#$&* course MTH 271 FEB. 19, 20124:22 pm 005. `query 5
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Given Solution: `a the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qexplain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: As another rate versus time graph, this too is relatively easy to translate into other forms. When you multiply height by width, you get the distance from one point to another. This could also be worded as average velocity times time change, and the result is the change in height. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time. When you multiply ave altitude by width you are representing ave vel * change in clock time, which gives change in position. This reasoning isn't confined to velocities. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qtext problem 0.5 #10 add x/(2-x) + 2/(x-2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x/ (2-x) + 2/ (x-2) = (x^2 -2x + 4 - 2x) / (x-2)(2-x) = (x-2)^2 / (-x^2 + 4x -4) = (x-2)^2 / -(x-2)^2 = 1/-1 = -1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a common denominator could be [ (2-x)(x-2) ]. In this case we have x / (2-x) + 2 / (x-2) = [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ] = x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ] = [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ] = [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ] = (x^2-4x+4) / [ -x^2+4x-4 ] = (x-2)^2 / [-(x-2)^2] = -1. NOTE however that there is a SIMPLER SOLUTION: We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qtext problem 0.5 #50 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6x + 900,000/x = [(6x^2) +900,000]/x When x= 240, [6(57600) + 900000] / 240 = 5,190 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a express with common denominator x: [x / x] * 6x + 900,000 / x = 6x^2 / x + 900,000 / x = (6x^2 + 900,000) / x so cost = (6x^2+900,000)/x Evaluating at x = 240 we get cost = (6 * 240^2 + 900000) / 240 = 5190. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " end document Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!