course mth 163 for quizExplain how the discriminant b2 - 4 a c tells us how many zeros a quadratic function has. if b2 - 4 a c equals a number less than zero there are no zeros, if it equals a number greater than zero then it has 2 zeros and if it is equal to zero then there is only one zero." "HԈ߄dN鰸ޥEγ
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12:48:08 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> f(-2)= -8 f(-a)= -a^3 f(x-4)=x^3-12x^2+48x-64 f(x) - 4=x^3-4 confidence assessment: 2
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12:49:05 ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. **
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RESPONSE --> self critique assessment: 1
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12:51:41 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> f(2)=4 f(-a)=2^(-a) f(x+3)=2^(x+3) f(x) + 3=2^x+3 confidence assessment: 2
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12:51:55 ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. **
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RESPONSE --> self critique assessment: 3
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12:52:56 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> you can visualize what your problem is about confidence assessment: 1
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12:53:04 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> confidence assessment: 1
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12:53:27 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.**
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RESPONSE --> self critique assessment: 3
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13:02:33 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> value(0)=1000(1.07)^0=1000 value(2)=1000(1.07)^2=1144.9 value(t+3)=1000(1.07)^(t+3) (t+3)/value(t)=[1000(1.07)^(t+3)]/(1000(1.07)^(t)) confidence assessment: 2
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13:03:52 ** Substitute very carefully and show your steps: value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. **
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RESPONSE --> self critique assessment: 3
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13:06:38 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> illumination(distance)/illumination(2*distance)= [50/distance^2]/[50/2distance^2] distance^2/2distance^2 =1/2 confidence assessment: 1
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13:09:12 ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. **
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RESPONSE --> ok is see my mistake i didn't flip and multiply. it should be [50 / distance^2] / [50 / (2*distance)^2] =[ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] =(2 * distance)^2 / distance^2 =4 * distance^2 / distance^2 = 4 self critique assessment: 2
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13:22:07 query #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph.
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RESPONSE --> 80=a4+b2+c 40=a25+b5+c 25=a100+b10+c 40= -21a-3b 15= -75a-5b 200= -105a -15b 45= -225 -15b 155= 120a a=1.29166666 40= -21(1.29166666)-3b b= -22.375 80=(1.29166666)4+(-22.375)2+c c=119.583333336 f(x)=1.29166666x^2-22.375x+119.583333336 confidence assessment: 1
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13:22:56 ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. **
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RESPONSE --> ok i worked the whole problem til i got the formula but i got a smooth curve with the given points self critique assessment: 2
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13:28:30 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> 1.2916666(60)^2-22.375(60)+119.58333335 =3427.08330935 confidence assessment: 1
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13:29:07 **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3**
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RESPONSE --> ?? self critique assessment: 0
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13:30:19 what is your estimate of the value f(7)?
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RESPONSE --> = about 30 confidence assessment: 3
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13:30:42 ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. **
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RESPONSE --> self critique assessment: 3
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13:31:04 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> about 3.5 confidence assessment: 2
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13:31:22 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> self critique assessment: 2
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13:31:47 what is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30?
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RESPONSE --> about 4.75 confidence assessment: 1
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13:31:56 ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. **
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RESPONSE --> self critique assessment: 3
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13:36:22 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> temperature confidence assessment: 2
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13:36:53 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 **
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RESPONSE --> i miss understood what you were asking for self critique assessment: 1
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13:40:01 What equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30?
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RESPONSE --> T(t)=150 t=150/T
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13:41:32 ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] **
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RESPONSE --> ok self critique assessment: 3
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13:44:05 query. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters?
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RESPONSE --> f(34)-f(47)= 20.71966 sec. confidence assessment: 2
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13:44:25 ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' **
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RESPONSE --> ok self critique assessment: 3
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13:45:55 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> f(23)-f(34)=19.0421 cm. confidence assessment: 2
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13:46:16 ** This would be f(34) - f(23). If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, you would find that f(34) = 50.6 and f(23) = 60.8 so f(34) - f(23) = 50.6 - 60.8 = -10.2. **
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RESPONSE --> self critique assessment: 0
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13:47:36 On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds?
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RESPONSE --> .577667 sec. per cm. confidence assessment: 2
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13:47:51 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in t / change in depth = 11 s / [ f(34) - f(23) ]. If f(t) gives depth in cm this result will be in seconds / centimeter, the ave number of seconds required for depth to change by 1 cm. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that 11 s / [ f(34) - f(23) ] = 11 s / (-10.2 cm) = -1.08 sec / cm, indicating that depth decreases by 1 cm in 1.08 seconds. **
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RESPONSE --> self critique assessment: 3
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13:48:15 On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds?
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RESPONSE --> 1.7311 cm. per sec. confidence assessment: 2
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13:48:28 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. **
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RESPONSE --> self critique assessment: 3
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13:54:24 query. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph.
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RESPONSE --> i drew the curve(which seemed to go down the middle of the given points) then I picked 3 points on the curve. I chose (0,96), (30,63) and (70,33) a= 96=a0^2+b0+c b=63=a30^2+b30+c c=33=a70^2+b70+c 33= -900a-30b 30= -4000a= 40b 42=8400a a=.005 33= -900(.005) -30b b= -1.25 96=0 +0+c c=96 y=.005x^2-1.25x+96 confidence assessment: 2
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13:54:40 ** STUDENT RESPONSE: I sketched a graph using the depth vs. time data:(0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). **
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RESPONSE --> self critique assessment: 3
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13:54:56 What 3 data point did you use as a basis for your model?
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RESPONSE --> I chose (0,96), (30,63) and (70,33) confidence assessment: 3
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13:55:05 ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.**
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RESPONSE --> self critique assessment: 3
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13:55:16 What was your function model?
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RESPONSE --> I chose (0,96), (30,63) and (70,33) a= 96=a0^2+b0+c b=63=a30^2+b30+c c=33=a70^2+b70+c 33= -900a-30b 30= -4000a= 40b 42=8400a a=.005 33= -900(.005) -30b b= -1.25 96=0 +0+c c=96 y=.005x^2-1.25x+96 confidence assessment: 2
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13:55:23 ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089x^2 - 1.4992x + 98.8544. **
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RESPONSE --> self critique assessment: 3
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13:55:36 What is the average deviation for your model?
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RESPONSE --> 4 confidence assessment: 2
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13:55:42 ** STUDENT RESPONSE CONTINUED: I added a column 2 columns to the chart given in our assignment one labeled 'Model data' and the other 'Deviation'. I then subtracted the model data readings from the corresponding given data to get the individual deviations. I then averaged out the numbers in the deviation column. The average deviation turned out to be 3.880975.**
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RESPONSE --> self critique assessment: 3
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13:55:56 How close is your model to the curve you sketched earlier?
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RESPONSE --> it was very close confidence assessment: 3
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13:56:08 ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.**
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RESPONSE --> self critique assessment: 3
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13:56:15 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 3
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13:56:21 ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation **
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RESPONSE --> self critique assessment: 3
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F~߆XKڻR assignment #004 004. Precalculus I 09-08-2008
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10:32:42 `q001. Note that this assignment has 4 questions If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.
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RESPONSE --> my answers are f(3)=13 f(7)=53 f(-5)=29 my graph is a prabola shape that opens upwards, and apears to be centered on the y axis confidence assessment: 2
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10:38:04 f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.
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RESPONSE --> ok i miss understood how much you wanted. I didn't work the whole problem out to get the x coordinates and the vertex and the points one space to the right and to the left but the should be b/(2 a) = -0/(2*1) = 0 f(0) = 0 ^ 2 + 4 = 0 + 4 = 4 vertex (0,4) 1,5) and (-1,5) are one space to the right and left and there are no x intercepts self critique assessment: 2
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10:48:59 `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.
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RESPONSE --> f(a)=a^2+4 f(x+h)=(x+h)^2+4 x2+4x+8 f(x+h)-f(x)=(x^2+2xh+h^2+4)- (x^2+4) 2xh+h^2 f(x+h) - f(x) ] / h= (x^2+2xh+h^2+4)- (x^2+4)/ (h^2+4) x^2+sxh+h^2+3-(x^2)/(h^2) confidence assessment: 2
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10:53:47 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.
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RESPONSE --> ok i see my mistake in[ f(x+h) - f(x) ] / h it should be = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h. i should have divided the whole problem by h but I only divided the f(x) self critique assessment: 2
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11:03:07 `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.
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RESPONSE --> f(x1)=5(x1)+7 f(x2)=5(x2)+7 [ f(x2) - f(x1) ] / ( x2 - x1 ) =[(5(x2)+7)-(5(x1)+7)]/(x2-x1) =(5x2-5x1)/(x2-x1) =5-5=0 confidence assessment: 2
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11:03:38 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.
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RESPONSE --> ok i see my mistake i should have factored the 5 out self critique assessment: 2
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11:06:23 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
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RESPONSE --> f(x) = 5x + 7 -3=5x + 7 -10=5x -2=x confidence assessment: 2
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11:07:26 If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.
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RESPONSE --> self critique assessment: 3
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11:07:45 end program
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RESPONSE --> self critique assessment: 3
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course mth 163 for quizExplain how the discriminant b2 - 4 a c tells us how many zeros a quadratic function has. if b2 - 4 a c equals a number less than zero there are no zeros, if it equals a number greater than zero then it has 2 zeros and if it is equal to zero then there is only one zero." "HԈ߄dN鰸ޥEγ
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12:48:08 Where f(x) = x^3, what are f(-2), f(-a), f(x-4) and f(x) - 4?
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RESPONSE --> f(-2)= -8 f(-a)= -a^3 f(x-4)=x^3-12x^2+48x-64 f(x) - 4=x^3-4 confidence assessment: 2
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12:49:05 ** COMMON ERROR WITH COMMENT: Where f(x) = x^3, f(-2) = -2^3 INSTRUCTOR CRITIQUE: Write that f(-2) = (-2)^3. By order of operations -2^3 = -(2^3) (you exponentiate first, before you apply the negative, which is effectively a multiplication by -1). For an odd power like 3 it makes no difference in the end---in both cases the result is -8 COMMON ERROR WITH COMMENT: f(-a) = -a^3. INSTRUCTOR COMMENT:f(-a) = (-a)^3, which because of the odd power is the same as -a^3. If g(x) = x^2, g(-a) would be (-a)^2 = a^2. ANSWERS TO THE REMAINING TWO QUESTIONS: f(x-4) = (x-4)^3. If you expand that you get (x^2 - 8 x + 16) ( x - 4) = x^3 - 12 x^4 + 48 x - 64. In more detail the expansion is as follows: (x-4)^3 = (x-4)(x-4)(x-4) = [ (x-4)(x-4) ] * (x-4) = [ x ( x - 4) - 4 ( x - 4) ] ( x - 4) = (x^2 - 4 x - 4 x + 16) ( x - 4) = (x^2 - 8x + 16) ( x - 4) = (x^2 - 8x + 16) * x - (x^2 - 8x + 16) * 4 = x^2 * x - 8 x * x + 16 * x - x^2 * 4 - (-8x) * 4 - 16 * 4 = x^3 - 8 x^2 + 16 x - 4 x^2 + 32 x - 64 = x^3 - 12 x^2 + 48 x - 64. f(x) - 4 = x^3 - 4. **
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RESPONSE --> self critique assessment: 1
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12:51:41 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> f(2)=4 f(-a)=2^(-a) f(x+3)=2^(x+3) f(x) + 3=2^x+3 confidence assessment: 2
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12:51:55 ** Where f(x) = 2^x we have: f(2)= 2^2 or 4; f(a) = 2^a; f(x+3) = 2^(x+3); and f(x) + 3 = 2^x + 3. **
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RESPONSE --> self critique assessment: 3
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12:52:56 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> you can visualize what your problem is about confidence assessment: 1
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12:53:04 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> confidence assessment: 1
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12:53:27 ** TWO STUDENT RESPONSES: Using the function notation gives more meaning to our equations. Example... 'depth(t) = ' is a lot more understandable than ' y = ' I read of the pros and cons in using names. I think using names helps to give meaning to the equation, especially if it's one you haven't looked at for a couple days, you know right away what you were doing by reading the 'meaningful' notation.**
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RESPONSE --> self critique assessment: 3
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13:02:33 What were your results for value(0), value(2), value(t+3) and value(t+3)/value(t)?
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RESPONSE --> value(0)=1000(1.07)^0=1000 value(2)=1000(1.07)^2=1144.9 value(t+3)=1000(1.07)^(t+3) (t+3)/value(t)=[1000(1.07)^(t+3)]/(1000(1.07)^(t)) confidence assessment: 2
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13:03:52 ** Substitute very carefully and show your steps: value(0) = $1000(1.07)^0 = $ 1000 value(2) = $1000(1.07)^2 = $1144.90 value(t + 3) = $1000(1.07)^(t + 3) value(t + 3) / value (t) = [$1000(1.07)^(t + 3)] / [ $1000(1.07)^t] , which we simplify. The $1000 in the numerator can be divided by the $1000 in the denominator to give us value(t+3) / value(t) = 1.07^(t+3) / [ 1.07^t]. By the laws of exponents 1.07^(t+3) = 1.07^t * 1.07^3 so we get value(t+3) / value(t) = 1.07^t * 1.07^3 / [ 1.07^t]. The 1.07^t divides out and we end up with value(t+3) / value(t) = 1.07^3. **
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RESPONSE --> self critique assessment: 3
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13:06:38 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> illumination(distance)/illumination(2*distance)= [50/distance^2]/[50/2distance^2] distance^2/2distance^2 =1/2 confidence assessment: 1
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13:09:12 ** We substitute carefully and literally to get illumination (distance) / illumination (2*distance) = [50 / distance^2] / [50 / (2*distance)^2] which is a complex fraction and needs to be simplified. You invert the denominator and multiply to get [ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] = (2 * distance)^2 / distance^2 = 4 * distance^2 / distance^2 = 4. **
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RESPONSE --> ok is see my mistake i didn't flip and multiply. it should be [50 / distance^2] / [50 / (2*distance)^2] =[ 50 / distance^2 ] * [ (2 * distance)^2 / 50 ] =(2 * distance)^2 / distance^2 =4 * distance^2 / distance^2 = 4 self critique assessment: 2
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13:22:07 query #3. Sketch a reasonable graph of y = f(x), if it is known that f(2) = 80, f(5) = 40 and f(10) = 25. Explain how you constructed your graph.
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RESPONSE --> 80=a4+b2+c 40=a25+b5+c 25=a100+b10+c 40= -21a-3b 15= -75a-5b 200= -105a -15b 45= -225 -15b 155= 120a a=1.29166666 40= -21(1.29166666)-3b b= -22.375 80=(1.29166666)4+(-22.375)2+c c=119.583333336 f(x)=1.29166666x^2-22.375x+119.583333336 confidence assessment: 1
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13:22:56 ** STUDENT ANSWER WITH INSTRUCTOR COMMENT: I used graph paper counted each small square as 2 units in each direction, I put a dot for the points (2,80), (5,40), & (10,25) and connected the dots with lines. INSTRUCTOR COMMENT: The points could be connected with straight lines, but you might also have used a smooth curve. **
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RESPONSE --> ok i worked the whole problem til i got the formula but i got a smooth curve with the given points self critique assessment: 2
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13:28:30 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> 1.2916666(60)^2-22.375(60)+119.58333335 =3427.08330935 confidence assessment: 1
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13:29:07 **If your graph was linear from the point (2, 80) to the point (5, 40) you would estimate x = 3.5, since 3.5 is halfway from 2 to 5 and 60 is halfway from 80 to 40. However with f(10) = 25 a straightforward smooth curve would be decreasing at a decreasing rate and the y = 60 value would probably occur just a bit earlier, perhaps somewhere around x = 3.3**
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RESPONSE --> ?? self critique assessment: 0
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13:30:19 what is your estimate of the value f(7)?
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RESPONSE --> = about 30 confidence assessment: 3
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13:30:42 ** STUDENT RESPONSE WITH INSTRUCTOR COMMENT: My estimation for the value of f(7) was f(7) = 34. A linear graph between (5, 40) and (10, 25) would have slope -3 and would therefore decline by 6 units between x = 5 and x = 7, giving your estimate of 34. However the graph is probably decreasing at a decreasing rate, implying a somewhat greater decline between 5 and 7 that would be implied by a straight-line approximation. A better estimate might be f(7) = 32 or 33. **
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RESPONSE --> self critique assessment: 3
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13:31:04 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> about 3.5 confidence assessment: 2
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13:31:22 ** If f(7) = 32 and f(9) = 27, not unreasonable estimates, then the difference is 5. **
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RESPONSE --> self critique assessment: 2
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13:31:47 what is your estimate of the difference in x values between the points where f(x) = 70 and where f(x) = 30?
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RESPONSE --> about 4.75 confidence assessment: 1
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13:31:56 ** f(x) will be 70 somewhere around x = 3 and f(x) will be 30 somewhere around x = 8 or 9. The difference in these x values is about 5 or 6. On your graph you could draw horizontal lines at y = 70 and at y = 30. You could then project these lines down to the x axis. The distance between these projection lines, which again should probably be around 5 or 6, can be estimated by the scale of the graph. **
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RESPONSE --> self critique assessment: 3
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13:36:22 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> temperature confidence assessment: 2
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13:36:53 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: The temperature at time t = 3; T(3)The temperature at time t = 5; T(5) The change in temperature between t = 3 and t = 5; T(3) - T(5) The order of the expressions is important. For example the change between 30 degrees and 80 degrees is 80 deg - 30 deg = 50 degrees; the change between 90 degrees and 20 degrees is 20 deg - 90 deg = -70 deg. The change between T(3) and T(5) is T(5) - T(3). When we specify the change in a quantity we subtract the earlier value from the later. INSTRUCTOR COMMENT: To average two numbers you add them and divide by 2. The average of the temperatures at t = 3 and t = 5 is therefore [T(3) + T(5)] /2 **
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RESPONSE --> i miss understood what you were asking for self critique assessment: 1
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13:40:01 What equation would we solve to find the clock time when the model predicts a temperature of 150? How would we find the length of time required for the temperature to fall from 80 to 30?
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RESPONSE --> T(t)=150 t=150/T
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13:41:32 ** GOOD STUDENT SOLUTION: To find the clock time when the model predicts a temperature of 150 we would solve the equation T(t) = 150. To find the length of time required for the temperature to fall from 80 to 30 we would first solve the equation T(t) = 80 to get the clock time at 80 degrees then find the t value or clok time when T(t) = 30. Then we could subtract the smaller t value from the larger t value to get the answer. [ value of t at T(t) = 30] - [ value of t at T(t) = 80)] **
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RESPONSE --> ok self critique assessment: 3
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13:44:05 query. use the f(x) notation at every opportunity:For how long was the depth between 34 and 47 centimeters?
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RESPONSE --> f(34)-f(47)= 20.71966 sec. confidence assessment: 2
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13:44:25 ** If depth = f(t) then you would solve f(t) = 34 to obtain t1 and f(t) = 47 to obtain t2. Then you would find abs(t2-t1). We use the absolute value because we don't know which is greater, t1 or t2, and the questions was 'how long' rather than 'what is the change in clock time ... ' **
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RESPONSE --> ok self critique assessment: 3
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13:45:55 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> f(23)-f(34)=19.0421 cm. confidence assessment: 2
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13:46:16 ** This would be f(34) - f(23). If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, you would find that f(34) = 50.6 and f(23) = 60.8 so f(34) - f(23) = 50.6 - 60.8 = -10.2. **
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RESPONSE --> self critique assessment: 0
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13:47:36 On the average, how many seconds did it take for the depth to change by 1 centimeter between t = 23 seconds and t = 34 seconds?
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RESPONSE --> .577667 sec. per cm. confidence assessment: 2
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13:47:51 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in t / change in depth = 11 s / [ f(34) - f(23) ]. If f(t) gives depth in cm this result will be in seconds / centimeter, the ave number of seconds required for depth to change by 1 cm. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that 11 s / [ f(34) - f(23) ] = 11 s / (-10.2 cm) = -1.08 sec / cm, indicating that depth decreases by 1 cm in 1.08 seconds. **
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RESPONSE --> self critique assessment: 3
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13:48:15 On the average, how by many centimeters did the depth change per second between t = 23 seconds and t = 34 seconds?
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RESPONSE --> 1.7311 cm. per sec. confidence assessment: 2
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13:48:28 ** Change in depth would be f(34) - f(23). Change in t would be 34 s - 23 s = 11 s. Thus change in depth / change in t = [ f(34) - f(23) ] / (11 s). If f(t) gives depth in cm this result will be in centimeters / second, the ave number of centimeters of depth change during each second. If for example if your model was f(t) = .01 t^2 - 1.5 t + 90, f(t) in cm, you would find that f(34) = 50.6 cm and f(23) = 60.8 cm so f(34) - f(23) = 50.6 cm - 60.8 cm = -10.2 cm so that [ f(34) - f(23) ] / (11 s) = (-10.2 cm) / (11 s) = -.92 cm / sec, indicating that between t = 23 sec and t = 34 sec depth decreases by -.92 cm in 1 sec, on the average. **
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RESPONSE --> self critique assessment: 3
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13:54:24 query. A hypothetical depth vs. time model based on three points, none of which are actual data points. Describe how you constructed your graph.
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RESPONSE --> i drew the curve(which seemed to go down the middle of the given points) then I picked 3 points on the curve. I chose (0,96), (30,63) and (70,33) a= 96=a0^2+b0+c b=63=a30^2+b30+c c=33=a70^2+b70+c 33= -900a-30b 30= -4000a= 40b 42=8400a a=.005 33= -900(.005) -30b b= -1.25 96=0 +0+c c=96 y=.005x^2-1.25x+96 confidence assessment: 2
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13:54:40 ** STUDENT RESPONSE: I sketched a graph using the depth vs. time data:(0, 96), (10, 89), (20, 68), (30, 65), (40, 48), (50, 49), (60, 36), & (70, 41). **
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RESPONSE --> self critique assessment: 3
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13:54:56 What 3 data point did you use as a basis for your model?
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RESPONSE --> I chose (0,96), (30,63) and (70,33) confidence assessment: 3
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13:55:05 ** STUDENT RESPONSE: After drawing a curved line through the scatter data I placed 3 dots on the graph at aproximately equal intervals. I obtained the x and y locations form the axis, thier locations were at points (4,93), (24, 68) & (60, 41). I used these 3 data points as a basis for obtaining the model equation.**
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RESPONSE --> self critique assessment: 3
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13:55:16 What was your function model?
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RESPONSE --> I chose (0,96), (30,63) and (70,33) a= 96=a0^2+b0+c b=63=a30^2+b30+c c=33=a70^2+b70+c 33= -900a-30b 30= -4000a= 40b 42=8400a a=.005 33= -900(.005) -30b b= -1.25 96=0 +0+c c=96 y=.005x^2-1.25x+96 confidence assessment: 2
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13:55:23 ** STUDENT RESPONSE CONTINUED: The function model obtained from points (4, 93), (24, 68), & (60, 41) is depth(t) = .0089x^2 - 1.4992x + 98.8544. **
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RESPONSE --> self critique assessment: 3
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13:55:36 What is the average deviation for your model?
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RESPONSE --> 4 confidence assessment: 2
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13:55:42 ** STUDENT RESPONSE CONTINUED: I added a column 2 columns to the chart given in our assignment one labeled 'Model data' and the other 'Deviation'. I then subtracted the model data readings from the corresponding given data to get the individual deviations. I then averaged out the numbers in the deviation column. The average deviation turned out to be 3.880975.**
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RESPONSE --> self critique assessment: 3
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13:55:56 How close is your model to the curve you sketched earlier?
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RESPONSE --> it was very close confidence assessment: 3
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13:56:08 ** STUDENT RESPONSE CONTINUED: I was really suprised at how close the model curve matched the curve that I had sketched earlier. It was very close.**
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RESPONSE --> self critique assessment: 3
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13:56:15 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> self critique assessment: 3
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13:56:21 ** STUDENT RESPONSE: I was surprized that the scattered data points on the hypothetical depth vs. time model would fit a curve drawn out through it. The thing that sank in for me with this was to not expect all data to be in neat patterns, but to look for a possible pattern in the data. INSTRUCTOR COMMENT: Excellent observation **
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RESPONSE --> self critique assessment: 3
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F~߆XKڻR assignment #004 004. Precalculus I 09-08-2008
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10:32:42 `q001. Note that this assignment has 4 questions If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.
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RESPONSE --> my answers are f(3)=13 f(7)=53 f(-5)=29 my graph is a prabola shape that opens upwards, and apears to be centered on the y axis confidence assessment: 2
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10:38:04 f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.
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RESPONSE --> ok i miss understood how much you wanted. I didn't work the whole problem out to get the x coordinates and the vertex and the points one space to the right and to the left but the should be b/(2 a) = -0/(2*1) = 0 f(0) = 0 ^ 2 + 4 = 0 + 4 = 4 vertex (0,4) 1,5) and (-1,5) are one space to the right and left and there are no x intercepts self critique assessment: 2
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10:48:59 `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.
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RESPONSE --> f(a)=a^2+4 f(x+h)=(x+h)^2+4 x2+4x+8 f(x+h)-f(x)=(x^2+2xh+h^2+4)- (x^2+4) 2xh+h^2 f(x+h) - f(x) ] / h= (x^2+2xh+h^2+4)- (x^2+4)/ (h^2+4) x^2+sxh+h^2+3-(x^2)/(h^2) confidence assessment: 2
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10:53:47 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.
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RESPONSE --> ok i see my mistake in[ f(x+h) - f(x) ] / h it should be = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h. i should have divided the whole problem by h but I only divided the f(x) self critique assessment: 2
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11:03:07 `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.
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RESPONSE --> f(x1)=5(x1)+7 f(x2)=5(x2)+7 [ f(x2) - f(x1) ] / ( x2 - x1 ) =[(5(x2)+7)-(5(x1)+7)]/(x2-x1) =(5x2-5x1)/(x2-x1) =5-5=0 confidence assessment: 2
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11:03:38 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.
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RESPONSE --> ok i see my mistake i should have factored the 5 out self critique assessment: 2
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11:06:23 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
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RESPONSE --> f(x) = 5x + 7 -3=5x + 7 -10=5x -2=x confidence assessment: 2
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11:07:26 If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.
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RESPONSE --> self critique assessment: 3
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11:07:45 end program
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RESPONSE --> self critique assessment: 3
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