course Mth 163 H?????X????G?assignment #004 004. `query 4 Precalculus I 01-31-2008 ?C?????????assignment #004 004. `query 4 Precalculus I 01-31-2008
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20:58:18 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) = -2^3 f(-2) = -8 f(-a) = -a^3 f(x-4) = (x-4)^3 = (x-4)(x-4)(x-4) f(x-4) = (x^2 - 8x + 16)(x-4) f(x-4) = x^3 -4x^2 -8x^2 + 32x + 16x - 64 f(x-4) = x^3 - 12x^2 + 48x - 64 f(x) - 4 = x^3 - 4 confidence assessment: 3
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21:02:19 Where f(x) = 2^x, find f(2), f(-a), f(x+3) and f(x) + 3?
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RESPONSE --> f(2) = 2^2 = 4 f(-a) = 2^-a f(x+3) = 2^(x+3) f(x) + 3 = 2^x +3 confidence assessment: 3
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21:07:50 query functions given by meaningful names. What are some of the advantages of using meaninful names for functions?
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RESPONSE --> Meaningful names for functions are very advantageous because it allows us to understand their purpose. It makes it easier to understand their value or whatever we may be using them for. confidence assessment: 3
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21:18:29 ** 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 --> 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} value(t+3) / value(t) = 1.07^(t+3) / 1.07^t self critique assessment: 2
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21:37:13 What did you get for illumination(distance)/illumination(2*distance)? Show your work on this one.
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RESPONSE --> illumination(1) = 50 / 1^2 = 50 illumination(2) = 50 / 2^2 = 12.5 illumination(3) = 50 / 3^2 = 5.56 illumination(distance)/illumination(2*distance) = 50 / {distance^2 / (2*distance)^2} = 50 / (distance^2 / 4*distance^2) = 50 / (1/4) = 200 confidence assessment: 2
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22:11:05 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 --> For f(2) = 80 I used 2 as my x coordinate and 80 as my y coordinate. I did the same for the rest. confidence assessment: 1
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22:12:06 what is your estimate of value of x for which f(x) = 60?
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RESPONSE --> x = 3.5 confidence assessment: 2
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22:13:10 what is your estimate of the value f(7)?
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RESPONSE --> f(7) = 29 confidence assessment: 2
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22:14:38 what is your estimate of the difference between f(7) and f(9)?
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RESPONSE --> Difference between f(7) and f(9) = 5 confidence assessment: 2
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22:16:44 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 --> An estimate for f(x) = 70 and f(x) = 30 is 5 confidence assessment: 2
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22:23:47 query #4. temperature vs. clock time function y = temperature = T(t), what is the symbolic expression for ...
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RESPONSE --> f(t) = 3 f(t) = 5 f(t) = 3-5 f(t) = (3-5)/2 confidence assessment: 1
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22:30:13 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 --> f(x) = 150 f(x) = 80 - 30 confidence assessment: 1
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22:36:56 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(t) = (f(t) = 34) - f(t) = 47) f(t) = 58-80 = 12 seconds confidence assessment: 1
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22:43:51 By how much did the depth change between t = 23 seconds and t = 34 seconds?
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RESPONSE --> f(t2-t1) / 2 = Depth (34-23) / 2 = 11/2 = 5.5cm confidence assessment: 1
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22:56:58 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 --> f(t = 34 - t = 23) / 2 = 1 (57cm - 67cm) / 2 = 1 -10cm / 2 = 1 -5cm = 1 confidence assessment: 0
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23:05:25 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 --> (f(34) - f(23)) / 2 = (56cm - 67cm) / 2 = -11cm / 2 = -5.5 cm/s confidence assessment: 1
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23:08:14 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 --> Time would be our x coordinate and Depth would be our y coordinate. confidence assessment: 2
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23:09:23 What 3 data point did you use as a basis for your model?
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RESPONSE --> (10, 80) (30, 65) (50, 50) confidence assessment: 3
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??????_???????assignment #004 004. `query 4 Precalculus I 02-02-2008
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11:39:56 What was your function model?
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RESPONSE --> y = -.292(x^2) + 10.9x + 0.8 confidence assessment: 2
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12:08:59 ** 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 --> My average deviation was a very high number that I believe is wrong. I did every step necessary to obtain the average, but obtained the number 117.1. My coordinates were (10, 80), (30, 65), (50, 50), and (70, 35) My model obtaind the data .6, 0, -234.2 and -702. Not sure if I have completed this part correctly. Need a little assistance here. Thanks. self critique assessment: 1
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12:12:30 How close is your model to the curve you sketched earlier?
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RESPONSE --> Fairly close other than the earlier curve had more of a curving pattern while my graph has more of a straight line pattern, which is because the depth and time is changing at a constant rate. confidence assessment: 2
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12:12:49 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Im OK self critique assessment: 3
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