course MTH 163 Assignment 8 q_a_ and query exercises and Week 4 Quiz 1. Week 4 Quizzes 2 and 3 will be logged to tomorrow's SEND file. vIRvМkassignment #008
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04:05:23 `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 --> To find the points for x=2 and x=9 we substitute these values into the equation: y = 1.1(2) + .8 y = 2.2 + .8 y = 3 (2 , 3) y = 1.1(9) + .8 y = 9.9 + .8 y = 10.7 (9 , 10.7) To find the rise we subtract the y-values in both points: 10.7 - 3 = 7.7 To find the run we subtract the x-values in both points: 9 - 2 = 7 To find the slope we divide the change in y by the change in x: 7.7 / 7 = 1.1 confidence assessment: 2
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04:05:36 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 --> I understand. self critique assessment: 2
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04:09:26 `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 --> To find the coordiantes for the point x=a we substitute the a-value (1.1) into the equation: y = 1.1(1.1) = .8 y = 1.1+.8 y = 1.9 (1.1 , 1.9) To find the coordiantes for the point x=b we substitute the b-value (.8) into the equation: y = 1.1(.8) + .8 y = .88 + .8 y = 1.68 (.8 , 1.68) confidence assessment: 2
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04:11:01 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 --> I went as far as to substitute the a- and b-values (per the generalized function for y=x) into the equation, but I understand how to find these points we are to substitute a and b, respectively, for x. self critique assessment: 2
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04:14:42 `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 --> To find the rise we subtract the change in the y-values: (1.1b + .8) - (1.1a + .8) = 1.1b - 1.1a To find the run we subtract the the change in the x-values: b - a confidence assessment: 2
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04:15:00 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 --> I understand. self critique assessment: 2
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04:24:44 `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 --> To find the average slope we divide the change in y by the change in x: (1.1b - 1.1a) / (b - a) 1.1b / b = 1.1 / 1 = 1.1 1.1a / a = 1.1 / 1 = 1.1 1.1 + 1.1 = 2.2 confidence assessment: 1
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04:26:32 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 made an error somewhere in my distributing leaving two sets of 1.1 to be added together. I understand how facotring out the (b - a) leaves (1.1 / 1), or 1.1 self critique assessment: 1
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fw _ױʹ assignment #008 008. `query 8 DERIVE exercise, Major Quiz ... Precalculus I 10-04-2007
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22:11:07 Were you able to complete the DERIVE exercise?
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RESPONSE --> Yes I was. I took a while to get the hang of how it all works, but I managed my way through it. confidence assessment: 2
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22:11:24 Did you understand everything in the DERIVE exercise?
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RESPONSE --> It all seems pretty clear to me yes. confidence assessment: 2
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22:12:04 Do you have any questions about the DERIVE exercise?
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RESPONSE --> Not at this point. I understand all the actions performed on it in this exercise. confidence assessment: 2
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Week 4 Quiz 1 - Version 9 To find the average rate of change in depth we divide the change in depth by the change in clock time. From time interval 17.8 to 26.7 seconds the the depth changes at an average rate of -0.742 cm/sec. From time interval 26.7 to 35.6 seconds the the depth changes at an average rate of -0.438 cm/sec. From time interval 35.6 to 44.5 seconds the the depth changes at an average rate of -0.135 cm/sec. Rate equals the change in depth divided by the change in clock time, which in notated on the graph as y and x, respectively. Therefore the rate equals the change in y divided by the change in x, or the rise divided by the run. Figuring out the rise and run between the two graph points representing the clock times 26.7 and 35.6 will give you an average rate of depth change for that time interval. After solving each equation for the function of x we find the x-value of the vertex by dividing our b-value by 2a. We find the y-value of the vertex by substituting the x-value of the vertex into the equation. The value to the right of the vertex is (xVertex+1, yVertex+a). The value to the left of the vertex is (xVertex-1,yVertex+a): f(x-1.75)=(x-1.75)^2 (x-1.75)(x-1.75) x(x-1.75) - 1.75(x-1.75) x^2 - 3.5x + 3.0625 xVertex = 3.5 / 2*1 = 1.75 yVertex = 1.75^2 - 3.5(1.75) + 3.0625 = 0 This gives us our three points: (1.75 , 0) , (2.75 , 1) , (0.75 , 1) f(x)-.85=(x^2)-.85 xVertex = 0 / 2*1 = 0 yVertex = 0^2 - .85 = -.85 This gives us our three points: (0 , -.85) , (1 , .15) , (-1 , .15) 2f(x)=2x^2 xVertex = 0 / 2*2 = 0 yVertex = 2(0)^2 = 0 This gives us our three points: (0 , 0) , (1 , 2) , (-1 , 2) 2f(x-1.75)+.85=2(x-1.75)^2+.85 2[(x-1.75)(x-1.75)]+.85 2[x(x-1.75) - 1.75(x-1.75)]+.85 2[x^2 - 3.5x + 3.0625]+.85 2x^2 - 7x + 6.125 + .85 2x^2 - 7x + 6.975 xVertex = 7 / 2*1 = 3.5 yVertex = 2(3.5)^2 - 7(3.5) + 6.975 = 31.475 This gives us our three points: (3.5 , 31.475) , (4.5 , 33.475) , (2.5 , 33.475)"