#$&* course Mth 271 Question: `explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval representedYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
.............................................
Given Solution: `a the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qexplain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When calculating the area of trapezoid we multiply the change in time, or the time interval (width) by average velocity (altitude),and we get the distance travelled or the change in position. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qtext problem 0.5 #10 add x/(2-x) + 2/(x-2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: [ x(x-2) + 2(2-x)]/[(2-x) (x-2)]= (X^2 - 4x +4) / - (x^2 - 4x + 4) = [(x-2)^2] / [-(x-2)^2] = -1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a common denominator could be [ (2-x)(x-2) ]. In this case we have x / (2-x) + 2 / (x-2) = [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ] = x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ] = [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ] = [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ] = (x^2-4x+4) / [ -x^2+4x-4 ] = (x-2)^2 / [-(x-2)^2] = -1. NOTE however that there is a SIMPLER SOLUTION: We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qtext problem 0.5 #50 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (6x^2 + 900 000) / x ( 6 * 240^2 + 900 000) / 240 = 5190 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!