course Mth 163 hئŋ
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12:51:55 Query problem 2. Describe the sum of the two graphs.
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RESPONSE --> The sum line on the graph would look much like the blue line but not exactly on it it is a little bit steeper confidence assessment: 3
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12:52:25 ** The 'black' graph takes values 8, 3, 0, -1, 0, 3, 8 at x = -3, -2, -1, 0, 1, 2, 3. The 'blue' graph takes approximate values 1.7, .8, .2, -.1, -.4, -.6, -.8 at the same x values. The 'blue' graph takes value zero at approximately x = -.4. The sum of the two graphs will coincide with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The sum will coincide with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. **
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RESPONSE --> ok self critique assessment: 3
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12:53:15 Where it is the sum graph higher than the 'black' graph, and where is it lower? Answer by giving specific intervals.
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RESPONSE --> it is higher until the point of about .5,-.5 and then it is lower confidence assessment: 3
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12:53:32 ** The sum of the graphs is higher than the 'black' graph where the 'blue' graph is positive, lower where the 'blue' graph is negative. The 'blue' graph is positive on the interval from x = -3 to x = -.4, approx.. This interval can be written [-3, -.4), or -3 <= x < -.4. **
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RESPONSE --> ok self critique assessment: 3
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12:55:07 Where it is the sum graph higher than the 'blue' graph, and where is it lower? Answer by giving specific intervals.
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RESPONSE --> when the blue graph is positive it is lower and when its negative it is higher confidence assessment: 3
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12:55:17 ** The sum of the graphs is higher than the 'blue' graph where the 'black' graph is positive, lower where the 'black' graph is negative. The 'black' graph is positive on the interval from x = -1 to x = 1, not including the endpoints of the interval. This interval can be written (-1, 1) or -1 < x < 1. **
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RESPONSE --> ok self critique assessment: 3
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12:55:48 Where does thus sum graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
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RESPONSE --> inbetween x=-1 and x=1 confidence assessment: 3
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12:55:55 ** The sum coincides with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. The coordinates would be about (-.4, -.7), on the 'black' graph. **
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RESPONSE --> ok self critique assessment: 3
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12:56:12 Where does thus sum graph coincide with the 'blue' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs.
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RESPONSE --> between x=-1 and x=1 confidence assessment: 3
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12:56:19 ** The sum coincides with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The coordinates would be about (-1, .2) and (1, -.4), on the 'blue' graph. **
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RESPONSE --> ok self critique assessment: 3
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13:03:54 Query problem 3 Describe the quotient graph obtained by dividing the 'black' graph by the 'blue' graph. You should answer the following questions: Where it is the quotient graph further from the x axis than the 'black' graph, and where is it closer? Answer by giving specific intervals, and explaining why you believe these to be the correct intervals. Where it is the quotient graph on the same side of the x axis as the 'black' graph, and where is it on the opposite side, and why? Answer by giving specific intervals. Where does thus quotient graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs. Where does the quotient graph have vertical asymptote(s), and why? Describe the graph at each vertical asymptote.
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RESPONSE --> confidence assessment:
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13:05:11 ** The 'black' graph is periodic, passing through 0 at approximately x = -3.1, 0, 3.1, 6.3. This graph has peaks with y = 1.5, approx., at x = 1.6 and 7.8, approx., and valleys with y = -1.5 at x = -1.6 and x = 4.7 approx. The 'blue' graph appears to be parabolic, passing thru the y axis at x = -1 and reaching a minimum value around y = -1.1 somewhere near x = 1. This graph passes thru the x axis at x = 5.5, approx., and first exceeds y = 1 around x = 7.5. The quotient will be further from the x axis than the 'black' graph wherever the 'blue' graph is within 1 unit of the origin, since division by a number whose magnitude is less than 1 gives a result whose magnitude is greater than the number being divided. This will occur to the left of x = 1, and between about x = 2 and x = 7.5. Between about x = 0 and x = 1 the 'blue' graph is more than 1 unit from the x axis and the quotient graph will be closer to the x axis than the 'black' graph. The same is true for x > 7.5, approx.. The 'black' graph is zero at or near x = -3.1, 0, 3.1, 6.3. At both of these points the 'blue' graph is nonzero so the quotient will be zero. The 'blue' graph is negative for x < 5.5, approx.. Since division by a negative number gives us the opposite sign as the number being divided, on this interval the quotient graph will be on the opposite side of the x axis from the 'black' graph. The 'blue' graph is positive for x > 5.5, approx.. Since division by a positive number gives us the same sign as the number being divided, on this interval the quotient graph will be on the same side of the x axis as the 'black' graph. The quotient graph will therefore start at the left with positive y values, about 3 times as far from the x axis as the 'black' graph (this since the value of the 'blue' graph is about -1/3, and division by -1/3 reverses the sign and gives us a result with 3 times the magnitude of the divisor). The quotient graph will have y value about 2.5 at x = -1.6, where the 'black' graph 'peaks', but the quotient graph will 'peak' slightly to the left of this point due to the increasing magnitude of the 'blue' graph. The quotient graph will then reach y = 0 / (-1) = 0 at x = 0 and, since the 'black' graph then becomes positive while the 'blue' graph remains negative, the quotient graph will become negative. Between x = 0 and x = 2 the magnitude of the 'blue' graph is a little greater than 1, so the quotient graph will be a little closer to the x axis than the 'black' graph (while remaining on the other side of the x axis). At x = 3.1 approx. the 'black graph is again zero, so the quotient graph will meet the x axis at this point. Past x = 3.1 the quotient graph will become positive, since the signs of both graphs are negative. As we approach x = 5.5, where the value of the 'blue' graph is zero, the quotient will increase more and more rapidly in magnitude (this since the result of dividing a negative number by a negative number near zero is a large positive number, larger the closer the divisor is to zero). The result will be a vertical asymptote at x = 5.5, with the y value approaching +infinity as x approaches 5.5 from the left. Just past x = 5.5 the 'blue' values become positive. Dividing a negative number by a positive number near zero results in a very large negative value, so that on this side of x = 5.5 the asymptote will rise up from -infinity. The quotient graph passes through the x axis near x = 6.3, where the 'black' graph is again zero. To the right of this point both graphs have positive values and the quotient graph will be positive. Around x = 7.5, where the 'blue' value is 1, the graph will coincide with the 'black' graph, giving us a point near (7.5, 1.3). Past this point the 'blue' value is greater than 1 so that the quotient graph will become nearer the x axis than the 'black' graph, increasingly so as x (and hence the 'blue' value) increases. This will result in a 'peak' of the quotient graph somewhere around x = 7.5, a bit to the left of the peak of the 'black' graph. **
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RESPONSE --> I kindof understand all this its a little hard to grasp just by looking at the pictures and reading this stuff its really confusing self critique assessment: 3
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13:06:56 Query problems 7-8 Sketch the graph of y = x^2 - 2 x^4 by first sketching the graphs of y = x^2 and y = -2 x^4. How does the result compare to the graph of y = x^2 - x^4, and how do you explain the difference?
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RESPONSE --> They both have the same basic shape but because of the 2 in front of the first x^4 it is a little wider than the second confidence assessment: 3
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13:07:08 ** At x = 0, 1/2, 1 and 2 we have x^2 values 0, 1/4, 1 and 4, while -x^4 takes values 0, -1/16, -1 and -16, and -2 x^4 takes values 0, -1/8, -2 and -32. All graphs clearly pass through the origin. The graphs of y = x^2 - x^4 and y = x^4 - 2 x^4 are both increasingly negative at far right and far left. Graphical addition will show that y = x^2 - x^4 takes value 0 and hence passes thru the x axis when the graphs have equal but opposite y values, which occurs at x = 1 and x = -1. To the left of x = -1 and to the right of x = 1 the negative values of -x^4 overwhelm the positive values of x^2 and the sum graph will be increasingly negative, with values dominated by -x^4. Near x = 0 the graph of y = -x^4 is 'flatter' than that of y = x^2 and the x^2 values win out, making the sum graph positive. y = x^2 - 2 x^4 will take value 0 where the graphs are equal and opposite in value; this occurs somewhere between x = .8 and x = .9, and also between x = -.9 and x = -.8, which places the zeros closer to the y axis than those of the graph of y = x^2 - x^4. The graph of y = -2 x^4 is still flatter near x = 0 than the graph of y = x^2, but not as flat as the graph of y = -x^4, so while the sum graph will be positive between the zeros the values won't be as great. Outside the zeros the sum graph will be increasingly negative, with values dominated by -2x^4. **
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RESPONSE --> ok self critique assessment: 3
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13:08:36 How does the shape of the graph change when you add x to get y = -2 x^4 + x^2 + x, and how do you explain this change?
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RESPONSE --> it still has the same basic shape but instead of the vertex being on the x axis as i would have expected the vertex is actually above it and is in the positive range confidence assessment: 3
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13:08:41 ** At x = 0 there is no change in the y value, so the graph still passes through (0, 0). As x increases through positive numbers we will have to increase the y values of y = x^2 -2 x^4 by greater and greater amounts. So it will take a little longer for the negative values of -2 x^4 to 'overwhelm' the positive values of x^2 + x than to overcome the positive values of x^2 and the x intercept will shift a bit to the right. As we move away from x = 0 through negative values of x we will find that the positive effect of y = x^2 is immediately overcome by the negative values of y = x, so there is no x intercept to the left of x = 0. The graph in fact stays fairly close to the graph of y = x near (0, 0), gradually moving away from that graph as the values of x^2 and -2 x^4 become more and more significant. **
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RESPONSE --> ok self critique assessment: 3
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13:10:14 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I am really surprised by that last graph I really dont understand why the vertex ended up like that...This whole idea of dividing graphs by each other and adding them is really confusing to me I think I just need some more hands on with it I will try to do some problems out of the book to see it I can learn it better that way
course Mth 163 dSÜ劒˽Ƹassignment #024
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13:12:18 Explain why, when either f(x) or g(x) is 0, then the product function also has a 0 for that value of x.
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RESPONSE --> because anything multiplied by 0 is zero confidence assessment: 3
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13:12:24 STUDENT RESPONSE: If you multiply any number by zero and you get zero.
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RESPONSE --> ok self critique assessment: 3
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13:12:55 Explain why, when the magnitude | f(x) | of f(x) is greater than 1 (i.e., when the graph of f(x) is more than one unit from the x axis), then the product function will be further from the x axis than the g(x) function.
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RESPONSE --> because it is automatically moved up 1 unit confidence assessment: 3
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13:13:04 STUDENT RESPONSE: If you multiply a number by another number greater than 1, the result is greater than the original number. If you multiply a number by another number whose magnitude is greater than 1, the result will have greater magnitude that the original number. If | f(x) | > 1 then the magnitude of f(x) * g(x) will be greater than the magnitude of g(x). The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude increases so does the distance from the x axis.
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RESPONSE --> oh ok self critique assessment: 3
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13:13:43 Explain why, when the magnitude | f(x) | of f(x) is less than 1 (i.e., when the graph of f(x) is less than one unit from the x axis), then the product function will be closer to the x axis than the g(x) function.
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RESPONSE --> because it would be like dividing if you multiply a number by .5 it is the same as dividing by 2 confidence assessment: 3
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13:13:51 STUDENT RESPONSE: If you multiply a number by another number less than 1, the result is less than the original number. If you multiply a number by another number whose magnitude is less than 1, the result will have a lesser magnitude that the original number. If | f(x) | < 1 then the magnitude of f(x) * g(x) will be less than the magnitude of g(x). The magnitude of g(x) at a given value of x is its distance from the x axis, so when the magnitude decreases so does the distance from the x axis.
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RESPONSE --> ok self critique assessment: 3
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13:14:46 Explain why, when f(x) and g(x) are either both positive or both negative, the product function is positive. When f(x) and g(x) have opposite signs the product function is negative.
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RESPONSE --> because 2 negatives make a positive but if you have -f(x)*g(x) you get -f(g(x)) confidence assessment: 3
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13:14:51 STUDENT RESPONSE: This is basic multiplication: + * + = +, - * - = -, + * - = -. The product of like signs is positive, the product of unlike signs is negative. Since the product function results from multiplication of the two functions, these rules apply.
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RESPONSE --> ok self critique assessment: 3
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13:15:27 Explain why, when f(x) = 1, the graph of the product function coincides with the graph of g(x).
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RESPONSE --> becuase the one wouldnt change anything confidence assessment: 3
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13:15:35 STUDENT RESPONSE: g(x) * 1 = g(x)
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RESPONSE --> ok self critique assessment: 3
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13:17:52 problem 4 Sketch graphs for y = f(x) = 2^x and y = g(x) = .5 x, for -2 < x < 2. Use your graphs to predict the shape of the y = g(x) * f(x) graph. Describe the graphs of the two functions, and explain how you used these graphs predict the shape of the graph of the product function.
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RESPONSE --> anything raised to the x is an exponential function and .5 x is representing the slope of a straight line this was an easy thing to predict confidence assessment: 3
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13:18:00 STUDENT RESPONSE: Where g(x) is - and f(x) is + graph will be -. where g(x) =0 graph will be at 0where both are + graph will be positive and rise more steeply. y=2^x asymptote negative x axis y intercept (0,1) y=.5x linear graph passing through (0,0) rising 1 unit for run of 2 units INSTRUCTOR COMMENT: It follows that since one function is negative for x < 0 while the other is always positive the product will be negative for x < 0, and since both functions are positive for x > 0 the product will be positive for x > 0. Since one function is zero at x=0 the product will be 0 at x = 0. For x > 0 the exponential rise of the one graph and the continuing rise of the other imply that the graph will rise more and more rapidly, without bound, for large positive x. For x < 0 one function is positive and the other is negative so the graph will be below the x axis. For large negative x, one graph approaches 0 while the other keeps increasing in magnitude; it's not immediatly clear which function 'wins'. However the exponential always 'beats' a fixed power so the graph will be asymptotic to the negative x axis. It will reach a minimum somewhere to the left of the x axis, before curving back toward the x axis and becoming asymptotic. **
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RESPONSE --> ok self critique assessment: 3
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13:20:09 problem 7 range(depth) = 2.9 `sqrt(depth) and depth(t) = t^2 - 40 t + 400. At what times is depth 0. How did you show that the vertex of the graph of depth vs. time coincides with these zeros?
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RESPONSE --> confidence assessment: 3
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13:20:50 ** The depth function is quadratic. Its vertex occurs at t = - b / (2 a) = - (-40) / (2 * 1) = 20. Its zeros can be found either by factoring or by the quadratic formula. t^2 - 40 t + 400 factors into (t - 20)(t - 20), so the only zero is at t = 20. This point (20, 0) happens to be the vertex, and the graph opens upward, so the graph never goes below the x axis. STUDENT QUESTION: When I simplified range(depth(t) = 2.9*'sqrt(t^2 - 40t +400) I got 2.9t - 58 which gives a negative range so I reversed it and got the correct results, what have I done wrong?
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RESPONSE --> i didnt understand how to show that the vertex was at the zeros self critique assessment: 3
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13:21:51 range(depth(t) = 2.9*'sqrt(t^2 - 40t +400) = 2.9*'sqrt( (t-20)^2) ) = 2.9( t - 20) = 2.9t - 58 I know it should be -2.9 + 58 I just don't understand how to get there. Thanks INSTRUCTOR RESPONSE TO QUESTION: This is a great question. What is sqrt( (-5) ^ 2)? `sqrt( (-5)^2 ) isn't -5, it's 5, since `sqrt(25) = 5. This shows that you have to be careful about possible negative values of t - 20. This is equivalent to saying that `sqrt( (-5)^2 ) = | -5 = 5|. `sqrt( (t-20) * (t-20) ) has to be positive. So `sqrt( (t-20) * (t-20) ) = | t - 20 |. If t < 20 then t - 20 is negative so that | t - 20 | = -(t - 20) = 20 - t. **
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RESPONSE --> ok self critique assessment: 3
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13:22:27 What two linear factors represent the depth function as their product?
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RESPONSE --> confidence assessment: 3
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13:22:37 depth(t) = (t-20)(t-20)
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RESPONSE --> oh ok confidence assessment: 3
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13:23:11 For t = 5, 10 and 15, what are the ranges?
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RESPONSE --> just plug all these numbers in for (t-20)(t-20) confidence assessment: 3
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13:23:52 ** depth(t) = t^2 - 40 t + 400 = (t-20)^2 so depth(5) = (5-20)^2 = (-15)^2 = 225 depth(10) = (10-20)^2 = (-10)^2 = 100 depth(15) = (15-20)^2 = (-5)^2 = 25. It follows that the ranges are range(depth(5)) = 2.9 sqrt(225) = 43.5 range(depth(10) = 2.9 sqrt(100) = 29 and range(depth(15) = 2.9 sqrt(25) = 14.5. **
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RESPONSE --> wow this just clicked completley to me it really just made since
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13:24:19 What is the function range(depth(t))? Show that its simplified form is linear in time.
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RESPONSE --> confidence assessment: 3
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13:24:48 ** range(depth(t) ) = 2.9 sqrt(depth(t)) = 2.9 sqrt(t^2 - 40 t + 400) = 2.9 sqrt( (t - 20)^2 ) = 2.9 | t - 20 |. **
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RESPONSE --> ok self critique assessment: 3
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13:28:03 problem 8 Illumination(r) = 40 / r^2; distance = 400 - .04 t^2. What is the composite function illumination(distance(t))?
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RESPONSE --> im not sure how to work with this one because of the division confidence assessment: 3
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13:28:20 ** Illumination(r) = 40 / r^2 so Illumination(distance(t)) = 40 / (distance(t))^2 = 40 / (400 - .04 t^2)^2. **
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RESPONSE --> oh ok self critique assessment: 3
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13:28:38 Give the illumination at t = 25, t = 50 and t = 75. At what average rate is illumination changing during the time interval from t = 25 to t = 50, and from t = 50 to t = 75?
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RESPONSE --> confidence assessment:
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13:28:58 ** illumination(distance(t)) = 40 / (400 - .04 t^2)^2 so illumination(distance(25)) = 40 / (400 - .04 * 25^2)^2 = .000284 illumination(distance(50)) = 40 / (400 - .04 * 50^2)^2 = .000444 illumination(distance(75)) = 40 / (400 - .04 * 75^2)^2 = .001306. from 25 to 50 change is .000444 - .000284 = .000160 so ave rate is .000160 / 25 = .0000064 from 50 to 75 change is .001306 - .000444 = .00086 so ave rate is .00086 / 25 = .000034 approx. **
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RESPONSE --> ok i understand the plugging in part well self critique assessment: 3
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13:33:05 problem 10 gradeAverage = -.5 + t / 10. t(Q) = 50 (1 - e ^ (-.02 (Q - 70) ) ). If the student's mental health quotient is an average 100, then what grade average should the student expect?
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RESPONSE --> i tried to work this out several differnet ways but appearantly they werent right because my numbers arent adding up at all confidence assessment: 3
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13:34:27 ** gradeAverage = -.5 + t / 10 = -.5 + 50 ( 1 - e^(-.02 (Q - 70) ) ) / 10 = -.5 + 5 ( 1 - e^(-.02 (Q - 70) ) ) So gradeAverage(t(100)) = -.5 + 5 ( 1 - e^(-.02 ( 100 - 70) ) = -.5 + 5( 1 - .5488 ) = -.5 + 5 ( .4522 ) = -.5 + 2.26 = 1.76. gradeAverage(t(110)) = -.5 + 5 ( 1 - e^(-.02 ( 110 - 70) ) = -.5 + 5( 1 - .4493 ) = -.5 + 5 ( .5517 ) = -.5 + 2.76 = 2.26. gradeAverage(t(120)) = -.5 + 5 ( 1 - e^(-.02 ( 120 - 70) ) = -.5 + 5( 1 - .3678 ) = -.5 + 5 ( .6322 ) = -.5 + 3.16 = 2.66. gradeAverage(t(130)) = -.5 + 5 ( 1 - e^(-.02 ( 130 - 70) ) = -.5 + 5( 1 - .3012 ) = -.5 + 5 ( .6988 ) = -.5 + 3.49 = 2.99. As Q gets larger and larger Q - 70 will get larger and larger, so -.02 ( Q - 70) will be a negative number with increasing magnitude; its magnitude increases without limit. It follows that e^(-.02 ( Q - 70) ) = will consist of e raised to a negative number whose magnitude increases without limit. As the magnitude of the negative exponent increases the result will be closer and closer to zero. So -.5 + 5 ( 1 - e^(-.02 ( Q - 70) ) ) will approach -.5 + 5 ( 1 - 0) = -.5 + 5 = 4.5. Side note: For Q = 100, 200 and 300 we would have grade averages 1.755941819, 4.128632108, 4.449740821. To get a 4-point Q would have to be close to 200. Pretty tough course
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RESPONSE --> ok I was actually doing this right I was just thinking that my grade average should be like a 90 or a high number not a 2.7 self critique assessment: 3
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13:34:49 What grade averages would be expected for mental health quotients of 110, 120 and 130?
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RESPONSE --> confidence assessment:
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13:34:54 110...2.2534, 120...2.66, 130...2.99
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RESPONSE --> ok self critique assessment: 3
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13:35:36 What is the upper limit on the expected grade average that can be achieved by this student?
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RESPONSE --> The limit would have to be how high they could actually score on tests confidence assessment: 3
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13:35:44 ** If Q is very, very large, e^-( .02(Q-70) ) would have a negative exponent with a very large magnitude and so would be very close to 0. In this case 50 ( 1-e^(-.02 (Q-70)) would be close to 50(1-0) = 50. Then the grade average would be -.5 + 50 / 10 = -.5 + 5 = 4.5 . DER [0.5488116360, 0.4493289641, 0.3678794411, 0.3011942119][1.755941819, 2.253355179, 2.660602794, 2.994028940]
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RESPONSE --> ok self critique assessment: 3
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13:35:55 What is the composite function gradeAverage( t(Q) )?
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RESPONSE --> confidence assessment: 3
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13:36:00 -.5+(50(1-e^(-.02(Q-70))/10
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RESPONSE --> ok self critique assessment: 3
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13:36:04 What do you get when you evaluate your composite function at t = 100, 110, 120 and 130?
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RESPONSE --> confidence assessment: 3
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13:36:09 t=100...9.5 t=110...10.5 t=120...11.5 t=130...12.5
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RESPONSE --> ok self critique assessment: 3
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13:36:14 ** you should get the same values you got before for these Q values. For example an approximate calculation for t = 130 is -.5 + 50(1-e^(-.02(130-70) ) / 10 = -.5 + 50 (1-e^-1.2) / 10 = -.5 + 50 (1 - .3) / 10 = -.5 + 35/10 = -.5 + 3.5 = 3, approx., pretty close to your 2.99 **
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RESPONSE --> ok self critique assessment:
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13:38:57 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I honestly learned that when I get ready to do these queries I think the worksheets made since but then the query comes and I completley forget everything I have found that I am really not good at learning things through the computer even though I really have been trying to make since of it all...I think after this course even though it is set up very well and probably works for most people I will stick with in class courses from now on self critique assessment: 3
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