course Mth 163 Making progress.... ְh벜}ҦxvStudent Name:
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13:59:33 `q001. Note that this assignment has 4 questions For the function y = 1.1 x + .8, what are the coordinates of the x = 2 and x = 9 points? What is the rise between these points and what is the run between these points? What therefore is the slope between these points?
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RESPONSE --> The coordinates for x=2 are (2,3) and the coordinates for x=9 are (9,10.7). The rise is 7.7 and the run is 7, so the slope is 1.1
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13:59:47 Evaluating y = 1.1 x +.8 for x = 2 and x = 9 we obtain y = 3 and y = 10.7. The graph points are therefore (2,3) and (9,10.7). The rise between these points is 10.7 - 3 = 7.7 and the run is 9-2 = 7. Thus the slope is 7.7 / 7 = 1.1.
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RESPONSE --> ok
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14:02:33 `q002. For the function y = 1.1 x + .8, what are the coordinates of the x = a point, in terms of the symbol a? What are the coordinates of the x = b point, in terms of the symbol b?
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RESPONSE --> I think it would be y=1.1a+.8 and y=1.1b+.8 or a=1.1 and b=.8
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14:04:41 If x = a, then y = 1.1 x + .8 gives us y = 1.1 a + .8. If x = b, then y = 1.1 x + .8 gives us y = 1.1 b + .8. Thus the coordinates of the x = a point are (a, 1.1 a + .8) and (b, 1.1 b + .8).
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RESPONSE --> O.K. I understand. If x=a then that would be the x cooordinate and the same for b.
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14:07:34 `q003. We see that the coordinates of the x = a point are (a, 1.1 a + .8) and (b, 1.1 b + .8). What therefore is the rise between these two points? What is the run between these two points?
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RESPONSE --> The rise would be 1.1b+.8-1.1a+.8, which would simplify to 1.1b-1.1a becasue .8-.8 is 0. and the run would be b-a.
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14:07:55 The rise between the points is the rise from y = 1.1 a + .8 to y = 1.1 b + .8, a rise of rise = (1.1 b + .8) -(1.1 a + .8) = 1.1 b + .8 - 1.1 a - .8 = 1.1 b - 1.1 a. The run is from x = a to x = b, a run of run = b - a.
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RESPONSE --> ok
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14:09:33 `q004. We see that the rise between the x = a and x = b points of the graph of y = 1.1x +.8 is 1.1 b + .8 - (1.1 a + .8), while the run is b - a. What therefore is the average slope of the graph between these points? Simplify your answer.
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RESPONSE --> I think the slope would be 0. The b's would cancel out and so would the a's. This would leave you with 1.1-1.1, which is 0.
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14:11:28 The slope is slope = rise / run = (1.1 b - 1.1 a) / (b - a) = 1.1 (b - a) / (b - a) = 1.1. The significance of this series of exercises is that the slope between any two points of the straight line y = 1.1 x + .8 must be 1.1, no matter whether the points are given by numbers (e.g., x = 2 and x = 9) or by symbols (x = a and x = b). Mostly
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RESPONSE --> I understand now that the b's and a's would not cancel. I see how you simplified it.
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꯲~ Student Name: assignment #009
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14:19:26 `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 --> y=1.1x1+.8 and y=1.1x2+.8 1 The coordinates of x=x1 are (x1, 1.1x1+.8) The coordinates of x=x2 are (x2,1.1x2+.8) The rise is 1.1x2+.8-1.1x1+.8 which is 1.1x2-1.1x1 The run is x2-x1 The slope is 1.1 I know this because you cannot cancel out the variables. 1.1(x2-x1)/(x2-x1)=1.1
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14:19:43 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 --> ok
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14:24:26 `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 coordinates of the x=x1 point in terms of the symbol x1 are (x1, 3.4x1+7). The coordinates of the x=x2 point in terms of the symbol x=x2 are (x2,3.4x2+7) The rise is 3.4x2-3.4x1 or 3.4(x2-x1). The run is (x2-x1) The slope is 3.4.
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14:24:32 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 --> ok
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S~ӷHbً Student Name: assignment #010
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14:30:12 `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 is a straight line. For every point of x, it is equal to that same point of y. The graph of y=.5x is also a straight line,less steep than the first graph and it crosses the first line. The graph of y=2x is very steep and crosses both lines.
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14:31:27 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 --> ok
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14:31:29 10-21-2006 14:31:29 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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14:36:46 `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 --> I think that the points of the graph would lie to the left of .5 on the x axis and below 2 on the y axis.
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14:43:18 10-21-2006 14:43:18 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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NOTES -------> I thought the coordinates would be (.5,1) I am confused on how to get the coordinates. I'm sure it's really simple and I feel like I should know it, but y would have a coeff. of 1 and .5 is the coeff. of x, right? No .5 is a, not the coeff. of x they are multiplied (kind of).
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14:58:46 `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 --> They all go in the same direction. The graph of y=x+3 lies three times above the graph of y=x and the graph of y=x-2 lies twice below the graph of y=x. And they are dotted, straight lines because the < represents less than not equal to.
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14:59:12 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 --> ok
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15:03:43 `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 --> The graph of y=2x has a slope twice as steep as the y=x graph and its points lie twice as far from the points of the y=x graph. The graph of y=2x crosses the y=x graph. The graph of y=2x-2 lies below and runs parallel to the y=2x graph.
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15:09:07 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'm still a little confused about the coordinates, but I understand the rest.
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15:10:35 `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 could graph a series of dotted lines between -2 and 3 on the x axis.
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15:11:01 10-21-2006 15:11:01 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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NOTES -------> ok
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15:13:37 `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 --> y2-y1/x2-x1
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15:14:04 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 --> Yes. I know that to be true.
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15:17:47 `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 is y-y1/x-x1.
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15:17:56 The slope from (x1, y1) to (x, y) is slope = rise/run = (y - y1) / (x - x1).
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RESPONSE --> ok
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15:19:08 `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 --> I say equal to because they all lie on the same line.
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15:19:17 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 --> ok
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15:25:09 `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 --> y-y1/x-x1=y2-y1/x2-x1
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15:25:33 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 --> ok
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15:39:17 `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 --> First I wanted to cross multiply and decided against it. Then I wanted to multiply each side by one of the x values to eliminate them. I'm just really not sure.
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15:42:48 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 --> O.K. My second instinct was right to muliply each side by the (x-x1). When we added y1 to each side, that eliminated the y1 from (y-y1). I understand. I was making it more difficult than it actually was.
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üwĮeݍߤʃywU Student Name: assignment #011
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16:54:54 `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 want to say two, but that seems too common sense. I also want to say eight, two one foot squares per side.
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16:55:55 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 got cha!
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16:56:56 `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 --> Is this a trick question? Just kidding. Is a cube the same thing as a square? If so, then it would take four.
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17:00:38 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 --> So here the difference in a cube and a square is that a cube is three dimensional?
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17:05:27 `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 --> Since it took four one foot squares to construct a square two feet on a side, I think it would take eight. Two rows with three and one between the rows on top to make three across and one between the rows at the bottom to make three across. This would leave the center empty.
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17:07:05 We would require three rows, each with 3 squares, for a total of 9 squares.
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RESPONSE --> O.K. I can't leave the center empty. Becasue it has to make a solid square when looking down on it?
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17:08:10 `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 --> Three layers of four, so 12.
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17:25:29 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 --> Confused. If you have three layers, then it's three feet high. Why does each layer have to have nine instead of four, because four makes a square. Unless by A side you mean EACH side, then it would have to be 9. I've got it, I think.
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17:31:32 `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 --> The weight would be double that of the origninal pyramid. Which would be 10,000 million tons.
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17:33:32 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 --> How do you know that it takes eight? I don't understand. Why eight instead of four or another even number?
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17:34:44 `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 --> Half as much to paint the larger pyramid because the cubes were larger?
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17:36:04 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 --> I still do not understand. If you have twice the dimension, why wouldn't it take twice the paint, instead of four times the paint.
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17:37:59 `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 --> 12=k(2)^2 12=4k 3=k
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17:38:04 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 --> ok
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17:38:47 `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 --> y=3x^2
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17:38:57 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 --> ok
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17:40:14 `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 --> y=3(5)^2 y=3(25) y=75
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17:40:18 If x = 5, then the equation y = 3 x^2 give us y = 3 (5)^2 = 3 * 25 = 75.
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RESPONSE --> ok
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17:49:42 `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 --> y=k(9)^3 y=k(729) y=729k 729k=(9)^3 729k=729 k=1 y=(1)*(9)^3 y=1*729 y=729 I've made a mistake. I just know it.
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17:51:42 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 --> Oh! I thought that all I could use was the x=9. I understand now. I was wondering how I was supposed to solve for k! Major blonde moment. I've got it.
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17:51:51 10-21-2006 17:51:51 `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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NOTES ------->
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18:05:32 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 meant to save that as notes, I hit enter by mistake. Here is my answer to number 11. I promise I haven't viewed the answer. 250=k(5) ^ -2 250=k(.04) 6250=k y=6250(12)^ -2 y=6250(.00694) y=43.375 Now I'm looking and I forgot the very important part of multiplying by the reciprical because the exponent is negative. I did obtain the correct value for k. Then I put that into the original equation with the new x value. I reworked it and obtained almost the same answer that I got the first time, y=43.4 Since the 4 is not a five or above, do we round the 3 to a 2?
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ďߖ߇ Student Name: assignment #012
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18:10:47 `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 --> Can you substitute the 7 where x is in the equation? I thought that x1 and x2 were different variable forms of x. y=k(7)^2 y=49k
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18:12:45 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 got the same answer in a much different way.
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18:32:52 `q002. If we know that y = k x^3, then if (x2/x1) = 7, what is (y2/y1)?
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RESPONSE --> y=k(7)^3 y=343k
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18:34:28 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 --> ok
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18:38:51 `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)?
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RESPONSE --> y=k(64)^ -2 y=k(1/64)^2 y=1/4096(k)?
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18:40:26 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 --> Yipee! I got it.
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昛fg Student Name: assignment #013
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18:41:35 `q001. Note that this assignment has 12 questions What does 2^5 mean?
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RESPONSE --> It means 2 raised to the fifth power.
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18:41:51 2^5 stands for 2 raised to the fifth power; i.e., 2^5 = 2*2*2*2*2. The result of this calculation is 2^5 = 32.
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RESPONSE --> Yes. Ok.
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18:45:00 `q002. What does 2^3 * 2^5 mean? Is the result of power of 2? If so, what power of 2 is it?
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RESPONSE --> It means 2 raised the third power or cubed, times 2 raised to the fifth power. I would solve the exponents first and then multiply the products according to the order of operations I got 8*32=256 If you added the exponents you would get 8. Is that what you are asking?
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18:45:31 2^3 * 2^5 means (2*2*2) * (2*2*2*2*2). This is the same as 2*2*2*2*2*2*2*2, or 2^8. When we multiply this number out, we obtain 256.
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RESPONSE --> ok, I thought that's what you wanted.
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18:46:17 `q003. Why do we say that a^b * a^c = a^(b+c)?
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RESPONSE --> because if the variable is the same you can add the exponents together, but if you have different variable you can't do this.
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18:47:03 We saw in the preceding example that 2^3 * 2^5 stood for a product of three 2's, multiply by a product of five 2's. We saw also that the result was identical to a product of eight 2's. This was one instance of the general rule that when we multiply to different powers of the same number, the result is that number raised to the sum of the two powers. One general way to state this rule is to let a stand for the number that is being raised to the different powers, and let b and c stand for those powers. Then we get the statement a^b * a^c = a^(b+c).
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RESPONSE --> yes I understand
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18:53:39 `q004. What does (2^3)^5 mean?
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RESPONSE --> It means that you work the problem in the parentheses first because that's what the order of operations says, then you raise that number to the fifth power. This would give you 8^5=32768 But the third rule of exponents says that if you have (a^m)^n you have a^mn, which means that you would have 2^15. The answer to this would be 32768, so this is just a shortcut, we got the same answer.
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18:53:51 Since 2^3 stands for 2*2*2, it follows that (2^3)^5 means (2^3)*(2^3)*(2^3)*(2^3)*(2^3) = (2*2*2)*(2*2*2)*(2*2*2)*(2*2*2)*(2*2*2) = 2*2*2*2*2*2*2*2*2*2*2*2*2*2*2 = 2^15.
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RESPONSE --> ok
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18:54:45 `q005. Why do we say that (a^b)^c = a^(b*c)?
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RESPONSE --> This is a shortcut. You can multiply the exponents and raise your number or value to that power.
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18:55:18 We saw in the last example how (2^3)^5 stands for the product of 5 quantities, each equal to the product of three 2's. We saw how this is equivalent to the product of fifteen 2's, and we saw how the fifteen was obtained by multiplying the exponents 3 and 5. In the present question a^b stands for the quantity a multiplied by itself b times. (a^b)^c stands for the quantity a^b multiplied by itself c times, which is equivalent to multiplying a by itself b * c times. Thus we say that (a^b)^c = a^(b * c).
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RESPONSE --> I understand.
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18:57:57 `q006. According to the law a^b * a^c = a*(b+c), if we multiply 2^5 by 2^-2 what power of 2 should we get?
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RESPONSE --> 2^5 * 2^ -2=2*(5+ -2) This would give us 2^3 which is 8.
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18:58:05 To according to the law, 2^5 * 2^-2 = 2^(5 + -2) = 2^(5-2) = 2^3.
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RESPONSE --> ok
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18:59:29 `q007. Since as we saw in the preceding question 2^5 * 2^-2 = 2^3, what therefore must be the value of 2^-2?
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RESPONSE --> 2^ -2 would be 1/2^2 which is 1/4 or .25
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18:59:51 One way of seeing this is to note that 2^5 = 32 and 2^3 = 8, so we have 32 * 2^-2 = 8. Dividing both sides by 32 we get 2^-2 = 8 / 32 = 1/4. We can learn something important if we keep the calculation in powers of 2. If 2^5 * 2^-2 = 2^3, then dividing both sides of the equation by 2^5 we obtain 2^-2 = 2^3/2^5, which is equal to 1/2^2. This shows us that 2^-2 = 1/2^2.
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RESPONSE --> ok
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19:07:23 `q008. Why must we say that 2^-n = 1 / 2^n, where n stands for any counting number?
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RESPONSE --> Because when you have a negitive exponent you have to flip it or multiply by the reciprocal. So then you have a positive exponent. You can't multiply a positive number by itself a negative amount of times.
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19:08:06 This is because for any number m, we have 2^m * 2^-n = 2^( m + -n) = 2^(m-n), and we also have 2^m * (1 / 2^n) = 2^m / 2^n = 2^(m-n). So whether we multiply 2^m by 2^-n or by 1 / 2^n we get the same result. This shows that 2^-n and 1 / 2^n are the same.
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RESPONSE --> Yes, I know this to be true, but I was trying to explain why you flip it. Ha!
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19:10:41 `q009. According to the law a^b * a^c = a*(b+c), if we multiply 2^3 by 2^-3 what power of 2 should we get? Since 2^-3 = 1 / 2^3, what number must we get when we multiply 2^3 by 2^-3?
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RESPONSE --> 2^3 * 2^ -3 = 2*(3+ -2)=2*1=2 We get a power of 2. 2
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19:12:08 2^3 * 2^-3 = 2^(3 + -3) = 2^(3-3) = 2^0. Since 2^-3 = 1 / 2^3 it follows that 2^3 * 2^-3 = 2^3 * ( 1 / 2^3) = 1.
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RESPONSE --> I'm sorry I made a mistake. I put (3+ -2) instead of (3+ -3). Opps. I understand thought that it would be 2^0.
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19:13:02 `q010. Continuing the last question, what therefore should be the value of 2^0?
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RESPONSE --> The zero exponent rule says that it will be 1. If we had 546^0, it would still be 1.
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19:13:05 Since 2^3 * 2^-3 = 2^0 and also 2^3 * 2^-3 = 1 we see that 2^0 must be 1.
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RESPONSE --> ok
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19:23:07 `q011. How do we solve the equation x^3 = 12?
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RESPONSE --> I first thought about subtracting 12 from each side to get x^3-12=0 and factoring it to x(x^2-12) and then to x(x-4)(x+3), but I'm not sure. Then I wondered if you could do this: x^3=12^3, but that wouldn't help would it? I know I should remember this from algebra. Help!
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19:25:07 We solve the equation by taking the 1/3 power of both sides: (x^3)^(1/3) = 12^(1/3), then by the law (a^b)^c = a^(bc) we have x^(3 * 1/3) = 12^(1/3), so that x^1 = 12^(1/3) or just x = 12^(1/3), which we can easily enough evaluate with a calculator. We obtain a result of approximately x = 2.29 .
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RESPONSE --> Ok. I was kind of on the right track with doing something to both sides. You take the reciprocal of the power?
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19:32:11 `q012. How do we solve the equation x^(5/2) = 44?
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RESPONSE --> (x^5/2)^(2/5)=44^(2/5) x=4.54 approx.
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19:32:17 If we take the 2/5 power of both sides we get (x^(5/2))^(2/5) = 44^(2/5) or x^(5/2 * 2/5) = 44^(2/5) or x^1 = 44^(2/5) so that x = 44^(2/5). Evaluating by calculator you should obtain approximately x = 4.54.
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RESPONSE --> ok
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