course Mth 163 end programZsǷqLL
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19:36:23 `q001. Note that this assignment has 2 questions For the function y = 1.1 x + .8, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2? What therefore is the rise between these two points, and what is the run? What is the average slope of the graph between these two points? Be sure to simplify your result.
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RESPONSE --> confidence assessment:
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HЯD̏θ䙣d assignment #009 009. Precalculus I 03-30-2008
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19:46:14 `q001. Note that this assignment has 2 questions For the function y = 1.1 x + .8, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2? What therefore is the rise between these two points, and what is the run? What is the average slope of the graph between these two points? Be sure to simplify your result.
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RESPONSE --> The coordinates of x=x1 points are x1, 1.1, x1+ .8. The coordinates of x=x2 points are x2, 1.1, x2 + .8. The rise of the two points are: (1.1 x2 + .8) - (1.1 x1 + .8) = 1.1 x2 + .8 - 1.1 x1 - .8 = 1.1 x2 -1.1 x1. The run of the two points are: x2 - x1. The slope is: (1.1 x2 - 1.1 x1) / (x2 - x1) = 1.1 (x2 - x1) / (x2-x1) = 1.1 confidence assessment: 2
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19:48:07 In terms of the symbol x1 the coordinates of the x = x1 point are ( x1, 1.1 x1 + .8) and the coordinates of the x = x2 point are ( x2, 1.1 x2 + .8). The rise between the two points is therefore rise = (1.1 x2 + .8) - (1.1 x1 + .8) = 1.1 x2 + .8 - 1.1 x1 - .8 = 1.1 x2 - 1.1 x1. The run is run = x2 - x1. The slope is therefore (1.1 x2 - 1.1 x1) / (x2 - x1) = 1.1 (x2 - x1) / (x2 - x1) = 1.1.
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RESPONSE --> I see how the coordinates of the points are found. The rise and run seem easy to find. It just a matter of putting everything in order. Finding the slope is also another thing that seems easy enough to find. You just need to place the rise over the run and divide it to find the slope. self critique assessment: 3
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19:57:29 `q002. For the function y = 3.4 x + 7, what are the coordinates of the x = x1 point, in terms of the symbol x1? What are the coordinates of the x = x2 point, in terms of the symbol x2? What therefore is the rise between these two points, and what is the run? What is the average slope of the graph between these two points? Be sure to simplify your result.
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RESPONSE --> The coordinate points of x=x1 are x1, 3.4 x1 + 7 and the corrdinante points of x=x2 are x2, 3.4 x2 + 7. The rise of the two points are: (3.4 x2 +7) - (3.4 x1 + 7) = 3.4 x2 + 7 - 3.4 x1 - 7 = 3.4 x2 - 3.4 x1. The run of the two points are: x2 - x1 The slope is: (3.4 x2 - 3.4 x1) / (x2 - x1) = 3.4 (x2 - x1) / (x2- x1) = 3.4 confidence assessment: 2
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19:58:22 In terms of the symbol x1 the coordinates of the x = x1 point are ( x1, 3.4 x1 + 7) and the coordinates of the x = x2 point are ( x2, 3.4 x2 + 7). The rise between the two points is therefore rise = (3.4 x2 + 7) - (3.4 x1 + 7) = 3.4 x2 + 7 - 3.4 x1 - 7 = 3.4 x2 - 3.4 x1. The run is run = x2 - x1. The slope is therefore (3.4 x2 - 3.4 x1) / (x2 - x1) = 3.4 (x2 - x1) / (x2 - x1) = 3.4.
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RESPONSE --> I understand how you find the coordinate points. I also see how you find the rise, run, and slope. It is just a matter of getting everything in the place that they belong. self critique assessment: 3
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לЎkޓαӒf~zՆ assignment #010 010. Precalculus I 03-30-2008
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20:13:14 `q001. Note that this assignment has 10 questions Sketch the function y = x and describe your graph. Describe how the graphs of y = .5 x and y = 2 x compare with the graph of y = x.
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RESPONSE --> The graph of y=x is a straight line that goes through (0,0) and has a slope of 1. It moves up the graph like (-1,-1) and (1,1). The graph of y=.5 lie closer to the x-axis. Some of the points are (0,0), (1, .5), and (-1, -.5) The graph y=2x are twice as far as the original graph. The points are (0,0), (1,2) and (-1,2). Depending on what is infront of the x is where the next point on the graph goes. All of these graphs start at (0,0). From there it just depends on what is with x as where the next point goes. confidence assessment: 2
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20:18:46 The graph of y = x consists of a straight line through the origin, having slope 1. This line has basic points (0,0) and (1,1). The points of the graph of y = .5 x all lie twice as close to the x-axis as the points of the graph of y = x. The point (0,0) of the y = x graph is already on the x-axis, so the corresponding point on the graph of y = .5 x is also (0,0). The point (1,1) of the y = x graph lies 1 unit above the x-axis, so the corresponding point on the graph of y = .5 x will lie twice as close, or .5 units above the x-axis, so that the corresponding point is (1, .5). The graph of y = .5 x Thus passes through the points (0,0) and (1,.5). Of course this result could have been found by simply plugging 0 and 1 into the function y = .5 x, but the point here is to see that we can get the same result if we think of moving all points twice as close. This order thinking will be useful when dealing with more complex functions. Thinking along similar lines we expect the points of the graph of y = 2 x to all lie twice as far from the x-axis as the points of the function y = x. Thus the two basic points (0,0) and (1,1) of the y = x graph will correspond to the points (0,0) and (1,2) on the graph of y = 2 x.
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RESPONSE --> I see what is being done. With all of these graphs you can visual what the graphs would look like even before you start. When you go to work them out and you put all of them on the graph and compare them you can see that they have the same origin but the points vary. It basically depends on what is infront of the variable x. When you have y=.5x then the grph is going to be closer to the origin. When you have y=2x then the graph is going to be twice as far from the origin. self critique assessment: 3
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20:26:44 `q002. If we were to sketch all the graphs of the form y = a x for which .5 < a < 2, what would our sketch look like?
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RESPONSE --> If you have y=.5x then the points in the graph will be (-1, -.5), (0,0), and (1, .5). If you have y=2x then the points will be (1,2), (0,0) and (-1, -2). With these figured out you would be able to figure out that graphs for .5 < a< 2 would like somewhere between y=.5x and y=2x. You would then have the slope of .5 and 2 so the graphs would lie between .5 and 2. confidence assessment: 2
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20:29:54 If a =.5 then our function is y = .5 x and the basic points will be (0,0) and (1,.5), as seen in the preceding problem. Similarly if a = 2 then our function is y = 2 x, with basic points (0,0) and (1,2). For .5 < a < 2, our functions will lie between the graphs of y = .5 x and y = 2 x. Since these two functions have slopes .5 and 2, the slopes of all the graphs will lie between .5 and 2. We could represent these functions by sketching dotted-line graphs of y = .5 x and y = 2 x (the dotted lines indicating that these graphs are not included in the family, because the < sign does not include equality). We could then sketch a series of several solid lines through the origin and lying between the two dotted-line graphs.
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RESPONSE --> I understand what is being done. If you need to find a graph that is between .5 < a< 2. You would need to figure out the points from the two main graphs. These would be y=.5x and y=2x. Once you have found these points then you can figure out that the graph would then lie somewhere between the graphs of .5 and 2. self critique assessment: 3
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20:44:07 `q003. Describe how the graphs of y = x - 2 and y = x + 3 compare with the graph of y = x. If we were to sketch all graphs of the form y = x + c for -2 < x < 3, what would our graph look like?
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RESPONSE --> The graph of y=x is a straight line. When you compare it to y=x-2 you see that it's points lie 2 places below thr origin. The graph y=x+3, the points are 3 units above the origin. To find y=x+c for -2 < x <3 it seems that you would taake the y=x-2 and y= x+3 and place those graph. You would see that the answer for y=x+c would lie somewhere between the other two. confidence assessment: 2
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20:51:45 The graph of y = x - 2 lies at every point 2 units below the corresponding point on the graph of y = x, so this graph is parallel to the graph of y = x and 2 units lower. Similarly the graph of y = x + 3 lies parallel to the graph of y = x and 3 units higher. To sketch the family y = x + c for -2 < x < 3, we first can draw dotted-line graphs of y = x - 2 and y = x + 3, then a series of several solid line graphs, all parallel to the graph of y = x, lying between the two dotted-line graphs. STUDENT COMMENT: I got a little confused with y = x + c part, but I understand the first part completely. ** The instructions said to sketch all graphs of the form y = x + c for -2 < x < 3. So for example c could be -1, 0, 1 or 2, giving us the functions y = x - 1, y = x, y = x + 1 andy x+ 2. c could also be -1.9, or .432, or 2.9, giving us functions y = x - 1.9, y = x + .432, y = x + 2.9. c can be any number between -2 and 3. These graphs are as described in the given solution. **
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RESPONSE --> I see how the other graps are done. I wasn't sure how to do the graph for y=x+c for -2
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21:01:26 `q004. Describe how the graph of y = 2 x compares with the graph of y = x. Describe how the graph of y = 2 x - 2 compares with the graph of y = 2 x.
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RESPONSE --> When you have graph y=2x every point on the graph will be twice as far from the origin. When this graph is compared to the graph of y=x you find that the points on the grpah y=2x will be twice as far from the origin than y=x/ When you have the graph of y=2x-2 every point on the graph will be 2 units below the graph y=2x. confidence assessment: 2
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21:07:04 The graph of y = 2 x lies at every point twice as far the x-axis as that of y = x. This graph passes through the points (0,0) and (1, 2), i.e., passing through the origin with slope 2. The graph of y = 2x - 2 will lie 2 units below the graph of y = 2 x. This graph will therefore have a slope of 2 and will pass-through the y axis at (0, -2).
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RESPONSE --> I understand how this is done. It is just a matter of understanding where the points will lie. For the graph y=2x-2 the coordinates will pass through (0,-2). It varies from the graph of y=2x because the first graph will be negative more while the other will be more positive. self critique assessment: 3
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21:14:55 `q005. Suppose we graph y = 2 x + c for all values of c for which -2 < c < 3. What with our graph look like?
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RESPONSE --> You would have to see what the points would be. One of the graphs would be y=x-2. The points on the graph would be two units below the origin. For the graph y=x + 3. The points on the graph would be three units above the origin. Therefore the graph for y= 2x +c for which -2 < c< 3 would lie somewhere between -2 and 3. confidence assessment: 2
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21:17:10 Each graph will lie c units vertically from the graph of y = 2 x, therefore having slope 2 the passing through the y-axis at the point (0, c). The family of functions defined by y = 2 x + c will therefore consist of a series of straight lines each with slope 2, passing through the y-axis between (0, -2) and (0, 3).
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RESPONSE --> When you place the points for the grpah the points would be (0,c), (0, -2), and (0,3). Therefore the points for the graph would basically lie somewhere between -2 and 3. self critique assessment: 3
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21:20:51 `q006. Sketch two points, not particularly close to one another, with one point in the second quadrant and the other in the first, with clearly different y values. Label the first point (x1, y1) and the second (x2, y2). Draw a straight line passing through both of these points and extending significantly beyond both. In terms of the symbols x1, x2, y1, and y2, what is slope of this straight line?
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RESPONSE --> To find the rise you would take y=y1 and y=y2 and the you would have a rise of y2-y1. The run is found the same way. The run would be x2-x1. The slope is then (y2-y1) / (x2-x1). confidence assessment: 2
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21:21:21 The rise of a line is from y = y1 to y = y2, a rise of y2-y1. The run is similarly found to be x2-x1. The slope is therefore slope = (y2-y1) / (x2-x1).
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RESPONSE --> I understand how it is done. It is easy to see and figure out how to find the rise, run, and slope. self critique assessment: 3
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21:24:07 `q007. On the sketch you made for the preceding problem, and add a point (x, y) on your straight line but not between the two points already labeled, and not too close to either. What is the slope from (x1, y1) to (x, y)?
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RESPONSE --> The slope looks to be from (x1, y1) to (x, y). The slope is then equal to the rise/run which is (y-y1)/ (x-x1). confidence assessment: 2
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21:24:43 The slope from (x1, y1) to (x, y) is slope = rise/run = (y - y1) / (x - x1).
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RESPONSE --> I see how it is done. When you write down the points and see the graph I can understand how to find the slope, rise, and run. self critique assessment: 3
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21:25:34 `q008. Should the slope from (x1, y1) to (x, y) be greater than, equal to or less than the slope from (x1, y1) to (x2, y2)?
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RESPONSE --> When you figure out the slope between both lines you can see the they are a straight line. That would make the slopes equal. confidence assessment: 2
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21:26:00 The slope between any two points of a straight line must be the same. The two slopes must therefore be equal.
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RESPONSE --> I understand it. If the slopes are the same then thay are both a straight line and at the same time they are equal. self critique assessment: 3
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21:28:10 `q009. The slope from (x1, y1) to (x, y) is equal to the slope from (x1, y1) to (x2, y2). If you set the expressions you obtained earlier for the slopes equal to one another, what equation do you get?
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RESPONSE --> The slopes you would have are (y2-y1) / (x2-x1). Also there would be (y-y1) / (x-x1). Then if you set the two slopes equal to each other you would then have (y2-y1)/ (x2-x1) = (y2-y1) / (x2-x1). confidence assessment: 2
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21:28:56 The slopes are (y2 - y1) / (x2 - x1) and (y - y1) / (x - x1). Setting the two slopes equal we obtain the equation (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1).
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RESPONSE --> I see. If one slope is one equation and they are eqaul to each other then the two slopes are the same. self critique assessment: 3
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21:32:53 `q010. Setting the two slopes equal you should have obtained (y - y1)/(x - x1) = (y2 - y1)/(x2 - x1). Solve this equation for y.
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RESPONSE --> You would take (y-y1) / (x-x1) = (y2-y1)/ (x2-x1). This would then equal (y2-y1)/(x2-x1). Then you have to multiply both sides by (x-x1) to get y by itself. When you do this you get (y-y1) = (y2-y1) / (x2-x1) * (x-x1). Then you add y1 to both sides. Now you have y= (y2-y1)/ (x2-x1) * (x-x1) + y1. confidence assessment: 2
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21:34:14 Starting with (y - y1) / (x - x1) = (y2 - y1) / (x2 - x1), we wish to isolate y on the left-hand side. We therefore begin by multiplying both sides by (x - x1) to obtain (y - y1) = (y2 - y1) / (x2 - x1) * (x - x1). We could then add y1 to both sides to obtain y = (y2 - y1) / (x2 - x1) * (x - x1) + y1.
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RESPONSE --> I see how this is done. In order to solve for y you have to have you equation out in front of you and then get y by itself on the left hand side. In order to do this you have to multiply both sides by (x-x1). After doing that you have to add y1 to both sides. when you do this you get the final equation with y= the answer. self critique assessment: 3
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ܷΜG{㚽 assignment #011 011. Precalculus I 03-30-2008
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21:39:15 `q001. Note that this assignment has 11 questions How many squares one foot on a side would it take to construct a square two feet on a side?
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RESPONSE --> I think it would take 2 one foot squares to make this complete. confidence assessment: 2
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21:41:48 A common response is that it takes 2 one-ft. squares to make a 2-foot square. However, below thought shows that this isn't the case. If we put 2 one foot squares side by side we get a one-foot by two-foot rectangle, not a square. If we put a second such rectangle together with the first, so that we have 2 rows with 2 squares in a row, then we have a two-foot square. Thus we see that it takes 4 squares one foot on a side to make a square 2 ft. on a side.
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RESPONSE --> I see that my answer wasn't correct. I see that when you place 2 one foot squares together you get a rectangle ad I need to get a square. If you add another 2 one foot square together you get a square. This square would consist of 2 rows with 2 squares. It would take 4 squares one foot on each side on order to make a square that is 2 feet on a side. self critique assessment: 3
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21:42:32 `q002. How many cubes one foot on a side would it take to construct a cube two feet on a side?
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RESPONSE --> It would take 3 rows with 3 squares on each side and this would be a total of 9 squares. confidence assessment: 2
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21:45:01 We could begin by constructing two rows with two cubes in a row, which would sit on a square two feet by two feet. However this would not give is a cube two feet on a side, because at this point the figure we have constructed is only one foot high. So we have to add a second layer, consisting of two more rows with two cubes a row. Thus we have 2 layers, each containing 2 rows with 2 cubes in a row. Each layer has 4 cubes, so our two layers contain 8 cubes.
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RESPONSE --> I think I read the question wrong. This would make sense though. In order to find this you would need 2 layers with 2 rows and 2 cubes on each. Each layer would then have 4 cubes and then would have two layers that contain 8 cubes. self critique assessment: 3
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21:45:45 `q003. How many squares one foot on a side would it take to construct a square three feet on a side?
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RESPONSE --> It would take 3 squares with 3 rows and this would have a total of 9 squares. confidence assessment: 2
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21:46:03 We would require three rows, each with 3 squares, for a total of 9 squares.
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RESPONSE --> I see how this is done and I was able to arrive at the same conclusion. self critique assessment: 3
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21:48:34 `q004. How many cubes one foot on a side would take to construct a cube three feet on a side?
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RESPONSE --> You would have to have cubes that are 3 feet high. Then each layer would have to have 3 rows with 3 cubes each. Then each layer would have 9 cubes and it would have a total of 27 cubes. confidence assessment: 2
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21:49:03 This would require three layers to make a cube three feet high. Each layer would have to contain 3 rows each with three cubes. Each layer would contain 9 cubes, so the three-layer construction would contain 27 cubes.
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RESPONSE --> I understand how this is done. I was able to come up with the same conclusion. It is like I can see how it would work in my head. self critique assessment: 3
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21:53:36 `q005. Suppose one of the Egyptian pyramids had been constructed of cubical stones. Suppose also that this pyramid had a weight of 100 million tons. If a larger pyramid was built as an exact replica, using cubical stones made of the same material but having twice the dimensions of those used in the original pyramid, then what would be the weight of the larger pyramid?
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RESPONSE --> Each of the stones would have double the size of each stone. It is when you compare the larger t the smaller. It takes 8 smaller cubes to construct a cube that is twice as big. The larger pyramid is 8 times heavier than the smaller. It would then weigh 800 millon tons. confidence assessment: 2
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21:56:00 Each stone of the larger pyramid has double the dimensions of each stone of the smaller pyramid. Since it takes 8 smaller cubes to construct a cube with twice the dimensions, each stone of the larger pyramid is equivalent to eight stones of the smaller. Thus the larger pyramid has 8 times the weight of the smaller. Its weight is therefore 8 * 100 million tons = 800 million tons.
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RESPONSE --> I understand how it is done. Also in order to find the weight I needed to mention that you would have to take 8 times 100 million tons in order to find the bigger pyramid from the smaller one. You would have 800 million tons. self critique assessment: 3
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21:59:11 `q006. Suppose that we wished to paint the outsides of the two pyramids described in the preceding problem. How many times as much paint would it take to paint the larger pyramid?
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RESPONSE --> The cubes on the second pyramid are twice as big as the first one. The faces have 4 times the area of the first one. You would have 4 times the area to paint, and the second would have 4 times the amount of paint. confidence assessment: 2
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22:02:17 The outside of each pyramid consists of square faces of uniform cubes. Since the cubes of the second pyramid have twice the dimension of the first, their square faces have 4 times the area of the cubes that make up the first. There is therefore 4 times the area to paint, and the second cube would require 4 times the paint
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RESPONSE --> When you take the larger pramid it is going to take more paint than the smaller one to paint the whole thing. When you work this out you find that it will take 4 more time the paint for the larger one than the smaller one. self critique assessment: 3
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22:04:16 `q007. Suppose that we know that y = k x^2 and that y = 12 when x = 2. What is the value of k?
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RESPONSE --> To find the value of k you have to take y = 12 and x = 2 into. Then you use the form form y = k x^2. You get: 12 = k * 2^2. You then have to simplify it and you get 12 = 4 * k. You then have to divide both sides by 410. When you do this you get k = 3. confidence assessment: 2
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22:06:23 To find the value of k we substitute y = 12 and x = 2 into the form y = k x^2. We obtain 12 = k * 2^2, which we simplify to give us 12 = 4 * k. The dividing both sides by 410 reversing the sides we easily obtain k = 3.
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RESPONSE --> I see how it is done. You have to substitute the variables into the formula. When you do this you get the equations and then divide both sides by the same number and then you find the value of k. self critique assessment: 3
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22:08:02 `q008. Substitute the value of k you obtained in the last problem into the form y = k x^2. What equation do you get relating x and y?
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RESPONSE --> You get k=3. You then substitute it into y= kx^2 and get the equation of y= 3x^2. confidence assessment:
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22:08:38 We obtained k = 3. Substituting this into the form y = k x^2 we have the equation y = 3 x^2.
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RESPONSE --> I understand how it is done. You have to substitute the form and when doing this you are able to get the correct equation. self critique assessment: 3
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22:10:11 `q009. Using the equation y = 3 x^2, determine the value of y if it is known that x = 5.
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RESPONSE --> You have x=5 and you use the equation y=3x^2. You then have the equation and answer of y=3(5)^2=3*25=75. confidence assessment: 2
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22:10:43 If x = 5, then the equation y = 3 x^2 give us y = 3 (5)^2 = 3 * 25 = 75.
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RESPONSE --> I understand how it is done. You take the equation and then substitute the variable into the equation in order to find the answer. self critique assessment: 3
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22:15:18 `q010. If it is known that y = k x^3 and that when x = 4, y = 256, then what value of y will correspond to x = 9? To determine your answer, first determine the value of k and substitute this value into y = k x^3 to obtain an equation for y in terms of x. Then substitute the new value of x.
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RESPONSE --> First you have to substitute x = 4, y = 256 into y = k x^3. You then get the equation of 256 = k * 4^3, and 256 = 64 k. You then divide both sides by 64. This is what you get k = 256 / 64 = 4. Then you Substitute k = 4 into the form y = k x^3, we get the equation of y = 4 x^3. You then have to find the value of y when x = 9. You can do this by substituting x=9 into the new equation: y = 4 * 9^3 = 4 * 729 = 2916. confidence assessment: 2
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22:15:56 To we first substitute x = 4, y = 256 into the form y = k x^3. We obtain the equation 256 = k * 4^3, or 256 = 64 k. Dividing both sides by 64 we obtain k = 256 / 64 = 4. Substituting k = 4 into the form y = k x^3, we obtain the equation y = 4 x^3. We wish to find the value of y when x = 9. We easily do so by substituting x equal space 9 into our new equation. Our result is y = 4 * 9^3 = 4 * 729 = 2916.
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RESPONSE --> I see how this is done. It is basically done by substituting and divide numbers and equations in order to get the final equation and get the final answer. self critique assessment: 3
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22:19:15 `q011. If it is known that y = k x^-2 and that when x = 5, y = 250, then what value of y will correspond to x = 12?
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RESPONSE --> You have to substitute x = 5 and y = 250 into the equation of y = k x^-2 and you get 250 = k * 5^-2. Then you have 5^-2 = 1 / 5^2 = 1/25, and you get 250 = 1/25 * k. You then have k = 250 * 25 = 6250. Then this equation would be y = k x^-2 and then you have y = 6250 x^-2. Then you have to use x = 12, and you get y = 6250 * 12^-2 = 6250 / 12^2 = 6250 / 144 = 42.6. confidence assessment: 2
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22:20:27 Substituting x = 5 and y = 250 into the form y = k x^-2 we obtain 250 = k * 5^-2. Since 5^-2 = 1 / 5^2 = 1/25, this becomes 250 = 1/25 * k, so that k = 250 * 25 = 6250. Thus our form y = k x^-2 becomes y = 6250 x^-2. When x = 12, we therefore have y = 6250 * 12^-2 = 6250 / 12^2 = 6250 / 144 = 42.6, approximately.
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RESPONSE --> I understand how this is done. You have to take the orginal formula and then substitute different variables in the equation. You then have to divide by the each side and then you are able to get the final equation and the answer. self critique assessment: 3
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ݶүZ|ԙ`ĵ assignment #012 012. Precalculus I 03-30-2008
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22:25:51 `q001. Note that this assignment has 3 questions If we know that y = k x^2, then if (x2/x1) = 7, what is (y2/y1)?
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RESPONSE --> You take y2 = k x2^2 and y1 = k x1^2. You then have y2 / y1 = (k x2^2) / ( k x1^2). With k / k = 1 you have the same answer. y2 / y1 = x2^2 / x1^2. This is also the same. y2 / y1 = (x2 / x1)^2.You then have (x2 / x1) = 7. You find y2 / y1 = (x2 / x1)^2 = 7^2 = 49. confidence assessment: 2
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22:27:46 If y2 = k x2^2 and y1 = k x1^2, then y2 / y1 = (k x2^2) / ( k x1^2). Since k / k = 1 this is the same as y2 / y1 = x2^2 / x1^2, which is the same as y2 / y1 = (x2 / x1)^2. In words this tells us if y to is proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x2 to x1. Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^2 = 7^2 = 49.
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RESPONSE --> I understand how this is done. You have to take the orginal formula and then you have to substitute the varibles and then you are able to have a new equation and you are able to find the answer. self critique assessment: 3
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22:30:27 `q002. If we know that y = k x^3, then if (x2/x1) = 7, what is (y2/y1)?
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RESPONSE --> You have y2 = k x2^3 and y1 = k x1^3. Then y2 / y1 = (k x2^3) / ( k x1^3). Since k / k = 1. y2 / y1 = x2^3 / x1^3. y2 / y1 = (x2 / x1)^3. Now you have (x2 / x1) = 7. This gives you y2 / y1 = (x2 / x1)^3 = 7^3 = 343. confidence assessment: 2
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22:32:15 If y2 = k x2^3 and y1 = k x1^3, then y2 / y1 = (k x2^3) / ( k x1^3). Since k / k = 1 this is the same as y2 / y1 = x2^3 / x1^3, which is the same as y2 / y1 = (x2 / x1)^3. In words this tells us if y to is proportional to the cube of x, then the ratio of y2 to y1 is the same as the cube of the ratio of x2 to x1. Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^3 = 7^3 = 343.
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RESPONSE --> I can see how this is done. You have the first equation and then you substitute the variables in. When you have done this you are able to find the final equation and then you have the answer. self critique assessment: 3
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22:35:21 `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)?
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RESPONSE --> You have the equations y2 = k x2^-2 and y1 = k x1^-2. y2 / y1 = (k x2^-2) / ( k x1^-2). k / k = 1. y2 / y1 = x2^-2 / x1^-2. y2 / y1 = (x2 / x1)^-2. You then take 1 / (x2 / x1)^2 and have (x1 / x2)^2. You then have (x2 / x1) = 64. You get y2 / y1 = (x1 / x2)^2 = (1/64)^2 = 1/ 4096. confidence assessment: 2
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22:36:26 If y2 = k x2^-2 and y1 = k x1^-2, then y2 / y1 = (k x2^-2) / ( k x1^-2). Since k / k = 1 this is the same as y2 / y1 = x2^-2 / x1^-2, which is the same as y2 / y1 = (x2 / x1)^-2, which is the same as 1 / (x2 / x1)^2, which gives us (x1 / x2)^2. So if y = k x^-2, then (y2 / y1) = (x1 / x2)^2.( In words this tells us if y to is inversely proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x1 to x2 (note that this is a reciprocal ratio). Now if (x2 / x1) = 64, we see that y2 / y1 = (x1 / x2)^2 = (1/64)^2 = 1/ 4096.
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RESPONSE --> I understand how this is done. You have to take the equation and then substitute the variables. In doing this you have to work your way in order to get to the final answer and equation. This may take some time but you are able to get the final answer. self critique assessment: 3
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