course mth 163 assignment #010010.
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18:32:37 `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 with a slope of 1. The graph for y = .5x is like the y = x graph, only vertically stretched by .5 The graph for y = 2x is also like the y = x graph just stretched vertically by 2. confidence assessment: 2
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18:36:58 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 described the graph of y=x but just left out the part that it passes through the origin and the basic points. I did not go into quite as much detail for the other two graphs, but do understand what they look like. self critique assessment: 2
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18:44:44 `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 --> Each graph would be slightly higher than the one before. confidence assessment: 1
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18:45:51 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 was not sure exactly how to word what I wanted to say about the graphs. I do understand what is being stated. self critique assessment: 2
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19:06:53 `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 y=x is a straight line with the slope of 1 with the basic point of (0, 0). The graph of y=x-2 lies 2 units to the right of the y=x graph. The basic point of this graph (0. -2) and (2, 0) The other graph y = x+4 lies four units to the left of the y=x graph. Basic points are (0, 3) and (-3, 0). For -2 < x < 3 graphs, they would lie between the two previously mentioned graphs. confidence assessment: 2
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19:08:32 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 was off on that. I mentioned that the shift was right or left when it should have been up or down. I was even looking at my graph. I do understand that the number are for vertical shift. self critique assessment: 2
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19:20:37 `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 --> On the y = 2x graph each point lies twice as high as on the y=x graph. On the graph y = 2x - 2 the points lie 2 units below those on the y = 2x graph. confidence assessment: 2
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19:21:32 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 did not include the slope in my description. self critique assessment: 2
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19:38:37 `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 --> From the previous question we know that y = 2x - 2 the slope is 2 and lies 2 units below the origin.. The graph y = 2x + 3 will have the same slope as the other, but will lie 3 units higher than the origin. Therefore, if we graph everything from -2 < x < 3 those graphs will lie between the two stated graphs. confidence assessment: 2
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19:38:49 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 --> self critique assessment: 3
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19:46:36 `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 --> The slope is (y2 - y1) / (x2 - x1) confidence assessment: 3
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19:46:48 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 --> self critique assessment: 3
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19:47:50 `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 --> (y1 - y) / (x1 - x) confidence assessment: 3
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19:48:20 The slope from (x1, y1) to (x, y) is slope = rise/run = (y - y1) / (x - x1).
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RESPONSE --> It would be the same as what I put. self critique assessment: 3
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19:50:22 `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 --> It should be the same as the slope between (x1, y1) and (x2, y2) confidence assessment: 3
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19:50:28 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 --> self critique assessment: 3
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19:59:32 `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 --> [ (y2 - y1) / (x2 - x1) } / [ (y1 - y) / (x1 - x) ] confidence assessment: 2
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19:59:51 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 --> self critique assessment: 3
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20:27:16 `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 --> In order to solve for y we must cross multiply giving us: (y2 - y1)(x1 - x) = (y1 - y) (x2 - x1) use foil x1y2 - xy2 - x1y1 + y1x=x2y1 - x2y - x1y1+ x1y x1y2 - xy2 + y1x = x2y1 - x2y + x1y x1y2 - xy2 + y1x + x2y1= - x2y + x1y x1y2 - xy2 + y1x + x2y1 = y (-x2 +x1) y = (x1y2 - xy2 + y1x + x2y1) / (-x2 +x1) confidence assessment: 2
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20:30:49 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 made that way too complicated. I don't know why I didn't think to clear the fraction. I know how to do this, but juts over thought it. self critique assessment: 2
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Ä|wÅ—®ÆÅËšIÀmw本èâç¸Îþ€× assignment #011 011. Precalculus I 02-15-2007
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20:45:33 `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 --> If you draw a model you would need 4 1ft squares confidence assessment: 2
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20:45:47 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 --> self critique assessment: 3
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20:51:13 `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 --> You would need 8 because if it takes four to make a 2 ft. square that is one foot high then you would need another row. confidence assessment: 3
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20:51:30 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 --> self critique assessment: 3
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20:55:08 `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 9 1 ft squares confidence assessment: 3
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20:55:16 We would require three rows, each with 3 squares, for a total of 9 squares.
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RESPONSE --> self critique assessment: 3
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21:06:06 `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 --> it would take 27 squares because you have 9 squares at one level, so to make it three feet high you need two more rows (3x3x3) confidence assessment: 3
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21:06:11 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 --> self critique assessment: 3
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21:17:21 `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 --> 100 million x 100 million confidence assessment: 1
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21:19:50 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 was way off, but not totally sure how to go about doing this one.. self critique assessment: 2
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21:28:14 `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 --> Wouldn't it take 8 times as much paiint for the same reason described in the last question? confidence assessment: 1
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21:31:37 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 should have went with my original instinct which was four times as much but I second guessed myself self critique assessment: 2
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21:33:17 `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 --> Substitute in y = k x^2: 12 = k (2^2) 12 = 4k divide 3 = k confidence assessment: 3
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21:33:31 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 --> self critique assessment: 3
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21:33:51 `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 confidence assessment: 3
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21:34:01 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 --> self critique assessment: 3
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21:37:11 `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 --> Substitute 5 in for x: y = 3 x^2 y = 3 (5)^2 y = 3 * 25 y = 75 confidence assessment: 3
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21:37:29 If x = 5, then the equation y = 3 x^2 give us y = 3 (5)^2 = 3 * 25 = 75.
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RESPONSE --> self critique assessment: 3
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21:44:18 `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 --> To find k: y = k x^3 256 = k (4)^3 256 = 64k k = 4 Then substitute back in: y = 4 (9)^3 y = 4(729) y = 2916 confidence assessment: 3
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21:44:25 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 --> self critique assessment: 3
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21:52:12 `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 --> Substitute: 250 = k (5)^-2 250 = k/25 ---multiply by 25 k = 6250 Substitute: y = 6250 (12)^-2 y = 6250 / 144 y = 43.40 confidence assessment: 3
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21:52:26 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 --> self critique assessment: 3
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