course Mth163 Submitted on Sunday, June 21,2009 at 11:32 pm. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If x = a & b the equations would look like this: y = 1.1a + .8 and y = 1.1b + .8 The coordinates of these points would be represented like this: (a, 1.1a + .8) and (b, 1.1b + .8) Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rise would be the change in y: (1.1b +.8) – (1.1a + .8) = 1.1b – 1.1a The run is the change in x: b - a Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope is found by rise / run and the rise was 1.1b – 1.1a and the run is b-a: (1.1b – 1.1a) / b – a = 1.1(b-a) / b – a = 1.1 is the slope. Confidence Assessment: positive ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 Open Query #2 008. `query 8 DERIVE exercise, Major Quiz ... Quiz#1 version 12 – finding the slope for each interval of time compared to depth: 7.1 / 6.7 = 1.06 cm / sec 4.9 / 6.7 = 0.73 cm /sec 4.9 / 6.7 = 0.39 cm / sec graphing the points and seeing the slope from the first two points shows this is the slope because it shows how much depth you get out of this much time Next after graphing each function this is what I have: f(x) = x^2 this is a parabola f(x)^2 = (x-.75)^2 this is shifted to the right of the original parabola f(x)^2 – 1.35 this shifted down 1.35 from the original parabola 5f(x) = 5(x)^2 this vertically stretched the parabola up 5 units making the graph thinner and longer 5f(x-.75) + 1.35 = 5(x-.75)^2 + 1.35 this shifted the original parabola right .75 and up 1.35 and stretched it 5 units up. Quiz#2 version 11 Pt.1 - find the depth of 86.19499cm 86.19499 = .02t^2 + -2.62t + 99 0 = .02t^2 + -2.62t + 12.80501 using quadratic formula 2.62 + `sqrt(-2.62^2 – 4(.02)(12.80501) / 2(.02) t = 5 , 126 Pt. 2 – f(t) = .02(13)^2 + -2.62*13 + 99 depth= 68.32cm Pt.3 - f(t) = 64.19499 64.19499 = .02t^2 + -2.62t + 99 0 = .02t^2 +-2.62t + 34.80501 using quadratic formula: 2.62 + `sqrt[-2.62^2 – 4(.o2)(34.80501)] / 2(.02) t = 120.475, 15.00248 Pt. 4 – f(t) = .02(31)^2 + -2.61*31 + 99 depth = 37cm Quiz # 3 version 9 Pt. 1 - plotting out coordinates (10, 83) and (28, 60) and finding the slope by the change in y divided by the change in x: (60 – 83) / (28 – 10) = 1.278 cm / sec for slope. Pt. 2 – t = 28 and t = 29 find the coordinates by finding the depth: f(t) = .019(28)^2 + - 2.01(28) + 102 depth = 60.616 coordinates ( 28, 60,616) f)t) = .019(29)^2 + -2/01(29) + 102 depth = 59.689 coordinates ( 29, 59.689) Pt 3 – find the slope of the last two coordinates: (59.689 – 60.616) / (29- 28) = -.927 cm / sec as your slope &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qWere you able to complete the DERIVE exercise? yes &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qDid you understand everything in the DERIVE exercise? Yes, If I read everything correctly I believe I did. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique Rating: ********************************************* Question: `qDo you have any questions about the DERIVE exercise? Only if I was wrong please explain where I misunderstood the directions. "