course MTH 163 Here are the q_a_ and query assignments for assignment 6 and the week 3 quizzes 1 and 2. 鼎C魨b飙朽C緔穙┞尽鄾狠禷ssignment #006
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00:41:32 `q001. Note that this assignment has 10 questions Recall that the graph of y = x^2 + 3 was identical to the graph of y = x^2, except that it was raised 3 units. This function is of the form y = x^2 + c. In the case of this specific function, c = 3. What function would this form give us for c = -1? How would the graph of this function compare with the graph of y = x^2?
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RESPONSE --> The function y = x^2 - 1 would be consistent with the form c = -1 The graph of this function would be positioned 1 unit lower than the function y = x^2 confidence assessment: 2
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00:41:47 If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2.
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RESPONSE --> I understand. self critique assessment: 2
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00:45:27 `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> All the graphs would have identical shapes; that of a parabola which is symmetrical about the y-axis. However, the vertex of each graph is positioned on the y-axis at whatever value c represents in the corresponding function. For example, for the function y = x^2 - 2 the vertex lies at (0 , -2) For the function y = x^2 + 2 the vertex lies at (0 , 2) confidence assessment: 2
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00:45:49 The graph of the c= -3 function y = x^2 - 3 will lie 3 units lower than the graph of y = x^2. The graph of the c= -2 function y = x^2 - 2 will lie 22222 units lower than the graph of y = x^2. The progression should be obvious. The graph of the c= 3 function y = x^2 + 3 will lie 3 units higher than the graph of y = x^2. The final graph will therefore show a series of 7 functions, with the lowest three units below the parabolic graph of y = x^2 and the highest three units above the graph of this function. Each graph will lie one unit higher than its predecessor.
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RESPONSE --> I understand. self critique assessment: 2
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00:52:59 `q003. The function y = (x -1)^3 is of the form y = (x -k)^3 with k = 1. What function would this form give us for k = 3? How would the graph of this function compare with that of y = x^3?
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RESPONSE --> The form k = 3 would give us the function y = (x-3)^3 The two graphs would have identical shapes; that of an anti-symmetric parabola. However, the point at which the function y = x^3 passes through the x-axis is three points to the right for the function y = (x-3)^3 confidence assessment: 2
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00:53:17 Recall how the graph of y = (x-1)^3 lies one unit to the right of the graph of y = x^3. The k = 3 function y = (x -3)^3 will lie 3 units to the right of the graph of y = x^3.
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RESPONSE --> self critique assessment: 2
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00:55:40 `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> The graph would have three functions with identical shapes. The graph for the form k = 2 would pass through the x-axis at x = 2 Each subsequent graph would be positioned one unit to the right. confidence assessment: 2
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00:55:53 The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right.
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RESPONSE --> I understand. self critique assessment: 2
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00:58:38 `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x?
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RESPONSE --> The form A = 2 would give us the function y = 2 * 2^x This graph would be position twice as far from the x-axis than the graph of y = 2^x would. confidence assessment: 2
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00:58:58 As we saw earlier, the graph of y = 3 * 2^x lies 3 times as far from the x-axis as a graph of y = 2^x and is at every point three times as steep. We would therefore expect the A = 2 function y = 2 * 2^x to lie 2 times is far from the x-axis as the graph of y = 2^x.
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RESPONSE --> I understand. self critique assessment: 2
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01:02:22 `q006. Suppose we wish to graph the functions y = A * 2^x for values A ranging from 2 to 5. If we graph all such functions on the same set of coordinate axes, what will the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
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RESPONSE --> The graph would contain four different functions, all having similar shapes. Each graph would pass through the y-axis at the point relative to their functions A value. For example, the function y = 4 * 2^x would pass through the y-axis at (0 , 4) confidence assessment: 2
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01:02:37 These graphs will range from 2 times as far to 5 times as far from the x-axis as the graph of y = 2^x, and will be from 2 to 5 times as steep. The y intercepts of these graphs will be (0,2), (0, 3), (0, 4), (0,5).
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RESPONSE --> I understand. self critique assessment: 2
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01:04:56 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
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RESPONSE --> The rise of this line would be 4, and the run of this line would be 6. Since slope is measured by rise/run, the slope of this line would be 4/6 or 2/3 confidence assessment: 2
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01:05:07 The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666....
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RESPONSE --> I understand. self critique assessment: 2
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01:09:31 `q008. What are the coordinates of the t = 5 and t = 9 points on the graph of y = 2 t^2 + 3? What is the slope of the straight line between these points?
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RESPONSE --> The function would yield the coordinates (5 , 53) and (9 , 165) The resulting straight line would have a rise of 110 and a run of 4, giving us a slope of 110/4 or 55/2 confidence assessment: 2
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01:11:26 The t = 5 value is y = 2 * 5^2 + 3 = 2 * 25 + 3 = 50 + 3 = 53. The t = 9 value is similarly calculated. We obtain y = 165. The rise between these points is therefore 165-53 = 112. The run is from t = 5 to t = 9, a run of 9 - 5 = 4. This slope of a straight line connecting these points is therefore rise/run = 112/4 = 28.
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RESPONSE --> I think in order to get my rise I accidentally typed 163 into the calculator instead of 165. I understand that the rise would actaully be 112, giving a slope of 112/4 or 28. self critique assessment: 1
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01:15:14 `q009. Suppose y = 2 t^2 + 3 represents the depth, in cm, of water in a container at clock time t, in seconds. At what average rate does the depth of water change between t = 5 seconds and t = 9 seconds?
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RESPONSE --> According to the results of the problem in the last question, the depth would change at a rate of 28 cm/second. confidence assessment: 2
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01:16:26 The t = 5 second and t = 9 second depths are easily calculated to be y = 53 cm and y = 165 cm. The depth therefore changes from 53 cm to 165 cm, a change of 165 cm - 53 cm = 112 cm, in the 4 seconds between the to clock times. The average rate of depth changes therefore 112 cm/(4 seconds) = 28 cm/second. We note that this calculation is identical to the calculation of the slope of the straight line connecting the t = 5 and t = 9 points of the graph y = 2 t^2 + 3.
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RESPONSE --> I understand. I didn't show any work because I immediately recognized that this problem, while worded differently, was essentially asking for the same thing as the last problem. self critique assessment: 1
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01:18:29 `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval?
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RESPONSE --> The slope of the straight line, which is rise/run, is essentailly the same as the average rate, which is depth/time. Both represent the change in y / change in t. confidence assessment: 2
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01:18:41 The rise of the graph represents the change in the depth y and the run represents the change clock time t. The slope, which is rise/run, therefore represents change in depth/change in clock time, which is the average rate at which the depth changes.
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RESPONSE --> I understand. self critique assessment: 2
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蕩栱欤鹌蟆~哂 憂~愇槼陮螂� assignment #006 006. `query 6 Precalculus I 09-29-2007
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13:46:15 Query 4 basic function families What are the four basic functions? What are the generalized forms of the four basic functions?
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RESPONSE --> The four basic functions are linear, quadratic, exponential, and power. The generalized form of the linear equation is y = f(x) = mx + b The generalized form of the quadratic equation is y = f(x) = ax^2 + bx + c , or alternatively, y = f(x) = a(x-h)^2 + c The generalized form of the exponential equation is y = f(x) = A b^x + c , or alternatively, y = f(x) = A ( 2 ^ (kx) ) + c The generalized form of the power equation is f(x) = A (x-h)^p + c confidence assessment: 2
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13:46:33 ** STUDENT RESPONSE: Linear is y=mx+b Quadratic is y=ax^2 + bx +c Exponential is y= A*2^ (kx)+c Power = A (x-h)^p+c INSTRUCTOR COMMENTS: These are the generalized forms. The basic functions are y = x, y = x^2, y = 2^x and y = x^p. **
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RESPONSE --> I understand. self critique assessment: 2
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13:53:04 For a function f(x), what is the significance of the function A f(x-h) + k and how does its graph compare to that of f(x)?
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RESPONSE --> The function A f(x-h) + k is a form of the generalized power function. A will determine it's vertical stretch, h will determine it's horizontal shift, and k will determine it's vertical shift. The function A f(x-h) + k could have a different vertical stretch, a different vertical shift, and a different horizontal shift than that of the function f(x). confidence assessment: 2
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13:53:27 ** STUDENT RESPONSE: A designates the x strectch factor while h affects a y shift & k affects an x shift INSTRUCTOR COMMENTS: k is the y shift for a simple enough reason -- if you add k to the y value it raises the graph by k units. h is the x shift. The reason isn't quite as simple, but not that hard to understand. It's because when x is replaced by x - h the y values on the table shift 'forward' by h units. A is a multiplier. When all y values are multiplied by A that moves them all A times as far from the x axis, which is what causes the stretch. Thus A f(x-h) + k is obtained from f(x) by vertical stretch A, horizontal shift h and vertical shift k. The two aren't the same, but of course they're closely related. **
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RESPONSE --> I understand. self critique assessment: 2
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14:00:06 query introduction to rates and slopes, problem 1 ave vel for function depth(t) = .02t^2 - 5t + 150 give the average rate of depth change from t = 20 to t = 40
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RESPONSE --> First we find the slope of the function with t = 20: .02(20^2) - 5(20) + 150 = 8 - 100 + 150 = 58 Then we find the slope of the function with t = 40 .02(40^2) - 50(40) + 150 = 32 - 200 + 150 = -18 We subtract our y-values and our t-values: -18 - 58 = -76 40 - 20 = 20 We then subtract our rise by our run to get our average: -76 / 20 = -(19 / 5) = -3.8 -3.8 cm / seconds NOTE: I can't seem to find a measurement of distance anywhere in the question. I'll be assuming it's centimeters. confidence assessment: 2
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14:00:26 ** depth(20) = .02(20^2) - 5(20) + 150 = 58 depth(40) = .02(40^2) - 5(40) + 150 = -18 change in depth = depth(40) - depth(20) = -18 - 58 = -76 change in clock time = 40 - 20 = 20. Ave rate of depth change = change in depth / change in clock time = -76 / 20 = -3.8 **
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RESPONSE --> I understand. self critique assessment: 2
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14:02:51 What is the average rate of depth change from t = 60 to t = 80?
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RESPONSE --> .02(60^2) - 5(60) + 150 = 72 - 300 + 150 = -78 .02(80^2) - 5(80) + 150 = 128 - 400 + 150 = -122 -122 - 78 = -200 80 - 60 = 20 -(200 / 20) = -10 -10 change in distance / change in time confidence assessment: 2
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14:15:39 ** depth(60) = .02(60^2) - 5(60) + 150 = -78 depth(80) = .02(80^2) - 5(80) + 150 = -122 change in depth = depth(80) - depth(60) = -122 - (-78) = -44 change in clock time = 40 - 20 = 20. Ave rate of depth change = change in depth / change in clock time = -44 / 20 = -2.2 **
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RESPONSE --> When I subtracted my depth averages I accidnetally stated ""-122 - 78"" when I should've stated :-122 - (-78)"" That would actually give a change of cepth of -44. That divided by 20 would give an average change in depth / change in clock time of -2.2. I understand now. self critique assessment: 1
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14:19:07 describe your graph of y = .02t^2 - 5t + 150
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RESPONSE --> It is a straight line with it's x-values increasing and it's y-values decreasing (in other words, it's positioned diagonally from top-left to bottom-right.) which passes through the y-axis at (0 , 150). confidence assessment: 2
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14:25:16 ** The graph is a parabola. y = .02t^2 - 5t + 150 has vertex at x = -b / (2a) = -(-5) / (2 * .02) = 125, at which point y = .02 (125^2) - 5(125) + 150 = -162.5. The graph opens upward, intercepting the x axis at about t = 35 and t = 215. Up to t = 125 the graph is decreasing at a decreasing rate. That is, it's decreasing but the slopes, which are negative, are increasing toward 0.**
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RESPONSE --> Not quite sure how I messed this one up. I suppose the graph I sketched was a little sloppy. Of course, I can look at the numbers and plainly see the averages are not constant and therefore couldn't possibly yield a straight line. Just wasn't thinking straight I suppose. self critique assessment: 1
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14:29:58 describe the pattern to the depth change rates
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RESPONSE --> Not sure what I'm missing, but I don't see any distinct pattern. confidence assessment: 2
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14:31:23 ** Rates are -3.8, -3 and -2.2 for the three intervals (20,40), (40,60) and (60,80). For each interval of `dt = 20 the rate changes by +.8. **
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RESPONSE --> Ok, I see that pattern. I was looking for a pattern amongst ALL the time intervals (0-20 , 20-40 , 40-60 , 60-80 , 80-100) The pattern between those three changes is clear to see. self critique assessment: 1
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14:35:17 query problem 2. ave rates at midpoint times what is the average rate of depth change for the 1-second time interval centered at the 50 sec midpoint?
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RESPONSE --> .02(49.5^2) - 5(49.5) + 150 = 49.005 - 247.5 + 150 = -48.495 .02(50.5^2) - 5(50.5) + 150 = 51.005 - 252.5 + 150 = -51.495 -51.495 - (-48.495) = -3 50.5 - 49.5 = 1 -3 / 1 = -3 -3 change in depth / change in clokc time is the average for the midpoint interval. confidence assessment: 2
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14:35:25 ** The 1-sec interval centered at t = 50 is 49.5 < t < 50.5. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(50.5) - depth(49.5)]/(50.5 - 49.5) = -3 / 1 = -3. **
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RESPONSE --> I understand. self critique assessment: 2
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14:37:59 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> .02(47^2) - 5(47) + 150 = 44.18 - 235 + 150 = -40.82 .02(53^2) - 5(53) + 150 = 56.18 - 265 + 150 = -58.82 -58.82 - (-40.82) = -18 53 - 47 = 6 -18 / 6 = -3 -3 change in depth / change in clock time confidence assessment: 2
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14:38:17 ** The 6-sec interval centered at t = 50 is 47 < t < 53. For depth(t) = .02t^2 - 5t + 150 we have ave rate = [depth(53) - depth(47)]/(53 - 47) = -18 / 6 = -3. **
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RESPONSE --> I understand. self critique assessment: 2
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14:38:32 What did you observe about your two results?
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RESPONSE --> They are identical. confidence assessment: 2
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14:38:47 ** The two rates match each other, and they also match the average rate for the interval 40 < t < 60, which is also centered at t = 50. For a quadratic function, and only for a quadratic function, the rate is the same for all intervals having the same midpoint. This is a property unique to quadratic functions. **
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RESPONSE --> I understand. self critique assessment: 2
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15:36:16 query problem 3. ave rates at midpt times for temperature function Temperature(t) = 75(2^(-.05t)) + 25. What is the average rate of depth change for the 1-second time interval centered at the 50 sec midpoint?
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RESPONSE --> 75(2^(-.05*49.5))+25 = 75(.18) + 25 = 38.49 75(2^(-.05*50.5))+25 = 75(.174)+25 = 38.03 38.03 - 38.49 = -.46 50.5 - 49.5 = 1 -.46 / 1 = -.46 -.46 change in temperature / change in clock time confidence assessment: 2
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15:39:45 STUDENT RESPONSE: .46 degrees/minute INSTRUCTOR COMMENT: More precisely .4595 deg/min, and this does not agree exactly with the result for the 6-second interval. Remember that you need to look at enough significant figures to see if there is a difference between apparently identical results.**
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RESPONSE --> I accidentally subtracted 38.49 from 38.03. I got it backwards, yielding a result of -.46. I see what I did wrong. self critique assessment: 1
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15:44:29 what is the average rate of change for the six-second time interval centered at the midpoint.
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RESPONSE --> 75(2^(-.05*47))+25 = 75(.196)+25 = 39.71 75(2^(-.05*53))+25 = 75(.159)+25 = 36.949 39.71 - 36.949 = 2.761 53 - 47 = 6 2.761 / 6 = .46 .46 deg/sec confidence assessment: 2
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15:45:02 STUDENT RESPONSE: .46 degrees/minute INSTRUCTOR COMMENT: The 1- and 6-second results might possibly be the same to two significant figures, but they aren't the same. Be sure to recalculate these according to my notes above and verify this for yourself. The average rate for the 6-second interval is .4603 deg/min. It differs from the average rate .4595 deg/min, calculated over the 1-second interval, by almost .001 deg/min. This differs from the behavior of the quadratic. For a quadratic that the results for all intervals centered at the same point will all agree. This is not the case for the present function, which is exponential. Exact agreement is a characteristic of quadratic functions, and of no other type. **
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RESPONSE --> I accidentally wrote minutes instead of seconds, but I uderstand. self critique assessment: 2
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Week 3 Quiz 1 Version 9 We stretch the first graph by a factor of 3 to obtain the second. For the first graph the vertex lies at (0 , 0) and the points one unit to the left and right of it are ((-1 , 1) and (1 , 1). For the second graph the vertex again lies at (0 , 0) and the points one unit to the left and right of it are ((-1 , 3) and (1 , 3). After shifting the graph -1.75 units in the x direction and .25 units in the y direction our three basic points are (-1.75 , .25) , (-2.75 , 3.25) and (-0.75 , 3.25) f(x-1.75) = (x-1.75)^2 f(x)+.25 = x^2 + .25 3f(x) = (x^2) / 3 3f(x-1.75) + .25 = [(x-1.75)^2 + .25] / 3 Week 3 Quiz 2 Version 9 For the graph y = x each function is doubled, positioning it twice as far from the x-axis. For the graph y = x^2 the distance of each point on the graph from the x-axis is doubled, making the parabola appear ""thinner"" and ""taller"". For the graph y = x^-1 the distance of each point on the graph from the x-axis is doubled, positioning the graph a little farther away from the y- and -x-axis'. For the graph y = 2^x the distance of each point on the graph from the x-axis is half of what it is in the original function, positioning the graph a little to the right of the original function ."