propeller-driven plane to travel from Glen Rock to Oakville. The jet plane averages 637 km/h while the propeller plane averages 273 km/h. How far is ... March 7, 2009 January 7, 2009 science thanks! December 7, 2008 science what could be a possible source of sulfates found in a supply of drinking water? December 7, 2008 math What is the pecent of change fo this problem... from 8 to 300 December 1, 2008 math So when 48.5 is rounded it equals 48.6? December 1, 2008 part 2 thank you Prepare a statement of cash flows, using the indirect method of presenting cash flows from operating activities a.Equipment and land were acquired for cash b.There were no disposal of equipment during the year c.The investments were sold for 45,000 cash d.The ... November 19, 2008 part 1 June 30, 2008/2007 Assets Cash- 34,700/23,500 accounts receivable- 101,600/92,300 inventory- 146,300/142,100 Investment-0/50,000 Land-145,000/0 equipment- 215,000/175,500 accumulated depreciation- (53,400)/ (41,300) 2008= 594,000 /2007=442,100 Liabilities and ... November 19, 2008 accounting 2 a.Equipment and land were acquired for cash b.There were no disposal of equipment during the year c.The investments were sold for 45,000 cash d.The common stock was issued for cash e.There was a 65,900 credit to retained earning for net income f.There was a 50,000 debit to ... November 19, 2008 accounting 2 June 30, 2008/2007 Assets Cash- 34,700/23,500 accounts receivable- 101,600/92,300 inventory- 146,300/142,100 Investment-0/50,000 Land-145,000/0 equipment- 215,000/175,500 accumulated depreciation- (53,400)/(41,300) 2008= 594,000 /2007=442,100 liabilities and stockholders ... November 19, 2008 microeconomic a. Calculate the total cost, the average variable cost, the aver-age total cost, and the marginal cost for each quantity of output. b. What is the break-even price? What is the shut-down price? c. Suppose that the price at which Kate can sell catered meals is \$21 per meal. In ... November 18, 2008 mircoeconomic Kate’s Katering provides catered meals, and the catered meals industry is perfectly competitive. Kate’s machinery costs \$100 per day and is the only fixed input. Her variable cost is com-prised of the wages paid to the cooks and the food ingredients. The variable ... November 18, 2008 medical coding and billing I`m doing this same eact researcgh project for Penn Foster and i`m haveing a really ough time with it. So if you could help meit would be really grateful! Thanks November 15, 2008 math riddle ALGEBRA WITH PIZZAZZ why did king kong eat a truck ? April 6, 2008 Science What volume of helium would be in a balloon the size of a soft drink bottle? March 13, 2008 MATH HELP!!!!!! i have a huge test on circles radius , circumference , diameter. Also Equivalent Fractions and Comparing and ordering fractions. January 16, 2008 science info on describing, Measuring, and graphing motion November 14, 2007 biology Explain how respiration and the circulatory system work together when an individual is exercising. Does tissue need oxygen at a higher rate? Does tissue need to void CO2 at a higher rate? Does the heart pulse speed up? Does the breathing rate increase? June 7, 2007 Government what are economic powers that are in the Legislative Branch? March 23, 2007 Government what are goals for American Government as Stated in the Preamble? March 23, 2007 health The nine essentialamino acids are A)NOT FOUND IN FOODS B)NOT MADE BY THE BODY C)ALL PRESENT IN INCOMPLETE PROTEIN D)MADE BY THE BODY not made by the body, so they must come from foods. January 5, 2007 College Algebra 1 help! 2x/1+x-1/1-x=2 Get a common denominator. As I see your problem (you did not use parenthesis), the common denominator will be (1+x)(1-x) Thank you! October 26, 2006 math A computer store sold a total of 300 items last month. The store sold six times as many hard drives as they did CD-ROM drives, and half as many floppy drives as hard drives. Based on this information, how many of each items were sold? August 19, 2006 1. Pages: 2. 1	4,657	15,445	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2016-44	longest	en	0.948332
https://cs.stackexchange.com/questions/111256/separating-numbers-with-a-minimal-difference	1,725,797,772,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651002.87/warc/CC-MAIN-20240908115103-20240908145103-00560.warc.gz	171,785,336	45,128	# Separating numbers with a minimal difference Given is a positive integer integer $$n$$, and integers $$a_1,b_1,\dots,a_n,b_n$$ with $$a_i\leq b_i$$ for each $$i$$. What is the complexity of deciding whether there exist integers $$c_1,\dots,c_n$$ such that $$a_i\leq c_i\leq b_i$$ for all $$i$$ and $$\|c_i-c_j\|\geq 2$$ for all $$i,j$$? What can be observed is that a greedy algorithm, wherein we assume that $$a_1\leq\dots\leq a_n$$ and choose $$c_i$$ according to this order, does not necessarily work. For example, we may have $$a_1=1, b_1=4, a_2=b_2=2$$. Putting $$c_1=1$$ does not work (it leaves no room for $$c_2$$), but what works is putting $$c_1=4$$ and $$c_2=2$$. Is the problem perhaps NP-hard? • Have you considered dynamic programming? Commented Jun 28, 2019 at 12:55 • Yes. The problem is that we don't know the order of the $c_i$'s, so we could not store, for example, the "best" solution using $c_1,\dots,c_k$ on the left-hand side. – pi66 Commented Jun 28, 2019 at 13:15 • Could you add a reference to the original problem? Commented Jun 28, 2019 at 20:52 • I don’t have a reference. I’d be interested if anyone knows one. – pi66 Commented Jun 28, 2019 at 21:45 As shown in the previous answer, this problem can be modeled as a scheduling problem with release and due dates. However, Schrage's heuristic works only for the case $$p_j=1$$ (all processing times are $$1$$) or when release and due dates agree (i.e. there is an order such that $$r_1\leq\cdots\leq r_n$$ and $$d_1\leq\cdots\leq d_n$$). For the case $$p_j=p$$ polynomial algorithms have been found by Simons (1978), Carlier (1981), and Garey et al. (1981), running in time $$O(n^2\log n)$$, $$O(n^2\log n)$$, and $$O(n\log n)$$, respectively. The algorithm of Garey et al. schedules unit-time tasks, but allows arbitrary release and due dates. The above problem can be reduced to this setting by dividing all dates and processing times by $$p$$. The main idea of the algorithm is to find a set of forbidden regions where no task is allowed to start. They show that Schrage's heuristic when respecting forbidden regions finds a feasible schedule of minimal makespan. When constructing the forbidden regions their algorithm may also detect infeasibility. To see how the algorithm works, first look at a simpler problem: schedule $$n$$ unit-time tasks between a release date $$r$$ and a deadline $$d$$ respecting forbidden regions $$F=\bigcup_i (a_i,b_i)$$, where each region $$(a_i,b_i)$$ is an open interval. (Note here we're not concerned with release and due dates of individual tasks.) This problem can be solved by Backscheduling: define a sentinel starting time $$s_{n+1}=d$$, and for $$i=n,n-1,\ldots,1$$ set starting time $$s_i$$ to the latest time not later than $$s_{i+1}-1$$ that is not forbidden. Write $$B(r,d,n,F)$$ for an application of Backscheduling for the above inputs, and define the return value as $$s_1$$, the latest possible starting date for scheduling the $$n$$ tasks. Now, if $$B(r,d,n,F) there is no feasible schedule for the $$n$$ tasks. Further, if $$r\leq B(r,d,n,F) no other task can start in $$(B(r,d,n,F)-1,r)$$ since otherwise there is again no feasible schedule for the $$n$$ tasks, and thus $$(B(r,d,n,F)-1,r)$$ can be declared a forbidden region. It turns out that this logic is sufficient to find all forbidden regions that are required to make Schrage's heuristic work. Assume that tasks are ordered such that $$r_1\leq r_2\leq\cdots\leq r_n$$ and write $$n(r,d)$$ for the number of tasks which are released and due in the closed interval $$[r,d]$$. 1. set $$F=\emptyset$$ 2. for tasks $$i=n,n-1,\ldots,1$$: 3. $$c=\min \{ B(r_i,d_j,n(r_i,d_j),F) \mid \forall j: d_j\geq d_i\}$$ 4. if $$c: return "no feasible schedule" 5. else if $$r_i\leq c: set $$F=F\cup\{(c-1,r_i)\}$$ A straightforward implementation as above would take time $$O(n^4)$$. Garey et al. show (besides correctness) that by updating the times obtained by Backscheduling, joining overlapping forbidden regions, and making "forbidden" queries $$O(1)$$ time can be brought down to $$O(n^2)$$, and by using even better datastructures to $$O(n\log n)$$. • Looks promising, but the first and last links go to paywalls, and the second leads to a broken certificate and "500 Internal Server Error" page for me. Could you summarise one of the algorithms, or give a publically accessible link? Thanks! Commented Jun 30, 2019 at 13:00 • Sorry, these articles seems to be hard to come by. Carlier's article is on Research Gate but is in French. I'll see if I can provide a brief summary. Commented Jun 30, 2019 at 13:13 Your problem is known as non-preemptive single-machine scheduling with release times and deadlines, with tasks of identical length, and can be solved efficiently using Schrage's greedy heuristic. Let us first describe the problem more formally. We are given a sequence of time intervals $$[r_i,d_i]$$ and a sequence of job lengths $$p_i$$. We want to schedule each job within its interval, so that no two jobs intersect. In our case $$r_i = a_i$$, $$d_i = b_i + 2$$, and $$p_i = 2$$. The starting time of each job $$c_i$$ thus satisfies $$a_i \leq c_i \leq b_i$$, and two jobs conflict if $$(c_i,c_i+2)$$ intersects $$(c_j,c_j+2)$$, which is the same as $$\|c_i - c_j\| < 2$$. (Without loss of generality, all jobs are scheduled at integer times.) Schrage's heuristic is a common heuristic which is optimal in some cases, though not necessarily this one. However, other algorithms exist in the literature which do solve this problem efficiently. • Schrage's heuristic is only optimal for $p_j=1$ not for $p_j=p$ (take $r=(0\:1)$, $d=(5\:3)$, $p=(2\:2)$). Polynomial algorithms for $p_j=p$ have been given by Simons (1978), Carlier (1981), and Garey et al. (1981), the latter in $O(n\log n)$. Commented Jun 29, 2019 at 1:44 • @MarcusRitt I agree that the heuristic doesn't work. Could you give the links (or at least the titles) for your references? – pi66 Commented Jun 29, 2019 at 8:56 • I don’t see how the value of $p$ enters the picture. You can divide everything by $p$ to get unit length jobs. Commented Jun 29, 2019 at 15:12 • @YuvalFilmus Yes, but then release and due dates may not be integer any more which makes the problem harder. Actually Garey et al. (1981) solve the unit-time problem with non-integer dates, and their algorithm can be applied after dividing everything by $p$. Commented Jun 29, 2019 at 15:37 Let A be the smallest of the $$a_i$$. Then there are two choices for the first $$c_i$$ to pick that cannot obviously be improved: One, find the i such that $$a_i = A$$ and $$b_i$$ is as small as possible, then let $$c_i = a_i$$. Two, pick j such that $$a_j = A+1$$, $$b_j, and $$b_j$$ as small as possible, then let $$c_j=b_j$$ (this may not be possible). Then remove that item from the list of intervals, update all $$a_i$$ to be greater by two than the $$c_i$$ that was picked, and check that no $$b_i$$ is too small. Depth first search and backtracking will find a solution. There is a good chance that this is fast, since the first choice is not bad as a heuristic. If there are exactly k intervals with $$a_i < A$$ for some A, and we find k values $$c_i <= A-2$$ then these $$c_i$$ are optimal and backtracking for these k items cannot help and will never be needed. On the other hand, if there are too many values $$b_i$$ too close together, that can be used to prove there is no solution. Finally, the problem can be approached from the smallest $$a_i$$, or equally well from the largest $$b_i$$.	2,171	7,547	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 84, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2024-38	latest	en	0.88271
https://gowers.wordpress.com/2011/10/13/domains-codomains-ranges-images-preimages-inverse-images/	1,686,103,318,000,000,000	text/html	crawl-data/CC-MAIN-2023-23/segments/1685224653501.53/warc/CC-MAIN-20230607010703-20230607040703-00126.warc.gz	325,433,234	62,165	## Domains, codomains, ranges, images, preimages, inverse images If I were writing a textbook, I would have discussed the basics of functions before talking about injections and surjections, but this is not a textbook — it is a series of blog posts that provide a kind of commentary on some of the lecture courses. However, now that I have got on to the subject of functions, it probably makes sense to discuss them a bit more, especially as I hear that they have made an appearance in Numbers and Sets. Let me start with the most basic question of all: what is a function? This is one of the first examples (of many, unfortunately) of a concept that you were probably reasonably happy with until your lecturer explained it to you. This is absolutely not a criticism of your lecturer (who is excellent, by the way). It’s more like a criticism of an entire mathematical tradition that goes back to the days when the foundations were laid for our subject in terms of set theory. Let’s have some examples of functions. The ones you are likely to have come across are ones that take real numbers to real numbers: things like $f(x)=x^2$, $g(x)=e^x$ or $h(x)=-2\sin(x+\pi/3)$. If someone asked, “Yes, but what are $f,g$ and $h$?” then an appropriate response might be, “Well, $f(x)$ is the result of doing something to $x$ that turns it into another real number.” Notice that in that last sentence I slightly avoided the issue. I didn’t say what $f$ itself was — I just said what $f(x)$ was. (It’s another real number.) Notice also that for a specific function we don’t feel quite as tempted to ask what it really is: we don’t say, “What is squared?” That’s because “squared” isn’t a noun, so it feels wrong to imagine that it must be a thing, just as you would expect strange looks if you went around asking, “What is and?” (That’s not to say that an answer isn’t possible: one could say that AND is a logical connective, and a logical connective is something that joins two statements to form a new statement, and it could, if you wanted, be regarded as a function from the set of all pairs of statements to the set of all statements, etc. etc.) What really matters about a function is not so much its essence as the following fact. • If $f$ is a function from $A$ to $B$ and $x$ is an element of $A$, then $f(x)$ is an element of $B$. • Given a function $f:A\to B$, we can define a natural notion of the graph of that function. It is the set of all points $(x,f(x))$ such that $x\in A$. To put it another way, it is the set of all points $(x,y)\in A\times B$ such that $y=f(x)$. This set has the property (discussed in the previous post) that for every $x\in A$ there is exactly one $y\in B$ such that $(x,y)$ belongs to the graph of $f$. And now the officially correct thing to do is to turn everything on its head and make the following definition. • Let $A$ and $B$ be sets. A function from $A$ to $B$ is a subset $f$ of $A\times B$ such that for every $x\in A$ there is exactly one $y\in B$ such that $(x,y)\in f$. • Then one goes on to say that it is traditional to write $y=f(x)$ instead of $(x,y)\in f$. Equivalently, we define $f(x)$ to be the unique $y$ such that $(x,y)\in f$. Why does anyone bother with this strange definition? One reason is that we sometimes want to talk about the set of all functions from $A$ to $B$, or, even more commonly, the set of all functions from $A$ to $B$ that satisfy certain conditions. It’s one thing to be able to recognise a function when you see one, but how do you say what counts as a function? We somehow want to capture the idea that any way of associating with each $x\in A$ some $y\in B$ counts as a function, even if that “way of associating” isn’t given by a rule of any kind. It may seem as though the “take any old subset of $A\times B$ as long as for each $x\in A$ there’s exactly one $y\in B$ such that $(x,y)$ belongs to that subset” is a pretty neat way of capturing this arbitrariness. And in a sense it is. I think psychologically we are happier with the idea of a completely arbitrary set than we are with the idea of a completely arbitrary “way of associating” elements of one set with elements of another. But just because we are happier with it, that doesn’t mean we should be happier with it. What is an “arbitrary” set of integers, say? Sets of integers defined by properties such as “the set of all $n$ such that $n$ is a prime greater than 1000″ are fine, but how can we capture the idea of a “completely arbitrary” set of integers that doesn’t have a definition of that kind? It’s more or less the same problem as that of capturing the idea of a completely arbitrary function that isn’t given by a rule. So, I respectfully submit, the subset-of-Cartesian-product definition of functions achieves virtually nothing. That will probably provoke a lot of disagreement, so let me qualify it slightly. It is useful for some purposes to “reduce everything to sets”. If I am working on the foundations of mathematics and I show that every statement to do with functions can be translated into an equivalent statement to do with sets, then I have shown that if I can sort sets out then I don’t have to do any further work to sort out functions as well. But in what one might call “everyday mathematics” I think that the definition of functions in terms of subsets of Cartesian products is of no use whatsoever. Hmm, I’m worried that I’m still exaggerating. Let me consider a statement that might make it seem important to answer the question, “What is a function?” It is the following. • There are uncountably many functions from $\mathbb{N}$to $\mathbb{N}$. • You haven’t yet been told what this means, but for now just think of “uncountably many” as meaning “not just infinitely many but an extra-specially big infinitely many” or something like that. We obviously can’t prove a statement like that by listing a whole bunch of functions, so doesn’t that force us to have some general idea of what a function is? Here are two arguments that it doesn’t. One is that I can prove this by reducing it to another problem. For each positive real number $\alpha$ I can define $f(n)=\lceil\alpha n\rceil$ (this means the smallest integer greater than $\alpha n$). It’s easy to check that these are all different functions. And then we can appeal to the fact that there are uncountably many positive real numbers. The second argument is closely related. It is also true that there are uncountably many subsets of $\mathbb{N}$, but nobody feels that we have to say what a subset of $\mathbb{N}$ is in order to make sense of this statement. We just need to know a few rules for dealing with sets: in particular, for defining new sets out of old ones. Just before I move on, let me express particular distaste for any definition that begins, “A function from $A$ to $B$ is a relation such that …” I absolutely hate this. The reason I hate it is that functions and relations are, to any reasonable person, different kinds of things, except that I don’t want to call them things at all, so what I really mean is that they have a different grammar. To illustrate what I mean by grammar, it’s rules like this. 1. If $f:A\to B$ and $x$ denotes an element of $A$, then $f(x)$ denotes an element of $B$. 2. If $P$ and $Q$ denote statements, then $P\vee Q$ denotes a statement. 3. If $\sim$ is a relation on $A\times B$, $x$ is an element of $A$, and $y$ is an element of $B$, then $x\sim y$ is a statement. 4. If $P$ is a property defined on a set $X$ and $x\in X$, then $P(x)$ is a statement. 5. If $P$ is a property defined on a set $X$, then $\{x\in X:P(x)\}$ is a subset of $X$. I’ll talk more about this kind of thing in a later post, but I hope these examples give you the idea. And 1 and 3 demonstrate that the grammar of functions is not the same as the grammar of relations. The fact that you can turn them into equivalent concepts that do have the same grammar is neither here nor there. You can do that with nouns and adjectives too. For example, I could decide that from now on I’m going to say, “There’s a red” where I used to say, “That’s red.” If I wanted to say, “My car is green,” I would say, “My car is a green,” just as I might say, “That dog is a Rottweiler.” It would be possible (basically, nouns and adjectives are both ways of picking out some subset of the set of all possible objects in the world) — but it would also be a bit weird. But none of this what I really wanted to talk about. Rather, I wanted to discuss a few confusing bits of terminology. To begin with, what are domains, ranges and codomains? What’s confusing about this is that there isn’t a standard terminology. The one I was taught as an undergraduate was this. If $f:A\to B$ is a function, then $A$ is called the domain of $f$ and $B$ is called the range of $f$. That’s the sense in which I’ve been using the words in these posts and I hope it agrees with what your lecturers have said. However, some people use the word “codomain” for $B$ instead of “range”. Worse still (from the point of view of communication between mathematicians), people who call $B$ the codomain often use the word “range” to refer to the set of values taken by $f$: that is, to the set $\{f(x):x\in A\}$. I call that the image of $f$, and I think that is probably the Cambridge standard. To give an example, let’s take the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x^2$. In my terminology, the domain and range are both $\mathbb{R}$ and the image is the set of non-negative real numbers. I’m also happy to call $\mathbb{R}$ the codomain if you want — for me, “codomain” and “range” mean the same thing. However, some people would say that the domain and codomain are $\mathbb{R}$ while the range is the set of non-negative reals. I think they would regard “range” as synonymous with “image”. So I suppose we would all get along fine if we just abolished the word “range”. Unfortunately, that isn’t the end of the confusion, since some people use the word “domain” to mean “set of points where $f$ makes sense”. To give an example, they might say this: “Let us define $f:\mathbb{R}\to\mathbb{R}$ by $f(x)=1/x(x+2)$. Then the domain of $f$ is the set of all real numbers apart from $0$ and $-2$.” Apparently, this use of the word “domain” is quite common in schools. According to my terminology, the function just defined was not a function from $\mathbb{R}$ to $\mathbb{R}$ at all. Why not? Because it doesn’t assign a real number to 0 or to -2. So if you want to be safe, then given a function $f:A\to B$ you can call $A$ the domain, $B$ the codomain, and $\{f(x):x\in A\}$ the image. That is still not the end of the potential confusion. I said earlier that the one thing you need to know about a function $f:A\to B$ is that if $x$ is an element of $A$ then $f(x)$ is an element of $B$. In a nice friendly world, you could deduce that whenever you see $f(*)$, then whatever $$ is must be an element of $A$. Unfortunately that’s not the case in the world we actually inhabit: we often write $f(C)$ when $C$ is a subset of $A$. Now in a sense that’s just plain incorrect. Functions from $A$ to $B$ are little machines that turn elements of $A$ into elements of $B$. So how can we write $f(C)$ if $C$ is a subset of $A$? The answer is that the wrongness of doing that tells us one of two things: (i) the writer has made a mistake; (ii) the writer means something different. You should have enough confidence in your lecturers and textbooks to assume that it is (ii) that holds and not (i). So what does $f(C)$ mean? It means the set of all $f(x)$ such that $x\in C$, or in symbols $\{f(x):x\in C\}$. For example, if $f$ is the function from $\mathbb{N}$ to $\mathbb{N}$ that takes $n$ to $n^2$ and $E$ is the set of all even numbers, then $f(E)=\{4,16,36,64,100,144,...\}$. The set $f(C)$ is a subset of $B$ and it is called the image of $C$. The alarm bells should be ringing again. Earlier, I defined the image of $f$ to be the set $\{f(x):x\in A\}$, which we now see that we can write as $f(A)$. So is the set $f(A)$ the image of $f$ (as I said earlier) or the image of $A$ (as would be consistent with the more recent definition)? You just have to be alert to the context. Functions have images, but if you’re talking about a given function that’s clear from the context, then you also talk about images of subsets. Oh, and while we’re at it, if $x$ is an element of $A$, then $f(x)$ is called the image of $x$. There is nothing for it but to get used to the fact that the same words and notation can be used for concepts with different — and confusable — meanings. If you see $f()$, then one of the first things you should do is look at what’s in those brackets and ask yourself what kind of object it is. If it’s an element of the domain (which will usually be indicated by a lower-case letter, which helps to reduce the confusion), then what you’ve got is an element of the codomain. If it’s a subset of the domain, then what you’ve got is a subset of the codomain. Let me give a different example of this kind of use of “element notation applied to subsets”. If $A$ and $B$ are sets of integers, we sometimes write $A+B$. You might object that this cannot be correct, on the grounds that $A$ and $B$ are sets and you don’t add sets together — you take things like unions and intersections. And that objection is right in the following sense: when we write $A+B$ we are not giving the usual meaning to the plus symbol. So what do we mean? We actually mean something fairly natural, which is the set of all numbers you can make by adding something in $A$ to something in $B$. In symbols, $A+B=\{x+y:x\in A, y\in B\}$. A quick example: if $A=\{1,3,5,10\}$ and $B=\{1,4,7\}$, then $A+B=\{2,4,5,6,7,8,9,10,11,12,14,17\}$. Here, I’m not really adding sets: I’m adding the elements and forming a set out of all possible results. In a similar way, when I take the set $f(A)$, I’m not applying the function $f$ to the set $A$: I’m applying $f$ to the elements of $A$ and forming a set out of all possible results. It’s a very important distinction. I’ve said that if $f:A\to B$ and $x$ is an element of $A$, then $f(x)$ is called the image of $x$. Something similar happens the other way round. If $f(x)=y$, then $x$ is called a preimage of $y$. Note a very important distinction between these two definitions: I talked about the image but a preimage. That’s because the definition of a function requires there to be exactly one image for each element $x\in A$, but if I pick $y\in B$ it might not have any preimages, and it might have more than one preimage. Finally, a rare situation where we don’t use the same word twice — however, we make up for this big time in our choice of symbolic notation. If we have a subset $D$ of $B$, then the inverse image of $D$, denoted $f^{-1}(D)$, is defined to be the set of all preimages of elements of $D$. Equivalently, it is the set of all $x\in A$ such that $f(x)\in D$. Equivalently again, it is the set $\{x\in A:f(x)\in D\}$. Here’s a quick example. Let $A=\{1,2,3,4,5\}$, let $B=\{1,2,3,4\}$ and define $f$ as follows: $f(1)=f(2)=1, f(3)=2, f(4)=f(5)=3$. Then the inverse image of the set $\{1,4\}$ is $\{1,2\}$. Why? Because 1 and 2 are the elements of $A$ that have images that belong to the set $\{1,4\}$. I hope you will have noticed something very important about that example, which is that the function $f$ in question does not have an inverse. In fact, it doesn’t even come close: the fact that $f(1)=f(2)$ shows that it isn’t an injection (which means that if we tried to form an inverse $g$ we wouldn’t be able to decide between setting $g(1)=1$ and $g(1)=2$) and the fact that 4 has no preimage shows that it isn’t a surjection either (we would have no idea what value to give to $g(4)$). And yet, I happily wrote $f^{-1}(\{1,4\})$. (Actually, I didn’t write it, but I’m writing it now. And I’m happy.) This is a very frequent source of confusion. Generation after generation of Cambridge undergraduates see an expression like $f^{-1}(D)$ and conclude, wrongly but not entirely unreasonably, that $f$ has an inverse. Indeed, it looks as though it must mean the image of $D$ under the inverse function of $f$. But it doesn’t (except that if $f$ does happen to have an inverse, then $f^{-1}(D)$ does happen to be the image of $D$ under that inverse). The best I can do to help you with understanding inverse images of sets is this. If you ever see a sentence of the form $x\in f^{-1}(D)$, then you are at liberty to translate it into the equivalent but more transparent sentence $f(x)\in D$. What’s more, I recommend doing so. Suppose, for example, that you are asked to prove the following simple fact. (At least, it’s very simple once you are used to the definitions and to standard techniques for writing proofs.) Fact. Let $X$ and $Y$ be sets and let $f:X\to Y$. Let $A$ be a subset of $X$ and $B$ be a subset of $Y$. Prove that $f(A)\subset B$ if and only if $A\subset f^{-1}(B)$. If you’re asked to prove an if and only if, then you start by assuming one side and deducing the other, and then you prove the implication in the opposite direction. So let’s begin by assuming that $f(A)\subset B$. What do we need to prove? We want to show that $A\subset f^{-1}(B)$. How do we show something like that? If you’ve learnt the definition of “is a subset of”, then you will call up to the front of your brain the following statement as what we want to prove. • For every $x\in A$, $x\in f^{-1}(B)$. • And if you have taken on board advice in the post on injections and surjections, you will now immediately write “Let $x\in A$.” The task is now to prove that $x\in f^{-1}(B)$. Aha! That is a sentence of exactly the form that allows us to get rid of that nasty and confusing $f^{-1}$, since it is equivalent to the statement $f(x)\in B$. So we know that $x\in A$ and we want to prove that $f(x)\in B$. What were we given? Oh yes, that $f(A)\subset B$. But $x\in A$, so $f(x)\in f(A)$. But if $f(x)\in f(A)$ and $f(A)\subset B$, it follows (directly from the definition of “is a subset of”) that $f(x)\in B$, just as we wanted. How about the other direction? This time we assume that $A\subset f^{-1}(B)$ and we want to prove that $f(A)\subset B$. So we want to prove that every element of $f(A)$ is an element of $B$. But every element of $f(A)$ is of the form $f(x)$ for some $x\in A$, so we can begin with, “Let $x\in A$,” and know that our target is to prove that $f(x)\in B$. (This is a slight elaboration of the “let” trick that is convenient for dealing with a situation where what we are sort of saying is, “For every $f(x)$ with $x\in A$, such-and-such happens.”) We know that $x\in A$, and that and the hypothesis that $A\subset f^{-1}(B)$ tell us that $x\in f^{-1}(B)$. Aha! We can get rid of that nasty and confusing $f^{-1}$: we now know that $f(x)\in B$. Better still, that’s exactly what we were trying to prove. Just in case I haven’t made it sufficiently clear, if $f:A\to B$, $y\in B$ and $f(x)=y$, then it is incorrect to say that $x$ is an inverse image of $y$ (it is a preimage) and it is incorrect to write $x=f^{-1}(y)$ (the function $f$ might not have an inverse, and if it doesn’t, then $f^{-1}()$ makes sense only if $*$ is a subset of the codomain of $f$). Similarly, if $D$ is a subset of $C$ then $f^{-1}(D)$ is not a preimage, or even the preimage, of $D$ (it is the inverse image). And the fact that we write $f^{-1}(D)$ does not mean that $f$ has an inverse. It only remains for me to apologize on behalf of the mathematical community for the historical accidents that have led to this jumble of overlapping terminology and notation. You have no choice but to learn it and be very careful about using it. The one thing I would say is that one gets used to it — so much so that it becomes hard to remember what it was like to find it confusing. That reminds me that I haven’t finished discussing non-standardness of terminology to do with functions. You will often see the following alternative terminology for injections, surjections and bijections. Injections are called one-to-one functions, surjections are called onto functions, and bijections are called one-to-one correspondences. This terminology is pretty confusing –see Terence Tao’s comment on one-to-one functions — but you probably have to learn it too. I’m mentioning this here because I want to recommend another of the quizzes, the one on functions. If you aren’t quite sure whether you have understood the material in this post and the previous one (and what your lecturers have said on similar topics), then trying out that quiz will soon tell you how you are doing. But it uses “one-to one function”, “onto function” and “one-to-one correspondence” instead of “injection”, “surjection” and “bijection”, so you’ll need to be ready to use that terminology. Added later. I received an email from Imre Leader expressing disagreement with my assertion that the set-theoretic definition of functions achieves nothing. Since he had some valid points to make, and since we ended up agreeing with each other completely (I think), I’d like to report on our exchange. Imre’s point is that, as I said above, people just are more comfortable with the idea of an arbitrary set not defined by a nice property than they are with the idea of an arbitrary function not defined by a nice rule. I maintain that this is irrational, but even if it is, I can’t deny that it is true. So if you give the set-theoretic definition of functions, then you make completely clear that functions can be just as arbitrary as sets. My eventual response (after a certain amount of thought, and an email that Imre disagreed with in a number of places) was that I agreed that the set-theoretic definition of functions helps one to understand how arbitrary functions can be, but that that benefit can be achieved in a different and better way. Instead of saying that a function from $A$ to $B$ is a subset $f$ of $A\times B$ such that for every $x\in A$ there is exactly one $y\in B$ such that $(x,y)\in f$, we should say that every function can be obtained from such a set. It turns out that Imre is entirely happy with that idea. (Also, he was at pains to stress that he is no fonder of turning everything into sets than I am.) Here in detail is what I would say. Let $G$ (to stand for “graph”) be a subset of $A\times B$ such that for every $x\in A$ there is exactly one $y\in B$ with $(x,y)\in G$. Then we can define a function $f:A\to B$ in terms of $G$ by letting $f(x)$ be the unique $y$ such that $(x,y)\in G$. Moreover, every function from $A$ to $B$ can be obtained this way, since if $f$ is such a function we can define $G$ to be the set of all $(x,y)$ such that $y=f(x)$. What I like about this approach is that it doesn’t feel unnecessarily paradoxical. I’m not saying, “Actually a function isn’t a function at all — it’s a funny kind of set.” Rather, I’m saying that there’s a one-to-one correspondence between functions and funny kinds of sets. This captures the arbitrariness of functions (if you believe that sets can be very arbitrary, then you can carry this arbitrariness over to functions), but it also preserves their function-like nature (given a set $G$ with certain properties, I then tell you a rule for associating elements of $B$ with elements of $A$). ### 153 Responses to “Domains, codomains, ranges, images, preimages, inverse images” 1. Qiaochu Yuan Says: I sympathize with your concerns about defining functions as relations, but I don’t think it’s completely fair to say that functions and relations don’t have the same grammar. After all, there is a category of sets and relations, and morphisms can be composed as usual in this category, and the category of sets and functions is a subcategory of this category, and so forth. What’s new about the category of relations, I guess, is that one can also define daggers (transposes) of relations. • gowers Says: Perhaps what I ought to have said more clearly is that the standard examples of functions (things like $f(x)=x^2$) and the standard examples of relations (things like “is congruent mod $m$ to”) have different grammar. One converts elements of one set into elements of another, and the other goes between elements of two (usually equal) sets and produces a statement. It’s certainly possible to conceive of ways of talking about relations that make them feel much more function-like, and in some contexts (particularly, as I’ve discussed in the comments on a different post, when one is checking that a function is well-defined), I think there might even be some benefit in doing so. • Qiaochu Yuan Says: And conversely I think it’s beneficial at times to think about functions in a way that is more relation-like. For example, here’s a small observation. Abstractly, the inverse image construction $f^{-1}(B)$ you describe in the post is a contravariant corepresentable functor $\text{Hom}(-, 2)$ from the category of sets and functions to itself (or the category of Boolean algebras, etc.). The image construction $f(A)$ you describe is a covariant functor, but it isn’t representable in the category of sets and functions by a straightforward cardinality argument. It is, however, representable in the category of sets and relations, by $1$! In fact inverse image and image for functions is a special case of inverse image and image for relations, which are related by transposition. Defining these constructions for functions only is, in a sense, breaking a natural symmetry. • Terence Tao Says: My basic rule of thumb to determine whether an object X is “truly” a set, as opposed to merely being modeled or represented by a set, is to ask oneself whether a statement such as “$x \in X$” could conceivably be useful for doing mathematics; similarly, I would only view $X$ as “truly” being a relation if sentences such as “$x X y$” could be viewed as useful mathematical statements. For instance, I do not consider $1 \in 2$ to be a useful statement, so I do not view the numbers 1 and 2 to be “equal” to sets such as the von Neumann ordinals $\{ \emptyset \}$ or $\{ \emptyset, \{\emptyset\} \}$, instead viewing the latter merely as one possible model or representation of the former. Similarly, “$(x,y) \in f$” and “$x f y$” do not seem like a useful mathematical sentence to me if $f$ is a function, so I do not view functions as “truly” equal to sets or relations, but instead merely being represented by them. In practice, these distinctions generally quite irrelevant, but occasionally it can be useful to maintain the distinction between a mathematical object and its representations, in order to be able to generalise the former in the absence of the latter. For instance, generalised functions such as distributions do not have any reasonable interpretations as sets or relations (at least, not in a fashion that recognisably corresponds to the analogous interpretations of classical functions), yet they still do many of the things that classical functions do (e.g. one can take linear combinations of generalised functions, limits of generalised functions, etc.). So this is one instance in which a dogmatic insistence that functions “are” sets or relations can get in the way of useful mathematical generalisation. • Scott Carnahan Says: I think functions as they are used in mathematical practice have a much more dynamic shape than relations and abstract sets. They have a strong directionality that suggests an irreversible process done by a machine, or perhaps the woes of time evolution. This is one reason why it may seem wrong to say that functions are relations or sets, but it seems okay to say that functions can be encoded as relations or sets. The notion of function as set seems to be an artifact of a historical choice of foundations of mathematics. One could conceivably hatch an alternative, more computationally oriented foundation like a lambda calculus, where one does the same everyday mathematics, but where functions and processes are more primitive than sets and relations. 2. plm Says: The “domain of a function not defined everywhere” can actually be called coimage. For 1/x as a (pre)function from R to R in the category of sets (with morphisms relations with at most one output per input), the coimage is R^ -up to bijection. • plm Says: Actually what I write is wrong according to standard terminology, coimage as I used it would only coincide with the category theory concept for injective (pre)functions. (Coimage is ok for 1/x, but not for 1/x^2 for instance.) 3. Colin Reid Says: I’m sure I’ve seen ‘preimage’ used to mean what you call the ‘inverse image’. The word ‘fibre’/’fiber’ is also popular in certain circles, though perhaps this means something more sophisticated than ‘inverse image’. 4. CD Says: I am enjoying these posts a lot. I wonder if you have encountered as much difficulty teaching “codomain” as I have. In my experience it is something of a pons asinorum for students in abstract mathematics: either they internalize it more or less immediately, or they never do. It presents many pedagogical difficulties. One is that much of our intuition for functions— and most use of functions outside of pure mathematics— simply does not require a precise choice of codomain. It’s quite common to have a definite, fixed choice of domain and something you want to do on it, but no fixed idea (or perhaps several distinct fixed ideas) of how you want to regard the result. Anybody who has ever worked with with a strongly typed programming language has, sooner or later, run up against a reminder that our pure-mathematics function concept contains just a little bit more structure than our intuition seems to want it to have. A second issue is that for many purposes the choice of codomain is arbitrary, subject to the restriction that it be a set containing all of the values f(x), for x in the domain. This is reflected in the “a function is a set of ordered pairs” definition, which popular in textbooks, despite the fact that it does not encode the codomain concept at all. By this definition, for example, the squaring function from real numbers to real numbers, is the same as the squaring function from real numbers to nonnegative real numbers. In many contexts this is innocuous or even desirable, for more or less the same reasons that the non-strict behavior of non-strongly-typed programming languages is innocuous or desirable. In other contexts this is unwanted— for example, textbooks using this definition of “function” will still insist on a difference between my examples, because one is surjective and the other isn’t. But the fact that this difference is not actually encoded in the definition is not that commonly noticed. For the reason, I’d say, that a lot of the time it just doesn’t matter. (And not encoding the difference serves a small but useful pedagogical purpose. I’m not sure that students have a better time with functions if, on the first go-round, they learned “functions are sets” not with the graph, but with an ordered triple consisting of two sets and then the graph. This draws almost too much attention to the arbitrary choices made in modeling mathematics within set theory.) The arbitrariness of the codomain seems related in my mind to the general idea that often we want to identify sets with sets containing those sets, whenever doing so simplifies the discussion and suppresses irrelevant choices in formalism. (A positive-integer-valued function is a real-valued function is a complex-valued function…) It takes experience and sensitivity to context— the context in which a function is being used, not just the context inherent in the definition of the function concept— to understand when the choice of codomain really matters and is not just an irrelevant choice of formalism. So: when teaching the subject, one is in the awkward position of drawing a useful technical distinction— between the image of a function and the function that the set is “to”— but then following this immediately with examples showing that the choice of the object that resolves this distinction is sometimes arbitrary, and sometimes crucially important. The best I have been able to do with teaching it, is to emphasize that the codomain is not really something intuitive, but extra structure we add to the intuition, to have useful notions available to us when we need them. 5. Sam Alexander Says: The way you defined “functions from A to B” in this post, you very deftly ducked under a very common problem. When “function” is naively defined as “set of pairs with the single-valued property”, then what you call the range (and what some call the codomain) becomes non-well-defined. For instance, using that naive definition, if I say f:R->R is defined by f(x)=x^2; and if I see g:R->[0,infty) is defined by g(x)=x^2; then f and g are actually the same set, and so if range/codomain could be well-defined, f and g would have to have the same one. Of course, the category theorists get around this by defining a function to be (say) a triple (f,A,B) where A and B are sets and f is what you’ve defined here as “a function from A to B”. • gowers Says: That was in fact deliberate. It was pointed out to me a year or two ago, and CD also mentions this above, that many textbooks get the definition of function wrong (or at any rate are inconsistent about it) and I wanted to be careful to avoid that mistake but without drawing attention to it too much. • Joel Says: I always find this worrying: morally speaking, should a function “know” what its codomain is? The category theory approach certainly solves that one, but it doesn’t seem to be a common approach when introducing functions. So, is the category theory approach the “right” one? If so, what would happen if we did try to teach that ordered triple version to starting undergraduates. CD says he isn’t sure this would be helpful. Has anyone tried it? If we do try it, we would have to warn students that they will find variants on this definition in many books. The “officially correct” definition in the original post says “A function from A to B is [a set of ordered pairs with certain properties] …”. Isn’t this in fact the “wrong” definition found in many textbooks, or is there a subtle difference, based on context, that I am missing? The “officially correct” definition is, of course, the one I was taught as an undergraduate, and which I have always worked with, but subject to the concerns expressed above. Joel 6. Thomas Says: From my topology teacher I adopted the square bracket notation for images and preimages of sets: $$f[A]$$ and $$f^{-1}[A]$$, respectively. Though it gets messy if A is an interval and written as such … • CD Says: Some textbooks call attention to the difference by decorating the f (e.g. if f is a function from A to B, then f^* is the function from the power set of A to the power set of B induced by f in the usual way). This convention has a lot to recommend it— among other things, it is not just a notational choice, but is functorial, and a subconscious introduction to the idea of a functor. And it allows one to enforce completely rigid interpretation of notation at an educational stage when some students might abuse and over-generalize the apparent flexibility of being able to write both f(element of domain) and f(subset of domain). On the downside, there is no standardized notation for the induced function on power sets— the standard is to write both f(x) and f(A)— and to any extent that confusion is avoided by not doing the standard thing, it is only postponed to the moment when students meet somebody who does do the standard thing. 7. JeffE Says: In my own personal notes, I always use “into” for injections, in contrast with “onto”. For example: Stereographically project the plane into the sphere. I haven’t quite convinced myself to use “unto” for bijections, though. (Map unto others as you would have them map unto you.) 8. Henning Makholm Says: From my vantage point in computer science, the underlying message of this post seems to be that the language of ordinary mathematics is a _typed_ language, and that the habitual tendency of mathematicians not to explain explicitly what the typing rules are, or even acknowledge that they exist, engenders confusion and mistakes. Expressing everything as sets is well and good for foundational purposes, but to insist that everything else is merely abbreviations for set-theoretical constructions amounts to denying every shred of intuition about what we’re doing — even though one often _cannot_ get from the symbols actually on the paper to their purportedly “real” set-theoretic meaning without knowing a lot of mostly unspoken rules of the subject area. It feels to me like insisting that all computer programming is just about bit patterns, so all the careful abstractions one builds when constructing a program are merely convenient (and ultimately irrelevant) abbreviations for particular bit manipulations. That’s just no way to get work done. There’s a large and refined body of work in computer science on how to design and describe rules like the ones explained in this post such in a rigorous and unambiguous, yet expressive and flexible, way. We do it in the context of programming languages, but that’s really just a constructive subset of mathematics, and easily extended to cover all of it. I’ve often wondered why mathematicians are not eagerly making use of these ideas to simplify and structure the exposition of mathematics. (Perhaps types got a bad rap after Russell’s theory of types turned out to be insufferably cumbersome to use relative to Zermelo’s untyped set theory? But that’s ignoring a century of later developments, where at least the last few decades have been keenly focused on practical convenience because programming languages are _tools_ rather than abstract objects of study). In a well-developed typeful formulation of mathematics, I expect that one would define “function from A to B” just as an abstract type with certain axiomatic properties, and then use the set-of-pairs definition simply as a “reference implementation”, serving as an existence proof which guarantees that the concept makes sense and is safe to use. • Qiaochu Yuan Says: I’m glad someone said something about typing. Ever since I attempted to teach myself Haskell, I’ve thought that a typed foundation of mathematics would be both easier to learn and ultimately more faithful to mathematical practice, and I often wonder why more mathematicians aren’t trying to write one down (or if they have, why I haven’t heard about it). • Terence Tao Says: Well, mainstream mathematics certainly has categories, which are very type-friendly (cf. also their philosophy that equality between objects is “evil” and should be replaced by isomorphism whenever possible). I would imagine therefore that category theoretic foundations of mathematics (such as ETCS) would be quite amenable to typing, but I am not an expert on these matters. In any event, once one moves away from foundations, I think most practicing mathematicians use some sort of informal typing when they discuss or do mathematics. For instance, I doubt mathematicians view a real number as an equivalence class of Cauchy sequences or as a Dedekind cut when manipulating such numbers. • gowers Says: I completely agree with this. In fact, I’ve been planning a post about parsing mathematical sentences, which will emphasize this point. • Uwe Stroinski Says: “I’ve often wondered why mathematicians are not eagerly making use of these ideas to simplify and structure the exposition of mathematics.” Good point. We do mathematics bottom-up in assembly language and should then present it top-down in some object oriented frame using information hiding, inheritance, polymorphism aso. I think, from all these concepts ‘top-down’ seems to be the hardest to achieve. • Henning Makholm Says: Terrence: the problem with category theory is that while it can express a lot of mathematics, to do so it translates everything into its own language and henceforth it is something completely different. Uwe, I have trouble discerning whether you’re being sarcastic. The goal of the program I’m imagining (which already makes it sound rather more grandiose than I think it should) would not be the recast mathematics in the mold of software development, but to vindicate the way everyday mathematics _already_ uses formulas. The most important thing to borrow from computer science might be a vocabulary for _describing_ systematically what must surely look to newcomers like an endless parade of disjoint ad-hoc notational conventions. In any case, I think you’re conflating software engineering with programming language research. The former, being both practically important and very, very difficult to make solid theoretical progress in, is mostly a collection glorified rules of thumb, quite susceptible to fads and populated in part by loud, dogmatic amateur propagandists for current and past fads. The latter is a proper quasimathematical field of study with rigorous definitions, theorems, and heavy use of Greek letters. I don’t really think OOP has much to offer mathematics. One of its ingredients, namely subtyping, is relevant on its own: as much as we don’t really want to identify the natural number 2 with the set {0,1}, we certainly do want it to be the same object as the integer 2, the rational number 2, the real number 2, and the complex number 2, which subtyping can help describe systematically (whereas a typical set-theoretic development would claim that there are “really” invisible injections being applied everywhere). Insisting on a strict top-down order of development would be almost as misplaced in mathematics as it is in programming. • Uwe Stroinski Says: Henning: Oops. Absolutely no sarcasm intended. Sorry for being so sloppy. I very much sympathize with your approach (albeit I probably do not completely understand it). Is software engineering really completely separable from programming language research ? I do not think so. There surely is a considerable overlap. At least there was, when I studied it. In any case, typed variables with restricted scopes (as discussed in this thread) are just the beginning. Once you have that, why not making explicit when you overload an operator like + or $f^{-1}$. Polymorphisms like that are everyday business for mathematicians and can be very confusing to students. In terms of programming languages such concepts are best described by OOP languages. At least, that is what I think. 9. J.P. McCarthy Says: Two minor points that I thought you were going to make. Firstly be careful to write $f^{-1}(\{x\})$ rather than $f^{-1}(x)$ so as not to imply that $f$ has an inverse. Secondly that the issue with codomains and images is related to the fact that all injections are invertible if we make no distinction between codomain and image. 10. Luqing Ye Says: Regard a function from set A to B as a subset of $A\times B$ which satisfy certain properties.This definition is also useful in such conditions: In Terence Tao ‘s book “Analysis”,the part of the book which involve set theory,there are two ways of presenting the power set axiom: (1)All the functions from set A from set B form a set. (2)All the subset of set A form a set. It seems only if we regard function as a subset of cartesian product can we prove that the two statements are equal.For more information,see “analysis” exercise 3.5.11 11. Chris Says: It might be interesting also to note that there are some instances where it matters a great deal which codomain is chosen for a function. For example, in general topology the concept of a function being $\theta$-continuous is not preserved under restricting the codomain. So even though we often don’t need to specify the codomain explicitly, there are cases where the choice matters (and not just any set containing the image will do). 12. Two-place functions aren’t one-place functions, are they? \| Logic Matters Says: […] Here’s a small niggle, that’s arisen rewriting a very early chapter of my Gödel book, and also in reading a couple of terrific blog posts by Tim Gowers (here and here). […] 13. Think of a function \| Explaining mathematics Says: […] (this appears to be the Nottingham standard), not the codomain of the function. Some, including Tim Gowers, prefer to use range to mean codomain (this appears to be the Cambridge standard). Perhaps we […] 14. Vania Mascioni Says: I feel that the dislike for the notion of “arbitrary set” as related to the notion of function might come from the fact that everyday practice makes us work with functions that not only have a name such as f(x), but an actual, working definition on how to get f(x) from x. In this sense, f(x) is the “name” standing for a specific operation, and we as a species are quite comfortable with naming things. Yet, when I think of real numbers I should feel the same way: most of us tend to come up with examples of real numbers who have names, such as -2, square root of 5, or pi. Yet the crushing majority of real numbers are transcendental and will never have a name of their own in our entire lifetime. So, when trying to answer the question “what is a real number?” one is necessarily forced to use the more abstract approach since there is no way to exhibit your general real number. unless dealing with random-like sequences of decimal digits should feel “concrete” to anyone. Similarly, going back to functions if we try to think of a random-valued real function it feels somewhat silly to think of it as an f(x) since we have nowhere near a formula for it. In this sense, I suspect that what Tim Gowers thinks about when speaking of the “grammar” of functions he is only referring to functions that actually can be meaningfully written as y = f(x) (with a “readable” f(x) part), and this seems quite analogous to the “grammar” associated to those real numbers that have a name: “see, square roof two is a real number.” But how can I reasonably fit in a sentence that forever unnamed (and rather unreal, in fact) transcendental “thing” that together with its brethren fills up most of the real line? • obryant Says: You’re using “transcendental” when you mean “computable”. As there are only countably many finite deterministic algorithms, there are only countably many reals whose decimal expansions we can compute by algorithm to arbitrary accuracy. We name transcendental numbers all the time ($\pi, e^{\sqrt{2}}$), but we never name noncomputable numbers. However, Chaitin would disagree. 15. rakayla Says: this really didnt help to me can you please give more details 16. JEENATH NISHA Says: i need theoretical proofs and examples for (a) f(AUB)=f(A)Uf(B) (b) f-1(AUB)=f-1(A)Uf-1(B) (c) f-1(A∩B)=f-1(A)∩f-1(B) 17. Anonymous Says: I’m puzzled by what appears to be a very elaborate and deliberate attempt to create a lot of ambiguity in the terminology associated with the theory of sets and functions. There’s a reason technical fields have an argot and it’s to minimize ambiguity, not create and justify a lot of it. • gowers Says: The one word I would dispute there is “deliberate”. It’s just an unfortunate example of a situation where by a historical accident the terminology didn’t standardize itself satisfactorily. 18. Optimization, Step 2 – geoff bourque's blog Says: […] me repeat that one more time for clarity (and I’m stealing this formulation from Sir Tim Gowers’ blog): If is a function from to and is an element of , then is an element of […] 19. Bram van Heuveln Says: Thank you for this post! I wish I had seen it at the time you wrote it, since I have been frustrated with the terminology for many years now. Here is one complaint I have in particular: going back to Russel and Whitehead, the word ‘co-domain’ seems to be short for ‘converse domain’, i.e. the domain of the converse or inverse of a function. But the way the terms ‘domain’, ‘co-domain’, and ‘range’ I most often see used (though there is certainly no consistency here!), it would be the ‘range’ (or ‘image’) that serves the role of the ‘converse domain’, and not the ‘co-domain’! For example, for a function like f(x)=1/x, the ‘domain’ is usually set to be the set of objects for which this function is defined, i.e. R/{0}, the ‘range’ as the set of values the function can take on, i.e. also R/{0}, but the co-domain would be R … Which is clearly not the ‘domain’ of the converse, or inverse, of this function. Another question I have here is where the R comes from: why would we think that we are initially/immediately talking about real numbers when talking about functions like 1/x? Couldn’t I be talking about complex numbers, say? Indeed, why do we indicate some kind of ‘intended’ set of ‘potential’ values of the function on the output side (the ‘co-domain’), but don’t do this on the input side? Indeed, I think it would make a lot of sense to first define a set of objects that the function is intended to work on (which we could call a ‘domain of discourse’), and then point out any values for which the function is actually defined (which we would call the ‘domain of definition’). And with that, we could have a ‘co-domain of discourse’ and a ‘co-domain of definition’ as the converse’s counterparts. Thus, for the function f(x)=1/x we could say that the domain and co-domain of discourse are both R, but the domain and co-domain of definition R/{0}. In fact, I have been taking the lack of standards (or logical consistency) in the terminology as justification to use exactly this terminology when I teach my students about functions. It may not be ‘standard’, but I think it is certainly a lot more consistent and coherent … What do you think? 20. Basics of Functions \| Part 1: Basic definitions, images, and pre-images – Nino's Notepad Says: […] you’re looking to learn about functions, here is a really good starting point. If you haven’t learned the basics of naive set theory and want to understand the […] 21. Prof. Tim Gowers’ on functions, domains, etc. \| For the love of Maths: the Queen of all Sciences Says: 22. Anonymous Says: f 23. Mary.Fannie Says: The only way to achieve happiness is to cherish what you have and forget what you don’t have 24. namraakhtar Says: It is the best information I ever read because it is very useful and informative. Keep doing the great work up. 25. namraakhtar Says: or label and patches visit my website https://austintrim.co/ 26. namraakhtar Says: for energy solar panels visit my website https://enss.pk/ 27. namraakhtar Says: continue blogging. 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http://docplayer.net/196134927-Drafting-your-panties-basic-block.html	1,604,001,975,000,000,000	text/html	crawl-data/CC-MAIN-2020-45/segments/1603107905777.48/warc/CC-MAIN-20201029184716-20201029214716-00257.warc.gz	32,768,844	26,070	# Drafting Your Panties Basic Block Size: px Start display at page: Transcription 1 Drafting Your Panties Basic Block Waist measurement: Hip measurement: Total rise measurement: Drafting Instructions Working it out Drafting Plan 1. Line A to B Draw a vertical line on your paper equal to the total rise measurement plus 4 cm and multiply this measurement by the Negative Ease Reduction Factor (NERF) NOTE: if your fabric is stable and does not stretch in both directions then you can omit multiplying by the NERF and just use the total rise plus 4 cm for line A to B, Total Rise = + 4 centimetres = Multiplied by NERF Line A to B = NOTES: Mark Point C Mark point C halfway between points A and B Line A-B = Divided by 2 = Measure down from point A by this quantity. 2 3. Mark points A to D and B to E Points A to D and points B to E are 1/6 of the total hip measurement plus 2 cm Total Hip = Divided by 6 = Plus 2 cm = Measure down from point A by this quantity and mark point D. Measure up from point B by this quantity and mark point E 4. Mark points D to F, G from C and C to H. D to F G from C C to H Are all equal to 1/3 of the measurement D to C Measurement A to C = Takeaway A to D = measurement D to C = Divided by 3 = Measure from point D to F G from C C to H by this quantity. 5. Square out from A By ¼ of the HIP measurement multiplied by the NERF. Mark this point A1 Hip = Divided by 4 = Repeat this for point B and Multiplied by NERF of 0.85 3 mark point B1 = 19.8 This creates the front and back waistline guidelines. 6. Square out from D By the same quantity used above Mark this point D1 Repeat this for point E and mark point E1 This creates the hip guidelines. 7. Square out from F By 4 cm - Square out from point A by this quantity and mark point A1 repeat for point B to B1 Square out from point D by this quantity and mark point D1 repeat for point E to E1 Set quantity for all sizes Notes: (if you change this set measurement after making your toile then use the notes space to record your personal preference) Mark point F1 4 8. Square out from G By 3 cm - Set quantity for all sizes 3 Notes: Mark point G1 9. Square out from C By 3.5cm - Set quantity for all sizes 3.5 Notes: Mark point C1 5 10. Square out from H By 1/12 of the total hip measurement. Hip= Divided by 12 = Square out from H by this ammount. Mark point H1 The basic draft is now complete now we just need to shape the panties! 11. Mark a point down from A by 1 to 1.5cm depending on how round your tummy is. (The rounder your tummy the less you need to drop this point you can also leave this flat) 12. Mark a point up from B by 1.5 to 2 cm depending on how much your back dips in. (The rounder your shape the less you should lower as you need the extra fabric to allow for your curves) Now draw in the waistline curves as illustrated keeping a right angle at centre back and centre front. (illustrated in blue) 13. Mark in from A1 and from B1 by 2 cm call these points A2 and B2 (if you don t have much difference between your hips and waist measurement then you can reduce this amount.) Please note that the above measurements are guidelines and you can experiment with them to achieve your perfect 6 fit. Connect point A2 to point D1 with a curved line for the hip do the same for B2 to E1 Shaping the leg curve 14. From point D1 mark a point 5 cm along call this D2 connect this point with a straight line to point F1 Find the centre of this new line (D2 to F1) and make a guide point 1.5 cm at a right angle upwards. Gently draw in the leg curve. 15. Connect F1 to G1 With a gentle curve 16. Connect G1 to C1 and C1 to H1 With a gentle curve. 17. Connect H1 to E1 With a straight line. Divide the straight line into quarters. Draw a curve to touch the centre as shown to follow the contours of the body. The flatter your bottom the flatter the line should be. 7 Planning the gusset 18. 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https://www.bankersadda.com/2018/06/reasoning-ability-for-sbi-clerk-prelims.html	1,563,945,857,000,000,000	text/html	crawl-data/CC-MAIN-2019-30/segments/1563195530385.82/warc/CC-MAIN-20190724041048-20190724063048-00124.warc.gz	611,989,228	56,264	Dear Aspirants, Bankersadda brings to you the SBI Clerk 20 Minutes Marathon of Reasoning Ability...its time to Chase your Success. This is a timer-based quiz of 20 minutes to help you practice for SBI Clerk Preliminary exam. You'll also get the full-length sectional test of English. so keep practicing on Bankersadda. You can also take up this challenge on Adda247 App. So, start practicing for the real examination right away. This will not only ensure your success in the exam but will also help you bag maximum marks in the Reasoning Ability Section with a planned strategy. Following is the Reasoning Ability Full-Length quiz to help you practice with the best of latest pattern questions. Directions (1-4): Study the information and answer the given questions: There are seven boxes viz. A, B, C, D, E, F and G. All the box are of different colour viz. Red, black, blue, yellow, cream, violet and orange but not necessarily in the same order. All the box are arranged from top to bottom. Box D is immediately above the yellow colour box. More than three boxes between cream colour box and orange colour box. There are two boxes between the box B and box E, which is of red colour. There is only one box between box E and box G. There are three box between box G and box A, which is of black colour. There are two boxes between box A and box C, which is of violet colour. The orange colour box is immediately above the box G. More than three boxes between the yellow colour box and the cream colour box. Q1. Which of the following box is at the top? A B C D E Solution: (1-4) Q2. What is the colour of box B ? violet black blue cream orange Q3. Which of the following is yellow colour box ? D E F G A Q4. How many box/es is/are there between D and the box which is of blue colour? Four Three Two One No one. Directions (5-9): In each of the questions below is given three statements followed by three conclusions numbered I, II and III. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts. Q5. Statements: Some light is pen. No pen is sun. Some sun is moon. Conclusions: I. Some moon is not light. II. Some moon is not pen. III. All light being moon is a possibility. None follows Only I follows Only II follows Only III follows Only II and III follow Solution: Q6. Statements: All star is water. No water is train. Some train is bus. Conclusions: I. All water being star is a possibility. II. All star being water is a possibility. III.No star is bus. None follows Only I follows Only II follows Only I and II follow Only II and III follow Solution: Q7. Statements: Some copy is pen. All pen is exam. No exam is test. Conclusions: I. All copy being test is a possibility. II. Some copy is test. III. Some pen is test. None follow Only I and II follow Only I and III follow Only II and III follow All follow Solution: Q8. Statements: All line is red. All red is black. No red is chair. Conclusions: I. No black is chair. II. No line is chair. III. All line being black is a possibility. Only I follows Only II follows Only III follows Only either I or III and II follow None follow Solution: Q9. Statements: Some night is day. All day is bright. All bright is star. Conclusions: I. Some night is bright. II. All day is star. III. Some star is night. All follows Only I follows Only II follows Only III follows Only I and II follow Solution: Directions (10-14): Study the information and answer the following questions: In a certain code language ‘he read very fast’ is written as ‘tx fs mp zq’, ‘fly bird very high’ is written as ‘nx tz zq bq’, ‘fast grow bird tree’ is written as ‘fs cz dv nx’, ‘he fly in tree’ is written as ‘mp bq hw cz’, Q10. What is the code for ‘tree’ in the given code language? mp bq cz hw None of these Solution: (10-14) Q11. What is the code for ‘read’ in the given code language? zq mp fs tx None of these Q12. What is the code for ‘grow’ in the given code language? fs cz dv nx None of these Q13. What is the code for ‘fast in’ in the given code language? fs cz fs hw fs bq hw nx None of these Q14. What is the code for ‘bird fly’ in the given code language? nx hw bq mp nx bq bq dv None of these Directions (15-19): Study the following information carefully and answer the given questions: Twelve persons are sitting in two parallel rows containing six persons each in such a way that there is an equal distance between adjacent persons. In the Row-1 A, B, C, D, E and F are sitting and all of them are facing towards north. In the Row-2 L, M, N, O, P and Q are sitting and all of them are facing towards south direction but not necessarily in the same order. In the given seating arrangement each member sitting in a row faces another member of the other row. O does not face E. Immediate neighbour of P faces B. M does not face D. Three persons sit between B and D. L sits second to right of Q. Only one person sits between L and P. P sits at one of the extreme end of the line. D does not sit on extreme left end. Three persons sit between N and M. C sits second to right of F. C does not face L. Q15. Who among the following sits between L and Q? M N O P None of these Solution: (15-19) Q16. Who among the following faces A? P Q L M O Q17. Who among the following sits second to the left of B? A C No one. D E Q18. How many persons sits between A and E? One Two Three Can’t be determined None of these Q19. Four of the following five are alike in a certain way based on the given seating arrangement and hence form a group. Which is the one that does not belong to that group? P B N D E Directions (20-21): Study the following information carefully to answer the given questions: There are six persons viz. A, B, C, D, E and F of different ages. C is older than only A and E. D is younger than only B. E is not youngest. The one who is third oldest, age is 81 year and E’s age is 62 years. Q20. Which of the following could be the possible age of C? 70 years. 94 years. 86 years. 61 years 81 years. Solution: (20-21) According to their age= B>D>F(81)>C>E(62)>A Q21. Which of the following is true with respect to the given information? D’s age is definitely less than 60. F is oldest. Only two people are older than C. There is a possibility that B’s age is 79 years None is true. Q22. Mohan walked 40km towards North and then he took a left turn and walked 20km. He again took a left turn and walked 40km. How far and in which direction is he from the starting point? 20m west 20km North 20km South 100km South 20km west Solution: 20km west Directions (23-25): These questions are based on the following set of numbers. 4384487737438833873338733837734783777374377833 Q23. how many 3’s in the above arrangement which is immediately preceded by even digit but not immediately followed by odd digit? One Three Two Four None of these Solution: 438, 438 Q24. How many 7’s in the above arrangements which is immediately followed by perfect square? Four Three Two One None of these Solution: 74, 74 Q25. Which of the following digit is 7th to left of 7th from right end in the above arrangements? 4 8 3 7 None of these Directions (26-28): Study the following information and answer the questions given below: There are AB axis in such a way that A is in north and B is in south direction. There is XY axis in such a way that X is in west direction and Y is in east direction. AB axis and XY axis intersect at a point Q in such a way that AQ is 13m, QB is 15m, QX is 10m, QY is 22 m. Sharvan starts walking from point X and walks 18m in south direction and then he takes a turn to his left and walks 30m. Siddharth starts walking from point A and walks 20m in east direction. Dipendra starts walking from point Y and walks 3m in north direction and then he takes a turn to his left and walks 2m and again he takes a turn to his left and walks 18m. Q26. Point B is in which direction with respect to Dipendra’s current position? south south-east south-west west north-west Solution: (26-28) Q27. Point Y is in which direction with respect to Sharvan’s current position? north east north-east north-west south Q28. What is distance between Siddharth’s current position and Sharvan’s current position? 31m 33m 11m 20m 25m Directions (29-32): Study the following information carefully and answer the questions given below: There are eight persons A, B, C, D, E, F, G and H are sitting around a circular table. All are facing towards the center of the table. No two successive persons are sitting together according to alphabetical order. For Example: A does not sit with B; similarly B does not sit with C and so on. All of them belong to different city viz. Agra, Noida, Pune, Varanasi, Delhi, Puri, Lucknow and Patna but not necessarily in the same order. There is only one person sits between the one who belongs to Puri and F, who belongs Patna. E is not immediate neighbor of G. The one who belongs to Lucknow sits second to right of E. There is only one person sits between the one who belongs to Lucknow and the one who belongs to Delhi. Immediate neighbor of E does not belong to Varanasi. B does not belong to Varanasi. G sits third to right of one who belongs to Puri. A belongs to Noida and he sits second to right of G.B is not immediate neighbor of E. Immediate neighbor of E does not belong to Pune. Q29. Who among following belongs to Pune? E A H G B Solution: (29-32) Q30. Who among following sits opposite to C? G F D B H Q31. Who among following sits second to left of D? A E C B G Q32. Four of the following five are alike in a certain way and hence they form a group. Which one of the following does not belong to that group? F H D E C Directions (33-35): In these questions, relationship between different elements is shown in the statements. These statements are followed by two conclusions: Q33.Statements: A ≤ D < C ≥ B < E Conclusion: I. C > A II. A ≥ C If only conclusion I follows. If only conclusion II follows. If either conclusion I or II follows. If neither conclusion I nor II follows. If both conclusion I and II follow. Solution: Conclusion: I. C > A ( True ) II. A ≥ C( Not True ) Q34.Statements: P > L ≤ M < N > Q Conclusion: I. P > Q II. Q > M If only conclusion I follows. If only conclusion II follows. If either conclusion I or II follows. If neither conclusion I nor II follows. If both conclusion I and II follow. Solution: Conclusion: I. P > Q( Not True ) II. Q > M( Not True ) Q35.Statement: S ≥ T = U < V ≥ X Conclusions: I. V > S II. V > T If only conclusion I follows. If only conclusion II follows. If either conclusion I or II follows. If neither conclusion I nor II follows. If both conclusion I and II follow. Solution: Conclusions: I. V > S( Not True ) II. V > T( True ) You may also like to read:	2,883	10,942	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5625	4	CC-MAIN-2019-30	latest	en	0.923435
http://math.stackexchange.com/questions/567054/getting-rid-of-the-square-roots-in-the-expression-sqrta-sqrtb-sqrtc	1,469,266,169,000,000,000	text/html	crawl-data/CC-MAIN-2016-30/segments/1469257821671.5/warc/CC-MAIN-20160723071021-00140-ip-10-185-27-174.ec2.internal.warc.gz	152,573,343	20,923	# Getting rid of the square roots in the expression $\sqrt{a} + \sqrt{b} + \sqrt{c} = d$ I need to find an alternate form of $\sqrt{a} + \sqrt{b} + \sqrt{c} = d$ without square roots for a problem that I'm working on, but it's rather complicated to do. What we can do is $\sqrt{a} + \sqrt{b} + \sqrt{c} = d \iff \sqrt{a} = d - \sqrt{b} - \sqrt{c} \iff a = (d - \sqrt{b} - \sqrt{c})^2$ and then calculate the right hand side, then iterate the process by putting a square root on one side, then proceed as above until all square roots are gone. However, this is a rather complicated process and I would like to do this with a computer but it exceeds WolframAlpha's server time. Can I get this done somewhere else? I would do this by hand myself but I will also need it for $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d} = e$ and perhaps even more complicated expressions, so if someone can show me an easy way to proceed with calculations - or point me to some computer program that can do this - I would appreciate it. Thanks in advance. (note: not sure how to tag my question, feel free to change the tag if you can find something more appropriate). Note: edited as per suggestions in the comments. - There are two problems: (1) the expressions are going to get large rapidly; a term with $n$ square roots will lead to a polynomial relation of degree $2^n$, and (2) your equivalences are running 'the wrong way' - it's not true that $a=b\Leftrightarrow a^2=\pm b^2$, but rather that $a^2=b^2\Leftrightarrow a=\pm b$, which is the opposite of what you want to make this work. In general, the polynomial relation will have $2^n$ distinct 'roots' for the one non-square-root variable of the original expression in terms of the other variables. – Steven Stadnicki Nov 14 '13 at 17:52 See the answers to this math StackExchange question: Rationalizing radicals – Dave L. Renfro Nov 14 '13 at 18:15 You are asking a special case of the following problem in abstract algebra: Suppose $x$ is a solution to $p(x) = 0$ and $y$ solves $q(y)=0$, for polynomials $p,q$ (with, say, integer coefficients); find a polynomial (with, again, integer or whatever coefficients) that has $x+y$ as a solution. In the "$\sqrt{a} + \sqrt{b} = c$" version of your question, you have $p(x) = x^2-a$ and $q(y) = y^2 - b$. There is a general way to do this. Unfortunately in the version I described, the degrees of the polynomials get large. This is because of the following. Suppose that $p$ is of degree $\deg p$, and $q$ has degree $\deg q$. Then generically $p$ has $\deg p$ many complex solutions and $q$ has $\deg q$ many complex solutions. Thus (unless these solutions happen to satisfy some coincidences) there are $(\deg p)(\deg q)$ many possible values of $x+y$ if all you know is that $x$ solves $p(x)=0$ and $y$ solves $p(y)=0$. Whatever the polynomial is that $x+y$ solves, it must be solved by all of possible of these $(\deg p)(\deg q)$ numbers, since you have no way of telling it which solution you want. Therefore, this polynomial must have degree $(\deg p)(\deg q)$. So in the example "$\sqrt a + \sqrt b = c$", we're looking for a polynomial solved by $c$ with "integer" (really, polynomial expressions in $a,b$) coefficients, and it necessarily will have degree $4$. Now adding on another square root means that in your case, your polynomial in $d$ will have degree $(4)(2) = 8$. In the next one, you get a polynomial of degree $16$. There's really nothing you can do about this. Of course, there will be some patterns. Let me focus on your case of sums of square roots. The two solutions to $x^2-a = 0$ are, of course, $\pm \sqrt a$, and the two solutions to $y^2-b$ are $\pm \sqrt b$. Thus the four possible values of $x+y$ are $\pm \sqrt a \pm \sqrt b$. Therefore the polynomial we want is the degree-$4$ polynomial vanishing on these four points, namely: $(z - \sqrt a - \sqrt b)(z - \sqrt a + \sqrt b)(z + \sqrt a - \sqrt b)(z + \sqrt a + \sqrt b) = ((z - \sqrt a)^2 - b)((z + \sqrt a)^2 - b) = (z^2 - a)^2 + b^2 - b((z - \sqrt a)^2 + (z + \sqrt a)^2) = z^4 - 2z^2a + a^2 + b^2 - 2z^2 b - 2ab = z^4 - 2(a+b)z^2 + (a-b)^2$ This illustrates, for example, that only even powers of the new variable (in your case, $d$) will appear — exactly because the set of solutions to this polynomial will necessarily be symmetric under $d \mapsto -d$. Such a polynomial is called "even". In all cases, let me henceforth call the new variable $z$ — so you asked about $\sqrt a + \sqrt b + \sqrt c = z$ or $\sqrt a + \sqrt b + \sqrt c + \sqrt d = z$, and so far I've discussed $\sqrt a + \sqrt b = z$. Let's say there are $n$ terms on the left, so that we're looking for an even degree-$2^n$ polynomial in $z$; and I will set $a = a_1$, $b = a_2$, $c = a_3$, and on up to $a_n$. The above calculation also illustrates that the coefficient on $z^{2^n - 2k}$ will be a symmetric polynomial in the $a_i$, homogeneous of degree $k$. That it's symmetric is clear: the problem as posed is symmetric in the $a_i$. That it is homogeneous of degree $k$ follows from rescaling all $a_i$ to $\lambda a_i$; then the solutions $z$ uniformly rescale to $\sqrt \lambda z$. So in the case that you originally asked about, with $a=3$, we're looking for: $z^8 + p_1(a_1,a_2,a_3) z^6 + p_2(a_1,a_2,a_3) z^4 + p_3(a_1,a_2,a_3) z^2 + p_4(a_1,a_2,a_3)$ where each $p_j$ is a homogeneous symmetric polynomial of degree $j$. It is well-known, then, that $p_j$ is a polynomial in the polynomials $s_1 = a_1 + a_2 + a_3$, $s_2 = a_1^2 + a_2^2 + a_3^2$, $s_3 = a_1^3 + a_2^3 + a_3^3$, $\dots$, $s_j = a_1^j + a_2^j + a_3^j$. For example, $p_1 = \alpha(a_1 + a_2 + a_3)$ for some coefficient $\alpha$, and $$p_2 = \beta(a_1 + a_2 + a_3)^2 + \gamma(a_1^2 + a_2^2 + a_3^2)$$ There are three as-yet undetermined coefficients in $p_3 = \delta s_1^3 + \epsilon s_1 s_2 + \zeta s_3$, and five coefficients in $p_4 = \eta s_1^4 + \theta s_1^2 s_2 + \iota s_2^2 + \kappa s_1 s_3 + \lambda s_4$. All together, we have reduced the problem from computing some arbitrary degree-$8$ polynomial with polynomial coefficients to computing $11$ rational numbers. Can we pair these down at all? When $a \neq 0$ but $b=c=0$, the two solutions need to be $z = \pm \sqrt a$. (Well, each of these is really for solutions, for "the two values of $\pm\sqrt{0}$".) So in this case our polynomial had better evaluate to $(z^2 - a)^4$. In particular, $\alpha = -4$, and $\beta + \gamma = 6$, $\delta + \epsilon + \zeta = -4$, and $\eta + \theta + \iota + \kappa + \lambda = 1$. This completely determines $\alpha$, and of the remaining $10$ unknowns, we have $3$ equations. When $a = b$ and $c=0$, the solutions should be $\pm 2\sqrt a$ and $0$. (Actually, two each of the first two and four of the last one.) Thus the polynomial should be $(z^2 - 4a)^2z^4$. This will determine $\alpha$ again, and give you three more equations, one of which is enough to determine $\beta,\gamma$. So we've paired the space of unknowns down to being $4$-dimensional. You can almost certainly continue in this way. For example, set $c=0$ and $b=4a$. Then $z = \pm3\sqrt a$ or $\pm \sqrt a$, each with multiplicity two, and so the polynomial in this case is $(z^2-a)^2(z^2-9a)^2$. I think this will give you two new equations when you combine it with what you already have, and I think it's enough to determine $p_3$. One final thing to mention: the coefficient on $z^0$ is always the product (up to a sign, which is $+1$ since there are an even number of solutions) of all the solutions. Well, the product of the solutions is $(\sqrt a + \sqrt b + \sqrt c)^2(\sqrt a - \sqrt b + \sqrt c)^2(\sqrt a - \sqrt b - \sqrt c)^2(\sqrt a + \sqrt b - \sqrt c)^2$, which is a version of the $n=2$ case. Doing this gives a different way to get $p_4$. Here's what I get, actually doing the arithmetic: $$z^8 - 4(a+b+c) z^6 + (2(a+b+c)^2 + 2(a^2+b^2+c^2))z^4 + (\frac43 (a+b+c)^3 - \frac83 (a^3+b^3+c^3))z^2 + (a^2 + b^2 + c^2 - 2(ab + bc + ac))^2$$ Note that the fractions in the coefficient on $z^2$ are strong evidence that I made an arithmetic mistake there, but I think the coefficients on $z^6$, $z^4$, and $z^0$ are correct. - I tried inserting the values $a=4$, $b=4$, $c=9$ and $z=7$ (as $\sqrt{4} + \sqrt{4} + \sqrt{9} = 7$) into your last equation, it did not result in $0$ so I think you are right that there is some arithmetic mistake (your other equation seems to be right though). But no worries I'll repeat the calculations myself, now that I understand your method. Thanks a lot for giving me a detailed answer. As $a$, $b$ and $c$ are just placeholders for more complicated expressions, I have a lot more work to do on my problem. – Sid Nov 15 '13 at 4:22 Just realized that I have a question: I am more interested in the order of $a, b$ and $c$ than in the order of $z$. Can we know what order we will get for those if we have $n$ terms on the left? – Sid Nov 15 '13 at 4:28 Or actually it seems to me like you have answered that question, sorry. – Sid Nov 15 '13 at 5:23	2,811	9,006	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.765625	4	CC-MAIN-2016-30	latest	en	0.928294
https://technodocbox.com/C_and_CPP/82077482-Upy14602-digital-electronics-and-microprocessors-lesson-plan.html	1,624,475,357,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623488539764.83/warc/CC-MAIN-20210623165014-20210623195014-00055.warc.gz	488,069,496	24,943	# UPY14602-DIGITAL ELECTRONICS AND MICROPROCESSORS Lesson Plan Size: px Start display at page: Transcription 1 UPY14602-DIGITAL ELECTRONICS AND MICROPROCESSORS Lesson Plan UNIT I - NUMBER SYSTEMS AND LOGIC GATES Introduction to decimal- Binary- Octal- Hexadecimal number systems-inter conversions-bcd code- Excess -3 code- Gray code One s complement and two s complements- Arithmetic operations- Addition- Subtraction- Multiplication and division- Basic and derived logic gates- Symbols and their truth tables- AND-OR-NOT- NAND- NOR- XOR- XNOR- Universal NAND and NOR gates-boolean algebra Basic laws of Boolean algebra De- Morgan s theorems- Reducing Boolean expressions using Boolean laws- SOP and POS forms of expressions- Min term and max terms- Karnaugh map simplification. 1 Introduction to decimal- Ability to understand the basic Binary- Octal- concepts of number systems. Hexadecimal number systems 2 Inter conversions Ability to understand the number system conversions 3 BCD code- Excess -3 Ability to understand the code code- Gray code Malvino A.P.and conversions 4 One s complement and Leach D.P.,Digital Ability to understand the arithmetic two s complements- Principles and operations Arithmetic operations- Applications, 4 th Addition- Subtraction Edition, McGraw 5 Multiplication and division Hill, Ability to understand the arithmetic operations 6 Basic and derived logic gates- Symbols and their various logic gates truth tables- AND-OR- NOT- NAND- NOR- XOR- XNOR. 7 Universal NAND and 2 NOR gates-boolean universal logic gates algebra Basic laws of Boolean algebra 8 De- Morgan s theorems theorems 9 Reducing Boolean various expressions using Boolean Boolean laws laws 10 SOP and POS forms of Ability to simplify logic expressions expressions- Min term and max terms 11 SOP and POS forms of expressions- Min term and max terms Ability to simplify logic expressions 12 Karnaugh map Ability to simplify logic expressions simplification. UNIT II - COMBINATIONAL LOGIC GATES Half and full adders- Half and full subtractors- Binary adders and subtractors- Two s complement adder/subtractor - Binary Coded Decimal (BCD) adder- Decoder-Encoder-Multiplexer- Demultiplexer-Analog to digital (A/D) conversion- Successive approximation- Digital to analog (D/A) conversion-r-2r ladder method. 13 Half and full adders 14 Half and full subtractors Malvino A.P.and 15 Binary adders and Leach D.P.,Digital subtractors Principles and 16 Two s complement Applications, 4 th adder/subtractor Edition, McGraw 17 Binary Coded Decimal Hill, (BCD) adder 18 Decoder-Encoder 19 Multiplexer 3 20 Demultiplexer 21 Analog to digital (A/D) Ability to understand ADC conversion 22 Successive approximation Ability to understand ADC 23 Digital to analog (D/A) Ability to understand DAC conversion 24 R-2R ladder method Ability to understand DAC UNIT III - SEQUENTIAL LOGIC SYSTEMS Flip flop-rs flip flop - Clocked RS flip flop-d flip flops JK flip flop - JK as master slave flip flops- Registers- Shift registers-shift left and Shift right registers- Counters-Synchronous and asynchronous counters-ripple counter-ring counter- Down counter Decade counter-.siso and SIPO Shift registers 25 Flip flop-rs flip flop - Clocked RS flip flop 26 D flip flops JK flip flop 27 JK as master slave flip flops 28 Registers- Shift registers- Shift left and Shift right Malvino A.P.and registers Leach D.P.,Digital 29 Counters-Synchronous Principles and counters Applications, 4 th 30 Asynchronous counters Edition, McGraw Hill, Ripple counter 32 Ring counter 33 Down counter 34 Decade Counter 4 35 SISO register 36 SIPO register UNIT IV - ARCHITECTURE AND PROGRAMMING OF MICROPROCESSOR Architecture of - Register organization of - Accumulator- General purpose Registers- Special purpose Registers - Bus structure (address, data and control buses) -Control signals-pin configuration of -Arithmetic and logic units-flags (zero, sign, parity, carry, auxiliary carry) - Addressing modes (register, Immediate, direct, indirect, implicit) of. 37 Architecture of various electrodes 38 Architecture of 39 Register organization of 40 Accumulator- General purpose Registers- Special purpose Registers 41 Bus structure (address, data and control buses) - Ramesh Goyankar, biosignal recording techniques 42 Control signals Pin configuration of Microprocessor Architecture- Programming and biosignal recording techniques 43 Pin configuration of Applications, 44 Arithmetic and logic units- Flags (zero, sign, parity, Prentice Hall, shock hazards carry, auxiliary carry) 45 Arithmetic and logic units- Flags (zero, sign, parity, carry, auxiliary carry) 46 Addressing modes (register, Immediate, direct, indirect, implicit) of 5 47 Addressing modes (register, Immediate, direct, indirect, implicit) of 48 Addressing modes (register, Immediate, direct, indirect, implicit) of UNIT V - INSTRUCTION SET OF MICROPROCESSOR Instruction Set-Types of instructions- Based on the number of bytes of operations-data transfer instructions - Arithmetic and logic instructions Branch instructions-subroutines-stack I/O instructions-machine cycle- Halt and Wait state- Timing diagram for opcode fetch- Memory read and write cycle Assembly language programming-simple programs using arithmetic and logic operations- Interrupts-Maskable and Non maskable interrupts. 49 Instruction Set-Types of instructions- Based on the various electrodes number of bytes of operations-data transfer instructions 50 Arithmetic and logic instructions Ramesh Goyankar, 51 Arithmetic and logic Microprocessor instructions Architecture- 52 Branch instructions Programming and Applications, 53 Subroutines Prentice Hall, biosignal recording techniques 54 Stack biosignal recording techniques 55 I/O instructions 56 Timing diagram for shock opcode fetch- Memory hazards read and write cycle 6 57 Assembly language programming-simple programs using arithmetic and logic operations 58 Assembly language programming-simple programs using arithmetic and logic operations 59 Interrupts-Maskable and Non maskable interrupts. 60 Interrupts-Maskable and Non maskable interrupts. Faculty Incharge HOD-Dept.Physics and Nanotechnology ### SRM ARTS AND SCIENCE COLLEGE SRM NAGAR, KATTANKULATHUR SRM ARTS AND SCIENCE COLLEGE SRM NAGAR, KATTANKULATHUR 603203 DEPARTMENT OF COMPUTER SCIENCE & APPLICATIONS LESSON PLAN (207-208) Course / Branch : B.Sc CS Total Hours : 50 Subject Name : Digital Electronics ### DIRECTORATE OF TECHNICAL EDUCATION DIPLOMA IN ELECTRICAL AND ELECTRONICS ENGINEERING II YEAR M SCHEME IV SEMESTER. DIRECTORATE OF TECHNICAL EDUCATION DIPLOMA IN ELECTRICAL AND ELECTRONICS ENGINEERING II YEAR M SCHEME IV SEMESTER 2015 2016 onwards DIGITAL ELECTRONICS CURRICULUM DEVELOPMENT CENTRE Curriculum Development ### END-TERM EXAMINATION (Please Write your Exam Roll No. immediately) END-TERM EXAMINATION DECEMBER 2006 Exam. Roll No... Exam Series code: 100919DEC06200963 Paper Code: MCA-103 Subject: Digital Electronics Time: 3 Hours Maximum ### CONTENTS CHAPTER 1: NUMBER SYSTEM. Foreword...(vii) Preface... (ix) Acknowledgement... (xi) About the Author...(xxiii) CONTENTS Foreword...(vii) Preface... (ix) Acknowledgement... (xi) About the Author...(xxiii) CHAPTER 1: NUMBER SYSTEM 1.1 Digital Electronics... 1 1.1.1 Introduction... 1 1.1.2 Advantages of Digital Systems... ### Principles of Digital Techniques PDT (17320) Assignment No State advantages of digital system over analog system. Assignment No. 1 1. State advantages of digital system over analog system. 2. Convert following numbers a. (138.56) 10 = (?) 2 = (?) 8 = (?) 16 b. (1110011.011) 2 = (?) 10 = (?) 8 = (?) 16 c. (3004.06) ### Scheme G. Sample Test Paper-I Sample Test Paper-I Marks : 25 Times:1 Hour 1. All questions are compulsory. 2. Illustrate your answers with neat sketches wherever necessary. 3. Figures to the right indicate full marks. 4. Assume suitable ### R10. II B. Tech I Semester, Supplementary Examinations, May SET - 1 1. a) Convert the following decimal numbers into an equivalent binary numbers. i) 53.625 ii) 4097.188 iii) 167 iv) 0.4475 b) Add the following numbers using 2 s complement method. i) -48 and +31 ### Injntu.com Injntu.com Injntu.com R16 1. a) What are the three methods of obtaining the 2 s complement of a given binary (3M) number? b) What do you mean by K-map? Name it advantages and disadvantages. (3M) c) Distinguish between a half-adder ### Digital logic fundamentals. Question Bank. Unit I Digital logic fundamentals Question Bank Subject Name : Digital Logic Fundamentals Subject code: CA102T Staff Name: R.Roseline Unit I 1. What is Number system? 2. Define binary logic. 3. Show how negative ### TEACHING & EXAMINATION SCHEME For the Examination COMPUTER SCIENCE. B.Sc. Part-I TEACHING & EXAMINATION SCHEME For the Examination -2015 COMPUTER SCIENCE THEORY B.Sc. 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(1M) With respect to digital signal explain the terms digits and bits.(2m) Also discuss active high and ### Philadelphia University Student Name: Student Number: Philadelphia University Student Name: Student Number: Faculty of Engineering Serial Number: Final Exam, First Semester: 2018/2019 Dept. of Computer Engineering Course Title: Logic Circuits Date: 03/01/2019 ### SIR C.R.REDDY COLLEGE OF ENGINEERING, ELURU DEPARTMENT OF INFORMATION TECHNOLOGY LESSON PLAN SIR C.R.REDDY COLLEGE OF ENGINEERING, ELURU DEPARTMENT OF INFORMATION TECHNOLOGY LESSON PLAN SUBJECT: CSE 2.1.6 DIGITAL LOGIC DESIGN CLASS: 2/4 B.Tech., I SEMESTER, A.Y.2017-18 INSTRUCTOR: Sri A.M.K.KANNA ### SUBJECT CODE: IT T35 DIGITAL SYSTEM DESIGN YEAR / SEM : 2 / 3 UNIT - I PART A (2 Marks) 1. Using Demorgan s theorem convert the following Boolean expression to an equivalent expression that has only OR and complement operations. Show the function can be implemented ### MGU-BCA-205- Second Sem- Core VI- Fundamentals of Digital Systems- MCQ s. 2. Why the decimal number system is also called as positional number system? MGU-BCA-205- Second Sem- Core VI- Fundamentals of Digital Systems- MCQ s Unit-1 Number Systems 1. What does a decimal number represents? A. Quality B. Quantity C. Position D. None of the above 2. Why the ### KING FAHD UNIVERSITY OF PETROLEUM & MINERALS COMPUTER ENGINEERING DEPARTMENT KING FAHD UNIVERSITY OF PETROLEUM & MINERALS COMPUTER ENGINEERING DEPARTMENT COE 202: Digital Logic Design Term 162 (Spring 2017) Instructor: Dr. Abdulaziz Barnawi Class time: U.T.R.: 11:00-11:50AM Class ### Hours / 100 Marks Seat No. 17333 13141 3 Hours / 100 Seat No. Instructions (1) All Questions are Compulsory. (2) Answer each next main Question on a new page. (3) Illustrate your answers with neat sketches wherever necessary. (4) ### (ii) Simplify and implement the following SOP function using NOR gates: DHANALAKSHMI COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING EE6301 DIGITAL LOGIC CIRCUITS UNIT I NUMBER SYSTEMS AND DIGITAL LOGIC FAMILIES PART A 1. How can an OR gate be ### Hours / 100 Marks Seat No. 17320 21718 3 Hours / 100 Seat No. Instructions (1) All Questions are Compulsory. (2) Answer each next main Question on a new page. (3) Figures to the right indicate full marks. (4) Assume suitable data, ### DIGITAL ELECTRONICS. P41l 3 HOURS UNIVERSITY OF SWAZILAND FACUL TY OF SCIENCE AND ENGINEERING DEPARTMENT OF PHYSICS MAIN EXAMINATION 2015/16 TITLE OF PAPER: COURSE NUMBER: TIME ALLOWED: INSTRUCTIONS: DIGITAL ELECTRONICS P41l 3 HOURS ANSWER ### R a) Simplify the logic functions from binary to seven segment display code converter (8M) b) Simplify the following using Tabular method SET - 1 1. a) Convert the decimal number 250.5 to base 3, base 4 b) Write and prove de-morgan laws c) Implement two input EX-OR gate from 2 to 1 multiplexer (3M) d) Write the demerits of PROM (3M) e) What ### 3. The high voltage level of a digital signal in positive logic is : a) 1 b) 0 c) either 1 or 0 1. The number of level in a digital signal is: a) one b) two c) four d) ten 2. A pure sine wave is : a) a digital signal b) analog signal c) can be digital or analog signal d) neither digital nor analog ### Code No: 07A3EC03 Set No. 1 Code No: 07A3EC03 Set No. 1 II B.Tech I Semester Regular Examinations, November 2008 SWITCHING THEORY AND LOGIC DESIGN ( Common to Electrical & Electronic Engineering, Electronics & Instrumentation Engineering, www.vidyarthiplus.com Question Paper Code : 31298 B.E./B.Tech. DEGREE EXAMINATION, NOVEMBER/DECEMBER 2013. Third Semester Computer Science and Engineering CS 2202/CS 34/EC 1206 A/10144 CS 303/080230012--DIGITAL ### SHRI ANGALAMMAN COLLEGE OF ENGINEERING. (An ISO 9001:2008 Certified Institution) SIRUGANOOR, TIRUCHIRAPPALLI SHRI ANGALAMMAN COLLEGE OF ENGINEERING AND TECHNOLOGY (An ISO 9001:2008 Certified Institution) SIRUGANOOR, TIRUCHIRAPPALLI 621 105 DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING EC1201 DIGITAL ### VALLIAMMAI ENGINEERING COLLEGE. SRM Nagar, Kattankulathur DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING EC6302 DIGITAL ELECTRONICS VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur-603 203 DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING EC6302 DIGITAL ELECTRONICS YEAR / SEMESTER: II / III ACADEMIC YEAR: 2015-2016 (ODD ### R07. Code No: V0423. II B. Tech II Semester, Supplementary Examinations, April SET - 1 II B. Tech II Semester, Supplementary Examinations, April - 2012 SWITCHING THEORY AND LOGIC DESIGN (Electronics and Communications Engineering) Time: 3 hours Max Marks: 80 Answer any FIVE Questions ### 1. Draw general diagram of computer showing different logical components (3) Tutorial 1 1. Draw general diagram of computer showing different logical components (3) 2. List at least three input devices (1.5) 3. List any three output devices (1.5) 4. Fill the blank cells of the ### Course Batch Semester Subject Code Subject Name. B.E-Marine Engineering B.E- ME-16 III UBEE307 Integrated Circuits Course Batch Semester Subject Code Subject Name B.E-Marine Engineering B.E- ME-16 III UBEE307 Integrated Circuits Part-A 1 Define De-Morgan's theorem. 2 Convert the following hexadecimal number to decimal ### UNIVERSITY POLYTECHNIC B.I.T., MESRA, RANCHI. COURSE STRUCTURE (W.E.F Batch Students) (Total Unit 7.5) Sessional Unit Code. Theory Unit Course COURSE STRUCTURE (W.E.F. 2011 Batch Students) (Total Unit 7.5) Course Theory Unit Course Sessional Unit Code Code DCP 4001 Data Structures 1.0 DCP 4002 Data Structures Lab. 0.5 DEC 4003 Electronics Circuits ### Code No: R Set No. 1 Code No: R059210504 Set No. 1 II B.Tech I Semester Regular Examinations, November 2006 DIGITAL LOGIC DESIGN ( Common to Computer Science & Engineering, Information Technology and Computer Science & Systems ### INTRODUCTION OF MICROPROCESSOR& INTERFACING DEVICES Introduction to Microprocessor Evolutions of Microprocessor Course Title Course Code MICROPROCESSOR & ASSEMBLY LANGUAGE PROGRAMMING DEC415 Lecture : Practical: 2 Course Credit Tutorial : 0 Total : 5 Course Learning Outcomes At end of the course, students will be ### B.Tech II Year I Semester (R13) Regular Examinations December 2014 DIGITAL LOGIC DESIGN B.Tech II Year I Semester () Regular Examinations December 2014 (Common to IT and CSE) (a) If 1010 2 + 10 2 = X 10, then X is ----- Write the first 9 decimal digits in base 3. (c) What is meant by don ### VALLIAMMAI ENGINEERING COLLEGE VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF INFORMATION TECHNOLOGY & COMPUTER SCIENCE AND ENGINEERING QUESTION BANK II SEMESTER CS6201- DIGITAL PRINCIPLE AND SYSTEM DESIGN ### Government of Karnataka Department of Technical Education Board of Technical Examinations, Bengaluru Government of Karnataka Department of Technical Education Board of Technical Examinations, Bengaluru Course Title: DIGITAL ELECTRONICS Course Code : 15EE34T Semester : III Course Group : Core Teaching INDEX Absorption law, 31, 38 Acyclic graph, 35 tree, 36 Addition operators, in VHDL (VHSIC hardware description language), 192 Algebraic division, 105 AND gate, 48 49 Antisymmetric, 34 Applicable input ### SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road QUESTION BANK (DESCRIPTIVE) SIDDHARTH GROUP OF INSTITUTIONS :: PUTTUR Siddharth Nagar, Narayanavanam Road 517583 QUESTION BANK (DESCRIPTIVE) Subject with Code : STLD(16EC402) Year & Sem: II-B.Tech & I-Sem Course & Branch: B.Tech ### INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad - 500 043 COMPUTER SCIENCE AND ENGINEERING TUTORIAL QUESTION BANK Name : DIGITAL LOGIC DESISN Code : AEC020 Class : B Tech III Semester ### II/IV B.Tech (Regular/Supplementary) DEGREE EXAMINATION. Answer ONE question from each unit. Hall Ticket Number: 14CS IT303 November, 2017 Third Semester Time: Three Hours Answer Question No.1 compulsorily. II/IV B.Tech (Regular/Supplementary) DEGREE EXAMINATION Common for CSE & IT Digital Logic ### Computer Logical Organization Tutorial Computer Logical Organization Tutorial COMPUTER LOGICAL ORGANIZATION TUTORIAL Simply Easy Learning by tutorialspoint.com tutorialspoint.com i ABOUT THE TUTORIAL Computer Logical Organization Tutorial Computer ### This tutorial gives a complete understanding on Computer Logical Organization starting from basic computer overview till its advanced architecture. About the Tutorial Computer Logical Organization refers to the level of abstraction above the digital logic level, but below the operating system level. At this level, the major components are functional ### VALLIAMMAI ENGINEERING COLLEGE VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203 DEPARTMENT OF INFORMATION TECHNOLOGY QUESTION BANK Academic Year 2018 19 III SEMESTER CS8351-DIGITAL PRINCIPLES AND SYSTEM DESIGN Regulation NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni-625531 Question Bank for the Units I to V SEMESTER BRANCH SUB CODE 3rd Semester B.E. / B.Tech. Electrical and Electronics Engineering ### Digital Logic Design Exercises. Assignment 1 Assignment 1 For Exercises 1-5, match the following numbers with their definition A Number Natural number C Integer number D Negative number E Rational number 1 A unit of an abstract mathematical system ### DHANALAKSHMI SRINIVASAN COLLEGE OF ENGINEERING AND TECHNOLOGY DHANALAKSHMI SRINIVASAN COLLEGE OF ENGINEERING AND TECHNOLOGY Dept/Sem: II CSE/03 DEPARTMENT OF ECE CS8351 DIGITAL PRINCIPLES AND SYSTEM DESIGN UNIT I BOOLEAN ALGEBRA AND LOGIC GATES PART A 1. How many ### Logic design Ibn Al Haitham collage /Computer science Eng. Sameer DEMORGAN'S THEOREMS One of DeMorgan's theorems stated as follows: The complement of a product of variables is equal to the sum of the complements of the variables. DeMorgan's second theorem is stated as ### COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: EC 1312 DIGITAL LOGIC CIRCUITS UNIT I KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING QUESTION BANK SUBJECT CODE & NAME: EC 1312 DIGITAL LOGIC CIRCUITS YEAR / SEM: III / V UNIT I NUMBER SYSTEM & BOOLEAN ALGEBRA ### ii) Do the following conversions: output is. (a) (101.10) 10 = (?) 2 i) Define X-NOR gate. (b) (10101) 2 = (?) Gray (2) /030832/31034 No. of Printed Pages : 4 Roll No.... rd 3 Sem. / ECE Subject : Digital Electronics - I SECTION-A Note: Very Short Answer type questions. Attempt any 15 parts. (15x2=30) Q.1 a) Define analog signal. b) ### 1. Mark the correct statement(s) 1. Mark the correct statement(s) 1.1 A theorem in Boolean algebra: a) Can easily be proved by e.g. logic induction b) Is a logical statement that is assumed to be true, c) Can be contradicted by another ### Code No: R Set No. 1 Code No: R059210504 Set No. 1 II B.Tech I Semester Supplementary Examinations, February 2007 DIGITAL LOGIC DESIGN ( Common to Computer Science & Engineering, Information Technology and Computer Science ### Code No: R Set No. 1 Code No: R059210504 Set No. 1 II B.Tech I Semester Regular Examinations, November 2007 DIGITAL LOGIC DESIGN ( Common to Computer Science & Engineering, Information Technology and Computer Science & Systems ### B.Sc II Year Computer Science (Optional) Swami Ramanand Teerth Marathwad University, Nanded B.Sc II Year Computer Science (Optional) (Semester Pattern) ( W.E.F. 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(b) Explain with an example the steps in subtraction of two n-digit unsigned numbers. ### Semester I. Semester I Code No. Name of the Paper Marks (Theory + CCE) FC Hindi 35+15 FC English 35+15 FC Development of Entrepreneurship 35+15 FC Udiyamita vikas 35+15 BCA 101 Computer Fundamentals 35+15 BCA 102 ### 1. INTRODUCTION TO MICROPROCESSOR AND MICROCOMPUTER ARCHITECTURE: 1. INTRODUCTION TO MICROPROCESSOR AND MICROCOMPUTER ARCHITECTURE: A microprocessor is a programmable electronics chip that has computing and decision making capabilities similar to central processing unit ### Architecture of 8085 microprocessor Architecture of 8085 microprocessor 8085 consists of various units and each unit performs its own functions. The various units of a microprocessor are listed below Accumulator Arithmetic and logic Unit ### Computer Architecture: Part III. 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What was ### APPENDIX A SHORT QUESTIONS AND ANSWERS APPENDIX A SHORT QUESTIONS AND ANSWERS Unit I Boolean Algebra and Logic Gates Part - A 1. Define binary logic? Binary logic consists of binary variables and logical operations. The variables are designated ### Written exam for IE1204/5 Digital Design Thursday 29/ Written exam for IE1204/5 Digital Design Thursday 29/10 2015 9.00-13.00 General Information Examiner: Ingo Sander. Teacher: William Sandqvist phone 08-7904487 Exam text does not have to be returned when ### ECE 331: N0. Professor Andrew Mason Michigan State University. Opening Remarks ECE 331: N0 ECE230 Review Professor Andrew Mason Michigan State University Spring 2013 1.1 Announcements Opening Remarks HW1 due next Mon Labs begin in week 4 No class next-next Mon MLK Day ECE230 Review	8,729	33,558	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.6875	4	CC-MAIN-2021-25	longest	en	0.781781
https://sciencedocbox.com/Physics/83447073-Algebra-i-staar-practice-test-a.html	1,632,310,230,000,000,000	text/html	crawl-data/CC-MAIN-2021-39/segments/1631780057347.80/warc/CC-MAIN-20210922102402-20210922132402-00230.warc.gz	555,611,752	27,206	# Algebra I STAAR Practice Test A Size: px Start display at page: Transcription 1 Algebra I STAAR Practice Test A 1 What is the value of if (, ) is a solution to the equation ? A C B 3 D 5 5 A plumber charges \$5 for a service call and \$15 per hour for repairs. She uses the graph below to help determine her costs for repairs that take varing amounts of time In the equation , what is the value for when equals 3? 0 0 Record our answer and fill in the bubbles on our answer document. O 6 8 Time (hours) Copright Pearson Education, Inc., or its affiliates. All rights reserved. 3 Which function best describes the data in the chart below? A 5 B C 5 D Sam mows lawns in the summer. His net profit for the summer, p, is represented b the equation p 5 15m 50, where m is the number of lawns he mows. Which is the best interpretation of this information? F Sam made a profit of \$35. G Sam made a profit of \$50. H Sam earns \$50 per lawn mowed. J It cost Sam \$50 to get his lawn mowing business started. After one ear, she decides to increase her initial service fee to \$30. She needs to update her graph. Which statement about her graph will be true? A The slope will be steeper. B The slope will be less steep. C The -intercept will move awa from the origin. D The -intercept will move toward the origin. 6 A dance studio charges \$35 per month plus an annual \$10 registration fee. Which equation represents the total cost, c, of the class for m months? F c 5 35m 10 G c 5 35m 1 10 H m 5 35c 10 J m 5 35c 1 10 STAAR Algebra I EOC Assessment Practice Algebra I STAAR Practice Test A 119 2 7 The quadratic function below models the profit a compan makes based on the quantit of items it sells to customers. Profit (hundreds) O 6 8 Number of Items Sold What is the greatest profit the compan can make based on the sale these items? A \$3.50 C \$3 B \$8 D \$0 10 Solve the equation e 5 1 e 5 e 1 3 e for e. F 7 H 8 7 G 7 8 J 5 11 Which of the following graphs would best represent the relationship between the diameter and the area of a circle? A Area Area vs. Diameter of a Circle Diameter 8 Tickets to a school soccer game are \$3 for adults and \$ for students. The equation models the number of students at the game when \$,050 was taken in from ticket sales. What does the slope of 3 mean in this situation? F The ratio of the number of adults at the game to the number of students at the game. G The ratio of the number of students at the game to the number of adults at the game. H The ratio of the cost of one adult ticket to the cost of one student ticket. J The ratio of the cost of one student ticket to the cost of one adult ticket. 9 The number of fish in a lake was 650 in Januar. In March, there were 570 fish. If the number of fish continues to decrease at this rate, what is a reasonable estimate for the number of fish remaining in September? A 90 C 330 B 0 D 90 B C D Area Area Area Area vs. Diameter of a Circle Diameter Area vs. Diameter of a Circle Diameter Area vs. Diameter of a Circle Diameter Copright Pearson Education, Inc., or its affiliates. All rights reserved. 10 Algebra I STAAR Practice Test A STAAR Algebra I EOC Assessment Practice 3 1 How will the graph of 5 5 be affected if the coefficient of becomes 5? F The graph will become wider. G The graph will become narrower. H The graph will be reflected across the -ais. J The graph will be reflected across the -ais. 13 At Colson s Discount Bookstore, paperbacks cost \$.50 each and hardbacks cost \$8.00 each. Which equation best describes the number of paperbacks, p, and hardbacks, h, that can be purchased for \$65.00? A 8p 1.5h 5 65 B.5p 1 8h 5 65 C 1.5(p 1 h) 5 65 D 65(p 1 h) A function is represented in the mapping diagram below Which of the following is NOT another correct representation of this function? A The range is {3, 5, 7, 9} and B The domain is {1,, 3, } and the range is {3, 5, 7, 9}. C is a positive integer less than 5 and D {(1, 3), (, 5), (3, 7), (, 9)} 16 Which describes the range of the function with the graph shown below? Copright Pearson Education, Inc., or its affiliates. All rights reserved. 1 The function f () represents a test score, f (), of a student studing hours for a test. The student studies for no more than 3 hours. What is the range of the function? F 0 # # 3 G 0 # # 60 H 0 # f () # 3 J 60 # f () # 90 F \$ 0 G \$ 5 H \$ 6 J \$ 6 6 O STAAR Algebra I EOC Assessment Practice Algebra I STAAR Practice Test A 11 4 17 Which equation best describes the graph below? A 5 3 B C 5 3 D O 18 What is the -intercept of the function f() 5 ( 5)? 0 For a fundraiser, the Math Club sold a ticket for each square ard of a football field. The randoml selected one square ard from the field to find out the winner of a \$5000 prize. If the function p() gives the profit from the sale, which statement best interprets the slope of this function? F The Math Club sold 5 tickets. G The Math Club sold 5000 tickets. H The Math Club sold tickets for \$5 each. J The Math Club sold tickets for \$5000 each. 1 A relation is given b the ordered pairs {(, 9), (, 0), (0, 3), (, ), (, 1)}. What is the domain of the relation? A {,,, 0,,, } B {,, 0,, } C all even integers and their opposites D all real numbers Record our answer and fill in the bubbles on our answer document. 19 Solve the equation for. A 5 B 5 C D 3 Copright Pearson Education, Inc., or its affiliates. All rights reserved. 1 Algebra I STAAR Practice Test A STAAR Algebra I EOC Assessment Practice 5 John plants trees for a living. He has weekl epenses of \$5 plus an additional \$35 for each tree. If he charges \$10 for each tree he plants, how man trees does he have to plant each week before he can make a profit? Record our answer and fill in the bubbles on our answer document. 3 The following scatter plot shows the relationship between the forearm length and shoe size of a group of mathematics students. Forearm Length vs. Shoe Size The function f () represents the total cost, f (), for tickets to a movie. Which statement is true if the function changes to f () ? F The total cost increased from \$9.75 to \$10.5. G The total cost decreased from \$10.5 to \$9.75. H The cost for each movie ticket increased from \$9.75 to \$10.5. J The cost for each movie ticket decreased from \$10.5 to \$9.75. Shoe Size (Men's US) Forearm Length (inches) 5 The line represented b the equation 5 1 is graphed below. O Copright Pearson Education, Inc., or its affiliates. All rights reserved. Based on these results, if Jim has a forearm length of 11 inches, which is the best estimate of his shoe size? A 1 C 6 B 8 D Which of the following best describes the change in the graph if the line becomes 5 1 1? A The -intercept increases. B The -intercept increases. C The -intercept decreases. D The -intercept decreases. STAAR Algebra I EOC Assessment Practice Algebra I STAAR Practice Test A 13 6 6 A model rocket was fired upward at an initial speed of 60 meters per second. The function h 5 60t.9t shows the relationship between the time elapsed and the rocket s height above the ground, where t is the time in seconds and h is the height in meters. The graph of this function is shown below. 7 A movie theater sold a total of 05 tickets to a movie, for a total of \$161. Adult tickets were \$9 and student tickets were \$5. What was the number of student tickets sold? Record our answer and fill in the bubbles on our answer document. Height (meters) Time (seconds) What is the best conclusion about the rocket s movement? F The speed of the rocket was greatest at about 6 seconds after it was fired. G The rocket traveled 180 meters. H The rocket reached its maimum height in about 6 seconds. J The rocket returned to the ground in about 6 seconds. 8 Which best describes the correlation shown in the scatterplot below? Liquid in Container (Gallons) O F absolute correlation G negative correlation H positive correlation J no correlation 6 8 Time (Das) Copright Pearson Education, Inc., or its affiliates. All rights reserved. 1 Algebra I STAAR Practice Test A STAAR Algebra I EOC Assessment Practice 7 9 Which equation best represents the area, A, of the triangle below? 1 31 What are the -intercepts of the graph of the equation ? A 5, 5 5 B 5, 5 5 C 5, 5 5 D 5, 5 5 A A 5 ( 1 ) 1 B A 5 ( 1 ) C A 5 1 ( 1 ) D A 5 ( 1 ) 3 Which best describes the effect of shifting the -intercept of this graph up 3 units, but keeping the -intercept the same? Copright Pearson Education, Inc., or its affiliates. All rights reserved. 30 A function is described b the equation f() The replacement set for the independent variable is {,, 6, 8}. Which set is the corresponding set for the dependent variable? F {, 8, 1, 16} G {,, 6, 8} H {1, 5, 9, 13} J {7, 11, 15, 19} O F The slope increases. G The slope decreases. H The slope does not change. J The slope changes sign. STAAR Algebra I EOC Assessment Practice Algebra I STAAR Practice Test A 15 8 33 The table below shows the relationship between the number of buses in the school parking lot and the number of students on those buses. Number of Number of Buses, Students, What does the slope of the line containing these points represent? A the number of students on an empt bus B the number of students each bus can hold C the number of buses used b the school D the number of students at the school 36 The quadratic epression 1 1 is modeled below using algebraic tiles. What are the solutions to the equation ? F 5 3 and 5 G 5 3 and 5 H 5 3 and 5 J 5 3 and 5 3 What are the roots of the quadratic equation ? F and 5 H and 5 G and 5 J and 5 35 Which tpe of function is graphed below? A quadratic B eponential C linear D logarithmic O 37 The distance Doug travels is directl related to the amount of time he travels. Doug travels 168 miles at a constant rate for 3 hours. How man hours will it take Doug to travel 39 miles at the same rate? Record our answer and fill in the bubbles on our answer document. Copright Pearson Education, Inc., or its affiliates. All rights reserved. 16 Algebra I STAAR Practice Test A STAAR Algebra I EOC Assessment Practice 9 38 What is the equation of the line that passes through the points (0, 3) and (, 0)? F G H J Jamie and Cnthia sold a total of 8 raffle tickets. Jamie sold 3 fewer than twice the number Cnthia sold. Which sstem of equations could be used to determine the number of raffle tickets sold b Jamie, j, and Cnthia, c? 1 Which equation best describes the relationship between hours of studing, h, and the grade on the test, g, in this table? Hours of Studing A g 5 6h 1 37 B h 5 6g 1 37 C g 5 1h 1 7 D h 5 1g 1 7 Grade on Test (%) Copright Pearson Education, Inc., or its affiliates. All rights reserved. A j 1 c 5 8 j 1 c 5 3 B j 1 c 5 8 j c 5 3 C j 1 c 5 8 j 1 3c 5 D j 1 c 5 8 j c If the domain of the function 5 1 is 0,, 5, what is the corresponding range? F 0,, 5 G,, 10 H 0,, 5 J,, 5 Simplif the algebraic epression ( 1 3)( 3) 1 3( 1 1 1). F G H J What is the solution to the sstem of equations shown below? A (1, 7) B (, 5) C (3, ) D (0, 1) STAAR Algebra I EOC Assessment Practice Algebra I STAAR Practice Test A 17 10 Which situation best describes the graph? 6 The graph below best represents which of the following situations? Distance Time F An airplane took off from the airport, then turned right, then turned left. G A car traveled at a constant rate, then stopped at a traffic light, then traveled at a constant rate. H Jake jogged at a constant rate, then stopped to take a break. J Ldia ascended up a mountain, then descended down the mountain. Altitude (miles) Time (hours) F An ant crawls into his anthill. G A person rides a roller coaster. H A famil drives through a mountainous area. J A racer runs a one-mile race around the track. 5 What is the first step in solving the inequalit 3 1 1, 11? A Add 11 to both sides of the inequalit. B Divide both sides of the inequalit b 3. C Subtract 1 from both sides of the inequalit. D Divide both sides of the inequalit b 3 and reverse the inequalit sign. 7 At a sporting goods store, the total cost of a baseball and a basketball is \$1.50. The cost of three baseballs and two basketballs is \$9.00. Which pair of equations can be used to determine the price of a baseball, b, and the price of a basketball, k? A b 1 k (b 1 k) B 3b 1 k b 1 k C b 1 k b 1 3k D b 1 k b 1 k Copright Pearson Education, Inc., or its affiliates. All rights reserved. 18 Algebra I STAAR Practice Test A STAAR Algebra I EOC Assessment Practice 11 8 Jane purchased cans of paint at \$ each and packs of markers at \$ each. She spent less than \$0, not including ta. Use the grid below to graph the inequalit 1, O 8 Which point represents a reasonable number of cans of paint and packs of markers that Jane purchased? 8 F (5, 3) H (3, 5) G (, 3) J (1, 5) 50 A function is represented in the mapping diagram below. 1 3 f() Which of the following is NOT another correct representation of this function? F The domain is {1,, 3, } and f() 5. G The domain is {1,, 3, } and f() 5. H is a negative integer greater than 5 and f() 5. J {(1, 1), (, ), (3, 9), (, 16)} Copright Pearson Education, Inc., or its affiliates. All rights reserved. 9 The equation represents the cost,, Jerem pas a plumber for hours of service. What function represents the cost, f() for hours of service? A f() B f() C f() D f() The quadratic function below models the path of a soccer ball as it is kicked up in the air. Height (Feet) O 1 3 Time (seconds) About how long does it take the ball to hit the ground? A 1.5 seconds B 3 seconds C 5 seconds D 0 seconds STAAR Algebra I EOC Assessment Practice Algebra I STAAR Practice Test A 19 12 5 What is m, the slope of the line that is shown in the graph below? 5 Which equation best represents the line on the graph? 8 6 O F 5 3 G 5 3 O 6 8 F 1 5 G 5 H 5 1 J 1 5 H 3 5 J The table shows the number of inches of accumulated snowfall, f(), after hours. Hours, Accumulated Snowfall (in inches), f() Can this situation be represented b a linear function? If so, what is the function? A no B es; f() C es; f() D es; f() Copright Pearson Education, Inc., or its affiliates. All rights reserved. STOP 130 Algebra I STAAR Practice Test A STAAR Algebra I EOC Assessment Practice ### Algebra I STAAR Practice Test B Algebra I STAAR Practice Test B 1 At a video store, used videos cost \$5.00 and new videos cost \$1.00. Which equation best describes the number of used, u, and new, n, videos that can be purchased for \$58.00? ### THIS IS A CLASS SET - DO NOT WRITE ON THIS PAPER THIS IS A CLASS SET - DO NOT WRITE ON THIS PAPER ALGEBRA EOC PRACTICE Which situation can be represented b =? A The number of eggs,, in dozen eggs for sale after dozen eggs are sold B The cost,, of buing ### REVIEW PACKET FOR END OF COURSE EXAM Math H REVIEW PACKET FOR END OF COURSE EXAM DO NOT WRITE ON PACKET! Do on binder paper, show support work. On this packet leave all fractional answers in improper fractional form (ecept where appropriate ### b(n) = 4n, where n represents the number of students in the class. What is the independent Which situation can be represented b =? A The number of eggs,, in dozen eggs for sale after dozen eggs are sold B The cost,, of buing movie tickets that sell for \$ each C The cost,, after a \$ discount, ### Standardized Test Practice Standardized Test Practice Name Date 1. Sallie makes candles using different scents. She sells each candle for \$8 on her website. She writes the equation 5 8 to determine how much mone she earns from candle ### A.7A Mini-Assessment LGER I (.) Linear Functions. The formulates equations and inequalities based on linear functions, uses a variet of methods to solve them, and analzes the solutions in terms of the situation. The student ### Evaluate: Domain and Range Name: Domain and Range Evaluate: Domain and Range 1. Kiera was driving in her neighborhood and approached a stop sign. When she applied the brakes, it took 4.5 seconds to come to a complete stop from a ### Algebra 1 Honors First Semester Review Permitted resources: Algebra 1 Honors First Semester Review TI-108 (or similar basic four function calculator) Algebra 1 and Geometr EOC Reference Sheet 4. Identif the mapping diagram that represents the ### Solve each system by graphing. Check your solution. y =-3x x + y = 5 y =-7 Practice Solving Sstems b Graphing Solve each sstem b graphing. Check our solution. 1. =- + 3 = - (1, ). = 1 - (, 1) =-3 + 5 3. = 3 + + = 1 (, 3). =-5 = - 7. = 3-5 3 - = 0 (1, 5) 5. -3 + = 5 =-7 (, 7). Released Form READY NCEXTEND2 End-of-Grade Alternate Assessment Mathematics Grade 8 Student Booklet Academic Services and Instructional Support Division of Accountabilit Services Copright 2013 b the North ### Ready To Go On? Skills Intervention 5-1 Using Transformations to Graph Quadratic Functions Read To Go On? Skills Intervention 5-1 Using Transformations to Graph Quadratic Functions Find these vocabular words in Lesson 5-1 and the Multilingual Glossar. Vocabular quadratic function parabola verte ### Algebra I Practice Exam Algebra I This practice assessment represents selected TEKS student expectations for each reporting category. These questions do not represent all the student expectations eligible for assessment. Copyright ### TAKS Mathematics. Test A GO ON. 1. Which of the following functions is not linear? A. f(x) 3x 4 B. f(x) 3 x 4 C. f(x) 3 4 TAKS Mathematics Test A. Which of the following functions is not linear? A. f() B. f() C. f() D. f(). Mia bought nuts and popcorn for a group of friends at the movie theater. A bag of nuts cost \$. and ### c. Find the slope and y-intercept of the graph of the linear equation. Then sketch its graph. Name Solve. End-of-Course. 7 =. 5 c =. One cell phone plan charges \$0 per month plus \$0.5 per minute used. A second cell phone plan charges \$5 per month plus \$0.0 per minute used. Write and solve an equation ### MATH GRADE 8 UNIT 4 LINEAR RELATIONSHIPS ANSWERS FOR EXERCISES. Copyright 2015 Pearson Education, Inc. 51 MATH GRADE 8 UNIT LINEAR RELATIONSHIPS FOR EXERCISES Copright Pearson Education, Inc. Grade 8 Unit : Linear Relationships LESSON : MODELING RUNNING SPEEDS 8.EE.. A Runner A 8.EE.. D sec 8.EE.. D. m/sec ### A calculator may be used on the exam. The Algebra Semester A eamination has the following tpes of questions: Selected Response Student Produced Response (Grid-in) Brief Constructed Response (BCR) Etended Constructed Response (ECR) Short Answer ### 3. Find the area of each rectangle shown below. 4. Simplify the expressions below. 5. If the expression 3a 2 9. is equal to 3, what is the value of d? Permitted resources: 2018 2019 Algebra 1 Midterm Review FSA Approved calculator Algebra 1 FSA Reference Sheet 1. The expression 13x + 5 represents the number of marbles you have after purchasing 13 bags ### Algebra I Final Study Guide 2011-2012 Algebra I Final Study Guide Short Answer Source: www.cityoforlando.net/public_works/stormwater/rain/rainfall.htm 1. For which one month period was the rate of change in rainfall amounts in Orlando ### Analytic Geometry 300 UNIT 9 ANALYTIC GEOMETRY. An air traffi c controller uses algebra and geometry to help airplanes get from one point to another. UNIT 9 Analtic Geometr An air traffi c controller uses algebra and geometr to help airplanes get from one point to another. 00 UNIT 9 ANALYTIC GEOMETRY Copright 00, K Inc. All rights reserved. This material ### Algebra 1 End-of-Course Assessment Practice Test with Solutions Algebra 1 End-of-Course Assessment Practice Test with Solutions For Multiple Choice Items, circle the correct response. For Fill-in Response Items, write your answer in the box provided, placing one digit ### Name. 1. Given the solution (3, y), what is the value of y if x + y = 6? 7. The graph of y = x 2 is shown below. A. 3 B. 4 C. 5 D. Name 1. Given the solution (, y), what is the value of y if x + y = 6? 7. The graph of y = x is shown below. 5 D. 6. What are the solutions to the equation x - x = 0? x = - or x = - x = - or x = 1 x = ### 26 Questions EOC Review #1 EOC REVIEW Name Period 6 Questions EOC Review # EOC REVIEW Solve each: Give the BEST Answer. You may use a graphing calculator.. Which quadrant contains the verte of the following: f ( ) 8 st nd rd d. 4th. What type ### ALGEBRA 1 SEMESTER 1 INSTRUCTIONAL MATERIALS Courses: Algebra 1 S1 (#2201) and Foundations in Algebra 1 S1 (#7769) Multiple Choice: Identify the choice that best completes the statement or answers the question. 1. Ramal goes to the grocery store and buys pounds of apples and pounds of bananas. Apples cost dollars per ### Lesson Master 6-1B. USES Objective E. Questions on SPUR Objectives. In 1 5, use the chart showing the percent of households that had a computer. Back to Lesson 6-1 6-1B USES Objective E In 1 5, use the chart showing the percent of households that had a computer. Year 1989 1993 1997 2001 Percent of Households 15.0 22.8 36.6 56.3 1. Make a line graph ### NAME DATE PERIOD. Graphing Equations in Slope-Intercept Form NAME DATE PERID 4-1 Skills Practice Graphing Equations in Slope-Intercept Form Write an equation of a line in slope-intercept form with the given slope and -intercept. 1. slope: 5, -intercept: -3. slope: ### Using Graphs to Relate Two Quantities - Think About a Plan Using Graphs to Relate Two Quantities Skiing Sketch a graph of each situation. Are the graphs the same? Explain. a. your speed as you travel from the bottom of a ski slope to the top ### Using Graphs to Relate Two Quantities - Using Graphs to Relate Two Quantities For Eercises, choose the correct letter.. The graph shows our distance from the practice field as ou go home after practice. You received a ride from a friend back ### An assessment of these skills will occur the first week of school. Fort Zumwalt West High School Pre-AP Algebra SUMMER REVIEW PACKET Name: * This packet is to be handed in to our Pre AP Algebra teacher on the first da of the school ear. *All work must be shown in the ### c. x x < 60 d. x x =9. What are the first four terms of the sequence? a. 12, 21, 30, 39 b. Algebra I Unit Reasoning with Linear Equations and Inequalities Post Test... A famil s cell phone plan costs \$ per month for, minutes and cents per minute over the limit. This month, the famil paid \$.. ### Semester 1 Final REVIEW Algebra Name s dc0p[8\ MKkuvtXaI esmoifntrw\araef GLSLFCA.l F pa\lglb ArNiigphHtsT qrievsneprqvlevdn. Semester Final REVIEW Evaluate each using the values given. ) z + ; use = -, and z = - ) ( - z) ; use ### 60 Minutes 60 Questions MTHEMTI TET 60 Minutes 60 Questions DIRETIN: olve each problem, choose the correct answer, and then fill in the corresponding oval on our answer document. Do not linger over problems that take too much ### Using a Graphing Calculator Using a Graphing Calculator Unit 1 Assignments Bridge to Geometry Name Date Period Warm Ups Name Period Date Friday Directions: Today s Date Tuesday Directions: Today s Date Wednesday Directions: Today ### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Eam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A ceramics workshop makes serving bowls, platters, and bread baskets to sell at its Winter ### 3. Find the area for each question below. a. (3x 2)(2x + 5) b. 4. Simplify the expressions below. is equal to 1, what is the value of a? Permitted resources: 2018 2019 Algebra 1 Midterm Review FSA Approved calculator Algebra 1 FSA Reference Sheet 1. The expression 13x + 5 represents the number of marbles you have after shopping at the game ### HW: Scatter Plots. 1. The scatter plot below shows the average traffic volume and average vehicle speed on a certain freeway for 50 days in 1999. Name: Date: 1. The scatter plot below shows the average traffic volume and average vehicle speed on a certain freeway for 50 days in 1999. 2. Ms. Ochoa recorded the age and shoe size of each student in ### Intermediate Algebra / MAT 135 Spring 2017 Master ( Master Templates) Test 1 Review #1 Intermediate Algebra / MAT 135 Spring 017 Master ( Master Templates) Student Name/ID: 1. Solve for. = 8 18. Solve for. = + a b 3. Solve for. a b = L 30. Two trains leave stations miles ### Test 1 Review #5. Intermediate Algebra / MAT 135 Fall 2016 Master (Prof. Fleischner) Test 1 Review #5 Intermediate Algebra / MAT 135 Fall 016 Master (Prof. Fleischner) Student Name/ID: 1. Solve for n. d = m + 9n. Solve for b. r = 5 b + a 3. Solve for C. A = 7 8 B + C ALEKS Test 1 Review ### (c) ( 5) 2. (d) 3. (c) 3(5 7) 2 6(3) (d) (9 13) ( 3) Question 4. Multiply using the distributive property and collect like terms if possible. Name: Chapter 1 Question 1. Evaluate the following epressions. (a) 5 (c) ( 5) (b) 5 (d) ( 1 ) 3 3 Question. Evaluate the following epressions. (a) 0 5() 3 4 (c) 3(5 7) 6(3) (b) 9 + (8 5) (d) (9 13) + 15 ### (TPP #3) Test Preparation Practice. Algebra Holt Algebra 1. Name Date Class Test Preparation Practice Algebra 1 Solve each problem. Choose the best answer for each question and record our answer on the Student Answer Sheet. Figures are not drawn to scale 1. Jack budgets \$35 for ### Statistics, Data Analysis, and Probability Statistics, Data Analsis, and Probabilit. The number of classic books Nanette sells in her bookshop varies according to the time of ear, as shown in the scatterplot below. Classic Book Sales Number of ### Math 1 Semester 1 Final Review Math Semester Final Review 0-0. Solve the equation (x + ) = 9. Write a justification for each step. Using these possible reasons: Substitution, Division, Subtraction, Addition, and Reflexive Properties ### CCSD Practice Proficiency Exam Fall B. c = E . Which epression represents the n th term of the sequence, 9, 7, 8,,? n n n 6. The chart below shows the number of passenger cars sold b two companies in June 008. Passenger Car Sales in June 008 Compan ### Name: Period: QVMS GTA FALL FINAL EXAM REVIEW PRE-AP ALGEBRA 1 Name: Period: QVMS GTA FALL FINAL EXAM REVIEW PRE-AP ALGEBRA ) When simplifing an epression, ou perform operations inside grouping smbols first. a. alwas b. sometimes c. never ) The opposite of a negative ### The semester A examination for Bridge to Algebra 2 consists of two parts. Part 1 is selected response; Part 2 is short answer. The semester A eamination for Bridge to Algebra 2 consists of two parts. Part 1 is selected response; Part 2 is short answer. Students ma use a calculator. If a calculator is used to find points on a graph, ### Bridge-Thickness Experiment. Student 2 Applications 1. Below are some results from the bridge-thickness eperiment. Bridge-Thickness Eperiment Thickness (laers) Breaking Weight (pennies) 15 5 5 a. Plot the (thickness, breaking weight) data. ### MATH 1710 College Algebra Final Exam Review MATH 7 College Algebra Final Eam Review MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. ) There were 80 people at a pla. The admission price was \$ ### A calculator may be used on the exam. The Algebra Semester A eamination will have the following tpes of questions: Selected Response Student Produced Response (Grid-in) Brief Constructed Response (BCR) Etended Constructed Response (ECR) Short ### Graph Quadratic Functions in Standard Form TEKS 4. 2A.4.A, 2A.4.B, 2A.6.B, 2A.8.A Graph Quadratic Functions in Standard Form Before You graphed linear functions. Now You will graph quadratic functions. Wh? So ou can model sports revenue, as in ### 4. Based on the table below, what is the joint relative frequency of the people surveyed who do not have a job and have a savings account? Name: Period: Date: Algebra 1 Common Semester 1 Final Review 1. How many surveyed do not like PS4 and do not like X-Box? 2. What percent of people surveyed like the X-Box, but not the PS4? 3. What is the ### EOC FSA Practice Test. Algebra 1. Calculator Portion EOC FSA Practice Test Algebra 1 Calculator Portion FSA Mathematics Reference Sheets Packet Algebra 1 EOC FSA Mathematics Reference Sheet Customary Conversions 1 foot = 12 inches 1 yard = 3 feet 1 mile ### 4. Based on the table below, what is the joint relative frequency of the people surveyed who do not have a job and have a savings account? Name: Period: Date: Algebra 1 Common Semester 1 Final Review Like PS4 1. How many surveyed do not like PS4 and do not like X-Box? 2. What percent of people surveyed like the X-Box, but not the PS4? 3. ### Skills Practice Skills Practice for Lesson 5.1 Skills Practice Skills Practice for Lesson. Name Date Widgets, Dumbbells, and Dumpsters Multiple Representations of Linear Functions Vocabular Write the term that best completes each statement.. A(n) is ### 8.4. If we let x denote the number of gallons pumped, then the price y in dollars can \$ \$1.70 \$ \$1.70 \$ \$1.70 \$ \$1. 8.4 An Introduction to Functions: Linear Functions, Applications, and Models We often describe one quantit in terms of another; for eample, the growth of a plant is related to the amount of light it receives, ### Intermediate Algebra. Exam 1 Review (Chapters 1, 2, and 3) Eam Review (Chapters,, and ) Intermediate Algebra Name. Epress the set in roster form. { N and 7}. Epress the set in set builder form. {-, 0,,,, }. Epress in set builder notation each set of numbers that ### Fair Game Review. Chapter of a mile the next day. How. far will you jog over the next two days? How many servings does the Name Date Chapter Evaluate the epression.. Fair Game Review 5 +. 3 3 7 3 8 4 3. 4 4. + 5 0 5 6 5. 3 6. 4 6 5 4 6 3 7. 5 8. 3 9 8 4 3 5 9. You plan to jog 3 4 of a mile tomorrow and 7 8 of a mile the net ### Part I. Hours Part I Answer all questions in this part. Each correct answer will receive credits. No partial credit will be allowed. For each question, write on the separate answer sheet the numeral preceding the word ### Name Date PD. Systems of Equations and Inequalities Name Date PD Sstems of Equations and Inequalities Sstems of Equations Vocabular: A sstem of linear equations is A solution of a sstem of linear equations is Points of Intersection (POI) are the same thing ### Fair Game Review. Chapter 5. Input, x Output, y. 1. Input, x Output, y. Describe the pattern of inputs x and outputs y. Name Date Chapter Fair Game Review Describe the pattern of inputs and outputs.. Input, utput,. 8 Input, utput,. Input, 9. utput, 8 Input, utput, 9. The table shows the number of customers in hours. Describe ### Name % Correct % MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1 Pre Test Unit Name % Correct % MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Decide if the given number is a solution to the given equation. 1) ### Summary and Vocabulary Chapter Chapter Summar and Vocabular Equations involving percents ma be written in the form p q = r, where p is the decimal form of the percent, q is the initial quantit, and r is the resulting quantit. ### Algebra 2 Semester Exam Review Algebra Semester Eam Review 7 Graph the numbers,,,, and 0 on a number line Identif the propert shown rs rs r when r and s Evaluate What is the value of k k when k? Simplif the epression 7 7 Solve the equation ### Algebra 2 Unit 1 Practice Algebra Unit Practice LESSON - Use this information for Items. Aaron has \$ to rent a bike in the cit. It costs \$ per hour to rent a bike. The additional fee for a helmet is \$ for the entire ride.. Write ### Name: Class: Date: Describe a pattern in each sequence. What are the next two terms of each sequence? Class: Date: Unit 3 Practice Test Describe a pattern in each sequence. What are the next two terms of each sequence? 1. 24, 22, 20, 18,... Tell whether the sequence is arithmetic. If it is, what is the ### Review: Expressions and Equations Review: Expressions and Equations Expressions Order of Operations Combine Like Terms Distributive Property Equations & Inequalities Graphs and Tables Independent/Dependent Variables Constant: a number ### Name Class Date. Pearson Education, Inc., publishing as Pearson Prentice Hall. All rights reserved. Travels in Air. Distance (miles) Time (seconds) Practice - Rate of Change and Slope Find the slope of each line.... O O O Find the slope of the line that passes through each pair of points.. (, ), (7, 5) 5. (8, ), (, ). (, 5), (, 7) 7. (-, 7), (, -) ### A C B D. 2. Which table corresponds to the equation y = 3x 2? A B C D. 3. Which function table represents the equation y = 2x + 1? 1. Jessie will be going on a hiking trip and plans to leave her cat at Pet Palace while she is awa. If Pet Palace charges an initial \$20 registration fee and \$7 per da of care, which table BEST represents ### PA CORE 8 UNIT 3 - FUNCTIONS FLEX Workbook PA CORE 8 UNIT - FUNCTIONS FLEX Workbook LESSON 9. INTRODUCTIONS TO FUNCTIONS 0. WORK WITH LINEAR FUNCTIONS. USE FUNCTIONS TO SOLVE PROBLEMS. USE GRAPHS TO DESCRIBE RELATIONSHIPS. COMPARE RELATIONSHIPS ### UNIT 5 INEQUALITIES CCM6+/7+ Name: Math Teacher: UNIT 5 INEQUALITIES 2015-2016 CCM6+/7+ Name: Math Teacher: Topic(s) Page(s) Unit 5 Vocabulary 2 Writing and Graphing Inequalities 3 8 Solving One-Step Inequalities 9 15 Solving Multi-Step Inequalities ### Algebra 1 STAAR Review Name: Date: Algebra 1 STAAR Review Name: Date: 1. Which graph does not represent y as a function of x? I. II. III. A) I only B) II only C) III only D) I and III E) I and II 2. Which expression is equivalent to? 3. ### Algebra - Chapter 5 Review Name Hour Algebra - Chapter 5 Review 1. Write an equation in slope-intercept form of the graph. 10 y 10 x 10 2. The cost of a school banquet is \$95 plus \$15 for each person attending. Write an equation ### CORE. Chapter 3: Interacting Linear Functions, Linear Systems. 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Property of the Virginia Department of Education VIRGINIA STANDARDS OF LEARNING Spring 009 Released Test GRADE 8 MATHEMATICS Form M09, CORE Propert of the Virginia Department of Education 009 b the Commonwealth of Virginia, Department of Education, P.O. ### Mini-Lecture 8.1 Solving Quadratic Equations by Completing the Square Mini-Lecture 8.1 Solving Quadratic Equations b Completing the Square Learning Objectives: 1. Use the square root propert to solve quadratic equations.. Solve quadratic equations b completing the square. ### West Campus State Math Competency Test Info and Practice West Campus State Math Competenc Test Info and Practice Question Page Skill A Simplif using order of operations (No grouping/no eponents) A Simplif using order of operations (With grouping and eponents) ### GRADE 8 TEKS/STAAR Spiraled Practice 1-20 Six Weeks 1 TEKSING TOWARD STAAR MATHEMATICS GRADE TEKS/STAAR Spiraled Practice -0 Si Weeks TEKSING TOWARD STAAR 0 TEKS/STAAR SPIRALED PRACTICE. If the area of a square is units, on which number line does S represents ### 1.1. Use a Problem Solving Plan. Read a problem and make a plan. Goal p Use a problem solving plan to solve problems. VOCABULARY. Formula. . Georgia Performance Standard(s) MMPd, MMPa Your Notes Use a Problem Solving Plan Goal p Use a problem solving plan to solve problems. VOCABULARY Formula A PROBLEM SOLVING PLAN Step Read the problem carefull. ### Ready To Go On? Skills Intervention 2-1 Solving Linear Equations and Inequalities A Read To Go n? Skills Intervention -1 Solving Linear Equations and Inequalities Find these vocabular words in Lesson -1 and the Multilingual Glossar. Vocabular equation solution of an equation linear ### 2014 TEXAS STAAR TEST END OF COURSE ALGEBRA I 0 TEXAS STAAR TEST END OF COURSE ALGEBRA I Total Possible Score: 5 Needed Correct to Pass: 0 Advanced Performance: Time Limit: Hours This file contains the State of Teas Assessments of Academic Readiness ### Practice EOC Questions Practice EOC Questions 1 One of the events at a high school track meet is the pole vault. The pole vaulter runs toward the crossbar and uses a pole to attempt to vault over the bar. Josh collects data Math Test Calculator 5 M INUTES, QUE S TI ON S Turn to Section of our answer sheet to answer the questions in this section. For questions -7, / +5 solve each problem, choose the best answer from the choices ### Algebra Final Exam Review Semester 1 Name: Hour: Algebra Final Exam Review Semester. Evaluate: when (P98 #7) 6. TOLL ROADS A toll road charges trucks a toll of \$ per axle. Write an expression for the total toll for a truck. (P #6). DVD STORAGE ### Algebra 1 Midterm Name Algebra 1 Midterm 01-013 Name Stud Guide Date: Period: THIS REVIEW WILL BE COLLECTED JANUARY 4 th or Januar 7 th. One problem each page will be graded! You ma not turn this eam review in after the due ### Key Focus #6 - Finding the Slope of a Line Between Two Points. Ke Focus #6 - Finding the Slope of a Line Between Two Points. Given the following equations of lines, find the SLOPES of the lines: = + 6... + 8 = 7 9 - = 7 - - 9 = 4.. 6. = 9-8 - = + 7 = 4-9 7. 8. 9.. ### Topic 1: Writing and Solving Equations and Inequalities Topic 1: Writing and Solving Equations and Inequalities In #1 3, solve each equation. Use inverse operations. 1. 8 21 5 = 15 2. 3 10 = 2(4 5) 3. 2( + 2) = 2 + 1 4. The rectangle and square have equivalent ### TEST REVIEW QUADRATICS EQUATIONS Name: 2. Which of the following statements is true about the graph of the function? Chapter MATHEMATICS 00 TEST REVIEW QUADRATICS EQUATIONS Name:. Which equation does not represent a quadratic function?. Which of the following statements is true about the graph of the function? it has 3. Graph Quadratic Functions in Standard Form Georgia Performance Standard(s) MMA3b, MMA3c Goal p Use intervals of increase and decrease to understand average rates of change of quadratic functions. Your ### MATH 103 Sample Final Exam Review MATH 0 Sample Final Eam Review This review is a collection of sample questions used b instructors of this course at Missouri State Universit. It contains a sampling of problems representing the material ### Algebra 1 Fall Semester Final Review Name It is very important that you review for the Algebra Final. Here are a few pieces of information you want to know. Your Final is worth 20% of your overall grade The final covers concepts from the entire ### Algebra 2 Unit 2 Practice Algebra Unit Practice LESSON 7-1 1. Consider a rectangle that has a perimeter of 80 cm. a. Write a function A(l) that represents the area of the rectangle with length l.. A rectangle has a perimeter of ### North Carolina Community College System Diagnostic and Placement Test Sample Questions North Carolina Communit College Sstem Diagnostic and Placement Test Sample Questions 0 The College Board. College Board, ACCUPLACER, WritePlacer and the acorn logo are registered trademarks of the College	10,654	39,540	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.90625	4	CC-MAIN-2021-39	latest	en	0.89597
http://mathhelpforum.com/number-theory/203024-prime-conjecture.html	1,529,865,350,000,000,000	text/html	crawl-data/CC-MAIN-2018-26/segments/1529267867050.73/warc/CC-MAIN-20180624180240-20180624200240-00444.warc.gz	204,327,100	14,479	1. ## Prime Conjecture Is the following conjecture true or false? If m is a positive odd integer such that 2m = 2 (mod. m(m-1)) then m is a prime. 27 = 2 (mod. 7x6). 243 = 2 (mod. 43x42). 2. ## Re: Prime Conjecture False. For example, let m = 9. EDIT: Sorry, m = 9 doesn't negate the statement. Ignore this... 3. ## Re: Prime Conjecture Originally Posted by richard1234 False. For example, let m = 9. 2^9 = 8 (mod.72). Hence your counterexample is false. Also, if 2m = 2 (mod. m(m-1)) then m = 3 (mod. 4). 4. ## Re: Prime Conjecture The conjecture is only true for m=3 and false for m<>3...even de Fermat was incompetent to discover it! 5. ## Re: Prime Conjecture Hence, there exists $\displaystyle k\in \mathbb{Z}$ such that $\displaystyle 2^m=2+m(m-1)k$. So consider the following manipulations: $\displaystyle 2(2^{m-1}-1)=m(m-1)k$ $\displaystyle 2^{m-1}-1 = \frac{m(m-1)}{2}k$ Therefore, this is true when: $\displaystyle \displaymode 2^{m-1}=1 (\text{mod } \sum_{i=1}^{m-1}(i))$ I dunno if that helps, but it seems like it might be a way to play with the problem. Edit: $\displaystyle \displaymode \sum_{i=0}^{m-2}2^i=0\text{ } (\text{mod }\sum_{j=0}^{m-1}{j})$ 6. ## Re: Prime Conjecture The conjecture is true for m = 7, 43 and more. But not for all primes = 3 (mod.4). For 231 = 2 (mod.31) but 231 = 8 (mod. 30). Is the following proof valid? Theorem. If m is a positive odd integer such that 2m ≡ 2 (mod. m(m-1)) then m ≡ 3 (mod. 4) and m is a prime. Proof. Let 2m ≡ 2 (mod. m(m-1)) then 2m - 2 =km(m-1) for some positive integer k. This gives 2m-1 – 1 = km(m-1)/2 and since 2m-1 – 1 is odd, (m-1)/2 is odd implying m ≡ 3 (mod. 4). Let a be a positive integer and p a prime such that (pa+1 , (m-1)/2) = pa. Then 2m-1 ≡ 1 (mod. pa). This implies pa divides phi(m). But this is true for all such primes p dividing (m-1)/2 and hece for their product. This implies (m-1)/2 divides phi(m) and therefore phi(m) = m - 1 Hence, m is a prime. 7. ## Re: Prime Conjecture Originally Posted by Stan [FONT=arial][SIZE=3]Is the following conjecture true or false? If m is a positive odd integer such that 2m = 2 (mod. m(m-1)) then m is a prime. m<184900 values for which the conjecture is true: m={3, 7, 19, 43, 127, 163, 379, 487, 883, 1459, 2647, 3079, 3943, 5419, 9199, 11827, 14407, 16759, 18523, 24967, 26407, 37339, 39367, 42463, 71443, 77659, 95923, 99079, 113779, 117307, 143263, 174763, 175447, 184843} 8. ## Re: Prime Conjecture Originally Posted by MaxJasper m<184900 values for which the conjecture is true: $\displaystyle m=\{3, 7, 19, 43, 127, 163, 379, 487, 883, 1459, 2647, 3079, 3943, 5419, 9199, 11827, 14407, 16759, 18523, 24967, 26407, 37339, 39367, 42463, 71443, 77659, 95923, 99079, 113779, 117307, 143263, 174763, 175447, 184843\}$ Does the theorem above hold true or is it erroneous? 9. ## Re: Prime Conjecture Originally Posted by Stan Does the theorem above hold true or is it erroneous? m<2000 for which the conjecture is FALSE: (resulting non-Primes) { 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, \ 69, 75, 77, 81, 85, 87, 91, 93, 95, 99, 105, 111, 115, 117, 119, 121, \ 123, 125, 129, 133, 135, 141, 143, 145, 147, 153, 155, 159, 161, 165, \ 169, 171, 175, 177, 183, 185, 187, 189, 195, 201, 203, 205, 207, 209, \ 213, 215, 217, 219, 221, 225, 231, 235, 237, 243, 245, 247, 249, 253, \ 255, 259, 261, 265, 267, 273, 275, 279, 285, 287, 289, 291, 295, 297, \ 299, 301, 303, 305, 309, 315, 319, 321, 323, 325, 327, 329, 333, 335, \ 339, 341, 343, 345, 351, 355, 357, 361, 363, 365, 369, 371, 375, 377, \ 381, 385, 387, 391, 393, 395, 399, 403, 405, 407, 411, 413, 415, 417, \ 423, 425, 427, 429, 435, 437, 441, 445, 447, 451, 453, 455, 459, 465, \ 469, 471, 473, 475, 477, 481, 483, 485, 489, 493, 495, 497, 501, 505, \ 507, 511, 513, 515, 517, 519, 525, 527, 529, 531, 533, 535, 537, 539, \ 543, 545, 549, 551, 553, 555, 559, 561, 565, 567, 573, 575, 579, 581, \ 583, 585, 589, 591, 595, 597, 603, 605, 609, 611, 615, 621, 623, 625, \ 627, 629, 633, 635, 637, 639, 645, 649, 651, 655, 657, 663, 665, 667, \ 669, 671, 675, 679, 681, 685, 687, 689, 693, 695, 697, 699, 703, 705, \ 707, 711, 713, 715, 717, 721, 723, 725, 729, 731, 735, 737, 741, 745, \ 747, 749, 753, 755, 759, 763, 765, 767, 771, 775, 777, 779, 781, 783, \ 785, 789, 791, 793, 795, 799, 801, 803, 805, 807, 813, 815, 817, 819, \ 825, 831, 833, 835, 837, 841, 843, 845, 847, 849, 851, 855, 861, 865, \ 867, 869, 871, 873, 875, 879, 885, 889, 891, 893, 895, 897, 899, 901, \ 903, 905, 909, 913, 915, 917, 921, 923, 925, 927, 931, 933, 935, 939, \ 943, 945, 949, 951, 955, 957, 959, 961, 963, 965, 969, 973, 975, 979, \ 981, 985, 987, 989, 993, 995, 999, 1001, 1003, 1005, 1007, 1011, \ 1015, 1017, 1023, 1025, 1027, 1029, 1035, 1037, 1041, 1043, 1045, \ 1047, 1053, 1055, 1057, 1059, 1065, 1067, 1071, 1073, 1075, 1077, \ 1079, 1081, 1083, 1085, 1089, 1095, 1099, 1101, 1105, 1107, 1111, \ 1113, 1115, 1119, 1121, 1125, 1127, 1131, 1133, 1135, 1137, 1139, \ 1141, 1143, 1145, 1147, 1149, 1155, 1157, 1159, 1161, 1165, 1167, \ 1169, 1173, 1175, 1177, 1179, 1183, 1185, 1189, 1191, 1195, 1197, \ 1199, 1203, 1205, 1207, 1209, 1211, 1215, 1219, 1221, 1225, 1227, \ 1233, 1235, 1239, 1241, 1243, 1245, 1247, 1251, 1253, 1255, 1257, \ 1261, 1263, 1265, 1267, 1269, 1271, 1273, 1275, 1281, 1285, 1287, \ 1293, 1295, 1299, 1305, 1309, 1311, 1313, 1315, 1317, 1323, 1325, \ 1329, 1331, 1333, 1335, 1337, 1339, 1341, 1343, 1345, 1347, 1349, \ 1351, 1353, 1355, 1357, 1359, 1363, 1365, 1369, 1371, 1375, 1377, \ 1379, 1383, 1385, 1387, 1389, 1391, 1393, 1395, 1397, 1401, 1403, \ 1405, 1407, 1411, 1413, 1415, 1417, 1419, 1421, 1425, 1431, 1435, \ 1437, 1441, 1443, 1445, 1449, 1455, 1457, 1461, 1463, 1465, 1467, \ 1469, 1473, 1475, 1477, 1479, 1485, 1491, 1495, 1497, 1501, 1503, \ 1505, 1507, 1509, 1513, 1515, 1517, 1519, 1521, 1525, 1527, 1529, \ 1533, 1535, 1537, 1539, 1541, 1545, 1547, 1551, 1555, 1557, 1561, \ 1563, 1565, 1569, 1573, 1575, 1577, 1581, 1585, 1587, 1589, 1591, \ 1593, 1595, 1599, 1603, 1605, 1611, 1615, 1617, 1623, 1625, 1629, \ 1631, 1633, 1635, 1639, 1641, 1643, 1645, 1647, 1649, 1651, 1653, \ 1655, 1659, 1661, 1665, 1671, 1673, 1675, 1677, 1679, 1681, 1683, \ 1685, 1687, 1689, 1691, 1695, 1701, 1703, 1705, 1707, 1711, 1713, \ 1715, 1717, 1719, 1725, 1727, 1729, 1731, 1735, 1737, 1739, 1743, \ 1745, 1749, 1751, 1755, 1757, 1761, 1763, 1765, 1767, 1769, 1771, \ 1773, 1775, 1779, 1781, 1785, 1791, 1793, 1795, 1797, 1799, 1803, \ 1805, 1807, 1809, 1813, 1815, 1817, 1819, 1821, 1825, 1827, 1829, \ 1833, 1835, 1837, 1839, 1841, 1843, 1845, 1849, 1851, 1853, 1855, \ 1857, 1859, 1863, 1865, 1869, 1875, 1881, 1883, 1885, 1887, 1891, \ 1893, 1895, 1897, 1899, 1903, 1905, 1909, 1911, 1915, 1917, 1919, \ 1921, 1923, 1925, 1927, 1929, 1935, 1937, 1939, 1941, 1943, 1945, \ 1947, 1953, 1955, 1957, 1959, 1961, 1963, 1965, 1967, 1969, 1971, \ 1975, 1977, 1981, 1983, 1985, 1989, 1991, 1995, 2001} 10. ## Re: Prime Conjecture Originally Posted by MaxJasper m<2000 for which the conjecture is FALSE: (resulting non-Primes) { 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, \ 69, 75, 77, 81, 85, 87, 91, 93, 95, 99, 105, 111, 115, 117, 119, 121, \ 123, 125, 129, 133, 135, 141, 143, 145, 147, 153, 155, 159, 161, 165, \ 169, 171, 175, 177, 183, 185, 187, 189, 195, 201, 203, 205, 207, 209, \ 213, 215, 217, 219, 221, 225, 231, 235, 237, 243, 245, 247, 249, 253, \ 255, 259, 261, 265, 267, 273, 275, 279, 285, 287, 289, 291, 295, 297, \ 299, 301, 303, 305, 309, 315, 319, 321, 323, 325, 327, 329, 333, 335, \ 339, 341, 343, 345, 351, 355, 357, 361, 363, 365, 369, 371, 375, 377, \ 381, 385, 387, 391, 393, 395, 399, 403, 405, 407, 411, 413, 415, 417, \ 423, 425, 427, 429, 435, 437, 441, 445, 447, 451, 453, 455, 459, 465, \ 469, 471, 473, 475, 477, 481, 483, 485, 489, 493, 495, 497, 501, 505, \ 507, 511, 513, 515, 517, 519, 525, 527, 529, 531, 533, 535, 537, 539, \ 543, 545, 549, 551, 553, 555, 559, 561, 565, 567, 573, 575, 579, 581, \ 583, 585, 589, 591, 595, 597, 603, 605, 609, 611, 615, 621, 623, 625, \ 627, 629, 633, 635, 637, 639, 645, 649, 651, 655, 657, 663, 665, 667, \ 669, 671, 675, 679, 681, 685, 687, 689, 693, 695, 697, 699, 703, 705, \ 707, 711, 713, 715, 717, 721, 723, 725, 729, 731, 735, 737, 741, 745, \ 747, 749, 753, 755, 759, 763, 765, 767, 771, 775, 777, 779, 781, 783, \ 785, 789, 791, 793, 795, 799, 801, 803, 805, 807, 813, 815, 817, 819, \ 825, 831, 833, 835, 837, 841, 843, 845, 847, 849, 851, 855, 861, 865, \ 867, 869, 871, 873, 875, 879, 885, 889, 891, 893, 895, 897, 899, 901, \ 903, 905, 909, 913, 915, 917, 921, 923, 925, 927, 931, 933, 935, 939, \ 943, 945, 949, 951, 955, 957, 959, 961, 963, 965, 969, 973, 975, 979, \ 981, 985, 987, 989, 993, 995, 999, 1001, 1003, 1005, 1007, 1011, \ 1015, 1017, 1023, 1025, 1027, 1029, 1035, 1037, 1041, 1043, 1045, \ 1047, 1053, 1055, 1057, 1059, 1065, 1067, 1071, 1073, 1075, 1077, \ 1079, 1081, 1083, 1085, 1089, 1095, 1099, 1101, 1105, 1107, 1111, \ 1113, 1115, 1119, 1121, 1125, 1127, 1131, 1133, 1135, 1137, 1139, \ 1141, 1143, 1145, 1147, 1149, 1155, 1157, 1159, 1161, 1165, 1167, \ 1169, 1173, 1175, 1177, 1179, 1183, 1185, 1189, 1191, 1195, 1197, \ 1199, 1203, 1205, 1207, 1209, 1211, 1215, 1219, 1221, 1225, 1227, \ 1233, 1235, 1239, 1241, 1243, 1245, 1247, 1251, 1253, 1255, 1257, \ 1261, 1263, 1265, 1267, 1269, 1271, 1273, 1275, 1281, 1285, 1287, \ 1293, 1295, 1299, 1305, 1309, 1311, 1313, 1315, 1317, 1323, 1325, \ 1329, 1331, 1333, 1335, 1337, 1339, 1341, 1343, 1345, 1347, 1349, \ 1351, 1353, 1355, 1357, 1359, 1363, 1365, 1369, 1371, 1375, 1377, \ 1379, 1383, 1385, 1387, 1389, 1391, 1393, 1395, 1397, 1401, 1403, \ 1405, 1407, 1411, 1413, 1415, 1417, 1419, 1421, 1425, 1431, 1435, \ 1437, 1441, 1443, 1445, 1449, 1455, 1457, 1461, 1463, 1465, 1467, \ 1469, 1473, 1475, 1477, 1479, 1485, 1491, 1495, 1497, 1501, 1503, \ 1505, 1507, 1509, 1513, 1515, 1517, 1519, 1521, 1525, 1527, 1529, \ 1533, 1535, 1537, 1539, 1541, 1545, 1547, 1551, 1555, 1557, 1561, \ 1563, 1565, 1569, 1573, 1575, 1577, 1581, 1585, 1587, 1589, 1591, \ 1593, 1595, 1599, 1603, 1605, 1611, 1615, 1617, 1623, 1625, 1629, \ 1631, 1633, 1635, 1639, 1641, 1643, 1645, 1647, 1649, 1651, 1653, \ 1655, 1659, 1661, 1665, 1671, 1673, 1675, 1677, 1679, 1681, 1683, \ 1685, 1687, 1689, 1691, 1695, 1701, 1703, 1705, 1707, 1711, 1713, \ 1715, 1717, 1719, 1725, 1727, 1729, 1731, 1735, 1737, 1739, 1743, \ 1745, 1749, 1751, 1755, 1757, 1761, 1763, 1765, 1767, 1769, 1771, \ 1773, 1775, 1779, 1781, 1785, 1791, 1793, 1795, 1797, 1799, 1803, \ 1805, 1807, 1809, 1813, 1815, 1817, 1819, 1821, 1825, 1827, 1829, \ 1833, 1835, 1837, 1839, 1841, 1843, 1845, 1849, 1851, 1853, 1855, \ 1857, 1859, 1863, 1865, 1869, 1875, 1881, 1883, 1885, 1887, 1891, \ 1893, 1895, 1897, 1899, 1903, 1905, 1909, 1911, 1915, 1917, 1919, \ 1921, 1923, 1925, 1927, 1929, 1935, 1937, 1939, 1941, 1943, 1945, \ 1947, 1953, 1955, 1957, 1959, 1961, 1963, 1965, 1967, 1969, 1971, \ 1975, 1977, 1981, 1983, 1985, 1989, 1991, 1995, 2001} But for each of the above non-prime values of m, 2^m - 1 is not zero (mod. m(m-1) and therefore the conjecture still holds. 11. ## Re: Prime Conjecture Originally Posted by Stan But for each of the above non-prime values of m, 2^m - 1 is not zero (mod. m(m-1) and therefore the conjecture still holds. Your original condition was: 2m = 2 (mod. m(m-1)) 12. ## Re: Prime Conjecture Originally Posted by MaxJasper Your original condition was: 2m = 2 (mod. m(m-1)) If you look at the original post, you will see the condition is 2^m = 2 (mod. m(m-1)). 13. ## Re: Prime Conjecture Sure 2^m was correct and I use it...I don't know what happened to ^ in my last post!	6,330	11,684	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.125	4	CC-MAIN-2018-26	latest	en	0.763416
https://discuss.codechef.com/t/sixfriends-editorial/100759	1,653,667,159,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00230.warc.gz	268,053,803	7,248	# SIXFRIENDS-Editorial Setter: Utkarsh Gupta Tester: Aryan, Satyam Editorialist: Devendra Singh 382 None # PROBLEM: Six friends went on a trip. They are looking for an accommodation. After struggling for hours, they found a hotel which has the following type of rooms: • Double Room at cost of X Rs. • Triple Room at cost of Y Rs. It is known that triple rooms are more expensive than double rooms, so X \lt Y. Output the minimum money required to accommodate all the six friends. # EXPLANATION: There are three ways to accommodate all six friends into hotel rooms: • Get three double rooms, money required = 3X • Get two triple rooms, money required = 2Y • Get two double rooms and a triple room, money required = 2X+Y Since X<Y we can ignore the third option as 3X<2X+Y \therefore\: Minimum money required = min(\:3X\: ,\: 2Y\:) # TIME COMPLEXITY: O(1) for each test case. # SOLUTION: Setter's Solution ``````//Utkarsh.25dec #include <bits/stdc++.h> #define ll long long int #define pb push_back #define mp make_pair #define mod 1000000007 #define vl vector <ll> #define all(c) (c).begin(),(c).end() using namespace std; ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=resa%mod;a=aa%mod;}return res;} ll modInverse(ll a){return power(a,mod-2);} const int N=500023; bool vis[N]; long long readInt(long long l,long long r,char endd){ long long x=0; int cnt=0; int fi=-1; bool is_neg=false; while(true){ char g=getchar(); if(g=='-'){ assert(fi==-1); is_neg=true; continue; } if('0'<=g && g<='9'){ x=10; x+=g-'0'; if(cnt==0){ fi=g-'0'; } cnt++; assert(fi!=0 \|\| cnt==1); assert(fi!=0 \|\| is_neg==false); assert(!(cnt>19 \|\| ( cnt==19 && fi>1) )); } else if(g==endd){ if(is_neg){ x= -x; } if(!(l <= x && x <= r)) { cerr << l << ' ' << r << ' ' << x << '\n'; assert(1 == 0); } return x; } else { assert(false); } } } string ret=""; int cnt=0; while(true){ char g=getchar(); assert(g!=-1); if(g==endd){ break; } cnt++; ret+=g; } assert(l<=cnt && cnt<=r); return ret; } long long readIntSp(long long l,long long r){ } long long readIntLn(long long l,long long r){ } } } void solve() { assert(x<y); cout<<min(3x,2y)<<'\n'; } int main() { #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif ios_base::sync_with_stdio(false); cin.tie(NULL),cout.tie(NULL); while(T--) solve(); assert(getchar()==-1); cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n"; } `````` Tester-1's Solution ``````/* in the name of Anton / / Compete against Yourself. Author - Aryan (@aryanc403) Atcoder library - https://atcoder.github.io/ac-library/production/document_en/ / #ifdef ARYANC403 #else #pragma GCC optimize ("Ofast") #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx") //#pragma GCC optimize ("-ffloat-store") #include<bits/stdc++.h> #define dbg(args...) 42; #endif // y_combinator from @neal template https://codeforces.com/contest/1553/submission/123849801 // http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2016/p0200r0.html template<class Fun> class y_combinator_result { Fun fun_; public: template<class T> explicit y_combinator_result(T &&fun): fun_(std::forward<T>(fun)) {} template<class ...Args> decltype(auto) operator()(Args &&...args) { return fun_(std::ref(this), std::forward<Args>(args)...); } }; template<class Fun> decltype(auto) y_combinator(Fun &&fun) { return y_combinator_result<std::decay_t<Fun>>(std::forward<Fun>(fun)); } using namespace std; #define fo(i,n) for(i=0;i<(n);++i) #define repA(i,j,n) for(i=(j);i<=(n);++i) #define repD(i,j,n) for(i=(j);i>=(n);--i) #define all(x) begin(x), end(x) #define sz(x) ((lli)(x).size()) #define pb push_back #define mp make_pair #define X first #define Y second #define endl "\n" typedef long long int lli; typedef long double mytype; typedef pair<lli,lli> ii; typedef vector<ii> vii; typedef vector<lli> vi; const auto start_time = std::chrono::high_resolution_clock::now(); void aryanc403() { #ifdef ARYANC403 auto end_time = std::chrono::high_resolution_clock::now(); std::chrono::duration<double> diff = end_time-start_time; cerr<<"Time Taken : "<<diff.count()<<"\n"; #endif } long long readInt(long long l, long long r, char endd) { long long x=0; int cnt=0; int fi=-1; bool is_neg=false; while(true) { char g=getchar(); if(g=='-') { assert(fi==-1); is_neg=true; continue; } if('0'<=g&&g<='9') { x=10; x+=g-'0'; if(cnt==0) { fi=g-'0'; } cnt++; assert(fi!=0 \|\| cnt==1); assert(fi!=0 \|\| is_neg==false); assert(!(cnt>19 \|\| ( cnt==19 && fi>1) )); } else if(g==endd) { if(is_neg) { x=-x; } assert(l<=x&&x<=r); return x; } else { assert(false); } } } string readString(int l, int r, char endd) { string ret=""; int cnt=0; while(true) { char g=getchar(); assert(g!=-1); if(g==endd) { break; } cnt++; ret+=g; } assert(l<=cnt&&cnt<=r); return ret; } long long readIntSp(long long l, long long r) { } long long readIntLn(long long l, long long r) { } string readStringLn(int l, int r) { } string readStringSp(int l, int r) { } assert(getchar()==EOF); } vi a(n); for(int i=0;i<n-1;++i) return a; } bool isBinaryString(const string s){ for(auto x:s){ if('0'<=x&&x<='1') continue; return false; } return true; } // #include<atcoder/dsu> // vector<vi> e(n); // atcoder::dsu d(n); // for(lli i=1;i<n;++i){ // e[u].pb(v); // e[v].pb(u); // d.merge(u,v); // } // assert(d.size(0)==n); // return e; // } const lli INF = 0xFFFFFFFFFFFFFFFL; lli seed; inline lli rnd(lli l=0,lli r=INF) {return uniform_int_distribution<lli>(l,r)(rng);} class CMP {public: bool operator()(ii a , ii b) //For min priority_queue . { return ! ( a.X < b.X \|\| ( a.X==b.X && a.Y <= b.Y )); }}; void add( map<lli,lli> &m, lli x,lli cnt=1) { auto jt=m.find(x); if(jt==m.end()) m.insert({x,cnt}); else jt->Y+=cnt; } void del( map<lli,lli> &m, lli x,lli cnt=1) { auto jt=m.find(x); if(jt->Y<=cnt) m.erase(jt); else jt->Y-=cnt; } bool cmp(const ii &a,const ii &b) { return a.X<b.X\|\|(a.X==b.X&&a.Y<b.Y); } const lli mod = 1000000007L; // const lli maxN = 1000000007L; lli T,n,i,j,k,in,cnt,l,r,u,v,x,y; lli m; string s; vi a; //priority_queue < ii , vector < ii > , CMP > pq;// min priority_queue . int main(void) { ios_base::sync_with_stdio(false);cin.tie(NULL); // freopen("txt.in", "r", stdin); // freopen("txt.out", "w", stdout); // cout<<std::fixed<<std::setprecision(35); while(T--) { cout<<min(3x,2y)<<endl; } aryanc403(); return 0; } `````` Tester-2's Solution ``````#include <bits/stdc++.h> using namespace std; #ifndef ONLINE_JUDGE #define debug(x) cerr<<#x<<" "; _print(x); cerr<<nline; #else #define debug(x); #endif #define ll long long / ------------------------Input Checker---------------------------------- / long long readInt(long long l,long long r,char endd){ long long x=0; int cnt=0; int fi=-1; bool is_neg=false; while(true){ char g=getchar(); if(g=='-'){ assert(fi==-1); is_neg=true; continue; } if('0'<=g && g<='9'){ x=10; x+=g-'0'; if(cnt==0){ fi=g-'0'; } cnt++; assert(fi!=0 \|\| cnt==1); assert(fi!=0 \|\| is_neg==false); assert(!(cnt>19 \|\| ( cnt==19 && fi>1) )); } else if(g==endd){ if(is_neg){ x= -x; } if(!(l <= x && x <= r)) { cerr << l << ' ' << r << ' ' << x << '\n'; assert(1 == 0); } return x; } else { assert(false); } } } string ret=""; int cnt=0; while(true){ char g=getchar(); assert(g!=-1); if(g==endd){ break; } cnt++; ret+=g; } assert(l<=cnt && cnt<=r); return ret; } long long readIntSp(long long l,long long r){ } long long readIntLn(long long l,long long r){ } } } /* ------------------------Main code starts here---------------------------------- / ll MAX=100000; ll tes_sum=0; void solve(){ assert(x<y); ll ans=min({3x,2y,2x+y}); cout<<ans<<"\n"; return; } int main(){ ios_base::sync_with_stdio(false); cin.tie(NULL); #ifndef ONLINE_JUDGE freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); freopen("error.txt", "w", stderr); #endif while(test_cases--){ solve(); } assert(getchar()==-1); return 0; } `````` Editorialist's Solution ``````#include "bits/stdc++.h" using namespace std; #define ll long long #define pb push_back #define all(_obj) _obj.begin(), _obj.end() #define F first #define S second #define pll pair<ll, ll> #define vll vector<ll> const int N = 1e5 + 11, mod = 1e9 + 7; ll max(ll a, ll b) { return ((a > b) ? a : b); } ll min(ll a, ll b) { return ((a > b) ? b : a); } void sol(void) { int x, y; cin >> x >> y; cout << min(3 * x, 2 * y) << '\n'; return; } int main() { ios_base::sync_with_stdio(false); cin.tie(NULL), cout.tie(NULL); int test = 1; cin >> test; while (test--) sol(); } ``````	2,770	8,679	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.65625	4	CC-MAIN-2022-21	latest	en	0.53493
https://gmatclub.com/forum/if-n-is-an-integer-between-2-and-100-and-if-n-is-also-the-square-of-114531.html?fl=similar	1,495,542,205,000,000,000	text/html	crawl-data/CC-MAIN-2017-22/segments/1495463607620.78/warc/CC-MAIN-20170523103136-20170523123136-00294.warc.gz	768,196,255	58,934	Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack It is currently 23 May 2017, 05:23 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If n is an integer between 2 and 100 and if n is also the square of Author Message TAGS: ### Hide Tags Intern Joined: 12 Sep 2010 Posts: 3 Followers: 0 Kudos [?]: 5 [0], given: 2 If n is an integer between 2 and 100 and if n is also the square of [#permalink] ### Show Tags 01 Jun 2011, 15:26 4 This post was BOOKMARKED 00:00 Difficulty: 65% (hard) Question Stats: 48% (02:01) correct 53% (00:48) wrong based on 80 sessions ### HideShow timer Statistics If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n? (1) n is even (2) The cube root of n is an integer [Reveal] Spoiler: 1) Not sufficient because n could be 4, 16, 36, 64, etc. 2) Doesn't between 2 and 100 mean 3 to 99? (I thought between means do not include the end points). Statement 2 says the cube of root n is an integer. Woudln't the cube root of 27 be included? as well as the cube root of 64? The OG guide says there is only one such value of n between 2 and 100, which is 64. OPEN DISCUSSION OF THIS QUESTION IS HERE: if-n-is-an-integer-between-2-and-100-and-if-n-is-also-the-sq-167846.html [Reveal] Spoiler: OA Math Forum Moderator Joined: 20 Dec 2010 Posts: 2013 Followers: 163 Kudos [?]: 1822 [1] , given: 376 Re: If n is an integer between 2 and 100 and if n is also the square of [#permalink] ### Show Tags 01 Jun 2011, 16:06 1 KUDOS KraZZyiE wrote: I have a question about statement 2..... If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n? 1) n is even 2) The cube root of n is an integer 1) Not sufficient because n could be 4, 16, 36, 64, etc. 2) Doesn't between 2 and 100 mean 3 to 99? (I thought between means do not include the end points). Statement 2 says the cube of root n is an integer. Woudln't the cube root of 27 be included? as well as the cube root of 64? The OG guide says there is only one such value of n between 2 and 100, which is 64. The question stem says that the "n" is square of an integer; 27 is not a square of any integer. Leaves us with just 64, which is 8^2. _________________ Director Joined: 01 Feb 2011 Posts: 755 Followers: 14 Kudos [?]: 124 [1] , given: 42 Re: If n is an integer between 2 and 100 and if n is also the square of [#permalink] ### Show Tags 01 Jun 2011, 16:49 1 KUDOS possibilities 4 9 16 25 36 49 64 81 1. Not sufficient 4 or 16 or 36.... 2. Sufficient as 64 is the only number is the above list thats also a cube of an integer. Intern Joined: 12 Sep 2010 Posts: 3 Followers: 0 Kudos [?]: 5 [0], given: 2 Re: If n is an integer between 2 and 100 and if n is also the square of [#permalink] ### Show Tags 02 Jun 2011, 14:53 thanks. great explanation Intern Status: arrogance and ignorance are infectious, thanks god that (true) laughing is too Joined: 15 Jun 2011 Posts: 9 Location: Vienna Followers: 1 Kudos [?]: 3 [0], given: 3 Re: If n is an integer between 2 and 100 and if n is also the square of [#permalink] ### Show Tags 16 Aug 2011, 05:49 thanks for the explanation according to the quant review and the gmat club error log it is question #110 in the DS section, quant review Senior Manager Joined: 16 Feb 2011 Posts: 259 Followers: 4 Kudos [?]: 161 [0], given: 9 Re: If n is an integer between 2 and 100 and if n is also the square of [#permalink] ### Show Tags 16 Aug 2011, 12:06 wow..i got it rite...B it is GMAT Club Legend Joined: 09 Sep 2013 Posts: 15394 Followers: 648 Kudos [?]: 204 [0], given: 0 Re: If n is an integer between 2 and 100 and if n is also the square of [#permalink] ### Show Tags 19 Jan 2016, 11:13 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Math Expert Joined: 02 Sep 2009 Posts: 38816 Followers: 7716 Kudos [?]: 105854 [0], given: 11593 Re: If n is an integer between 2 and 100 and if n is also the square of [#permalink] ### Show Tags 19 Jan 2016, 12:11 Expert's post 3 This post was BOOKMARKED KraZZyiE wrote: If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n? (1) n is even (2) The cube root of n is an integer [Reveal] Spoiler: 1) Not sufficient because n could be 4, 16, 36, 64, etc. 2) Doesn't between 2 and 100 mean 3 to 99? (I thought between means do not include the end points). Statement 2 says the cube of root n is an integer. Woudln't the cube root of 27 be included? as well as the cube root of 64? The OG guide says there is only one such value of n between 2 and 100, which is 64. If n is an integer between 2 and 100 and if n is also the square of an integer, what is the value of n? Given: n is a perfect square between 2 and 100 (a perfect square is an integer that can be written as the square of some other integer, for example 16=4^2, is a perfect square). (1) n is even --> n can be any even perfect square in the given range: 4, 16, 36, ... Not sufficient. (2) The cube root of n is an integer --> so n is also a perfect cube between 2 and 100. There are 4 perfect cubes in this range: 2^3=8, 3^3=27 and 4^3=64 but only one of them namely 64 is also a perfect square, so n=64=8^2=4^3. Sufficient. OPEN DISCUSSION OF THIS QUESTION IS HERE: if-n-is-an-integer-between-2-and-100-and-if-n-is-also-the-sq-167846.html _________________ Re: If n is an integer between 2 and 100 and if n is also the square of [#permalink] 19 Jan 2016, 12:11 Similar topics Replies Last post Similar Topics: 1 If n is an integer, and n ≠ 0, is (100 + n^2)/n a positive integer? 2 01 Sep 2016, 22:55 4 If y is an integer such that 2 < y < 100 and if y is also the square 5 19 May 2015, 10:24 19 If n is an integer between 2 and 100 and if n is also the sq 14 08 Feb 2017, 21:12 6 If n is an integer between 2 and 100 and if n is also the 3 23 Oct 2015, 11:21 10 If n is an integer between 2 and 100 and if n is also the 9 19 Sep 2014, 07:31 Display posts from previous: Sort by	2,180	6,985	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.09375	4	CC-MAIN-2017-22	latest	en	0.910395
https://serokell.io/blog/incomplete-and-utter-introduction-to-modal-logic	1,723,241,329,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640782236.55/warc/CC-MAIN-20240809204446-20240809234446-00704.warc.gz	411,916,603	106,022	Incomplete and Utter Introduction to Modal Logic, Pt. 1 Foreword Modal logic is one of the most popular branches of mathematical logic. Modal logic covers such areas of human knowledge as mathematics (especially, topology and graph theory), computer science, linguistics, artificial intelligence, and philosophy. Moreover, modal logic in itself still attracts by its mathematical beauty. We would like to introduce the reader to modal logic, fundamental technical tools, and connections with other disciplines. Our introduction is not so comprehensive indeed, the title is an allusion to the wonderful book by Stephen Fry. History and background The grandfather of modal logic is the 17th-century German mathematician and philosopher Gottfried Leibniz, one of the founders of calculus and mechanics. From his point of view, there are two kinds of truth. The statement is called necessarily true if it’s true for any state of affair. Note that, one also has possibly true statements. That is, such statements that could be false, other things being equal. For instance, the proposition “sir Winston Churchill was a prime minister of Great Britain in 1940” is possibly true. As we know, Edward Wood, 1st Earl of Halifax, also could be a prime minister at the same time. Of course, one may provide any other example from the history of the United Kingdom and all of them would be valid to the same degree. In other words, any factual statement might be possibly true since “the trueness” of any factual statement depends strongly on circumstances and external factors. The example of a statement which is necessarily true is $1 + 1 = 2$, where signs $1$, $2$, $+$ and $=$ are understood in the usual sense. Modal logic became a part of contemporary logic at the beginning of the 20th century. In the 1910s, Clarence Lewis, American philosopher, was the first who proposed to extend the language of classical logic with two modal connectives $\Box$ and $\Diamond$. Informally, one may read $\Box \varphi$ as $\varphi$ is necessary. $\Diamond \varphi$ has a dual interpretation as $\varphi$ is possible. Lewis sought to analyse the notion of necessity from a logical point of view. Thus, modalities in mathematical logic arose initially as a tool for philosophical issue analysis. Lewis understood those modalities intuitively rather than formally. In other words, modal operators had no mathematically valid semantical interpretation. In the year 1944, Alfred Tarski and John McKinsey wrote a paper called The Algebra of Topology. In this work, the strong connection between modal logic ${\bf S}4$ and topological and metric spaces was established. Historically, topological semantics is the very first mathematical semantics of modal logic. Those results marked the beginning of a separate branch of modal logic that studies the topological aspects of modalities. In the 1950s and 1960s, Saul Kripke formulated relational semantics for modal logic. Note that we will mostly consider Kripke semantics in this post. Later, in the 1970s, such mathematicians as Dov Gabbay, Krister Segerberg, Robert Goldblatt, S. K. Thomason, Henrik Sahlqvist, Johan van Benthem provided the set of technical tools to investigate systems of modal logic much more deeply and precisely. So, modal logic continues its development in the direction that was established by these researchers. Let us briefly overview some areas of modal logic use. Topology and graphs Modal logic provides a compact and laconic language to characterise some properties of directed graphs and topological spaces. In this blog post, we study Kripke frames as the underlying structures. Without loss of generality, one may think of Kripke frames as directed graphs. It means that formal definitions of a Kripke frame and a directed graph are precisely the same. Here, modal language is a powerful tool in representing first-order adjacency properties. We realise the fact that not every reader has a background in first-order logic. In order to keep the post self-contained, we remind the required notions from first-order logic to describe this connection quite clearly. As we will see, modal logic is strongly connected with binary relations through Kripke frames. The properties of binary relation in a Kripke frame are mostly first-order. We consider examples of first-order relation properties expressed in modal language. Moreover, we formulate and discuss the famous Salqvist’s theorem that connects modal formulas of the special kind with binary relation properties that are first-order definable. Anyway, in such a perspective, one may consider modal logic as a logic of directed graphs since there is no formal difference between a binary relation and edges in a graph. That is, one may use modal logic in directed graph characterisation in a limited way. You may draw any directed graph you prefer on a plane. A graph is a combinatorial object rather than a geometrical one. A topological space is closer to geometry. Although, there’s a topological way of graph consideration, however. We also discuss how exactly modal logic has a topological interpretation. In the author’s opinion, topological and geometrical aspects of modal logic are one of the most astonishing and beautiful. It is a well-known result proved by Alfred Tarski and John McKinsey that the modal logic ${\bf S}4$ is the logic of all topological spaces. Similarly to directed graphs, modal logic might be used in classifying topological spaces and other spatial constructions in a restricted way. We will discuss the topological and spatial aspects of modal logic in the second part. The foundations of mathematics The foundation of mathematics is the branch of mathematical logic that studies miscellaneous metamathematical issues. In mathematics, we often consider some abstract structure and formulate a theory based on some list axioms that describe primitive properties of considered constructions. The examples are group theory, elementary geometry, graph theory, arithmetics, etc. In metamathematics, we are interested in what a mathematical theory is in itself. That is, metamathematics arose at the beginning of the previous century to answer philosophical questions mathematically. The main interest was to prove the consistency of formal arithmetics with purely formal methods. As it’s well known, Gödel’s famous incompleteness theorems place limits the formal proof of arithmetics consistency within arithmetics itself. However, modal logic also allows one to study the properties of provability in formal arithmetical theories. Moreover, the second Gödel’s incompleteness theorem has a purely modal formulation! We discuss that topic in more detail in the follow-up of the series. Basic definitions and constructions Let us define the modal language. That’s the starting point for every logic which we take into consideration. In our case, a modal language is a grammar according to rules of which we write logical formulas enriched with modalities. Let $\operatorname{PV} = \{ p_0, p_1, \dots \}$ be a countably infinite set of propositional variables, or atomic propositions, or propositional letters. Informally, a propositional variable ranges over such atomic statements as “it is raining” or something like that. A modal language as the set of modal formulas $\operatorname{Fm}$ is defined as follows: 1. Any $p_i \in \operatorname{PV}$ is a formula. 2. If $\varphi$ is a formula, then $\neg \varphi$ is a formula. 3. If $\varphi$ and $\psi$ are formulas, then $(\varphi \circ \psi)$ is formula, where $\circ \in \{ \to, \land, \lor\}$. 4. If $\varphi$ is a formula, then $\Box \varphi$ is a formula. Note that the possibility operator $\Diamond$ is expressed as $\Diamond$ = $\neg \Box \neg$. In our approach, the statement $\varphi$ is possibly true if it’s false that necessiation of $\neg \varphi$ is true. Also, one may introduce $\Diamond$ as the primitive one, that is $\Box = \neg \Diamond \neg$. Also, it is useful to have constants $\top$ and $\bot$ that we define as $\neg p_0 \lor p_0$ and $\neg \top$ correspondingly. As we said, a modal language extends the classical one. One may ask how we can read $\Box$. Here are several ways to understand this modal connective: 1. $\Box \varphi$ as $\varphi$ is necessary. This is historically the very first way of the modal connective reading. Dually, one may read $\Diamond \varphi$ as $\varphi$ is possible. Modalities of this kind are alethic modalities. 2. $\Box \varphi$ as it is known that $\varphi$. In this case, we call $\Box$ an epistemic modality. 3. $\Box \varphi$ as $\varphi$ is a statement provable in formal arithmetics. 4. $\Box$ as an interior in a topological space. Here, we call such modality a topological one. 5. $\Box \varphi$ as there will be $\varphi$ always in the future, a temporal modality, in other words. We will discuss items 3-5 in the next part of the series as promised. We defined the syntax of modal logic above. But syntax doesn’t provide logic, only grammar. In logic, one has inference rules and the definition of proof. We also want to know what kind of modal statements we need to prove basically. Firstly, let us describe a more basic notion like normal modal logic. By normal modal logic, we mean a set of modal formulas $\mathcal{L}$ that contains all Boolean tautologies; ${\bf AK}$-axiom (named after Kripke) $\Box (p \to q) \to (\Box p \to \Box q)$ and closed under the following three inference rules: 1. If $\varphi \in \mathcal{L}$ and $\varphi \to \psi \in \mathcal{L}$, then $\psi \in \mathcal{L}$ (Modus ponens). That is the basic logical rule that allows you to get a consequence from an implication if you have a premise. 2. If $\varphi \in \mathcal{L}$, then $\Box \varphi \in \mathcal{L}$ (Necessiation rule). If $\varphi$ belongs to a logic, then $\Box \varphi$ belongs to $\mathcal{L}$ too. Informally, if $\varphi$ is provable, then $\varphi$ is necessary. 3. If $\varphi ( p ) \in \mathcal{L}$, then $\varphi (p := \psi) \in \mathcal{L}$. If formula $\varphi$ with variable $p$ belongs to a logic, then its instances belong to the same logic. Here, the instance of the formula is the result of a substitution of another formula instead of a variable. In other words, any normal modal logic is closed under substitution. The fact that any normal modal logic contains all Boolean tautologies tell us that modal logic extends the classical one. Literally, the Kripke axiom ${\bf AK}$-axiom denotes that $\Box$ distributes over implication. Such an explanation doesn’t help us so much, indeed. It’s quite convenient to read this axiom in terms of necessity. Then, if the implication $\varphi \to \psi$ is necessary and the premise is necessary, then the consequence is also necessary. For instance, it is necessary that if the number is divided by four then this number is divided by two. Then, if it is necessary that the number is divided by four, then it is necessary that it’s divided by two. Of course, within the natural language that we use in everyday speech the last two sentences sound quite monotonous, but we merely illustrated how this logical form works by example. In such a situation, it’s crucially important to follow the formal structure even if this structure looks wordy and counterintuitive. Of course, this sometimes runs counter to our preferences in linguistic aesthetics, although structure following allows analysing modal sentences much more precisely. If we take into consideration some logical system represented by axioms and inference rules, then one needs to determine what derivation is in an arbitrary normal modal logic. A derivation of formula $\varphi$ in normal modal logic $\mathcal{L}$ is a sequence of formulas, each element of which is either axiom or some formula obtained from the previous ones via inference rules. The last element of such a sequence is a formula $\varphi$ itself. Here is an example of derivation in normal modal logic: 1. $\varphi \to (\psi \to (\varphi \land \psi))$ Boolean tautology 2. $\Box (\varphi \to (\psi \to (\varphi \land \psi)))$ Necessiation rule 3. $\Box (\varphi \to (\psi \to (\varphi \land \psi))) \to (\Box \varphi \to \Box (\psi \to (\varphi \land \psi)))$ ${\bf AK}$-axiom. 4. $\Box \varphi \to \Box (\psi \to (\varphi \land \psi))$ Modus ponens, (2), (3) 5. $\Box (\psi \to (\varphi \land \psi)) \to (\Box \psi \to \Box (\varphi \land \psi))$ ${\bf K}$-axiom 6. $\Box \varphi \to (\Box \psi \to \Box (\varphi \land \psi))$ Transitivity of implication, (4), (5) 7. $(\Box \varphi \to (\Box \psi \to \Box (\varphi \land \psi))) \to (\Box \varphi \land \Box \psi \to \Box (\varphi \land \psi))$ The instance of a Boolean tautology 8. $(\Box \varphi \land \Box \psi) \to \Box (\varphi \land \psi)$ Modus ponens, (6), (7) We leave the converse implication as an exercise to the reader. The minimal normal modal logic ${\bf K}$ is defined via the following list of axioms and inference rules: 1. $(p \to (q \to r)) \to ((p \to q) \to (p \to r))$ 2. $p \to (q \to p)$ 3. $p \to (q \to (p \land q))$ 4. $(p_1 \land p_2) \to p_i$, $i = 1,2$ 5. $p_i \to (p_1 \lor p_2)$, $i = 1,2$ 6. $(q \to p) \to ((r \to p) \to (q \lor r) \to p)$ 7. $(p \to q) \to ((p \to \neg q) \to \neg p)$ 8. $\neg \neg p \to p$ 9. $\Box (p \to q) \to (\Box p \to \Box q)$ 10. Inference rules: Modus Ponens, Necessiation, Substitution. The axioms (1)-(8) are exactly axioms of the usual classical propositional logic that axiomatise the set of Boolean tautologies together with Modus Ponens and Substitution rule. The axiom (9) is a Kripke axiom ${\bf AK}$. One may also claim that ${\bf K}$ is the smallest normal modal logic. In the definition of normal modal logic, we just require that the set of formulas should contain Boolean tautologies, etc. By the way, there might be something different from the required list of formulas. For instance, the set of all formulas is also a normal modal logic, the trivial one. The minimal normal modal logic is a normal modal logic that contains only Boolean tautologies, ${\bf AK}$-axiom and closed under the inference rules, no less no more. The logic ${\bf K}$ is the underlying logic for us. Other modal logics are solely extensions of ${\bf K}$. Note that modal logic is not needed to be a normal one. Weaker modal logics are studied via so-called neighbourhood semantics that we drop in our introductory post. Kripke semantics We have defined the syntax of the modal logic above introducing the grammar of a modal language. Let us define Kripke frames and models to have the truth definition in modal logic. Definition A Kripke frame is a pair $\mathbb{F} = \langle W, R \rangle$, where $W$ is a non-empty set, a set of so-called possible worlds, informally. Note that we will call an element of $W$ a world or a point. World and point are synonyms for us. $R \subseteq W \times W$ is a binary relation on $W$. For example, the set $\{0, 1, ..., n\}$ with strict order relation $<$ is a Kripke frame. Moreover, any directed graph might be considered as a Kripke frame, where the set of vertices is a set of possible worlds and the set of edges is a binary relation. Definition A Kripke model is a pair $\mathcal{M} = \langle \mathcal{F}, \vartheta \rangle$, where $\mathbb{F} = \langle W, R \rangle$ is a Kripke frame and $\vartheta : \operatorname{PV} \to 2^W$ is a valuation function that maps any proposition variable to some subset of possible words. Informally, we should match any atomic statement with corresponding states of affairs in which this atomic statement is true. For complex formulas, we introduce the truth definition inductively. We denote this relation as $\mathcal{M}, w \models \varphi$ that should be readen as “in model $\mathcal{M}$, in world $w$ the statement $\varphi$ is true”. 1. $\mathcal{M}, w \models p$ iff $w \in v ( p )$ 2. $\mathcal{M}, w \not\models \bot$, that is, there is no point in which false could be true. 3. $\mathcal{M}, w \models \varphi \to \psi$ iff $\mathcal{M}, w \models \varphi$ implies $\mathcal{M}, w \models \psi$ 4. $\mathcal{M}, w \models \Box \varphi$ iff for all $v$ such that $w R v$, $\mathcal{M}, v \models \varphi$ It is not so difficult to obtain the truth conditions for other connectives. One needs to keep in mind that the other Boolean connectives are introduced via implication and bottom as 1. $\neg \varphi \leftrightharpoons \varphi \to \bot$ 2. $\varphi \lor \psi \leftrightharpoons \neg (\varphi \to \neg \psi)$ 3. $\varphi \land \psi \leftrightharpoons \neg (\neg \varphi \lor \neg \psi)$ We also know that $\Diamond = \neg \Box \neg$, so the truth definition for $\Diamond \varphi$ is the following one: $\mathcal{M}, w \models \Diamond \varphi$ iff there exists $v$ such that $w R v$ and $\mathcal{M}, v \models \varphi$. There are a lot of examples of Kripke models, indeed. Here, we refer the reader to the book Modal Logic of Open Minds by Johan van Benthem to study miscellaneous cases in depth. Let us consider briefly the following graph as a Kripke frame with the valuation map $\vartheta$: As an exercise, the reader might determine which formulas are true in every point of the model above. Also, we will use the following notions: 1. $\mathcal{M} \models \varphi$ iff $\mathcal{M}, w \models \varphi$ for each $w$. That is, a formula $\varphi$ is true in a Kripke model $\mathcal{M}$ if and only if it’s true in each possible world $w$ 2. $\mathcal{F} \models \varphi$ iff for each valuation $v$ and for all $w$, $\langle \mathcal{F}, w \rangle \models \varphi$. One should read $\mathcal{F} \models \varphi$ as $\varphi$ is valid in a frame $\mathcal{F}$ In other words, a formula $\varphi$ is true in a Kripke frame $\mathcal{F}$ if and only if it’s true in every Kripke model on this frame as an underlying one. 3. $\operatorname{Log}(\mathcal{F}) = \{ \varphi \in Fm \: \| \: \mathcal{F} \models \varphi \}$, where $\mathcal{F}$ is a Kripke frame. We will call the set $\operatorname{Log} (\mathcal{F})$ the logic of frame $\mathcal{F}$ 4. Let $\mathbb{F}$ be a class of Kripke frames, then $\operatorname{Log}(\mathbb{F}) = \bigcap \limits_{\mathcal{F} \in \mathbb{F}} \operatorname{Log}(\mathcal{F})$. More simply, the logic of class is a set of formulas each of which is true in every Kripke frame that belongs to this class. In logic, we often require that any true formula should be provable. If some logic satisfies this requirement, then we call that the logic is complete. Here, a modal logic $\mathcal{L}$ is a Kripke complete if and only if there exists some class of Kripke frames $\mathbb{F}$ such that $\mathcal{L} = \operatorname{Log}(\mathbb{F})$. That is, any valid formula in this class of frames is provable in $\mathcal{L}$. So, we’re going to solve the following issue. What’s the most general logic which is valid in an arbitrary Kripke frame? In other words, we are going to characterise the logic of all possible Kripke frames. We defined above what minimal normal modal logic is. One may show that this logic is the logic of all Kripke frames: Theorem ${\bf K} = \operatorname{Log} (\mathbb{F})$, where $\mathbb{F}$ is the class of all Kripke frames. The left inclusion ${\bf K} \subseteq \operatorname{Log} (\mathcal{F})$ called soundness is more or less obvious. The soundness theorem claims that the logic of all Kripke frames is the normal modal one. We will prove only that any formula provable in ${\bf K}$ is valid in any Kripke frame and show that $\operatorname{Log}(\mathcal{F})$ is closed under substitution, where $\mathcal{F} \in \mathbb{F}$ is an arbitrary Kripke frame. Let us show that the normality axiom is valid in an arbitrary Kripke frame. Let $\mathbb{F} = \langle W, R \rangle$ be a Kripke frame and $\vartheta : \operatorname{PV} \to \mathcal{P}(W)$ a valuation map. So, we have a Kripke model $\mathcal{M} = \langle \mathbb{F}, \vartheta \rangle$. Let $\mathcal{M}, a \models \Box (\varphi \to \psi)$ and $\mathcal{M}, a \models \Box \varphi$. Let $a R b$, then $\mathcal{M}, b \models \varphi \to \psi$ and $\mathcal{M}, b \models \varphi$ by the truth definition for $\Box$. So $\mathcal{M}, b \models \varphi$, so far as Modus Ponens holds in every Kripke model. Thus, $\mathcal{M}, a \models \Box \varphi$. Now we show that $\operatorname{Log}(\mathbb{F})$ is closed under substitution. We provide only a sketch since the full proof is quite technical. Let $\vartheta : \operatorname{PV} \to \mathcal{P}(W)$ be a valuation and $\mathcal{M} = \langle \mathcal{F}, \vartheta \rangle$ a Kripke model. Let us put $\|\|\varphi\|\| = \{ w \in W \: \| \: \mathcal{M}, w \models \varphi \}$, the set of all points in a Kripke model, where the formula $\varphi$ is true. Let $\mathcal{F} \models \varphi ( p )$ and let $\psi$ be an arbitrary formula. We build a Kripke model $\mathcal{M} = \langle \mathcal{F}, \vartheta' \rangle$ such that $\vartheta' ( p ) = \|\|\psi\|\|$. Then, one may show by induction that $\mathcal{M}, x \models \varphi ( p := \psi ) \Leftrightarrow \mathcal{M}', x \models \varphi ( p )$. The right inclusion (${\bf K} \supseteq \operatorname{Log}(\mathbb{F})$) called completeness is quite non-trivial, one needs to build the so-called canonical model. We haven’t defined what a canonical frame and model are. The idea of a canonical frame is that an observed logic itself forms a Kripke frame (and Kripke model too). Canonical frames and models often allow us to prove the fact that some normal modal logic is complete with respect to some class of Kripke frames. We provide only the main proof sketch. The following construction is very and very abstract, but sometimes one has to be patient to produce fruits. Let $\Gamma$ be a set of formulas and $\mathcal{L}$ a normal modal logic. $\Gamma$ is $\mathcal{L}$-inconsistent, if there exist some formulas $\varphi_1, \dots, \varphi_n$ such that $\neg (\varphi_1 \land \dots \land \varphi_n) \in \mathcal{L}$. That is, the set of formulas $\Gamma$ is $\mathcal{L}$-inconsistent, if there exists a finite subset such that negation of its conjuction is provable in an observed logic. $\Gamma$ is $\mathcal{L}$-consistent, if $\Gamma$ is not inconsistent. A $\mathcal{L}$-consistent set $\Gamma$ is maximal if it doesn’t have any non-trivial extensions. In other words, if $\Gamma$ is a subset of $\Gamma'$, where $\Gamma'$ is a $\mathcal{L}$-consistent, then $\Gamma = \Gamma'$. Now we are ready to define a canonical frame. Let $\mathcal{L}$ be a normal modal logic, then a canonical frame is a pair $\mathcal{F}^{\mathcal{L}} = \langle W^{\mathcal{L}}, R^{\mathcal{L}} \rangle$, where $W^{\mathcal{L}}$ is the set of all maximal $\mathcal{L}$-consistent set. $R$ is a canonical relation such that: $\Gamma R^{\mathcal{L}} \Delta \Leftrightarrow \Box \varphi \in \Gamma \Rightarrow \varphi \in \Delta$. That is, any boxed formula from $\Gamma$ can be unboxed in $\Delta$. A canonical model is a canonical frame equipped with the canonical valuation $\vartheta^{\mathcal{L}} ( p ) = \{ \Gamma \in W^{\mathcal{L}} \: \| \: p \in \Gamma \}$. This valuation maps each variable to set of all maximal $\mathcal{L}$-consistent sets that contain a given variable. Here comes the theorem: Theorem 1. $\mathcal{M}^{\mathcal{L}}, \Gamma \models \varphi \Leftrightarrow \varphi \in \Gamma$, i.e., the truth definition is generated by membership to some maximal consistent set. 2. $\varphi$ is provable in $\mathcal{L}$ iff $\mathcal{M}^{\mathcal{L}} \models \varphi$, i.e. any formula is provable iff it is true at every point of a canonical model. 3. If $\varphi \in \operatorname{Log}(\mathcal{F}^{\mathcal{L}})$, then $\varphi$ is provable in $\mathcal{L}$, i.e. any formula that valid in a canonical frame is provable. The completeness theorem for ${\bf K}$ is a simple corollary from the theorem above. Any formula that provable in ${\bf K}$ is valid in every frame. If it is valid in every frame, then it is also valid in the canonical frame. Thus, a formula provable in ${\bf K}$. That’s it. The construction above allow us to claim that any formula that valid in all possible Kripke frames is provable in ${\bf K}$. The reader might read the complete proof here. As we told above, modal logic is strictly connected with first-order logic. First-order logic extends classical logic with quantifiers as follows. Classical propositional logic deals with statements and the ways of their combinations with such connectives as a conjunction, disjunction, negation, and implication. For example, if Neil Robertson will make a maximum 147 break, then he will win the current frame (a snooker frame, not Kripke frame). This statement has the form $p \to q$. $p$ denotes Neil Robertson will make a maximum 147 break. $q$ denotes he will win the current frame. In first-order logic, one may analyse formally universal and existential statements which tell about properties of objects. More strictly, we add to the list of logical connectives quantifiers $\forall$ (for all …) and $\exists$ (there exists …). We extended the set of connectives, so one needs to redefine the notion of formula. Suppose also we have a countably infinite set of relation symbols (or letters) of arbitrary finite arity. We well denote them as $P$. Also one has a countably infinite set of individual variables $x, y, z, \dots$. The notion of formula with such set of predicate symbols: 1. If $x_1, \dots, x_n$ are variables and $P$ is a relation symbol of an arity $n$, then $P(x_1, \dots, x_n)$ is a formula 2. $\bot$ is a formula 3. If $A, B$ are formulas, then $(A \to B)$ is a formula 4. If $x$ is a variable and $\forall x \: A$ is a formula The other connectives and quantifiers are expressed as follows: 1. $\neg A := A \to \bot$ 2. $(A \land B) := \neg (A \to \neg B)$ 3. $(A \lor B) := \neg (\neg A \land \neg B)$ 4. $(\exists x \: A) := \neg \forall x \: \neg A$ Note that there is a need to distinguish free and bound variables in a first-order formula. A variable is bounded if it’s captured by a quantifier. Otherwise, a variable is called free. For instance, the variable $x$ is bounded in the formula $\exists x \: R(x, y, z)$. One may read this formula, for instance, as there is exists a point $x$ such that $x$ lies between points $y$ and $z$. Variables $y$ and $z$ are free in this formula since there are no quantifiers that bound them. Now we would like to realise what truth for a first-order formula is. Unfortunately, we don’t have truth tables with 0 and 1 as in classical propositional logic. The definition of truth for the first-order case is much more sophisticated. An interpretation of the first-order language (as the infinite set of relation symbols) is a pair $\langle A, I \rangle$, where $A$ is a non-empty set (a domain) and $I$ is an interpretation function. $I$ maps every relation letter $P$ of an arity $n$ to the function of type $A^{n} \to \{ 0, 1 \}$, i.e., some $n$-ary predicate on a domain $A$. To define truth conditions for first-order formulas, we suppose that we have an arbitrary variable assignment function $v$ such that $v$ maps every variable to some element of our domain $M$. An interpretation and variable assignment give us a first-order model, the definition of truth is the following one: 1. $\mathcal{M} \models_{\operatorname{FO}} P(x_1, \dots, x_n) \Leftrightarrow I ( P ) (v(x_1), \dots, v(x_n)) = 1$ 2. $\mathcal{M} \models_{\operatorname{FO}} A \to B \Leftrightarrow \mathcal{M} \models_{\operatorname{FO}} A \Rightarrow \mathcal{M} \models_{\operatorname{FO}} B$ 3. $\mathcal{M} \nvDash_{\operatorname{FO}} \bot$ 4. $\mathcal{M} \models_{\operatorname{FO}} \forall x \: A(x) \Leftrightarrow \mathcal{M} \models_{\operatorname{FO}} A(a)$ for each $a \in A$ Note that the truth definition for existential formulas might be expressed as follows: $\mathcal{M} \models \exists x \: A(x) \Leftrightarrow \mathcal{M} \models_{\operatorname{FO}} A(a)$ for some $a \in A$. A first-order formula is satisfiable if it’s true in some model and it’s valid if it’s true in every interpretation. We use $\models_{\operatorname{FO}}$ in the same sense as in Kripke models, but with the index $\operatorname{FO}$ to distinguish both relations that denoted equally. Let us comment on the conditions briefly. The first condition is the underlying one. As we said above any $n$-ary relation symbol maps to $n$-ary relation on observed domain, that is, $n$-ary truth function $A^n \to \{ 0,1 \}$. On the other hand, any variable (which should be free, indeed) maps to some element of a domain. After that, we apply the obtained truth function to the result of the variable assignment. We obviously require that an elementary formula $P(x_1, \dots, x_n)$ is true in the model $\mathcal{M}$ if the result of that application equals $1$. The last fact mean that a predicate is true on elements $v(x_1), \dots, v(x_n)$. For example, suppose we have a ternary relation symbol $R$ such that we have interpret $R(x,y,z)$ as $y$ lies between points $x$ and $z$. Suppose also that our domain is the real line $\mathbb{R}$. We map our ternary symbol to the truth function $\pi$ such that $\pi(a,b,c) = 1$ if and only if $a \leq b$ and $b \leq c$. We also define the variable assignment as $v(x) = \sqrt{2}$, $v(y) = \sqrt{3}$, and $v(z) = \sqrt{5}$. It is clear that in the model on real numbers $\mathcal{M}$ the formula $R(x,y,z)$ is true since $I ( R ) (v(x), v(y), v(z)) = \pi(\sqrt{2}, \sqrt{3}, \sqrt{5}) = 1$. The last equation follows from the obvious fact that $\sqrt{2} \leq \sqrt{3} \leq \sqrt{5}$. The items 2 and 3 are agreed with our understanding of what false and implication are. The last condition is the truth condition for quantified formulas. This condition describes our intuition about the circumstance when a universal statement is true. Let us consider an example. Let us assume that we want to read the formula $A(x)$ as $x$ has a father. Our domain $M$ is the set of all people who have ever lived on the planet. Then the statement $\forall x \: A(x)$ is true since every human has a father, as the reader already knows even regardless of this post. Now let us return to modal logic and Kripke models. Let us take a look at truth defitions for $\Box$ and $\Diamond$ one more time: 1. $\mathcal{M}, w \vdash \Diamond \varphi \Leftrightarrow \exists u \:\: w R u \: \& \: \mathcal{M}, u \models \varphi$, i.e., a formula $\Diamond \varphi$ is true at the point $w$, if there exists $R$-successor of $w$, namely, $u$ such that a formula $\varphi$ is true. 2. $\mathcal{M}, w \vdash \Box \varphi \Leftrightarrow \forall u \:\: w R u \Rightarrow \mathcal{M}, u \models \varphi$, i.e., a formula $\Box \varphi$ is true at the point $w$, if at every $R$-successor of $w$ a formula $\varphi$ is true. In those explanations, we used first-order quantifiers over points in Kripke models informally. But we may build the bridge between Kripke models and first-order one more precisely. Let us assume that we have the following list of predicate and relation symbols: 1. A binary relation symbols $R$ 2. A countable infinite set of unary predicate letters $P_i$, where $i = 0, 1, 2 \dots$ The translation $t$ from modal formulas to first-order ones is defined inductively 1. $t(p_i)(w) = P_i (a)$, where $x$ is a variable 2. $t(\varphi \to \psi)(w) = t(\varphi)(w) \to t(\psi)(w)$ 3. $t(\bot)(w) = \bot$ 4. $t(\Box \varphi)(w) = \forall v \: (R(w, v) \to t(\varphi)(v))$ The translation for diamonds is the following one: $t(\Diamond \phi)(w) = \exists v \: (R(w, v) \land t(\phi)(v))$ In other words, every modal formula has its first-order counterpart, a formula with one free variable $w$. The reader can see that we just mapped modalised formulas the first-order one with respect to their truth definition. But there’s the question: is the truth of formula really preserved with such a translation? Here is the lemma: Lemma Let $\mathcal{M} = \langle W, R, \vartheta \rangle$ be a Kripke model and $a \in W$, then $\mathcal{M}, a \models \varphi$ if and only if $\mathcal{M}' \models_{\operatorname{FO}} t(\varphi)(a)$. Here $\mathcal{M}'$ is the first-order model such that its domain is $W$, and interpretation for the predicate letter is agreed with the relation on an observed Kripke frame. An interpretation of unary predicate letters is agreed with the evaluation in a Kripke model as follows. $\mathcal{M}' \models P_i(w)$ if and only if $w \in \vartheta(p_i)$. Here $p_i$ is a propositional variable with the same index $i$. In other words, those unary predicate letters allow us to encode an evaluation via the first-order language. The lemma claims that a modal formula is true at some point $w$ if only if its translation (in which a point $w$ is a parameter) in the first-order language is true in the corresponding model and, hence, satisfiable. That is, one has a bridge between truth in a Kripke modal and first-order satisfiability. If a formula is true in some model at some point, it doesn’t imply its provability in the minimal normal modal logic. We would like to connect provability in ${\bf K}$ and in first-order predicate calculus. Here is the theorem proved by Johan van Benthem in the 1970s: Theorem A formula $\phi$ is provable in ${\bf K}$ iff and only the universal closure of its standard translation $\forall w \: t(\phi)(w)$ is provable in the first-order predicate calculus. That is, there’s an embedding of ${\bf K}$ to first-order logic. The lemma and the theorem above gives more precise meaning of the analogy between modalities and quantifiers which looks informal prima facie. Modal formulas are closely connected with the special properties of relations on Kripke frames. We mean a modal formula is able to express the condition on a relation in a Kripke frame. Let us consider a simple example. Suppose we have a set $W$ with transitive relation $R$. It means that for all $a, b, c \in W$, if $a R b$ and $b R c$, then $a R c$. Let us show that formula $\Box p \to \Box \Box p$ is valid on frame $\langle W, R \rangle$ if and only if this frame is transitive. For simplicity, we will use the equivalent form written in diamonds: $\Diamond \Diamond p \to \Diamond p$. It is very easy to show that these formulas are equivalent to each other: Lemma Let $\mathcal{F} = \langle W, R \rangle$, then $\mathcal{F} \models \Diamond \Diamond p \to \Diamond p$ iff $R$ is transitive. Let $\langle W, R \rangle$ be a transitive frame, $\vartheta$ be a valutation and $w \in W$. So, we have a Kripke model $\mathcal{M} = \langle W, R, \vartheta \rangle$. Suppose $\mathcal{M}, w \models \Diamond \Diamond p$. We need to show that $\mathcal{M}, w \models \Diamond p$. If $\mathcal{M}, w \models \Diamond \Diamond p$, then there exists $v \in W$ such that $w R v$ and $\mathcal{M}, v \models \Diamond p$. But, there exists $u \in W$ such that $v R u$ and $\mathcal{M}, u \models p$. Well, we have $w R v$ and $v R u$, so we also have $w R u$ by transitivity. Hence, $\mathcal{M}, w \models \Diamond p$. Consequently, $\mathcal{M}, w \models \Diamond \Diamond p \to \Diamond p$. The converse implication is harder a little bit. Let $\mathbb{F} = \langle W, R \rangle$ be a Kripke frame such that $\mathcal{F} \models \Diamond \Diamond p \to \Diamond p$, that is, this formula is true for each valuation map. Let $w, v, u \in W$ such that $w R v$ and $v R u$. Let us show that $w R u$. Suppose, we have the valuation such that $\vartheta ( p ) = \{ w \}$. So, $\langle \mathcal{F}, \vartheta \rangle, w \models \Diamond \Diamond p$. On the other hand, $\langle \mathcal{F}, \vartheta \rangle, w \models \Diamond \Diamond p \to \Diamond p$, so $\langle \mathcal{F}, \vartheta \rangle, w \models \Diamond p$. But we have only one point, where $p$ is true, $u$. So, $w R u$. Also we may make this table that describes the correspondence between modal formulas: Name Modal formula Relation ${\bf AT}$ $\Box p \to p$ for all $x \in W$, $x R x$ ${\bf A4}$ $\Box p \to \Box \Box p$ for all $x, y, z \in W$, $x R y$ and $y R z$ implies $x R z$ ${\bf AD}$ $\Box p \to \Diamond p$ Seriality: for all $x \in W$ there exists $y \in W$ such that $x R y$ ${\bf ACR}$ $\Diamond \Box p \to \Box \Diamond p$ Church-Rosser property (confluence): if $x R y$ and $x R z$, then there exists $x_1$ such that $y R x_1$ and $z R x_1$ ${\bf AB}$ $p \to \Box \Diamond p$ Symmetry: for all $x, y \in W$, $x R y$ implies $y R x$ Let us take a look at the corresponding frame properties on a picture: More formally and generally, a modal formula $\varphi$ defines (or characterises) the class of frames $\mathbb{F}$, if for each frame $\mathcal{F}$,$\mathcal{F} \models \varphi \Leftrightarrow F \in \mathcal{F}$. That is, a formula $\Box p \to p$ characterises the class of all reflexive frames, etc. It is clear that our list is incomplete as this blog post itself, but we have described the most popular formulas and the corresponding relation properties. By the way, one may obtain new modal logics adding one of these formulae as axioms and get a botanic garden of modal logics. There are three columns in the following table. The first column is the name of the concrete logic, a sort of identification mark. The second column describes how the given logic is axiomatised. Generally, ${\bf K} \oplus \Gamma$ denotes that we extend minimal normal modal logic with formulas from $\Gamma$ as the additional axioms. The third column is the completeness theorem that claims that a given logic is the set of formulas that are valid in some class of Kripke frames. Name Logic Frames ${\bf T}$ ${\bf K} \oplus {\bf AT}$ The logic of all reflexive frames ${\bf K}4$ ${\bf K} \oplus {\bf A}4$ The logic of all transitive frames ${\bf D}$ ${\bf K} \oplus {\bf AD} = {\bf K} \oplus \Diamond \top$ The logic of all serial frames ${\bf S}4$ ${\bf T} \oplus {\bf A}4$ = ${\bf K}4 \oplus {\bf AT}$ The logic of all preorders (reflexive and transitive relations) ${\bf S}4.2$ ${\bf S}4 \oplus {\bf ACR}$ The logic of all confluent preorders (preorders that satisfy the Church-Rosser property) ${\bf S}5$ ${\bf S}4 \oplus {\bf AB}$ The logic of all equivalence relations The fact that a given logic is the logic of some class of frames tells us that this logic is complete with respect to this class. For instance, when we tell that ${\bf T}$ is the logic of all reflexive frames, it means that any formula which is valid in an arbitrary reflexive frame is provable in ${\bf T}$. One may prove the corresponding completeness theorem ensuring that the formulas from the table are canonical ones. A modal formula $\varphi$ is called canonical, if the logic $\mathcal{L} = {\bf K} \oplus \varphi$ is the logic of its canonical frame. It’s not difficult to ensure that if logic has an axiomatisation with canonical formulas, then this logic is Kripke complete. For example, if we want to prove that ${\bf S}4$ is the logic of all preorders, it’s enough to check that the relation on the ${\bf S}4$-canonical frame is a preorder. Let us remember first-order logic once more. The first is to rewrite the first table above changing the properties for the third column with the relevant first-order formulas as follows: Name Modal formula Property of a frame written in the first-order language ${\bf AT}$ $\Box p \to p$ $\forall x \: (x R x)$ ${\bf A4}$ $\Box p \to \Box \Box p$ $\forall x \: \forall y \: \forall z \: (x R y \land y R z \to x R z)$ ${\bf AD}$ $\Box p \to \Diamond p$ $\forall x \: \exists y \:\: (x R y)$ ${\bf ACR}$ $\Diamond \Box p \to \Box \Diamond p$ $\forall x \: \forall y \: \forall z \:\: (x R y \land x R z \to \exists z_1 \:\: (y R z_1 \land z R z_1))$ ${\bf AB}$ $p \to \Box \Diamond p$ $\forall x \: \forall y \:\: (x R y \to y R x)$ The current question we would like to ask is there some bridge between modal formulas and first-order properties of relation in a Kripke frame. Before that, we introduce a fistful of definitions. A modal formula $\varphi$ is called elementary if the class of frames which this formula characterises is defined by some first-order formula. For example, the formulas from the table above are elementary since the corresponding frame properties might be arranged as well-formed first-order formulas. Let us define the special kind of formulas that we call Sahlqvist formulas: A box-atom is a formula of the form $\nabla p$ or $\nabla \neg p$, where $\nabla$ is a finite (possibly, empty) sequence of boxes. A Sahlqvist formula is a modal formula of the form $\Box \dots \Box (\varphi \to \psi)$. $\Box \dots \Box$ is a $n$-sequence of boxes, $n \geq 0$. $\varphi$ is formula that contains $\land$, $\lor$, $\Box$, $\Diamond$, perhaps, $0$ times. $\phi$ is constructed from box-atoms, $\land$ and $\Diamond$. For instance, any formula from our table is a Sahlqvist formula. The example of a non-Sahlqvist formula is the McKinsey formula $\Box \Diamond p \to \Diamond \Box p$, which is also non-elementary. This formula should be a separate topic for an extensive discussion. The reader may continue such a discussion herself with the classical paper by Robert Goldblatt called The McKinsey Axiom is not Canonical. Theorem (Sahlqvist’s theorem) Let $\varphi$ be a Sahlqvist formula, then $\varphi$ is canonical and elementary. Sahlqvist’s theorem is extremely non-trivial to be proved accurately in detail. However, this theorem gives us crucially significant consequences. If some normal modal logic is defined with Sahlqvist formulas as axioms, then it’s automatically Kripke complete since every axiom is canonical and elementary. Moreover, any Sahlqvist formula defines some property of a Kripke frame which is definable in the first-order language. The modal definability of first-order properties in itself has certain advantages. Let us observed them concisely and slightly philosophically. As we said at the beginning, a modal language extends the propositional one with unary operators $\Box$ and $\Diamond$. As the reader could have seen, necessity and possibility have connections with universality and existence. We established this connection defining the standard translation and embedding the minimal normal modal logic into the first-order predicate logic. On the one hand, it is a well-known result that first-order logic is undecidable. Thus, we don’t have a general procedure that defines whether a given first-order formula is true or not (alternatively, provable or not). On the other hand, we have already observed the analogy between quantifiers and modalities. On the third hand (yes, I have three hands, it’s practically useful sometimes), modal logics are mostly decidable as we will see a little bit later. That is, one has a method to check the validity of some binary relation properties by encoding them in modal logic. Such a way doesn’t work for an arbitrary property, indeed. But a large class of such characteristics might be covered via modal language. In the second part, we also take a look at the example of a formula whose class of Kripke frames is not first-order definable. In order to remain intellectually honest, we should note quite frankly and openly that there also exist first-order properties of Kripke frames which are undefinable in a modal language. A relation is called irreflexive, if it is false that $a R a$ for each $a$. The example of an irreflexive relation is the set of natural numbers with $<$, just because there is no natural number that could be less than itself, indeed. Let us define the notion of $p$-morphism, a natural homomorphism between Kripke frames. By natural homomorphism, we mean a map that preserves a structure of Kripke frame and validness at the same time. Let us explain the idea with the precise definition and related lemma. Definition Let $\mathcal{F}_1 = \langle W_1, R_1 \rangle$, $\mathcal{F}_2 = \langle W_2, R_2 \rangle$ be Kripke frames, then $p$-morphism is a map $f : \mathcal{F}_1 \to \mathcal{F}_2$ such that: 1. $f$ is monotone, i. e., $a R_1 b$ implies $f(a) R_2 f(b)$: 1. $f$ has a lifting property, i. e., $f(a) R_2 c$ implies that there exists $b \in W_1$ $a R b$ and $f(b) = c$ A $p$-mophism between Kripke models $\mathcal{M}_1 = \langle \mathcal{F}_1, v_1 \rangle$, $\mathcal{M}_2 = \langle \mathcal{F}_2, v_2 \rangle$ is a $p$-morphism $f : \mathcal{F}_1 \to \mathcal{F}_2$ such that: $\mathcal{M}_1, w \models p \Leftrightarrow \mathcal{M}_2, f(w) \models p$ for every variable $p$. If $f : \mathcal{F}_1 \to \mathcal{F}_2$ is a surjective $p$-morphism, then we will write $f : \mathcal{F}_1 \twoheadrightarrow \mathcal{F}_2$. By surjection, we mean a map $f : \mathcal{F}_1 \to \mathcal{F}_2$ such that for each $y \in \mathcal{F}_2$ there exists $x \in \mathcal{F}_1$ such that $f(x) = y$. The following lemma describes a $p$-morphism behaviour with formulas that are true in models and valid in frames. Lemma 1. If $f : \mathcal{M}_1 \to \mathcal{M}_2$, then $\mathcal{M}_1, w \models \varphi$ iff $\mathcal{M}_2, f(w) \models \varphi$ 2. If $f : \mathcal{F}_1 \twoheadrightarrow \mathcal{F}_2$, then $\operatorname{Log}(\mathcal{F}_1) \subseteq \operatorname{Log}(\mathcal{F}_2)$ Proof We prove only the first part of the lemma. Let us suppose that $\Diamond$ is a primitive modality for a technical simplicity. The base case with variables is already proved since the condition of the lemma for variables is the part of model $p$-morphism definition. Let us assume that $\varphi \eqcirc \Diamond \psi$. We prove that $\mathcal{M}_1, w \models \Diamond \psi$ iff $\mathcal{M}_2, f(w) \models \Diamond \psi$. In other words, one needs to prove two implications: 1. If $\mathcal{M}_1, w \models \Diamond \psi$, then $\mathcal{M}_2, f(w) \models \Diamond \psi$. Let $\mathcal{M}_1, w \models \Diamond \psi$, then there exists $v \in R_1(w)$ such that $\mathcal{M}_1, v \models \psi$. By induction hypothesis, $\mathcal{M}_2, f(v) \models \psi$. $f$ is a $p$-morphism, hence $f$ is monotone and $w R_1 v$ implies $f(w) R_2 f(v)$. Thus, $\mathcal{M}_2, f(w) \models \Diamond \psi$. We visualise the reasoning above as follows: 1. If $\mathcal{M}_2, f(w) \models \Diamond \psi$, then $\mathcal{M}_1, w \models \Diamond \psi$ Let $\mathcal{M}_2, f(w) \models \Diamond \psi$. Then there exists $x \in R_2 (f(w))$ such that $f(w) R_2 x$. $f$ is a $p$-morphism and $f(w) R_2 x$, then there exists $v \in W_1$ such that $w R_1 v$ and $f(v) = x$. By induction hypothesis, $\mathcal{M}_1, v \models \psi$. But $w R_1 v$, so $\mathcal{M}_1, w \models \Diamond \psi$. Take a look at the picture. Now we show that there doesn’t exist a modal formula that expresses irreflexivity of a relation. Lemma The class of all irreflexive frame is not modal definable. Proof Let $\mathcal{F}_1 = \langle \mathbb{N}, < \rangle$ be a Kripke frame, a natural numbers with less-than relation, which is obviously irreflexive. Let $\mathcal{F}_2 = \langle \{ * \}, R = \{ (, ) \} \rangle$. Let us put $f : \mathbb{N} \to \{ * \}$ as $f(x) = $ for all $x \in \mathbb{N}$. It is easy to see that this map is monotone and surjective. Let us check the lifting property. Let $f(x) R $. Let $y := x + 1$, then $x < y$ and $f(y) = *$. Then $f$ is a $p$-morphism, but $\mathcal{F}_1$ is an irreflexive frame and $\mathcal{F}_2$ is merely reflexive point. If the reader is interested in modal definability in more detail, then she might take into consideration the Goldblatt-Thomason theorem that connects modal definability, first-order definability with $p$-morphisms and some other operations on Kripke frames for further study. Decidability in modal logic The decidability of a formal system allows one to establish whether a formula is provable or not algorithmically. In modal logic, the famous Harrop’s theorem provides the most widespread method of decidability proving. Let us introduce some definitions to formulate this theorem. Definition A normal modal logic $\mathcal{L}$ is finitely axiomatisable if $\mathcal{L} = {\bf K} \oplus \Gamma$ for some $\Gamma$, where $\Gamma$ is a finite set of formulae. In other words, $\mathcal{L}$ is finitely axiomatisable if this logic extends minimal normal modal logic with some finite set of axioms. Definition A normal modal logic $\mathcal{L}$ has a finite model property (or finitely approximable), if $\mathcal{L} = \operatorname{Log}(\mathbb{F})$, where $\mathbb{F}$ is some class of finite frames. That is, a finite model property is a Kripke completeness with respect to the class of finite Kripke frames. Now we formulate the famous Harrop’s theorem: Theorem (Harrop) Let $\mathcal{L}$ be a normal modal logic such that $\mathcal{L}$ is finitely axiomatisable and has a finite model property. Then $\mathcal{L}$ is decidable. Proof We provide only a quite brief proof sketch. It is clear that the set of formulas provable in $\mathcal{L}$ is enumerable since this logic merely extends ${\bf K}$ with the finite set of axioms. On the other hand, let $\varphi \notin \mathcal{L}$. Then the set $\operatorname{Bad} = \{ \varphi \in Fm \: \| \: \varphi \notin \mathcal{L} \}$ is also enumerable, so far as one may enumerate finite frames such that $\mathcal{F} \models \mathcal{L}$ and $\mathcal{F} \nvDash \varphi$, where $\varphi \in \operatorname{Bad}$. So, if we have some finitely axiomatisable logic, then one needs to can show that this logic is complete with respect to some class of finite frames. Here, we will assume that $\Diamond$ is a primitive modality for simplicity. Definition Let $\mathcal{M} = \langle W, R, \vartheta \rangle$ be a Kripke model and $\Gamma$ a set of formulas closed under subformulas (that is, if $\varphi \in \Gamma$ and $\psi$ is a subformula of $\varphi$, then $\psi \in \Gamma$). We put the following equivalence relation: $x \sim_{\Gamma} y \Leftrightarrow \mathcal{M}, x \models \varphi \Rightarrow \mathcal{M}, y \models \varphi$, where $\varphi \in \Gamma$ Then, a filtration of a model $\mathcal{M}$ through a set $\Gamma$ is a model $\overline{M} = \langle \overline{W}, \overline{R}, \overline{\vartheta} \rangle$, where 1. $\overline{W} = W / \sim_{\Gamma}$. In other words, $\overline{W}$ is a quotient by relation $\sim_{\Gamma}$, the set of equivalence classes. 2. Let $\bar{x}, \bar{y} \in \overline{W}$, then $\bar{x} \overline{R} \bar{y} \Leftrightarrow \exists w \in \bar{x} \: \exists v \in \bar{y} \:\: w R y$ 3. $\overline{\vartheta} ( p ) = \{ \bar{x} \in \bar{W} \: \| \: \exists w \in \bar{x} \: \mathcal{M}, w \models p \}$ Here is a quite obvious observation. Suppose, one have a model $\mathcal{M} = \langle W, R, \vartheta \rangle$ and its filtration $\overline{M} = \langle \overline{W}, \overline{R}, \overline{\vartheta} \rangle$ through the set of subformulas $\operatorname{Sub}(\varphi)$, where $\varphi$ is a arbitrary formula. Then $\|\overline{W}\| \leq 2^{\|\operatorname{Sub}(\varphi)\|}$. Harrop’s theorem provides the uniform method to prove that some normal modal logic is decidable if this logic is finitely axiomatisable and has finite model property. We need filtration to prove that a given logic is finitely approximable. Here comes the first lemma about minimal filtration. Lemma Let $\mathcal{M}$ be a Kripke model and $\Gamma$ a set of formulas closed under subformulas, then $\mathcal{M}, x \models \varphi \Leftrightarrow \overline{\mathcal{M}}, \bar{x} \models \varphi$, where $\varphi \in \Gamma$. Proof Quite simple induction on $\varphi$. Let us check this statement for $\varphi \eqcirc \Diamond \psi$. At first, let us check the only if part. Let $\mathcal{M}, x \models \Diamond \psi$. Then there exists $y \in R(x)$ such that $\mathcal{M}, y \models \psi$. By induction hypothesis, $\overline{\mathcal{M}}, \bar{y} \models \psi$. But $\bar{x} \overline{R} \bar{y}$ by the definition of $\overline{R}$. This, $\overline{\mathcal{M}}, \bar{x} \models \Diamond \psi$. The converse implication has the same level of completexity. Let $\overline{\mathcal{M}}, \bar{x} \models \Diamond \psi$. Then there exists $c \in \overline{R}(\bar{x})$ such that $\overline{\mathcal{M}}, c \models \psi$. Consequently, there exist $w \in \bar{x}$ and $v \in c$ such that $w R v$. By assumption, $\mathcal{M}, v \models \psi$. Thus, $\mathcal{M}, w \models \Diamond \psi$. Now we may formulate the theorem which claims that the minimal normal modal logic is the logic of all finite Kripke frames. Theorem 1. ${\bf K} = \operatorname{Log}(\mathbb{F})$, where $\mathbb{F}$ is the class of all finite Kripke frames. 2. ${\bf T} = \operatorname{Log}(\mathbb{F})$, where $\mathbb{F}$ is the class of all finite reflexive frames. 3. ${\bf D} = \operatorname{Log}(\mathbb{F})$, where $\mathbb{F}$ is the class of all finite serial frames. 4. ${\bf B} = {\bf K} \oplus p \to \Box \Diamond p = \operatorname{Log}(\mathbb{F})$, where $\mathbb{F}$ is the class of all finite symmetric frames. Proof Let ${\bf K} \not\vdash \varphi$. It means that there exists a model $\mathcal{M}$ and $x \in \mathcal{M}$ such that $\mathcal{M}, x \not\models \psi$ according to the completeness theorem. Let $\overline{\mathcal{M}}$ be a minimal filtration of a model $\mathcal{M}$ through $\operatorname{Sub}(\psi)$, the set of subformulas of $\psi$. By the previous lemma, $\overline{\mathcal{M}}, \bar{x} \not\models \psi$. It is clear that this model is finite, since $\|\overline{W}\| \leq 2^{\|\operatorname{Sub}(\varphi)\|}$, as we observed above. (2), (3), (4) are proved similarly, but one needs to check that a minimal filtration preserves reflexivity, seriality, and symmetry. However, we are in trouble. We haven’t already discussed a finite model property of logics that contain transitivity as an axiom. Unfortunately, a minimal filtration doesn’t have to preserve the transitivity. Let $\mathcal{M} = \langle W, R, \vartheta \rangle$ be a transitive model and $\overline{M} = \langle \overline{W}, \overline{R}, \overline{\vartheta} \rangle$ a minimal filtration of $\mathcal{M}$ through $\Gamma$. Let $\bar{w}, \bar{v}, \bar{u} \in \overline{W}$ such that $\bar{w} \overline{R} \bar{v}$ and $\bar{v} \overline{R} \bar{u}$. If $\bar{w} \overline{R} \bar{v}$, then there exists $x \in \bar{w}$ and $y \in \bar{v}$ such that $x R y$. From the other hand, if $\bar{v} \overline{R} \bar{u}$, then there exists $y' \in \bar{v}$ and $z \in \bar{u}$ such that $y' R z$. It is clear that $y$ doesn’t have to see $y'$ within the equivalence class $\bar{u}$. Thus, a relation in minimally filtrated model isn’t necessarily transitive, even if the original relation is transitive. A solution to that problem is the transitive closure of a relation in a minimal filtration. Let us discuss what a transitive closure is. Suppose we have some set $W$ with a binary relation $R$. Generally, this relation is not transitive, but we’d like to extend it to the transitive one. What should we make? Suppose we have $x, y, z \in W$ such that $x R y$ and $y R z$. We add $x R z$ to the extended relation which we denote as $R^{+}$. We perform this action for each situation. That gives us a transitive version of a relation. Here we going to extend a relation in a minimal filtration. This solution was proposed initially by Dov Gabbay. We will denote this closure $(\overline{R})^{+}$. A transitive closure of a relation in a minimal filtration allows us to prove that ${\bf K}4$ is the logic of finite transitive frames. Firstly, we formulate the lemma that explains why a transitive closure is a good idea: Lemma Let $\mathcal{M} = \langle W, R, \vartheta \rangle$ be a transitive model and $\overline{M} = \langle \overline{W}, \overline{R}, \overline{\vartheta} \rangle$ a minimal filtration through $\Gamma$, then the following conditions hold: 1. If $a \overline{R} b$, then $a (\overline{R})^{+} b$, where $a, b \in \overline{W}$. This statement claims that a relation in a minimal filtration is a subrelation if its transitive closure. 2. If $w R v$, then $\bar{w} \overline{R} \bar{v}$. If two relation are connected with the underlying relation, then a transitive closure preserve this connection. 3. Let $\bar{w} (\overline{R})^{+} \bar{v}$ and $\mathcal{M}, w \models \Box \varphi$, then $\mathcal{M}, v \models \varphi$. Here we claim that a transitive closure should respect the truth condition for $\Box$. Using the lemma above, it is not so hard to obtain the following theorem: Theorem 1. ${\bf K}4 = \operatorname{Log}(\mathbb{F})$, where $\mathbb{F}$ is the class of all finite transitive frames 2. ${\bf S}4 = \operatorname{Log}(\mathbb{F})$, where $\mathbb{F}$ is the class of all finite preorders. 3. ${\bf S}5 = \operatorname{Log}(\mathbb{F})$, where $\mathbb{F}$ is the class of all finite equivalence relations. The theorem above allows us to claim that the logics ${\bf K}4$, ${\bf S}4$, and ${\bf S}5$ have a finite model property. All these logics are finitely axiomatisable. Then, by Harrop’s theorem, ${\bf K}4$, ${\bf S}4$, and ${\bf S}5$ are decidable. That is, one has a uniform method to provide a countermodel in which every unprovable formula fails. Summary We discussed a brief history of modal logic and its mathematical and philosophical roots. After that, we introduced the grammar of modal logic to define what modal formula is. As we have already told, the definition of a modal language is not enough to deal with modal logic. From a syntactical perspective, we have formal proofs as derivation with axioms and inference rules. We defined normal modal logic as a set of formulas with specific limitations. As an underlying logic, we fixed minimal normal logic ${\bf K}$. Here, the other modal logics merely extend the minimal normal modal one with the additional axioms. Note that syntax doesn’t answer the question about the truth of a formula. The distinction between proof and truth is a foundation of the logical culture in itself. Truth is a semantical concept. To define the truth condition for modal formulas, we defined Kripke frames and Kripke models. After that, we formulated the completeness theorems for the list of modal logics. As usual, completeness tells us that, very roughly speaking, any valid formula is provable. The underlying modal logic ${\bf K}$ is a logic of all possible Kripke frames. Other modal logics are complete with respect to a narrower class of frames with the relation in which is restricted somehow. Alright, we have the completeness theorem for some logic, but we also would like to define whether a given formula is provable or not algorithmically. For this purpose, we took a look at methods of proving the fact that a given normal modal logic has finite model property. Finite model property with finite axiomatisation gives us decidability according to the Harrop’s theorem. We studied the background of modal logic so that we could continue our journey in more concrete case studies. In the second part, we will overview concrete branches of modal logic, such as topological semantics, provability logic, and temporal logic. We will discuss how exactly modal logic is connected with geometry, the foundations of mathematics, and computer science. I sincerely hope the reader managed to survive in this landscape of such an abstract and theoretically saturated material.	16,266	59,255	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 692, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.8125	4	CC-MAIN-2024-33	latest	en	0.947238
https://www.r-bloggers.com/2024/01/probably-more-than-chance-a-beginners-guide-to-probability-distributions-in-r/	1,726,283,522,000,000,000	text/html	crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00449.warc.gz	880,164,978	25,505	Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. ### Introduction Welcome to “Probably More Than Chance: A Beginner’s Guide to Probability Distributions in R.” In this journey, we’ll explore the intriguing world of probability distributions, an essential concept in statistics and data analysis. These distributions are the backbone of understanding how data behaves under various conditions. R, a powerful tool for statistical computing, offers us an intuitive way to explore and visualize these distributions. Whether you’re a beginner or an enthusiast in the realm of data analysis, this guide will help you grasp the basics and see how R can bring statistical theories to life. Now, let’s dive into the fascinating world of probability distributions, starting with understanding what they are and why they’re crucial in data analysis. ### Understanding the Basics: What Are Probability Distributions? At the heart of statistical analysis lies a fundamental question: “How likely is it?” This is where probability distributions come into play. Imagine probability distributions as a roadmap, guiding us through the landscape of potential outcomes in a random event. They tell us not just what could happen, but how likely each outcome is. In a simple analogy, think of a probability distribution as a recipe book for a grand feast of data. Each recipe (distribution) has its unique ingredients (parameters) and preparation methods (formulas), leading to a variety of delightful dishes (outcomes). Whether it’s the bell curve of a normal distribution or the discrete bars of a binomial outcome, each distribution serves up insights into the nature of our data. Now, with our foundational understanding set, let’s step into the world of specific distributions, starting with the classic: the Normal distribution. ### The Normal Distribution: The Bell Curve in R The Normal distribution, frequently encountered in real-world scenarios like IQ scores or heights in a population, exemplifies a symmetric bell-shaped curve. It symbolizes situations where most observations cluster around a central mean, with fewer occurrences as we move away from the center. Here’s how you can generate and visualize a Normal distribution in R: ```# Generating a normal distribution in R normal_data <- rnorm(1000, mean = 50, sd = 10) hist(normal_data, main = “Normal Distribution”, xlab = “Values”, breaks = 30, col = “blue”)``` In this snippet rnorm() is the function for generating normally distributed random numbers: • The first argument, 1000, indicates the number of observations to generate. • mean = 50 sets the average value around which data points are centered. • sd = 10 specifies the standard deviation, reflecting how spread out the values are around the mean. Running this code in R produces a histogram that visually represents the Normal distribution, showing the characteristic bell curve. ### The Poisson Distribution: Predicting Rare Events The Poisson distribution is your statistical crystal ball for predicting the frequency of rare events. It’s like observing shooting stars on a dark night; you know they’re scarce, but you want to predict their occurrence. Commonly, it’s used for counting events like the number of emails you receive in a day or cars passing through an intersection. In R, simulating a Poisson distribution is straightforward: ```# Generating a Poisson distribution in R poisson_data <- rpois(1000, lambda = 3) hist(poisson_data, main = “Poisson Distribution”, xlab = “Occurrences”, breaks = 20, col = “green”)``` The key function here is rpois, which generates random numbers following a Poisson distribution: • The first argument, 1000, is the number of random values we want to generate. • lambda = 3 represents the average rate of occurrence (λ). In this example, it’s as if we expect, on average, 3 events (like emails or cars) in a given time frame. By running this code, you create a histogram that illustrates how often different counts of an event occur, showcasing the unique characteristics of the Poisson distribution. ### The Uniform Distribution: Equal Probability Events Imagine you’re picking a card at random from a well-shuffled deck. Each card has an equal chance of being selected — this scenario exemplifies the Uniform distribution. It’s the go-to model when every outcome has the same likelihood, whether it’s rolling a fair dice or selecting a random number between 0 and 1. In R, creating a Uniform distribution is as simple as: ```# Generating a Uniform distribution in R uniform_data <- runif(1000, min = 0, max = 1) hist(uniform_data, main = “Uniform Distribution”, xlab = “Values”, breaks = 25, col = “red”)``` Let’s break down the function runif() generating uniformly distributed numbers: • 1000 specifies the quantity of random numbers to generate. • min = 0 and max = 1 set the lower and upper limits between which the numbers will be uniformly distributed. This code creates a histogram showing a flat, even distribution of values, which is the hallmark of the Uniform distribution in its purest form. ### The Exponential Distribution: Time Between Events The Exponential distribution is akin to timing the unpredictable yet inevitable — like waiting for a meteor to shoot across the sky. It’s primarily used to model the time elapsed between events, such as the lifespan of a machine part or the interval between bus arrivals. Simulating an Exponential distribution in R is quite straightforward: ```# Generating an Exponential distribution in R exponential_data <- rexp(1000, rate = 0.2) hist(exponential_data, main = “Exponential Distribution”, xlab = “Time”, breaks = 30, col = “purple”)``` In this snippet, rexp is the function at play. It generates random numbers following an Exponential distribution: • The first argument, 1000, is the number of random values to generate. • rate = 0.2 sets the rate parameter, which is the inverse of the mean. In this case, it implies an average waiting time of 5 units (since 1/0.2 = 5). Running this code in R produces a histogram that visualizes the Exponential distribution, showcasing how the frequency of occurrences decreases as time increases. ### The Binomial Distribution: Success or Failure Outcomes The Binomial distribution is the statistical equivalent of a coin flip experiment, but with more coins and more flips. It’s perfect for scenarios with two possible outcomes: success or failure, win or lose, yes or no. For instance, it can be used to predict the number of heads in 100 coin tosses or the likelihood of a certain number of successes in a series of yes/no experiments. Generating a Binomial distribution in R is quite intuitive: ```# Generating a Binomial distribution in R binomial_data <- rbinom(1000, size = 10, prob = 0.5) hist(binomial_data, main = “Binomial Distribution”, xlab = “Number of Successes”, breaks = 10, col = “orange”)``` Here’s a look at the function and its arguments. rbinom is the function for generating binomially distributed numbers: • 1000 is the number of experiments or trials to simulate. • size = 10 specifies the number of trials in each experiment (like flipping a coin 10 times per experiment) • prob = 0.5 sets the probability of success on each trial (similar to a fair coin having a 50% chance of landing heads). This code snippet creates a histogram that illustrates the distribution of successes across multiple experiments, providing a visual representation of the Binomial distribution. ### Practical Applications: Using Distributions in Real-World Data Analysis The power of probability distributions extends far beyond academic exercises; they are vital tools in making informed decisions from real-world data. For instance, an environmental scientist might use the Exponential distribution to model the time until the next significant weather event, or a retail analyst might employ the Normal distribution to understand customer spending behaviors. Let’s explore a more complex example using R, involving customer purchase behavior: ```# Example: Analyzing Customer Purchase Behavior set.seed(123) # For reproducibility purchase_times <- rexp(200, rate = 1/45) purchase_amounts <- rnorm(200, mean = 200, sd = 50) plot(purchase_times, purchase_amounts, main = “Customer Purchase Behavior”, xlab = “Time Between Purchases (days)”, ylab = “Purchase Amount (\$)”, pch = 19, col = “brown”)``` In this scenario we simulate the time between purchases (purchase_times) using rexp, assuming an average of 45 days between purchases. This reflects the Exponential distribution’s ability to model waiting times. purchase_amounts represents the amount spent on each purchase, modeled with rnorm to reflect a Normal distribution with an average purchase of \$200 and a standard deviation of \$50. The plot function creates a scatter plot, allowing us to visualize the relationship between the time intervals and purchase amounts. This example offers a glimpse into how real-world phenomena, like customer behavior, can be modeled and analyzed using different probability distributions in R. The insights drawn from such analysis can significantly influence business strategies and decision-making. ### Conclusion Our exploration of probability distributions in R, “Probably More Than Chance: A Beginner’s Guide to Probability Distributions in R,” has taken us through a variety of statistical landscapes. From the ubiquitous bell curve of the Normal distribution to the event-counting Poisson, each distribution offers unique insights into data. We’ve seen how these distributions are not just theoretical abstractions but are deeply embedded in the fabric of everyday data analysis. By using R, we’ve brought these concepts to life, offering both visual and quantitative understanding. Whether it’s predicting customer behavior or analyzing environmental patterns, these tools empower us to make data-driven decisions with greater confidence. As you continue your journey in data analysis with R, remember that each dataset tells a story, and probability distributions are key to unlocking their meanings. Embrace the power of R and these statistical techniques to uncover the hidden narratives in your data. But our journey doesn’t end here. Stay tuned for our next article, where we’ll dive into the world of hypothesis testing. We’ll unravel how to make definitive statements about your data, moving from probability to certainty. From setting up hypotheses to interpreting p-values, we’ll demystify the process, making it accessible for beginners and a refresher for seasoned practitioners. Get ready to test your assumptions and validate your theories with R in our upcoming guide! Probably More Than Chance: A Beginner’s Guide to Probability Distributions in R was originally published in Numbers around us on Medium, where people are continuing the conversation by highlighting and responding to this story.	2,257	10,975	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.53125	4	CC-MAIN-2024-38	latest	en	0.855273
http://docplayer.net/24823234-14-2-the-mean-value-and-the-root-mean-square-value-introduction-prerequisites-learning-outcomes.html	1,545,005,692,000,000,000	text/html	crawl-data/CC-MAIN-2018-51/segments/1544376828018.77/warc/CC-MAIN-20181216234902-20181217020902-00011.warc.gz	73,895,065	26,898	# 14.2. The Mean Value and the Root-Mean-Square Value. Introduction. Prerequisites. Learning Outcomes Save this PDF as: Size: px Start display at page: Download "14.2. The Mean Value and the Root-Mean-Square Value. Introduction. Prerequisites. Learning Outcomes" ## Transcription 1 he Men Vlue nd the Root-Men-Squre Vlue 4. Introduction Currents nd voltges often vry with time nd engineers my wish to know the men vlue of such current or voltge over some prticulr time intervl. he men vlue of time-vrying function is defined in terms of n integrl. An ssocited quntity is the root-men-squre (r.m.s). For exmple, the r.m.s. vlue of current is used in the clcultion of the power dissipted by resistor. Prerequisites Before strting this Section you should... Lerning Outcomes On completion you should be ble to... be ble to clculte definite integrls be fmilir with tble of trigonometric identities clculte the men vlue of function clculte the root-men-squre vlue of function HELM (8): Workbook 4: Applictions of Integrtion 2 . Averge vlue of function Suppose time-vrying function f(t) is defined on the intervl t b. he re, A, under the grph of f(t) is given by the integrl A = f(t) f(t) dt. his is illustrted in Figure 5. f(t) m b t b t () the re under the curve from t = to t = b Figure 5 (b) the re under the curve nd the re of the rectngle re equl On Figure 3 we hve lso drwn rectngle with bse spnning the intervl t b nd which hs the sme re s tht under the curve. Suppose the height of the rectngle is m. hen re of rectngle = re under curve m(b ) = f(t) dt m = b he vlue of m is the men vlue of the function cross the intervl t b. f(t) dt Key Point he men vlue of function f(t) in the intervl t b is b f(t) dt he men vlue depends upon the intervl chosen. If the vlues of or b re chnged, then the men vlue of the function cross the intervl from to b will in generl chnge s well. Exmple Find the men vlue of f(t) = t over the intervl t 3. Solution Using Key Point with = nd b = 3 nd f(t) = t men vlue = b f(t) dt = 3 3 t dt = [ t 3 3 ] 3 = 3 3 HELM (8): Section 4.: he Men Vlue nd the Root-Men-Squre Vlue 3 sk Find the men vlue of f(t) = t over the intervl t 5. Use Key Point with = nd b = 5 to write down the required integrl: Your solution men vlue = Answer 5 5 t dt Now evlute the integrl: Your solution men vlue = Answer 5 5 t dt = 3 [ t 3 3 ] 5 = 3 [ ] = = 3 Engineering Exmple Sonic boom Introduction Impulsive signls re described by their pek mplitudes nd their durtion. Another quntity of interest is the totl energy of the impulse. he effect of blst wve from n explosion on structures, for exmple, is relted to its totl energy. his Exmple looks t the clcultion of the energy on sonic boom. Sonic booms re cused when n ircrft trvels fster thn the speed of sound in ir. An idelized sonic-boom pressure wveform is shown in Figure 6 where the instntneous sound pressure p(t) is plotted versus time t. his wve type is often clled n N-wve becuse it resembles the shpe of the letter N. he energy in sound wve is proportionl to the squre of the sound pressure. P p(t) t P Figure 6: An idelized sonic-boom pressure wveform HELM (8): Workbook 4: Applictions of Integrtion 4 Problem in words Clculte the energy in n idel N-wve sonic boom in terms of its pek pressure, its durtion nd the density nd sound speed in ir. Mthemticl sttement of problem Represent the positive pek pressure by P nd the durtion by. he totl coustic energy E crried cross unit re norml to the sonic-boom wve front during time is defined by E = <p(t) > /ρc where ρ is the ir density, c the speed of sound nd the time verge of [p(t)] is <p(t) > = p(t) dt () Find n pproprite expression for p(t). () () (b) Hence show tht E cn be expressed in terms of P,, ρ nd c s E = P 3ρc. Mthemticl nlysis () he intervl of integrtion needed to compute () is [, ]. herefore it is necessry to find n expression for p(t) only in this intervl. Figure 6 shows tht, in this intervl, the dependence of the sound pressure p on the vrible t is liner, i.e. p(t) = t + b. From Figure 6 lso p() = P nd p( ) = P. he constnts nd b re determined from these conditions. At t =, + b = P implies tht b = P. At t =, + b = P implies tht = P /. Consequently, the sound pressure in the intervl [, ] my be written p(t) = P (b) his expression for p(t) my be used to compute the integrl () p(t) dt = = = P ( P t + P ) dt = [ 4P 3 t3 P ( ] t + P t ) = P /3. t + P. ( 4P t 4P ) t + P dt Hence, from Eqution (), the totl coustic energy E crried cross unit re norml to the sonicboom wve front during time is E = P 3ρc. Interprettion he energy in n N-wve is given by third of the sound intensity corresponding to the pek pressure multiplied by the durtion. HELM (8): Section 4.: he Men Vlue nd the Root-Men-Squre Vlue 3 5 Exercises. Clculte the men vlue of the given functions cross the specified intervl. () f(t) = + t cross [, ] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(t) = t cross [, ] (e) f(z) = z + z cross [, 3]. Clculte the men vlue of the given functions over the specified intervl. () f(x) = x 3 cross [, 3] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(z) = z 3 cross [, ] 3. Clculte the men vlue of the following: () f(t) = sin t cross [ ], π (b) f(t) = sin t cross [, π] (c) f(t) = sin ωt cross [, π] (d) f(t) = cos t cross [ ], π (e) f(t) = cos t cross [, π] (f) f(t) = cos ωt cross [, π] (g) f(t) = sin ωt + cos ωt cross [, ] 4. Clculte the men vlue of the following functions: () f(t) = t + cross [, 3] (b) f(t) = e t cross [, ] (c) f(t) = + e t cross [, ] Answers. () (b) (c) (d) 4 (e) () (b).693 (c).948 (d) 3. () (b) (c) [ cos(πω)] π π πω (d) + sin ω cos ω (g) ω 4. () 4 (b).75 (c).75 9 π (e) (f) sin(πω) πω 4 HELM (8): Workbook 4: Applictions of Integrtion 6 . Root-men-squre vlue of function If f(t) is defined on the intervl t b, the men-squre vlue is given by the expression: b [f(t)] dt his is simply the men vlue of [f(t)] over the given intervl. he relted quntity: the root-men-squre (r.m.s.) vlue is given by the following formul. Key Point 3 Root-Men-Squre Vlue b r.m.s vlue = [f(t)] b dt he r.m.s. vlue depends upon the intervl chosen. If the vlues of or b re chnged, then the r.m.s. vlue of the function cross the intervl from to b will in generl chnge s well. Note tht when finding n r.m.s. vlue the function must be squred before it is integrted. Exmple 3 Find the r.m.s. vlue of f(t) = t cross the intervl from t = to t = 3. Solution r.m.s = b [f(t)] dt = 3 3 [t ] dt = 3 t 4 dt = [ t 5 5 ] HELM (8): Section 4.: he Men Vlue nd the Root-Men-Squre Vlue 5 7 Exmple 4 Clculte the r.m.s vlue of f(t) = sin t cross the intervl t π. Solution Here = nd b = π so r.m.s = π π sin t dt. he integrl of sin t is performed by using trigonometricl identities to rewrite it in the lterntive form ( cos t). his technique ws described in 3.7. π [ ] π ( cos t) sin t r.m.s. vlue = dt = t = π 4π 4π (π) = =.77 hus the r.m.s vlue is.77 to 3 d.p. In the previous Exmple the mplitude of the sine wve ws, nd the r.m.s. vlue ws.77. In generl, if the mplitude of sine wve is A, its r.m.s vlue is.77a. Key Point 4 he r.m.s vlue of ny sinusoidl wveform tken cross n intervl of width equl to one period is.77 mplitude of the wveform. Engineering Exmple 3 Electrodynmic meters Introduction A dynmometer or electrodynmic meter is n nlogue instrument tht cn mesure d.c. current or.c. current up to frequency of khz. A typicl dynmometer is shown in Figure 7. It consists of circulr dynmic coil positioned in mgnetic field produced by two wound circulr sttor coils connected in series with ech other. he torque on the moving coil depends upon the mutul inductnce between the coils given by: = I I dm 6 HELM (8): Workbook 4: Applictions of Integrtion 8 where I is the current in the fixed coil, I the current in the moving coil nd θ is the ngle between the coils. he torque is therefore proportionl to the squre of the current. If the current is lternting the moving coil is unble to follow the current nd the pointer position is relted to the men vlue of the squre of the current. he scle cn be suitbly grduted so tht the pointer position shows the squre root of this vlue, i.e. the r.m.s. current. Pointer Scle Moving coil Spring Fixed sttor coils Figure 7: An electrodynmic meter Problem in words A dynmometer is in circuit in series with 4 Ω resistor, rectifying device nd 4 V r.m.s lternting sinusoidl power supply. he rectifier resists current with resistnce of Ω in one direction nd resistnce of kω in the opposite direction. Clculte the reding indicted on the meter. Mthemticl Sttement of the problem We know from Key Point 4 in the text tht the r.m.s. vlue of ny sinusoidl wveform tken cross n intervl equl to one period is.77 mplitude of the wveform. Where.77 is n pproximtion of. his llows us to stte tht the mplitude of the sinusoidl power supply will be: V pek = V rms = V rms In this cse the r.m.s power supply is 4 V so we hve V pek = 4 = V During the prt of the cycle where the voltge of the power supply is positive the rectifier behves s resistor with resistnce of Ω nd this is combined with the 4 Ω resistnce to give resistnce of 6 Ω in totl. Using Ohm s lw V = IR I = V R As V = V pek sin(θ) where θ = ωt where ω is the ngulr frequency nd t is time we find tht during the positive prt of the cycle Irms = π ( ) sin(θ) π 6 HELM (8): Section 4.: he Men Vlue nd the Root-Men-Squre Vlue 7 9 During the prt of the cycle where the voltge of the power supply is negtive the rectifier behves s resistor with resistnce of kω nd this is combined with the 4 Ω resistnce to give 4 Ω in totl. So we find tht during the negtive prt of the cycle Irms = π π π ( ) sin(θ) 4 herefore over n entire cycle Irms = π π ( sin(θ) 6 ) + π π π We cn clculte this vlue to find I rms nd therefore I rms. Mthemticl nlysis Irms = π π ( sin(θ) 6 ( π Irms = sin (θ) π 36 ) + π π π + π π ( ) sin(θ) 4 ( ) sin(θ) 4 sin (θ) 96 ) Substituting the trigonometric identity sin (θ) cos(θ) we get I rms = = = ( π cos(θ) 4π π ( [ θ 36 sin(θ) 7 ] π + π ( π 4π 36 + π ) = A 96 + π cos(θ) 96 [ θ 96 sin(θ) 39 ] π π ) ) I rms =.3 A to d.p. Interprettion he reding on the meter would be.3 A. 8 HELM (8): Workbook 4: Applictions of Integrtion 10 Exercises. Clculte the r.m.s vlues of the given functions cross the specified intervl. () f(t) = + t cross [, ] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(t) = t cross [, ] (e) f(z) = z + z cross [, 3]. Clculte the r.m.s vlues of the given functions over the specified intervl. () f(x) = x 3 cross [, 3] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(z) = z 3 cross [, ] 3. Clculte the r.m.s vlues of the following: [ () f(t) = sin t cross, π ] (b) f(t) = sin t cross [, π] (c) f(t) = sin ωt cross [, π] (d) f(t) = cos t cross [ ], π (e) f(t) = cos t cross [, π] (f) f(t) = cos ωt cross [, π] (g) f(t) = sin ωt + cos ωt cross [, ] 4. Clculte the r.m.s vlues of the following functions: Answers () f(t) = t + cross [, 3] (b) f(t) = e t cross [, ] (c) f(t) = + e t cross [, ]. ().87 (b).575 (c).447 (d).7889 (e) ().4957 (b).77 (c) (d) ().77 (b).77 (c) (d).77 (e).77 (f) 4. ().58 (b).3466 (c).74 sin πω cos πω πω sin πω cos πω + πω (g) + sin ω ω HELM (8): Section 4.: he Men Vlue nd the Root-Men-Squre Vlue 9 ### 15.6. The mean value and the root-mean-square value of a function. Introduction. Prerequisites. Learning Outcomes. Learning Style The men vlue nd the root-men-squre vlue of function 5.6 Introduction Currents nd voltges often vry with time nd engineers my wish to know the verge vlue of such current or voltge over some prticulr time ### Arc Length. P i 1 P i (1) L = lim. i=1 Arc Length Suppose tht curve C is defined by the eqution y = f(x), where f is continuous nd x b. We obtin polygonl pproximtion to C by dividing the intervl [, b] into n subintervls with endpoints x, x,...,x ### Graphs on Logarithmic and Semilogarithmic Paper 0CH_PHClter_TMSETE_ 3//00 :3 PM Pge Grphs on Logrithmic nd Semilogrithmic Pper OBJECTIVES When ou hve completed this chpter, ou should be ble to: Mke grphs on logrithmic nd semilogrithmic pper. Grph empiricl ### 5.6 Substitution Method 5.6 Substitution Method Recll the Chin Rule: (f(g(x))) = f (g(x))g (x) Wht hppens if we wnt to find f (g(x))g (x) dx? The Substitution Method: If F (x) = f(x), then f(u(x))u (x) dx = F (u(x)) + C. Steps: ### 2 DIODE CLIPPING and CLAMPING CIRCUITS 2 DIODE CLIPPING nd CLAMPING CIRCUITS 2.1 Ojectives Understnding the operting principle of diode clipping circuit Understnding the operting principle of clmping circuit Understnding the wveform chnge of ### Sect 8.3 Triangles and Hexagons 13 Objective 1: Sect 8.3 Tringles nd Hexgons Understnding nd Clssifying Different Types of Polygons. A Polygon is closed two-dimensionl geometric figure consisting of t lest three line segments for its ### Area Between Curves: We know that a definite integral Are Between Curves: We know tht definite integrl fx) dx cn be used to find the signed re of the region bounded by the function f nd the x xis between nd b. 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Evlute the double integrl x dx d. Solution. We cn either slice the region verticll or horizontll. ( x x Slicing ### 1+(dy/dx) 2 dx. We get dy dx = 3x1/2 = 3 x, = 9x. Hence 1 + Mth.9 Em Solutions NAME: #.) / #.) / #.) /5 #.) / #5.) / #6.) /5 #7.) / Totl: / Instructions: There re 5 pges nd totl of points on the em. You must show ll necessr work to get credit. You m not use our ### 2.6.3 Characteristic of Compound-wound motor connection common use aids speed falls opposes speed increases ratio series-to-shunt field ampere-turns .6.3 Chrcteristic of Compound-wound motor A compound-wound motor hs both series nd shunt field winding, (i.e. one winding in series nd one in prllel with the rmture circuit), by vrying the number of turns ### 4.0 5-Minute Review: Rational Functions mth 130 dy 4: working with limits 1 40 5-Minute Review: Rtionl Functions DEFINITION A rtionl function 1 is function of the form y = r(x) = p(x) q(x), 1 Here the term rtionl mens rtio s in the rtio of two ### AREA OF A SURFACE OF REVOLUTION AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7. ### 11. Fourier series. sin mx cos nx dx = 0 for any m, n, sin 2 mx dx = π. . Fourier series Summry of the bsic ides The following is quick summry of the introductory tretment of Fourier series in MATH. We consider function f with period π, tht is, stisfying f(x + π) = f(x) for ### Pythagoras theorem and trigonometry (2) HPTR 10 Pythgors theorem nd trigonometry (2) 31 HPTR Liner equtions In hpter 19, Pythgors theorem nd trigonometry were used to find the lengths of sides nd the sizes of ngles in right-ngled tringles. These ### Section 4.3. By the Mean Value Theorem, for every i = 1, 2, 3,..., n, there exists a point c i in the interval [x i 1, x i ] such that Difference Equtions to Differentil Equtions Section 4.3 The Fundmentl Theorem of Clculus We re now redy to mke the long-promised connection between differentition nd integrtion, between res nd tngent lines. ### 9 CONTINUOUS DISTRIBUTIONS 9 CONTINUOUS DISTIBUTIONS A rndom vrible whose vlue my fll nywhere in rnge of vlues is continuous rndom vrible nd will be ssocited with some continuous distribution. Continuous distributions re to discrete ### Chapter G - Problems Chpter G - Problems Blinn College - Physics 2426 - Terry Honn Problem G.1 A plne flies horizonlly t speed of 280 mês in position where the erth's mgnetic field hs mgnitude 6.0µ10-5 T nd is directed t n ### Treatment Spring Late Summer Fall 0.10 5.56 3.85 0.61 6.97 3.01 1.91 3.01 2.13 2.99 5.33 2.50 1.06 3.53 6.10 Mean = 1.33 Mean = 4.88 Mean = 3. The nlysis of vrince (ANOVA) Although the t-test is one of the most commonly used sttisticl hypothesis tests, it hs limittions. The mjor limittion is tht the t-test cn be used to compre the mens of only ### Example A rectangular box without lid is to be made from a square cardboard of sides 18 cm by cutting equal squares from each corner and then folding 1 Exmple A rectngulr box without lid is to be mde from squre crdbord of sides 18 cm by cutting equl squres from ech corner nd then folding up the sides. 1 Exmple A rectngulr box without lid is to be mde ### Binary Representation of Numbers Autar Kaw Binry Representtion of Numbers Autr Kw After reding this chpter, you should be ble to: 1. convert bse- rel number to its binry representtion,. convert binry number to n equivlent bse- number. In everydy ### Name: Lab Partner: Section: Chpter 4 Newton s 2 nd Lw Nme: Lb Prtner: Section: 4.1 Purpose In this experiment, Newton s 2 nd lw will be investigted. 4.2 Introduction How does n object chnge its motion when force is pplied? A force ### Mathematics Higher Level Mthemtics Higher Level Higher Mthemtics Exmintion Section : The Exmintion Mthemtics Higher Level. Structure of the exmintion pper The Higher Mthemtics Exmintion is divided into two ppers s detiled below: ### Section 5-4 Trigonometric Functions 5- Trigonometric Functions Section 5- Trigonometric Functions Definition of the Trigonometric Functions Clcultor Evlution of Trigonometric Functions Definition of the Trigonometric Functions Alternte Form ### Polynomial Functions. 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Since ### 1 Numerical Solution to Quadratic Equations cs42: introduction to numericl nlysis 09/4/0 Lecture 2: Introduction Prt II nd Solving Equtions Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mrk Cowlishw Numericl Solution to Qudrtic Equtions Recll ### CHAPTER 5a. SIMULTANEOUS LINEAR EQUATIONS CHAPTER 5. SIMULTANEOUS LINEAR EQUATIONS A. J. Clrk School of Engineering Deprtment of Civil nd Environmentl Engineering by Dr. Ibrhim A. Asskkf Spring 00 ENCE 0 - Computtion Methods in Civil Engineering ### Memorial University of Newfoundland Department of Physics and Physical Oceanography. Department of Physics and Physical Oceanography Memoril University of Newfoundlnd Deprtment of Physics nd Physicl Ocenogrphy Memoril Physics University 2055 Lbortory of Newfoundlnd Deprtment of Physics nd Physicl Ocenogrphy Introduction Physics 2055 ### m, where m = m 1 + m m n. Lecture 7 : Moments nd Centers of Mss If we hve msses m, m 2,..., m n t points x, x 2,..., x n long the x-xis, the moment of the system round the origin is M 0 = m x + m 2 x 2 + + m n x n. The center of ### addition, there are double entries for the symbols used to signify different parameters. These parameters are explained in this appendix. APPENDIX A: The ellipse August 15, 1997 Becuse of its importnce in both pproximting the erth s shpe nd describing stellite orbits, n informl discussion of the ellipse is presented in this ppendix. The ### Basic Analysis of Autarky and Free Trade Models Bsic Anlysis of Autrky nd Free Trde Models AUTARKY Autrky condition in prticulr commodity mrket refers to sitution in which country does not engge in ny trde in tht commodity with other countries. Consequently ### The Quadratic Formula and the Discriminant 9-9 The Qudrtic Formul nd the Discriminnt Objectives Solve qudrtic equtions by using the Qudrtic Formul. Determine the number of solutions of qudrtic eqution by using the discriminnt. Vocbulry discriminnt ### Applications to Physics and Engineering Section 7.5 Applictions to Physics nd Engineering Applictions to Physics nd Engineering Work The term work is used in everydy lnguge to men the totl mount of effort required to perform tsk. In physics ### PROBLEMS 13 - APPLICATIONS OF DERIVATIVES Page 1 PROBLEMS - APPLICATIONS OF DERIVATIVES Pge ( ) Wter seeps out of conicl filter t the constnt rte of 5 cc / sec. When the height of wter level in the cone is 5 cm, find the rte t which the height decreses. ### MATH 150 HOMEWORK 4 SOLUTIONS MATH 150 HOMEWORK 4 SOLUTIONS Section 1.8 Show tht the product of two of the numbers 65 1000 8 2001 + 3 177, 79 1212 9 2399 + 2 2001, nd 24 4493 5 8192 + 7 1777 is nonnegtive. Is your proof constructive ### Theory of Forces. Forces and Motion his eek extbook -- Red Chpter 4, 5 Competent roblem Solver - Chpter 4 re-lb Computer Quiz ht s on the next Quiz? Check out smple quiz on web by hurs. ht you missed on first quiz Kinemtics - Everything ### Tests for One Poisson Mean Chpter 412 Tests for One Poisson Men Introduction The Poisson probbility lw gives the probbility distribution of the number of events occurring in specified intervl of time or spce. The Poisson distribution ### Rotating DC Motors Part II Rotting Motors rt II II.1 Motor Equivlent Circuit The next step in our consiertion of motors is to evelop n equivlent circuit which cn be use to better unerstn motor opertion. The rmtures in rel motors ### Mathematics. Vectors. hsn.uk.net. Higher. Contents. Vectors 128 HSN23100 hsn.uk.net Higher Mthemtics UNIT 3 OUTCOME 1 Vectors Contents Vectors 18 1 Vectors nd Sclrs 18 Components 18 3 Mgnitude 130 4 Equl Vectors 131 5 Addition nd Subtrction of Vectors 13 6 Multipliction by ### α Must use radians. τ = F trnsltionl nd rottionl nlogues trnsltionl ( liner ) motion rottionl motion trnsltionl displcement d A = r A Δθ ngulr displcement Δx Δy Must use rdins. 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Lvl Kennesw Stte University August, 8 Abstrct This hndout contins mteril on very importnt integrtion method clled integrtion by substitution. Substitution is ### Electric Circuits. Simple Electric Cell. Electric Current Electric Circuits Count Alessndro olt (745-87) Georg Simon Ohm (787-854) Chrles Augustin de Coulomb (736 806) André Mrie AMPÈRE (775-836) Crbon Electrode () Simple Electric Cell wire Zn Zn Zn Zn Sulfuric ### Reasoning to Solve Equations and Inequalities Lesson4 Resoning to Solve Equtions nd Inequlities In erlier work in this unit, you modeled situtions with severl vriles nd equtions. For exmple, suppose you were given usiness plns for concert showing ### 2.2. Volumes. A(x i ) x. Once again we recognize a Riemann sum at the right. In the limit as n we get the so called Cavalieri s principle: V = 2.2. VOLUMES 52 2.2. Volumes 2.2.1. Volumes by Slices. First we study how to find the volume of some solids by the method of cross sections (or slices ). The ide is to divide the solid into slices perpendiculr ### Lecture 3 Gaussian Probability Distribution Lecture 3 Gussin Probbility Distribution Introduction l Gussin probbility distribution is perhps the most used distribution in ll of science. u lso clled bell shped curve or norml distribution l Unlike ### Distributions. (corresponding to the cumulative distribution function for the discrete case). Distributions Recll tht n integrble function f : R [,] such tht R f()d = is clled probbility density function (pdf). The distribution function for the pdf is given by F() = (corresponding to the cumultive ### Random Variables and Cumulative Distribution Probbility: One Rndom Vrible 3 Rndom Vribles nd Cumultive Distribution A probbility distribution shows the probbilities observed in n experiment. The quntity observed in given tril of n experiment is number ### Education Spending (in billions of dollars) Use the distributive property. 0 CHAPTER Review of the Rel Number System 96. An pproximtion of federl spending on eduction in billions of dollrs from 200 through 2005 cn be obtined using the e xpression y = 9.0499x - 8,07.87, where ### 1. 1 m/s m/s m/s. 5. None of these m/s m/s m/s m/s correct m/s Crete ssignment, 99552, Homework 5, Sep 15 t 10:11 m 1 This print-out should he 30 questions. Multiple-choice questions my continue on the next column or pge find ll choices before nswering. The due time ### 1 PRECALCULUS READINESS DIAGNOSTIC TEST PRACTICE PRECALCULUS READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the smples, work the problems, then check your nswers t the end of ech topic. If you don t get the nswer given, check your work nd look ### Ae2 Mathematics : Fourier Series Ae Mthemtics : Fourier Series J. D. Gibbon (Professor J. D Gibbon, Dept of Mthemtics [email protected] http://www.imperil.c.uk/ jdg These notes re not identicl word-for-word with my lectures which will ### Two special Right-triangles 1. The Mth Right Tringle Trigonometry Hndout B (length of ) - c - (length of side ) (Length of side to ) Pythgoren s Theorem: for tringles with right ngle ( side + side = ) + = c Two specil Right-tringles. The ### Use Geometry Expressions to create a more complex locus of points. Find evidence for equivalence using Geometry Expressions. Lerning Objectives Loci nd Conics Lesson 3: The Ellipse Level: Preclculus Time required: 120 minutes In this lesson, students will generlize their knowledge of the circle to the ellipse. The prmetric nd ### Chapter 6 Solving equations Chpter 6 Solving equtions Defining n eqution 6.1 Up to now we hve looked minly t epressions. An epression is n incomplete sttement nd hs no equl sign. Now we wnt to look t equtions. An eqution hs n = sign ### SPECIAL PRODUCTS AND FACTORIZATION MODULE - Specil Products nd Fctoriztion 4 SPECIAL PRODUCTS AND FACTORIZATION In n erlier lesson you hve lernt multipliction of lgebric epressions, prticulrly polynomils. In the study of lgebr, we come ### PHY 222 Lab 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS PHY 222 Lb 8 MOTION OF ELECTRONS IN ELECTRIC AND MAGNETIC FIELDS Nme: Prtners: INTRODUCTION Before coming to lb, plese red this pcket nd do the prelb on pge 13 of this hndout. From previous experiments, ### Factoring Polynomials Fctoring Polynomils Some definitions (not necessrily ll for secondry school mthemtics): A polynomil is the sum of one or more terms, in which ech term consists of product of constnt nd one or more vribles ### 4. DC MOTORS. Understand the basic principles of operation of a DC motor. Understand the operation and basic characteristics of simple DC motors. 4. DC MOTORS Almost every mechnicl movement tht we see round us is ccomplished by n electric motor. Electric mchines re mens o converting energy. Motors tke electricl energy nd produce mechnicl energy. ### On the Meaning of Regression Coefficients for Categorical and Continuous Variables: Model I and Model II; Effect Coding and Dummy Coding Dt_nlysisclm On the Mening of Regression for tegoricl nd ontinuous Vribles: I nd II; Effect oding nd Dummy oding R Grdner Deprtment of Psychology This describes the simple cse where there is one ctegoricl ### The Velocity Factor of an Insulated Two-Wire Transmission Line The Velocity Fctor of n Insulted Two-Wire Trnsmission Line Problem Kirk T. McDonld Joseph Henry Lbortories, Princeton University, Princeton, NJ 08544 Mrch 7, 008 Estimte the velocity fctor F = v/c nd the ### Basic Math Review. Numbers. Important Properties. Absolute Value PROPERTIES OF ADDITION NATURAL NUMBERS {1, 2, 3, 4, 5, } ƒ Bsic Mth Review Numers NATURAL NUMBERS {1,, 3, 4, 5, } WHOLE NUMBERS {0, 1,, 3, 4, } INTEGERS {, 3,, 1, 0, 1,, } The Numer Line 5 4 3 1 0 1 3 4 5 Negtive integers Positive integers RATIONAL NUMBERS All ### So there are two points of intersection, one being x = 0, y = 0 2 = 0 and the other being x = 2, y = 2 2 = 4. y = x 2 (2,4) Ares The motivtion for our definition of integrl ws the problem of finding the re between some curve nd the is for running between two specified vlues. We pproimted the region b union of thin rectngles ### Notes #5. We then define the upper and lower sums for the partition P to be, respectively, U(P,f)= M k x k. k=1. L(P,f)= m k x k. Notes #5. The Riemnn Integrl Drboux pproch Suppose we hve bounded function f on closed intervl [, b]. We will prtition this intervl into subintervls (not necessrily of the sme length) nd crete mximl nd ### Written Homework 6 Solutions Written Homework 6 Solutions Section.10 0. Explin in terms of liner pproximtions or differentils why the pproximtion is resonble: 1.01) 6 1.06 Solution: First strt by finding the liner pproximtion of f ### Answer, Key Homework 4 David McIntyre Mar 25, Answer, Key Homework 4 Dvid McIntyre 45123 Mr 25, 2004 1 his print-out should hve 18 questions. Multiple-choice questions my continue on the next column or pe find ll choices before mkin your selection. ### A5682: Introduction to Cosmology Course Notes. 4. Cosmic Dynamics: The Friedmann Equation. = GM s R 2 s(t). 4. Cosmic Dynmics: The Friedmnn Eqution Reding: Chpter 4 Newtonin Derivtion of the Friedmnn Eqution Consider n isolted sphere of rdius R s nd mss M s, in uniform, isotropic expnsion (Hubble flow). The ### 6.2 Volumes of Revolution: The Disk Method mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of ### Using Definite Integrals Chpter 6 Using Definite Integrls 6. Using Definite Integrls to Find Are nd Length Motivting Questions In this section, we strive to understnd the ides generted by the following importnt questions: How	11,176	37,221	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.859375	4	CC-MAIN-2018-51	latest	en	0.792954
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Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Senior Manager Joined: 31 Aug 2009 Posts: 417 Location: Sydney, Australia If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 10 Oct 2009, 06:51 1 KUDOS 7 This post was BOOKMARKED 00:00 Difficulty: 95% (hard) Question Stats: 46% (02:50) correct 54% (02:07) wrong based on 317 sessions ### HideShow timer Statistics If a and b are integers, and \|a\| > \|b\|, is a·\|b\| < a – b? (1) a < 0 (2) ab >= 0 [Reveal] Spoiler: OA Manager Joined: 02 Jan 2009 Posts: 94 Location: India Schools: LBS Re: Absolute values [#permalink] ### Show Tags 10 Oct 2009, 07:45 yangsta8 wrote: If a and b are integers, and \|a\| > \|b\|, is a·\|b\| < a – b? 1. a < 0 2. ab >= 0 I managed to solve this by taking values for a and b that were both positive and negative. I managed to get the correct answer based on this approach but it took me almost 4 minutes, roughly 2 mins to solve for each of the clues. Is there any faster way to solve this? I think this is a 700+ question. 1. a < 0 .... a is -ve \|b\| is always +ve so a.\|b\| is always negative also \|a\|>\|b\| so even if b is +ve (a-b) will always be -ve. But a.\|b\| will always give a much smaller value than a-b. So suff. 2. ab >= 0 . here a and b will always have the same sign unless and until one of them is 0. so if we have a and b both to be -ve, our problem boils down to the one in case 1. i.e. the equation holds good. But consider a = 0, and b to be +ve and we see that it contradicts our previous result. So 2 is insuff. So A. _________________ The Legion dies, it does not surrender. Senior Manager Joined: 31 Aug 2009 Posts: 417 Location: Sydney, Australia Re: Absolute values [#permalink] ### Show Tags 10 Oct 2009, 07:56 jax91 wrote: 1. a < 0 .... a is -ve \|b\| is always +ve so a.\|b\| is always negative also \|a\|>\|b\| so even if b is +ve (a-b) will always be -ve. The bolded statement above is true. But of b is negative then (a-b) can potentially be > 0 hence Statement 1 can be insuff. Manager Joined: 02 Jan 2009 Posts: 94 Location: India Schools: LBS Re: Absolute values [#permalink] ### Show Tags 10 Oct 2009, 08:22 yangsta8 wrote: jax91 wrote: 1. a < 0 .... a is -ve \|b\| is always +ve so a.\|b\| is always negative also \|a\|>\|b\| so even if b is +ve (a-b) will always be -ve. The bolded statement above is true. But of b is negative then (a-b) can potentially be > 0 hence Statement 1 can be insuff. \|a\| > \|b\| so if be is -ve, like a -11 then \|b\| = 11 so \|a\| has to be > 11 say \|a\| = 12 if a is -ve that makes a = -12 so (a-b) = -12 - (-11) = -12 +11 = -1 and a.\|b\| = -12 . \|-11\| = -12.(11) = -132 please correct me if i missed something. _________________ The Legion dies, it does not surrender. Senior Manager Joined: 31 Aug 2009 Posts: 417 Location: Sydney, Australia Re: Absolute values [#permalink] ### Show Tags 10 Oct 2009, 08:27 1 KUDOS jax91 wrote: so (a-b) = -12 - (-11) = -12 +11 = -1 and a.\|b\| = -12 . \|-11\| = -12.(11) = -132 Your working for the above is correct. So in the above you prove that a.\|b\| < a-b is true. However if you take a smaller number Let a = -2 let b = 1 a.\|b\| = -2 . \|1\| = -2.1 = -2 (a-b) = -2 - 1 = -3 a.\|b\| < a-b is false. Hence A is insufficient. Manager Joined: 02 Jan 2009 Posts: 94 Location: India Schools: LBS Re: Absolute values [#permalink] ### Show Tags 10 Oct 2009, 08:35 yangsta8 wrote: jax91 wrote: so (a-b) = -12 - (-11) = -12 +11 = -1 and a.\|b\| = -12 . \|-11\| = -12.(11) = -132 Your working for the above is correct. So in the above you prove that a.\|b\| < a-b is true. However if you take a smaller number Let a = -2 let b = 1 a.\|b\| = -2 . \|1\| = -2.1 = -2 (a-b) = -2 - 1 = -3 a.\|b\| < a-b is false. Hence A is insufficient. Missed that condition. Thanks for pointing it out. _________________ The Legion dies, it does not surrender. Math Expert Joined: 02 Sep 2009 Posts: 39672 Re: Absolute values [#permalink] ### Show Tags 10 Oct 2009, 11:54 3 KUDOS Expert's post 1 This post was BOOKMARKED If a and b are integers, and \|a\| > \|b\|, is a·\|b\| < a – b? 1. a < 0 2. ab >= 0 The key to solve this problem is to determine the relationship between the signs of a-b and a·\|b\|, with the given condition that \|a\|>\|b\|. When determined the rest of problem will go smoothly. So, NOTE that when \|a\|>\|b\|: a-b>0 IF and ONLY a>0, no matter what possible values will take b (possible means not violating the given condition \|a\|>\|b\|) a-b<0 IF and ONLY a<0, no matter what possible values will take b (possible means not violating the given condition \|a\|>\|b\|) Above conclusion means that: when a>0 --> a·\|b\|>0 and a – b>0 AND when a<0 --> a·\|b\|<0 and a – b<0 Also NOTE that even knowing the signs of LHS and RHS of our inequality its not possible to determine LHS<RHS or not. Generally speaking even not considering the statements, we can conclude, that if they are giving us ONLY the info about the signs of a and b, it won't help us to answer the Q. (in our case we can even not consider them separately or together, we know answer would be E, as the statements are only about the signs of the variables) But still let's look at the statement: (1) a<0 --> a\|b\|<0 (a negative \|b\| positive) and we already determined that when a<0 a-b<0 --> so both are negative but we can not determine is a · \|b\| < a – b or not. Not Sufficient (2) ab >= 0 a>0 b=>0 (a can not be zero as \|a\|>\|b\|) or a<0, b<=0 a>0 b=>0 --> a\|b\|>0 and a – b>0 --> both are positive but we can not determine is a · \|b\| < a – b or not. a<0, b<=0 --> a\|b\|<0 and a-b<0 --> both are negative but we can not determine is a · \|b\| < a – b or not. Not Sufficient (1)+(2) --> a<0, b<0 same thing --> a\|b\|<0 and a-b<0 we can not determine is a · \|b\| < a – b or not. Hope this helps. _________________ SVP Joined: 29 Aug 2007 Posts: 2473 Re: Absolute values [#permalink] ### Show Tags 10 Oct 2009, 23:19 yangsta8 wrote: If a and b are integers, and \|a\| > \|b\|, is a·\|b\| < a – b? 1. a < 0 2. ab >= 0 I managed to solve this by taking values for a and b that were both positive and negative. I managed to get the correct answer based on this approach but it took me almost 4 minutes, roughly 2 mins to solve for each of the clues. Is there any faster way to solve this? I think this is a 700+ question. 1. If a < 0, b could be -ve or 0 or +ve. If b = -ve, (a·\|b\|) and (a – b) both are -ve however the relationship between (a·\|b\|) and (a – b) cannot be established. If b = 0, (a·\|b\|) = 0 but (a – b) is still -ve. so (a·\|b\|) < (a – b) is not true. If b = +ve, (a·\|b\|) and (a – b) bot hare -ve however the relationship between (a·\|b\|) and (a – b) cannot be established. NSF... 2. If ab >= 0, a and b could be both -ve or +ve or 0 or only one of either is 0 and the other is non-zero. The relationship between (a·\|b\|) and (a – b) cannot be established here too. NSF... Togather also the relationship between (a·\|b\|) and (a – b) cannot be established. NSF...... Thats E. _________________ Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html GT Intern Joined: 02 May 2009 Posts: 22 Re: Absolute values [#permalink] ### Show Tags 11 Oct 2009, 08:38 i would go for c, pls post OA Manager Joined: 05 Jun 2009 Posts: 77 Re: Absolute values [#permalink] ### Show Tags 11 Oct 2009, 09:20 E for me Condtion 2 does not bind anything, if B equals 0 or less than 1 and if B equals a large number with A being a bigger abs valued negative number it does not prove sufficient. Senior Manager Joined: 13 Aug 2012 Posts: 464 Concentration: Marketing, Finance GPA: 3.23 Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 17 Jan 2013, 00:18 1 KUDOS yangsta8 wrote: If a and b are integers, and \|a\| > \|b\|, is a·\|b\| < a – b? 1. a < 0 2. ab >= 0 I noticed that these kinds of questions are important on the GMAT... I've seen many of these types that initially threw me off until I faced this type and practiced as much... My solution: 1. Test a is (-) and b is (-) : -2\|-1\| < -2 + 1 ==> -2 < -1 YES! Test a is (-) and b is (+): -2 \|1\| < -2 - 1 ==> -2 < -3 NO! We stop here since we know that the information is INSUFFICIENT. 2. The info meant a and b must have the same sign or one of them is 0. Test a is (-) and b is (-): From Statement 1 we know this is YES! Test a is (+) and b is (+): 2\|1\| < 2 - 1 ==> 2 < 1 NO! We stop here and we know that the information is INSUFFICIENT. Together: We know a is (-) and be is either (-) or (0) Test a is (-) and b is (-): From Statement 1 we know this is YES! Test a is (-) and b is 0: 0 < -2 NO! Still information together is INSUFFICIENT! _________________ Impossible is nothing to God. Manager Joined: 12 Sep 2010 Posts: 233 Concentration: Healthcare, General Management Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 21 Jan 2013, 16:48 mbaiseasy wrote: yangsta8 wrote: If a and b are integers, and \|a\| > \|b\|, is a·\|b\| < a – b? 1. a < 0 2. ab >= 0 I noticed that these kinds of questions are important on the GMAT... I've seen many of these types that initially threw me off until I faced this type and practiced as much... My solution: 1. Test a is (-) and b is (-) : -2\|-1\| < -2 + 1 ==> -2 < -1 YES! Test a is (-) and b is (+): -2 \|1\| < -2 - 1 ==> -2 < -3 NO! We stop here since we know that the information is INSUFFICIENT. 2. The info meant a and b must have the same sign or one of them is 0. Test a is (-) and b is (-): From Statement 1 we know this is YES! Test a is (+) and b is (+): 2\|1\| < 2 - 1 ==> 2 < 1 NO! We stop here and we know that the information is INSUFFICIENT. Together: We know a is (-) and be is either (-) or (0) Test a is (-) and b is (-): From Statement 1 we know this is YES! Test a is (-) and b is 0: 0 < -2 NO! Still information together is INSUFFICIENT! This is the type of question that I have the most trouble with. Part of my problem is to think about the numbers that would prove the statement sufficient/insufficient. For example, in statement 1, if a=-5 and b=2, then -5 l 2 l < -5 - (2) Yes. So at this point, I think to myself what numbers can I plug-in to prove that statement 1 insufficient. This is when I get in trouble because I spend too much time thinking about what numbers do I need to plug in next. So I just pick another random number: a=-5 and b=-2, then -5 l -2 l < -5 - (-2) Yes. By this time I'm probably close to the 2-minute mark. How do you know what numbers to test efficiently? Can anybody suggest more of this type of questions or any materials that I can practice on? Any advice will be greatly appreciated. Thanks. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7440 Location: Pune, India Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 21 Jan 2013, 21:34 Samwong wrote: This is the type of question that I have the most trouble with. Part of my problem is to think about the numbers that would prove the statement sufficient/insufficient. For example, in statement 1, if a=-5 and b=2, then -5 l 2 l < -5 - (2) Yes. So at this point, I think to myself what numbers can I plug-in to prove that statement 1 insufficient. This is when I get in trouble because I spend too much time thinking about what numbers do I need to plug in next. So I just pick another random number: a=-5 and b=-2, then -5 l -2 l < -5 - (-2) Yes. By this time I'm probably close to the 2-minute mark. How do you know what numbers to test efficiently? Can anybody suggest more of this type of questions or any materials that I can practice on? Any advice will be greatly appreciated. Thanks. Such questions are best solved using part logic and part number plugging. When you plug numbers, keep in mind that you have to try to get the answer but keep life as simple as possible for yourself. Let me explain what I mean: If a and b are integers, and \|a\| > \|b\|, is a·\|b\| < a – b? 1. a < 0 2. ab >= 0 Read the question stem: "If a and b are integers" - very good .. no fractions to worry about "and \|a\| > \|b\|" - absolute value of a is greater than absolute value of b.. we don't know anything about their signs "is a·\|b\| < a – b" There is no obvious relation between the left hand side (LHS) and the right hand side(RHS) so I need to move on and look at the statements. 1. a < 0 a is negative means LHS is negative or 0 (if b is 0). RHS could be negative or positive depending on value of b. Assume b = 0 (makes life simple) We get 0 < a. Inequality does not hold since a is negative. Assume b is a large negative number say, -10 and a is a number a little smaller than b. LHS is a large negative number and RHS is a small negative number. e.g. a = -11, b = -10 -110 < -1 Inequality holds. Insufficient. 2. ab >= 0 This means EITHER a is negative, b is negative (or one or both are 0) OR a is positive, b is positive (or one or both are 0) Now notice that even with both statements together, you cannot say whether the inequality holds. With statement 1, we saw two cases - b large negative, a is a little smaller than b (inequality holds) a negative, b is 0 (inequality does not hold) According to this statement as well, both cases are possible. Hence this statement alone is not sufficient and both together are also not sufficient. Hence answer must be (E) Check out a related post that discusses how to choose the numbers you should plug: http://www.veritasprep.com/blog/2012/12 ... game-plan/ (Edited) _________________ Karishma Veritas Prep \| GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 12 Sep 2010 Posts: 233 Concentration: Healthcare, General Management Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 01 Feb 2013, 16:25 VeritasPrepKarishma wrote: Samwong wrote: This is the type of question that I have the most trouble with. Part of my problem is to think about the numbers that would prove the statement sufficient/insufficient. For example, in statement 1, if a=-5 and b=2, then -5 l 2 l < -5 - (2) Yes. So at this point, I think to myself what numbers can I plug-in to prove that statement 1 insufficient. This is when I get in trouble because I spend too much time thinking about what numbers do I need to plug in next. So I just pick another random number: a=-5 and b=-2, then -5 l -2 l < -5 - (-2) Yes. By this time I'm probably close to the 2-minute mark. How do you know what numbers to test efficiently? Can anybody suggest more of this type of questions or any materials that I can practice on? Any advice will be greatly appreciated. Thanks. Such questions are best solved using part logic and part number plugging. When you plug numbers, keep in mind that you have to try to get the answer but keep life as simple as possible for yourself. Let me explain what I mean: If a and b are integers, and \|a\| > \|b\|, is a·\|b\| < a – b? 1. a < 0 2. ab >= 0 Read the question stem: "If a and b are integers" - very good .. no fractions to worry about "and \|a\| > \|b\|" - absolute value of a is greater than absolute value of b.. we don't know anything about their signs "is a·\|b\| < a – b" There is no obvious relation between the left hand side (LHS) and the right hand side(RHS) so I need to move on and look at the statements. 1. a < 0 a is negative means LHS is negative or 0 (if b is 0). RHS could be negative or positive depending on value of b. Assume b = 0 (makes life simple) We get 0 < a. Inequality does not hold since a is negative. Assume b is a large negative number say, -10 and a is a small negative number. LHS is negative and RHS is positive. Inequality holds. Insufficient. 2. ab >= 0 This means EITHER a is negative, b is negative (or one or both are 0) OR a is positive, b is positive (or one or both are 0) Now notice that even with both statements together, you cannot say whether the inequality holds. With statement 1, we saw two cases - a small negative, b large negative (inequality holds) a negative, b is 0 (inequality does not hold) According to this statement as well, both cases are possible. Hence this statement alone is not sufficient and both together are also not sufficient. Hence answer must be (E) Check out a related post that discusses how to choose the numbers you should plug: http://www.veritasprep.com/blog/2012/12 ... game-plan/ Hi Karishma, Thank you for answering my question. However, in statement 1, I don't see how the RHS can be positive when "a" is negative with the given condition \|a\| > \|b\|. If b = -10 then "a" has to be less than -10 (ie -11, -12, -13...) Thus, -11 - ( -10) = -1 RHS = negative. Please check whether I miss something. Thanks. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7440 Location: Pune, India Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 03 Feb 2013, 23:09 Samwong wrote: Hi Karishma, Thank you for answering my question. However, in statement 1, I don't see how the RHS can be positive when "a" is negative with the given condition \|a\| > \|b\|. If b = -10 then "a" has to be less than -10 (ie -11, -12, -13...) Thus, -11 - ( -10) = -1 RHS = negative. Please check whether I miss something. Thanks. Yes, you are right. \|a\| cannot be smaller than \|b\| so we should take numbers where a is a little less than b to get a small negative on RHS. The LHS will be negative with a greater absolute value. a = -11, b = -10 -110 < -11 - (-10) -110 < -1 (inequality holds) _________________ Karishma Veritas Prep \| GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199 Veritas Prep Reviews Current Student Joined: 11 Apr 2013 Posts: 53 Schools: Booth '17 (M) Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 16 Apr 2013, 14:39 OK, my process for solving this was way simpler than all this number plugging and stuff I'm seeing, but maybe I'm doing something wrong and I just happened to get the right answer. Here's my thought process. 1) Manipulate the question stem is a - \|b\| < a - b ? - subtract 'a' from both sides to get is -\|b\| < -b ? the answer is 'yes' if 'b' is negative, and 'no' if 'b' is non-negative, so in order to be sufficient the data needs to tell us definitively the sign of 'b' 1) a < 0 tells us nothing about the sign of 'b', INSUFFICIENT 2) ab >= 0 tells us that 'a' and 'b' have the same sign, or that 'a' or 'b' or both are zero. From the original question we know that \|a\| > \|b\|, so 'a' cannot be zero, which means this statement is telling us that either 'a' and 'b' have the same sign, or b=0. This does nothing to establish the sign of 'b', INSUFFICIENT 1 & 2 together: applying " a < 0 " to "ab >= 0 " tells us that 'b' must non-positive, but can still equal zero or any negative number greater than 'a'. Since there are multiple possibilities for the sign of 'b', we cannot answer the original question of " is -\|b\| < -b ?", so the answer is E, statements 1 and 2 are INSUFFICIENT Is this valid logic? I think it hinges on the question of whether or not you can manipulate the question stem by subtracting 'a' from both sides. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7440 Location: Pune, India Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 16 Apr 2013, 22:33 dustwun wrote: OK, my process for solving this was way simpler than all this number plugging and stuff I'm seeing, but maybe I'm doing something wrong and I just happened to get the right answer. Here's my thought process. 1) Manipulate the question stem is a - \|b\| < a - b ? - subtract 'a' from both sides to get is -\|b\| < -b ? the answer is 'yes' if 'b' is negative, and 'no' if 'b' is non-negative, so in order to be sufficient the data needs to tell us definitively the sign of 'b' 1) a < 0 tells us nothing about the sign of 'b', INSUFFICIENT 2) ab >= 0 tells us that 'a' and 'b' have the same sign, or that 'a' or 'b' or both are zero. From the original question we know that \|a\| > \|b\|, so 'a' cannot be zero, which means this statement is telling us that either 'a' and 'b' have the same sign, or b=0. This does nothing to establish the sign of 'b', INSUFFICIENT 1 & 2 together: applying " a < 0 " to "ab >= 0 " tells us that 'b' must non-positive, but can still equal zero or any negative number greater than 'a'. Since there are multiple possibilities for the sign of 'b', we cannot answer the original question of " is -\|b\| < -b ?", so the answer is E, statements 1 and 2 are INSUFFICIENT Is this valid logic? I think it hinges on the question of whether or not you can manipulate the question stem by subtracting 'a' from both sides. You got the question wrong. It is Is a * \|b\| < a - b ? (there is a dot there, not a minus sign) _________________ Karishma Veritas Prep \| GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for \$199 Veritas Prep Reviews Current Student Joined: 11 Apr 2013 Posts: 53 Schools: Booth '17 (M) Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 16 Apr 2013, 22:56 VeritasPrepKarishma wrote: You got the question wrong. It is Is a * \|b\| < a - b ? (there is a dot there, not a minus sign) aaaaand that's why I have a bag over my head. See you at Harvard next year! GMAT Club Legend Joined: 09 Sep 2013 Posts: 15966 Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 10 Feb 2015, 02:18 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ GMAT Club Legend Joined: 09 Sep 2013 Posts: 15966 Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] ### Show Tags 28 Aug 2016, 05:46 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a [#permalink] 28 Aug 2016, 05:46 Similar topics Replies Last post Similar Topics: 16 If a and b are positive integers such that a < b, is b even? 12 01 Jan 2017, 10:22 3 If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a 7 23 Sep 2013, 02:13 If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a 6 18 Mar 2011, 04:20 1 If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a 6 18 Mar 2012, 14:24 37 If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a 27 19 Jun 2017, 10:08 Display posts from previous: Sort by # If a and b are integers, and \|a\| > \|b\|, is a \|b\| < a new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	7,317	24,092	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2017-26	latest	en	0.906289
https://statkat.com/stattest.php?t=9&t2=18&t3=39&t4=6	1,652,736,939,000,000,000	text/html	crawl-data/CC-MAIN-2022-21/segments/1652662512249.16/warc/CC-MAIN-20220516204516-20220516234516-00286.warc.gz	593,065,040	8,308	Two sample t test - equal variances not assumed - overview This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table Two sample $t$ test - equal variances not assumed Spearman's rho McNemar's test One sample $t$ test for the mean Independent/grouping variableVariable 1Independent variableIndependent variable One categorical with 2 independent groupsOne of ordinal level2 paired groupsNone Dependent variableVariable 2Dependent variableDependent variable One quantitative of interval or ratio levelOne of ordinal levelOne categorical with 2 independent groupsOne quantitative of interval or ratio level Null hypothesisNull hypothesisNull hypothesisNull hypothesis H0: $\mu_1 = \mu_2$ Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2. H0: $\rho_s = 0$ Here $\rho_s$ is the Spearman correlation in the population. The Spearman correlation is a measure for the strength and direction of the monotonic relationship between two variables of at least ordinal measurement level. In words, the null hypothesis would be: H0: there is no monotonic relationship between the two variables in the population. Let's say that the scores on the dependent variable are scored 0 and 1. Then for each pair of scores, the data allow four options: 1. First score of pair is 0, second score of pair is 0 2. First score of pair is 0, second score of pair is 1 (switched) 3. First score of pair is 1, second score of pair is 0 (switched) 4. First score of pair is 1, second score of pair is 1 The null hypothesis H0 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) = P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is the same as the probability that a pair of scores switches from 1 to 0. Other formulations of the null hypothesis are: • H0: $\pi_1 = \pi_2$, where $\pi_1$ is the population proportion of ones for the first paired group and $\pi_2$ is the population proportion of ones for the second paired group • H0: for each pair of scores, P(first score of pair is 1) = P(second score of pair is 1) H0: $\mu = \mu_0$ Here $\mu$ is the population mean, and $\mu_0$ is the population mean according to the null hypothesis. Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis H1 two sided: $\mu_1 \neq \mu_2$ H1 right sided: $\mu_1 > \mu_2$ H1 left sided: $\mu_1 < \mu_2$ H1 two sided: $\rho_s \neq 0$ H1 right sided: $\rho_s > 0$ H1 left sided: $\rho_s < 0$ The alternative hypothesis H1 is that for each pair of scores, P(first score of pair is 0 while second score of pair is 1) $\neq$ P(first score of pair is 1 while second score of pair is 0). That is, the probability that a pair of scores switches from 0 to 1 is not the same as the probability that a pair of scores switches from 1 to 0. Other formulations of the alternative hypothesis are: • H1: $\pi_1 \neq \pi_2$ • H1: for each pair of scores, P(first score of pair is 1) $\neq$ P(second score of pair is 1) H1 two sided: $\mu \neq \mu_0$ H1 right sided: $\mu > \mu_0$ H1 left sided: $\mu < \mu_0$ AssumptionsAssumptionsAssumptionsAssumptions • Within each population, the scores on the dependent variable are normally distributed • Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another Note: this assumption is only important for the significance test, not for the correlation coefficient itself. The correlation coefficient itself just measures the strength of the monotonic relationship between two variables. • Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another • Scores are normally distributed in the population • Sample is a simple random sample from the population. That is, observations are independent of one another Test statisticTest statisticTest statisticTest statistic $t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}}$ Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s^2_1$ is the sample variance in group 1, $s^2_2$ is the sample variance in group 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis. The denominator $\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1 - \bar{y}_2$ is removed from 0. Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$. $t = \dfrac{r_s \times \sqrt{N - 2}}{\sqrt{1 - r_s^2}}$ Here $r_s$ is the sample Spearman correlation and $N$ is the sample size. The sample Spearman correlation $r_s$ is equal to the Pearson correlation applied to the rank scores. $X^2 = \dfrac{(b - c)^2}{b + c}$ Here $b$ is the number of pairs in the sample for which the first score is 0 while the second score is 1, and $c$ is the number of pairs in the sample for which the first score is 1 while the second score is 0. $t = \dfrac{\bar{y} - \mu_0}{s / \sqrt{N}}$ Here $\bar{y}$ is the sample mean, $\mu_0$ is the population mean according to the null hypothesis, $s$ is the sample standard deviation, and $N$ is the sample size. The denominator $s / \sqrt{N}$ is the standard error of the sampling distribution of $\bar{y}$. The $t$ value indicates how many standard errors $\bar{y}$ is removed from $\mu_0$. Sampling distribution of $t$ if H0 were trueSampling distribution of $t$ if H0 were trueSampling distribution of $X^2$ if H0 were trueSampling distribution of $t$ if H0 were true Approximately the $t$ distribution with $k$ degrees of freedom, with $k$ equal to $k = \dfrac{\Bigg(\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}\Bigg)^2}{\dfrac{1}{n_1 - 1} \Bigg(\dfrac{s^2_1}{n_1}\Bigg)^2 + \dfrac{1}{n_2 - 1} \Bigg(\dfrac{s^2_2}{n_2}\Bigg)^2}$ or $k$ = the smaller of $n_1$ - 1 and $n_2$ - 1 First definition of $k$ is used by computer programs, second definition is often used for hand calculations. Approximately the $t$ distribution with $N - 2$ degrees of freedom If $b + c$ is large enough (say, > 20), approximately the chi-squared distribution with 1 degree of freedom. If $b + c$ is small, the Binomial($n$, $P$) distribution should be used, with $n = b + c$ and $P = 0.5$. In that case the test statistic becomes equal to $b$. $t$ distribution with $N - 1$ degrees of freedom Significant?Significant?Significant?Significant? Two sided: Right sided: Left sided: Two sided: Right sided: Left sided: For test statistic $X^2$: • Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2}$ or • Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$ If $b + c$ is small, the table for the binomial distribution should be used, with as test statistic $b$: • Check if $b$ observed in sample is in the rejection region or • Find two sided $p$ value corresponding to observed $b$ and check if it is equal to or smaller than $\alpha$ Two sided: Right sided: Left sided: Approximate $C\%$ confidence interval for $\mu_1 - \mu_2$n.a.n.a.$C\%$ confidence interval for $\mu$ $(\bar{y}_1 - \bar{y}_2) \pm t^ \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}$ where the critical value $t^$ is the value under the $t_{k}$ distribution with the area $C / 100$ between $-t^$ and $t^$ (e.g. $t^$ = 2.086 for a 95% confidence interval when df = 20). The confidence interval for $\mu_1 - \mu_2$ can also be used as significance test. --$\bar{y} \pm t^* \times \dfrac{s}{\sqrt{N}}$ where the critical value $t^$ is the value under the $t_{N-1}$ distribution with the area $C / 100$ between $-t^$ and $t^$ (e.g. $t^$ = 2.086 for a 95% confidence interval when df = 20). The confidence interval for $\mu$ can also be used as significance test. n.a.n.a.n.a.Effect size ---Cohen's $d$: Standardized difference between the sample mean and $\mu_0$: $$d = \frac{\bar{y} - \mu_0}{s}$$ Cohen's $d$ indicates how many standard deviations $s$ the sample mean $\bar{y}$ is removed from $\mu_0.$ Visual representationn.a.n.a.Visual representation -- n.a.n.a.Equivalent ton.a. -- - Example contextExample contextExample contextExample context Is the average mental health score different between men and women?Is there a monotonic relationship between physical health and mental health?Does a tv documentary about spiders change whether people are afraid (yes/no) of spiders?Is the average mental health score of office workers different from $\mu_0 = 50$? SPSSSPSSSPSSSPSS Analyze > Compare Means > Independent-Samples T Test... • Put your dependent (quantitative) variable in the box below Test Variable(s) and your independent (grouping) variable in the box below Grouping Variable • Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow • Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2 • Continue and click OK Analyze > Correlate > Bivariate... • Put your two variables in the box below Variables • Under Correlation Coefficients, select Spearman Analyze > Nonparametric Tests > Legacy Dialogs > 2 Related Samples... • Put the two paired variables in the boxes below Variable 1 and Variable 2 • Under Test Type, select the McNemar test Analyze > Compare Means > One-Sample T Test... • Put your variable in the box below Test Variable(s) • Fill in the value for $\mu_0$ in the box next to Test Value JamoviJamoviJamoviJamovi T-Tests > Independent Samples T-Test • Put your dependent (quantitative) variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable • Under Tests, select Welch's • Under Hypothesis, select your alternative hypothesis Regression > Correlation Matrix • Put your two variables in the white box at the right • Under Correlation Coefficients, select Spearman • Under Hypothesis, select your alternative hypothesis Frequencies > Paired Samples - McNemar test • Put one of the two paired variables in the box below Rows and the other paired variable in the box below Columns T-Tests > One Sample T-Test • Put your variable in the box below Dependent Variables • Under Hypothesis, fill in the value for $\mu_0$ in the box next to Test Value, and select your alternative hypothesis Practice questionsPractice questionsPractice questionsPractice questions	3,132	11,287	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.78125	4	CC-MAIN-2022-21	latest	en	0.877248
https://numeracycontinuum.com/math-tips/how-to-find-the-partial-sum-of-an-arithmetic-sequence-solution-found.html	1,660,033,535,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882570913.16/warc/CC-MAIN-20220809064307-20220809094307-00163.warc.gz	386,364,583	20,811	# How To Find The Partial Sum Of An Arithmetic Sequence? (Solution found) An arithmetic series is the sum of the terms of an arithmetic sequence. The nth partial sum of an arithmetic sequence can be calculated using the first and last terms as follows: Sn=n(a1+an)2. ## What is a partial sum arithmetic? The simple answer is that a partial sum is actually just the sum of part of a sequence. You can find a partial sum for both finite sequences and infinite sequences. A partial sum, on the other hand, is just the sum of part of a sequence. ## What is the partial sum of a sequence? A Partial Sum is the sum of part of the sequence. The sum of infinite terms is an Infinite Series. And Partial Sums are sometimes called “Finite Series”. ## How do you find the nth partial sum of a sequence? The nth partial sum of a geometric sequence can be calculated using the first term a1 and common ratio r as follows: Sn=a1(1−rn)1−r. The infinite sum of a geometric sequence can be calculated if the common ratio is a fraction between −1 and 1 (that is \|r\|<1) as follows: S∞=a11−r. ## What is the formula of the sum of arithmetic sequence? The sum of the arithmetic sequence can be derived using the general arithmetic sequence, an n = a1 1 + (n – 1)d. ## Introduction to Arithmetic Progressions Generally speaking, a progression is a sequence or series of numbers in which the numbers are organized in a certain order so that the relationship between the succeeding terms of a series or sequence remains constant. It is feasible to find the n thterm of a series by following a set of steps. There are three different types of progressions in mathematics: 1. Arithmetic Progression (AP), Geometric Progression (GP), and Harmonic Progression (HP) are all types of progression. AP, also known as Arithmetic Sequence, is a sequence or series of integers in which the common difference between two subsequent numbers in the series is always the same. As an illustration: Series 1: 1, 3, 5, 7, 9, and 11. Every pair of successive integers in this sequence has a common difference of 2, which is always true. Series 2: numbers 28, 25, 22, 19, 16, 13,. There is no common difference between any two successive numbers in this series; instead, there is a strict -3 between any two consecutive numbers. ### Terminology and Representation • Common difference, d = a 2– a 1= a 3– a 2=. = a n– a n – 1 • A n= n thterm of Arithmetic Progression • S n= Sum of first n elements in the series • A n= n ### General Form of an AP Given that ais treated as a first term anddis treated as a common difference, the N thterm of the AP may be calculated using the following formula: As a result, using the above-mentioned procedure to compute the n terms of an AP, the general form of the AP is as follows: Example: The 35th term in the sequence 5, 11, 17, 23,. is to be found. Solution: When looking at the given series,a = 5, d = a 2– a 1= 11 – 5 = 6, and n = 35 are the values. As a result, we must use the following equations to figure out the 35th term: n= a + (n – 1)da n= 5 + (35 – 1) x 6a n= 5 + 34 x 6a n= 209 As a result, the number 209 represents the 35th term. ### Sum of n Terms of Arithmetic Progression The arithmetic progression sum is calculated using the formula S n= (n/2) ### Derivation of the Formula Allowing ‘l’ to signify the n thterm of the series and S n to represent the sun of the first n terms of the series a, (a+d), (a+2d),., a+(n-1)d S n = a 1 plus a 2 plus a 3 plus .a n-1 plus a n S n= a + (a + d) + (a + 2d) +. + (l – 2d) + (l – d) + l. + (l – 2d) + (l – d) + l. + (l – 2d) + (l – d) + l. (1) When we write the series in reverse order, we obtain S n= l + (l – d) + (l – 2d) +. + (a + 2d) + (a + d) + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a + d + a … (2) Adding equations (1) and (2) results in equation (2). + (a + l) + (a + l) + (a + l) +. (3) As a result, the formula for calculating the sum of a series is S n= (n/2)(a + l), where an is the first term of the series, l is the last term of the series, and n is the number of terms in the series. d S n= (n/2)(a + a + (n – 1)d)(a + a + (n – 1)d) S n= (n/2)(2a + (n – 1) x d)(n/2)(2a + (n – 1) x d) Observation: The successive terms in an Arithmetic Progression can alternatively be written as a-3d, a-2d, a-d, a, a+d, a+2d, a+3d, and so on. ### Sample Problems on Arithmetic Progressions Problem 1: Calculate the sum of the first 35 terms in the sequence 5,11,17,23, and so on. a = 5 in the given series, d = a 2–a in the provided series, and so on. The number 1 equals 11 – 5 = 6, and the number n equals 35. S n= (n/2)(2a + (n – 1) x d)(n/2)(2a + (n – 1) x d) S n= (35/2)(2 x 5 + (35 – 1) x 6)(35/2)(2 x 5 + (35 – 1) x 6) S n= (35/2)(10 + 34 x 6) n= (35/2)(10 + 34 x 6) n= (35/2)(10 + 34 x 6) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) S n= (35/2)(10 + 204) A = 35214/2A = 3745S n= 35214/2A = 3745 Find the sum of a series where the first term of the series is 5 and the last term of the series is 209, and the number of terms in the series is 35, as shown in Problem 2. Problem 2. S n= (35/2)(5 + 209) S n= (35/2)(5 + 209) S n= (35/2)(5 + 209) A = 35214/2A = 3745S n= 35214/2A = 3745 Problem 3: A amount of 21 rupees is divided among three brothers, with each of the three pieces of money being in the AP and the sum of their squares being the sum of their squares being 155. You might be interested: Explain How A Geometric Sequence And An Arithmetic Sequence Are The Same How Are They Different? (Solution) Solution: Assume that the three components of money are (a-d), a, and (a+d), and that the total amount allocated is in AP. 155 divided by two equals 155 Taking the value of ‘a’ into consideration, we obtain 3(7) 2+ 2d. 2= 4d = 2 = 2 The three portions of the money that was dispersed are as follows:a + d = 7 + 2 = 9a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5a = 7a – d = 7 – 2 = 5 As a result, the most significant portion is Rupees 9 million. ## Finite Sequence: Definition & Examples – Video & Lesson Transcript When there is a finite series, the first term is followed by a second term, and so on until the last term. In a finite sequence, the letternoften reflects the total number of phrases in the sequence. In a finite series, the first term may be represented by a (1), the second term can be represented by a (2), etc. Parentheses are often used to separate numbers adjacent to thea, but parentheses will be used at other points in this course to distinguish them from subscripts. This terminology is illustrated in the following graphic. ## Examples Despite the fact that there are many other forms of finite sequences, we shall confine ourselves to the field of mathematics for the time being. The prime numbers smaller than 40, for example, are an example of a finite sequence, as seen in the table below: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, Another example is the range of natural numbers between zero and one hundred. Due to the fact that it would be tedious to write down all of the terms in this finite sequence, we will demonstrate it as follows: 1, 2, 3, 4, 5,., 100 are the numbers from 1 to 100. Alternatively, there might be alternative sequences that begin in the same way and end in 100, but which do not contain the natural numbers less than or equal to 100. ## Finding Patterns Let’s take a look at some finite sequences to see if there are any patterns. #### Example 1 Examine a few finite sequences to see if any patterns can be discovered. ## Finding Sums of Finite Arithmetic Series – Sequences and Series (Algebra 2) The sum of all terms for a finitearithmetic sequencegiven bywherea1 is the first term,dis the common difference, andnis the number of terms may be determined using the following formula: wherea1 is the first term,dis the common difference, andnis the number of terms Consider the arithmetic sum as an example of how the total is computed in this manner. If you choose not to add all of the words at once, keep in mind that the first and last terms may be recast as2fives. This method may be used to rewrite the second and second-to-last terms as well. There are 9fives in all, and the aggregate is 9 x 5 = 9. expand more Due to the fact that the difference between each term is constant, this sequence from 1 through 1000 is arithmetic. The first phrase, a1, is one and the last term, is one thousand thousand. The total number of terms is less than 1000. The sum of all positive integers up to and including 1000 is 500 500. }}}!}}}Test}}} arrow back} arrow forwardarrow leftarrow right ## Arithmetic Series It is the sum of the terms of an arithmetic sequence that is known as an arithmetic series. A geometric series is made up of the terms of a geometric sequence and is represented by the symbol. You can work with other sorts of series as well, but you won’t have much experience with them until you get to calculus. For the time being, you’ll most likely be collaborating with these two. How to deal with arithmetic series is explained and shown on this page, among other things. You can only take the “partial” sum of an arithmetic series for a variety of reasons that will be explored in greater detail later in calculus. The following is the formula for the firstnterms of the anarithmeticsequence, starting with i= 1, and it is written: Content Continues Below The “2” on the right-hand side of the “equals” sign may be converted to a one-half multiplied on the parenthesis, which reveals that the formula for the total is, in effect,n times the “average” of the first and final terms, as seen in the example below. The summation formula may be demonstrated via induction, by the way. #### Find the35 th partial sum,S 35, of the arithmetic sequence with terms The first thirty-five terms of this sequence are added together to provide the 35th partial sum of the series. The first few words in the sequence are as follows: Due to the fact that all of the words share a common difference, this is in fact an arithmetic sequence. The final term in the partial sum will be as follows: Plugging this into the formula, the 35 th partial sum is:Then my answer is:35 th partial sum:Then my answer is:35 th partial sum: S 35 = 350 S 35 = 350 If I had merely looked at the formula for the terms in the series above, I would have seen the common difference in the above sequence. If we had used a continuous variable, such as the “x” we used when graphing straight lines, rather than a discrete variable, then ” ” would have been a straight line that rose by one-half at each step, rather than the discrete variable. #### Find the value of the following summation: It appears that each term will be two units greater in size than the preceding term based on the formula ” 2 n– 5 ” for the then-thirteenth term. (If I wasn’t sure about something, I could always plug in some values to see if they were correct.) As a result, this is a purely arithmetic sum. However, this summation begins at n= 15, not at n= 1, and the summation formula is only applicable to sums that begin at n=1. So, how am I supposed to proceed with this summation? By employing a simple trick: The quickest way to determine the value of this sum is to first calculate the 14th and 47th partial sums, and then subtract the 14th from the 47th partial sum. By performing this subtraction, I will have subtracted the first through fourteenth terms from the first through forty-seventh terms, and I will be left with the sum of the fifteenth through forty-seventh terms, as shown in the following table. These are the fourteenth and forty-seventh terms, respectively, that are required: a14= 2(14) – 5 = 23a47= 2(47) – 5 = 89a14= 2(14) – 5 = 23a47= 2(47) – 5 = 89 With these values, I now have everything I need to find the two partial sums for my subtraction, which are as follows: I got the following result after subtracting: Then here’s what I’d say: As a side note, this subtraction can also be written as ” S 47 – S 14 “. Don’t be surprised if you come across an exercise that employs this notation and requires you to decipher its meaning before you can proceed with your calculations; this is common. If you’re dealing with something more complex, however, it may be necessary to group symbols together in order to make the meaning more obvious. In order to do so correctly, the author of the previous exercise should have formatted the summation with grouping symbols in the manner described below: #### Find the value ofnfor which the following equation is true: Knowing that the first term has the value a1= 0.25(1) + 2 = 2.25, I may proceed to the second term. It appears from the formula that each term will be 0.25 units larger than the preceding term, indicating that this is an arithmetical series withd= 0.25, as shown in the diagram. The summation formula for arithmetical series therefore provides me with the following results: The number n is equal to 2.25 + 0.25 + 2 = 42n is equal to 0.25 + 4.25 + 42 = 420.25 n2+ 4.25 n– 42 = 0n2+ 17 n– 168 = 0(n+ 24)(n– 7 = 0n2+ 17 n– 168 = 0(n+ 24)(n– 7). You might be interested: Why Is Geometric Mean Less Than Arithmetic? (Solution found) Then n= 7 is the answer. However, your instructor may easily assign you a summation that needs you to use, say, eighty-six words or a thousand terms in order to arrive at the correct total. As a result, be certain that you are able to do the calculations from the formula. #### Find the sum of1 + 5 + 9 +. + 49 + 53 After looking through the phrases, I can see that this is, in fact, an arithmetic sequence: The sum of 5 and 1 equals 49 and 5 equals 453 and 49 equals 4. The reason for this is that they won’t always inform me, especially on the exam, what sort of series they’ve given me. (And I want to get into the habit of checking this way.) They’ve provided me the first and last terms of this series, however I’m curious as to how many overall terms there are in this series. This is something I’ll have to sort out for myself. After plugging these numbers into the algorithm, I can calculate how many terms there are in total: a n=a1+ (n–1) d 53 = 1 + (n–1) a n=a1+ (n–1) (4) 53 = 1 + 4 n– 453 = 4 n– 356 = 4 n– 14 =n 53 = 1 + 4 n– 453 = 4 n– 356 = 4 n– 14 =n There are a total of 14 words in this series. + 49 + 53 = 1 + 5 + 9 Then I’ll give you my answer: partial sum S 14 = 378 S 14= 378 After that, we’ll look at geometric series. ## Partial Sums of an Arithmetic Sequence MathJax is the software we utilize. To obtain the sum of an arithmetic series, a finite number of terms from the sequence can be added together. As a result of the unlimited long-term behavior of an arithmetic sequence, we are always limited to adding a finite number of terms to it. ### Specific Numerical Results Take the number$8+13+18+23+ldots+273$ as an example. In a short time, we realize that the terms have a common difference of 5, and that this total is the product of an arithmetic series whose explicit formula is$a n=5n+3$. As a result, the sequence of partial sums is defined by$s n=sumlimits_ n (5k+3)$, where $n$ is any positive integer. When we solve the equation$5n+3=273$, we find out that 273 is the 54th term in the sequence. We might proceed in the following manner if we used a little imagination. Utilizing the characteristics of sums is an additional method of solving the problem. (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 (5)frac+54(3) = 7587 ### Specific Partial Sum Formulas It is possible to use either strategy to generate a formula for the series of partial sums$s n=sumlimits_ n (5k+3)$, just by leaving $n$ as a variable in the equation. As a result, we get the explicit formulae for the sequence of partial sums that are shown below. • $s n=dfrac(a 1+a n)=dfrac(8+5n+3)=dfrac$ • _$sumlimits_ _$sumlimits_ _$sumlimits_ n (5k+3)=5sumlimits_ n k+sumlimits_ n 3=(5)dfrac+3n=dfrac n 5=(5)dfrac+3n=dfrac n 3=(5)dfrac+3n=dfrac$ ### Applications Various applications for the arithmetic sequence may be found on the internet. Examples include the following: • The seats in an auditorium are arranged in a semicircular configuration, similar to that of a theater. There are 30 rows in total, with the first row having 24 seats and each successive row having an extra 2 seats each row after that. In the explicit formula, the last row will have$a_ =2(30)+22=82$seats, and there will be$s n=dfrac(24+82)=1590$seats in the auditorium • Campbell County’s grain production was 150 million bushels in 1990, and it has been rising at a rate of 3 million bushels per year since then. With the explicit formula ($a n=3n+147$), the county produced$a_ =3(21)+147=210$million bushels in the year 2010, and a total of$s_ =dfrac(150+210)=3780$million bushels throughout the course of those twenty-one years. ## Partial Sums An example of a Partial Sum is the sum of a portion of a sequence. ### Example: This is the sequence of even integers starting with 2 and going up to infinity: There are four words in that sequence, and this is the partial sum of those four terms: 2 + 4 + 6 + 8 = 20 Now, let us define the terms a bit more precisely: A sequence is a collection of items (typically numbers) that are arranged in a specific order. A Partial Sum is the sum of a portion of a sequence of numbers. An Infinite Series is made up of the sum of all infinite terms. Partial sums are also referred to as “Finite Series” in some circles. ## Sigma Partial sums are frequently written with the phrase “add them all up” in mind: This sign (calledSigma) is derived from the Greek word for “sum up,” which implies to bring everything together. ### Sum What? Thevalues are shown belowand above the Sigma: 4Σn=1n it saysngoes from 1 to 4,which is1,2,3and4 ### OK, Let’s Go. So now we add up 1,2,3 and 4: 4Σn=1n = 1 + 2 + 3 + 4 =10 It is common to see partial sums printed with the phrase “add them all up” in the text: So, to summarize, this sign (calledSigma) indicates “add up” or “compile.” ## More Powerful But you have the ability to accomplish far more powerful things than that! We may square each time and add the results to get the following: The number n 2 equals one plus two plus three plus four plus five equals thirty. We can put the first four words in this sequence together to get the total. 2n+1: The sum of (2n+1) is 3 + 5 + 7 + 9 = 24. We may also use other letters; for example, we can useiand sum upi (i+1), going from 1 to 3: the product of i(i+1) = 12 plus 23 plus 34 = 20 Furthermore, we can begin and conclude with any number. ## Properties Partial Sums have a number of beneficial qualities that may be used to assist us in doing the computations. ### Multiplying by a Constant Property Suppose we have a summarization we want to make, let’s say we name ita ka kcould be 2 or k(k-7)+2 or anything really, and cis any constant number (like 2, or-9.1, or anything really), then: In other words, if every term we are summing is multiplied by a constant, we may “pull” the constant outside thesigma by multiplying the constant by the total of the terms. ### Example: So, let’s say there’s something we want to sum up, and we’ll name ita ka, where ka may be anything from 2 to 7+2, or anything else at all. If andc is a constant value (such as 2, or -9.1, or something else), then: As a result, if every term we are summing is multiplied by a constant, we may “pull” the constant outside of a square root of the total of its squares. Which means that when two terms are put together and we wish to sum them up, we may actually sum them individually and then add the results together, which is a valuable knowledge. ### Example: It will be simpler to do the two sums separately and then combine them at the end. It should be noted that this also works for subtraction: ## Useful Shortcuts Lastly, here are some helpful shortcuts that make the sums a whole lot easier. In each example, we are attempting to sum from 1 to some valuen. Summing1equalsn Summing the constantcequalsctimesn A shortcut when summingk A shortcut when summingk 2 A shortcut when summingk 3 Also true when summingk 3 Summing odd numbers Let’s have a look at some of them: ### Example 1: You sell concrete blocks for landscaping. A buyer has expressed interest in purchasing the full “pyramid” of blocks that you keep out front. The stack is 14 blocks in height. What is the total number of blocks in there? Because each layer is a square, the computation goes as follows: 1 2+ 2 2+ 3 2+ 1 2+ 2 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2+ 1 2 However, the following is a far more straightforward way to write it: We may utilize the following formula fork 2 from the previous section: That was a lot less difficult than adding 1 2+ 2 2+ 3 2+. + 14 2 to get a total of 1 2+ 2 2+. ### Example 2: The customer wants a better price. According to the client, the bricks on the outside of the pyramid should be less expensive because they need to be cleaned. You acknowledge that you have read and understand the following: What is the overall cost of the project? You may use this formula to figure out how many “inner” and “outer” blocks are in any layer (except the first). As a result, the cost per layer is as follows: • Cost (outer blocks) = $7 x 4 (size-1) • Cost (inner blocks) =$11 x (size-2) 2 As a result, all layers (except from the first) will cost: Now that we know the total, let’s see if we can make the computations a little easier! Using the “Addition Property” from the previous section: Using the “Multiply by Constant Property” from before, we can create the following: That’s a nice thing. However, because we are starting from i=2 instead of i=1, we are unable to take any shortcuts. IF, ON THE OTHER HAND, WE INVENT TWO NEW VARIABLES: We have the following:(I removed the k=0 example because I am aware that 0 2 =0) And now we may take advantage of the shortcuts: After a quick computation, we arrive at: $7 364 plus$11 650 equals \$9,698.00. And don’t forget about the top layer (size=1), which is made up of only one single block. Keep in mind that when we combine the “outer” and “inner” blocks together together with the one on top, we obtain the sum of 364 + 650 + 1 = 1015. yay! ## Arithmetic Sequences and Sums A sequence is a collection of items (typically numbers) that are arranged in a specific order. Each number in the sequence is referred to as aterm (or “element” or “member” in certain cases); for additional information, see Sequences and Series. ## Arithmetic Sequence An Arithmetic Sequence is characterized by the fact that the difference between one term and the next is a constant. In other words, we just increase the value by the same amount each time. endlessly. ### Example: 1, 4, 7, 10, 13, 16, 19, 22, and 25 are the numbers 1 through 25. Each number in this series has a three-digit gap between them. Each time the pattern is repeated, the last number is increased by three, as seen below: As a general rule, we could write an arithmetic series along the lines of • There are two words: Ais the first term, and dis is the difference between the two terms (sometimes known as the “common difference”). ### Example: (continued) 1, 4, 7, 10, 13, 16, 19, 22, and 25 are the numbers 1 through 25. Has: • In this equation, A = 1 represents the first term, while d = 3 represents the “common difference” between terms. And this is what we get: ### Rule It is possible to define an Arithmetic Sequence as a rule:x n= a + d(n1) (We use “n1” since it is not used in the first term of the sequence). ### Example: Write a rule, and calculate the 9th term, for this Arithmetic Sequence: 3, 8, 13, 18, 23, 28, 33, and 38 are the numbers three, eight, thirteen, and eighteen. Each number in this sequence has a five-point gap between them. The values ofaanddare as follows: • A = 3 (the first term) • D = 5 (the “common difference”) • A = 3 (the first term). In this equation, A = 3 (the first term), d = 5 (the “common difference”), and e = 3. ## Advanced Topic: Summing an Arithmetic Series To summarize the terms of this arithmetic sequence:a + (a+d) + (a+2d) + (a+3d) + (a+4d) + (a+5d) + (a+6d) + (a+7d) + (a+8d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + (a+9d) + ( make use of the following formula: What exactly is that amusing symbol? It is referred to as The Sigma Notation is a type of notation that is used to represent a sigma function. Additionally, the starting and finishing values are displayed below and above it: “Sum upnwherengoes from 1 to 4,” the text states. 10 is the correct answer. ### Example: Add up the first 10 terms of the arithmetic sequence: The values ofa,dandnare as follows: • A = 1 (which is the first term) • D = 3 (the “common difference” between two words) • D = 3 (the “common difference” between two terms) • N = 10 (number of words to add together) To summarize, the equation becomes:= 2+9 + 3 = 5(29). Why don’t you try adding the terms yourself and check whether it comes out to 145? ## Footnote: Why Does the Formula Work? Let’s take a look at why the formula works because we’ll be employing an unusual “technique” that’s worth understanding. First, we’ll refer to the entire total as “S”: S = a + (a + d) +. + (a + (n2)d) +(a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n1)d) + (a + (n2)d) + (a + (n1)d) + (a + (n1)d) + (a + (n1)d) + After that, rewrite S in the opposite order: S = (a + (n1)d)+ (a + (n2)d)+. +(a + d)+a. +(a + d)+a. +(a + d)+a. Now, term by phrase, add these two together: S = a + (a+d) + . + (a + (n-2)d) + (a + (n-1)d) S = (a + (n-1)d) + (a + (n-2)d) + . + (a + d) + a 2S = (2a + (n-1)d) + (2a + (n-1)d) + . + (2a + (n-1)d) + (2a + (n-1)d) Each and every term is the same! Furthermore, there are “n” of them. 2S = n (2a + (n1)d) = n (2a + (n1)d) Now, we can simply divide by two to obtain the following result: The function S = (n/2) (2a + (n1)d) is defined as This is the formula we’ve come up with:	7,651	26,366	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.6875	5	CC-MAIN-2022-33	longest	en	0.945332
http://www.corestandards.org/Math/Content/MD/	1,477,152,434,000,000,000	text/html	crawl-data/CC-MAIN-2016-44/segments/1476988719027.25/warc/CC-MAIN-20161020183839-00404-ip-10-171-6-4.ec2.internal.warc.gz	407,277,104	13,384	# Measurement & Data ## Kindergarten #### Describe and compare measurable attributes. CCSS.Math.Content.K.MD.A.1 Describe measurable attributes of objects, such as length or weight. Describe several measurable attributes of a single object. CCSS.Math.Content.K.MD.A.2 Directly compare two objects with a measurable attribute in common, to see which object has "more of"/"less of" the attribute, and describe the difference. For example, directly compare the heights of two children and describe one child as taller/shorter. #### Classify objects and count the number of objects in each category. CCSS.Math.Content.K.MD.B.3 Classify objects into given categories; count the numbers of objects in each category and sort the categories by count.1 #### Measure lengths indirectly and by iterating length units. CCSS.Math.Content.1.MD.A.1 Order three objects by length; compare the lengths of two objects indirectly by using a third object. CCSS.Math.Content.1.MD.A.2 Express the length of an object as a whole number of length units, by laying multiple copies of a shorter object (the length unit) end to end; understand that the length measurement of an object is the number of same-size length units that span it with no gaps or overlaps. Limit to contexts where the object being measured is spanned by a whole number of length units with no gaps or overlaps. #### Tell and write time. CCSS.Math.Content.1.MD.B.3 Tell and write time in hours and half-hours using analog and digital clocks. #### Represent and interpret data. CCSS.Math.Content.1.MD.C.4 Organize, represent, and interpret data with up to three categories; ask and answer questions about the total number of data points, how many in each category, and how many more or less are in one category than in another. #### Measure and estimate lengths in standard units. CCSS.Math.Content.2.MD.A.1 Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes. CCSS.Math.Content.2.MD.A.2 Measure the length of an object twice, using length units of different lengths for the two measurements; describe how the two measurements relate to the size of the unit chosen. CCSS.Math.Content.2.MD.A.3 Estimate lengths using units of inches, feet, centimeters, and meters. CCSS.Math.Content.2.MD.A.4 Measure to determine how much longer one object is than another, expressing the length difference in terms of a standard length unit. #### Relate addition and subtraction to length. CCSS.Math.Content.2.MD.B.5 Use addition and subtraction within 100 to solve word problems involving lengths that are given in the same units, e.g., by using drawings (such as drawings of rulers) and equations with a symbol for the unknown number to represent the problem. CCSS.Math.Content.2.MD.B.6 Represent whole numbers as lengths from 0 on a number line diagram with equally spaced points corresponding to the numbers 0, 1, 2, ..., and represent whole-number sums and differences within 100 on a number line diagram. #### Work with time and money. CCSS.Math.Content.2.MD.C.7 Tell and write time from analog and digital clocks to the nearest five minutes, using a.m. and p.m. CCSS.Math.Content.2.MD.C.8 Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have? #### Represent and interpret data. CCSS.Math.Content.2.MD.D.9 Generate measurement data by measuring lengths of several objects to the nearest whole unit, or by making repeated measurements of the same object. Show the measurements by making a line plot, where the horizontal scale is marked off in whole-number units. CCSS.Math.Content.2.MD.D.10 Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems1 using information presented in a bar graph. #### Solve problems involving measurement and estimation. CCSS.Math.Content.3.MD.A.1 Tell and write time to the nearest minute and measure time intervals in minutes. Solve word problems involving addition and subtraction of time intervals in minutes, e.g., by representing the problem on a number line diagram. CCSS.Math.Content.3.MD.A.2 Measure and estimate liquid volumes and masses of objects using standard units of grams (g), kilograms (kg), and liters (l).1 Add, subtract, multiply, or divide to solve one-step word problems involving masses or volumes that are given in the same units, e.g., by using drawings (such as a beaker with a measurement scale) to represent the problem.2 #### Represent and interpret data. CCSS.Math.Content.3.MD.B.3 Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step "how many more" and "how many less" problems using information presented in scaled bar graphs. For example, draw a bar graph in which each square in the bar graph might represent 5 pets. CCSS.Math.Content.3.MD.B.4 Generate measurement data by measuring lengths using rulers marked with halves and fourths of an inch. Show the data by making a line plot, where the horizontal scale is marked off in appropriate units— whole numbers, halves, or quarters. #### Geometric measurement: understand concepts of area and relate area to multiplication and to addition. CCSS.Math.Content.3.MD.C.5 Recognize area as an attribute of plane figures and understand concepts of area measurement. CCSS.Math.Content.3.MD.C.5.a A square with side length 1 unit, called "a unit square," is said to have "one square unit" of area, and can be used to measure area. CCSS.Math.Content.3.MD.C.5.b A plane figure which can be covered without gaps or overlaps by n unit squares is said to have an area of n square units. CCSS.Math.Content.3.MD.C.6 Measure areas by counting unit squares (square cm, square m, square in, square ft, and improvised units). CCSS.Math.Content.3.MD.C.7 Relate area to the operations of multiplication and addition. CCSS.Math.Content.3.MD.C.7.a Find the area of a rectangle with whole-number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths. CCSS.Math.Content.3.MD.C.7.b Multiply side lengths to find areas of rectangles with whole-number side lengths in the context of solving real world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning. CCSS.Math.Content.3.MD.C.7.c Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Use area models to represent the distributive property in mathematical reasoning. CCSS.Math.Content.3.MD.C.7.d Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems. #### Geometric measurement: recognize perimeter. CCSS.Math.Content.3.MD.D.8 Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters. #### Solve problems involving measurement and conversion of measurements. CCSS.Math.Content.4.MD.A.1 Know relative sizes of measurement units within one system of units including km, m, cm; kg, g; lb, oz.; l, ml; hr, min, sec. Within a single system of measurement, express measurements in a larger unit in terms of a smaller unit. Record measurement equivalents in a two-column table. For example, know that 1 ft is 12 times as long as 1 in. Express the length of a 4 ft snake as 48 in. Generate a conversion table for feet and inches listing the number pairs (1, 12), (2, 24), (3, 36), ... CCSS.Math.Content.4.MD.A.2 Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale. CCSS.Math.Content.4.MD.A.3 Apply the area and perimeter formulas for rectangles in real world and mathematical problems. For example, find the width of a rectangular room given the area of the flooring and the length, by viewing the area formula as a multiplication equation with an unknown factor. #### Represent and interpret data. CCSS.Math.Content.4.MD.B.4 Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Solve problems involving addition and subtraction of fractions by using information presented in line plots. For example, from a line plot find and interpret the difference in length between the longest and shortest specimens in an insect collection. #### Geometric measurement: understand concepts of angle and measure angles. CCSS.Math.Content.4.MD.C.5 Recognize angles as geometric shapes that are formed wherever two rays share a common endpoint, and understand concepts of angle measurement: CCSS.Math.Content.4.MD.C.5.a An angle is measured with reference to a circle with its center at the common endpoint of the rays, by considering the fraction of the circular arc between the points where the two rays intersect the circle. An angle that turns through 1/360 of a circle is called a "one-degree angle," and can be used to measure angles. CCSS.Math.Content.4.MD.C.5.b An angle that turns through n one-degree angles is said to have an angle measure of n degrees. CCSS.Math.Content.4.MD.C.6 Measure angles in whole-number degrees using a protractor. Sketch angles of specified measure. CCSS.Math.Content.4.MD.C.7 Recognize angle measure as additive. When an angle is decomposed into non-overlapping parts, the angle measure of the whole is the sum of the angle measures of the parts. Solve addition and subtraction problems to find unknown angles on a diagram in real world and mathematical problems, e.g., by using an equation with a symbol for the unknown angle measure. #### Convert like measurement units within a given measurement system. CCSS.Math.Content.5.MD.A.1 Convert among different-sized standard measurement units within a given measurement system (e.g., convert 5 cm to 0.05 m), and use these conversions in solving multi-step, real world problems. #### Represent and interpret data. CCSS.Math.Content.5.MD.B.2 Make a line plot to display a data set of measurements in fractions of a unit (1/2, 1/4, 1/8). Use operations on fractions for this grade to solve problems involving information presented in line plots. For example, given different measurements of liquid in identical beakers, find the amount of liquid each beaker would contain if the total amount in all the beakers were redistributed equally. #### Geometric measurement: understand concepts of volume. CCSS.Math.Content.5.MD.C.3 Recognize volume as an attribute of solid figures and understand concepts of volume measurement. CCSS.Math.Content.5.MD.C.3.a A cube with side length 1 unit, called a "unit cube," is said to have "one cubic unit" of volume, and can be used to measure volume. CCSS.Math.Content.5.MD.C.3.b A solid figure which can be packed without gaps or overlaps using n unit cubes is said to have a volume of n cubic units. CCSS.Math.Content.5.MD.C.4 Measure volumes by counting unit cubes, using cubic cm, cubic in, cubic ft, and improvised units. CCSS.Math.Content.5.MD.C.5 Relate volume to the operations of multiplication and addition and solve real world and mathematical problems involving volume. CCSS.Math.Content.5.MD.C.5.a Find the volume of a right rectangular prism with whole-number side lengths by packing it with unit cubes, and show that the volume is the same as would be found by multiplying the edge lengths, equivalently by multiplying the height by the area of the base. Represent threefold whole-number products as volumes, e.g., to represent the associative property of multiplication. CCSS.Math.Content.5.MD.C.5.b Apply the formulas V = l × w × h and V = b × h for rectangular prisms to find volumes of right rectangular prisms with whole-number edge lengths in the context of solving real world and mathematical problems. CCSS.Math.Content.5.MD.C.5.c Recognize volume as additive. Find volumes of solid figures composed of two non-overlapping right rectangular prisms by adding the volumes of the non-overlapping parts, applying this technique to solve real world problems. ## High School: Statistics & Probability #### Calculate expected values and use them to solve problems CCSS.Math.Content.HSS.MD.A.1 (+) Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space; graph the corresponding probability distribution using the same graphical displays as for data distributions. CCSS.Math.Content.HSS.MD.A.2 (+) Calculate the expected value of a random variable; interpret it as the mean of the probability distribution. CCSS.Math.Content.HSS.MD.A.3 (+) Develop a probability distribution for a random variable defined for a sample space in which theoretical probabilities can be calculated; find the expected value. For example, find the theoretical probability distribution for the number of correct answers obtained by guessing on all five questions of a multiple-choice test where each question has four choices, and find the expected grade under various grading schemes. CCSS.Math.Content.HSS.MD.A.4 (+) Develop a probability distribution for a random variable defined for a sample space in which probabilities are assigned empirically; find the expected value. For example, find a current data distribution on the number of TV sets per household in the United States, and calculate the expected number of sets per household. How many TV sets would you expect to find in 100 randomly selected households? #### Use probability to evaluate outcomes of decisions CCSS.Math.Content.HSS.MD.B.5 (+) Weigh the possible outcomes of a decision by assigning probabilities to payoff values and finding expected values. CCSS.Math.Content.HSS.MD.B.5.a Find the expected payoff for a game of chance. For example, find the expected winnings from a state lottery ticket or a game at a fast-food restaurant. CCSS.Math.Content.HSS.MD.B.5.b Evaluate and compare strategies on the basis of expected values. For example, compare a high-deductible versus a low-deductible automobile insurance policy using various, but reasonable, chances of having a minor or a major accident. CCSS.Math.Content.HSS.MD.B.6 (+) Use probabilities to make fair decisions (e.g., drawing by lots, using a random number generator). CCSS.Math.Content.HSS.MD.B.7 (+) Analyze decisions and strategies using probability concepts (e.g., product testing, medical testing, pulling a hockey goalie at the end of a game).	3,366	15,141	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.953125	4	CC-MAIN-2016-44	longest	en	0.86625
https://mathematica.stackexchange.com/questions/60960/integers-which-are-the-sum-of-both-two-and-three-consecutive-squares/60975	1,579,715,369,000,000,000	text/html	crawl-data/CC-MAIN-2020-05/segments/1579250607314.32/warc/CC-MAIN-20200122161553-20200122190553-00185.warc.gz	539,011,442	39,081	Integers which are the sum of both two and three consecutive squares This is a math problem I came across the other day: $365$ can be written as a sum of two and also three consecutive perfect squares: $$365=14^2+13^2=12^2+11^2+10^2$$ What is the next number with this property? Give the last 4 digits of the number. The perfect squares cannot be zero. I would like to know what would be a good way (especially performance-wise) to check the first, let's say, $1000000$ natural numbers if they can be represented in the way above, using Mathematica? • This problem can be recast as the Pell equation $v^2-24w^2=1$, whose integral solutions $(v,w)$ come from coefficients of $(5+\sqrt{24})^k=v_k+w_k\sqrt{24}$ for integers $k$. This was done more generally for the problem of describing the integers that are a sum of $n$ and $n+1$ consecutive squares almost 50 years ago, by Alder and Simons. See "$n$ and $n + 1$ Consecutive Integers with Equal Sums of Squares," Amer. Math. Monthly 74 (1967), 28–30. Writing $n(n+1)=a^2b$ where $b$ is squarefree, the task amounts to solving $v^2-4bw^2=1$ in integers $v$ and $w$. – KConrad Oct 1 '14 at 12:14 • @KConrad Why dont you put together an answer similar to this comment, just a little bit extended? It looks extremely interesting. – VividD Oct 1 '14 at 12:51 • @KConrad, thank you for pointing that paper out; it was a fun read! Thanks to your suggestion, I was able to come up with an efficient routine. – J. M. will be back soon Oct 22 '15 at 15:31 • Incidentally, sequences associated with this problem involve nice relations between second, third, and fifth powers. See math.stackexchange.com/questions/952216 – Tito Piezas III Jan 13 '16 at 2:12 You can just step through $i$ and $j$ while trying to simultaneously satisfy $$i^2+(i+1)^2=j^2+(j+1)^2+(j+2)^2$$ Just loop and if the inequality is too small on the left, increment $i$. If it's too small on the right, increment $j$. That looks like this: Clear[f, g, i, j]; f[i_] = i^2 + (i + 1)^2; g[j_] = j^2 + (j + 1)^2 + (j + 2)^2; max = 10^6; i = 1; j = 1; While[f[i] <= max && g[i] <= max, If[f[i] == g[j], Print[{i, j, f[i]}]; i++;]; If[f[i] < g[j], i++]; If[f[i] > g[j], j++]; ]; (Output: {13, 10, 365} {133, 108, 35645} ) This executes almost instantaneously. So $133^2+134^2 = 108^2 + 109^2 + 110^2 = 35645$. You can increase max to find more, like these: {13, 10, 365} {133, 108, 35645} {1321, 1078, 3492725} {13081, 10680, 342251285} {129493, 105730, 33537133085} That's up to $10^{12}$, which takes about 10 seconds. Further discussion Any useful algorithm here will focus on the $i$ and $j$, rather than the $n$, from $$i^2+(i+1)^2=j^2+(j+1)^2+(j+2)^2=n$$ If you are searching in a straightforward way, with all things equal, checking all possible $i$ and $j$ (keeping in mind that you iterate them together) takes about $\sqrt{n}$ time (whereas checking all possible $n$ takes, well, $n$ time). You can try something using FindInstance, but even the following: Timing[FindInstance[i^2 + (i + 1)^2 == j^2 + (j + 1)^2 + (j + 2)^2 && i > 0 && j > 0, {i, j}, Integers]] will still take about ten times as long as the code above. http://oeis.org/A007667 Clear[fa, ga]; fa = Total[{#, # + 1}^2] &; ga = Total[{#, # + 1, # + 2}^2] &; Update: A closed form function soln = Assuming[{C[1] ∈ Integers && C[1] >= 0 && x > 0 && y > 0}, Simplify@ Reduce[Total[{x, x + 1}^2] == Total[{y, y + 1, y + 2}^2], {x, y}, Integers]] /. C[1] -> n; fa@{ToRules[soln]}[[All, All, -1]][[1, 1]] // FullSimplify ClearAll[genOEISa007667]; genOEISa007667[n_Integer] := 1/8 (10 + 3 (5 - 2 Sqrt[6])^(2 n) (5 + 2 Sqrt[6]) + (15 - 6 Sqrt[6]) (5 + 2 Sqrt[6])^(2 n)) Finding all in the range $1$ to $10^{197}$ in less than 0.005 seconds: genOEISa007667/@ Range[100]; // AbsoluteTiming (* {0.005990,Null} ) The first 50 numbers: genOEISa007667/@ Range[50] // Simplify // Column Previous version: ClearSystemCache[] (res = Reduce[{fa[x] == ga[y] && 1 < x < 100000000000000000 && 1 < y < 100000000000000000}, {x, y}, Integers]); // AbsoluteTiming ({0.102073, Null} ) initials = (res /. Or \| And -> List)[[All, All, -1]]; {Row[#, ","], fa[First@#]} & /@ initials // TableForm • Where is the next set? – Kellen Myers Sep 30 '14 at 8:53 • @kguler very nice and instructive – ubpdqn Sep 30 '14 at 12:20 • Is there any reason, mathematically easy to explain, why the factor between i_n and i_(n+1) seems to converge to 9.8989..? – Guntram Blohm supports Monica Sep 30 '14 at 12:31 • @GuntramBlohm, good observation. – kglr Sep 30 '14 at 15:00 • Limit[genOEISa007667[n + 1]/genOEISa007667[n], n -> Infinity] gives 49 + 20 Sqrt[6] or 97.98979485566356 – Bob Hanlon Sep 30 '14 at 15:52 Update: the lhs of $i^2+(i+1)^2=j^2+(j+1)^2+(j+2)^2$ is always odd so $j$ should be even. The fastest solution I found: max = 10^12; {(Sqrt[3 + 6 (# + 1)^2] - 1)/2, #, #^2 + (# + 1)^2 + (# + 2)^2} & @@@ (2 Position[#, 0]) &@UnitStep[Abs[# - Round[#]] - 1.^-10] &[ Sqrt[0.75 + 1.5 #^2] - 0.5] &@ Range[3., Sqrt[max/3.], 2] // AbsoluteTiming (* {0.023773, {{13, 10, 365}, {133, 108, 35645}, {1321, 1078, 3492725}, {13081, 10680, 342251285}, {129493, 105730, 33537133085}}} ) Only 0.02 sec for max = 10^12! It fast because it checks all number simultaneously and uses PackedArrays. I use floating point numbers, but it is enough to obtain the correct result. I'm surprised no one bothered to pursue the method given in Simons and Alder's paper, as commented by KConrad. A Mathematica implementation of their idea is surprisingly short: n = 10; MapThread[With[{z = (#1 + 1)/2}, 3 #2 + 2 z - 2 + {z, 0}] &, LinearRecurrence[{10, -1}, #, {1, n} + 1] & /@ {{1, 5}, {0, 2}}] {{13, 10}, {133, 108}, {1321, 1078}, {13081, 10680}, {129493, 105730}, {1281853, 1046628}, {12689041, 10360558}, {125608561, 102558960}, {1243396573, 1015229050}, {12308357173, 10049731548}} 2 (# + 1) # + 1 & /@ %[[All, 1]] {365, 35645, 3492725, 342251285, 33537133085, 3286296790925, 322023548377445, 31555021444198565, 3092070077983081805, 302991312620897818205} As mentioned, the problem can be reduced to solving a certain Pell equation. I used LinearRecurrence[] to generate the Pell solutions, after which they are transformed to the first terms of the sum of consecutive squares. To check, say, the second term of the sequence: 133^2 + 134^2 == 108^2 + 109^2 + 110^2 == 35645 True I recommend reading that fascinating paper if you can. Not being contented with what I've written above, I've decided to write a full Mathematica implementation of the Simons and Alder method. The function consecutiveSquareSum[n, k], given below, will list the first k triples consisting of the first term of the sums of n and n + 1 consecutive squares, and the sum itself: SquarefreePart[n_Integer?Positive] := Times @@ Power @@@ MapAt[Mod[#, 2] &, FactorInteger[n], {All, 2}] ( generate k solutions to the Diophantine equation x^2 - n y^2 == 1 ) pellSolutions[n_Integer, k_Integer?Positive] := Module[{p, q, r, u, v}, r = Length[Last[ContinuedFraction[Sqrt[n]]]]; If[OddQ[r], r = 2]; {p, q} = Through[{Numerator, Denominator}[Last[Convergents[Sqrt[n], r]]]]; {u, v} = {2 p, n q^2 - p^2}; Transpose[LinearRecurrence[{u, v}, #, {2, k + 1}] & /@ {{1, p}, {0, q}}]] consecutiveSquareSum[n_Integer?Positive, k_Integer?Positive] := Module[{a, a2b, b, cQ, d}, cQ = MatchQ[Mod[n, 4], 1 \| 2]; a2b = If[cQ, n (n + 1)/2, n (n + 1)/4]; b = SquarefreePart[a2b]; a = Sqrt[a2b/b]; d = If[cQ, 2 b, b]; Block[{z = (#1 + 1)/2, y}, y = a b #2 + n (z - 1); {y + z, y, ((n/3 + 1/2) n + 1/6) n + y (y + n) (n + 1)}] & @@@ pellSolutions[d, k]] As an example, generate the first five sums of $100$ and $101$ consecutive squares: consecutiveSquareSum[100, 5] {{20201, 20100, 41008358350}, {8140601, 8100200, 6627019056398350}, {3272521201, 3256280300, 1070939533497408458350}, {1315545402001, 1309016600400, 173065970485621168960538350}, {528845979103001, 526221417100500, 27967806961346412612849840638350}} Taking the first triple as an example, this says that $20201^2 + \cdots + 20300^2 = 20100^2 + \cdots + 20200^2 = 41008358350$. This is not particularly fast but works: f = (-1 + Sqrt[3 + 6 #^2])/2 &; q = IntegerQ /@ (f /@ Range[1000000]); yields the triples : {1, 11, 109, 1079, 10681, 105731} fanswer = f /@ answer {1, 13, 133, 1321, 13081, 129493} Dropping first case which is 0^2+1^2+2^2=1^2+2^2: trip = Rest[{(# - 1), #, # + 1} & /@ answer]; ftrip = Rest[{#, (# + 1)} & /@ fanswer]; s = StringTemplate["\!$$\SuperscriptBox[$$$, $$2$$]$"]; func = Function[x, StringJoin @@ Riffle[TemplateApply[s, #] & /@ x, "+"]]; Grid[MapThread[{#1, "=", #2, "=", ToExpression[#2]} &, {func /@ trip, func /@ ftrip}]] yields: Solutions can be greatly simplified simply computing the next number. For the equation: $$n^2+(n+1)^2=k^2+(k+1)^2+(k+2)^2$$ Using the first number. $(p_1 ; s_1) - (1 ; 0 )$ Let's use these numbers. Which are the sequence. The following is found using the previous value according to the formula. $$p_2=5p_1+12s_1$$ $$s_2=2p_1+5s_1$$ Using the numbers $p,s$ - you can find solutions on formulas. $$n=p^2\pm6ps+12s^2$$ $$k=6s(2s\pm{p})$$ $$$$ $$n=-(2p^2\pm6ps+6s^2)$$ $$k=-2p(p\pm3s)$$ So the formula looked compact, you can do the replacement. $$x=p^2\pm6ps+6s^2$$ $$y=p^2\pm4ps+6s^2$$ Then the solutions are. $$n=4x^2\pm6xy+3y^2$$ $$k=2x(2x\pm3y)$$ $$$$ $$n=-(2x^2\pm6xy+6y^2)$$ $$k=-6y(y\pm{x})$$ Or such replacement. $$z=2ps$$ $$q=p^2+6s^2$$ Then the solutions are. $$n=-(2q^2\pm6qz+6z^2)$$ $$k=-2q(q\pm3z)$$ $$**$$ $$n=q^2\pm6qz+12z^2$$ $$k=6z(2z\pm{q})$$ Interestingly, all this variety of formulas give the same solution. So that one can restrict the upper formula. The rest of the formula was drawn to show what interesting patterns there. • Hi ! Can you make this post self contained, e.g. add a few lines of code demonstrating the method outlined here. – Sektor Dec 24 '14 at 8:22 • @Sektor I tried it as easy as possible to write. So you can even use a calculator. I wrote not that interesting calculation, and to show how different calculations can lead to the same result. – individ Dec 24 '14 at 8:38 • I can see that you are trying to write in a clear fashion. I just wanted you to demonstrate that this method actually solves the problem at hand. – Sektor Dec 24 '14 at 8:40 You could always use brute force: max = 10^12; ( sq2 = Range[Floor[Sqrt[max/2]]]^2; sq2 = sq2[[2 ;;]] + sq2[[ ;; -2]]; sq3 = Range[Floor[Sqrt[max/3]]]^2; sq3 = sq3[[3 ;;]] + sq3[[2 ;; -2]] + sq3[[;; -3]]; Intersection[sq2, sq3] ) // Timing {0.046800, {365, 35645, 3492725, 342251285, 33537133085}} Not as elegant as the others, but quite fast: Finding all numbers < 10^18 takes about 22 seconds. (It needs a lot of RAM, though!) {21.793340, {365, 35645, 3492725, 342251285, 33537133085, 3286296790925, 322023548377445, 31555021444198565}}	3,987	10,869	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 2, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.96875	4	CC-MAIN-2020-05	latest	en	0.911866
http://blog.themathmom.com/2009/08/who-is-your-role-model-explained-by.html?showComment=1465019119571	1,660,699,284,000,000,000	text/html	crawl-data/CC-MAIN-2022-33/segments/1659882572833.78/warc/CC-MAIN-20220817001643-20220817031643-00726.warc.gz	6,901,458	43,611	### Who is your role model - explained by math. This fun trick came in the email from my friend. Try it and see how math could explain this mystery. Do this without looking at the answers 1. Pick your favorite number between 1 - 9 2. Multiply your favorite number by 3 4. Then multiply by 3 (I will wait while you get the calculator) 5. You should have a 2 digit number Now scroll down......... Now match your number with the number below to see who is your ROLE MODEL on the list below: 1. Albert Einstein 2. Nelson Mandela 3. Margaret Thatcher 4. Tom Cruise 5. Bill Gates 6. Gandhi 8. Queen Elizabeth II 9. The Math Mom 10. George Washington I know... I just have that effect on people... one day you too can be like me. Do not aspire to anything less... PS... Stop picking different numbers. I am your idol, just deal with it... Curious to know how this "magic" works? Let's say you picked number N. Following the instructions you multiply it by 3, getting 3N. And multiply by 3: 3 x (3N+3) = 9N + 9 Now you are instructed in (6) to add two digits that compound your number. How could this be that the sum of two digits in your number is exactly 9? Well, here is secret: Number 9 has a magical property. If you multiply any number by it, in the resulting number the sum of digits is divisible by 9. Your number is 9N+9, for sure is divisible by 9. This means that in your number, the sum of its digits is divisible by 9. This sum could be 9, 18, 27, 36 etc. But no matter what N you choose you always get a sum of 9 in this trick. Why? Here is the second secret: Your number N was between 1 and 9, so your result is between 9x1+9=18 and 9x9+9=90 For a number between 18 and 90, the largest sum of digits is 8+9=17. 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Great topics share through this post and really this is very nice post so many thanks to all online bookkeeping 18. شركة زهرة العلا للخرمات المنزلية بالسعودية واحدة من افضل الشركات التي تعمل بالمجال يرجع كل هذا الي الثقة والسرعة والدقة وجودة الخدمة التي تقدم للعميل ولافضل الاسعار الممكنة مع الحفاظ علي الجودة ونوعية المبيدات ومواد النظافة واثرها علي البيئة وصحة الاسرة كما ان الموارد البشرية الخاصة بنا علي اعلي دراية وفهم بالامور وذو خبرات طويلة في مكافحة الحشرات,النظافة العامة مثل الفلل,الشقق,الواجهات الزجاجية والحجرية,البلاط والرخام,الحدائق,المسابح وغيرها من امور النظافة كما انهم يجيدون التعامل مع كافة الاثاث الخشبي والزجاجي وغيرها والقيام باعمال النقل والفك والتركيب والتغليف بواسطة النجارين المحترفين كما يوجد لدينا افخك مستودعات تخزين الاثاث بالرياض والمتميز بالصفات القياسية للحفاظ علي العفش من الاتربة والرطوبة وغيرها من الامور التي تؤدي الي التلف فقط للحصول علي الخدمات المنفردة اتصل بنا علي الارقام الموضحة بالموقع ليصلك مندوبنا للمعاينة المجانية واعطاء النصائح والارشادات شركة رش مبيدات بالدمام شركة مكافحة حشرات بالخبر شركة مكافحة حشرات بالطائف شركة مكافحة حشرات ببريدة شركة مكافحة حشرات بنجران القضاء علي النمل الابيض شركة مكافحة حشرات بجازان شركة مكافحة حشرات بالخرج شركة مكافحة حشرات بالمدينة محلات الباركية بالرياض تركيب غرف نوم ايكيا مستودعات تخزين الاثاث بالرياض شركة تنظيف مساجد بالرياض 19. ان كنتم تملكون شقة او مجموعة من المنازل وتبحثون لها عن من يقوم بتنظيفها شركة تنظيف منازل بالمدينة المنورة ستتكفل بكل جوانب التنظيف على اساس قويم من الخدمة العصرية حيث ان التنظيف لدينا تنظيف منازل بالمدينة المنورة يجعلنا نتحكم في كل اجزاء الغرفة ونأتي على كل جوانبها بدءا من الارضية الى الفرش الى الجدران الى الاسقف الى كل ما من شأنه ان يجعلها نظيفة بالكامل وهذا بفضل الامكانات التي نمتلكها تنظيف منازل بالمدينة المنورة نمتلك ارقى اليات التنظيف التي تجعلنا نتحكم تحكما كاملا في تنظيف المنازل ونتصدر هذا الجانب على مستوى المدينة المنورة شركة تنظيف منازل بالمدينة المنورة فلا تتوانوا ان تطلبوا خدمة غسيل منازل بالمدينة المنورة لأنكم بالفعل سترون وتدركون بصمة التنظيف التي سنتركها على شقتكم مثل سابقيكم من العملاء الذين كانوا قد طلبوا خدمة تنظيف المنازل لدى شركتنا شركة تنظيف المنازل بالمدينة المنورة . عن طريق استخدام أفضل عمال موجودين فى شركة تنظيف منازل والذين ستجدهم متخصصين فى عملهم والذين ستجدهم لدى شركة غسيل منازل بالمدينة المنورة يقومون بعملهم فى تنظيف منازل ويقومون بتنظيف الحمامات والمراتب والرخام والبيارات والخزانات والارضيات والزجاج تنظيف منازل بالمدينه المنوره كل أنواع النظافة التى ذكرتها جعلت منا افضل شركة تنظيف منازل بالمدينة المنورة من وسط كل شركات تنظيف المنازل بالمدينة المنورة لأننا لا نترك أى شىء دون أن ينعم بالنظافة والجمال حتى تحصل أنت على ما تريده من منظر جمالى داخل شقتك ولنشعرك بكل سبل السعادة والراحة التى نريدك أن تشعر بها دائما طوال مدة تعاملك معنا.	4,309	14,913	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.375	4	CC-MAIN-2022-33	latest	en	0.940461
http://mathhelpforum.com/algebra/199448-just-simple-order-operations-question-2-print.html	1,524,439,200,000,000,000	text/html	crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00112.warc.gz	207,318,711	5,996	# Just a simple order of operations question Show 40 post(s) from this thread on one page Page 2 of 2 First 12 • May 30th 2012, 07:49 AM Reckoner Re: Just a simple order of operations question Quote: Originally Posted by SplashDamage In this example (which is actually from a previous test that i failed hard) ... I am not sure what to do. Is that your work there? Because that looks good to me (so long as we agree that x is positive and that y and z are nonzero). You are correct when you say that raising a product to an nth power is the same as multiplying that product by itself n times. The result of this is that the power gets "distributed" to each factor. That is, in general, $\displaystyle (ab)^n=\overbrace{(ab)(ab)(ab)\cdots(ab)}^{n\ \mathrm{factors}}=a^nb^n$. Another useful property is $\displaystyle (a^m)^n = a^{m\cdot n} = a^{mn}$. So, $\displaystyle \frac{27z^9\left(y^2x^{-1/2}\right)^2}{\left(3z^3x^{-1/3}\right)^3y^4}$ $\displaystyle =\frac{27z^9\left(y^2\right)^2\left(x^{-1/2}\right)^2} {\left(3\right)^3\left(z^3\right)^3\left(x^{-1/3}\right)^3y^4}$ $\displaystyle =\frac{27z^9y^4x^{-1}}{27z^9x^{-1}y^4} = 1$, again with $\displaystyle x>0$ and $\displaystyle y,z\neq0$. Edit: But be sure you see that we cannot distribute powers over addition or subtraction. That is, $\displaystyle (a+b)^n\neq a^n+b^n$ for arbitrary $\displaystyle a$ and $\displaystyle b$ (though there are specific values for which this will be true). In this case, we would have to multiply $\displaystyle (a+b)$ by itself $\displaystyle n$ times. • May 30th 2012, 07:54 AM Reckoner Re: Just a simple order of operations question Quote: Originally Posted by SplashDamage I mean, if you had something like (x + 3) (y + 4) (x + y)... How do you multiply them all together? Like you would with just two. Multiply any two factors together. This will give you a new expression with one fewer factor than the previous one. Then repeat the process until everything is multiplied out and you are left with only one factor. $\displaystyle (x+3)(y+4)(x+y)$ $\displaystyle =\left[(x+3)(y+4)\right](x+y)$ $\displaystyle =(xy+4x+3y+12)(x+y)$ Now we distribute, $\displaystyle =xy(x+y)+4x(x+y)+3y(x+y)+12(x+y)$ $\displaystyle =x^2y + xy^2 + 4x^2 + 4xy + 3xy + 3y^2 + 12x + 12y$ and we could combine like terms. Using this procedure, you can multiply any finite number of polynomials together. • May 30th 2012, 08:00 AM SplashDamage Re: Just a simple order of operations question Wow that is amazingly helpful. You're epic. What would happen if I ended up doing the ( ) ( ) ( ) thing depending on ^n? And would you recommend the way you did it over that method? Also another thing. This concerns getting the lowest common denominator in order to -/+ fractions together. For example I am confronted with this. (1/2) - (2/x+4) The way I was taught to find the LCD was split the lower halves into their prime factors, which I thought was 2 and 2^2. Although I think I got that wrong. Can you explain to me how to properly find the LCD as my lecturer clearly fails at explaining these things. • May 30th 2012, 08:19 AM Reckoner Re: Just a simple order of operations question Quote: Originally Posted by SplashDamage Wow that is amazingly helpful. You're epic. What would happen if I ended up doing the ( ) ( ) ( ) thing depending on ^n? And would you recommend the way you did it over that method? Yes, if you have an expression with more than one term, you don't have much choice other than multiplying the expression by itself over and over again. $\displaystyle (a+b)^n = \overbrace{(a+b)(a+b)(a+b)\cdots(a+b)}^{n\ \mathrm factors}$, and then multiply everything out. Quote: Originally Posted by SplashDamage For example I am confronted with this. (1/2) - (2/x+4) The way I was taught to find the LCD was split the lower halves into their prime factors, which I thought was 2 and 2^2. Although I think I got that wrong. Can you explain to me how to properly find the LCD as my lecturer clearly fails at explaining these things. A simple procedure for finding the LCD is to completely factor each denominator (so that you have prime factors), and then take the highest power of each factor from either denominator. For example, to find the LCD of $\displaystyle \frac1{3a^2+6ab+3b^2},\quad\frac1{4a^2+4ab},\quad \mathrm{and}\quad\frac1{2a^3}$ Factor each denominator completely: $\displaystyle 3a^2+6ab+3b^2 = 3\left(a^2+2ab+b^2\right) = 3(a+b)^2$ $\displaystyle 4a^2+4ab=4a(a+b)=2^2a(a+b)$ $\displaystyle 2a^3 = 2a^3$ To find the LCD, multiply the highest power of each factor: $\displaystyle \mathrm{LCD} = 2^2\cdot3^1\cdot a^3\cdot(a+b)^2 = 12a^3(a+b)^2$. For your example, I assume you mean $\displaystyle \frac12 - \frac2{x+4}$ and not $\displaystyle \frac12 - \left(\frac2x + 4\right)$? In this case the denominators are already factored completely. And since there are no common factors, our LCD would simply be the product: $\displaystyle \mathrm{LCD}=2(x+4)$. • May 30th 2012, 08:28 AM SplashDamage Re: Just a simple order of operations question Quote: Originally Posted by Reckoner To find the LCD, multiply the highest power of each factor: $\displaystyle \mathrm{LCD} = 2^2\cdot3^1\cdot a^3\cdot(a+b)^2 = 12a^3(a+b)^2$. Not sure what you did/meant here and yeah I meant 2 / (x + 4) • May 30th 2012, 08:40 AM SplashDamage Re: Just a simple order of operations question Also I figured at if we had 2 and (x + 4)... wouldnt the factors be 2, x, 4 and then wouldnt it be 2 * x * 4, making 8x? • May 30th 2012, 08:40 AM Reckoner Re: Just a simple order of operations question Quote: Originally Posted by SplashDamage Not sure what you did/meant here The first denominator is: $\displaystyle 3\cdot(a+b)^2$ The second: $\displaystyle 2^2\cdot a\cdot(a+b)$ And the third: $\displaystyle 2\cdot a^3$ In all of these we have four different factors: 2, 3, $\displaystyle a$, and $\displaystyle (a+b)$. So we first look for the highest power of 2. The first denominator has zero factors of 2 ($\displaystyle 2^0$), the second one has two factors ($\displaystyle 2^2$), and the third denominator has one factor ($\displaystyle 2^1$). In our LCD, we use the highest power of 2 among these three, so we take $\displaystyle 2^2$. Similarly, $\displaystyle 3^1$ is the highest power of 3 (the other denominators have zero factors of three), $\displaystyle a^3$ is the highest power of $\displaystyle a$, and $\displaystyle (a+b)^2$ is the highest power of $\displaystyle (a+b)$. We take each of these and multiply them together. • May 30th 2012, 08:43 AM SplashDamage Re: Just a simple order of operations question OMG i get it. That makes the other question I just made clear. • May 30th 2012, 08:43 AM Reckoner Re: Just a simple order of operations question Quote: Originally Posted by SplashDamage Also I figured at if we had 2 and (x + 4)... wouldnt the factors be 2, x, 4 No, factors are the things that are being multiplied to form the final product. $\displaystyle x$ and 4 are being added, not multiplied. So $\displaystyle (x+4)$, in its entirety, is a single factor, and cannot be split further. • May 30th 2012, 09:11 AM SplashDamage Re: Just a simple order of operations question So something like 5x can be split into 5 and x, or something like 25x could be split into 5^2 and x. And two factors can be grouped as one factor if there is parenthesis around them for example (x + 4)? and if that parenthesis was not there, would the LCD indeed be 4x? actually no it would still be 2 (x + 4) wouldnt it. • May 30th 2012, 09:15 AM Reckoner Re: Just a simple order of operations question Quote: Originally Posted by SplashDamage So something like 5x can be split into 5 and x, or something like 25x could be split into 5^2 and x. Yes Quote: Originally Posted by SplashDamage And two factors can be grouped as one factor if there is parenthesis around them for example (x + 4)? I'm not sure I understand what you mean. We can group factors as we please, for multiplication is associative: $\displaystyle abc=(ab)c=a(bc)$. $\displaystyle x+4$ is a single factor, however (though it consists of two terms). • May 30th 2012, 09:17 AM SplashDamage Re: Just a simple order of operations question So with or without the brackets, it would still be 2(x + 4)? • May 30th 2012, 09:29 AM Reckoner Re: Just a simple order of operations question Quote: Originally Posted by SplashDamage So with or without the brackets, it would still be 2(x + 4)? Right. When $\displaystyle x+4$ is by itself, we don't need to write the parentheses. But it's still a single factor, and we can't factor it further. • May 30th 2012, 09:54 AM SplashDamage Re: Just a simple order of operations question What would happen if we had (x + 8) instead? since 8 can be factored. also if we had x + 8 • May 30th 2012, 09:59 AM Reckoner Re: Just a simple order of operations question Quote: Originally Posted by SplashDamage What would happen if we had (x + 8) instead? since 8 can be factored. 8 can be factored as $\displaystyle 2^3$, yes. But even though 2 is a factor of 8, it is not a factor of $\displaystyle (x+8)$ (because it is not a factor of $\displaystyle x$). Since $\displaystyle x$ and 8 have no common factors, the overall expression $\displaystyle (x+8)$ cannot be factored any further. If, on the other hand, we had something like $\displaystyle (2x+8)$, we could factor out a 2 from each term: $\displaystyle 2x+8 = 2\cdot x+2\cdot4 = 2(x+4)$. Show 40 post(s) from this thread on one page Page 2 of 2 First 12	2,845	9,543	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.0625	4	CC-MAIN-2018-17	latest	en	0.868703
https://de.scribd.com/document/137501645/Matematik-Pengurusan-Bab-6	1,576,034,728,000,000,000	text/html	crawl-data/CC-MAIN-2019-51/segments/1575540529745.80/warc/CC-MAIN-20191211021635-20191211045635-00540.warc.gz	334,516,515	80,290	Sie sind auf Seite 1von 5 # Matematik Pengurusan Matematik Pengurusan BAB 6 : APLIKASI PEMBEZAAN Latihan 6.1 1) Find the second degree differentiation a) y = 4 x 3 12 x 2 + 6 x + 2 y(x) = 3( 4 x 31 ) 2(12 x 21 ) + 6 x11 + 2 2) ## Find the third degree differentiation a) y = 4x 2 y(x) = 2( 4 x 21 ) = 8 x1 1 y(x) = 8 x1 =8 y(x) = 0 b) y = 4 x 12 x + 6 x + 24 y(x) = 3 2 Latihan 6.2 1) If the production cost for one unit product are RM10, while its fixed cost are RM5,000 a) Find the cost function = Fixed Cost + x (Per Unit Cost) = 5,000 + 10x b) What is the total cost for producing 200 units of the above product? = 5,000 + 10(200) = 5,000 + 2,000 = RM7,000 Derive the average cost function, C ( x ) C ( x) 5,000 +10 x = = = x x 5,000 +10 x ## = 12 x 2 24 x1 + 6 y(x) = 2(12 x 21 ) 24 x11 +6 = 24 x 24 b) y= = 2 x 2 y(x) 1 2)( 2 x 2 ) = ( 3( 4 x 31 ) 2(12 x 21 ) + 6 x11 + 0 2 x2 ## = 12 x 2 24 x1 + 6 y(x) = 2(12 x 21 ) 24 x11 +6 = 24 x 24 y(x) = 24 x11 0 = 24 ) c) 4 x 3 ) = ( 3)( 4x y(x) = ( 3 1 c) y(x) = 12 x 12 = 4 x y = 2 x + x +1 = 1(2 x 11 ) + 2( x 21 ) + 0 2 d) = 2x + 2x ## Determine the ultimate total cost function (pembezaan 1 darjah), C ' ( x) 1 2)( 2 x 21 ) +2 x1 y(x) = ( = 4 x 3 + 2 3)( 4 x 31 ) +0 y(x) = ( = 12 x 4 12 = 4 x [email protected] 014-3580486 C ( x )( x ) ## 5000 +10 ( x ) x = 5000 +10 x = 0 +10 x11 = 10 Matematik Pengurusan Matematik Pengurusan ## c) 3) 2) a) = Suppose the average total cost function 100000 +1500 + 0.2q is C (q ) = q 2 ## Given the total cost function is C(q) = q + 3q + 400 4 What is the quantity which has to be produced so that the average total cost in minimised? C '( q ) C ( q )(q ) a) ## Obtain the function, C ( q ) C (q) (q) q2 + 3q + 400 = 4 q average cost = 4 + q2 1 400 400 ## 100000 +1500 + 0.2q )(q ) =( q = 100000 +1500q + 0.2q 2 0 = 4 + q2 400 1 = q2 4 q 2 = 1600 b) Derive the marginal total cost function, C ' ( q ) = 0 +1500q 11 + 2(0.2q 21 ) = 1500 +0.4q Determine the rate of change for cost of producing 10 units of the product, C ' ( q ) = 1500 + 0.4(10) = 1500 + 4 = 1504 q = 1600 = 40 unit c) q 2 1 3q 400 = + 4 xq + q q q 400 = 4 +3 + q b) ## Derive the ultimate total cost function, C ' ( x ) q + 3 + 400q 1 4 q 11 + 0 + 400q 11 = 4 1 2 = + 400q 4 1 400 = 4 + q2 [email protected] 014-3580486 Matematik Pengurusan Matematik Pengurusan b) b) 4) Elyana trading supplies attire to the supermarkets in North Peninsula. The company annual cost is given by a function C= 15 + 0.15q + 200 , where q is quantity q (in dozen) and C (in RM thousand) is the total cost in a year a) What is the quantity which minimses the total cost? C ( x) = 15q 1 + 0.15q + 200 C ' ( x) = ( 1)(15q 11 ) +0.15q 11 +0 = ( 15q 2 ) + 0.15 0 = ( 15q 2 ) + 0.15 2 15q -0.15 = What is the minimum total cost = ( 15q 2 ) + 0.15 1 2)( 15q 2 ) = ( 3 = 30q 300 300 = q3 = >0 10 3 Oleh itu, total cost have a minimun value wen q = 10 dozen C ' ( x) C ' ' ( x) What the quantity which has to be produced so that the average cost is minimsed? C (q) ## 2500 + 75 + 0.25q q = 2500q 1 + 75 + 0.25q C ' (q) = 1( 2500q 11 ) + 0 + 0.25q 11 ## = 2500q 2 +0.25 0 = 2500q 2 +0.25 -0.25 = 2500 q2 2500 q2 = 0.25 ## C (10) = q + 0.15q + 200 = 15 + 0.15(10) + 200 10 15 = 1.5 + 1.5 + 200 = 203 = RM203,000 5) The total production cost of a cosmetics product is C = 2500 + 75q + 0.25q 2 a) Find the average total cost quantity, C ( q ) = = C (q) (q) 2500 + 75q + 0.25q 2 q ## = 10,000 q = 10,000 = 100 unit -0.15 = q 2 q2 = 15 c) = 100 q = 100 15 0.15 = 10 dozen What is total cost at the production level which minimises the average total cost? C = 2500 + 75q + 0.25q 2 = 2500 + 75(100) + 0.25(100) 2 = 2500 + 7500 + 2,500 = RM12,500 ## 2500 75q 0.25q 2 + + q q q 2500 = q + 75 + 0.25q Latihan 6.3 [email protected] 014-3580486 Matematik Pengurusan Matematik Pengurusan ## 2) 1) The demand function of a health product is given a function p = 0.001q 2 +840 a) ## Obtain the total function, R(q) = pq = (0.001q 2 +840)(q ) = 0.001q 3 +840q R(q) q 3 0.001q +840q = q 3 0.001q 840q + = q q revenue Given the total demand function, p(x) = 2 0.01x , where p is the unit price in thousand RM and x is the quantity of the item, a) Find the total revenue function, R(x) = px = (2 0.01x)( x ) = 2 x 0.01x 2 b) Determine the price mazimises the total revenue which Latihan 6.4 1) A company has an average total cost function of K = 4 +100q 1 . The demand equation is given by function p = 54 q , where p is the unit price (RM) and q is the quantity a) Derive the revenue function, R(x) = pq = (54 q )(q ) = 54q q 2 b) Determine the cost function, K ( q ) = K ( q )(q ) = (4 +100q 1 )(q ) = 4q +100 (q ) c) Obtain the profit function, = R(x) - K ( q ) = (54q q 2 ) - (4q +100) = 54q q 2 4q 100 = 50q q 2 100 b) ## Derive the average total revenue cost function, Step 1 : Find the ultimate total revenue function, R(x) = 2 x11 (2)(0.01x 21 ) = 2 0.02x Step 2 : Find value when R(x) = 0 0 = 2 0.02x 0.02x = 2 x = 2 0.02 = 100 Step 3 : Find the price = 2 0.01x = 2 (0.01 x 100) =21 =1 = RM1,000 = 0.001q 2 +840 c) Determine the marginal total revenue function, R(q) = 3(0.001q 31 ) +840q 11 = 0.003q 2 +840 [email protected] 014-3580486 Matematik Pengurusan Matematik Pengurusan d) Find the price for which the company will maximise its profit, by using differentiaiton method ( q ) = 50q q 2 100 d = 50q 11 2( q 21 ) 0 dq = 50 2q ## (300 x x 2 ) (0.1x 2 +14 x +100) = 300 x x 2 0.1x 2 14 x 100 = 286 x 1.1x 2 100 c) Determine the quantity which maximise the profit ( q ) = 286 x 1.1x 2 100 So, the quantity is d = 286 x11 2(1.1x 21 ) 0 dq When dq = 0 0 = 50 2q 2q = 50 q = 50 2 = 25 So, p = 54 q = 54 25 = RM29 2) The demand function pf a local product is p = 300 x and the cost function is C(x) = 0.1x 2 +14 x +100 a) Obtain the total revenue function, R(x) = px = (300 x)(x) = 300 x x 2 = 286 2.2 x 0 = 286 2.2 x 2.2x = 286 x = 286 2.2 = 130 unit d) Calculate the price at which the profit is macimised So, the price is p = 300 x = 300 130 = RM170 e) Find the value of the maximum profit ( q ) = 286 x 1.1x 2 100 = 286(130) 1.1(130)2 100 = RM18,490 ## (q ) b) Derive the total profit function, = R(x) - C(x) [email protected] 014-3580486	2,766	6,287	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.21875	4	CC-MAIN-2019-51	latest	en	0.588647
https://www.jiskha.com/members/profile/posts.cgi?name=STEVE&page=14	1,498,690,535,000,000,000	text/html	crawl-data/CC-MAIN-2017-26/segments/1498128323807.79/warc/CC-MAIN-20170628222452-20170629002452-00274.warc.gz	898,153,220	9,381	# Posts by STEVE Total # Posts: 51,767 math correct. a person can determine the miles walked. After that, the calories follow. Math .03 * 27700 math there are only 2 colors. 3 socks must have two of the same color... math correct Math - Calculus Actually, the shape does not matter for this question because it asked about the rate of flow through the channel. I suspect there were other parts where we'd have to work with the volume or depth in the bay, but not here. Math - Calculus just find the derivative dd/dt the shape of the bay does not matter for this question. math just divide and you get x+4 + 9/(x-2) So, as x gets huge, the 9/(x-2) vanishes, and the graph approaches the line y = x+4 Math(Calculus) well, x^2-4 has a minimum at x=0, so e^(x^2-4) will as well. since x^2-4 has an absolute max at x=±2, so does e^(x^2-4) But, as long as this is calculus, dh/dx = 2x e^(x^2-4) dh/dx=0 at x=0 (minimum) Algebra Since F=ma, a = 8x/m. So, we have a(x) = -8/m x v(x) = -4/m x^2 + c1 s(x) = -4/3 x^3 + c1x + c2 v(0) = 0 and s(0) = 10, so s(x) = -4/3 x^3 + 10 For the damping, just work with v(x) = -4/m x^2 + 8v and follow through to get s(x) Calculus Since the y-axis is also the axis of symmetry for the curves, we can just take one side and revolve it. This will be easiest using shells of thickness dx: v = ?[0,2] 2?rh dx where r=x and h=(8-x^2)-x^2=8-2x^2 v = ?[0,2] 2?x(8-2x^2) dx = 16? Math x = 3+4cos? y = 2+4sin? since cos^2? + sin^2? = 1, (x-3)^2 + (y-2)^2 = 4^2 no way to tell all we know is that 15 students got above 74% and 15 got below 74% math x grows by 2 the slope is 2, so y grows by 4. so, y=2+4=6 or, more formally, (y-2)/(3-1) = 2 Math each term is calculated using the previous term, not the next one! Calculus - Fundamental thm of Calc that would be ((1/4)(x^2)^2?1)^15 2x = 2x(x^4/4 - 1)^15 Calculus - Related Rates Let the distance on the road be x. Then the distance z to the house is z^2 = x^2+25 z dz/dt = x dx/dt z(1) = ?26 ?26 dz/dt = 1 * 65 ... Calculus Help google hole in sphere. You will find many excellent discussions. Math well, you have a base perimeter of 30. So, the base of each triangular lateral face is 30/4 = 7.5 Now you can see what the side edges of the pyramid are, and there are four of them. let 'er rip... math reimburses? Did the doctor pay the patient something? maths L = km/h^2 So, Lh^2/m = k is constant You want L such that L50^2 = 40h^2/175 If you will supply the original height, then you can solve for L. math I don't see any graph ... algebra #1 The line of symmetry of y=x^2 is the y-axis (x=0). Similarly, for a translated parabola y=a(x-h)^2 + k the axis of symmetry is the line x=h. So, for y = -2x^2+3 the axis is x=0 #2 same for \|x\|. So, for y = 1/3 \|x\| - 2 the axis is x=0 #3 Huh? This stumps you? P(4,-4)-->(-... math I suspect there was meant to be a question in there somewhere, but I couldn't see it ... Math not mixed numbers. As a fraction, 87.3/8.73 = 108.73/8.73 = 10/1 = 10 In case you wanted individual fractions, 87.3 = 873/10 8.73 = 873/100 Math I think Natavia is right. If P = 1/26 that is the theoretical probability. No mention is made of actually performing the experiment. Math recall that alternate interior angles between parallel lines are congruent. Draw a line through one vertex parallel to the opposite base. the rest is easy. Math 4 1/4 = 1 Math 6 * 2/3 = 4 Math 4/12 = 2/6 = 1/3 Intermediate Maths (a) 12(1/2)^(t/5700) = 10 t = 1499.3 years (b) 12(1/2)^(1599.3/5700) = 9.879 grams The region is a semi-circle of radius 6, so the area ranges from 0 to 18? The integral is the area moving from the left side to some point x in [-6,6]. Pre calculus set this up as a geometric series and then just apply the sum formula. a = 2 r = 0.9 allow for round trips up and down Probability to start out, there are 13 diamonds in 52 cards. As you draw the cards, the probabilities change, due to the makeup of the remaining cards. The result is 13/52 * 12/51 * 11/50 * 10/49 = (13121110)/(52515049) = 13P4/52P4 Calculus A,C,D are true. The Intermediate Value theorem says that there is a zero in (-2,1) and (1,4) Rolle's Theorem (or MVT) says that f'(x)=0 somewhere in (-2,4) Mean Value Theorem says D is true Math 1 I must say, these problems of yours are some cumbersome. I'm not sure they are teaching much about factoring, since most of them only have a single monomial common factor. Like this one: 48x^3 - 128x^2 -56x -16 8(6x^3-16x^2-7x-2) Math 1 392n^4 + 168n^3 + 168n^2 +72n 8n(49n^3+21n^2 + 21n+9) grouping gives 8n(7n^2(7n+3)+3(7n+3)) 8n(7n+3)(7n^2+3) Math 1 5x(16x^3-12x^2-12x-9) 5x(2x-3)(8x^2+6x+3) the discriminant of the quadratic is negative, so it has no real factors. Math1 The x^2 is easy: 195x^5 + 168x^4 - 28x^3 - 24x^2 x^2(195x^3+168x^2-28x-24) Now things get tough. If there are any rational roots of the form p/q, then we know that p divides 24 q divides 195 Some time and effort should convince you that there are no other rational roots, so no... Maths ( Higher Level) Simultaneous Equations. the method is just to write the words as math: mother is x years old, her son is y years old and the sum of theirs ages together is 58 x+y = 58 Five years ago, the mother was five times as old as the son (x-5) = 5(y-5) Now it's easy! x+y=58 x-5y = -20 subtract and you get ... algebra 2, check my answer plzzzzzz x = 5y^3 x/5 = y^3 y = (x/5)^(1/3) Or, if you insist on rational denominators, y = (25x)^(1/3)/5 your notation would seem to indicate 3?(25x)/5 rather than cubrt(25x)/5 = ∛(25x)/5 Math 5/8 Math well 45/3 = 15 = one-third of the bag. So, ... Math well, each foot of rope has 3 pieces, right ? 1 foot = 3/3 So, how many in 8 feet? Math 42 Math If b is breaths, then 10^9 b * 1min/16b * 1day/1440min * 1yr/365day = 118.9 years This makes sense, since 1 billion seconds is about 30 years. Functions (Math) Draw the angle. It should be clear that y = -5 x = 2 r = ?29 Now just recall that sin? = y/r cos? = x/r tan? = y/x now let 'er rip MATHS Note that ?STP is similar to ?RTS and angle SRT = angle TPQ (alternate interior angles) Calculus dy/dt = k/y y dy = k dt 1/2 y^2 = kt + c y^2 = 2kt + c y = ?(2kt+c) E does not work, as you can easily see: y = ?(2kt)+4 dy/dt = 2k/(2?(2kt)) = k/?(2kt) but that is not k/y = k/(?(2kt)+4) calculus just plug and chug. The surface can be thought of as a stack of thin rings. A = ?2?r ds where r=y and ds=?(1+y'^2) dx so, y' = (9x^4-4)/(12x^2) A = 2??[1/2,1] (x^3/4+1/(3x))((9x^4+4)/(12x^2)) dx = 1981?/3072 Calculus (trig derivatives) Draw a diagram. It is clear that when the distance of the beam from the point P on shore nearest the lighthouse is x, tan? = x/2 sec^2? d?/dt = 1/2 dx/dt when x=4, we have (1+(4/2)^2)(2?/20) = 1/2 dx/dt dx/dt = ? mi/s MATHS Since the bisector meets BC at a 45° angle, CB=PB = 15 so AP=25-15=10 Mathematics on each trip, walk = 1/3 bus = 2 3/4 - 1/3 = 2 5/12 math the hexagon has an apothem a=9. so, its sides are s=6?3 Now, you know how to find the area of a hexagon of side s. The area A of the pyramid is just A=6(sa/2) The volume is 1/3 (area of hexagon)42 Math Let Ø = angle to bottom of picture ? = angle subtended by the picture x = distance from wall to observer tanØ = 4/x tan(?+Ø) = 9/x so (tan?+tanØ)/(1-tan?tanØ) = 9/x (tan?+(4/x))/(1-(4/x)tan?) = 9/x see what you can do with that. or, if y=... Calculus g'(x) = ?((x^2)^3+2) * 2x now just plug in x=2 Alg II COT cot ?/6 = ?3 Alg II COT Math true programming there are lots of sort routines you can find online. So, this will accept an arbitrary number of names: n=0 name="x" while name ? "" read name if name?"" then names[n++] = name end while snames = sort names,n while n>0 print names[n--] trig (4cosx+1)(2cosx+1) = 0 now it's a cinch... maths There are 11P4 ways to pick the pained boats. Maths well, all the 400's and all the 900's and all the 40's & 90's of the other 8 hundreds and all the other 8 numbers ending in 9 for the other 8 hundreds maths a set of n elements has 2^n-1 non-empty subsets Algebra correct Math The way you have written it is incorrect. It should be 4tanx(1-tan^2(x)) ----------------------- (1+tan^2(x))^2 = 4tanx/(1+tanx^2) * (1-tanx^2)/(1+tanx^2) = [4sinx/cosx * secx^2][(1-tanx^2)/(1+tanx^2)] = (4sinx cosx)(1-tanx^2)/secx^2 = 2sin2x (cosx^2-sinx^2) = 2sin2x cos2x = ... Math take sin of both sides and use your sum and double-angle formulas to get a polynomial in x sin(2arcsin(x/?6)+arcsin(4x)) = 1 x = ?39 - 6 see http://www.wolframalpha.com/input/?i=sin(2arcsin(x%2F%E2%88%9A6)%2Barcsin(4x))+%3D+1 Maths well, 22 tens is 220 ... algerbra looks ok to me, aside from using "square of" to mean "square root of" Math use a common denominator. Then you have 12/28 + 7/28 not so hard now, eh? algerbra right on. geometry their areas are in the ratio 2:3:4 Math wrong in so many ways. First, 1.00 is 100% .609 = 60.9% so, 2.608 = 260.9% But you read the question wrong. You know that 2 is 50% of 4, right? So, you divide 2/4 = 0.5 = 50% But you divided 321 by 123. Surely you could see that 2.6% is a totally unreasonable answer. Actually... maths 380 They have no common factors LCM(x,y) = xy/GCD(x,y) maths I guess he is then lost. Or, try the law of cosines. Maths A. usingsymmetry, the area is just ?[0,?/4] 2cosx - secx dx = ?2 - 2tanh-1 ?/8 Using discs of thickness dx for the volume, v = 2?[0,?/4] ?(R^2-r^2) dx where R=2cosx and r=secx v = 2?[0,?/4] ?((2cosx)^2-(secx)^2) dx = ?^2 Using shells of thickness dy, we have to split the ... Maths sides have shrunk by ?4 = 2 Math well, just add up the numbers! math (8/6) * 48 = 64 math well, the sum of all 9 angles is 7180 = 1260 (1260-462)/6 = ? math x+x+9 < 24 2x < 15 x < 7.5 so, x <= 7 The sides of length x must each be > 9/2, so x >= 5 5 <= x <= 7 Maths well, 05 is half of 10, so ... Math If R is the radius at the equator, then let r be the radius at 45°. That is Rcos45° = R/?2 So, the ratio is R/r = ?2 Algebra 6(j + j+10) = 780 maths (A) subtract each number from 1000. Which difference is smallest in size, + or -? (B) 46+2x+96+x+62 = 306 (C) add up all the rentals and divide by the number of days geometry it can only be angle PKE=16° That means angle PKN=32° and KPM=148° MK and PN bisect the angles of the rhombus. Now you can get all the angles you need. Note that diagonals are perpendicular. Math do you not have a formula for the payments? Just plug your numbers into the formula. Math There are 10C3 ways to pick the females, and 10C2 ways to pick the males. So, that makes 10C3 10C2 ways to form the committee. Geometry If the area is 10, then 1/2(5+15)h = 10 1/2(20)h = 10 10h = 10 h = 1 If the radius is 7.5 (diameter=15) then 3.147.5^2/2 = 3.1456.25/2 = 176.625/2 = 88.3125 math 6/(6+8+6) * 30 = 9 math 4 esses in 13 letters, so p = 4/13 * 3/12 Math 3/4 * 8 * 9? = 54? cm^3 Math Not quite. There are 4 possible outcomes Coin1 Coin2 H H H T T H T T Two of those result in a tail and a head. So, the probability is 2/4 = 1/2 algebra p = s + 12 + 2s+3 = 3s+15 expression, not equation 7(a-b)-8(a-2b) 7a-7b-8a+16b now you can finish it off Math (3/r)^2 + (4/r)^2 = 1 25/r^2 = 1 r = ±1/5 tanx = 4/3 x is in QI or QIII matha they are planed in a pentagram figure https://richardwiseman.wordpress.com/2013/07/29/answer-to-the-friday-puzzle-216/ Trigonometry Draw a diagram. If the distance is x, then (h-85)/x = tan11°6' h/x = tan26°7' So, (h-85)cot 11°6' = h*cot 26°7' Use that to find h, and then you can get x. t If the sides are a and b, and the included angle is C, then c^2 = a^2+b^2-2ab cosC sinA/a = sinC/c A+B+C = 180 Trigonometry These are all just about the basic trig functions. Draw a diagram and just decide which function to use. Post a New Question	4,386	11,743	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.953125	4	CC-MAIN-2017-26	longest	en	0.918908
http://en.wikipedia.org/wiki/Discrete_Fourier_Transform	1,432,627,321,000,000,000	text/html	crawl-data/CC-MAIN-2015-22/segments/1432207928817.29/warc/CC-MAIN-20150521113208-00213-ip-10-180-206-219.ec2.internal.warc.gz	79,976,744	45,665	# Discrete Fourier transform (Redirected from Discrete Fourier Transform) Fourier transforms Continuous Fourier transform Fourier series Discrete-time Fourier transform Discrete Fourier transform Fourier analysis Related transforms Relationship between the (continuous) Fourier transform and the discrete Fourier transform. Left column: A continuous function (top) and its Fourier transform (bottom). Center-left column: Periodic summation of the original function (top). Fourier transform (bottom) is zero except at discrete points. The inverse transform is a sum of sinusoids called Fourier series. Center-right column: Original function is discretized (multiplied by a Dirac comb) (top). Its Fourier transform (bottom) is a periodic summation (DTFT) of the original transform. Right column: The DFT (bottom) computes discrete samples of the continuous DTFT. The inverse DFT (top) is a periodic summation of the original samples. The FFT algorithm computes one cycle of the DFT and its inverse is one cycle of the DFT inverse. Depiction of a Fourier transform (upper left) and its periodic summation (DTFT) in the lower left corner. The spectral sequences at (a) upper right and (b) lower right are respectively computed from (a) one cycle of the periodic summation of s(t) and (b) one cycle of the periodic summation of the s(nT) sequence. The respective formulas are (a) the Fourier series integral and (b) the DFT summation. Its similarities to the original transform, S(f), and its relative computational ease are often the motivation for computing a DFT sequence. In mathematics, the discrete Fourier transform (DFT) converts a finite list of equally spaced samples of a function into the list of coefficients of a finite combination of complex sinusoids, ordered by their frequencies, that has those same sample values. It can be said to convert the sampled function from its original domain (often time or position along a line) to the frequency domain. The input samples are complex numbers (in practice, usually real numbers), and the output coefficients are complex as well. The frequencies of the output sinusoids are integer multiples of a fundamental frequency, whose corresponding period is the length of the sampling interval. The combination of sinusoids obtained through the DFT is therefore periodic with that same period. The DFT differs from the discrete-time Fourier transform (DTFT) in that its input and output sequences are both finite; it is therefore said to be the Fourier analysis of finite-domain (or periodic) discrete-time functions. The DFT is the most important discrete transform, used to perform Fourier analysis in many practical applications.[1] In digital signal processing, the function is any quantity or signal that varies over time, such as the pressure of a sound wave, a radio signal, or daily temperature readings, sampled over a finite time interval (often defined by a window function[2]). In image processing, the samples can be the values of pixels along a row or column of a raster image. The DFT is also used to efficiently solve partial differential equations, and to perform other operations such as convolutions or multiplying large integers. Since it deals with a finite amount of data, it can be implemented in computers by numerical algorithms or even dedicated hardware. These implementations usually employ efficient fast Fourier transform (FFT) algorithms;[3] so much so that the terms "FFT" and "DFT" are often used interchangeably. Prior to its current usage, the "FFT" initialism may have also been used for the ambiguous term "finite Fourier transform". ## Definition The sequence of N complex numbers $x_0, x_1, \ldots, x_{N-1}$ is transformed into an N-periodic sequence of complex numbers: $X_k\ \stackrel{\text{def}}{=}\ \sum_{n=0}^{N-1} x_n \cdot e^{-2 \pi i k n / N}, \quad k\in\mathbb{Z}\,$ (integers) [note 1] (Eq.1) Each $X_k$ is a complex number that encodes both amplitude and phase of a sinusoidal component of function $x_n$. The sinusoid's frequency is k cycles per N samples. Its amplitude and phase are: $\|X_k\|/N = \sqrt{\operatorname{Re}(X_k)^2 + \operatorname{Im}(X_k)^2}/N$ $\arg(X_k) = \operatorname{atan2}\big( \operatorname{Im}(X_k), \operatorname{Re}(X_k) \big)=-i\operatorname{ln}\left(\frac{X_k}{\|X_k\|}\right),$ where atan2 is the two-argument form of the arctan function. Assuming periodicity (see Periodicity), the customary domain of k actually computed is [0N-1]. That is always the case when the DFT is implemented via the Fast Fourier transform algorithm. But other common domains are [-N/2, N/2-1] (N even) and [-(N-1)/2, (N-1)/2] (N odd), as when the left and right halves of an FFT output sequence are swapped.[4] The transform is sometimes denoted by the symbol $\mathcal{F}$, as in $\mathbf{X} = \mathcal{F} \left \{ \mathbf{x} \right \}$ or $\mathcal{F} \left ( \mathbf{x} \right )$ or $\mathcal{F} \mathbf{x}$.[note 2] Eq.1 can be interpreted or derived in various ways, for example: $x_n = \frac{1}{N} \sum_{k=0}^{N-1} X_k \cdot e^{i 2 \pi k n / N}, \quad n\in\mathbb{Z}\,$ (Eq.2) which is also N-periodic. In the domain $\scriptstyle n\ \in\ [0,\ N-1],$ this is the inverse transform of Eq.1. The normalization factor multiplying the DFT and IDFT (here 1 and 1/N) and the signs of the exponents are merely conventions, and differ in some treatments. The only requirements of these conventions are that the DFT and IDFT have opposite-sign exponents and that the product of their normalization factors be 1/N. A normalization of $\scriptstyle \sqrt{1/N}$ for both the DFT and IDFT, for instance, makes the transforms unitary. In the following discussion the terms "sequence" and "vector" will be considered interchangeable. ## Properties ### Completeness The discrete Fourier transform is an invertible, linear transformation $\mathcal{F}\colon\mathbb{C}^N \to \mathbb{C}^N$ with $\mathbb{C}$ denoting the set of complex numbers. In other words, for any N > 0, an N-dimensional complex vector has a DFT and an IDFT which are in turn N-dimensional complex vectors. ### Orthogonality The vectors $u_k=\left[ e^{ \frac{2\pi i}{N} kn} \;\|\; n=0,1,\ldots,N-1 \right]^T$ form an orthogonal basis over the set of N-dimensional complex vectors: $u^T_k u_{k'}^* = \sum_{n=0}^{N-1} \left(e^{ \frac{2\pi i}{N} kn}\right) \left(e^{\frac{2\pi i}{N} (-k')n}\right) = \sum_{n=0}^{N-1} e^{ \frac{2\pi i}{N} (k-k') n} = N~\delta_{kk'}$ where $~\delta_{kk'}$ is the Kronecker delta. (In the last step, the summation is trivial if $k=k'$, where it is 1+1+⋅⋅⋅=N, and otherwise is a geometric series that can be explicitly summed to obtain zero.) This orthogonality condition can be used to derive the formula for the IDFT from the definition of the DFT, and is equivalent to the unitarity property below. ### The Plancherel theorem and Parseval's theorem If Xk and Yk are the DFTs of xn and yn respectively then the Parseval's theorem states: $\sum_{n=0}^{N-1} x_n y^_n = \frac{1}{N} \sum_{k=0}^{N-1} X_k Y^_k$ where the star denotes complex conjugation. Plancherel theorem is a special case of the Parseval's theorem and states: $\sum_{n=0}^{N-1} \|x_n\|^2 = \frac{1}{N} \sum_{k=0}^{N-1} \|X_k\|^2.$ These theorems are also equivalent to the unitary condition below. ### Periodicity The periodicity can be shown directly from the definition: $X_{k+N} \ \stackrel{\mathrm{def}}{=} \ \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} (k+N) n} = \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} k n} \underbrace{e^{-2 \pi i n}}_{1} = \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} k n} = X_k.$ Similarly, it can be shown that the IDFT formula leads to a periodic extension. ### Shift theorem Multiplying $x_n$ by a linear phase $e^{\frac{2\pi i}{N}n m}$ for some integer m corresponds to a circular shift of the output $X_k$: $X_k$ is replaced by $X_{k-m}$, where the subscript is interpreted modulo N (i.e., periodically). Similarly, a circular shift of the input $x_n$ corresponds to multiplying the output $X_k$ by a linear phase. Mathematically, if $\{x_n\}$ represents the vector x then if $\mathcal{F}(\{x_n\})_k=X_k$ then $\mathcal{F}(\{ x_n\cdot e^{\frac{2\pi i}{N}n m} \})_k=X_{k-m}$ and $\mathcal{F}(\{x_{n-m}\})_k=X_k\cdot e^{-\frac{2\pi i}{N}k m}$ ### Circular convolution theorem and cross-correlation theorem The convolution theorem for the discrete-time Fourier transform indicates that a convolution of two infinite sequences can be obtained as the inverse transform of the product of the individual transforms. An important simplification occurs when the sequences are of finite length, N. In terms of the DFT and inverse DFT, it can be written as follows: $\mathcal{F}^{-1} \left \{ \mathbf{X\cdot Y} \right \}_n \ = \sum_{l=0}^{N-1}x_l \cdot (y_N)_{n-l} \ \ \stackrel{\mathrm{def}}{=} \ \ (\mathbf{x * y_N})_n\ ,$ which is the convolution of the $\mathbf{x}$ sequence with a $\mathbf{y}$ sequence extended by periodic summation: $(\mathbf{y_N})_n \ \stackrel{\mathrm{def}}{=} \ \sum_{p=-\infty}^{\infty} y_{(n-pN)} = y_{n (mod N)}. \,$ Similarly, the cross-correlation of $\mathbf{x}$ and $\mathbf{y_N}$ is given by: $\mathcal{F}^{-1} \left \{ \mathbf{X^* \cdot Y} \right \}_n = \sum_{l=0}^{N-1}x_l^* \cdot (y_N)_{n+l} \ \ \stackrel{\mathrm{def}}{=} \ \ (\mathbf{x \star y_N})_n\ .$ When either sequence contains a string of zeros, of length L, L+1 of the circular convolution outputs are equivalent to values of $\mathbf{x * y}.$ Methods have also been developed to use this property as part of an efficient process that constructs $\mathbf{x * y}$ with an $\mathbf{x}$ or $\mathbf{y}$ sequence potentially much longer than the practical transform size (N). Two such methods are called overlap-save and overlap-add.[5] The efficiency results from the fact that a direct evaluation of either summation (above) requires $\scriptstyle O(N^2)$ operations for an output sequence of length N. An indirect method, using transforms, can take advantage of the $\scriptstyle O(N\log N)$ efficiency of the fast Fourier transform (FFT) to achieve much better performance. Furthermore, convolutions can be used to efficiently compute DFTs via Rader's FFT algorithm and Bluestein's FFT algorithm. ### Convolution theorem duality It can also be shown that: $\mathcal{F} \left \{ \mathbf{x\cdot y} \right \}_k \ \stackrel{\mathrm{def}}{=} \sum_{n=0}^{N-1} x_n \cdot y_n \cdot e^{-\frac{2\pi i}{N} k n}$ $=\frac{1}{N} (\mathbf{X * Y_N})_k, \,$ which is the circular convolution of $\mathbf{X}$ and $\mathbf{Y}$. ### Trigonometric interpolation polynomial $p(t) = \frac{1}{N} \left[ X_0 + X_1 e^{2\pi it} + \cdots + X_{N/2-1} e^{(N/2-1)2\pi it} + X_{N/2} \cos(N\pi t) + X_{N/2+1} e^{(-N/2+1)2\pi it} + \cdots + X_{N-1} e^{-2\pi it} \right]$ for N even , $p(t) = \frac{1}{N} \left[ X_0 + X_1 e^{2\pi it} + \cdots + X_{\lfloor N/2 \rfloor} e^{\lfloor N/2 \rfloor 2\pi it} + X_{\lfloor N/2 \rfloor+1} e^{-\lfloor N/2 \rfloor 2\pi it} + \cdots + X_{N-1} e^{-2\pi it} \right]$ for N odd, where the coefficients Xk are given by the DFT of xn above, satisfies the interpolation property $p(n/N) = x_n$ for $n=0,\ldots,N-1$. For even N, notice that the Nyquist component $\frac{X_{N/2}}{N} \cos(N\pi t)$ is handled specially. This interpolation is not unique: aliasing implies that one could add N to any of the complex-sinusoid frequencies (e.g. changing $e^{-it}$ to $e^{i(N-1)t}$ ) without changing the interpolation property, but giving different values in between the $x_n$ points. The choice above, however, is typical because it has two useful properties. First, it consists of sinusoids whose frequencies have the smallest possible magnitudes: the interpolation is bandlimited. Second, if the $x_n$ are real numbers, then $p(t)$ is real as well. In contrast, the most obvious trigonometric interpolation polynomial is the one in which the frequencies range from 0 to $N-1$ (instead of roughly $-N/2$ to $+N/2$ as above), similar to the inverse DFT formula. This interpolation does not minimize the slope, and is not generally real-valued for real $x_n$; its use is a common mistake. ### The unitary DFT Another way of looking at the DFT is to note that in the above discussion, the DFT can be expressed as a Vandermonde matrix, introduced by Sylvester in 1867, $\mathbf{F} = \begin{bmatrix} \omega_N^{0 \cdot 0} & \omega_N^{0 \cdot 1} & \ldots & \omega_N^{0 \cdot (N-1)} \\ \omega_N^{1 \cdot 0} & \omega_N^{1 \cdot 1} & \ldots & \omega_N^{1 \cdot (N-1)} \\ \vdots & \vdots & \ddots & \vdots \\ \omega_N^{(N-1) \cdot 0} & \omega_N^{(N-1) \cdot 1} & \ldots & \omega_N^{(N-1) \cdot (N-1)} \\ \end{bmatrix}$ where $\omega_N = e^{-2 \pi i/N}\,$ is a primitive Nth root of unity. The inverse transform is then given by the inverse of the above matrix, $\mathbf{F}^{-1}=\frac{1}{N}\mathbf{F}^$ With unitary normalization constants $1/\sqrt{N}$, the DFT becomes a unitary transformation, defined by a unitary matrix: $\mathbf{U}=\mathbf{F}/\sqrt{N}$ $\mathbf{U}^{-1}=\mathbf{U}^$ $\|\det(\mathbf{U})\|=1$ where det() is the determinant function. The determinant is the product of the eigenvalues, which are always $\pm 1$ or $\pm i$ as described below. In a real vector space, a unitary transformation can be thought of as simply a rigid rotation of the coordinate system, and all of the properties of a rigid rotation can be found in the unitary DFT. The orthogonality of the DFT is now expressed as an orthonormality condition (which arises in many areas of mathematics as described in root of unity): $\sum_{m=0}^{N-1}U_{km}U_{mn}^=\delta_{kn}$ If X is defined as the unitary DFT of the vector x, then $X_k=\sum_{n=0}^{N-1} U_{kn}x_n$ and the Plancherel theorem is expressed as $\sum_{n=0}^{N-1}x_n y_n^ = \sum_{k=0}^{N-1}X_k Y_k^$ If we view the DFT as just a coordinate transformation which simply specifies the components of a vector in a new coordinate system, then the above is just the statement that the dot product of two vectors is preserved under a unitary DFT transformation. For the special case $\mathbf{x} = \mathbf{y}$, this implies that the length of a vector is preserved as well—this is just Parseval's theorem, $\sum_{n=0}^{N-1}\|x_n\|^2 = \sum_{k=0}^{N-1}\|X_k\|^2$ A consequence of the circular convolution theorem is that the DFT matrix F diagonalizes any circulant matrix. ### Expressing the inverse DFT in terms of the DFT A useful property of the DFT is that the inverse DFT can be easily expressed in terms of the (forward) DFT, via several well-known "tricks". (For example, in computations, it is often convenient to only implement a fast Fourier transform corresponding to one transform direction and then to get the other transform direction from the first.) First, we can compute the inverse DFT by reversing the inputs (Duhamel et al., 1988): $\mathcal{F}^{-1}(\{x_n\}) = \mathcal{F}(\{x_{N - n}\}) / N$ (As usual, the subscripts are interpreted modulo N; thus, for $n=0$, we have $x_{N-0}=x_0$.) Second, one can also conjugate the inputs and outputs: $\mathcal{F}^{-1}(\mathbf{x}) = \mathcal{F}(\mathbf{x}^)^* / N$ Third, a variant of this conjugation trick, which is sometimes preferable because it requires no modification of the data values, involves swapping real and imaginary parts (which can be done on a computer simply by modifying pointers). Define swap($x_n$) as $x_n$ with its real and imaginary parts swapped—that is, if $x_n = a + b i$ then swap($x_n$) is $b + a i$. Equivalently, swap($x_n$) equals $i x_n^$. Then $\mathcal{F}^{-1}(\mathbf{x}) = \textrm{swap}(\mathcal{F}(\textrm{swap}(\mathbf{x}))) / N$ That is, the inverse transform is the same as the forward transform with the real and imaginary parts swapped for both input and output, up to a normalization (Duhamel et al., 1988). The conjugation trick can also be used to define a new transform, closely related to the DFT, that is involutory—that is, which is its own inverse. In particular, $T(\mathbf{x}) = \mathcal{F}(\mathbf{x}^) / \sqrt{N}$ is clearly its own inverse: $T(T(\mathbf{x})) = \mathbf{x}$. A closely related involutory transformation (by a factor of (1+i) /√2) is $H(\mathbf{x}) = \mathcal{F}((1+i) \mathbf{x}^) / \sqrt{2N}$, since the $(1+i)$ factors in $H(H(\mathbf{x}))$ cancel the 2. For real inputs $\mathbf{x}$, the real part of $H(\mathbf{x})$ is none other than the discrete Hartley transform, which is also involutory. ### Eigenvalues and eigenvectors The eigenvalues of the DFT matrix are simple and well-known, whereas the eigenvectors are complicated, not unique, and are the subject of ongoing research. Consider the unitary form $\mathbf{U}$ defined above for the DFT of length N, where $\mathbf{U}_{m,n} = \frac1{\sqrt{N}}\omega_N^{(m-1)(n-1)} = \frac1{\sqrt{N}}e^{-\frac{2\pi i}N (m-1)(n-1)}.$ This matrix satisfies the matrix polynomial equation: $\mathbf{U}^4 = \mathbf{I}.$ This can be seen from the inverse properties above: operating $\mathbf{U}$ twice gives the original data in reverse order, so operating $\mathbf{U}$ four times gives back the original data and is thus the identity matrix. This means that the eigenvalues $\lambda$ satisfy the equation: $\lambda^4 = 1.$ Therefore, the eigenvalues of $\mathbf{U}$ are the fourth roots of unity: $\lambda$ is +1, −1, +i, or −i. Since there are only four distinct eigenvalues for this $N\times N$ matrix, they have some multiplicity. The multiplicity gives the number of linearly independent eigenvectors corresponding to each eigenvalue. (Note that there are N independent eigenvectors; a unitary matrix is never defective.) The problem of their multiplicity was solved by McClellan and Parks (1972), although it was later shown to have been equivalent to a problem solved by Gauss (Dickinson and Steiglitz, 1982). The multiplicity depends on the value of N modulo 4, and is given by the following table: Multiplicities of the eigenvalues λ of the unitary DFT matrix U as a function of the transform size N (in terms of an integer m). size N λ = +1 λ = −1 λ = -i λ = +i 4m m + 1 m m m − 1 4m + 1 m + 1 m m m 4m + 2 m + 1 m + 1 m m 4m + 3 m + 1 m + 1 m + 1 m Otherwise stated, the characteristic polynomial of $\mathbf{U}$ is: $\det (\lambda I - \mathbf{U})= (\lambda-1)^{\left\lfloor \tfrac {N+4}{4}\right\rfloor} (\lambda+1)^{\left\lfloor \tfrac {N+2}{4}\right\rfloor} (\lambda+i)^{\left\lfloor \tfrac {N+1}{4}\right\rfloor} (\lambda-i)^{\left\lfloor \tfrac {N-1}{4}\right\rfloor}.$ No simple analytical formula for general eigenvectors is known. Moreover, the eigenvectors are not unique because any linear combination of eigenvectors for the same eigenvalue is also an eigenvector for that eigenvalue. Various researchers have proposed different choices of eigenvectors, selected to satisfy useful properties like orthogonality and to have "simple" forms (e.g., McClellan and Parks, 1972; Dickinson and Steiglitz, 1982; Grünbaum, 1982; Atakishiyev and Wolf, 1997; Candan et al., 2000; Hanna et al., 2004; Gurevich and Hadani, 2008). A straightforward approach is to discretize an eigenfunction of the continuous Fourier transform, of which the most famous is the Gaussian function. Since periodic summation of the function means discretizing its frequency spectrum and discretization means periodic summation of the spectrum, the discretized and periodically summed Gaussian function yields an eigenvector of the discrete transform: • $F(m) = \sum_{k\in\mathbb{Z}} \exp\left(-\frac{\pi\cdot(m+N\cdot k)^2}{N}\right)$. A closed form expression for the series is not known, but it converges rapidly. Two other simple closed-form analytical eigenvectors for special DFT period N were found (Kong, 2008): For DFT period N = 2L + 1 = 4K +1, where K is an integer, the following is an eigenvector of DFT: • $F(m)=\prod_{s=K+1}^L\left[\cos\left(\frac{2\pi}{N}m\right)- \cos\left(\frac{2\pi}{N}s\right)\right]$ For DFT period N = 2L = 4K, where K is an integer, the following is an eigenvector of DFT: • $F(m)=\sin\left(\frac{2\pi}{N}m\right)\prod_{s=K+1}^{L-1}\left[\cos\left(\frac{2\pi}{N}m\right)- \cos\left(\frac{2\pi}{N}s\right)\right]$ The choice of eigenvectors of the DFT matrix has become important in recent years in order to define a discrete analogue of the fractional Fourier transform—the DFT matrix can be taken to fractional powers by exponentiating the eigenvalues (e.g., Rubio and Santhanam, 2005). For the continuous Fourier transform, the natural orthogonal eigenfunctions are the Hermite functions, so various discrete analogues of these have been employed as the eigenvectors of the DFT, such as the Kravchuk polynomials (Atakishiyev and Wolf, 1997). The "best" choice of eigenvectors to define a fractional discrete Fourier transform remains an open question, however. ### Uncertainty principle If the random variable Xk is constrained by $\sum_{n=0}^{N-1}\|X_n\|^2=1 ~,$ then $P_n=\|X_n\|^2$ may be considered to represent a discrete probability mass function of n, with an associated probability mass function constructed from the transformed variable, $Q_m=N\|x_m\|^2 ~.$ For the case of continuous functions $P(x)$ and $Q(k)$, the Heisenberg uncertainty principle states that $D_0(X)D_0(x)\ge\frac{1}{16\pi^2}$ where $D_0(X)$ and $D_0(x)$ are the variances of $\|X\|^2$ and $\|x\|^2$ respectively, with the equality attained in the case of a suitably normalized Gaussian distribution. Although the variances may be analogously defined for the DFT, an analogous uncertainty principle is not useful, because the uncertainty will not be shift-invariant. Still, a meaningful uncertainty principle has been introduced by Massar and Spindel.[6] However, the Hirschman entropic uncertainty will have a useful analog for the case of the DFT.[7] The Hirschman uncertainty principle is expressed in terms of the Shannon entropy of the two probability functions. In the discrete case, the Shannon entropies are defined as $H(X)=-\sum_{n=0}^{N-1} P_n\ln P_n$ and $H(x)=-\sum_{m=0}^{N-1} Q_m\ln Q_m ~,$ and the entropic uncertainty principle becomes[7] $H(X)+H(x) \ge \ln(N) ~.$ The equality is obtained for $P_n$ equal to translations and modulations of a suitably normalized Kronecker comb of period $A$ where $A$ is any exact integer divisor of $N$. The probability mass function $Q_m$ will then be proportional to a suitably translated Kronecker comb of period $B=N/A$.[7] There is also a well-known deterministic uncertainty principle that uses signal sparsity (or the number of non-zero coefficients).[8] Let $\\|x\\|_0$ and $\\|X\\|_0$ be the number of non-zero elements of the time and frequency sequences $x_0,x_1,\ldots,x_{N-1}$ and $X_0,X_1,\ldots,X_{N-1}$, respectively. Then, $N\leq \\|x\\|_0 \cdot \\|X\\|_0.$ As an immediate consequence of the inequality of arithmetic and geometric means, one also has $2\sqrt{N}\leq\\|x\\|_0+\\|X\\|_0$. Both uncertainty principles were shown to be tight for specifically-chosen "picket-fence" sequences (discrete impulse trains), and find practical use for signal recovery applications.[8] ### The real-input DFT If $x_0, \ldots, x_{N-1}$ are real numbers, as they often are in practical applications, then the DFT obeys the symmetry: $X_{N-k} \equiv X_{-k} = X_k^,$ where $X^\,$ denotes complex conjugation. It follows that X0 and XN/2 are real-valued, and the remainder of the DFT is completely specified by just N/2-1 complex numbers. ## Generalized DFT (shifted and non-linear phase) It is possible to shift the transform sampling in time and/or frequency domain by some real shifts a and b, respectively. This is sometimes known as a generalized DFT (or GDFT), also called the shifted DFT or offset DFT, and has analogous properties to the ordinary DFT: $X_k = \sum_{n=0}^{N-1} x_n e^{-\frac{2 \pi i}{N} (k+b) (n+a)} \quad \quad k = 0, \dots, N-1.$ Most often, shifts of $1/2$ (half a sample) are used. While the ordinary DFT corresponds to a periodic signal in both time and frequency domains, $a=1/2$ produces a signal that is anti-periodic in frequency domain ($X_{k+N} = - X_k$) and vice versa for $b=1/2$. Thus, the specific case of $a = b = 1/2$ is known as an odd-time odd-frequency discrete Fourier transform (or O2 DFT). Such shifted transforms are most often used for symmetric data, to represent different boundary symmetries, and for real-symmetric data they correspond to different forms of the discrete cosine and sine transforms. Another interesting choice is $a=b=-(N-1)/2$, which is called the centered DFT (or CDFT). The centered DFT has the useful property that, when N is a multiple of four, all four of its eigenvalues (see above) have equal multiplicities (Rubio and Santhanam, 2005)[9] The term GDFT is also used for the non-linear phase extensions of DFT. Hence, GDFT method provides a generalization for constant amplitude orthogonal block transforms including linear and non-linear phase types. GDFT is a framework to improve time and frequency domain properties of the traditional DFT, e.g. auto/cross-correlations, by the addition of the properly designed phase shaping function (non-linear, in general) to the original linear phase functions (Akansu and Agirman-Tosun, 2010).[10] The discrete Fourier transform can be viewed as a special case of the z-transform, evaluated on the unit circle in the complex plane; more general z-transforms correspond to complex shifts a and b above. ## Multidimensional DFT The ordinary DFT transforms a one-dimensional sequence or array $x_n$ that is a function of exactly one discrete variable n. The multidimensional DFT of a multidimensional array $x_{n_1, n_2, \dots, n_d}$ that is a function of d discrete variables $n_\ell = 0, 1, \dots, N_\ell-1$ for $\ell$ in $1, 2, \dots, d$ is defined by: $X_{k_1, k_2, \dots, k_d} = \sum_{n_1=0}^{N_1-1} \left(\omega_{N_1}^{~k_1 n_1} \sum_{n_2=0}^{N_2-1} \left( \omega_{N_2}^{~k_2 n_2} \cdots \sum_{n_d=0}^{N_d-1} \omega_{N_d}^{~k_d n_d}\cdot x_{n_1, n_2, \dots, n_d} \right) \right) \, ,$ where $\omega_{N_\ell} = \exp(-2\pi i/N_\ell)$ as above and the d output indices run from $k_\ell = 0, 1, \dots, N_\ell-1$. This is more compactly expressed in vector notation, where we define $\mathbf{n} = (n_1, n_2, \dots, n_d)$ and $\mathbf{k} = (k_1, k_2, \dots, k_d)$ as d-dimensional vectors of indices from 0 to $\mathbf{N} - 1$, which we define as $\mathbf{N} - 1 = (N_1 - 1, N_2 - 1, \dots, N_d - 1)$: $X_\mathbf{k} = \sum_{\mathbf{n}=0}^{\mathbf{N}-1} e^{-2\pi i \mathbf{k} \cdot (\mathbf{n} / \mathbf{N})} x_\mathbf{n} \, ,$ where the division $\mathbf{n} / \mathbf{N}$ is defined as $\mathbf{n} / \mathbf{N} = (n_1/N_1, \dots, n_d/N_d)$ to be performed element-wise, and the sum denotes the set of nested summations above. The inverse of the multi-dimensional DFT is, analogous to the one-dimensional case, given by: $x_\mathbf{n} = \frac{1}{\prod_{\ell=1}^d N_\ell} \sum_{\mathbf{k}=0}^{\mathbf{N}-1} e^{2\pi i \mathbf{n} \cdot (\mathbf{k} / \mathbf{N})} X_\mathbf{k} \, .$ As the one-dimensional DFT expresses the input $x_n$ as a superposition of sinusoids, the multidimensional DFT expresses the input as a superposition of plane waves, or multidimensional sinusoids. The direction of oscillation in space is $\mathbf{k} / \mathbf{N}$. The amplitudes are $X_\mathbf{k}$. This decomposition is of great importance for everything from digital image processing (two-dimensional) to solving partial differential equations. The solution is broken up into plane waves. The multidimensional DFT can be computed by the composition of a sequence of one-dimensional DFTs along each dimension. In the two-dimensional case $x_{n_1,n_2}$ the $N_1$ independent DFTs of the rows (i.e., along $n_2$) are computed first to form a new array $y_{n_1,k_2}$. Then the $N_2$ independent DFTs of y along the columns (along $n_1$) are computed to form the final result $X_{k_1,k_2}$. Alternatively the columns can be computed first and then the rows. The order is immaterial because the nested summations above commute. An algorithm to compute a one-dimensional DFT is thus sufficient to efficiently compute a multidimensional DFT. This approach is known as the row-column algorithm. There are also intrinsically multidimensional FFT algorithms. ### The real-input multidimensional DFT For input data $x_{n_1, n_2, \dots, n_d}$ consisting of real numbers, the DFT outputs have a conjugate symmetry similar to the one-dimensional case above: $X_{k_1, k_2, \dots, k_d} = X_{N_1 - k_1, N_2 - k_2, \dots, N_d - k_d}^ ,$ where the star again denotes complex conjugation and the $\ell$-th subscript is again interpreted modulo $N_\ell$ (for $\ell = 1,2,\ldots,d$). ## Applications The DFT has seen wide usage across a large number of fields; we only sketch a few examples below (see also the references at the end). All applications of the DFT depend crucially on the availability of a fast algorithm to compute discrete Fourier transforms and their inverses, a fast Fourier transform. ### Spectral analysis When the DFT is used for spectral analysis, the $\{x_n\}\,$ sequence usually represents a finite set of uniformly spaced time-samples of some signal $x(t)\,$, where t represents time. The conversion from continuous time to samples (discrete-time) changes the underlying Fourier transform of x(t) into a discrete-time Fourier transform (DTFT), which generally entails a type of distortion called aliasing. Choice of an appropriate sample-rate (see Nyquist rate) is the key to minimizing that distortion. Similarly, the conversion from a very long (or infinite) sequence to a manageable size entails a type of distortion called leakage, which is manifested as a loss of detail (a.k.a. resolution) in the DTFT. Choice of an appropriate sub-sequence length is the primary key to minimizing that effect. When the available data (and time to process it) is more than the amount needed to attain the desired frequency resolution, a standard technique is to perform multiple DFTs, for example to create a spectrogram. If the desired result is a power spectrum and noise or randomness is present in the data, averaging the magnitude components of the multiple DFTs is a useful procedure to reduce the variance of the spectrum (also called a periodogram in this context); two examples of such techniques are the Welch method and the Bartlett method; the general subject of estimating the power spectrum of a noisy signal is called spectral estimation. A final source of distortion (or perhaps illusion) is the DFT itself, because it is just a discrete sampling of the DTFT, which is a function of a continuous frequency domain. That can be mitigated by increasing the resolution of the DFT. That procedure is illustrated at Sampling the DTFT. • The procedure is sometimes referred to as zero-padding, which is a particular implementation used in conjunction with the fast Fourier transform (FFT) algorithm. The inefficiency of performing multiplications and additions with zero-valued "samples" is more than offset by the inherent efficiency of the FFT. • As already noted, leakage imposes a limit on the inherent resolution of the DTFT. So there is a practical limit to the benefit that can be obtained from a fine-grained DFT. ### Data compression The field of digital signal processing relies heavily on operations in the frequency domain (i.e. on the Fourier transform). For example, several lossy image and sound compression methods employ the discrete Fourier transform: the signal is cut into short segments, each is transformed, and then the Fourier coefficients of high frequencies, which are assumed to be unnoticeable, are discarded. The decompressor computes the inverse transform based on this reduced number of Fourier coefficients. (Compression applications often use a specialized form of the DFT, the discrete cosine transform or sometimes the modified discrete cosine transform.) Some relatively recent compression algorithms, however, use wavelet transforms, which give a more uniform compromise between time and frequency domain than obtained by chopping data into segments and transforming each segment. In the case of JPEG2000, this avoids the spurious image features that appear when images are highly compressed with the original JPEG. ### Partial differential equations Discrete Fourier transforms are often used to solve partial differential equations, where again the DFT is used as an approximation for the Fourier series (which is recovered in the limit of infinite N). The advantage of this approach is that it expands the signal in complex exponentials einx, which are eigenfunctions of differentiation: d/dx einx = in einx. Thus, in the Fourier representation, differentiation is simple—we just multiply by i n. (Note, however, that the choice of n is not unique due to aliasing; for the method to be convergent, a choice similar to that in the trigonometric interpolation section above should be used.) A linear differential equation with constant coefficients is transformed into an easily solvable algebraic equation. One then uses the inverse DFT to transform the result back into the ordinary spatial representation. Such an approach is called a spectral method. ### Polynomial multiplication Suppose we wish to compute the polynomial product c(x) = a(x) · b(x). The ordinary product expression for the coefficients of c involves a linear (acyclic) convolution, where indices do not "wrap around." This can be rewritten as a cyclic convolution by taking the coefficient vectors for a(x) and b(x) with constant term first, then appending zeros so that the resultant coefficient vectors a and b have dimension d > deg(a(x)) + deg(b(x)). Then, $\mathbf{c} = \mathbf{a} * \mathbf{b}$ Where c is the vector of coefficients for c(x), and the convolution operator $\,$ is defined so $c_n = \sum_{m=0}^{d-1}a_m b_{n-m\ \mathrm{mod}\ d} \qquad\qquad\qquad n=0,1\dots,d-1$ But convolution becomes multiplication under the DFT: $\mathcal{F}(\mathbf{c}) = \mathcal{F}(\mathbf{a})\mathcal{F}(\mathbf{b})$ Here the vector product is taken elementwise. Thus the coefficients of the product polynomial c(x) are just the terms 0, ..., deg(a(x)) + deg(b(x)) of the coefficient vector $\mathbf{c} = \mathcal{F}^{-1}(\mathcal{F}(\mathbf{a})\mathcal{F}(\mathbf{b})).$ With a fast Fourier transform, the resulting algorithm takes O (N log N) arithmetic operations. Due to its simplicity and speed, the Cooley–Tukey FFT algorithm, which is limited to composite sizes, is often chosen for the transform operation. In this case, d should be chosen as the smallest integer greater than the sum of the input polynomial degrees that is factorizable into small prime factors (e.g. 2, 3, and 5, depending upon the FFT implementation). #### Multiplication of large integers The fastest known algorithms for the multiplication of very large integers use the polynomial multiplication method outlined above. Integers can be treated as the value of a polynomial evaluated specifically at the number base, with the coefficients of the polynomial corresponding to the digits in that base. After polynomial multiplication, a relatively low-complexity carry-propagation step completes the multiplication. #### Convolution When data is convolved with a function with wide support, such as for downsampling by a large sampling ratio, because of the Convolution theorem and the FFT algorithm, it may be faster to transform it, multiply pointwise by the transform of the filter and then reverse transform it. Alternatively, a good filter is obtained by simply truncating the transformed data and re-transforming the shortened data set. ## Some discrete Fourier transform pairs Some DFT pairs $x_n = \frac{1}{N}\sum_{k=0}^{N-1}X_k e^{i 2 \pi kn/N}$ $X_k = \sum_{n=0}^{N-1}x_n e^{-i 2 \pi kn/N}$ Note $x_n e^{i 2 \pi n\ell/N} \,$ $X_{k-\ell}\,$ Shift theorem $x_{n-\ell}\,$ $X_k e^{-i 2 \pi k\ell/N} \,$ $x_n \in \mathbb{R}$ $X_k=X_{N-k}^\,$ Real DFT $a^n\,$ $\left\{ \begin{matrix} N & \mbox{if } a = e^{i 2 \pi k/N} \\ \frac{1-a^N}{1-a \, e^{-i 2 \pi k/N} } & \mbox{otherwise} \end{matrix} \right.$ from the geometric progression formula ${N-1 \choose n}\,$ $\left(1+e^{-i 2 \pi k/N} \right)^{N-1}\,$ from the binomial theorem $\left\{ \begin{matrix} \frac{1}{W} & \mbox{if } 2n < W \mbox{ or } 2(N-n) < W \\ 0 & \mbox{otherwise} \end{matrix} \right.$ $\left\{ \begin{matrix} 1 & \mbox{if } k = 0 \\ \frac{\sin\left(\frac{\pi W k}{N}\right)} {W \sin\left(\frac{\pi k}{N}\right)} & \mbox{otherwise} \end{matrix} \right.$ $x_n$ is a rectangular window function of W points centered on n=0, where W is an odd integer, and $X_k$ is a sinc-like function (specifically, $X_k$ is a Dirichlet kernel) $\sum_{j\in\mathbb{Z}} \exp\left(-\frac{\pi}{cN}\cdot(n+N\cdot j)^2\right)$ $\sqrt{cN} \cdot \sum_{j\in\mathbb{Z}} \exp\left(-\frac{\pi c}{N}\cdot(k+N\cdot j)^2\right)$ Discretization and periodic summation of the scaled Gaussian functions for $c>0$. Since either $c$ or $\frac{1}{c}$ is larger than one and thus warrants fast convergence of one of the two series, for large $c$ you may choose to compute the frequency spectrum and convert to the time domain using the discrete Fourier transform. ## Generalizations ### Representation theory The DFT can be interpreted as the complex-valued representation theory of the finite cyclic group. In other words, a sequence of n complex numbers can be thought of as an element of n-dimensional complex space Cn or equivalently a function f from the finite cyclic group of order n to the complex numbers, ZnC. So f is a class function on the finite cyclic group, and thus can be expressed as a linear combination of the irreducible characters of this group, which are the roots of unity. From this point of view, one may generalize the DFT to representation theory generally, or more narrowly to the representation theory of finite groups. More narrowly still, one may generalize the DFT by either changing the target (taking values in a field other than the complex numbers), or the domain (a group other than a finite cyclic group), as detailed in the sequel. ### Other fields Many of the properties of the DFT only depend on the fact that $e^{-\frac{2 \pi i}{N}}$ is a primitive root of unity, sometimes denoted $\omega_N$ or $W_N$ (so that $\omega_N^N = 1$). Such properties include the completeness, orthogonality, Plancherel/Parseval, periodicity, shift, convolution, and unitarity properties above, as well as many FFT algorithms. For this reason, the discrete Fourier transform can be defined by using roots of unity in fields other than the complex numbers, and such generalizations are commonly called number-theoretic transforms (NTTs) in the case of finite fields. For more information, see number-theoretic transform and discrete Fourier transform (general). ### Other finite groups The standard DFT acts on a sequence x0, x1, …, xN−1 of complex numbers, which can be viewed as a function {0, 1, …, N − 1} → C. The multidimensional DFT acts on multidimensional sequences, which can be viewed as functions $\{0, 1, \ldots, N_1-1\} \times \cdots \times \{0, 1, \ldots, N_d-1\} \to \mathbb{C}.$ This suggests the generalization to Fourier transforms on arbitrary finite groups, which act on functions GC where G is a finite group. In this framework, the standard DFT is seen as the Fourier transform on a cyclic group, while the multidimensional DFT is a Fourier transform on a direct sum of cyclic groups. ## Alternatives There are various alternatives to the DFT for various applications, prominent among which are wavelets. The analog of the DFT is the discrete wavelet transform (DWT). From the point of view of time–frequency analysis, a key limitation of the Fourier transform is that it does not include location information, only frequency information, and thus has difficulty in representing transients. As wavelets have location as well as frequency, they are better able to represent location, at the expense of greater difficulty representing frequency. For details, see comparison of the discrete wavelet transform with the discrete Fourier transform. ## Notes 1. ^ In this context, it is common to define $\omega$ to be the Nth primitive root of unity, $\omega = e^{-2 \pi i / N}$, to obtain the following form: $X_k = \sum_{n=0}^{N-1} x_n \cdot \omega^{k n}$ 2. ^ As a linear transformation on a finite-dimensional vector space, the DFT expression can also be written in terms of a DFT matrix; when scaled appropriately it becomes a unitary matrix and the Xk can thus be viewed as coefficients of x in an orthonormal basis. ## Citations 1. ^ Strang, Gilbert (May–June 1994). "Wavelets". American Scientist 82 (3): 253. Retrieved 8 October 2013. This is the most important numerical algorithm of our lifetime... 2. ^ Sahidullah, Md.; Saha, Goutam (Feb 2013). "A Novel Windowing Technique for Efficient Computation of MFCC for Speaker Recognition". IEEE Signal Processing Letters 20 (2): 149–152. doi:10.1109/LSP.2012.2235067. 3. ^ Cooley et al., 1969 4. ^ "Shift zero-frequency component to center of spectrum - MATLAB fftshift". http://www.mathworks.com/. Natick, MA 01760: The MathWorks, Inc. Retrieved 10 March 2014. 5. ^ T. G. Stockham, Jr., "High-speed convolution and correlation," in 1966 Proc. AFIPS Spring Joint Computing Conf. Reprinted in Digital Signal Processing, L. R. Rabiner and C. M. Rader, editors, New York: IEEE Press, 1972. 6. ^ Massar, S.; Spindel, P. (2008). "Uncertainty Relation for the Discrete Fourier Transform". Physical Review Letters 100 (19). arXiv:0710.0723. Bibcode:2008PhRvL.100s0401M. doi:10.1103/PhysRevLett.100.190401. 7. ^ a b c DeBrunner, Victor; Havlicek, Joseph P.; Przebinda, Tomasz; Özaydin, Murad (2005). "Entropy-Based Uncertainty Measures for $L^2(\mathbb{R}^n),\ell^2(\mathbb{Z})$, and $\ell^2(\mathbb{Z}/N\mathbb{Z})$ With a Hirschman Optimal Transform for $\ell^2(\mathbb{Z}/N\mathbb{Z})$" (PDF). IEEE Transactions on Signal Processing 53 (8): 2690. Bibcode:2005ITSP...53.2690D. doi:10.1109/TSP.2005.850329. Retrieved 2011-06-23. 8. ^ a b Donoho, D.L.; Stark, P.B (1989). "Uncertainty principles and signal recovery". SIAM Journal on Applied Mathematics 49 (3): 906–931. doi:10.1137/0149053. 9. ^ Santhanam, Balu; Santhanam, Thalanayar S. "Discrete Gauss-Hermite functions and eigenvectors of the centered discrete Fourier transform", Proceedings of the 32nd IEEE International Conference on Acoustics, Speech, and Signal Processing (ICASSP 2007, SPTM-P12.4), vol. III, pp. 1385-1388. 10. ^ Akansu, Ali N.; Agirman-Tosun, Handan "Generalized Discrete Fourier Transform With Nonlinear Phase", IEEE Transactions on Signal Processing, vol. 58, no. 9, pp. 4547-4556, Sept. 2010. ## References • Brigham, E. Oran (1988). The fast Fourier transform and its applications. Englewood Cliffs, N.J.: Prentice Hall. ISBN 0-13-307505-2. • Oppenheim, Alan V.; Schafer, R. W.; and Buck, J. R. (1999). Discrete-time signal processing. Upper Saddle River, N.J.: Prentice Hall. ISBN 0-13-754920-2. • Smith, Steven W. (1999). "Chapter 8: The Discrete Fourier Transform". The Scientist and Engineer's Guide to Digital Signal Processing (Second ed.). San Diego, Calif.: California Technical Publishing. ISBN 0-9660176-3-3. • Cormen, Thomas H.; Charles E. Leiserson; Ronald L. Rivest; Clifford Stein (2001). "Chapter 30: Polynomials and the FFT". Introduction to Algorithms (Second ed.). MIT Press and McGraw-Hill. pp. 822–848. ISBN 0-262-03293-7. esp. section 30.2: The DFT and FFT, pp. 830–838. • P. Duhamel, B. Piron, and J. M. Etcheto (1988). "On computing the inverse DFT". IEEE Trans. Acoust., Speech and Sig. Processing 36 (2): 285–286. doi:10.1109/29.1519. • J. H. McClellan and T. W. Parks (1972). "Eigenvalues and eigenvectors of the discrete Fourier transformation". IEEE Trans. Audio Electroacoust. 20 (1): 66–74. doi:10.1109/TAU.1972.1162342. • Bradley W. Dickinson and Kenneth Steiglitz (1982). "Eigenvectors and functions of the discrete Fourier transform". IEEE Trans. Acoust., Speech and Sig. Processing 30 (1): 25–31. doi:10.1109/TASSP.1982.1163843. (Note that this paper has an apparent typo in its table of the eigenvalue multiplicities: the +i/−i columns are interchanged. The correct table can be found in McClellan and Parks, 1972, and is easily confirmed numerically.) • F. A. Grünbaum (1982). "The eigenvectors of the discrete Fourier transform". J. Math. Anal. Appl. 88 (2): 355–363. doi:10.1016/0022-247X(82)90199-8. • Natig M. Atakishiyev and Kurt Bernardo Wolf (1997). "Fractional Fourier-Kravchuk transform". J. Opt. Soc. Am. A 14 (7): 1467–1477. Bibcode:1997JOSAA..14.1467A. doi:10.1364/JOSAA.14.001467. • C. Candan, M. A. Kutay and H. M.Ozaktas (2000). "The discrete fractional Fourier transform". IEEE Trans. on Signal Processing 48 (5): 1329–1337. Bibcode:2000ITSP...48.1329C. doi:10.1109/78.839980. • Magdy Tawfik Hanna, Nabila Philip Attalla Seif, and Waleed Abd El Maguid Ahmed (2004). "Hermite-Gaussian-like eigenvectors of the discrete Fourier transform matrix based on the singular-value decomposition of its orthogonal projection matrices". IEEE Trans. Circ. Syst. I 51 (11): 2245–2254. doi:10.1109/TCSI.2004.836850. • Shamgar Gurevich and Ronny Hadani (2009). "On the diagonalization of the discrete Fourier transform". Applied and Computational Harmonic Analysis 27 (1): 87–99. arXiv:0808.3281. doi:10.1016/j.acha.2008.11.003. preprint at. • Shamgar Gurevich, Ronny Hadani, and Nir Sochen (2008). "The finite harmonic oscillator and its applications to sequences, communication and radar". IEEE Transactions on Information Theory 54 (9): 4239–4253. arXiv:0808.1495. doi:10.1109/TIT.2008.926440. preprint at. • Juan G. Vargas-Rubio and Balu Santhanam (2005). "On the multiangle centered discrete fractional Fourier transform". IEEE Sig. Proc. Lett. 12 (4): 273–276. Bibcode:2005ISPL...12..273V. doi:10.1109/LSP.2005.843762. • J. Cooley, P. Lewis, and P. Welch (1969). "The finite Fourier transform". IEEE Trans. Audio Electroacoustics 17 (2): 77–85. doi:10.1109/TAU.1969.1162036. • F.N. Kong (2008). "Analytic Expressions of Two Discrete Hermite-Gaussian Signals". IEEE Trans. Circuits and Systems –II: Express Briefs. 55 (1): 56–60. doi:10.1109/TCSII.2007.909865.	13,155	46,207	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 219, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2015-22	latest	en	0.877631
http://gmatclub.com/forum/1-what-is-a-the-greatest-possible-area-of-a-triangular-40489.html?fl=similar	1,467,097,604,000,000,000	text/html	crawl-data/CC-MAIN-2016-26/segments/1466783396459.32/warc/CC-MAIN-20160624154956-00149-ip-10-164-35-72.ec2.internal.warc.gz	131,526,323	42,003	Find all School-related info fast with the new School-Specific MBA Forum It is currently 28 Jun 2016, 00:06 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track Your Progress every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # 1) What is a the greatest possible area of a triangular post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message Manager Joined: 24 Jun 2006 Posts: 62 Followers: 1 Kudos [?]: 3 [0], given: 0 1) What is a the greatest possible area of a triangular [#permalink] ### Show Tags 25 Dec 2006, 16:59 This topic is locked. If you want to discuss this question please re-post it in the respective forum. 1) What is a the greatest possible area of a triangular region with the vertex at the center of a circle with radius 1 and other two vertices on the circle 1) (sqrt 3)/4 2) 1/2 3) pi/4 4) 1 5) sqrt 2 2) A certain circular area has its center at point P and has radius 4 and Points X & Y lie in the same plane as the circular area. Does point y lie outside the circular area. (Data Sufficiency) 1) Dist between X & P = 4.5 2) Disat between X & Y = 9 VP Joined: 28 Mar 2006 Posts: 1381 Followers: 2 Kudos [?]: 26 [0], given: 0 Re: GCD [#permalink] ### Show Tags 25 Dec 2006, 17:40 mitul wrote: 1) What is a the greatest possible area of a triangular region with the vertex at the center of a circle with radius 1 and other two vertices on the circle 1) (sqrt 3)/4 2) 1/2 3) pi/4 4) 1 5) sqrt 2 It should be a isoceles right triangle with base=ht=1 So area =1/2 2) A certain circular area has its center at point P and has radius 4 and Points X & Y lie in the same plane as the circular area. Does point y lie outside the circular area. (Data Sufficiency) 1) Dist between X & P = 4.5 NOT SUFF. Is radius =4 and P is 4.5 from centre then Y should be between 0.5 and 8.5 ('cos thrid side is always less than sum of other 2 sides and greater than the difference). 2) Distance between X & Y = 9 SUFF. The third side length is between 5(9-4) and 13(9+4) which is outside the circle So B Director Joined: 28 Dec 2005 Posts: 921 Followers: 2 Kudos [?]: 43 [0], given: 0 Re: GCD [#permalink] ### Show Tags 25 Dec 2006, 17:48 mitul wrote: 1) What is a the greatest possible area of a triangular region with the vertex at the center of a circle with radius 1 and other two vertices on the circle 1) (sqrt 3)/4 2) 1/2 3) pi/4 4) 1 5) sqrt 2 2) A certain circular area has its center at point P and has radius 4 and Points X & Y lie in the same plane as the circular area. Does point y lie outside the circular area. (Data Sufficiency) 1) Dist between X & P = 4.5 2) Disat between X & Y = 9 1) I am not sure this is so, can someone please confirm, that such a triangle of largest area is an equilateral one? If that is the case, the answer is sqrt(3)/4. Using special triangle properties the side of the triangle are in the ratio: 1:sqrt(3):2. [30:60:90]. Thus hyp is 1, so the height is sqrt(3)/2 and the base of the complete triangle is 1/22 = 1. Thus the area is 1/21sqrt(3)/2 = sqrt(3)/4. 2) The only way to do this is by drawing the circle, the center and the points X and Y. 1) X is obviously outside the circle because it is more than the radius away from the center. But doesnt say anything about Y. INSUFF. 2)Distance between X and Y is given, but this also insufficient since we have no indication about the location wrt the circle. INSUFF. Together, it becomes clear that Y will always be outside the circle since the only way Y could have been at least on the circle would have been if X and Y were (4.5+4 = 8.5). So, my answer is C. Senior Manager Joined: 23 Jun 2006 Posts: 387 Followers: 1 Kudos [?]: 318 [0], given: 0 [#permalink] ### Show Tags 25 Dec 2006, 18:08 for question 1 answer is 1/2 (C)... i.e. a triangle with 90 degree angle. an alternative formula for a triangle's area is absin(x)/2 where a and b are two sides of the triangle and x is the angle between them. in our case, a=b=1 regardless of angle. so maximum is attained when sin(x) is maximal which happens when x is 90 degrees. area is 0.5 in that case. question 2 is indeed C. Director Joined: 28 Dec 2005 Posts: 921 Followers: 2 Kudos [?]: 43 [0], given: 0 [#permalink] ### Show Tags 25 Dec 2006, 18:15 hobbit wrote: for question 1 answer is 1/2 (C)... i.e. a triangle with 90 degree angle. an alternative formula for a triangle's area is absin(x)/2 where a and b are two sides of the triangle and x is the angle between them. in our case, a=b=1 regardless of angle. so maximum is attained when sin(x) is maximal which happens when x is 90 degrees. area is 0.5 in that case. question 2 is indeed C. Thanks Hobbit. Makes sense. Intern Joined: 13 Jun 2005 Posts: 28 Followers: 0 Kudos [?]: 26 [0], given: 0 [#permalink] ### Show Tags 25 Dec 2006, 18:20 Prob 1. If one vertex is on the circle, and the other two sides are on the circle ( two sides are equal). Isosceles triangle. If it is right angled at the center, area = 1/2 x 1 x1 = 1/2 Tried..... The third side is >0 and <2 and tried to find the hyp but it works out to be less 1/2 I'm going with (2) Ans 1/2 Prob 2. 1) X is 4.5 from P, outside the circle but where is Y (insuff) 2) distance b/w X & Y is 9 X can be inside the circle or Y can be inside the circle or both can be outside and we can still have a distance of 9 (insuff) together diameter of circle is 8 . X is 4.5 from center and is outside the circle Since distance between X and Y is 9, Y needs to be outside (SUFF) Ans: C VP Joined: 28 Mar 2006 Posts: 1381 Followers: 2 Kudos [?]: 26 [0], given: 0 [#permalink] ### Show Tags 25 Dec 2006, 19:14 hobbit wrote: for question 1 answer is 1/2 (C)... i.e. a triangle with 90 degree angle. an alternative formula for a triangle's area is absin(x)/2 where a and b are two sides of the triangle and x is the angle between them. in our case, a=b=1 regardless of angle. so maximum is attained when sin(x) is maximal which happens when x is 90 degrees. area is 0.5 in that case. question 2 is indeed C. Why isnt it B enough for question (2)? Please see my reasoning above. Thanks! Intern Joined: 24 Feb 2006 Posts: 48 Followers: 1 Kudos [?]: 1 [0], given: 0 [#permalink] ### Show Tags 25 Dec 2006, 19:19 2) A certain circular area has its center at point P and has radius 4 and Points X & Y lie in the same plane as the circular area. Does point y lie outside the circular area. (Data Sufficiency) 1) Dist between X & P = 4.5 2) Disat between X & Y = 9 1) insuff as no info about Y 2) insuff as no info relates to the circle. together : 1 gives X is outside of the circle by (4.5 - 4) = 0.5 If Y were in the circle max distance between X and Y will be diameter + 0.5 = 42 + 0.5 = 8.5 but dist between X & Y = 9 >8.5 so Y is outside of the circle. ans is (C) [#permalink] 25 Dec 2006, 19:19 Display posts from previous: Sort by # 1) What is a the greatest possible area of a triangular post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	2,315	7,704	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.796875	4	CC-MAIN-2016-26	longest	en	0.863799
http://www.markedbyteachers.com/as-and-a-level/science/investigation-on-whether-rubber-obeys-hooke-s-rule-1.html	1,607,078,686,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141735600.89/warc/CC-MAIN-20201204101314-20201204131314-00094.warc.gz	143,325,246	25,594	• Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month Page 1. 1 1 2. 2 2 3. 3 3 4. 4 4 5. 5 5 6. 6 6 7. 7 7 8. 8 8 Investigation on whether Rubber obeys Hooke's Rule Extracts from this document... Introduction Physics AT1 – Hooke’s Rule Investigation on whether Rubber obeys Hooke’s Rule Plan Introduction Hooke's Rule states that extension of a material is proportional to the tension force applied to it unless the elastic limit is reached, which is the point at which the material no longer obeys Hooke's Rule. There are only a few materials that obey this rule. In this investigation, we will find out whether rubber obeys Hooke’s Rule. We will measure in detail the way in which the extension of a rubber band depends on the tension in the band. This will be done by applying various amounts of weights, as it is a continual variation. Hooke’s Rule = F = ke • F = Force in Newtons • k = Spring constant • e = Extension in Centimetres Rubber is a natural polymer which is made up of long chains of molecules which are bent back and forth with weak forces acting between them. As the rubber band is stretched, molecules straighten out and allow the rubber band to become larger. Eventually, as the molecules become fully stretched, the long chains will become parallel to each other and can stretch up to ten times its original length. Extra force will make the rubber band break. Middle 800 8 900 9 1000 10 Obtaining Evidence Method 1. First a boss and clamp were fixed together on a stand. 2. The rubber band was suspended off the boss clamp and was held with a piece of plastecine. 3. A second boss and clamp was fixed to the stand above the rubber band. This boss clamp was used to hold the meter rule in place, but in such a way that the length of the rubber band could be measured. 4. The original length of the rubber band (without extension) was then measured in cm and recorded in a results table. This would be important when calculating the extension. 5. Then a weight of 5N was added to the rubber band. The length of the rubber band increased and the new length was measured using the metre rule. 6. The extension was then calculated by subtracting the original length from the new stretched length. (Extension = New Length – Original Length). 7. This experiment was repeated with varied weights i.e. 10N, 15N, 20N etc. The same equation to calculate extension was used for each experiment. 8. The results were recorded in a table and a graph was created. Results 1st Set Mass/g Weight/N Original Length/cm New Length/cm Extension/cm 0 0 7.2 7.2 0.0 100 1 7.2 8.6 1.4 200 2 7.2 9.1 2.6 300 3 7.2 10.3 4.4 400 4 7.2 11.1 5.3 500 5 7.2 12.3 6.6 600 6 7.2 13.5 7.2 700 7 7.2 14.5 8.1 800 8 7.2 15.3 9.0 900 9 7.2 16.0 9.6 1000 10 7.2 16.7 9.9 Conclusion However if the experiment was to be carried out again changes could be made to reduce experimental error. A pointer on the rubber band would have helped in gathering the information more accurately. Also advanced equipment could be used to make the measurements even more accurate, for example an ultra sonic measuring device to measure the extension to a very accurate degree or a special laser. Computer technology could assist by creating graphs as the experiment is happening. Also a tensomometer may have been used as an accurate way of applying tension to the rubber. These improvements would increase the accuracy of the experiment vastly. Also more results could have been taken to increase the reliability of the evidence. If the investigation was extended further, other variables in the experiment could be measured, for example, the temperature which could have been measured as it was observed in the preliminary work that the temperature of the rubber increases as the extension increases and this factor could be investigated more thoroughly. Also different materials could have been investigated, for example, plastic, metal etc. and the results of the extension of these materials will be compared with the findings of the extension of rubber. Therefore, it can be seen how different types of cross-linking and bonding between molecules affects the extension This student written piece of work is one of many that can be found in our AS and A Level Waves & Cosmology section. Found what you're looking for? • Start learning 29% faster today • 150,000+ documents available • Just £6.99 a month Here's what a star student thought of this essay 3 star(s) Response to the question The author has provided a good response to the question, as they have carried out a successful experiment to investigate the relationship between the stress applied to the rubber band and the extension, showing that the rubber band appears to ... Response to the question The author has provided a good response to the question, as they have carried out a successful experiment to investigate the relationship between the stress applied to the rubber band and the extension, showing that the rubber band appears to be following HookeÃ¢â‚¬â„¢s Law. They have also considered the science behind the experiment, including a good explanation of elasticity. Level of analysis The author has analysed their results and compared them to the expected trend predicted in the hypothesis. However, there were many opportunities to extend their analysis slightly, for example using the gradient of their graph to calculate the Ã¢â‚¬Ëœspring constantÃ¢â‚¬â„¢ in the equation given in the hypothesis. They could also have calculated the uncertainties in their measurements and plotted uncertainty bars on the graph, and, if they had repeated their experiment more times, any anomalies in their results. These may be essential to gain high marks, and provide the opportunity to show good mathematical skills. Occasionally they have not shown a thorough understanding of the physics behind the experiment Ã¢â‚¬â€œ for example their explanation of the energy transfers when rubber is stretched is flawed. They need to think more logically about the molecules, in order to realise that the rubber gains elastic potential energy as the intermolecular bonds/ cross-links are stretched. The energy is not used to increase the temperature (although a slight increase in temperature could be due to friction as the molecules slide past each other), and the work done is only to pull the molecules in order to stretch the bonds. Less force could be required to produce a relatively larger extension due to the intermolecular bonds breaking/ weakening after a certain stress has been applied. They have also conflated the terms Ã¢â‚¬Ëœelastic limitÃ¢â‚¬â„¢ and Ã¢â‚¬Ëœlimit of proportionalityÃ¢â‚¬â„¢ Ã¢â‚¬â€œ the latter is the point after which the material stops obeying HookeÃ¢â‚¬â„¢s Law, whereas the elastic limit is the point beyond which the material stops behaving elastically i.e. undergoes permanent deformation under stress. Despite these small mistakes, they have analysed their results well and used them to support their conclusion. Quality of writing Their quality of written communication is good, with no obvious spelling or grammatical mistakes. However, they could present their work better by including more clear headings to break up the text and show to the marker that they can do an Ã¢â‚¬ËœanalysisÃ¢â‚¬â„¢ of their Ã¢â‚¬ËœresultsÃ¢â‚¬â„¢ and come to a Ã¢â‚¬ËœconclusionÃ¢â‚¬â„¢. They also occasionally use more colloquial language such as Ã¢â‚¬Ëœloads of varied massesÃ¢â‚¬â„¢, which could be better expressed as Ã¢â‚¬Ëœa variety of masses ranging from 100g to 1000g, increasing in 100g amountsÃ¢â‚¬â„¢ Ã¢â‚¬â€œ this is a bit clearer and less informal. However, the majority of their report is very well done, with only minor problems in the layout. Reviewed by dragonkeeper13 23/04/2012 Not the one? Search for your essay title... • Join over 1.2 million students every month • Accelerate your learning by 29% • Unlimited access from just £6.99 per month Related AS and A Level Waves & Cosmology essays 1. Experiment B11: Measuring focal length of lenses cm Method (c): Lenses-mirror method (for a concave lens) Image distance v/cm Separation between lenses t/cm Focal length f = v ? t/cm u ??2fx 20 3 17 u ??1.9fx 21 5 16 u ? 1.8fx 22 7 15 u ??1.7fx 23 9 14 u ??1.6fx 24 11 13 2. Investigate any relationship present between the distance between a solar cell and a lamp, ... Power Experiment Bulb Output Solar Cell Output Current (A) Voltage (V) Power (W) Current (mA) 1.94 11.46 22.2324 92 1.73 9.38 16.2274 52 1.5 7.38 11.07 22 1.32 5.98 7.8936 11 1.52 7.78 11.8256 28 1.76 9.59 16.8784 52 1.66 8.63 14.3258 38 1.45 6.94 10.063 19 1.2 5.32 6.384 1. Determine the value of 'g', where 'g' is the acceleration due to gravity. Preliminary Experiment Mass (g) Length of spring (m) Extension (m) Time taken for 10 oscillations (s) Time Period (s) 0 0 0 0 100 0.046 0.023 3.69 0.369 200 0.087 0.064 5.13 0.513 300 0.130 0.107 6.51 0.651 400 0.181 0.149 7.91 0.791 500 0.265 0.193 9.29 0.929 The table above shows the results obtained during the preliminary experiment. 2. Hooke's Law / Young's Modulus - trying to find out what factors effect the ... This shows that Hookes Law does have a limit as a spring does and does have limitations and if kept within these boundaries will provide reliable accurate results. Hooke�s law can also be explained by the molecular structure of the substance the force is acting on: F Sep 1. 1. Stretching Rubber Bands In conclusion the rubber band does behave a lot like the steel spring, in that, the coil of a spring can become elongated because of the stretching force we place on it. However, because a metal's structure is not like a polymer's, this does not happen until you have reached 2. Hooke's Law A second boss clamp will hold in place a 1 metre ruler starting from the bottom of the spring to measure how much it extends in mm. I will then add weights to the spring and measure extension. Before I did the real thing I did a practice experiment to find the elastic limit of the springs we had. 1. Waves and Cosmology - AQA GCE Physics Revision Notes * Quarks are assigned baryon numbers to explain which particle can exist and outcome of interactions. All quarks have a baryon number of 1 / 3, whereas all anti quarks has a baryon number of -1 / 3. * Strangeness explains the behaviour of massive particles such as kaons, they 2. I intend to investigate whether any correlation exists between the wavelength of light exerted ... Instantly this dictates that the investigation can?t have any human input, instead requiring specially configured logging equipment. When evaluating what would be required, there were two routes I could have chosen: The first involved triggering a light source and starting a voltage logger simultaneously. • Over 160,000 pieces of student written work • Annotated by experienced teachers • Ideas and feedback to	2,780	10,978	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2020-50	latest	en	0.935899
https://byjus.com/maths/skip-counting/	1,656,484,698,000,000,000	text/html	crawl-data/CC-MAIN-2022-27/segments/1656103624904.34/warc/CC-MAIN-20220629054527-20220629084527-00621.warc.gz	192,817,494	160,750	Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis : # Skip Counting Skip counting deals with counting by a number, other than 1. For example, skip count by 2, skip count by 3, skip count by 5, 10 or 100. Skip counting is taught to us in our primary classes while learning counting numbers. It helps kids to count the objects quickly. The major application of skip counting is the multiplication tables or multiples of numbers, where we skip count each number to get its multiple. For example, to find the table of 5, we have to skip count by 5 to get its multiples such as 5, 10, 15, 20, 25, 30, and so on. Let us learn here the examples of skip counting. ## What is Skip Counting? Skip counting is a method of counting numbers by adding a number every time to the previous number. For example, skip counting by 2, we get 2, 4, 6, 8, 10, 12 and so on. Hence we get the series of even numbers here. Note: Skip count by any number is possible. It helps in counting the number quickly Skip counting has a huge application in multiplication tables. ## Forward Skip Counting Here, we count the number in the forward direction of a number. It means we skip count for positive numbers. Skip counting has a major application in real life. Suppose we have to count marbles which are in 100s of numbers. Counting one-one marble will consume more time. Hence, if we skip count them by any big number such as 10 or 20, then we can quickly count them. Let us learn here how to skip count by different numbers. ### Skip Counting By 2 Here, we add the number 2 for each counting, in such a manner that the counting is represented in alternate numbers. Suppose we start counting from 2, then, skipping the count by 2 for each step we can write as; 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, …., so on ### Skip Counting By 3 Here, we add the number 3 for each step of counting. 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60 and so on. ### Skip Counting By 4 When we skip count natural numbers by 4 then we add 4 at each step. Here it is how to do: Skip Counting by 4 Result 0+4 4 4+4 8 8+4 12 12+4 16 16+4 20 20+4 24 24+4 28 28+4 32 32+4 36 36+4 40 So we can skip count 4 we got a table of 4. ### Skip Counting By 5 When we count skipping 5 numbers in between, then we skip count by 5. Let us start counting by skipping 5 and write the succeeding numbers. 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, … ### Skip Counting By 10 When we skip by10 in between while counting the numbers, then it is said to skip counting by 10. Skip count by 10 is very easy and quick to do. See the below-given example to understand it better. 0+10 = 10 10+10=20 20+10=30 30+10=40 40+10=50 50+10=60 60+10=70 70+10=80 80+10=90 90+10=100 100+10=110 110+10=120 120+10=130 130+10=140 140+10=150 And so on ## Skip Counting by Bigger Numbers Let us learn here to skip count by some bigger numbers such as 25, 50, 100. ### Skip Counting by 25 When we have hundreds of things to count we can also skip count them by 25. It will help in counting fast. Skip count by 25 is done as: 25, 50, 75, 100, 125, 150, 175, 200, 225, 250, 275, 300, …. You can practice writing more numbers in forward skip counting by 25. ### Skip Counting by 50 Skip counting by 50 means counting the objects after every 50 objects. It can be done as: 50, 100, 150, 200, 250, 300, 350, 400, 450, 500, …. ### Skip Counting By 100 If we have to count objects in a very large number, say 1000, then we can skip count by 100. Let us understand counting by 100 with the help of few examples, given below; 100, 200, 300, 400, 500, 600, 700, 800, 900, 1000, …. In the above counting, you can see the digit at the hundredth place changes after each count. ## Skip Counting Table (2 to 10) Let us prepare a table where we will see the skip counted numbers from 2 to 10. Count By Skip Counting 2’s 2 4 6 8 10 12 14 16 18 20 22 24 3’s 3 6 9 12 15 18 21 24 27 30 33 36 4’s 4 8 12 16 20 24 28 32 36 40 44 48 5’s 5 10 15 20 25 30 35 40 45 50 55 60 6’s 6 12 18 24 30 36 42 48 54 60 66 72 7’s 7 14 21 28 35 42 49 56 63 70 77 84 8’s 8 16 24 32 40 48 56 64 72 80 88 96 9’s 9 18 27 36 45 54 63 72 81 90 99 108 10’s 10 20 30 40 50 60 70 80 90 100 110 120 This table resembles the multiplication table from 2 to 10. ## Backward Skip Counting Students can practice backward counting also, towards negative numbers. Suppose we want to skip count by -2 then, -2, -4, -6, -8, -10, …. Mathematically, it is possible to do, but in real life, we don’t skip count by negative numbers. But for the practice, you can skip count by -3,-4,-5 or any negative number. ## Skip Counting Examples Q.1: If there are 20 marbles, then which of the following will help to count fast? • Skip count by 1 • Skip count by 2 Solution: Skip count by 2 will help to count fast than skip count by 1. It will take half of the time to count the marbles rather than counting one by one. Q.2: If there are 40 balls in a basket, and we need to skip count it by 5. Then how many times do we need to count the balls? Solution: Given, number of balls = 40 If we skip count by 5, then we have, 5, 10, 15, 20, 25, 30, 35, 40 Thus, we will skip count by 5, 8 times here. ### Practise Questions 1. How to skip count by 9? 2. Skip count by 15. 3. Suppose there are 100 marbles. Skip count these marbles by 20. 4. Skip count by -10. ## Frequently Asked Questions – FAQs ### What is skip counting? Skip counting is a method of counting forward by any number apart from 1. If we skip count by a number that means we are adding that number at each step to get another number. ### How do we skip count? Skip counting is a method of counting numbers by skipping them with a certain number. For example, if we skip count by 1 and start counting from 0, then we get a sequence of even numbers such as, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, and so on. ### How to skip count by 10? Skip count by 10 gives a multiplication table of 10 if we start counting from 10. Hence we get: 10 10+10 = 20 20+10 = 30 30+10 = 40 40+10 = 50 50+10 = 60 60+10 = 70 70+10 = 80 80+10 = 90 90+100 = 100 And so on. But remember, if we start counting from 1 and skip count by 10, then we don’t get the table of 10. ### How to skip count by 6? Skip counting by 6 means we have to skip 6 numbers in forward count and jump to next one. 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, …. You can count as far as you want. ### How to skip count by 7? Skip counting by 7 numbers means at each step we skip seven numbers. Hence, if we subtract the succeeding number by preceding one, then we get 7 as the result. It is done as: 7,14,21,28,35,42,49,56,63,… Check: 14-7 = 7 21-14=7 28-21=7 Quiz on Skip Counting	2,262	6,815	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.875	5	CC-MAIN-2022-27	longest	en	0.940004
https://junewalkeronline.com/id/231acd-properties-of-eigenvalues-and-eigenvectors-ppt	1,611,332,345,000,000,000	text/html	crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00164.warc.gz	414,089,853	10,333	# properties of eigenvalues and eigenvectors ppt Solution: Example 5. 1. is diagonalizable. [3]). numerically different eigenvalues. sree2728. Lecture 11: Eigenvalues and Eigenvectors De &nition 11.1. The next matrix R (a reï¬ection and at the same time a permutation) is also special. SECTION 7B Properties of Eigenvalues and Eigenvectors 31st March 08. 4. Control theory, vibration analysis, electric circuits, advanced dynamics and quantum mechanics are just a few of the application areas. We call such a basis an eigenvector basis of . Thus, given a general polynomial p, we can form a matrix A Eigenvalues and Eigenvectors Questions with Solutions Examples and questions on the eigenvalues and eigenvectors of square matrices along with their solutions are presented. The set of all the eigenvalues of . The second postulate states that the possible values of the physical properties are given by the eigenvalues of the operators. n . Permutations have all j jD1. Î» =2, 2 , 3. that we found for the matrix A = â¡ â£ 2 â 36 05 â 6 01 0 â¤ â¦. Yet again . of . Uploaded by. The a. n EIGENVECTORS AND EIGENVALUES Proposition 9.2. Sep 25, 2020 - Properties of Eigenvalues and Eigenvectors: A Review Engineering Mathematics Notes \| EduRev is made by best teachers of Engineering Mathematics . no degeneracy), then its eigenvectors form a `complete setâ of unit vectors (i.e a complete âbasisâ) âProof: M orthonormal vectors must span an M-dimensional space. A . A given nth-degree polynomial p(c) is the characteristic polynomial of some matrix. Key Terms. Interpret the matrix and eigenvalues geometrically. Eigenvalues and Eigenvectors 22.2 Introduction Many applications of matrices in both engineering and science utilize eigenvalues and, sometimes, eigenvectors. Eigenvalue problems. Completeness of Eigenvectors of a Hermitian operator â¢THEOREM: If an operator in an M-dimensional Hilbert space has M distinct eigenvalues (i.e. A. Eigenvalues: Each n x n square matrix has n eigenvalues that are real or complex numbers. If is an ð×ð symmetric matrix, then the following properties are true. The determinant of a triangular matrix is the product of the elements at the diagonal. Step 1: Find the eigenvalues for A. 286 Chapter 6. Arpit Srivastava. Find the eigenvalues of A = [01 â 10]. are often thought of as superpositions of eigenvectors in the appropriate function space. Eigenvalues and Eigenvectors EXAMPLE 1 (continued 5) Determination of Eigenvalues and Eigenvectors 1 1 1 1 1 1 5 2 1 1, Check: ( 1) . Note: Here we have two distinct eigenvalues and three linearly independent eigenvectors. That is, the eigenspace of ð has dimension . Evaluation of Eigenvalues and Eigenvectors Before we discuss methods for computing eigenvalues, we mention an inter-esting observation. WhenAhas eigenvalues 1 and 2 , its inverse has eigenvalues. Viewing the matrix as a linear transformation, the eigenvectors indicate directions of pure stretch and the eigenvalues the degree of stretching. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P. In other words, A is diagonalizable if and only if there are enough eigenvectors to form a basis of . In this lesson we explore the properties of eigenvectors and how to use them to solve a system of linear differential equations. 3. spectrum . Example 4. Theorem Thus, the characteristic equation of A is A 3 4 0 0 3 0 0 0 1 13. InstituteofofScience The eigenvalues of a diagonal or triangular matrix are its diagonal elements. 2 Find the eigenvalues and the eigenvectors of these two matrices: AD 14 23 and ACID 24 24 : ACIhas the eigenvectors asA. Solution: Solve det(AâÎ»I)= 6.3 Finding eigenvectors To find the eigenvectors â¦ This document is highly rated by Computer Science Engineering (CSE) students and has been viewed 4747 times. Eigenvalues and Eigenvectors Projections have D 0 and 1. Hence, in this case there do not exist two linearly independent eigenvectors for the two eigenvalues 1 and 1 since and are not linearly independent for any values of s and t. Symmetric Matrices There is a very important class of matrices called symmetric matrices that have quite nice properties concerning eigenvalues and eigenvectors. Uploaded by. Numercal Anlys & Finit. Eigenvalues, Eigenvectors and Their Uses 1 Introduction 2 De ning Eigenvalues and Eigenvectors 3 Key Properties of Eigenvalues and Eigenvectors 4 Applications of Eigenvalues and Eigenvectors 5 Symmetric Powers of a Symmetric Matrix 6 Some Eigenvalue-Eigenvector Calculations in R James H. Steiger (Vanderbilt University) Eigenvalues, Eigenvectors and Their Uses 2 / 23 The companion matrix of equation (3.177) is one such matrix. If $$A$$ is a square matrix, its eigenvectors $$X$$ satisfy the matrix equation $$AX = \lambda X$$, and the eigenvalues $$\lambda$$ are determined by the characteristic equation Theorem If A is an matrix with , then. All eigenvalues of are real. the three eigenvectors onto a unit vector, v,inthe chosen direction (38, 41). Check the trace! The important properties of a positive semi-deï¬nite matrix is that its eigenvalues are always positive or null, and that its eigen-vectors are pairwise orthogonal when their eigenvalues are differ-ent. First, we need to consider the conditions under which we'll have a steady state. MATH 685/ CSI 700/ OR 682 Lecture Notes Lecture 6. EXAMPLE 2 Example 2: Find the eigenvalues A. Let A be an n â¥ n matrix over a ï¬eld K and assume that all the roots of the charac-teristic polynomial A(X)=det(XIA) of A belong to K. For every eigenvalue i of A, the geometric multiplicity of i is always less than or equal to its If so, the solutions of partial differential equations (e.g., the physics of Maxwell's equations or Schrodinger's equations, etc.) Uploaded by. NationalInstitute D: Eigenvalues and eigenfunctions . Using eigenvalues and eigenvectors to calculate the final values when repeatedly applying a matrix. Vectors that map to their scalar multiples, and the associated scalars In linear algebra, an eigenvector or characteristic vector of a linear transformation is a nonzero vector that changes by a scalar factor when that linear transformation is applied to it. eigenvectors.Forexample,thepowermethod(apartialmethod,seeSection ... Remark5.1 Properties 5.1 and 5.2 do not exclude the possibility that there exist circles containing no eigenvaluesâ¦ $\begingroup$ Are you interested in eigenvalues and eigenvectors in a finite dimensional linear algebra sense? Î». Let A be a square matrix (or linear transformation). Or are infinite dimensional concepts acceptable? Ppt Evaluation - Free download as Powerpoint Presentation (.ppt / .pptx), PDF File (.pdf), Text File (.txt) or view presentation slides online. We note that in the above example the eigenvalues for the matrix are (formally) 2, 2, 2, and 3, the elements along the main diagonal. That is a major theme of this chapter (it is captured in a table at the very end). 2. Properties of Eigen values and Vectors Technology Science&&Technology A square matrix A and its transpose (AT) have the same eigenvalues.However the matrices A and AT will usually have different eigenvectors. Computing Eigenvalues and Eigenvectors Problem Transformations Power Iteration and Variants Other Methods Example: Similarity Transformation From eigenvalues and eigenvectors for previous example, 3 1 1 3 1 1 1 1 = 1 1 1 1 2 0 0 4 and hence 0:5 0:5 0:5 0:5 3 1 1 3 1 1 1 1 = 2 0 0 4 matrix Face Recognition. The largest of the absolute values of the eigenvalues of . The properties of the eigenvalues and their corresponding eigenvectors are also discussed and used in solving questions. Each projection is given by the dot product between the eigenvector and v (an example of a dot product will be shown in Eq. Special properties of a matrix lead to special eigenvalues and eigenvectors. Lecture 13: Eigenvalues and eigenfunctions An operator does not change the âdirectionâ of its eigenvector In quantum mechanics: An operator does not change the state of its eigenvectors (âeigenstatesâ, These special 'eigen-things' are very useful in linear algebra and will let us examine Google's famous PageRank algorithm for presenting web search results. 3 Compute the eigenvalues and eigenvectors ofAandA 1. Eigenvalues and Eigenvectors Matrices: Eigenvalues and Eigenvectors Matrices: Eigenvalues and Eigenvectors A number âis called an eigenvalue of A if there exists a non-zero vector ~u such that corresponding to that eigenvalue . eigenvectors, characteristic vectors . We shall see that the spectrum consists of at least one eigenvalue and at most of . â¦ 2 2 2 2 2 xxO ª º ª ºª º ª º « » « »« » « » ¬ ¼ ¬ ¼¬ ¼ ¬ ¼ x Ax A . Eigenvalue problems Eigenvalue problems occur in many areas of science and engineering, such as structural analysis Eigenvalues are also important in analyzing numerical methods Theory and algorithms apply to complex matrices as well as real matrices With complex matrices, we use conjugate transpose, AH, instead of â¦ The eigenvectors are also composed of real values (these last two properties are a consequence of the symmetry of the matrix, one or more complex scalars called eigenvalues and associated vectors, called eigenvectors. Check these properties for the eigenvalues. If there is no change of value from one month to the next, then the eigenvalue should have value 1. AD 02 11 and A 1 D 1=2 1 1=2 0 : A 1 has the eigenvectors asA. is called the . This document is highly rated by Engineering Mathematics students and has been viewed 695 times. Eigenvectors are special vectors associated with a matrix. Its eigenvalues are by 1. If ð is an eigenvalue of with algebraic multiplicity , then ð has linearly independent eigenvectors. independent eigenvectors of A. This is no accident. of . Eigenvalues and Eigenvectors: Practice Problems. Eigenvectors are particular vectors that are unrotated by a transformation matrix, and eigenvalues are the amount by which the eigenvectors are stretched. Eigenvalues, eigenvectors and applications Dr. D. Sukumar Department of Mathematics Indian Institute of Technology Hyderabad Recent Trends in Applied Sciences with Engineering Applications June 27-29, 2013 Department of Applied Science Government Engineering College,Kozhikode, Kerala Dr. D. Sukumar (IITH) Eigenvalues Nov 21, 2020 - Eigenvalues and Eigenvectors Computer Science Engineering (CSE) Notes \| EduRev is made by best teachers of Computer Science Engineering (CSE). of an operator are deï¬ned as the solutions of the eigenvalue problem: A[u. n (rx)] = a. n. u. n (rx) where n = 1, 2,... indexes the possible solutions.	2,633	10,722	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2021-04	latest	en	0.787338
https://math.stackexchange.com/questions/2869913/use-de-moivre-laplace-to-approximate-1-sum-k-0n-n-choose-k-pk1-p	1,582,020,818,000,000,000	text/html	crawl-data/CC-MAIN-2020-10/segments/1581875143646.38/warc/CC-MAIN-20200218085715-20200218115715-00442.warc.gz	465,611,648	37,789	# Use De Moivre–Laplace to approximate $1 - \sum_{k=0}^{n} {n \choose k} p^{k}(1-p)^{n-k} \log\left(1+\left(\frac{p}{1-p}\right)^{n-2k}\right)$ I am trying to use De Moivre–Laplace theorem to approximate $$1 - \sum_{k=0}^{n} {n \choose k} p^{k}(1-p)^{n-k} \log\left(1+\left(\frac{p}{1-p}\right)^{n-2k}\right)$$ The idea of an approximation is that we don't have the sum term which is difficult to calculate if $n$ is high. Using the De Moivre–Laplace theorem gets us that: $${n \choose k} p^{k}(1-p)^{n-k} \approx \frac{1}{\sqrt{2 \pi np(1-p)}}e^{-\frac{(k-np)^2}{2np(1-p)}}$$ Now we see that \begin{align} F &= 1 - \sum_{k=0}^{n} {n \choose k} p^{k}(1-p)^{n-k} \log\left(1+\left(\frac{p}{1-p}\right)^{n-2k}\right) \\&\approx 1 - \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi np(1-p)}}e^{-\frac{(x-np)^2}{2np(1-p)}}\log_2\left(1+\left(\frac{p}{1-p}\right)^{n-2x}\right) dx \end{align} my calculation is inspired by Entropy of a binomial distribution If one has an other suggestion to approximate $F$ or get a closed for i would like to hear those. So far i've tried approximating $F$ with a least squares method using a tanh function as the fit function. • Missed $dp$ in the integral – Yuri Negometyanov Aug 4 '18 at 20:22 • I didnt get it. There is something seriously wrong here. The sum formula is eventually independent of $k$ but dependent on $p$ and the final integral is independent of $p$ but dependent on $k$. What does $k$ mean after the integration? sounds strange right? – Seyhmus Güngören Aug 4 '18 at 21:06 • I guess $k$ is eventually $x$ and the integration over $p$ is probably a typo. I do not get the limit $-\infty..+\infty$ though when originally we have $0..n$. – Diger Aug 4 '18 at 21:12 • Actually it doesnt matter. So for $p<0.5$ use the approximation $\log(1+y)\approx y$ and for large $y$, we have $\log(1+y)\approx \log(y)$. One more thing you have $y=a^{(f(x))}$ and you can write this as $\exp(a,f(x))$, then in the integral you have the multiplication of two exponentials. I think these forms are well studied. – Seyhmus Güngören Aug 4 '18 at 22:07 • take $p$ small enough and $n$ large enough. I think it should work. Btw. It will be very good, if you could put the results that you already have to the question description. Plus: the answer of wolfram.alpha for the evaluation of this integral, just the link. – Seyhmus Güngören Aug 4 '18 at 22:28 $$\color{brown}{\textbf{Transformations}}$$ Let WLOG the inequality $$q=\dfrac p{1-p}\in(0,1)\tag1$$ is valid. Otherwise, the corresponding opposite events can be reversed. This allows to present the issue expression in the form of \begin{align} &S(n,p)=1 - (1-p)^n\sum_{k=0}^{n} {n \choose k} q^k\log\left(1+q^{n-2k}\right),\tag2\\[4pt] \end{align} or \begin{align} &=1 - (1-p)^n\sum_{k=0}^{n} {n \choose k}q^kq^{n-2k} - (1-p)^n\sum_{k=0}^{n} {n \choose k}q^k\left(\log\left(1+q^{n-2k}\right)-q^{n-2k}\right)\\[4pt] &=1 - (1-p)^n(1+q)^n - (1-p)^n\sum_{k=0}^{n} {n \choose k}q^k\left(\log\left(1+q^{n-2k}\right)-q^{n-2k}\right)\\[4pt] &S(n,p)= - (1-p)^n\sum_{k=0}^{n} {n \choose k}q^k\left(\log\left(1+q^{n-2k}\right)-q^{n-2k}\right).\tag3\\[4pt] \end{align} Formula $(3)$ can simplify the calculations, because it does not contain the difference of the closed values. $$\color{brown}{\textbf{How to calculate this.}}$$ Note that the sum of $(3)$ contains both the positive and the negative degrees of $q.$ This means that in the case $n\to \infty$ the sum contains the terms of the different scale. The calculations in the formula $(3)$ can be divided on the two parts. $\color{green}{\textbf{The Maclaurin series.}}$ The Maclaurin series for the logarithmic part converges when the term $\mathbf{\color{blue}{q^{n-2k} < 1}}.$ This corresponds with the values $k<\frac n2$ in the case $\mathbf{q<1}$ and with the values $k>\frac n2$ in the case $\mathbf{q>1}.$ Then the Maclaurin series in the form of $$\log(1+q^{n-2k}) = \sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}q^{(2n-k)i}\tag4$$ can be used. If $\mathbf{\color{blue}{q^{n-2k} > 1}},$ then $$\log(1+q^{n-2k}) = \log(q^{2n-k}(1+q^{k-2n})) = (2n-k)\log q + \log(1+q^{k-2n}).\tag5$$ If $\mathbf{\color{blue}{q^{n-2k} = 1}},$ then $LHS(4) = \log2.$ If $\mathbf{\color{blue}{q^{n-2k} \lesssim 1}},$ then $$\log(1+q^{2n-k}) = \log\frac{1+r}{1-r} = 2r\sum_{i=0}^\infty\frac{(-1)^i}{2i+1}r^{2i},\quad \text{ where } r=\frac{q^{2n-k}}{2+q^{2n-k}}\approx\frac{q^{2n-k}}3,\tag6$$ and can be used some terms of the series. $\color{green}{\textbf{The double summations.}}$ After the substitution of the $(4)$ or $(5)$ to $(3)$ the sums can be rearranged. For example, $$\sum_{k=0}^{L}{n \choose k}q^k\sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}q^{(2n-k)i}= \sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}\sum_{k=0}^{L}{n \choose k}q^kq^{(2n-k)i}$$ $$= q^{n+1}\sum_{i=1}^\infty\frac{(-1)^{i+1}}{i}\sum_{k=0}^{L}{n \choose k}\left(q^{i+1}\right)^{n-k},$$ wherein the order of the summation can be chosen, taking in account the given data. • hmmmm, i am quite curious now. Does the limit converge to 0 as $n \rightarrow \infty?$ – Kees Til Aug 11 '18 at 0:15 • @KeesTil If $k<\frac n2,$ then we have suitable Maclaurin series. If $k>\frac n2,$ then the factor $(1-p)^n$ provides the convergence. – Yuri Negometyanov Aug 11 '18 at 0:37 • @KeesTil Yes, that's valid. Fortunately, $p$ and $1-p$ are the possibilities of the opposite events, whose designations can be changed for this task. – Yuri Negometyanov Aug 11 '18 at 19:10 • @KeesTil I see that $S(p)=S(1-p).$ Please check this. Thanks for the useful comments. – Yuri Negometyanov Aug 11 '18 at 19:29 • I don't get the last sentence where you day that the inner sums can be calculated via the de Moivre Laplace theorem. How can we do this, $q+1 \neq 1$. I use this definition btw: en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem – Kees Til Aug 12 '18 at 10:33 The expression looks very much like the Bernstein approximation of a function ($1-f(x)$) on $[0,1]$. But the argument (in fact the degree $n-2k$) of $log$ function ruins everything. Here is a quick idea. Denote $$y(p)=\sum_{k=0}^{n} {n \choose k} p^{k}(1-p)^{n-k} \log\left(1+\left(\frac{p}{1-p}\right)^{n-2k}\right).$$ Let us assume that we can represent $y(p)$ in the form $y(p)=\sum_{m=0}^\infty y_m p^m$, where $y_m$ are constants not depending on $p$. Note that $y(p)=y(1-p)$. Let us consider the equation $$y(p)+y(1-p)=f(p). \tag{eq1}\label{eq1}$$ Although we can write out the expression for $f(p)$, let us think that we don't know how $f(p)$ looks like. But for sure, $f(p)$ must satisfy $f(p)=f(1-p)$. It is know (see for example http://eqworld.ipmnet.ru/en/solutions/fe/fe1116.pdf) that equations like \eqref{eq1} have a solution, for example, $$\tag{eq2}\label{eq2} y(p)=f(p) \sin^2({\pi p \over 2}).$$ By expanding $\sin^2({\pi p \over 2})$ into the Maclaurin series we get $$y(p)=f(p) \sum_{m=1}^\infty {(-1)^{m+1} 2^{2m-1} \over (2m)!} {p^{2m} \pi^{2m} \over 2^{2m}}.$$ Let us assume that $f(p)$ is an analytic function i.e. $f(p)=\sum_{m=0}^\infty {f^{(m)}(0)\over m!} p^m$. By writing \eqref{eq2} in the series form we have: $$\sum_{m=0}^\infty y_m p^m = \left ( \sum_{m=0}^\infty {f^{(m)}(0)\over m!} p^m \right ) \left ( \sum_{m=1}^\infty {(-1)^{m+1} \over (2m)!} {p^{2m} \pi^{2m} \over 2} \right ).$$ From this relation it may be possible to find the expressions for $f^{(m)}(0)$ through $y_m$ by equating the coefficients at $p^m$. If this works out, we go back to the right part of \eqref{eq2} and try to find how many terms in the product $$\left ( f^{(0)}(0) + f^{(1)}(0) p + f^{(2)}(0) {p \over 2} + \dots \right ) \left ( \sum_{m=1}^\infty {(-1)^{m+1} \over (2m)!} {p^{2m} \pi^{2m} \over 2} \right ).$$ yield the approximate value. • so i can't use this because of the "n-2k" term? – Kees Til Aug 7 '18 at 18:07 • It is impossible (at least I don't see how) to use Bernstein approximation because of the (n-2k) degree. But the idea above is the one you can try. – rrv Aug 8 '18 at 5:21 • I tried this method, however my terms were very complicated and i could not see something that was simplified. My Mathemathica code: A = FullSimplify[ Series[Sum[ Binomial[n, k]p^k (1 - p)^(n - k) Log[2, 1 + (p/(1 - p))^(n - 2 k)], {k, 0, n}] , {p, 0, 2}]] B = FullSimplify[ Series[Sum[((-1)^(m + 1))/2 m!(p^(2m)Pi^(2*m)/(2)), {m, 1, Infinity}], {p, 0, 2}]] – Kees Til Aug 8 '18 at 9:30 We can apply a method similar to this. Since the summand has a sharp peak around $k = n/2$, we can take an expansion valid for large $n$ and for $k$ close to $n/2$ and then, also due to the tails being small, extend the summation range indefinitely: $$a_k = \binom n k p^k q^{n - k} \ln \left( 1 + \left( \frac p q \right)^{n - 2 k} \right), \quad q = 1 - p, \\ a_{n/2 + i} \sim \sqrt {\frac 2 {\pi n}} \left( 2 \sqrt {p q} \right)^n \left( \frac p q \right)^i \ln \left( 1 + \left( \frac q p \right) ^{2 i} \right), \\ \sum_{k = 0}^n a_k \sim \sqrt {\frac 2 {\pi n}} \left( 2 \sqrt {p q} \right)^n \sum_{i = -\infty}^\infty \left( \frac p q \right)^i \ln \left( 1 + \left( \frac q p \right) ^{2 i} \right), \\ n \to \infty, p \text{ fixed}, 0 < p < 1, p \neq \frac 1 2.$$ • i don't really get it right now, but you now have a summation term from $-\infty$ to $\infty$, does that not make the computation expensive? – Kees Til Aug 11 '18 at 18:08 • Depends on what your goal is. If $p$ is fixed, the sum over $i$ is a constant. – Maxim Aug 11 '18 at 18:17 • can you write out the sum for fixed $p$, i don't see the solution directly – Kees Til Aug 11 '18 at 18:30 • Do you mean a closed form? This infinite sum probably doesn't have one. – Maxim Aug 11 '18 at 18:45 • too bad my p is not fixed so i dont think this will work for me :( – Kees Til Aug 11 '18 at 18:53	3,715	9,719	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 3, "equation": 0, "x-ck12": 0, "texerror": 0}	4.09375	4	CC-MAIN-2020-10	latest	en	0.849103
https://www.ukessays.com/essays/mathematics/applications-of-mathematics-in-real-life-situations.php	1,623,659,798,000,000,000	text/html	crawl-data/CC-MAIN-2021-25/segments/1623487611641.26/warc/CC-MAIN-20210614074543-20210614104543-00285.warc.gz	966,162,448	15,994	Disclaimer: This is an example of a student written essay. Any opinions, findings, conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of UKEssays.com. # Applications of Mathematics in Real Life Info: 2547 words (10 pages) Essay Published: 2nd Oct 2017 1.0 Application of Matrices Matrix concepts can be applied in various fields such as: • Quantum Mechanics • 3D Games • Animations • Cryptography and Others. We shall analyze the last one which is Encryption in further details. Encryption is indeed an important measure of security when there are transactions of data between parties. Firstly, we shall make use of the AB = X and B = A-1X concept, where the matrix A is the Encoder and the inverse of A is the Decoder. As messages are mainly sent in letters, we have a set a relationship between letters from the message and numbers in our matrix. For our example we will take the following relationship. A B C D E F G H I J K L M N 2 3 4 5 6 7 8 9 10 11 12 13 14 15 O P Q R S T U V W X Y Z Space 16 17 18 19 20 21 22 23 24 25 26 1 0 Now let’s encode our message which is “BAD”. Referring to our relationship table, it becomes “3, 2, 5”. We decide for a coding matrix A, which can be 5 2 4 6 1 0 2 7 2 As it is a 3*3 matrix, we can encode 3 numbers at a time. Encrypting the first 3 numbers, 10, 0, 13 (Matrix B) using matrix multiplications: 5 2 4 3 39 6 1 0 2 = 20 2 7 2 5 30 39, 20, 30 becomes the Encrypted Message which should be decoded now using inverse of matrix A. Inverse of A = 1/73 12/73 -2/73 39 3 -6/73 1/73 12/73 20 = 2 20/73 -31/146 -7/146 30 5 Decoding the message gives us back 3, 2, 5 which can be referred back to the relationship table to get the message sent. Note: For this example • We need to have a matrix A which does have an inverse, we need to cater for the blank – space hence we can allocate a “0” for it • The coding matrix as well as the number letter relationship is independent that is for some other encryption A can be 3 while B is 4 and so on. Conclusion: Matrices do play a major role in not only Encryption of Messages to avoid misuse of data, but in other fields mentioned above as well. ## 2.0 Application of Statistics We live in the Information Age…” is a common saying in today’s world. It is a true fact as in our era as we make use of information in every field to be able to get an idea of what is actually happening and what we can do to not only to reduce problems of the past but devise ways and techniques for much proper less time consuming, less complicated, less costly and more beneficial processes to obtain output Statistics are the ways we can achieve the above by manipulating the data in their own ways to obtain a set of conclusions which will help us take many crucial decisions. We can also agree to the fact that indeed statistics are being used in many vast and important fields which will be illustrated below: 1. Weather Forecast/Emergencies Precautions We make use of statistics to a very great extent in weather forecasting. This is so as almost every forecasting is based on data and information gathered from previous ones and in addition with other related data. For example: In reference to the amount of rainfall or cyclones obtained from previous years, we can have a close overview for the current year and hence take necessary precautions 1. Medical Studies/Prediction of Diseases/Genetics Statistics have a huge role in medical fields nowadays. By referring to data and information on other past patients, we can learn more and extract new remedies and treatment not only to make the medical field prospect but help needy peoples. Also, using genetic data, some parents might come to know well before the coming problems with their expected children hence can prevent the possible problems accordingly. For example: Using information recorded from past patients suffering from a particular disease in a particular season, expected patients with the same disease can be protected using vaccinations well before. 1. Politics To achieve a well organized political structure for the country’s well being, the potential people needs to be elected and based on their contribution, their work and their reputation on previous data, the obtain their chances to be candidates in next elections. 1. Consumer Goods/Stock Market/Quality Testing Many wholesalers, retailers and even small scaled businessmen nowadays do keep tracks of both their purchases and sales which constantly being referred to be able to extract many important conclusions for better decision making and productivity. For example: If a certain commodity’s sales rises during a particular festive season, the businessman will know from statistics that he should have the item in stock for his own profit. In addition, it is crucial acknowledge what is happening in the economy of the country, hence statistics is a blessing here also as it gives us feedback and predictions for the future. Companies too use statistics to test whether their products are as per their customer’s wants and needs. To do so, they make use of product batches and hence it is time consuming for them to check for all the products. Our academic experts are ready and waiting to assist with any writing project you may have. From simple essay plans, through to full dissertations, you can guarantee we have a service perfectly matched to your needs. Conclusion: In the light of the above, we have seen that statistics are merely predictions; hence we cannot rely completely on statistics. However, they allow us to have better overviews of what is expected and hence we prepare accordingly to eliminate errors and mistakes. ## 3.0 Application of Regression and Correlation Regression and Correlation is used with sets of data, most commonly 2 sets of data to conclude about one main point, the relationship between the data. Regression deals mainly with the graphs of best fits for the data to be able to obtain the Correlation between them. Example: The Correlation: Positive Linear Correlation The study also includes the Correlation degree or measure namely the Pearson Product Moment Correlation Coefficient, which lies between -1 and 1. Regression and Correlation is used in everyday life in various situations namely: Firstly, to compare the previous sale figures such as to have better understandings for future sales. Moreover, companies are able to see how the varied prices on commodities have an effect on sales and clients requirements. Also, also the regression predictions will allow the companies to eliminate future problems and risk hence obtaining better business models with proper decision makings. 1. In Regression Testing As defined by the Internet, Regression Testing is to verify that modified code does not break the existing functionality of the application and works within the requirements of the system. Therefore, it is a much easier and quicker way to find mistakes in systems to be able to implement new designed and modified ones. 1. In Medical Fields According to Paul I. M. Schmitz (1970-1986), in the biomedical field, data in binary form such as disease/no disease or survival/death are very common. In these applications a multivariate normal distribution for the x-variables in both the disease group and the non-disease group was assumed. In pregnancy cases, the IQ of the unborn child is mainly based on predictions from regressions. Hence it is clear that regression played an important part in medical fields. 1. In Education / Candidates Selection / World War Many institutions such as Harvard use regression models to be able to select the students that are eligible. Students also turn to counselors which uses the same techniques to predict the best school for the latter. For selecting best candidates for employment, companies do make use of regression methods. In the same way, during world wars, regression made it simple to pick the most capable soldiers to increase their winning chances. Conclusion: Like Statistics, Regression is of great importance when it comes to have predictions for the future to be able to make better decisions. ## 4.0 References Matrix Encryption [Online] Available from http://www.austincc.edu/lrosen/1314/webact2/webact2.htm [Accessed: 26th – 29th November 2014]. Maths Worksheet Center [Online] Available from http://www.mathworksheetscenter.com/mathtips/statsareimportant.html [Accessed: 30th November – 3rd December 2014] What Are Some Ways Linear Regression Can Be Applied in Business Settings? [Online] Available from http://smallbusiness.chron.com/ways-linear-regression-can-applied-business-settings-35431.html [Accessed: 26th – 05th November 2014]. Pearson’s Product Moment Correlation Coefficient [Online] Available from http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient [Accessed: 5th November 2014]. Numbers Numerical Methods For Bioscience Students [Online] Available from http://web.anglia.ac.uk/numbers/graphsCharts.html [Accessed: 7th December 2014] Testing Basic Interview Questions [Online] Available from http://testingbasicinterviewquestions.blogspot.com/2012/01/what-is-regression-testing-explain-it.html [Accessed: 7th December 2014] Regression Models [Online] Available from http://www.psychstat.missouristate.edu/introbook/sbk16m.htm [Accessed: 7th December 2014] Paul I. M. Schmitz. (1970-1986) Developments In Logistic Regression Methodology. 1.p.2.1 [Online] Available from file:///C:/Users/User/Downloads/860423_SCHMITZ,%20Paulus%20Ignatius%20Maria.pdf [Accessed: 7th December 2014] View all ## DMCA / Removal Request If you are the original writer of this essay and no longer wish to have your work published on UKEssays.com then please: Related Services	2,207	9,865	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.3125	4	CC-MAIN-2021-25	latest	en	0.891232
https://www.mapleprimes.com/users/mmcdara/answers	1,701,875,790,000,000,000	text/html	crawl-data/CC-MAIN-2023-50/segments/1700679100599.20/warc/CC-MAIN-20231206130723-20231206160723-00836.warc.gz	965,129,627	425,495	## 17 Badges 7 years, 185 days ## Three ways... Leading idea: find the zeroes of diff(s, phi): > restart; W:=fmuM*2(1-sqrt(1+2phi/(muM*2)))+(1-f)(1+3beta-(1+3beta-3betaphi+betaphi2)exp(phi))+gnuM*2(1-sqrt(1-2phi/(nuM*2)))+(1-g)(1+3beta-(1+3beta+3betaphi+betaphi2)exp(-phi)): s:= subs(g=0.95,mu=1,beta=0.19,nu=0.5,M=2.277, f=0.4490, W): plot(s,phi=-2.3..0.6); # As it is seen for beta=0.19 and f=0.449 we have three consecutive local extrema. How can I find these critical values of (beta,f)? > ds := diff(s, phi): omega := -2.3..0.6: plot(ds, phi=omega); > # Using solve: one solution is missing fnormal~({solve(ds)}); Locations := select((x -> verify(x, omega, 'interval')), %); (1) > # Using fsolve: the method is governed by what you see on the plot above loc_1 := fsolve(ds, phi=omega); loc_2 := fsolve(ds, phi=loc_1..op(2, omega)); loc_3 := fsolve(ds, phi=op(1, omega)..loc_1); loc_4 := fsolve(ds, phi=op(1, omega)..loc_3-1e-6); Locations := fnormal~({loc_1, loc_2, loc_3, loc_4}) (2) > # Using RootFinding:-NextZero: fds := unapply(ds, phi): Locations := [ RootFinding:-NextZero(phi -> fds(phi), -2.30) ]; for i from 1 to 3 do Locations := [ Locations[], RootFinding:-NextZero(phi -> fds(phi), Locations[-1]) ]; end do: Locations := fnormal~(Locations) (3) > Download plot_mmcdara.mw ## Suggestion: try a numerical approach... dsolve does not return a formal solution. I don't lnow if it is possible to get one but I suggest you to dsolve numerically. At the end of the attached fie you will find an attempt to solve formally using the option type='series'. Nevertheless this seems a dead end unless you use a very large number of terms in the series expansion. > > > > (1) > # Solve numerically: soln := dsolve({eq1, eq2, eq3, ic1, ic2, ic3}, numeric) (2) > pn := plots:-display( plots:-odeplot(soln, [t, U(t)], t=0..10, color=red , legend=typeset(U(t))), plots:-odeplot(soln, [t, V(t)], t=0..10, color=blue , legend=typeset(V(t))), plots:-odeplot(soln, [t, w(t)], t=0..10, color=green, legend=typeset(w(t))) ): display(pn) > Order := 20: sols:= dsolve({eq1, eq2, eq3, ic1, ic2, ic3}, {U(t), V(t), w(t)}, type='series'): sols := map(convert, sols, polynom): (3) > ps := plots:-display( plot(eval(U(t), sols), t=0..10, color=red , style=point, symbol=circle), plot(eval(V(t), sols), t=0..10, color=blue , style=point, symbol=circle), plot(eval(w(t), sols), t=0..10, color=green, style=point, symbol=circle) ): display(pn, ps, view=[default, 0..4]) Do_It_Numerically.mw ## I believe this can be done more simply... There is no need to build the Rough (Rough-Hurwitz) matrix for a polynomial of order 3. The conditions for its 3 roots have a strictly negative real parts are very simple. I propose you do the things this way > restart > with(LinearAlgebra): > J := Matrix(3, 3, [[0, 0, -m+Rstar/m], [xi, -1, m-Rstar/m], [0, 1, -m]]) (1) > char_poly := CharacteristicPolynomial(J, lambda) (2) > # Coefficients of lambda in char_poly sorted by decreasing powers of lambda c := [coeffs(char_poly, lambda, 'p')]; sort([p]); c := c[sort([p], output=permutation)] (3) > # Condiitons for char_poly to be a Hurwitz polynomial HurwitzConditions := { (c >~ 0)[], c[2]c[3] - c[1]c[4] > 0}; (4) > # char_poly is a Hurwitz polynomial iif sol := {solve(HurwitzConditions)}: print~(%): (5) > # Check that random choices of (m, xi, Rstar) which verifie sol[2] lead # to characteristic polynomials whose roots have a negative real part. ind := indets(sol): K := 1: sol[K]; V := {m = rand(0.0 .. 1.)()}: W := select(has, eval(sol[K], V), ind): V := V union select(type, W, `=`): W := select(has, eval(sol[K], V), ind): V := V union {Rstar = rand(lhs(W[1])..2lhs(W[1]))()}; eval(char_poly, V): evalf([solve(%)]): Re~(%); (6) > # Check that random choices of (m, xi, Rstar) which verifie sol[2] lead # to characteristic polynomials whose roots have a negative real part. ind := indets(sol): K := 2: sol[K]; V := {m = rand(0.0 .. 1.)(), xi = rand(0.0 .. 1.)()}: W := select(has, eval(sol[K], %), ind): V := V union {Rstar = rand(lhs(W[1])..rhs(W[2]))()}; eval(char_poly, V): evalf([solve(%)]): Re~(%); (7) > # Check that random choices of (m, xi, Rstar) which verifie sol[3] lead # to characteristic polynomials whose roots have a negative real part. ind := indets(sol): K := 3: sol[K]; V := {m = rand(0.0 .. 1.)()}: W := select(has, eval(sol[K], V), ind): V := V union {xi = rand(2rhs(W[3])..rhs(W[3]))()}: W := select(has, eval(sol[K], V), ind): V := V union {Rstar = rand(lhs(W[1])..rhs(W[2]))()}; eval(char_poly, V): evalf([solve(%)]): Re~(%); (8) > # Check that random choices of (m, xi, Rstar) which verifie sol[4] lead # to characteristic polynomials whose roots have a negative real part. ind := indets(sol): K := 4: sol[K]; V := {m = rand(0.0 .. 1.)()}: W := select(has, eval(sol[K], V), ind): V := V union {xi = rand(lhs(W[1])..rhs(W[3]))()}: W := select(has, eval(sol[K], V), ind): V := V union {Rstar = rand(lhs(W[1])..2lhs(W[1]))()}; eval(char_poly, V): evalf([solve(%)]): Re~(%); (9) > # Check that random choices of (m, xi, Rstar) which verifie sol[5] lead # to characteristic polynomials whose roots have a negative real part. ind := indets(sol): K := 5: sol[K]; V := {m = rand(-1.0 .. 0.)()}; W := select(has, eval(sol[K], V), ind); V := V union {xi = rand(lhs(W[1])..rhs(W[3]))()}; W := select(has, eval(sol[K], V), ind); V := V union {Rstar = rand(2rhs(W[1])..rhs(W[1]))()}; eval(char_poly, V): evalf([solve(%)]): Re~(%); (10) Download Hurwitz_mmcdara.mw ## One way... ```restart: lst := [24300, 18907875, 151200, 147000]: ilst := map(ifactor, lst); [ 2 5 2 2 3 5 5 3 2 3 3 2] [(2) (3) (5) , (3) (5) (7) , (2) (3) (5) (7), (2) (3) (5) (7) ] xlst := eval(ilst, ``(2)=``(x)) [ 2 5 2 2 3 5 5 3 2 3 3 2] [(x) (3) (5) , (3) (5) (7) , (x) (3) (5) (7), (x) (3) (5) (7) ] ``` One_way.mw To understand why this works use lprint: ```lprint(ilst[1]) ``(2)^2``(3)^5``(5)^2 ``` As (2)^p is ``(2)^p `` prevents the evaluation), just replace ``(2) by ``(x) ( ``(x) will preserve the representation (x)^p) To retrieve the numbers: ```expand(xlst) [ 2 5 3] [6075 x , 18907875, 4725 x , 18375 x ] ``` ## Check your indices... Three errors: 1. You define delta(k) at the top of your worksheet but use delta[k] further on. 2. You use sum instead of add. 3. You did some mistake in the indices of Theta in the loop. > > > > > > > # For k=0 the computation of T2 is responsible of the error. # When r=0 you invoke Theta[r+2]= Theta[2]=Theta[k+2] (being processed) for k=2. # # Check your indices. for k from 0 to 7 do Theta_Indices_in_T1 := seq([k-r+1, r+1], r = 0 .. k); T1 := betaadd((k-r+1)Theta[k-r+1](r+1)Theta[r+1], r = 0 .. k); Theta_Indices_in_T2 := seq([k-r, r+2], r = 0 .. k); T2 := betaadd(Theta[k-r](r+1)(r+2)Theta[r+2], r = 0 .. k); T3 := betaadd(add(Theta[r-m](k-r+1)Theta[k-r+1]/(1+delta[m-1]), m = 0 .. r), r = 0 .. k); T4 :=add((k-r+1)Theta[k-r+1]/(1+delta[r-1]), r = 0 .. k)-2lambdaTheta[k]; Theta[k+2] := -1/((k+1)(k+2))(T1+T2+T3+T4) end do; > Download AF_mmcdara.mw My advice: Verify the expressions of Theta[k+2] ## Non convergent Taylor expansion in the r... The radius of convergence of function tanh being Pi/2, there necessarily exists a finite range for xx and yy outside of which the Taylor expansion doesn't converge whatever the nuimber of terms you use. This is reason why your plot (and @dharr's) exhibit sush large values. More of this, given that the range for yy where h (@dharr's g) has significant variations (obviously has values in (0, 1); along xx the variations of h are smaller) is 0.5512527188e-1 .. 0.5569069537e-1 (definition given in the file [BL, BR]), you can simplify f by taking only a 6 terms, not the one hundred you manipulate. All the details are in the attached file. So, to answeer your question "Why Taylor series can not estimate my function in desired interval [-1<x,y<1]?" " Because your are not in the range where this expansion converges" The convergence range is `yy = 0.55291125973999346567e-1 .. 0.55524841270134109157e-1` The xx convergence range is likely larger for the reason I gave below (slow variations in the xx direction). A rough estimation is ```fsolve(g(eval(f0, yy=AT))=0.001) .. fsolve(g(eval(f0, yy=AT))=0.999); -3.3962547 .. 0.1126052 ``` > > with(Student[MultivariateCalculus]): > > # Preliminaries f0 := evalf(f): g := phi -> .5(1+tanh(phi)): # note that 0 <= g(..) <= 1 Tf := (n, p, q) -> mtaylor(f0, [xx=p, yy=q], n); Tg := (m, n, p, q) -> mtaylor(g(Tf(n, p, q)), [xx=p, yy=q], m); # examples Tf(5, 0, 0): Tg(4, 5, 0, 0): (1) > # Observe expression g(f0)): # Observation 1: # # As the expression below is numerical equal to 0, and because the min value # of g(any_function) is 0 too, xx=yy=0 is already extremely far from the # region where g(..) takes values significantly larger than 0 and lower then 1. eval(f0, [xx=0, yy=0]); g(%); (2) > # Observation 2: # # Can we get an approximation of where the previous region takes place? # To do this we search(arbitrarily on the direction and get BL := fsolve(eval(g(f0), yy=xx)=0.001); BR := fsolve(eval(g(f0), yy=xx)=0.999); plot3d( g(f0) , xx=BL..BR, yy=BL..BR , color = blue , style = surface ); > # Consider now the taylor expansion of g(f0) around xx = yy = AT # where AT stands for (BL+BR)/2). # # Comparing the values of g(g0) and TG(5, AT, AT) shows a good agreement. AT := (BL+BR)/2; eval(g(f0), [xx=AT, yy=AT]); Tg(5, (BL+BR)/2, (BL+BR)/2): eval(%, [xx=AT, yy=AT]); (3) > # Before computing the Taylor expansion of g(f0), let us # begin observing how close could be g(f0) and g(Tf(n, AT, AT)). # # Taking arbitrarily n=2 already gives a good agreement. # For more security I will nevertheless use the value n = 3. plot3d( [g(f0), g(Tf(3, AT, AT))] , xx=BL..BR, yy=BL..BR , color = [blue, red] , style = [surface\$2] ); > # Finally ask yourself theis simple question: # "Can the Taylor expansion of tanh(x) around x=0 be reasonably acute # in the range, let's say -4..4, where on considers than tanh(x) # takes values in its almost complete rnge of variation 1..1?@ # # Let us see: plot( [tanh(x), seq(mtaylor(tanh(x), x, k), k=4..16, 4)] , x=-4..4 , color=[blue, red, orange, cyan, green] , legend=[typeset(tanh(x)), seq(cat("k=", k), k=4..16, 4)] , view=[default, -2..2] , gridlines=true ) > # The conclusion is obvious: as no Taylor expansion of tanh(x) gives # an acceptable representation of tanh(x) outside of a limited range # (here around -1..1), there is absolutely no chance at all that # a Taylor expansion og g(f0) will give you a reliable representation # in a quite limited range around yy=AT (as you see the variation of # g(f0) is far more rapid along yy than along xx). # Defining the range where tanh(x) has significant variations this way SignificantRange := fsolve(tanh(x)=-0.999)..fsolve(tanh(x)=0.999); # one observes this range extends far from 0 for about 2.4 times the radius of # convergence of its Taylor expansion. # Converting this radius of convergence in terms of BR and BL then gives: xi := (Pi/2) / op(2, SignificantRange); # and then ConvergenceRange := AT-(AT-BL)xi .. AT+(BR-AT)xi; g35f3 := Tg(35, 3, AT, AT): plot3d( [g(f0), g35f3] , xx=ConvergenceRange, yy=ConvergenceRange , color = [blue, red] , style = [surface\$2] ); > # But if push further out from this range PossibleRange := AT-(AT-BL)(xi1.1) .. AT+(BR-AT)(xi1.1); plot3d( [g(f0), g35f3] , xx=PossibleRange, yy=PossibleRange , color = [blue, red] , style = [surface\$2] ); Download taylorProblem_mmcdara.mw ## Generally speaking things are done in th... Generally speaking things are done in the other way: starting from an ODE/PDE one derive a numerical scheme which is convergent and consistent with this equation (a fex other properties are required). The only situation where we start from a discrete equaation (or a set of) and want to buid a continuous ODE/PDE is when you want to answer this question "I have a continuous ODE/PDE C and its discrete version D, I know that D doesn't represents exactly C and because of some truncation error I would like to tknow what is the continuous ODE/PDE Ctrue that D represents withe a null truncation error See here for a Maple example and the references within. If you are in this framework please let me know. Otherwise here is a step by step explanation on how to find a continuous PDE that z11 would discretize: (do not use "h" and "1" to denote the same space step, firstly it's confusing, and secondly one can get nothing wifh that: use "h" only) > > > > > > (1) > > > > > > > (2) > R0, NoR0 := selectremove(has, [op(expand(lhs(z11)))], r); R0, NoR0 := map(add, [R0, NoR0])[] (3) > R0 := factor(R0) (4) > # Remark: # # The notation f(i+(1/2)h, t) is commonly ised to represent symbolically the # mean value of fome function f(x, t) at time t within the interval [i, i+h]. # # More precisely, in discretization scheme f(i+(1/2)h, t) is th approximation # of int(r(x, t), x=i..i+h) got by the mid-point rule. # # This strategy, when it comes to a function representing a mass is named # "mass lumping". This name is used whatever the function f represents # # Thus here MassLumping := r(i+(1/2)h, t) = ``(r(i+h, t)+r(i, t))/2 (5) > # Generally one starts from the continuous equation and then derive a discretization # scheme from it. # For instance, assuming is mass-liming is used, a term such as r(x, t)diff(r(x, t), x) # could be discretized as: r(x, t)diff(r(x, t), x) = r(i+(1/2)h, t) . ``((r(i+h, t)-r(i, t)) / h); (6) > # Now let's try to buid the continuous PDE from its discretzation # ASSUMING THE MASS-LUMPING STRATEGY IS USED # # Step 1 , for convenience set: I/h = C R0 := algsubs(I/h = C, R0); (7) > # Set 2: FromMassLumping := 2~(rhs=lhs)(value(MassLumping)); R := subsop(2=rhs(FromMassLumping), R0) (8) > # Set 3: identify the second term in R using a 2nd order Taylor expansion MyRule := op(2, R) = convert(convert(taylor(op(2, R), h, 2), polynom), diff) (9) > # Set 4: use MyRule to rewrite R R := eval(eval(R, C=I/h), MyRule) (10) > # Set 5: use a 1st order expansion of r(i+(1/2)h, t) as h --> 0 UnlumpedMass := op(-1, R) = convert(taylor(op(-1, R), h, 1), polynom) (11) > # Set 6: use a 1st order expansion of r(i+(1/2)h, t) as h --> 0 R := eval(R, UnlumpedMass) (12) > # Set 7: finally replace "i" by "x" R := eval(R, i=x) (13) > # Step 8: focus on NoR: # # Basically I apply here some of the same steps as above NoR := algsubs(I/h = C, NoR0); NoR := eval(NoR, i=x); NoR := convert(convert(taylor(NoR, h, 2), polynom), diff); NoR := eval(NoR, C=I/h) (14) > # Step 9: assembling ContinuousPDE := R + NoR (15) > Download ContinuousEquation.mw ## I've been stuck several hours until ... I've been stuck several hours until I saw that the expression of eq I copid-pasted from your file 123.mw was WRONG. You wrote indeed ```eq := a^3bT^2(diff(V(xi), `\$`(xi, 2)))+a^3bT(diff(V(xi), xi))+3a^2bV(xi)^2-(3ac+2bomega)V(xi) ``` instead of ```eq := a^3bxi^2(diff(V(xi), `\$`(xi, 2)))+a^3bxi(diff(V(xi), xi))+3a^2bV(xi)^2-(3ac+2bomega)V(xi) ``` ! Once the corrections are done one can get several solutions among them the two solutions given at formula (3.4) in the image you provide: > restart From equation (3.4) > `eq (3.4)` := a^3bxi^2(diff(V(xi), `\$`(xi, 2)))+a^3bxi(diff(V(xi), xi))+3a^2bV(xi)^2-(3ac+2bomega)V(xi) (1) The "constant V" case: > # The "constant V" case: # Let K a non null constant Cst := factor(eval(`eq (3.4)`, V(xi)=K)): # K being non nul by definition one keeps onlu Cst := Cst/K: # Then Cst is null if isolate(Cst, omega): # The special case displayed in the excerpt you reproduce corresponds to K=a/3 # (why this choice?). map(factor, eval(%, K=a/3)) (2) The "non constant V" case: > verbose := false: > Guess := xi -> (p[0]+p[1]xi+p[2]xi^(2)+p[3]xi^(3))/(q[0]+q[1]xi+q[2]xi^(2)+q[3]xi^(3)) (3) > # Plug this guess into `eq (3.4)` f := eval(`eq (3.4)`, V=(xi -> Guess(xi))): if verbose then print(f): end if: > # Assuming that the denominator doesn't vanish one can concentrate on the numerator of f. # This numerator is to be seen seen like a polynom with indeterminate xi (9th degree)? # For it to be identically null whatever xi, it is necessary that all the coefficients # of this polynom are equal to 0 collect(numer(f), xi): coefficients := [coeffs(%, xi)]: if verbose then print(%): end if: SOL := solve(coefficients): if verbose then print~({SOL}): end if: > SOL := allvalues([SOL]): > AllPossibleSolutions := NULL: for s in map(op, {SOL}) do try eval(Guess(xi), s): AllPossibleSolutions := AllPossibleSolutions, simplify(%); catch: end try end do: AllNonConstantSolutions := {AllPossibleSolutions}: # eliminate potential multiple occurrences AllNonConstantSolutions := select(has, AllNonConstantSolutions, xi): # select non constant guesses AllNonConstantSolutions := remove(has, AllNonConstantSolutions, I): # remove possibly complex solutions print~(AllNonConstantSolutions): (4) > # Check the last solution # # Note the result is obviously not 0 eval(`eq (3.4)`, diff=Diff); eval(%, V(xi) = AllNonConstantSolutions[-1]); simplify(value(%)); (5) > # A few words # # The first solution is generic and not interesting: # # The last solution corresponds to the solution given in equation (3.5) # once done the transformation {aq[1] -> q[1], p[2] -> q[2]} (remember a <> 0). # The previous one is also the solution given in equation (3.5) # under the transformation p[1] --> q[1] and p[2] --> -q[2]). # # I didn't dig into these solutions to understand the meaning of # AllPossibleSolutions[2..4]. > # Give a look now to constant solutions AllConstantSolutions := {AllPossibleSolutions}: # eliminate potential multiple occurrences AllConstantSolutions := remove(has, AllConstantSolutions, xi): # select non constant guesses AllConstantSolutions := remove(has, AllConstantSolutions, I): # remove possibly complex solutions print~(AllConstantSolutions): (6) > # A few words # # The first solution is obvious # # The last solution corresponds to the constant olution given in equation (3.5) > Download 123_mmcdara_2.mw It remains a point which puzzles me look to the text in red in the attached file) It would be interesting to have more information on the paper you cite. It looks like the non constant solution is the eigen function of the operator which describes lhs(`eq (3.4)`) and the constant solution is the associated eigen value. ## Analysis... I don't understand your "can you check please...look like need to determine.constraints..more effectvely?" sentence. The attached file presents a quick analysis of your problem. I'm convinced that either your model, or your data, the units you use, are not correct. By the way, what is this model? It looks like a kind of viscoelasticity model. > > > > > > > > > > > > > > > > (1) > (2) A SIMPLE ANALYSIS > # Let's tke for instance stress[5]: s5 := stress[5] (3) > # You impose quite large values for the owar bounds of tau. # So s5 is almost equal to s5_1 := eval(s5, {seq(tau[i]=+infinity, i=1..9)}) (4) > # Assuming that the Gshave almost the same value g one gets s5_2 := normal(eval(s5_1, {seq(G[i]=g, i=1..9)})) (5) > # s5_2 is singular at singular(s5_2); plot(s5_2, g=-1e10..-1e8) > # What is the range of positive values for s5_2? # As you impose the Gs are negative: solve(s5_2 > 0) assuming g < 0; domains := solve({s5_2 > 0, g < 0}) (6) > # In the two domains 5_2 is contiously increasing: normal(diff(s5_2, g)) (7) > # The minium value of s5_2 in domains[1] is s5_2_min_1 := limit(s5_2, g=-infinity); # a value 10^6 times larger than true_stress[5] (8) > # The minium value of s5_2 in domains[1] is s5_2_min_2 := 0; # So if you expect than s5 canbe close to true_stress[5]=191, # it seems reasonable to impose that g verifies domains[2] (9) > # Note that the value of s5_2 are rapidly much higher than 191 in the # range of interest. plot(s5_2, g=lhs(op(1, domains[2]))..0, axis[2]=[mode=log]) > # When is s5_2 equal to true_stress[5]? solve({s5_2 = true_stress[5], domains[2][]}) (10) > # Now, assuming each tau is large enough for the limit tau -> +oo # is reasonable, lets define this simplified problem ObjLimit := eval(obj, {seq(tau[i]=+infinity, i=1..9)}): # And, all the G[1],.., G[9] only intervene through their sum SG, set: ObjSimple := algsubs(add(1.G[i], i=1..9) = SG, ObjLimit): ObjSimple := simplify(ObjSimple, size); (11) > # Try to minimize ObjSimple without taking any precautions at all: minimize(ObjSimple, location=true) (12) > # Even for its minimizer ObjSimple gets the amazing high value 10^9. # Might one reason be the crude approximation tau = +oo? # # Let's look to the values of exp(-strain/(strain_ratetau[a])) for the # constraints you impose to the taus. exp(-strain/(strain_ratetau[a])); TauMin := {1540 = tau[1], 155260 = tau[2], 10546880 = tau[3], 6.00000000010^8 = tau[4], 1.08000000010^10 = tau[5], 4.04000000010^11 = tau[6], 1.10000000010^13 = tau[7], 1.08000000010^14 = tau[8], 2.04000000010^16 = tau[9]}; eval( [seq([seq(exp(-x/(strain_ratetau[i])), i=1..9)], x in varepsilon)], (rhs=lhs)~(TauMin)): print~(%): (13) > # In ObjSimple I used the value "1" instead of the values above. # As you can see my approximation is not that crude and cannot be responsible # of the extremely high value of the minimum of ObjSimple. # # So the problem is elsewhere, on the side of your model certainly, # maybe you should also look on the side of the units you use (are they consistent?) Download 9terms_mmcdara.mw LAST COMMENT: As your model is governed by the sum SG = G[1]+..+G[7] when parameters tau[1]..tau[7] are "large enough" (which is your case),this means your model is not identifiable in the sense that an infinity of 7-uples (G[1], .., G[7]) can give the same value of SG. Note that tau values are not identifiable neither when the (positive) lower bounds they lust verify are "large enough". ## First: convert solution into hyperbolic ... Next arrange the result the way it suits you (an example is provided at the end of the attached file): > restart > V := exp(lambdaS) = S^4a4 + S^3a3 + S^2a2 + Sa1 + a0; > V1 := subs(S = 2, V); > V2 := subs(S = 1, V); > V3 := subs(S = 0, V); > V4 := subs(S = -1, V); > V5 := subs(S = -2, V); (1) > sol := solve(subs(a0 = 1, {V1, V2, V4, V5}), {a1, a2, a3, a4}); (2) > trigsol := convert~(sol, trigh) (3) > rel := { sinh(2lambda)=2sinh(lambda)cosh(lambda) , cos(2lambda)=cos(lambda)^2+sin(lambda)^2 }: map(factor, simplify(trigsol, rel)) (4) > Download trigh.mw The key is to define the rewritting rules rel that will lead you to the desire result. ## Some results... The first point is that your couple of equations (eq1, eq2) cannot be solved formally. THIS SECTION IS OBSOLETE GIVEN @Christian Wolinski's remark I kept it however tio justify what @Christian Wolinski wrote Having said that the natural approach is to give x a numerical value and solve (eq1, eq2) numerically, which will provide you a single solution (if any). I adopted a different way based on the observation that eq1-eq2 is a polynomial equation which doesn't contain x. Its solution is easily computable, even by hand, and iis of the form y=f(z) where z can be coosed arbitrarily. Things are a little bit more delicate for eq1-eq2 can be seen as a polynomial of degree 2 and y: then you can get 2 real solutions, a double real solution or 2 conjugate solutions depending on the value of z. In the attached file I considered that you would be interested by real solutions only. This constraints implies z is to be choosen in some subdomain of the real line (denoted Omega in the worksheet). Each of the two relations y=f(z) is aimed to be plugged into equation eq1 (the simplest one). Once done the resulting equation, which depends only on z, is formally solved. The results are of small direct interest as they are represented by a RootOf of a complex expression. Using the classical command allvalues to force the evaluations of the roots (if any) of this equation still gives extremely complex, and IMO useless, expressions... but evaluating the output of allvalues as floats give, for each relation y=f(z), a collection of numeric z-solutions. The last step is to compute the corresponding y-solutions by applying f to each z-solution. An attempt to represent the (y, z)-solutions for different values of x in the range 0..1 is also proposed. xyz.mw UPDATED ANSWER The attached file contains a procedure which is aimed to find (I write this way for it seems that a few solutions are missed, likely for numerical reasons) all the solutions (y(x), z(x)) in a square domain -L <= y <= L, -L <= z <= L with L strictly positive. Note that this domain could probably be reduced by ignoring negative y values (I didn't look into this point, but it seems that all solutions verify y > 0... something that is maybe easy to prove). I only focused here on your claim "I need to find the first 3 or 5 solutions" (in fact this > restart > eq1 := z(1/150-2y)exp(-2Pixy)+1/5((z^2-y^2)sin(2Pixz)-2yzcos(2Pixz))=0 (1) > eq2 := (z^2-y^2+1/150y-1)exp(-2Pixy)+1/5((z^2-y^2)sin(2Pixz)+2yzcos(2Pix*z))=0 (2) How do I proceed (1) Principles > # As no formal solution (y(x), z(x)) can be found a good start is to # do some graphics, to understand what happens. Digits := 15: X := 0.1; # my arbitrary choice in [0, 1] L := 20: NG := 1000: pp := plots:-implicitplot( eval([eq1, eq2], x=X) , y=-L..L, z=-L..L , color=[red, blue] , legend=["eq1=0", "eq2=0"] , grid=[NG, NG] ): (3) > # 10 solutions are likely to be seen here plots:-display(pp) > NC1 := nops~([op(1, pp)])[]-2; # NC1 red curves NC2 := nops~([op(2, pp)])[]-2; # NC2 blue curves # Matrix representation of each curve AllCurves := map2(op, -1, [plottools:-getdata(pp)]): (4) > # Let J the objective function which is equal to 0 iif eq1 and eq2 areboth # verified (log[10] is an artifact to lessen the range of variation of J) J := log[10](add(lhs~([eq1, eq2])^~2)): > # Solution search algorithm Digits := 15: SignChanges := NULL: minima := NULL: SolutionsChecked := NULL: # For each curve eq1=0 for ac in AllCurves[1..NC1] do pts1 := convert(ac, listlist): # Find the points on this curve where lhs(eq2) changes its sign # (points where a red and a blue curve will tangent will be missed) StartSign := signum(eval(eval(lhs(eq2), x=X), [y, z] =~ pts1[1])); for i from 2 to numelems(pts1) do EndSign := signum(eval(eval(lhs(eq2), x=X), [y, z] =~ pts1[i])); if EndSign <> StartSign then StartSign := EndSign; SignChanges := SignChanges, [pts1[max(1, i-2)], pts1[i]] end if: end do: end do: # Each couple of points (y, z) within which a sign change occurs defines # a domain where to search a solution where J is minimal (theoritically null). for s in [SignChanges] do dom := y = (min..max)(map2(op, 1, s)), z = (min..max)(map2(op, 2, s)); try minima := minima, Optimization:-Minimize(eval(J, x=X), dom, optimalitytolerance=1e-8, iterationlimit=1000): catch: printf("No solution found in domain %a\n", [dom]) end try: end do: # Evaluate eq1 and eq2 at the minima found. for sol in [minima] do SolutionsChecked := SolutionsChecked, eval(eval(lhs~([eq1, eq2]), x=X), sol[2]) end do: #print~(map2(op, 2, [minima]) implies~ [SolutionsChecked]): # Due to numerical precision some minima appear to give significantly non nul solution. # I arbitrarily discard them. AlmostSolutions := zip((u, v) -> if `and`(is~(abs~(u) <=~ 1e-5)[]) then v end if, [SolutionsChecked], map2(op, 2, [minima])): print~(AlmostSolutions): plots:-display( pp, seq( plot([eval([y, z], AlmostSolutions[i])], style=point, symbol=circle, symbolsize=20, color=black), i = 1..numelems(AlmostSolutions) ) ) How do I proceed (2) The solution procedure > SearchSolutions := proc(X, L, {see::boolean:=false, verbose::boolean:=false}) local NG, pp, NC1, NC2, AllCurves, SignChanges, minima, SolutionsChecked, ac, pts1, StartSign, i, EndSign, s, dom, sol, AlmostSolutions: uses plots: NG := 1000: pp := implicitplot( eval([eq1, eq2], x=X) , y=-L..L, z=-L..L , color=[red, blue] , legend=["eq1=0", "eq2=0"] , grid=[NG, NG] , title=typeset('x'=X) ): NC1 := nops~([op(1, pp)])[]-2; NC2 := nops~([op(2, pp)])[]-2; AllCurves := map2(op, -1, [plottools:-getdata(pp)]): SignChanges := NULL: minima := NULL: SolutionsChecked := NULL: for ac in AllCurves[1..NC1] do pts1 := convert(ac, listlist): StartSign := signum(eval(eval(lhs(eq2), x=X), [y, z] =~ pts1[1])); for i from 2 to numelems(pts1) do EndSign := signum(eval(eval(lhs(eq2), x=X), [y, z] =~ pts1[i])); if EndSign <> StartSign then StartSign := EndSign; SignChanges := SignChanges, [pts1[max(1, i-2)], pts1[i]] end if: end do: end do: for s in [SignChanges] do dom := y = (min..max)(map2(op, 1, s)), z = (min..max)(map2(op, 2, s)); try minima := minima, Optimization:-Minimize(eval(J, x=X), dom, optimalitytolerance=1e-8, iterationlimit=1000): catch: if verbose then printf("X = %3a No solution found in domain %a\n", X, [dom]) end if; end try: end do: for sol in [minima] do SolutionsChecked := SolutionsChecked, eval(eval(lhs~([eq1, eq2]), x=X), sol[2]) end do: AlmostSolutions := zip((u, v) -> if `and`(is~(abs~(u) <=~ 1e-5)[]) then v end if, [SolutionsChecked], map2(op, 2, [minima])): if see then print( plots:-display( pp, seq( plot([eval([y, z], AlmostSolutions[i])], style=point, symbol=circle, symbolsize=20, color=black), i = 1..numelems(AlmostSolutions) ) ) ); print(): end if: return AlmostSolutions end proc: > SearchSolutions(0.2, 20, see=true) (5) > # Sweep the range 0..1 for x and collect all the solutions found in this range Xs := [seq](0.1..1, 0.1): Sol_wrt_X := table([ seq(xs = SearchSolutions(xs, 20, verbose=true), xs in Xs) ]): X = .4 No solution found in domain [y = 15.6066929420051 .. 15.6457631113523, z = 15.6245382892261 .. 15.6255081599189] X = .4 No solution found in domain [y = 11.8689358202139 .. 11.9077186173179, z = 11.8748079235297 .. 11.8760651664658] X = .5 No solution found in domain [y = 12.4853581013289 .. 12.5244290234156, z = 12.499626883656 .. 12.5005960016093] X = .8 No solution found in domain [y = 16.5565565565565 .. 16.5965965965966, z = 16.5624419179027 .. 16.5629167275484] X = .8 No solution found in domain [y = 12.7927927927928 .. 12.8328328328328, z = -12.8127796536036 .. -12.8121695242539] X = .8 No solution found in domain [y = 5.3053053053053 .. 5.3453453453453, z = 5.31222489410908 .. 5.31366125682572] X = .8 No solution found in domain [y = .314327082446639 .. .356413032179948, z = -2.59865527442221 .. -2.59660936472894] X = 1.0 No solution found in domain [y = .4604604604604 .. .5005005005005, z = -10.0192011299064 .. -10.0185580322442] X = 1.0 No solution found in domain [y = 13.2332332332332 .. 13.2732732732732, z = 13.2497897695701 .. 13.2502623826246] X = 1.0 No solution found in domain [y = 2.23548973487505 .. 2.27305935450115, z = 2.24899474960935 .. 2.25146517002328] > # Plot all the solutions found for each value of X in Xs plots:-display( seq( plot( [seq](eval([y, z], Sol_wrt_X[k][i]), i = 1..numelems(Sol_wrt_X[k])) , style=point , symbol=solidcircle , symbolsize=10 , color=ColorTools:-Color([rand()/10^12, rand()/10^12, rand()/10^12]) , legend=typeset('x'=k) , legendstyle=[location=right] , labels=[y, z] ) , k in Xs[2..-1] ) , gridlines=true , view=[-0.1..20, -20..20] , size=[500, 400] ) > yz_sol_1.mw ## Alternative... If you don't mind having r and t dimensionless: ```r := t -> max(t, 0): plot(r(t), t=-1..2, useunits = [Unit('s'), Unit(('W')/'m'^2)]) ``` ## solve and plots:-inequal... Finding the domains where f(x)^2-g(y)^2 > 0 requires knowing explicitely the expressions of f and g, if not you can simply consider the expression a^2-b^2 > 0 instead. Examples: > restart > expr := a^2-b^2; (1) > solve(expr >= 0) (2) > plots:-inequal(expr >= 0, a=-3..3, b=-3..3 ); > f := u -> u^2: g := u -> u/(1+u^2): expr := f(x)^2 - g(y)^2; solve(expr >= 0) (3) > plots:-inequal(expr >= 0, x=-3..3, y=-3..3 ); > Download inequality.mw ## Boring that you change the answer that d... This is an answer to your previous question "why solve gives noolution?" or something like that. The answer is simple: solve can solve systems of 3 polynomial equations in 3 unknowns if each equation contans relatively low degrees of unwkowns combinations. Beforetryink to use solve, give a look to your equations: it should be clear to you that yoour system cannot be solved in a formal way, that's it. I end my worksheet with an advice and an example of a numeric solution. It's up to you to take inspiration of this, or not and keep asking while solve cannot solve your equations. 141123_Problem_mmcdara.mw About your "reformulated" question: You write: "Instead of solving my original equations, which are convoluted and not in polynomial form, I try to solve for their numerators first (since their numerators are polynomials). Broadly speaking, such solutions should also solve the original non-polynomial system" I "solved for the numerators" in the attached file and no formal solution (using solve) could be got. If you say that the system in your new file can be "solved for the numerators", isn't it because you have simplified your initial problem in somesense (isn't it the meaning of "no correlation" in the name of your new file?). You write "They need to be verified": What does "verifying a formal [thus exact] solution" mean to you? Checking that the roots ofthe numerator do not nullify the denominators? if it is so do that: ```num := numer~ (Eqs); sol := solve(Eqs, Vars,...): # or SolveTools:-PolynomialSystem maybe den := denom~(Eqs): eval(den, sol); # or eval(den, Vars=~sol) # or seq(eval(den, Vars=~s), s in [sol] # or somethingelse depending on the structure of sol ``` Here is an example where the olution (named sol_solve) has been obtained by using simply solve (no need to use SolveTools:-PolynomialSystem). You will see at the end of the file that the zeroes of the numerators are also those of the denominators. 141123_Problem_NoCorrelation_mmcdara.mw ## Right, use entries and indices... ```T := table([1=1, 2=4, 3=9]): max(indices(T)); 3 max(entries(T)); 9 ``` Note that for more complex forms of the indices the above can give unexpected results ```T := table([[1, 3]=4, [3, 2]=5]) max(indices(T)) 3 # You can use sort to find the largest index corresponding to a given order ind := indices(T, nolist); [1, 3], [3, 2] max_index = sort([ind], key=(x -> x[2]))[-1] max_index = [3, 2] ``` @Carl Love pointed out the incompleteness ogf my reply, here is a way to get the index of the maximal entrie ```T := table([ 1=23, 2=36, 3=14 ]);: max_index = sort([entries(T)], output=permutation)[-1] max_index = 2 ``` @Carl Love : using the 'pairs' option seems to do lead to something more complex. Maybe this can be made simpler? ```max_index = lhs(sort([indices(T,'pairs')], key=(x -> rhs(x)))[-1]); max_index = 2 ``` I don't see here any advantage in using 'pairs' instead of op(op(T)). ```max_index = lhs(sort(op(op(T)), key=(x -> rhs(x)))[-1]); max_index = 2``` 1 2 3 4 5 6 7 Last Page 1 of 49	11,997	36,152	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2023-50	longest	en	0.581524
https://physics.info/newton-first/	1,723,047,133,000,000,000	text/html	crawl-data/CC-MAIN-2024-33/segments/1722640694594.35/warc/CC-MAIN-20240807143134-20240807173134-00183.warc.gz	365,858,016	19,723	The Physics Hypertextbook Opus in profectus Forces Discussion introduction The first chapter of this book dealt with the topic of kinematics — the mathematical description of motion. With the exception of falling bodies and projectiles (which involve some mysterious thing called gravity) the factors affecting this motion were never discussed. It is now time to expand our studies to include the quantities that affect motion — mass and force. The mathematical description of motion that includes these quantities is called dynamics. Many introductory textbooks often define a force as "a push or a pull". This is a reasonable informal definition to help you conceptualize a force, but it is a terrible operational definition. What is "a push or a pull"? How would you measure such a thing? Most importantly, how does "a push or a pull" relate to the other quantities already defined in this book? Physics, like mathematics, is axiomatic. Each new topic begins with elemental concepts, called axioms, that are so simple that they cannot be made any simpler or are so generally well understood that an explanation would not help people to understand them any better. The two quantities that play this role in kinematics are distance and time. No real attempt was made to define either of these quantities formally in this book (so far) and none was needed. Nearly everyone on the planet knows what distance and time mean. examples How about we build up the concept of force with real world examples? Here we go… • Forces that act on all objects. • Weight (W, Fg) The force of gravity acting on an object due to its mass. An object's weight is directed down, toward the center of the gravitating body; like the Earth or moon, for example. • Forces associated with solids. • Normal (N, Fn) The force between two solids in contact that prevents them from occupying the same space. The normal force is directed perpendicular to the surface. A "normal" in mathematics is a line perpendicular to a planar curve or surface; thus the name "normal force". • Friction (f, Ff) The force between solids in contact that resists their sliding across one another. Friction is directed opposite the direction of relative motion or the intended direction of motion of either of the surfaces. • Tension (T, Ft) The force exerted by an object being pulled upon from opposite ends like a string, rope, cable, chain, etc. Tension is directed along the axis of the object. (Although normally associated with solids, liquids and gases can also be said exert tension in some circumstances.) • Elasticity (Fe, Fs) The force exerted by an object under deformation (typically tension or compression) that will return to its original shape when released like a spring or rubber band. Elasticity, like tension, is directed along an axis (although there are exceptions to this rule). • Forces associated with fluids. Fluids include liquids (like water) and gases (like air). • Buoyancy (B, Fb) The force exerted on an object immersed in a fluid. Buoyancy is usually directed up (although there are exceptions to this rule). • Drag (R, D, Fd) The force that resists the motion of an object through a fluid. Drag is directed opposite the direction of motion of the object relative to the fluid. • Lift (L, F) The force that a moving fluid exerts as it flows around an object; typically a wing or wing-like structure, but also golf balls and baseballs. Lift is generally directed perpendicular to the direction of fluid flow (although there are exceptions to this rule). • Thrust (T, Ft) The force that a fluid exerts when expelled by a propeller, turbine, rocket, squid, clam, etc. Thrust is directed opposite the direction the fluid is expelled. • Forces associated with physical phenomena. • Electrostatic Force (FE) The attraction or repulsion between charged bodies. Experienced in everyday life through static cling and in school as the explanation behind much of elementary chemistry. • Magnetic Force (FB) The attraction or repulsion between charged bodies in motion. Experienced in everyday life through magnets and in school as the explanation behind why a compass needle points north. • Fundamental forces. All the forces in the universe can be explained in terms of the following four fundamental interactions. • Gravity The interaction between objects due to their mass. Weight is a synonym for the force of gravity. • Electromagnetism The interaction between objects due to their charge. All the forces discussed above are electromagnetic in origin except weight. • Strong Nuclear Interaction The interaction between subatomic particles with "color" (an abstract quantity that has nothing to do with human vision). This is the force that holds protons and neutrons together in the nucleus and holds quarks together in the protons and neutrons. It cannot be felt outside of the nucleus. • Weak Nuclear Interaction The interaction between subatomic particles with "flavor" (an abstract quantity that has nothing to do with human taste). This force, which is many times weaker than the strong nuclear interaction, is involved in certain forms of radioactive decay. • Fictitious forces. These are apparent forces that objects experience in an accelerating coordinate system like an accelerating car, airplane, spaceship, elevator, or amusement park ride. Fictitious forces do not arise from an external object like genuine forces do, but rather as a consequence of trying to keep up with an accelerating environment. • Centrifugal Force The force experienced by all objects in a rotating coordinate system that seems to pull them away from the center of rotation. • Coriolis Force The force experienced by moving objects in a rotating coordinate system that seems to deflect them at right angles to their direction of motion. • "G Force" Not really a force (or even a fictitious force) but rather an apparent gravity-like sensation experienced by objects in an accelerating coordinate system. • Generic forces. When you don't know what to call a force, you can always give it a generic name like… • Push • Pull • Force • Applied Force free body diagrams Physics is a simple subject taught by simpleminded folk. When physicists look at an object, their first instinct is to simplify that object. A book isn't made up of pages of paper bound together with glue and twine, it's a box. A car doesn't have rubber tires that rotate, six-way adjustable seats, ample cup holders, and a rear window defogger; it's a box. A person doesn't have two arms, two legs, and a head; they aren't made of bone, muscle, skin, and hair; they're a box. This is the beginning of a type of drawing used by physicists and engineers called a free body diagram. Physics is built on the logical process of analysis — breaking complex situations down into a set of simpler ones. This is how we generate our initial understanding of a situation. In many cases this first approximation of reality is good enough. When it isn't, we add another layer to our analysis. We keep repeating the process until we reach a level of understanding that suits our needs. Just drawing a box is not going to tell us anything. Objects don't exist in isolation. They interact with the world around them. A force is one type of interaction. The forces acting on an object are represented by arrows coming out of the box — out of the center of the box. This means that in essence, every object is a point — a thing with no dimensions whatsoever. The box we initially drew is just a place to put a dot and the dot is just a place to start the arrows. This process is called point approximation and results in the simplest type of free body diagram. Let's apply this technique to a series of examples. Draw a free body diagram of… • a book lying on a level table • a person floating in still water • a wrecking ball hanging vertically from a cable • a helicopter hovering in place • a child pushing a wagon on level ground a book lying on a level table First example: Let's start with the archetypal example that all physics teachers begin with — a demonstration so simple it requires no preparation. Reach into the drawer, pull out the textbook, and lay it on top in a manner befitting its importance. Behold! A book lying on a level table. Is there anything more grand? Now watch as we reduce it to its essence. Draw a box to represent the book. Draw a horizontal line under the box to represent the table if you're feeling bold. Then identify the forces acting on it. Something keeps the book down. We need to draw an arrow coming out of the center pointing down to represent that force. Thousands of years ago, there was no name for that force. "Books lie on tables because that's what they do," was the thinking. We now have a more sophisticated understanding of the world. Books lie on tables because gravity pulls them down. We could label this arrow Fg for "force of gravity" or W for it's more prosaic name, weight. (Prosaic means non-poetic, by the way. Prosaic is a poetic way to say common. Prosaic is a non-prosaic word. Back to the diagram.) Gravity pulls the book down, but it doesn't fall down. Therefore there has to be some force that also pushes the book up. What do we call this force? The "table force"? No that sounds silly and besides, it's not the act of being a table that makes the force. It's some characteristic the table has. Place a book in water or in the air and down it goes. The thing about a table that makes it work is that it's solid. So what do we call this force? The "solid force"? That actually doesn't sound half bad, but it's not the name that's used. Think about it this way. Rest on a table and there's an upward force. Lean against a wall and there's a sideways force. Jump on a trampoline high enough to hit your head on the ceiling and you'll feel a downward force. The direction of the force always seems to be coming out of the solid surface. A direction which is perpendicular to the plane of a surface is said to be normal. The force that a solid surface exerts on anything in the normal direction is called the normal force. Calling a force "normal" may seem a little odd since we generally think of the word normal as meaning ordinary, usual, or expected. If there's a normal force, shoudn't there also be an abnormal force? The origin of the Modern English word normal is the Latin word for a carpenter's square — norma. The word didn't acquire its current meaning until the 19th century. Normal force is closer to the original meaning of the word normal than normal behavior (behavior at a right angle?), normal use (use only at a right angle?), or normal body temperature (take your temperature at a right angle?). Are we done? Well in terms of identifying forces, yes we are. This is a pretty simple problem. You've got a book, a table, and the Earth. The Earth exerts a force on the book called gravity or weight. The table exerts a force on the book called normal or the normal force. What else is there? Forces come from the interaction between things. When you run out of things, you run out of forces. The last word for this simple problem is about length. How long should we draw the arrow representing each force. There are two ways to answer this question. One is, "Who cares?" We've identified all the forces and got their directions right, let's move on and let the algebra take care of the rest. This is a reasonable reply. Directions are what really matter since they determine the algebraic sign when we start combining forces. The algebra really will take care of it all. The second answer is, "Who cares is not an acceptable answer." We should make an effort and determine which force is greater given the situation described. Knowing the relative size of the forces may tell us something interesting or useful and help us understand what's going on. So what is going on? In essence, a whole lot of nothing. Our book isn't going anywhere or doing anything physically interesting. Wait long enough and the paper will decompose (that's chemistry) and decomposers will help decompose it (that's biology). Given the lack of any activity, I think it's safe to say that the downward gravitational force is balanced by the upward normal force. W = N In summary, draw a box with two arrows of equal lengths coming out of the center, one pointing up and one pointing down. Label the one pointing down weight (or use the symbol W or Fg) and label the one pointing up normal (or use the symbol N or Fn). It may seem like I've said a lot for such a simple question, but I rambled with a reason. There were quite a few concepts that needed to be explained: identifying the forces of weight and normal, determining their directions and relative sizes, knowing when to quit drawing, and knowing when to quit adding forces. a person floating in still water Second example: a person floating in still water. We could draw a stick figure, but that has too much unnecessary detail. Remember, analysis is about breaking up complex situations into a set of simple things. Draw a box to represent the person. Draw a wavy line to represent water if you feel like being fancy. Identify the forces acting on the person. They're on Earth and they have mass, therefore they have weight. But we all know what it's like to float in water. You feel weightless. There must be a second force to counteract the weight. The force experienced by objects immersed in a fluid is called buoyancy. The person is pulled down by gravity and buoyed up by buoyancy. Since the person is neither rising nor sinking nor moving in any other direction, these forces must cancel W = B In summary, draw a box with two arrows of equal lengths coming out of the center, one pointing up and one pointing down. Label the one pointing down weight (or W or Fg) and the one pointing up buoyancy (or B or Fb). Buoyancy is the forced that objects experience when they are immersed in a fluid. Fluids are substances that can flow. All liquids and gases are fluids. Air is a gas, therefore air is a fluid. But wait, wasn't the book in the previous example immersed in the air. I said there were only three objects in that problem: the book, the table, and the Earth. What about the air? Shouldn't we draw a second upward arrow on the book to represent the buoyant force of the air on the book? The air does indeed exist and it does indeed exert an upward force on the book, but does adding an extra arrow to the previous example really help us understand the situation in any way? Probably not. People float in water and even when they sink they feel lighter in water. The buoyant force in this example is significant. It's what the problem's probably all about. Books in the air just feel like books. Whatever buoyant force is exerted on them is imperceptible and quite difficult to measure. Analysis is a skill. It isn't a set of procedures one follows. When you reduce a situation to its essence you have to make a judgment call. Sometimes small effects are worth studying and sometimes they aren't. An observant person deals with the details that are significant and quietly ignores the rest. An obsessive person pays attention to all details equally. The former are mentally healthy. The latter are mentally ill. a wrecking ball hanging vertically from a cable Third example: a wrecking ball hanging vertically from a cable. Start by drawing a box. No wait, that's silly. Draw a circle. It's a simple shape and it's the shape of the actual thing itself. Draw a line coming out the top if you feel so inclined. Keep it light, however. You don't want to be distracted by it when you add in the forces. The wrecking ball has mass. It's on the Earth (in the Earth's gravitational field to be more precise). Therefore it has weight. Weight points down. One vector done. The wrecking ball is suspended. It isn't falling. Therefore something is acting against gravity. That thing is the cable which suspends the ball. The force it exerts is called tension. The cable is vertical. Therefore the force is vertical. Gravity down. Tension up. Size? Nothing's going anywhere. This sounds like the previous two questions. Tension and weight cancel. W = T In summary, draw a circle with two arrows of equal length coming out of the center, one pointing up and one pointing down. Label the one pointing down weight (or W or Fg) and the one pointing up tension (or T or Ft). a helicopter hovering in place Fourth example: a helicopter hovering in place. How do you draw a helicopter? A box. What if you're tired of drawing boxes? A circle is a good alternative. What if even that's too much effort? Draw a small circle, I suppose. What if I want to try drawing a helicopter? Extra credit will not be awarded. You know the rest of the story. All objects have weight. Draw an arrow pointing down and label it. The helicopter is neither rising nor falling. What keeps it up? The rotor. What force does the rotor apply? A rotor is a kind of wing and wings provide lift. Draw an arrow pointing up and label it. The helicopter isn't sitting on the ground, so there is no normal force. It's not a hot air balloon or a ship at sea, so buoyancy isn't significant. There are no strings attached, so tension is nonexistent. In other words, stop drawing forces. Have I mentioned that knowing when to quit is an important skill? If not, I probably should have. Once again, we have an object going nowhere fast. When this happens it should be somewhat obvious that the forces must cancel. W = L In summary, draw a rectangle with two arrows of equal lengths coming out of the center, one pointing up and one pointing down. Label the one pointing down weight (or W or Fg) and the one pointing up lift (or L or F). and now… the law Let's do one more free body diagram for practice. a child pushing a wagon on level ground First, establish what the problem is about. This is somewhat ambiguous. Are we being asked to draw the child or the wagon or both? The long answer is, "it depends." The short answer is, "I am telling you that I want you to deal with the wagon." Draw a rectangle to represent the wagon. Next, identify the forces. Gravity pulls everything down, so draw an arrow pointing down and label it weight (or W or Fg according to your preference). It is not falling, but lies on solid ground. That means a normal force is present. The ground is level (i.e., horizontal), so the normal force points up. Draw an arrow pointing up and label it normal (or N or Fn). The wagon is not moving vertically so these forces are equal. Draw the arows representing normal and weight with equal length. W = N The child is pushing the wagon. We have to assume he's using the wagon for its intended purpose and is pushing it horizontally. I read left to right, which means I prefer using right for the forward direction on paper, blackboards, whiteboards, and computer displays. Draw an arrow to the right coming out of the center of the block. I see no reason to give this force a technical name so let's just call it push (P). If you disagree with me, there is an option. You could call it the applied force (Fa). That has the benefit of making you sound well-educated, but also has the drawback of being less precise. Calling a force an applied force says nothing about it since all forces have to be applied to exist. The word push is also a bit vague since all forces are a kind of push or pull, but pushing is something we generally think of as being done by hands. Since there is no benefit to using technobabble and the plain word push actually describes what the child is doing, we'll use the word push. Motion on the Earth does not take place in a vacuum. When one thing moves, it moves through or across another. When a wheel turns on an axle, the two surfaces rub against one another. This is called dry friction. Grease can be used to separate the solid metal parts, but this just reduces the problem to layers within the grease sliding past one another. This is called viscous friction. Pushing a wagon forward means pushing the air out of the way. This is another kind of viscous friction called drag. Round wheels sag when loaded, which makes them difficult to rotate. This is called rolling resistance. These resistive forces are often collectively called friction and they are everywhere. A real world analysis of any situation that involves motion must include friction. Draw an arrow to the left (opposite the assumed direction of motion) and label it friction (or f or Ff). Now for the tricky part. How do the horizontal forces compare? Is the push greater than or less than the friction? To answer this question, we first need to do something that physicists are famous for. We are going to exit the real world and enter a fantasy realm. We are going to pretend that friction doesn't exist. Watch the swinging pendulum. Your eyes are getting heavy. You are getting sleepy. Sleepy. I am going to count to three. When I say the word three you will awake in a world without friction. One. Two. Three. Welcome to the real world. No wait, that's a line from the Matrix. Assuming hypnosis worked, you should now slide off whatever it is your sitting on and fall to the ground. While you're down there I'd like you to answer this seemingly simple question. What does it take to make something move? More precisely, what does it take to make something move with a constant velocity? In the real world where friction is everywhere, motion winds down. Hit the brakes of your car and you'll come to a stop rather quickly. Turn the engine of your car off and you'll come to a stop gradually. Bowl a bowling ball down your lane and you probably won't perceive much of a change in speed. (If you're a good bowler, however, you're probably used to seeing the ball curve into the pocket. Remember, velocity is speed plus direction. Whenever either one changes, velocity changes.) Slap a hockey puck with a hockey stick and you'll basically see it move with one speed in one direction. I've chosen these examples and presented them in this order for a reason. There's less friction in coasting to a stop than braking to a stop. There's less friction in a hockey puck on ice than a bowling ball on a wooden lane. How about an example that's a little less everyday? Push a railroad car on a level track. Think you can't do it? Well think again. I'm not asking you push an entire train or even a locomotive — just a nice empty boxcar or subway car. I'm also not saying it's going to be easy. You may need a friend or two to help. This is something that is routinely done by railroad maitenance crews. Workers moving a subway car. Source: 所さんの目がテン！ MORE TEXT FINISH THIS WITH A GALILEO REFERENCE Heaven is a place where nothing ever happens. Isaac Newton (1642–1727) England. Did most of the work during the plague years of 1665 & 1666. Philosophiæ Naturalis Principia Mathematica (The Mathematical Principles of Natural Philosophy) published in 1687 (20+ year lag!) at Halley's expense. Lex. I. Law I. Corpus omne perſeverare in ſtatu ſuo quieſcendi vel movendi uniformiter in directum, niſi quatennus illud a viribus impreſſi cogitur ſtatum suummutare. Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon. Projectilia perſeverant in motibus ſuis, niſi quatenus a reſiſtentia aëris retardantur, & vi gravitatis impelluntur deorſum. Trochus, cujus partes cohærendo perpetuo retrahunt ſeſe a motibus rectilineis, non ceſſat rotari, niſi quatenus ab aëre retardantur. Majora autem planetarum & cometarum corpora motus ſuos & progreſſivos & circulares in ſpatiis minus reſiſtentibus factos conſervant diutius. Projectiles continue in their motions, so far as they are not retarded by the resistance of the air, or impelled downwards by the force of gravity. A top, whose parts by their cohesion are continually drawn aside from rectilinear motions, does not cease its rotations, otherwise than it is retarded by the air. The greater bodies of the planets and comets, meeting with less resistance in freer spaces, persevere in their motions both progressive and circular for a much longer time. (Newton, interpreted by Elert) An object at rest tends to remain at rest and an object in motion tends to continue moving with constant velocity unless compelled by a net external force to act otherwise. This rather complicated sentence says quite a bit. A common misconception is that moving objects contain a quantity called "go" (or something like that — in the old days they called it "impetus") and they eventually stop since they run out of "go". If no forces act on a body, its speed and direction of motion remain constant. Motion is just as natural a state as is rest. Motion (or the lack of motion) doesn't need a cause, but a change in motion does. Definitio. III. Definition III. Materiæ vis insita est potentia resistendi, qua corpus unumquodque, quantum in se est, perseverat in statu suo vel quiescendi vel movendi uniformiter in directum. The vis insita, or innate force of matter, is a power of resisting, by which every body endeavours to persevere in its present state, whether it be of rest, or of moving uniformly forward in a right line. … … Definitio. IV. Definition IV. Vis impressa est actio in corpus exercita, ad mutandum ejus statum vel quiescendi vel movendi uniformiter in directum. An impressed force is an action exerted upon a body, in order to change its state, either of rest, or of moving uniformly forward in a right line. Consistit hæc vis in actione sola, neque post actionem permanet in corpore. Perserverat enim corpus in statu omni novo per solam vim inertiæ. Est autem vis impresa diversarum originum, ut ex ictu, ex pressione, ex vi centripeta. This force consists in the action only; and remains no longer in the body when the action is over. For a body maintains every new state it acquires, by its vis inertiæ only. Impressed forces are of different origins as from percussion, from pressure, from centripetal force. In general, inertia is resistance to change. In mechanics, inertia is the resistance to change in velocity or, if you prefer, the resistance to acceleration. In general, a force is an interaction that causes a change. In mechanics, a force is that which causes a change in velocity or, if you prefer, that which causes an acceleration. When more than one force acts on an object it is the net force that is important. Since force is a vector quantity, use geometry instead of arithmetic when combining forces. External force: For a force to accelerate an object it must come from outside it. You can't pull yourself up by your own bootstraps. Anyone who says you can is literally wrong.	5,929	26,828	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.515625	4	CC-MAIN-2024-33	latest	en	0.957935
http://docplayer.net/401947-I-introduction-mpri-cours-2-12-2-lecture-iv-integer-factorization-what-is-the-factorization-of-a-random-number-ii-smoothness-testing-f.html	1,548,229,353,000,000,000	text/html	crawl-data/CC-MAIN-2019-04/segments/1547584203540.82/warc/CC-MAIN-20190123064911-20190123090911-00024.warc.gz	63,229,094	26,728	# I. Introduction. MPRI Cours Lecture IV: Integer factorization. What is the factorization of a random number? II. Smoothness testing. F. Save this PDF as: Size: px Start display at page: Download "I. Introduction. MPRI Cours 2-12-2. Lecture IV: Integer factorization. What is the factorization of a random number? II. Smoothness testing. F." ## Transcription 1 F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 MPRI Cours I. Introduction Input: an integer N; logox F. Morain logocnrs logoinria Output: N = k i=1 pα i i with p i (proven) prime. Lecture IV: Integer factorization Major impact: estimate the security of RSA cryptosystems. Also: primitive for a lot of number theory problems. I. Introduction. II. Smoothness testing. III. Pollard s RHO method. IV. Pollard s p 1 method. 2013/10/14 How do we test and compare algorithms? Cunningham project, RSA Security (partitions, RSA keys) though abandoned? Decimals of π. F. Morain École polytechnique MPRI cours /22 What is the factorization of a random number? F. Morain École polytechnique MPRI cours /22 II. Smoothness testing N = N 1 N 2 N r with N i prime, N i N i+1. Prop. r log 2 N; r = log log N. Size of the factors: D k = lim N + log N k / log N exists and On average k D k N 1 N 0.62, N 2 N 0.21, N 3 N an integer has one large factor, a medium size one and a bunch of small ones. Def. a B-smooth number has all its prime factors B. B-smooth numbers are the heart of all efficient factorization or discrete logarithm algorithms. De Bruijn s function: ψ(x, y) = #{z x, z is y smooth}. Thm. (Candfield, Erdős, Pomerance) ε > 0, uniformly in y (log x) 1+ε, as x with u = log x/ log y. ψ(x, y) = x u u(1+o(1)) 2 B-smooth numbers (cont d) A) Trial division Prop. Let L(x) = exp ( log x log log x ). For all real α > 0, β > 0, as x ψ(x α, L(x) β x α ) = L(x). α 2β +o(1) Ordinary interpretation: a number x α is L(x) β -smooth with probability ψ(x α, L(x) β ) x α = L(x) α 2β +o(1). Algorithm: divide x X by all p B, say {p 1, p 2,..., p m }, m = π(b) = O(B). Cost: m divisions or p B T(x, p) = O(m lg X lg B). Implementation: use any method to compute and store all primes 2 32 (one char per (p i+1 p i )/2; see Brent). Useful generalization: given x 1, x 2,..., x n X, can we find the B-smooth part of the x i s more rapidly than repeating the above in O(nm lg B lg X)? F. Morain École polytechnique MPRI cours /22 B) Product trees Algorithm: Franke/Kleinjung/FM/Wirth improved by Bernstein 1. [Product tree] Compute z = p 1 p m. p 1 p 2 p 3 p 4 p 1 p 2 p 3 p 4 p 1 p 2 p 3 p 4 2. [Remainder tree] Compute z mod x 1,..., z mod x n. z mod (x 1 x 2 x 3 x 4 ) z mod (x 1 x 2 ) z mod (x 3 x 4 ) z mod x 1 z mod x 2 z mod x 3 z mod x 4 3. [explode valuation] For each k {1,..., n}, compute y k = z 2e mod x k with e s.t. 2 2e x k ; print gcd(x k, y k ). F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 Validity and analysis Validity: let y k = z 2e mod x k. Suppose p x k. Then ν p (x k ) 2 e, since 2 ν p ν 2 2e. Therefore ν p (y k ) 2 e ν and the gcd will contain the right valuation. Division: If A has r + s digits and B has s digits, then plain division requires D(r + s, s) = O(rs) word operations. In case r s, break into r/s divisions of complexity M(s). Step 1: M((m/2) lg B). Step 2: D(m lg B, n lg X) (m/n)m(n) lg B lg X. Ex. B = 2 32, m , X = Rem. If space is an issue, do this by blocks. Rem. For more information see Bernstein s web page. F. Morain École polytechnique MPRI cours /22 3 F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 III. Pollard s RHO method Epact Prop. Let f : E E, #E = m; X n+1 = f (X n ) with X 0 E. X 0 X 1 X 2 X µ 1 X µ X µ+1 X µ+λ 1 Thm. (Flajolet, Odlyzko, 1990) When m πm λ µ m. Prop. There exists a unique e > 0 (epact) s.t. µ e < λ + µ and X 2e = X e. It is the smallest non-zero multiple of λ that is µ: if µ = 0, e = λ and if µ > 0, e = µ λ λ. Floyd s algorithm: X <- X0; Y <- X0; e <- 0; repeat X <- f(x); Y <- f(f(y)); e <- e+1; until X = Y; Thm. e π 5 m m. F. Morain École polytechnique MPRI cours /22 Application to the factorization of N F. Morain École polytechnique MPRI cours /22 Practice Idea: suppose p N and we have a random f mod N s.t. f mod p is random. function f(x, N) return (x 2 + 1) mod N; end. function rho(n) 1. [initialization] x:=1; y:=1; g:=1; 2. [loop] while (g = 1) do x:=f(x, N); y:=f(f(y, N), N); g:=gcd(x-y, N); endwhile; 3. return g; Choosing f : some choices are bad, as x x 2 et x x 2 2. Tables exist for given f s. Trick: compute gcd( i (x 2i x i ), N), using backtrack whenever needed. Improvements: reducing the number of evaluations of f, the number of comparisons (see Brent, Montgomery). Conjecture. RHO finds p N using O( p) iterations. 4 F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 Theoretical results (1/4) Bach (2/4) Thm. (Bach, 1991) Proba RHO with f (x) = x finding p N after k iterations is at least ( k 2) when p goes to infinity. Sketch of the proof: define p + O(p 3/2 ) f 0 (X, Y) = X, f i+1 (X, Y) = f 2 i + Y, f 1 (X, Y) = X 2 + Y, f 2 (X, Y) = (X 2 + Y 2 ) + Y,... Divisor of N found by inspecting gcd(f 2i+1 (x, y) f i (x, y), N) for i = 0, 1, 2,.... Prop. (a) deg X (f i ) = 2 i ; (b) f i X 2i mod Y; (c) f i Y 2i Y mod X; (d) f i+j (X, Y) = f i (f j (X, Y), Y). (e) f i is absolutely irreducible (Eisentein s criterion). (f) f j f i = fj 1 2 f i 1 2 = (f j 1 + f i 1 )(f j 2 + f i 2 ) (f j i + X)(f j i X). (g) For all k 1 f l+n ± f l f l+kn ± f l. F. Morain École polytechnique MPRI cours /22 Bach (3/4) For i < j, ρ i,j is the unique poly in Z[X, Y] s.t. (a) ρ i,j is a monic (in Y) irreducible divisor of f j f i. (b) Let ω i,j denote a primitive (2 j 2 i ) root of unity. Then ρ i,j (ω i,j, 0) = 0. Proof: and and for i 1: f j f i X 2i (X 2j 2 i 1) mod Y, X 2j 2 i 1 = µ 2 j 2 i Φ µ (X). f j f 0 ρ 0,j d j,d j ρ 0,d f j 1 + f i 1 ρ i,j d j i,d j i ρ. i,i+d Conj. we have in fact equalities instead of divisibility., F. Morain École polytechnique MPRI cours /22 Bach (4/4) Thm1. f k f l factor over Z[X, Y]. Moreover, they are squarefree. Proof uses projectivization of f. Weil s thm: if f Z[X, Y] is absolutely irreducible of degree d, then N p the number of projective zeroes of f is s.t. ( ) d 1 p. N p (p + 1) 2 2 Thm2. Fix k 1. Choose x and y at random s.t. 0 x, y < p. Then, proba for some i, j < k, i j, f i (x, y) = f j (x, y) mod p is at least ( k ) 2 /p + O(1/p 3/2 ) as p tends to infinity. Proof: same as i < j < k and ρ i,j (x, y) 0 mod p. Use inclusion-exclusion, Weil s inequality and Bézout s theorems. Same result for RHO to find p N. 5 F. Morain École polytechnique MPRI cours /22 F. Morain École polytechnique MPRI cours /22 IV. Pollard s p 1 method First phase Idea: assume p N and a is prime to p. Then Invented by Pollard in Williams: p + 1. Bach and Shallit: Φ k factoring methods. Shanks, Schnorr, Lenstra, etc.: quadratic forms. Lenstra (1985): ECM. Rem. Almost all the ideas invented for the classical p 1 can be transposed to the other methods. (p a p 1 1 and p N) p gcd(a p 1 1, N). Generalization: if R is known s.t. p 1 R, will yield a factor. gcd((a R mod N) 1, N) How do we find R? Only reasonable hope is that p 1 B! for some (small) B. In other words, p 1 is B-smooth. Algorithm: R = p α B 1 p α = lcm(2,..., B 1 ). Rem. (usual trick) we compute gcd( k ((ar k 1) mod N), N). F. Morain École polytechnique MPRI cours /22 Second phase: the classical one F. Morain École polytechnique MPRI cours /22 Second phase: BSGS Let b = a R mod N and gcd(b, N) = 1. Hyp. p 1 = Qs with Q R and s prime, B 1 < s B 2. Test: is gcd(b s 1, N) > 1 for some s. s j = j-th prime. In practice all s j+1 s j are small (Cramer s conjecture implies s j+1 s j (log B 2 ) 2 ). Precompute c δ b δ mod N for all possible δ (small); Compute next value with one multiplication b s j+1 = b s j c sj+1 s j mod N. Cost: O((log B 2 ) 2 ) + O(log s 1 ) + (π(b 2 ) π(b 1 )) multiplications +(π(b 2 ) π(b 1 )) gcd s. When B 2 B 1, π(b 2 ) dominates. Rem. We need a table of all primes < B 2 ; memory is O(B 2 ). Record. Nohara (66dd of , 2006; see Select w B 2, v 1 = B 1 /w, v 2 = B 2 /w. Write our prime s as s = vw u, with 0 u < w, v 1 v v 2. Lem. gcd(b s 1, N) > 1 iff gcd(b wv b u, N) > 1. Algorithm: 1. Precompute b u mod N for all 0 u < w. 2. Precompute all(b w ) v for all v 1 v v For all u and all v evaluate gcd(b vw b u, N). Number of multiplications: w + (v 2 v 1 ) + O(log 2 w) = O( B 2 ) Memory: O( B 2 ). Number of gcd: π(b 2 ) π(b 1 ). 6 Second phase: using fast polynomial arithmetic Algorithm: 1. Compute h(x) = 0 u<w (X bu ) Z/NZ[X] 2. Evaluate all h((b w ) v ) for all v 1 v v Evaluate all gcd(h(b wv ), N). Analysis: Step 1: O((log w)m pol (w)) operations (using a product tree). Step 2: O((log w)m int (log N)) for b w ; v 2 v 1 for (b w ) v ; multi-point evaluation on w points takes O((log w)m pol (w)). Rem. Evaluating h(x) along a geometric progression of length w takes O(w log w) operations (see Montgomery-Silverman). ( ) Total cost: O((log w)m pol (w)) = O. B 0.5+o(1) 2 Trick: use gcd(u, w) = 1 and w = F. Morain École polytechnique MPRI cours /22 Second phase: using the birthday paradox Consider B = b mod p ; s := #B. If we draw s elements at random in B, then we have a collision (birthday paradox). Algorithm: build (b i ) with b 0 = b, and { b 2 b i+1 = i mod N with proba 1/2 b 2 i b mod N with proba 1/2. We gather r s values and compute where r i=1 j i (b i b j ) = Disc(P(X)) = i P(X) = r (X b i ). i=1 use fast polynomial operations again. P (b i ) F. Morain École polytechnique MPRI cours /22 ### FACTORING. n = 2 25 + 1. fall in the arithmetic sequence FACTORING The claim that factorization is harder than primality testing (or primality certification) is not currently substantiated rigorously. As some sort of backward evidence that factoring is hard, ### Arithmetic algorithms for cryptology 5 October 2015, Paris. Sieves. Razvan Barbulescu CNRS and IMJ-PRG. R. Barbulescu Sieves 0 / 28 Arithmetic algorithms for cryptology 5 October 2015, Paris Sieves Razvan Barbulescu CNRS and IMJ-PRG R. 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http://gmatclub.com/forum/2-integers-will-be-randomly-selected-from-the-sets-above-58629.html	1,369,031,851,000,000,000	text/html	crawl-data/CC-MAIN-2013-20/segments/1368698493317/warc/CC-MAIN-20130516100133-00035-ip-10-60-113-184.ec2.internal.warc.gz	110,873,556	33,426	Find all School-related info fast with the new School-Specific MBA Forum It is currently 19 May 2013, 23:37 # 2 integers will be randomly selected from the sets above Author Message TAGS: Manager Joined: 10 Dec 2007 Posts: 52 Followers: 0 Kudos [?]: 5 [1] , given: 0 2 integers will be randomly selected from the sets above [#permalink] 17 Jan 2008, 14:50 1 KUDOS 00:00 Question Stats: 88% (01:51) correct 11% (01:30) wrong based on 1 sessions A = {2,3,4,5} B = {4,5,6,7,8} 2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 [Reveal] Spoiler: OA CEO Joined: 17 Nov 2007 Posts: 3591 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Followers: 230 Kudos [?]: 1298 [0], given: 346 Re: PS - probability [#permalink] 17 Jan 2008, 14:59 B p=\frac{C^4_1}{C^4_1C^5_1}=\frac{4}{45}=0.20 or p=\frac{4}{4}\frac{1}{5}=0.20 or p=\frac{4}{5}\frac{1}{4}+\frac{1}{5}\frac{0}{4}=0.20 _________________ iPhone/iPod/iPad: GMAT ToolKit - The bestselling GMAT prep app \| GMAT Club (free) \| PrepGame \| GRE ToolKit \| LSAT ToolKit Android: GMAT ToolKit (NEW!). POLL: What tool do you need next? Math: GMAT Math Book \|\|\| General: GMATTimer \|\|\| Chicago Booth: Slide Presentation The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do. Find out what's new at GMAT Club - latest features and updates Director Joined: 01 Jan 2008 Posts: 635 Followers: 3 Kudos [?]: 114 [2] , given: 1 Re: PS - probability [#permalink] 17 Jan 2008, 15:26 2 KUDOS You don't need to perform any calculations in this example. Regardless of the number you choose from set A, there is only one number (out of 5) that makes the sum equal 9. Therefore, the probability is 1/5. The answer is B. Manager Joined: 04 Jan 2008 Posts: 86 Followers: 1 Kudos [?]: 10 [0], given: 0 Re: PS - probability [#permalink] 22 Jan 2008, 17:10 B total 20 ways to sum, 4 of the sums equals 9 so, prob is 4/20= 1/5= .20 SVP Joined: 07 Nov 2007 Posts: 1841 Location: New York Followers: 20 Kudos [?]: 289 [0], given: 5 Re: PS - probability [#permalink] 24 Aug 2008, 14:52 vermatanya wrote: Quote: A = {2,3,4,5} B = {4,5,6,7,8} 2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 -thanx p= 4/(4c15c1)=1/5=0.2 _________________ Smiling wins more friends than frowning Manager Joined: 10 Jul 2009 Posts: 136 Location: Ukraine, Kyiv Followers: 2 Kudos [?]: 14 [1] , given: 60 Re: PS - probability [#permalink] 07 Sep 2009, 11:29 1 KUDOS Maybe I am late with response, but anyway. We have two sets with 4 and 5 integers included in each one respectively. Thus, we have 45=20 possible ways of selecting two integers from these sets. The requirement is that we need the sum of 9. Here we have 4 possible choices: 2 and 7 3 and 6 4 and 5 5 and 4 Probability = favourable / total 4/20 = 0.2 _________________ Never, never, never give up Manager Joined: 27 Oct 2008 Posts: 188 Followers: 1 Kudos [?]: 42 [0], given: 3 Re: PS - probability [#permalink] 27 Sep 2009, 06:17 A = {2,3,4,5} B = {4,5,6,7,8} Total number of possibilities of choosing two numbers is = 4 5 = 20 ways No. of possibilities of choosing numbers such that their sum is 9 is: (2,7),(3,6),(4,5),(5,4) thus 4 ways So probability is = 4/20 = 1/5 = 0.2 Senior Manager Joined: 22 Dec 2009 Posts: 368 Followers: 9 Kudos [?]: 136 [0], given: 47 Re: PS - probability [#permalink] 14 Feb 2010, 13:24 vermatanya wrote: A = {2,3,4,5} B = {4,5,6,7,8} 2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 -thanx Total ways of choosing 1 number from Set A and 1 from Set B.... 4c1 x 5c1 = 20 Comb which yield 9 as the sum - 2,7.....3,6.....4,5....5,4 = 4 comb... Probability = 4 / 20 = .20 = B _________________ Cheers! JT........... If u like my post..... payback in Kudos!! \|For CR refer Powerscore CR Bible\|For SC refer Manhattan SC Guide\| ~~Better Burn Out... Than Fade Away~~ Senior Manager Joined: 01 Feb 2010 Posts: 275 Followers: 1 Kudos [?]: 30 [0], given: 2 Re: PS - probability [#permalink] 15 Feb 2010, 00:24 vermatanya wrote: Quote: A = {2,3,4,5} B = {4,5,6,7,8} 2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 -thanx P(A) = 1 P(B) = 1/5 P(A&B) = 1/5 = 0.2 hence B Intern Joined: 24 Feb 2012 Posts: 36 Followers: 0 Kudos [?]: 3 [0], given: 18 Re: PS - probability [#permalink] 24 Feb 2012, 21:54 walker wrote: B p=\frac{C^4_1}{C^4_1C^5_1}=\frac{4}{45}=0.20 or p=\frac{4}{4}\frac{1}{5}=0.20 or p=\frac{4}{5}\frac{1}{4}+\frac{1}{5}\frac{0}{4}=0.20 Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please? GMAT Club team member Joined: 02 Sep 2009 Posts: 11518 Followers: 1792 Kudos [?]: 9538 [3] , given: 826 Re: PS - probability [#permalink] 24 Feb 2012, 22:49 3 KUDOS fortsill wrote: walker wrote: B p=\frac{C^4_1}{C^4_1C^5_1}=\frac{4}{45}=0.20 or p=\frac{4}{4}\frac{1}{5}=0.20 or p=\frac{4}{5}\frac{1}{4}+\frac{1}{5}\frac{0}{4}=0.20 Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please? I'd offer another approach which I hope will also clarify the above. A = {2,3,4,5} B = {4,5,6,7,8} 2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 Rearrange the first set: A = {5, 4, 3, 2} B = {4, 5, 6, 7, 8} As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 45=20. P=favorable/total=4/20=0.2. Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=11/5. Hope it's clear. _________________ Manager Joined: 08 Jun 2011 Posts: 98 Followers: 1 Kudos [?]: 11 [0], given: 65 Re: PS - probability [#permalink] 03 May 2012, 05:11 Bunuel wrote: fortsill wrote: walker wrote: B p=\frac{C^4_1}{C^4_1C^5_1}=\frac{4}{45}=0.20 or p=\frac{4}{4}\frac{1}{5}=0.20 or p=\frac{4}{5}\frac{1}{4}+\frac{1}{5}\frac{0}{4}=0.20 Although the calculation seems overkill, do you mind going over the logic behind the fractions above, please? I'd offer another approach which I hope will also clarify the above. A = {2,3,4,5} B = {4,5,6,7,8} 2 integers will be randomly selected from the sets above , one integer forms set A and one integer form set B. What is the Probability that the sum of 2 integers will equal 9? A. 0.15 B. 0.20 C. 0.25 D. 0.30 E. 0.33 Rearrange the first set: A = {5, 4, 3, 2} B = {4, 5, 6, 7, 8} As you can see numbers in each column (the numbers of the same color) give the sum of 9. So there are 4 such pares possible, total # of pairs is 45=20. P=favorable/total=4/20=0.2. Or: we can select ANY number from set A (4/4=1) but in this case we must select its matching pair from set B (the number of the same color) and since there are only one matching pair of this particular number in B then the probability of this is 1/5. So, overall: P=1*1/5. Hope it's clear. Bunuel, I am confused. We are selecting a pair from a total of 9 numbers or 7 numbers (if we eliminate duplicates). So we will be getting for total number of ways, in either case 2 from 9 is 36 and 2 from 7 is 21 So we have [ (2, 7),(3,6), (4,5) ] 3 pairs or [ (2, 7),(3,6), (4,5) (5,4) ] 4 pairs if we don't eliminate duplicates Either way, I am getting the wrong answer 3/36 and 4/21 are both wrong. Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same. How is it different from this one (the general concept)? if-two-of-the-four-expressions-x-y-x-5y-x-y-5x-y-are-92727.html#p713823 Thank you Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 3105 Location: Pune, India Followers: 567 Kudos [?]: 1997 [1] , given: 92 Re: PS - probability [#permalink] 03 May 2012, 09:29 1 KUDOS 2 from 9 is 36 and 2 from 7 is 21 You are not selecting 2 numbers from 9 numbers. You are selecting 1 number from 4 numbers and 1 number from 5 numbers. Mind you, they are different. While selecting 2 numbers from {1, 2, 3, 4, 5, 6, 7, 8, 9}, you can select (3, 4) While selecting 1 from 4 and 1 from 5: {1, 2, 3, 4} and {5, 6, 7, 8, 9}, you cannot select (3, 4). That's the problem number 1. Why is that wrong? and why aren't we eliminating duplicates because essentially (4,5) and (5,4) are the same same. Secondly, the two given sets are distinct: A = {2,3,4,5} B = {4,5,6,7,8} When we pick a number from A and from B, we get {4, 5} or we can do it in this way {5, 4}. These are two different cases. In the first case, the 4 comes from A and in the second case, it comes from B. _________________ Karishma Veritas Prep \| GMAT Instructor My Blog Save 10% on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 3105 Location: Pune, India Followers: 567 Kudos [?]: 1997 [0], given: 92 Re: 2 integers will be randomly selected from the sets above [#permalink] 20 Aug 2012, 04:12 Responding to a pm: No, there will not be 8 different pairs. A = {2,3,4,5} B = {4,5,6,7,8} If you select 2 from A, from B you must select 7. If you select 3 from A, from B you must select 6. If you select 4 from A, from B you must select 5. If you select 5 from A, from B you must select 4. The total number of ways in which the sum can be 9 is 4. The question was whether the last two cases are distinct or not. Is 4 from A and 5 from B the same as 4 from B and 5 from A. Since 4 and 5 come from different sets in the two cases, they represent two different cases and should be counted as such. _________________ Karishma Veritas Prep \| GMAT Instructor My Blog Save 10% on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Intern Joined: 12 Oct 2012 Posts: 18 WE: General Management (Hospitality and Tourism) Followers: 0 Kudos [?]: 5 [0], given: 38 Re: 2 integers will be randomly selected from the sets above [#permalink] 15 Nov 2012, 03:25 HI Karishma, I understood the question, got 0.20, but I have a doubt. At first, I got 6 pairs - (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options. How do we come too know that we need not include duplicate pairs?? I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 3105 Location: Pune, India Followers: 567 Kudos [?]: 1997 [0], given: 92 Re: 2 integers will be randomly selected from the sets above [#permalink] 15 Nov 2012, 03:33 HI Karishma, I understood the question, got 0.20, but I have a doubt. At first, I got 6 pairs - (2,7);(3,6);(4,5);(5,4);(6,3);(7,2), leading, P= 6/20= 0.3 which is also one of the options. How do we come too know that we need not include duplicate pairs?? I understand from the explanations that 6,3 & 7,2 are not counted, but how can i figure out looking at the q? You need to select a pair {a, b} such that their sum is 9. a is selected from set A and b is selected from set B. a can take a value from set A only i.e. one of 2/3/4/5 b can take a value from set B only i.e. one of 4/5/6/7/8 How do you select {6, 3}? a cannot be 6. A selection is different from another when you select different numbers from each set. One selection is {4, 5} - 4 from A, 5 from B Another selection is {5, 4} - 5 from A, 4 from B One selection is {3, 6} - 3 from A, 6 from B There is no such selection {6, 3} - A has no 6. _________________ Karishma Veritas Prep \| GMAT Instructor My Blog Save 10% on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Intern Joined: 12 Oct 2012 Posts: 18 WE: General Management (Hospitality and Tourism) Followers: 0 Kudos [?]: 5 [0], given: 38 Re: 2 integers will be randomly selected from the sets above [#permalink] 15 Nov 2012, 03:40 when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it cannot be (b,a) and it will always be (a,b) in terms of order?? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 3105 Location: Pune, India Followers: 567 Kudos [?]: 1997 [0], given: 92 Re: 2 integers will be randomly selected from the sets above [#permalink] 15 Nov 2012, 03:46 when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it cannot be (b,a) and it will always be (a,b) in terms of order?? It doesn't matter how you write it. One number is from A and one is from B. You know which number is from A and which is from B. You can write it is {a, b} or {b, a}, it doesn't matter. {4, 5} - 4 is from A and 5 is from B {5, 4} - 5 is from A and 4 is from B These two are different. {2, 6} - 2 is from A and 6 is from B {6, 2} - 2 is from A and 6 is from B These two are same. _________________ Karishma Veritas Prep \| GMAT Instructor My Blog Save 10% on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Intern Joined: 12 Oct 2012 Posts: 18 WE: General Management (Hospitality and Tourism) Followers: 0 Kudos [?]: 5 [0], given: 38 Re: 2 integers will be randomly selected from the sets above [#permalink] 15 Nov 2012, 03:51 VeritasPrepKarishma wrote: when Q says that you need to select one integer from Set A & one from Set B, going by the explanation, does it mean that it cannot be (b,a) and it will always be (a,b) in terms of order?? It doesn't matter how you write it. One number is from A and one is from B. You know which number is from A and which is from B. You can write it is {a, b} or {b, a}, it doesn't matter. {4, 5} - 4 is from A and 5 is from B {5, 4} - 5 is from A and 4 is from B These two are different. {2, 6} - 2 is from A and 6 is from B {6, 2} - 2 is from A and 6 is from B These two are same. Thanks karishma. Hope i master these concepts soon. Re: 2 integers will be randomly selected from the sets above [#permalink] 15 Nov 2012, 03:51 Similar topics Replies Last post Similar Topics: If two different numbers are randomly selected from a set 9 02 Jun 2005, 09:08 If two numbers are randomly selected from set (1, 2, 3, 4, 1 23 Oct 2007, 12:32 2 If an integer is to be randomly selected from set M above 9 12 Jan 2008, 05:29 2 Two integers will be randomly selected from sets A and B 5 24 Oct 2012, 22:05 Two integers will be randomly selected from the sets above, 2 02 Dec 2012, 07:51 Display posts from previous: Sort by # 2 integers will be randomly selected from the sets above Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.	5,536	16,010	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.5	4	CC-MAIN-2013-20	latest	en	0.705749
https://alloutlotto.com/toto-latest-result-latest-4d-result.html	1,560,661,904,000,000,000	text/html	crawl-data/CC-MAIN-2019-26/segments/1560627997731.69/warc/CC-MAIN-20190616042701-20190616064701-00145.warc.gz	349,162,979	6,489	The probability and odds can be taken into a mathematical perspective: The probability of winning the jackpot (through October 27, 2017) was 1:(75C5) x (15); that is: 75 ways for the first white ball times 74 ways for the second times 73 for the third times 72 for the fourth times 71 for the last white ball divided by 5 x 4 x 3 x 2 x 1, or 5!, and this number is then multiplied by 15 (15 possible numbers for the "Megaball"). Therefore, (75 x 74 x 73 x 72 x 71)/ (5 x 4 x 3 x 2 x 1) x 15 = 258,890,850, which means any combination of five white balls plus the Megaball has a 1:258,890,850 chance of winning the jackpot. Similarly, the odds for second prize are 1:(75C5) x (15/14) = 1: 18,492,204 chance of winning. 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https://www.jiskha.com/search?query=a+density+of+1.0+M+solution+of+NaCL&page=40	1,534,521,148,000,000,000	text/html	crawl-data/CC-MAIN-2018-34/segments/1534221212598.67/warc/CC-MAIN-20180817143416-20180817163416-00680.warc.gz	929,927,524	15,352	# a density of 1.0 M solution of NaCL 29,843 results, page 40 1. ## Physics The zeppelin Hindenburg exploded on May 6, 1937, in Lakehurst, New Jersey. The accident was due to this type of dirigible using hydrogen (density of H2, M = 2g/mol) as a gas. It carried up to 70 passengers and crew. Estimate how many passengers and crew 2. ## Physics The zeppelin Hindenburg exploded on May 6, 1937, in Lakehurst, New Jersey. The accident was due to this type of dirigible using hydrogen (density of H2, M = 2g/mol) as a gas. It carried up to 70 passengers and crew. Estimate how many passengers and crew 3. ## Geography (Ms. Sue) Which has the higher population density: The United States or Russia and the Republic? Why? A: I believe the United States has the higher population density because ? 4. ## Earth Science Silver has a density of 10.5 g/cm cubed, and gold has a density of 19.3 g/cm cubed. Which would have the greater mass, 5cm cubed of silver or 5 cm cubed of gold? 5. ## Physics A gold prospector finds a solid rock that is composed solely of quartz and gold. The mass and volume of the rock are, respectively, 14.8 kg and 1.89 × 10-3 m3. The density of gold is 19 300 kg/m3, and the density of quartz is 2660 kg/m3. Find the mass of 6. ## Math a 25.00 gram sample is placed in a graduated cylinder and then the cylinder is filled to the 50.0 mL mark with benzene. the mass of benzene and solid together is 58.80 gram. assuming that the solid is insoluble in benzene and that the density of benzene is 7. ## physics - density A person got a metal rock, he knows this rock density is ρm, but he is worried that there might be a hole within the rock, how he found there is a hole or not? And how he found the fraction of the hole size to total rock size? write your experimental 8. ## PHYSICS A simple open U-tube contains mercury. When 11.9 cm of water is poured into the right arm of the tube, how high above its initial level does the mercury rise in the left arm? Take the density of mercury to be 13.6 g/cm3 and the density of water to be 1.00 9. ## Chemistry Concept The iron in steel has a density of 7.86 g/cm3. When iron rusts, it forms hydrated Fe2O3 that has a density of 5.12 g/cm3. What happens to the volume as iron rusts? Like does it expand? contract? stay the same? expand then contract? <-- that is my 10. ## physics A gold prospector finds a solid rock that is composed solely of quartz and gold. The mass and volume of the rock are, respectively, 13.3 kg and 3.56 × 10-3 m3. The density of gold is 19 300 kg/m3, and the density of quartz is 2660 kg/m3. Find the mass of 11. ## Chemistry A. The forensic technician at a crime scene has just prepared a luminol stock solution by adding 13.0 g of luminol into a total volume of 75.0 mL of H2O. -What is the molarity of the stock solution of luminol? B. Before investigating the scene, the 12. ## chemistry You are tasked with preparing a 500mL nutrient solution containing 5mg/L of phosphorous (P), 3mg/L nitrate (NO3 -) and 100 mg/L chloride (Cl-). To make this solution, you have powdered sodium chloride, a 100 mg/L solution of phosphorus, a 1000 mg/L of 13. ## Simple Chemistry A.Write a brief procedure outling how would you prepare 250mL of a 0.150M solution of a CoCl2 from solid CoCl2 and distilled water b. Using the solution from part a and distilled water, how would you prepare 100ml of a 0.0600 M solution of CoCl2 C. What A solution contains one or more of the following ions: Hg2 2+,Ba2+ ,Fe2+ and . When potassium chloride is added to the solution, a precipitate forms. The precipitate is filtered off and potassium sulfate is added to the remaining solution, producing no 15. ## Chemistry 1. Sugar can be dissolved in water because of the presence of bonds between _______ in its molecule. 2. If the maximum amount of a compound that can dissolve in 100 mL of water is 22 g, then a solution of 20 g of that substance in 100 mL of water is said 16. ## Chemistry 1. Sugar can be dissolved in water because of the presence of bonds between _______ in its molecule. 2. If the maximum amount of a compound that can dissolve in 100 mL of water is 22 g, then a solution of 20 g of that substance in 100 mL of water is said 17. ## chemistry The reaction between sodium hydroxide and hydrobromic acid is represented in the equation below: OH- + H+ -> H2O A 62.5 mL sample of 1.86 M NaOH at 22.5 deg Celsius was mixed with 62.5 mL sample of 1.86 M of HBr at 22.5 deg Celsius in a coffee cup 18. ## chemistry! URGENT A standard solution is prepared for the analysis of fluoxymesterone (C20H29FO3), an anabolic steroid. A stock solution is first prepared by dissolving 10.0 mg of fluoxymestrone in enough water to give a total volume of 500.0 ml. A 100.0 micro liter aliquot 19. ## chemistry Gas X has a density of 2.60g/L at STP. Determine the molar mass of this gas. Sorry there was some typos on the previous question. please show me again steps by steps. thank you =] Same answer. You just left the g off the density. One mole of ANY (ideal) 20. ## Chemistry-Question 3 A 4.86 kg sample of a solution of the solute chloroform in the solvent (C2H5)2O that has a concentration of 1.91 pph by mass chloroform is available. Calculate the amount (g) of CHCl3 that is present in the sample. Molar Mass (g/mol) CHCl3 119.38 (C2H5)2O 21. ## Chemistry Hi, I'm really having trouble this this chemistry problem. I have the answers; however I don't have extensive knowledge about titrations and I'm struggling with working it out. If anyone could suggest how to work this out (or even where to start) that 22. ## AP Chemistry A 0.3500 MHCl solution was used to titrate a 21.60 mL sodium acetate solution. The endpoint was reached after 25.75 mL of titrant were added. Find the molar concentration of sodium acetate in the original solution. 23. ## ap chemistry Which of the following statements about the pH scale is not true? A. In a pH expression, the hydronium ions, H3O+, can be abbreviated simply as H+. B. A solution with a pH of 4 has twice the [H+] of a solution with a pH of 2. C. The pH scale is based on a 24. ## Differential equations,Calculus So I have the following differential equation. The general solution I have is: t=k(-1/r)+c I now need to find the particular solution when t=0 and the radius (r) = 1cm. So k is a constant which is approx 3.9 (5/4pi) So for the particular solution im really 25. ## Chemistry titrations 25.00mL of an unknown sulfuric acid solution is titrated to the second equivalence point with 21.02mL of 0.420M potassium hydroxide solution. What is the concentration of the sulfuric acid solution? 26. ## chemistry 35.00mL of 0.250M KOH solution is required to completely neutralize 10.50mL of H2SO4 solution. what is the molarity of the sulfuric acid solution KOH(aq)+ H2SO4 -------> K2SO4(aq) +H2O(l 27. ## math Solve this system using elimination. If a single solution exists, write the solution as an ordered pair. Your answer will be an ordered pair, no solution, or infinitely many solutions 3a+5b=-19 5a-27b=145 28. ## Physics ( Thickness of a Spherical Shell) Aluminum floater: A spherical shell with an outer radius R=2 cm is made of aluminum (mass density of Al is 2.7 g/cm3). How thick is the shell if the average density of the spherical object is exactly 0.5 g/cm3 ? Assume there is a vacuum in the hollow 29. ## Physics An electromagnetic wave has an electric field strength of 145 V/m at a point P in space at time t. A parallel-plate capacitor whose plates have area 0.109 m2 has a uniform electric field between its plates with the same energy density as the answer to part 30. ## Chemistry In a thermochemistry experiment the reaction of 0.0857 g of an unknown metal and an excess of 4.0 M HCl resulted in a 0.210 mL decrease in the volume of an ice/water mixture. Calculate the enthalpy of this reaction per 1 g of metal reacted. Needed data: 31. ## Chemistry If some NH4Cl is added to an aqeous solution of NH3: A. pH of solution will decrease B. PH will not change C. pH will increase D. NH4Cl cannot be added to NH3 E. The solution will not have pH I think the answer is B, because if NH4Cl is added to the 32. ## Chemistry If 43.2 mL of 2.46 M HClO4 solution completely reacts with a 75.0 mL sample of a sodium hydroxide solution, what is the molarity of the sodium hydroxide solution? (hint: write the complete balanced rxn) 33. ## Chemistry A 230.-mL sample of a 0.280 M CaCl2 solution is left on a hot plate overnight; the following morning, the solution is 1.10 M. What volume of water evaporated from the 0.280 M CaCl2 solution? 34. ## intro to chem The molarity of an aqueous solution of potassium hydroxide, KOH, is determined by titration against a 0.197 M HI solution. If 28.4 mL of the base are required to neutralize 18.6 mL of HI, what is the molarity of the KOH solution? 35. ## ap chemistry A 37 mL sample of a solution of sulfuric acid is neutralized by 42 mL of a 0.113 M sodium hydroxide solution. Calculate the molarity of the sulfuric acid solution. Answer in units of mol/L 36. ## Chemistry a 20.0ml sample of HCL solution is placed in a flask with a few drops of indicator. If 30.0ml of a 0.240 M NaOH solution is needed to reach the endpoint, what is the molarity of the HCL solution? 37. ## Chemistry A 69 mL sample of a solution of sulfuric acid is neutralized by 41 mL of a .161 M sodium hydroxide solution. Calculate the molarity of the sulfuric acid solution. Answer in units of mol/L. A 11 mL sample of a solution of AlBr3 was diluted with water to 23 mL. A 15 mL sample of the dilute solution was found to contain 0.045 moles of Br-. What was the concentration of AlBr3 in the original undiluted solution? answer is 2.09 but i cant get it. 39. ## math algebra Solve the following system using elimination. If a single solution exists, write the solution as an ordered pair. Your answer will be an ordered pair, no solution, or infinitely many solutions. -a+b=-4 2a-3b=-2 40. ## Math: Differential Equations Solve the initial value problem y' = y^2, y(0) = 1 and determine the interval where the solution exists. I understand how the final solution comes to y = 1/(1-x), but do not understand how the solution is defined from (-infinity, 1) and not (1, infinity). 41. ## Physics The density of copper is 9.8 kilograms per meter cube and the density of gold is 1.9 kilograms per meter cube. When two wires of these metals are held under the same tension, the wave speed in the gold is found to be half that in the copper wire. What is 42. ## math 1 litre of coffee at 80oC need to be cooled to 40oC. Assuming that all the ice melts and no heat is lost to the environment. Take the specific heat capacity of coffee to be 4200J/kgK, density of coffee to be 1000kg/m3 and latent heat of fusion of water to 43. ## chem True/false 1. will addding a strong acid (HNO3) to a solution that is 0.010M in Ag(NH3)2(^+) will tend to dissociate the complex ion into Ag+ and NH4+ 2.adding a strong acid (HNO3)to a solution that is 0.010M in AgBr2(^-) will tend to dissociate the 44. ## physics Challenger Deep in the Marianas Trench of the Pacific Ocean is the deepest known spot in the Earth's oceans, at 10.922 km below sea level. Taking density of seawater at atmospheric pressure (p0 = 101.3 kPa) to be 1024 kg/m3 and its bulk modulus to be B(p) 45. ## Math The pH of a solution ranges from 0 to 14. An acid has a pH less than 7. Pure water is neutral and has a pH of 7. The pH of a solution is given by pH = -logx where x represents the concentration of the hydrogen ions in the solution in moles per liter. Find 46. ## Chemistry A solution contains Cr+3 ion and Mg+2 ion. The addition of 1.00 L of 1.55 M NaF solution is required to cause the complete precipitation of these ions as CrF3(s) and MgF2(s). The total mass of the precipitate is 49.8 g. Find the mass of Cr+3 in the How would you prepare 100.0 mL of 0.07500 M AgNO3 solution starting with pure solute? 2. An experiment calls for you to use 250 mL of 1.0 M HNO3 solution. All you have to available is a bottle of 6.0 M HNO3. How would you prepare the desired solution? 48. ## Chemistry A solution containing 22.18 mL of a .1212 M H2CO3 solution is mixed with 17.22 mL of a .1234 M Na2CO3 solution. What is the pH of the final mixture? Ka1 = 6.2x10-7 Ka2 = 4.4x10-11. If you diluted the mixture up to 100 mL, would the pH change? If it does 49. ## Chemistry An experiment calls for 100ml of 0.1M copper(II) sulfate solution. Write a step-by-step procedure for a science teacher to make 100ml of 5M stock solution and the required diluted solution. 50. ## chemistry 15.0 mL of an iodine solution of unknown molarity was titrated against a .001 M solution of S2o32-. The blue black iodine-starch complex just disappeared when 20.2 mL of thiosulfate was added. What is the exact molarity of the Ia solution 51. ## Math The pH of a solution ranges from 0 to 14. An acid has a pH less than 7. Pure water is neutral and has a pH of 7. The pH of a solution is given by pH = -logx where x represents the concentration of the hydrogen ions in the solution in moles per liter. Find 52. ## chem You wish to prepare an aqueous solution of glycerol, C3H5(OH)3, in which the mole fraction of the solute is 0.093. How many grams of glycerol must you combine with 425 grams of water to make this solution? What is the molality of the solution? PLEASE HELP! 53. ## Chemistry If 20.00mL of a solution that contains oxalate ion requires 10.00mL of 0.020M MnO4- solution in a titration, what was the molarity of the oxalate solution? 5C2O42- + 2MnO4- + 12H+ --> 2Mn2+ + 10CO2 + 8 H2O 54. ## Chemistry A) Determine the pH of a 0.98 x 10^-2 mol L solution of hydrocyanic acid (HCN) Ka = 4.0 x 10^-10 B) What is the pH of a 0.243mol L solution of methylamine? (pKb for CH3NH2 = 3.30) C) determine the pH of a buffer solution of Na2CO3 (pkb = 3.68, 0.125 M) and 55. ## analytical chemistry An EDTA solution is standardized against a solution of primary standard CaCO3 (0.5622 g dissolve in 1000 ml of solution) and titrating aliquots of it with the EDTA. If a 25.00 ml aliquot required 21.88 ml of the EDTA, what is the concentration of the EDTA? 56. ## Chemistry Calculate the concentration, in molarity, of a solution prepared by adding 9 mL of water to 1 mL of 0.1 M HCl solution. If 2.0 mL of 0.010 M NaOH is mixed with enough water to make the total volume 8 mL, what is the molarity of the resulting solution? 57. ## AP Chemistry A 0.2500 M NaOH solution was used to titrate a 16.85 mL hydrofluoric acid solution. The endpoint was reached after 31.20mL of titrant were added. Find the molar concentration of hydrofluoric acid in the original solution. 58. ## Chemistry What is the molarity of a 5.00 x 102 ml solution containing 2490 g of KI? How many moles of LiF would be required to produce a 2.5 M solution with a volume of 10.5 L? How many grams of NaI would be used to produce a 2.0 M solution with a volume of 10.00 L? 59. ## chemistry A standard sodium carbonate solution is prepared in a 250 cm^3 volumetric flask during a titration,20 cm^3 of a 0,1 mol.dm^-3 nitric acid solution neutralises 25 cm^3 of the above standard solution according to the following balanced equation 2HNO3(aq) 60. ## Science What is anything that has mass and takes up space? 1) Weight 2) Volume 3) Density 4) Matter When you measure the boiling point of mercury, you are investigating a ____. 1) Chemical change 2) Chemical property 3) Physical change 4) Physical property Ab 61. ## Chemistry A total charge of 96.5 kC is passed through an electrolytic cell. Determine the quantity of substance produced in each of the following cases. (a) the mass (in grams) of silver metal from a silver nitrate solution (b) the volume (in liters at 273 K and 62. ## chemistry A chemist added 3.15 grams of HNO3 to a solution of the same acid with concentration of 9 mol/L. dilutes the resulting solution up to the 100ml volume. The resulting solution will have what concentration? 63. ## math a chemistry experiment calls for 30% sulfuric acid solution. if the supply room has only 50% and 20% sulfuric acid solution on hand,how much of each should be mixed to obtain 12 litres of 30% solution? 64. ## chemistry What is the mass of BaCl2 contained in 250 mL of solution since after the addition of 40.00 ml of AgNO3 0.1020M to 25.00 mL of the BACl2 solution, the excess solution of AgNO3 consumed 15.00 ml of NH4SCN 0.09800 M 65. ## Chemistry A solution is made by mixing 90.6 g of magnesium sulfate heptahydrate with 2320.0 g of water. 690.00 mL of the solution has a mass of 745.20 g. What is the molarity of the magnesium sulfate solution? 66. ## Chemistry A 1.0m solution of LiCl decreases the freezing point of water more than a .10 m solution of LiCl. Explain why based on colligative properties. I know that the 1.0m solution would decrease it more, but cannot explain why. Any help is appreciated! 67. ## Chemistry How many grams of nickel (II) sulfate hexahydrate would be present in one liter of a 0.40 molar solution? (Hint: a 1.0 molar solution would have 1 mole of NiSO4*6H2O per liter of solution) 68. ## Math chandra has 3 liters of a 28% solution of sodium hydroxide in a container. What is the amount and concentration of sodium peroxide solution she must add to this in order to end up with 7 liters of 48% solution? 69. ## math Solve the following system using elimination. If a single solution exists, write the solution as an ordered pair. Your answer will be an ordered pair, no solution, or infinitely many solutions. x-3y=5 2x-6y=10 70. ## chemistry prepare M/20 solution of oxalic acid. Find out the molarity and strength in gram per litre of the given solution of potassium permanganate with the help of the prepared solution of oxalic acid 71. ## chemistry Sulfuric was titrated with NaOH solution. 208 mL of .250 M NaOH solution was needed to completely neutralize 65.0 mL of H2SO4. What is the concentration of the sulfuric acid solution? 72. ## Math solve 3(2x-2)=18 select the correct choice below& if necessary fill in the answer box to complete your choice. a. the solution is x= simplify answer. b. the solution is all real numbers. c. there is no solution. 73. ## MATH Solve this system using elimination. If a single solution exists, write the solution as an ordered pair. Your answer will be an ordered pair, no solution, or infinitely many solutions 2x-y=0 3x+y=5 74. ## Chemistry A 53 mL sample of a solution of sulfuric acid is neutralized by 37 mL of a 0.18 M sodium hydroxide solution. Calculate the molarity of the sulfuric acid solution. Answer in units of mol/L 75. ## Chemistry 1.How many grams of KCl is in 225 mL of a 0.452 M solution? 2.What volume in mL of 0.555 M solution contains 85.0g Na2SO4? 3.How many grams of CaCl2 would be dissolved in 2.0L of a 0.10 M solution of CaCl2? Can you please show me how to do these problems? 76. ## precalc I'm terrible at pre calc, could someone tell me if the solution set for log subscript 2 x + log sub 2 (x-1)=1 is the solution -1? Using either the elimination or substitution method is the solution set for 4x + y =11 and 2x -y = 7 is the answer (3, -1)? 77. ## chemistry 1. what mass of sodium hydroxide is required to make up 100 mL of 0.5000M solution 2.What volume of this solution must be diluted down to 100 mL if 0.100M solution of NaOH is required? 78. ## Math A doctor needs to administer 2 mL of a solution to a 200 pound patient. However, the doctor has the solution availible only in 20 mg/mL strength. What volume of the 20 mg/mL solution would be needed to administer the desired dose? 79. ## Chemistry A 14.7 mL sample of 0.0550 M KOH solution required 55.5 mL of aqueous acetic acid solution in a titration experiment. Calculate the molarity of the acetic acid solution. 80. ## math Solve the following system using elimination. If a single solution exists, write the solution as an ordered pair. Your answer will be an ordered pair, no solution, or infinitely many solutions x-y=4 3x-5=2 81. ## Buffer Solutions A buffered solution is made by adding 75.0g sodium acetate to 500.0 mL of a .64 M solution of acetic acid. What is the pH of the final solution? (Ka for acetic acid, CH3COOH, is 1.8 x 10^-5) 82. ## math a chemistry experiment calls for 30% sulfuric acid solution. if the supply room has only 50% and 20% sulfuric acid solution on hand,how much of each should be mixed to obtain 12 litres of 30% solution? The freezing point of an aqueous solution that contains a non-electrolyte is -9.0 degrees celcious. A. What is the freezing-point depression of the solution? B. What is the molal concentration of the solution? 84. ## algebra 1 Solve this system using elimination. If a single solution exists, write the solution as an ordered pair. Your answer will be an ordered pair, no solution, or infinitely many solutions. 5x=y=15 3x+2y=9 85. ## AP Chemistry A 65 mL sample of a solution of sulfuric acid is neutralized by 46 mL of a 0.14 M sodium hydroxide solution. Calculate the molarity of the sulfuric acid solution. Answer in units of mol/L 86. ## math A doctor needs to administer 2 mL of a 100 mg/mL solution to a 200-pound patient. However, the doctor has the solution available only in 20 mg/mL strength. What volume of the 20 mg/mL solution would be needed to administer the desired dose? 87. ## physics A rectangular block of ice 8 m on each side and 1.5 m thick floats in sea water. The density of the sea water is 1025 kg/m3. The density of ice is 917 kg/m3. a) How high does the top of the ice block float above the water level? b) How many penguins of 88. ## Chemisty How many grams of sodium bicarbonate mut be decomposed by: HCl (ag) + NaHCO3 (ag) ----> NaCl (ag) + CO2 (g) + H2O (l) to yield 75.0 mL of carbon dioxide at STP? 89. ## Science quick help needed identify the following formulas as acids, bases or salts: HCH3COO NaCL H2CO3 C6H8)7 Mg(OH) NaOH KBr Sr(OH)2 HNO3 please help :) thanks :) 90. ## chemistry How many liters of 0.101 m NaCl contains 1.70 mol of this salt? Assume this solvent is H2O. Answer in units of L I do not understand what i am looking for or how to solve for it. Steps would be appreciated. 91. ## chemistry 2 CaCl2, NaCl, and sugar derived from beet juice are all commonly used in winter to melt ice on streets. At the same molality, which of these would be most effective at this task and why? 92. ## chemistry name each of the following ionic compounds by stock system... couldn't find these in the list. a)NaCl b) KF c) CaS d) Co(NO3)2 e)FePO4 f)Hg2SO4 g) Hg3(PO4)2 93. ## chemistry Arrange the following aqueous solutions in order of increasing freezing points (lowest to highest temperature): 0.10 m glucose, 0.10 m BaCl2, 0.20 m NaCl, and 0.20 m Na2SO4. 94. ## Chemistry Which compound would you expect to have the lowest melting point? Base your answer off of the bonding theory presented in this course. MgO BeS NaCl KCl 95. ## Algebra How do you know if a value is a solution for an inequality? How is this different from determining if a value is a solution to an equation? If you replace the equal sign of an equation and put an inequality sign in its place, is there ever a time when the 96. ## math 116 How do you know if a value is a solution for an inequality? How is this different from determining if a value is a solution to an equation? If you replace the equal sign of an equation and put an inequality sign in its place, is there ever a time when the 97. ## math 116 How do you know if a value is a solution for an inequality? How is this different from determining if a value is a solution to an equation? If you replace the equal sign of an equation and put an inequality sign in its place, is there ever a time when the 98. ## chemistry A solution of PbI2 has [Pb2+] = 4.5 x 10-5 and [I-] = 6.5 x 10-4. PbI2 has Ksp = 8.7 x 10-9. Write down the reaction that is taking place. Calculate Q. Is Q larger or smaller than the Ksp? Is the solution unsaturated, saturated, or supersaturated? Will a 99. ## chemistry A solution of PbI2 has [Pb2+] = 4.5 x 10-5 and [I-] = 6.5 x 10-4. PbI2 has Ksp = 8.7 x 10-9. Write down the reaction that is taking place. Calculate Q. Is Q larger or smaller than the Ksp? Is the solution unsaturated, saturated, or supersaturated? Will a 100. ## Chemistry A solution contains 0.010 M BaCl2. Calculate the molarity of the ions in solution, the ionic strength of the solution, the activity coefficient of the barium ion and its activity. Repeat for the chloride ion. Show all calculations please.	6,860	24,583	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.71875	4	CC-MAIN-2018-34	latest	en	0.92233
https://mathoverflow.net/questions/30381/definition-of-function/30397	1,606,510,410,000,000,000	text/html	crawl-data/CC-MAIN-2020-50/segments/1606141194171.48/warc/CC-MAIN-20201127191451-20201127221451-00625.warc.gz	393,583,922	33,097	# Definition of Function What is authoritative canonical formal definition of function? For example, According to Wolfram MathWorld, $$isafun_1(f)\;\leftrightarrow\; \forall a\in f\;(\exists x\exists y \;\langle x,y\rangle = a) \; \wedge \; \forall x\forall y_1\forall y_2\;((\langle x,y_1\rangle\in f\wedge\langle x,y_2\rangle \in f)\rightarrow y_1=y_2))$$ According to Bourbaki "Elements de Mathematiques, Theorie des Ensembles", $$isafun_2(f)\;\leftrightarrow\; \exists d\exists g\exists c\;(\langle d,g,c\rangle=f \;\wedge\;isafun_1(g)\;\wedge$$ $$\;\wedge\; \forall x(x \in d\rightarrow \exists y(\langle x,y\rangle \in g)) \;\wedge\; \forall x\forall y(\langle x,y\rangle \in g\rightarrow (x \in d\wedge y\in c)))$$ How to make agree definition of function as triple with extensional equality $$\forall f\forall g\;[\;(isafun(f)\wedge isafun(g)) \; \rightarrow \; [\;(\forall x(\;f(x)=g(x)\;))\leftrightarrow f=g\;]\;]$$ ? Why such divergences in definitions exist? 1. Why function is not a pair in $isafun_2$? First component of triple is perfectly derivable from the second. 2. What word function exactly means if no underlying theory is specified in context? If I build fully formal knowledge base about mathematics for automated reasoning and want to add notion of contextless function -- how I must describe it? • I meant "equivalent" in the spirit of category theory. The answer, why there are different definitions, is simple: In analytical number theory for example, nobody cares for a fixed codomain for a function. But in other areas it may matter a lot: algebraic geometry, category theory, mathematical logic. In fact, in some definitions of a category, it is required that $hom(x,y)$ are pairwise disjoint for different $(x,y)$. – Martin Brandenburg Jul 3 '10 at 8:56 • To Martin Brandenburg: Give me a couple of examples, please, when fixed codomain matters a lot and just working with function range is not enough. – Vag Jul 3 '10 at 9:06 • 1) Well, extensiona equality has then the form: $f=g$ iff $cod(f)=cod(g), dom(f)=dom(g)$ and $f(x)=g(x)$ for all $x \in dom(f)$. I don't think this causes any problems. 2) Try to define "surjective" with the first definition. – Martin Brandenburg Jul 3 '10 at 9:16 • To Dror Speiser: I've seen numerous sources that define function as in 1st case, not only MathWorld. For instance, us.metamath.org/mpegif/df-fun.html. – Vag Jul 3 '10 at 12:23 • You're right, the definition in mathworld is just not the same as Bourbaki. Maybe you can send an email to update it, so that it would contain a formal paragraph with the second definition you wrote above,. – Dror Speiser Jul 3 '10 at 13:53 The fact is that different subject areas of mathematics use different definitions for this basic concept. The Bourbaki definition is quite common, particularly in many of the areas well-represented here on MO, but other areas use the ordered-pair definition. For example, if you open any set-theory text, you will find that a function $f$ is a set of ordered pairs having the functional property that any $x$ is paired with at most one $y$, denoted $f(x)$. This definition, which is completely established and much older than the Bourbaki definition, makes a function a special kind of binary relation, which is any set of ordered pairs. The domain of a function is the set of $x$ for which $f(x)$ exists. The range is the set of all such $f(x)$, and so on. The assertion $f:A\to B$ is a statement about the three objects, $f$, $A$ and $B$, that $f$ is a function with domain $A$ having its range a subset of $B$. In particular, the same function $f$ can have many different codomains. Another useful variation of the function concept is the concept of a partial function, common in many parts of logic, particularly set theory and computability theory. A partial function on $A$ is simply a function whose domain is included in $A$. In this case, we write $f:A\to B$, but with with three dots (my MO tex ability can't seem to do it), to mean that $f$ is a function with $dom(f)\subset A$ and $ran(f)\subset B$. This notion is particularly usefful in computability theory, where one has functions that might not produce an output on all input. But it also arises in set theory, where one often build partial orders consisting of small partial functions from one set to another. The union of a chain of such functions is a function again. It would be silly to insist in the Bourbaki style that there are really invisible functors running through this construction adjusting the domains and co-domains. One could object that the set-theorists could use the Bourbaki definition, if only they prepared better: in any context where many functions are treated, they should simply delimit an upper bound for the co-domains under consideration and use that co-domain for all the functions. But this proposal bumps into set-theoretic issues. For example, if I consider the class of all functions from an ordinal to the ordinals, then the only common co-domain is the class of all ordinals. But as this is a proper class, it isn't available if I want to consider only set functions. So there are good set-theoretic reasons not to use the Bourbaki definition. There are numerous other basic concepts that are given different precise meanings in different subjects of mathematics. For example, the concept of tree. In graph theory, it is a graph with no loops, whereas in set theory, it is a kind of partial order. In finite combinatorics, it might be a finitte partial order having no diamonds, but in the infinitary theory, one often means a partial order such that the predecessors of every node are well-ordered (making the levels of the tree form a well-ordered hierarchy). The graph-theoretic definition does not allow for the cases of Souslin trees and Kurepa trees, which are central in the other theory. There are surely numerous other examples where terminology differs. • The term "function" is not meaningless, but the precise definition has to be established by context. The same is true for many other words: "ring" (does it have a unity?), "poset" (weak, strong, or the same as a preorder?), "graph" (is it a simple graph?), etc. We lie to students by telling them that each term in mathematics has an authoritative definition, which doesn't stand up to close scrutiny. But it has an inner truth: the precise definition of "graph" may change, but it's not going to change into the definition of a ring. – Carl Mummert Jul 3 '10 at 13:24 • Vag, there are of course degrees of precision in mathematics, each serving a different kind of purpose. At the bottom underlying level, there is the hyper-precise accounts arising in the various formal systems, which would be more precise even than the definitions you and I mentioned, but few would argue that we should all operate only in that formal language just because we believe that it is possible in principle to do so. – Joel David Hamkins Jul 3 '10 at 14:41 • Vag, you are asking how we manage to teach the concept of function... just go into any class where it is taught, and you'll see that these problems you seem to be talking about simply do not exist. – Mariano Suárez-Álvarez Jul 3 '10 at 15:02 • Many words lack meaning out of context, while becoming precise in a context. Why should you expect that there is a meaning for this word outside of any context? Our various mathematical contexts are set up specifically to give words precise meanings, and there seems to be little reason why these contexts must agree. In the case of function, I took as my answer task to explain that indeed there is a difference in meaning for different subject areas. But we can all translate easily from one context to the other, and any amibguity is therefore quickly resolved. Is there any issue left? – Joel David Hamkins Jul 6 '10 at 16:54 • But the function concept has been made absolutely precise. In fact, it has been made fully precise twice, in two different ways. Each group prefers to use their own precise definition, for sound reasons, and I don't see how it would help anyone for you to give a third, incompatible definition, that was different than either of these. Rather than giving a "contextless" definition, you would instead merely be providing a third, unneeded, context. – Joel David Hamkins Jul 16 '10 at 20:00 When we use the Bourbaki definition of function as a triple $(domain, codomain, graph)$, then two functions are usually defined to be equal iff their domains and graphs are equal. So equal functions can have different codomains. The problem is that the same sign "=" is used both for the the equality of functions and the "universal" equality (In ZFC, for example, the "universal" equality is defined for all sets). That is, the sign "=" is overloaded. Normally, from the context one can determine what is the intended meaning. But there is a more serious trouble (as Vag pointed out early) with the Bourbaki definition, when a function is an element of a set. So it seems that the definition of function as a set of ordered pairs having the functional property is more preferable. • Let me devise some theory with contradiction in it. You'll come and see and state: "there is contradiction in it!". I'll reply: "No. This is misunderstanding on your side, because there is hidden patch that must be implicitly applied mentally while reading my theory, this patch removes contradiction completely." – Vag Jul 5 '10 at 19:55 • OK, I agree with your previous comment which you deleted. Then you asked about the case when a function is an element of a set. Indeed, in this case there is a serious trouble with the Bourbaki definition. So now in my work I decided not to use the Bourbaki definition. – Victor Makarov Jul 6 '10 at 12:32 • Could you explain the serious trouble you mention? Why can't we have Bourbaki triples being elements of a set? This seems to happen quite often. – Joel David Hamkins Jul 6 '10 at 17:08 • That sounds like a problem with calling that relation equality, rather than a problem with the Bourbaki definition of function itself. Of course, it isn't really equality, but just an equivalence relation, whose equivalence classes are canonically indexed by the ordered-pair definition of function. But we have no reason to expect equality of sets to respect that relation. – Joel David Hamkins Jul 6 '10 at 23:18 • Moschovakis uses the set-theoretic definition of function---the set of ordered pairs definition---not the Bourbaki definition. And for that definition, this notion of equivalence is equality. – Joel David Hamkins Jul 8 '10 at 20:48	2,572	10,614	{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.609375	4	CC-MAIN-2020-50	latest	en	0.638314
https://gomathanswerkey.com/texas-go-math-grade-3-lesson-10-5-answer-key/	1,718,739,073,000,000,000	text/html	crawl-data/CC-MAIN-2024-26/segments/1718198861773.80/warc/CC-MAIN-20240618171806-20240618201806-00155.warc.gz	254,567,261	56,425	# Texas Go Math Grade 3 Lesson 10.5 Answer Key Model with Arrays Refer to our Texas Go Math Grade 3 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 3 Lesson 10.5 Answer Key Model with Arrays. ## Texas Go Math Grade 3 Lesson 10.5 Answer Key Model with Arrays Essential Question How can you use arrays to solve division problems? An arrangement of objects, pictures, or numbers in columns and rows is called an array, and division of the objects or pictures in equal group solve the problem. Investigate Materials square tiles You can use arrays to model division and find equal groups. A. Count out 30 tiles. Make an array to find how many rows of 5 are in 30. 6 rows Explanation: 30 tiles are grouped 5 of 6 rows 30Ã·5 = 6 B. Make a row of 5 tiles. Explanation: The array contains 30 tiles which are grouped 5 of 6 rows. 30Ã·5 = 6 C. Continue to make as many rows of 5 tiles as you can. How many rows of 5 did you make? ____ 6 rows Explanation: 30Ã·5 = 6 So, we can make 6 rows. Make Connections You can write a division equation to show how many rows of 5 are in 30. Show the array you made in Investigate by completing the drawing below. 30 Ã· 5 = ___ There are ___ rows of 5 tiles in 30. So, 30 Ã· 5 = _____ completing the drawing: 30 Ã· 5 = 6 There are 6 rows of 5 tiles in 30. So, 30 Ã· 5 = 6 Math Idea You can divide to find the number of equal rows or to find the number in each row. n x n 1 x 1 2 x 2 3 x 3 Share and Show Use square tiles to make an array. Solve. Question 1. How many rows of 3 are in 18? 6 rows Explanation: Array has 18 tiles arranged in rows. Each row has 3 tiles. Total rows in array 18 Ã· 3 = 6 How many rows of 6 are in 12? 2 rows Explanation: Array has 12 tiles arranged in rows. Each row has 6 tiles. Total rows in array 18 Ã· 6 = 2 Question 3. How many rows of 7 are in 21? 3 rows Explanation: Array has 21 tiles arranged in rows. Each row has 7 tiles. Total rows in array 21 Ã· 7 = 3 Question 4. How many rows of 8 are in 32? 4 rows Explanation: Array has 32 tiles arranged in rows. Each row has 8 tiles. Total rows in array 32 Ã· 8 = 4 Make an array. Then write a division equation. Question 5. 25 tiles in 5 rows 5 tiles in a row 25Ã·5 = 5 Explanation: Array has 25 tiles arranged in rows. Total 5 rows in array Each row has 25 Ã· 5 = 5 Question 6. 14 tiles in 2 rows 7 tiles in a row 14Ã·2 = 7 Explanation: Array has 14 tiles arranged in rows. Each row has 7 tiles Total rows in array 14 Ã· 2 = 7 Question 7. 28 tiles in 4 rows 7 tiles in a row 28Ã·4 = 7 Explanation: Array has 28 tiles arranged in rows. Total 4 rows in array Each row has 28 Ã· 7 = 7 27 tiles in 9 rows 3 tiles in a row 27Ã·9=3 Explanation: Array has 27 tiles arranged in rows. Total 9 rows in array Each row has 27 Ã· 9 = 3 Question 9. How many rows of 3 are in 15? 5 rows 15 Ã· 5 = 3 Explanation: Array has 15 tiles arranged in rows. Each row has 3 tiles Total rows in array 27 Ã· 9 = 3 Question 10. How many rows of 8 are in 24? 3 rows 24 Ã· 8 = 3 Explanation: Array has 24 tiles arranged in rows. Each row has 8 tiles Total rows in array 24 Ã· 8 = 3 Math Talk Mathematical Processes Explain when you Count the number of rows to find the answer and when you count the number of tiles in each row to find the answer. To find the number of tiles in a row,Â total tiles Ã· number of row is calculated. To find the number of rows,Â total tiles Ã· number of tiles in row is calculated. Question 11. Write Math Show two ways you could make an array with tiles for 18 Ã· 6. Shade squares on the grid to record the arrays. 18 Ã· 6 = 3 6 rows of 3 tiles or 3 rows of 6 tiles Explanation: count the number of tiles on the grid and shade as, 6 rows of 3 tiles 3 rows of 6 tiles Problem Solving Question 12. Thomas has 28 tomato seedlings to plant in his garden. He wants to plant 7 seedlings in each row. How many rows of tomato seedlings will Thomas plant? 4 rows of tomato seedlings. Explanation: Thomas has 28 tomato seedlings to plant in his garden. He wants to plant 7 seedlings in each row. Total rows of tomato seedlings Thomas plant 28 Ã· 7 = 4 Question 13. H.O.T. Use Math Language Tell how to use an array to find how many rows of 8 are in 40. An array is an orderly arrangement (often in rows, columns or a matrix) that is most commonly used as a visual tool for demonstrating multiplication and division . 40Â Ã· 8 = 5 Explanation: An array refers to a set of numbers or objects that will follow a specific pattern. 40Â Ã· 8 = 5 Faith plants 36 flowers in 6 equal rows. How many flowers are in each row? 6 flowers in each row. Explanation: Faith plants 36 flowers in 6 equal rows. Total flowers in each row 36 Ã· 6 = 6 Question 15. H.O.T. Multi-Step There were 20 plants sold at a store on Saturday and 30 plants sold at the store on Sunday. Customers bought 5 plants each. How many customers in all bought the plants? 10 customers bought the plants. Explanation: There were 20 plants sold at a store on Saturday and 30 plants sold at the store on Sunday. Total plants sold on both days 20 + 30 = 50 Customers bought 5 plants each. Number of customers bought the plants 50 Ã· 5 =10 Question 16. H.O.T. Multi-Step Lionel made an array with 24 tiles. The number of rows is 5 more than the number of tiles in each row. How many rows are in Lionelâ€™s array? 3 rows Explanation: Lionel made an array with 24 tiles. The number of rows is 5 more than the number of tiles in each row. Number of rows in Lionel’s array is 3 Fill in the bubble for the correct answer choice. You may use objects or models to solve. Question 17. Multi-Step Mrs. Weston is baking 24 oatmeal raisin cookies. There will be 4 rows of cookies on the cookie tray. She has already put one row of cookies on the tray. How many more cookies does she have left to put on the tray? (A) 6 (B) 20 (C) 18 (D) 3 Mrs. Weston is baking 24 oatmeal raisin cookies. 24 Ã· 4 = 6 4 – 1 = 3 rows left 3 x 6 = 18 18 more cookies she have left to put on the tray. Question 18. Representations Which division equation is shown by the array? (A) 24 Ã· 3 = 8 (B) 24 Ã· 4 = 6 (C) 24 Ã· 2 = 12 (D) 21 Ã· 3 = 7 Option (A) Explanation: The array contains 24 tiles, arranged as 3 rows of 8 tiles. Question 19. Multi-Step Hajune finds 9 black rocks one week and 9 black rocks the next week. He wants to keep the rocks in a box with 3 rows. How many black rocks can he put in each row? (A) 18 (B) 6 (C) 12 (D) 3 6 black rocks in a box. Explanation: Hajune finds 9 black rocks one week and 9 black rocks the next week. Total black rocks 9 + 9 = 18 He wants to keep the rocks in a box with 3 rows. Total black rocks he put in each row 18 Ã· 3 = 6 Texas Test Prep Question 20. Sally makes an array using 36 coins. If she puts 4 coins in each row, how many rows does she make? (A) 40 (B) 32 (C) 8 (D) 9 Option(D) Explanation: Sally makes an array using 36 coins. If she puts 4 coins in each row, Total rows she make 36 Ã· 4 = 9 ### Texas Go Math Grade 3 Lesson 10.5 Homework and Practice Answer Key Make an array. Then write a division equation. Question 1. 21 tiles in 3 rows 7 tiles in each row 21 Ã· 3 = 7 Explanation: Array has 21 tiles arranged in 3 rows. Each row has 27 Ã· 9 = 3 tiles Question 2. 36 tiles in 6 rows 6 tiles in each row 36 Ã· 6 = 6 Explanation: Array has 36 tiles arranged in 6 rows. Each row has 36 Ã· 6 = 6 tiles Question 3. 16 tiles in 8 rows 2 tiles in each row 16 Ã· 8 = 2 Explanation: Array has 16 tiles arranged in 8 rows. Each row has 16 Ã· 8 = 2 tiles Question 4. 18 tiles in 3 rows 6 tiles in each row 18 Ã· 3 = 6 Explanation: Array has 18 tiles arranged in 3 rows. Each row has 18 Ã· 3 = 6 tiles Question 5. How many rows of 4 are in 32? 8 tiles in each row 32 Ã· 4 = 8 Explanation: Array has 21 tiles arranged. Each row has 4 tiles number of rows 27 Ã· 9 = 3 Question 6. How many rows of 4 are in 16? 4 tiles in each row 16 Ã· 4 = 4 Explanation: Array has 21 tiles arranged. Each row has 4 tiles number of rows 27 Ã· 9 = 3 Problem Solving Question 7. Hannah has a collection of 27 seashells. She wants to put 9 shells on each shelf. How many shelves does Hannah need? 3 shelves 27 Ã· 9 = 3 Explanation: Hannah has a collection of 27 seashells. She wants to put 9 shells on each shelf. Number of shelves Hannah needs 27 Ã· 9 = 3 Go Math Grade 3 Practice and Homework Lesson 10.5 Answer Key Question 8. Dexter buys 18 fish. He wants to put an equal number of fish into 3 fishbowls. How many fish will he put in each fishbowl? 6 fish in each bowl. Explanation: He wants to put an equal number of fish into 3 fishbowls. Number of fish he put in each fishbowl 18 Ã· 3 = 6 Question 9. Quentin is making a tile design with 35 tiles. He wants to put the tiles in rows. How can he set up the tiles so there are an equal number of tiles in each row? 35Ã·7 =5 or 35Ã·5 =7 Explanation: Quentin is making a tile design with 35 tiles. He wants to put the tiles in rows. He can set up the tiles in 2 ways, an equal number of tiles in each row 35Ã·7 =5 or 35Ã·5 =7 Texas Test Prep Lesson Check Question 10. Marla makes this array. What division equation can she write? (A) 14 Ã· 3 = 7 (B) 14 Ã· 2 = 7 (C) 2 Ã— 7 = 14 (D) 7 + 7 = 14 Option(B) Explanation: Marla makes array of 14 tiles, arranged in 2 equal rows. Number of tiles in each row 14 Ã· 2 = 7 Question 11. Claude makes this array. What division equation can he write? (A) 24 Ã· 4 = 6 (B) 24 Ã· 3 = 8 (C) 4 Ã— 6 = 24 (D) 6 Ã— 4 = 20 Option(A) Explanation: Claude makes array of 24 tiles. Each row contains 4 tiles. Total tiles arranged in equal rows 24Â Ã· 4 = 6 Question 12. Sondra runs for 32 minutes. If she runs 4 miles, how many minutes does it take her to run one mile? (A) 6 minutes (B) 7 minutes (C) 8 minutes (D) 28 minutes Option(C) Explanation: Sondra runs for 32 minutes. If she runs 4 miles, Total minutes it take her to run one mile 32 Ã· 4 = 8 Question 13. There are 25 students on the playground. The students are in teams of 5. How many teams are on the playground? (A) 6 (B) 5 (C) 4 (D) 7 Option(B) Explanation: There are 25 students on the playground. The students are in teams of 5. Total teams on the playground 25 Ã·Â 5 = 5 Question 14. Multi-Step Jennifer collects 9 animal cards and 21 nature cards. She wants to arrange them on a bulletin board in 6 rows. How many cards will be in each row? (A) 5 (B) 2 (C) 3 (D) 7 Option(A) Explanation: Jennifer collects 9 animal cards and 21 nature cards. 9 + 21 = 30 She wants to arrange them on a bulletin board in 6 rows. Number of cards in each row 30 Ã· 6 = 5 Question 15. Multi-Step Zach is planting a garden of 27 tomato plants. He puts the plants into rows of 9. If he has already planted one row, how many more rows does he need to plant? (A) 3 (B) 2 (C) 18 (D) 4	3,330	10,808	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	4.84375	5	CC-MAIN-2024-26	latest	en	0.891899
https://nrich.maths.org/public/leg.php?code=-68&cl=3&cldcmpid=1576	1,508,818,904,000,000,000	text/html	crawl-data/CC-MAIN-2017-43/segments/1508187828134.97/warc/CC-MAIN-20171024033919-20171024053919-00183.warc.gz	790,679,442	10,200	# Search by Topic #### Resources tagged with Visualising similar to Logoland - Sequences: Filter by: Content type: Stage: Challenge level: ### Triangles Within Pentagons ##### Stage: 4 Challenge Level: Show that all pentagonal numbers are one third of a triangular number. ### Triangles Within Triangles ##### Stage: 4 Challenge Level: Can you find a rule which connects consecutive triangular numbers? ### Triangles Within Squares ##### Stage: 4 Challenge Level: Can you find a rule which relates triangular numbers to square numbers? ### Three Frogs ##### Stage: 4 Challenge Level: Three frogs hopped onto the table. A red frog on the left a green in the middle and a blue frog on the right. Then frogs started jumping randomly over any adjacent frog. Is it possible for them to. . . . ### Natural Sum ##### Stage: 4 Challenge Level: The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . . ### Jam ##### Stage: 4 Challenge Level: A game for 2 players ### Yih or Luk Tsut K'i or Three Men's Morris ##### Stage: 3, 4 and 5 Challenge Level: Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . . ### Coordinate Patterns ##### Stage: 3 Challenge Level: Charlie and Alison have been drawing patterns on coordinate grids. Can you picture where the patterns lead? ### Hypotenuse Lattice Points ##### Stage: 4 Challenge Level: The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN? ### Rotating Triangle ##### Stage: 3 and 4 Challenge Level: What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle? ### Problem Solving, Using and Applying and Functional Mathematics ##### Stage: 1, 2, 3, 4 and 5 Challenge Level: Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information. ### Jam ##### Stage: 4 Challenge Level: To avoid losing think of another very well known game where the patterns of play are similar. ### More Pebbles ##### Stage: 2 and 3 Challenge Level: Have a go at this 3D extension to the Pebbles problem. ### Icosagram ##### Stage: 3 Challenge Level: Draw a pentagon with all the diagonals. This is called a pentagram. How many diagonals are there? How many diagonals are there in a hexagram, heptagram, ... Does any pattern occur when looking at. . . . ### Building Gnomons ##### Stage: 4 Challenge Level: Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible. ### Hidden Rectangles ##### Stage: 3 Challenge Level: Rectangles are considered different if they vary in size or have different locations. How many different rectangles can be drawn on a chessboard? ### Dice, Routes and Pathways ##### Stage: 1, 2 and 3 This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . . ### Cuboids ##### Stage: 3 Challenge Level: Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all? ### Proximity ##### Stage: 4 Challenge Level: We are given a regular icosahedron having three red vertices. Show that it has a vertex that has at least two red neighbours. ### Seven Squares - Group-worthy Task ##### Stage: 3 Challenge Level: Choose a couple of the sequences. Try to picture how to make the next, and the next, and the next... Can you describe your reasoning? ### AMGM ##### Stage: 4 Challenge Level: Can you use the diagram to prove the AM-GM inequality? ### Seven Squares ##### Stage: 3 Challenge Level: Watch these videos to see how Phoebe, Alice and Luke chose to draw 7 squares. How would they draw 100? ### Intersecting Circles ##### Stage: 3 Challenge Level: Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have? ### Sliding Puzzle ##### Stage: 1, 2, 3 and 4 Challenge Level: The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves. ### Steel Cables ##### Stage: 4 Challenge Level: Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions? ### The Triangle Game ##### Stage: 3 and 4 Challenge Level: Can you discover whether this is a fair game? ### Picture Story ##### Stage: 4 Challenge Level: Can you see how this picture illustrates the formula for the sum of the first six cube numbers? ### Changing Places ##### Stage: 4 Challenge Level: Place a red counter in the top left corner of a 4x4 array, which is covered by 14 other smaller counters, leaving a gap in the bottom right hand corner (HOME). What is the smallest number of moves. . . . ### Mystic Rose ##### Stage: 4 Challenge Level: Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes. ### Königsberg ##### Stage: 3 Challenge Level: Can you cross each of the seven bridges that join the north and south of the river to the two islands, once and once only, without retracing your steps? ### Cubes Within Cubes Revisited ##### Stage: 3 Challenge Level: Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need? ### Clocked ##### Stage: 3 Challenge Level: Is it possible to rearrange the numbers 1,2......12 around a clock face in such a way that every two numbers in adjacent positions differ by any of 3, 4 or 5 hours? ### John's Train Is on Time ##### Stage: 3 Challenge Level: A train leaves on time. After it has gone 8 miles (at 33mph) the driver looks at his watch and sees that the hour hand is exactly over the minute hand. When did the train leave the station? ### Tilting Triangles ##### Stage: 4 Challenge Level: A right-angled isosceles triangle is rotated about the centre point of a square. What can you say about the area of the part of the square covered by the triangle as it rotates? ### Painting Cubes ##### Stage: 3 Challenge Level: Imagine you have six different colours of paint. You paint a cube using a different colour for each of the six faces. How many different cubes can be painted using the same set of six colours? ### Pattern Power ##### Stage: 1, 2 and 3 Mathematics is the study of patterns. Studying pattern is an opportunity to observe, hypothesise, experiment, discover and create. ### Convex Polygons ##### Stage: 3 Challenge Level: Show that among the interior angles of a convex polygon there cannot be more than three acute angles. ### Cubes Within Cubes ##### Stage: 2 and 3 Challenge Level: We start with one yellow cube and build around it to make a 3x3x3 cube with red cubes. Then we build around that red cube with blue cubes and so on. How many cubes of each colour have we used? ### Coloured Edges ##### Stage: 3 Challenge Level: The whole set of tiles is used to make a square. This has a green and blue border. There are no green or blue tiles anywhere in the square except on this border. How many tiles are there in the set? ### Muggles Magic ##### Stage: 3 Challenge Level: You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area. ### Cutting a Cube ##### Stage: 3 Challenge Level: A half-cube is cut into two pieces by a plane through the long diagonal and at right angles to it. Can you draw a net of these pieces? Are they identical? ### Flight of the Flibbins ##### Stage: 3 Challenge Level: Blue Flibbins are so jealous of their red partners that they will not leave them on their own with any other bue Flibbin. What is the quickest way of getting the five pairs of Flibbins safely to. . . . ### Reflecting Squarely ##### Stage: 3 Challenge Level: In how many ways can you fit all three pieces together to make shapes with line symmetry? ### Squares, Squares and More Squares ##### Stage: 3 Challenge Level: Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares? ##### Stage: 3 Challenge Level: Can you mark 4 points on a flat surface so that there are only two different distances between them? ### Playground Snapshot ##### Stage: 2 and 3 Challenge Level: The image in this problem is part of a piece of equipment found in the playground of a school. How would you describe it to someone over the phone? ### Threesomes ##### Stage: 3 Challenge Level: Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw? ### Rolling Triangle ##### Stage: 3 Challenge Level: The triangle ABC is equilateral. The arc AB has centre C, the arc BC has centre A and the arc CA has centre B. Explain how and why this shape can roll along between two parallel tracks. ### Tourism ##### Stage: 3 Challenge Level: If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.	2,271	9,809	{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}	3.828125	4	CC-MAIN-2017-43	latest	en	0.826331

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