course mth271 I hope I didn't put the same stuff on there twice, if so I am extremely sorry about that. ????????C???b?assignment #001001. `query 1
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16:50:11 Section 0.1.26 solve x/2-x/3>5
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RESPONSE --> (3x/6 - 2x/6)6 > (5)6 = 3x-2x>30 x>30 so -3>x<3 confidence assessment: 3
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16:52:07 It's easiest to avoid denominators where possible. So the preferred first step is to multiply both sides of the original equation by the common denominator 6: 6(x/2) - 6(x/3) = 6 * 5, which gives you3x - 2x = 6 * 5 which gives you x > 6 * 5 which simplifies to x > 30. The interval associated with this solution is 30 < x < infinity, or (30, infinity). To graph you would make an arrow starting at x = 30 and pointing to the right, indicating by an open circle that x = 30 isn't included.**
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RESPONSE --> I should have had 30
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16:54:09 Section 0.1.28 solve 2x^2+1<9x-3
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RESPONSE --> 2x^2-9x+3+1<0 then 2x^2 -9x+4<0 then (2x-1)(x-4)<0 then I got a -2
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16:56:13 The given inequality rearranges to give the quadratic 2x^2 - 9 x + 4 < 0. The left-hand side has zeros at x = .5 and x = 4, as we see by factoring [ we get (2x-1)(x-4) = 0 which is true if 2x-1 = 0 or x - 4 = 0, i.e., x = 1/2 or x = 4. ] The left-hand side is a continuous function of x (in fact a quadratic function with a parabola for a graph), and can change sign only by passing thru 0. So on each interval x < 1/2, 1/2 < x < 4, 4 < x the function must have the same sign. Testing an arbitrary point in each interval tells us that only on the middle interval is the function negative, so only on this interval is the inequality true. Note that we can also reason this out from the fact that large negative or positive x the left-hand side is greater than the right because of the higher power. Both intervals contain large positive and large negative x, so the inequality isn't true on either of these intervals. In any case the correct interval is 1/2 < x < 4. ALTERNATE BUT EQUIVALENT EXPLANATION: The way to solve this is to rearrange the equation to get 2 x^2 - 9 x + 4< 0. The expression 2 x^2 - 9 x + 4 is equal to 0 when x = 1/2 or x = 4. These zeros can be found either by factoring the expression to get ( 2x - 1) ( x - 4), which is zero when x = 1/2 or 4, or by substituting into the quadratic formula. You should be able to factor basic quadratics or use the quadratic formula when factoring fails. The function can only be zero at x = 1/2 or x = 4, so the function can only change from positive to negative or negative to positive at these x values. This fact partitions the x axis into the intervals (-infinity, 1/2), (1/2, 4) and (4, infinity). Over each of these intervals the quadratic expression can't change its sign. If x = 0 the quadratic expression 2 x^2 - 9 x + 4 is equal to 4. Therefore the expression is positive on the interval (-infinity, 1/2). The expression changes sign at x = 1/2 and is therefore negative on the interval (1/2, 4). It changes sign again at 4 so is positive on the interval (4, infinity). The solution to the equation is therefore the interval (1/2, 4), or in inequality form 1/2 < x < 4. **
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RESPONSE --> I should have solved for x equal to 0 and then I would have gotten 1/2 and 4. self critique assessment: 2
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?m????????? assignment #002 002. `Query 2 Applied Calculus I 08-03-2007 ?z??}????Sx????assignment #002 002. `Query 2 Applied Calculus I 08-03-2007
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17:03:32 What were temperature and time for the first, third and fifth data points (express as temp vs clock time ordered pairs)?
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RESPONSE --> um confidence assessment: 1
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17:05:04 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> ok, I don't think I did this part. I did the stuff in the book. confidence assessment: 1
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17:07:18 A good choice of points `spreads' the points out rather than using three adjacent points. For example choosing the t = 10, 20, 30 points would not be a good idea here since the resulting model will fit those points perfectly but by the time we get to t = 60 the fit will probably not be good. Using for example t = 10, 30 and 60 would spread the three points out more and the solution would be more likely to fit the data. The solution to this problem by a former student will be outlined in the remaining `answers'. STUDENT SOLUTION (this student probably used a version different from the one you used; this solution is given here for comparison of the steps) For my quadratic model, I used the three points (10, 75) (20, 60) (60, 30). **
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RESPONSE --> So I should spread out the points i choose to help make the solution more plausible and to fit the data. self critique assessment: 2
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17:08:50 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> -b+or - the sqt of b^2-4ac all over 2*a confidence assessment: 1
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17:09:44 STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
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RESPONSE --> I should have chosen 10,75 and plug it into the a+b+c=0 equation to get 100a+10b+c+75 self critique assessment: 3
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17:10:05 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> an equation of a+b+c=0 confidence assessment: 2
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17:10:35 STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE --> It should have been 400a+20b+c+60 self critique assessment: 3
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17:10:55 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> a+b+c=0 confidence assessment: 2
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17:11:45 STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE --> it should be 3600+60b+c=30 the 60 is sqared to get 3600 self critique assessment: 2
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17:12:48 What multiple of which equation did you first add to what multiple of which other equation to eliminate c, and what is the first equation you got when you eliminated c?
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RESPONSE --> I would have subtracted the first equation from the second equation and would be able to eliminate c leaving a+b=0 confidence assessment: 2
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17:13:52 STUDENT SOLUTION CONTINUED: First, I subtracted the second equation from the third equation in order to eliminate c. By doing this, I obtained my first new equation 3200a + 40b = -30. **
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RESPONSE --> if I had used the 2nd and 3rd equation then I would have gotten the equation 3200a+40b+-30 self critique assessment: 2
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17:15:28 To get the second equation what multiple of which equation did you add to what multiple of which other quation, and what is the resulting equation?
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RESPONSE --> I would have done the same thing with the other equation i did not use and one of the other ones that i did use to get another equation of a+b=0 confidence assessment: 2
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17:15:53 STUDENT SOLUTION CONTINUED: This time, I subtracted the first equation from the third equation in order to again eliminate c. I obtained my second new equation: 3500a + 50b = -45**
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RESPONSE --> The equation i would have gotten was 3500a+50b=-45 self critique assessment: 2
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17:16:25 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> I eliminated b in order to get the answer of a confidence assessment: 2
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17:17:24 STUDENT SOLUTION CONTINUED: In order to solve for a and b, I decided to eliminate b because of its smaller value. In order to do this, I multiplied the first new equation by -5 -5 ( 3200a + 40b = -30) and multiplied the second new equation by 4 4 ( 3500a + 50b = -45) making the values of -200 b and 200 b cancel one another out. The resulting equation is -2000 a = -310. **
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RESPONSE --> To eliminate b i would need a common number in both equations so that it could be eliminated from the equations thus giving the answer to a self critique assessment: 2
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17:20:21 What equation did you get when you substituted this value into one of the 2-variable equations, and what did you get for the other variable?
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RESPONSE --> I plugged in the answer that I got for a into an equation that was a+b=0 and was able to find the answer to b confidence assessment: 2
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17:20:46 STUDENT SOLUTION CONTINUED: After eliminating b, I solved a to equal .015 a = .015 I then substituted this value into the equation 3200 (.015) + 40b = -30 and solved to find that b = -1.95. **
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RESPONSE --> That is what I said I think, just without the numbers self critique assessment: 2
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17:21:18 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> I plugged in a and b into one of the original equations in order to find what c is confidence assessment: 2
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17:21:32 STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE --> c would equal 93 self critique assessment: 3
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17:22:00 What is the resulting quadratic model?
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RESPONSE --> plug in to the quadratic formula? confidence assessment: 2
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17:23:11 STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE --> oh yeah, make an equation with the a,b , and c that i would find and put it into the equation y=ax^2+bx+c and you would get y=(.015)x^2-(1.95)x+93 self critique assessment: 2
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17:23:44 What did your quadratic model give you for the first, second and third clock times on your table, and what were your deviations for these clock times?
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RESPONSE --> plug in the x value in order to obtain the y and get a point confidence assessment: 2
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17:24:28 STUDENT SOLUTION CONTINUED: This model y = (.015) x^2 - (1.95)x + 93 evaluated for clock times 0, 10 and 20 gave me these numbers: First prediction: 93 Deviation: 2 Then, since I used the next two ordered pairs to make the model, I got back }the exact numbers with no deviation. So. the next two were Fourth prediction: 48 Deviation: 1 Fifth prediction: 39 Deviation: 2. **
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RESPONSE --> plug in all 3 ordered pairs, one at a time, and solve self critique assessment: 2
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17:24:44 What was your average deviation?
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RESPONSE --> 2.5 confidence assessment: 1
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17:25:00 STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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RESPONSE --> the deviation should have been .6 self critique assessment: 2
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17:25:12 Is there a pattern to your deviations?
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RESPONSE --> i believe so confidence assessment: 1
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17:25:34 STUDENT SOLUTION CONTINUED: There was no obvious pattern to my deviations. INSTRUCTOR NOTE: Common patterns include deviations that start positive, go negative in the middle then end up positive again at the end, and deviations that do the opposite, going from negative to positive to negative. **
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RESPONSE --> There was not any pattern self critique assessment: 2
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17:26:50 Have you studied the steps in the modeling process as presented in Overview, the Flow Model, Summaries of the Modeling Process, and do you completely understand the process?
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RESPONSE --> I believe I understand the process, however, I forgot to do the flow model, thus not being able to give number answers but I understand the concept of how to get a,b ,and c and then plugging those into an equation and then putting the 3 ordered pairs into the equation and such. confidence assessment: 3
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17:26:58 STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE --> yes i do self critique assessment: 3
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17:30:12 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> I probably will remember them forever. First you get three equations of ax^2+b+c=0 and then subtract 2 of them to obtain an equation of a+b=0 to eliminate the c and then do the same thing for one of the same equations used and the other that was not used. Then find a common number and either eliminate a or b to find a or b and then plug that number back into one of the a+b=0 equations to find b or a and then plug both of those back into one of the original equations to find c. Then you make an equation of y= a^2x+ bx+c by using those numbers and then plug in the ordered pairs that were chosen from the chart. confidence assessment: 3
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17:30:39 STUDENT SOLUTION CONTINUED: Yes, sir, I have memorized the steps of the modeling process at this point. I also printed out an outline of the steps in order to refresh my memory often, so that I will remember them forever!!! INSTRUCTOR COMMENT: OK, I'm convinced. **
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RESPONSE --> I will even print out an outline self critique assessment: 3
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17:32:09 Query Completion of Model first problem: Completion of model from your data.Give your data in the form of depth vs. clock time ordered pairs.
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RESPONSE --> The ordered pairs are:(14,0), (10,10), (7,20) confidence assessment: 1
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17:33:11 STUDENT SOLUTION: Here are my data which are from the simulated data provided on the website under randomized problems. (5.3, 63.7) (10.6. 54.8) (15.9, 46) (21.2, 37.7) (26.5, 32) (31.8, 26.6). **
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RESPONSE --> They should have been (5.3, 63.7) (10.6, 54.8) (15.9, 46) (21.2,37.7) (26.5, 32) (31.8,26.6) self critique assessment: 2
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17:34:07 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (37.7, 46), (15.9, 17) confidence assessment: 1
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17:34:51 STUDENT SOLUTION CONTINUED: As the basis for my graph, I used ( 5.3, 63.7) (15.9, 46) (26.5, 32)**
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RESPONSE --> (15.9, 46) (5.3,63.7) and (26.5, 32) in order to spread it out self critique assessment: 2
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17:36:46 Give the first of your three equations.
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RESPONSE --> 28.09a+5.3b+c= 252.81a+15.9b+c=46 691.69a+2.3b+c= confidence assessment: 1
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17:37:12 STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **
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RESPONSE --> I forgot to put 63.7 in self critique assessment: 2
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17:37:44 Give the second of your three equations.
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RESPONSE --> 252.81a+15.9b+c=46 confidence assessment: 2
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17:37:58 STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE --> i believe that is what i got self critique assessment: 3
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17:38:17 Give the third of your three equations.
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RESPONSE --> i forgot the numbers to that one confidence assessment: 1
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17:38:58 STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE --> it should have been 702.25a+26.5b+c=32 the 702.25 is 26.5 squared. self critique assessment: 3
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17:39:34 Give the first of the equations you got when you eliminated c.
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RESPONSE --> it should be an equation of a+b=# confidence assessment: 2
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17:39:55 STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE --> it should be 449.44a+ 10.6b=-14 self critique assessment: 3
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17:40:05 Give the second of the equations you got when you eliminated c.
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RESPONSE --> a+b=# confidence assessment: 2
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17:40:29 ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE --> it should be 674.16a+21.2b=-31.7 self critique assessment: 2
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17:41:33 Explain how you solved for one of the variables.
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RESPONSE --> you subtract from both equations either a or b and if necessary multiply one equation by a number that will get the same number as the other equation in order to eliminate it. then you can solve for a or b confidence assessment: 3
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17:41:50 STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated b by multiplying the first equation by 21.2, which was the b value in the second equation. Then, I multiplied the seond equation by -10.6, which was the b value of the first equation, only I made it negative so they would cancel out. **
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RESPONSE --> i believe i said that, only without the numbers self critique assessment: 3
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17:42:04 What values did you get for a and b?
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RESPONSE --> a=# and b=# confidence assessment: 2
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17:42:22 STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE --> a=.0165 and b=-2 self critique assessment: 3
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17:42:42 What did you then get for c?
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RESPONSE --> i plugged in a and b into one of the original equations and solved for c confidence assessment: 3
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17:42:51 STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE --> c= 73.4 self critique assessment: 3
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17:43:16 What is your function model?
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RESPONSE --> y=#a^2x+bx+c confidence assessment: 2
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17:43:41 STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE --> y=(.0165)x^2+(-2)x+73.4 self critique assessment: 3
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17:43:56 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE --> i am not sure confidence assessment: 1
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17:44:22 STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE --> the clock time was 46sec and prediction was 16.314cm self critique assessment: 2
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17:44:33 What clock time corresponds to the given depth (give depth also)?
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RESPONSE --> not sure confidence assessment: 1
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17:46:03 The specifics will depend on your model and the requested depth. For your model y = (.0165)x^2 + (-2)x + 73.4, if we wanted to find the clock time associated with depth 68 we would note that depth is y, so we would let y be 68 and solve the resulting equation: 68 = .01t^2 - 1.6t + 126 using the quadratic formula. There are two solutions, x = 55.5 and x = 104.5, approximately. **
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RESPONSE --> oh, if you wanted to find the clock time for 68 we would plug that in as y and get 68=.01t^2-1.6t+126 and then solve by using the quadratic formula -b+ or - the sqrt of b^2-4ac all over 2a self critique assessment: 3
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17:50:09 Completion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.
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RESPONSE --> i dont know confidence assessment: 1
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17:50:28 STUDENT SOLUTION: Grade vs. percent of assignments reviewed (0, 1) (10, 1.790569) (20, 2.118034) (30, 2.369306) (40, 2.581139) (50, 2.767767) (60, 2.936492) (70, 3.09165) (80, 3.236068) (90, 3.371708) (100, 3.5). **
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RESPONSE --> it should be all those numbered pairs self critique assessment: 2
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17:50:48 What three points on your graph did you use as a basis for your model?
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RESPONSE --> three points on the graph confidence assessment: 0
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17:51:21 STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE --> (20, 2.110834) (50, 2.767767) and (100,3.5) self critique assessment: 2
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17:51:33 Give the first of your three equations.
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RESPONSE --> a+b+c=# confidence assessment: 1
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17:51:51 STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE --> 400a+20b+c=2.110834 self critique assessment: 2
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17:52:02 Give the second of your three equations.
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RESPONSE --> a+b+c=# confidence assessment: 2
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17:52:20 STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE --> 2500a+50b+c=2.767767 self critique assessment: 2
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17:52:41 Give the third of your three equations.
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RESPONSE --> 10000a+100b+c=3.5 confidence assessment: 3
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17:52:49 STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE --> that is what i got self critique assessment: 3
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17:52:58 Give the first of the equations you got when you eliminated c.
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RESPONSE --> a+b=# confidence assessment: 2
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17:53:15 STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE --> 7500a+50b=.732233 self critique assessment: 2
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17:53:27 Give the second of the equations you got when you eliminated c.
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RESPONSE --> a+b=# confidence assessment: 2
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17:53:44 STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE --> 9600a+80b=1.381966 self critique assessment: 2
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17:55:19 Explain how you solved for one of the variables.
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RESPONSE --> I would subtract a or b and if necessary get a common number with both equations and then subtract and that will eliminate either a or b and will get you a or b confidence assessment: 3
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17:55:36 STUDENT SOLUTION CONTINUED: In order to solve for a, I eliminated the variable b. In order to do this, I multiplied the first new equation by 80 and the second new equation by -50. **
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RESPONSE --> that is what i said only without the numbers self critique assessment: 3
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17:55:59 What values did you get for a and b?
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RESPONSE --> a=3 and b=# confidence assessment:
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17:56:29 STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE --> a=-.0000876638 and b = .01727 self critique assessment: 3
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17:56:51 What did you then get for c?
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RESPONSE --> i plugged in a and b into one of the original equations and solved for c confidence assessment: 3
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17:57:01 STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE --> c=1.773 self critique assessment: 3
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17:57:18 What is your function model?
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RESPONSE --> y=ax^2+bx+c confidence assessment: 3
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17:57:44 STUDENT ANSWER: y = (0) x^2 + (.01727)x + 1.773 **
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RESPONSE --> y=.01727x+1.773 self critique assessment: 3
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17:57:53 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> no idea confidence assessment: 1
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17:58:57 The precise solution depends on the model desired average. For example if the model is y = -.00028 x^2 + .06 x + .5 (your model will probably be different from this) and the grade average desired is 3.3 we would find the percent of review x corresponding to grade average y = 3.3 then we have 3.3 = -.00028 x^2 + .06 x + .5. This equation is easily solved using the quadratic formula, remembering to put the equation into the required form a x^2 + b x + c = 0. We get two solutions, x = 69 and x = 146. Our solutions are therefore 69% grade review, which is realistically within the 0 - 100% range, and 146%, which we might reject as being outside the range of possibility. To get a range you would solve two equations, on each for the percent of review for the lower and higher ends of the range. In many models the attempt to solve for a 4.0 average results in an expression which includes the square root of a negative number; this indicates that there is no real solution and that a 4.0 is not possible. **
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RESPONSE --> it depends on the average. but you would put in percent for a,b,c and the grade average as y self critique assessment: 3
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17:59:06 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> no idea confidence assessment: 1
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17:59:36 Here you plug in your percent of review for the variable. For example if your model is y = -.00028 x^2 + .06 x + .5 and the percent of review is 75, you plug in 75 for x and evaluate the result. The result gives you the grade average corresponding to the percent of review according to the model. **
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RESPONSE --> you plug in the percent of review for x and solve for y self critique assessment: 3
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17:59:45 How well does your model fit the data (support your answer)?
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RESPONSE --> it does confidence assessment: 1
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18:00:12 You should have evaluated your function at each given percent of review-i.e., at 0, 10, 20, 30, . 100 to get the predicted grade average for each. Comparing your results with the given grade averages shows whether your model fits the data. **
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RESPONSE --> it depends on the model self critique assessment: 2
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18:01:21 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> (#,#) (# #) (# #) (# #) confidence assessment: 1
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18:05:27 STUDENT SOLUTION: (1, 935.1395) (2, 264..4411) (3, 105.1209) (4, 61.01488) (5, 43.06238) (6, 25.91537) (7, 19.92772) (8, 16.27232) (9, 11.28082) (10, 9.484465)**
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RESPONSE --> they should be 1-10 and then numbers from 9 to 935 self critique assessment: 2
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18:05:58 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (2, #) (4,#) (7,#) confidence assessment: 1
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18:06:39 STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE --> (2, 264.4411) (4, 61.01488) and (8, 1.27232) self critique assessment: 3
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18:07:02 Give the first of your three equations.
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RESPONSE --> a+b+c=# confidence assessment: 2
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18:07:45 STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE --> 4a+2b+c=264.4411 self critique assessment: 3
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18:08:03 Give the second of your three equations.
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RESPONSE --> a+b+c=# confidence assessment: 2
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18:08:38 STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE --> 16a+4b+c=61.01488 self critique assessment: 3
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18:08:50 Give the third of your three equations.
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RESPONSE --> a+b+c=# confidence assessment: 3
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18:09:26 STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE --> 64a+8b+c=16.27232 self critique assessment: 3
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18:10:37 Give the first of the equations you got when you eliminated c.
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RESPONSE --> 12a+2b= -203.42622 confidence assessment: 3
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18:11:04 STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE --> they probably chose different equations than i did self critique assessment: 3
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18:12:31 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 60a+6b= -248.16878 confidence assessment: 3
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18:12:47 STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE --> thats what i got self critique assessment: 3
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18:13:27 Explain how you solved for one of the variables.
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RESPONSE --> multiply the first equation by3 to get 6b so that it can be subtracted from the equations to solve for a. confidence assessment: 3
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18:15:02 STUDENT SOLUTION CONTINUED: I solved for a by eliminating the variable b. I multiplied the first new equation by 4 and the second new equation by -6 **
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RESPONSE --> i did it slightly differently self critique assessment: 3
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18:16:21 What values did you get for a and b?
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RESPONSE --> a= 15.08791167 and b=-192.24058 confidence assessment: 2
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18:16:31 STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE --> thats what i got self critique assessment: 3
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18:17:35 What did you then get for c?
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RESPONSE --> c= -59.688 confidence assessment: 2
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18:18:04 STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE --> i must have done the math wrong but it should be 588.5691 self critique assessment: 3
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18:19:18 What is your function model?
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RESPONSE --> 15.088x^2-192.24x+588.5691=y confidence assessment: 2
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18:19:27 STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE --> thats what i got self critique assessment: 3
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18:19:34 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> no idea confidence assessment: 1
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18:20:37 STUDENT SOLUTION CONTINUED: The given distance was 1.6 Earth distances from the sun. My illumination prediction was 319.61 w/m^2, obtained by evaluating my function model for x = 1.6. **
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RESPONSE --> it should be 1.6=x and the illumination prediction should be 319.61w/m^2 self critique assessment: 2
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18:20:44 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE --> not sure confidence assessment: 1
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18:21:32 The precise solution depends on the model and the range of averages. For example if the model is y =9.4 r^2 - 139 r + 500 and the illumination range is 25 to 100 we would find the distance r corresponding to illumination y = 25, then the distance r corresponding to illumination y = 100, by solving the equations 25=9.4 r^2 - 139 r + 500 and 100 =9.4 r^2 - 139 r + 500 Both of these equations are easily solved using the quadratic formula, remembering to put both into the required form a r^2 + b r + c = 0. Both give two solutions, only one solution of each having and correspondence at all with the data. The solutions which correspond to the data are r = 3.9 when y = 100 and r = 5.4 when y = 25. So when the distance x has range 50% - 69% if the illumination range is 25 to 100. Note that a quadratic model does not fit this data well. Sometimes data is quadratic in nature, sometimes it is not. We will see as the course goes on how some situations are accurately modeled by quadratic functions, while others are more accurately modeled by exponential or power functions. **
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RESPONSE --> it depends on the model and the averages. but you would plug in the numbers in the correct places and use the quadratic formula to get the answer. self critique assessment: 2
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18:23:39 ppCal1 Section 0.2 EXTRA QUESTION. What is the midpoint between two points What are your points and what is the midpoint? How did you find the midpoint?{}{}What is the midpoint between the points (3, 8) and (7, 12)?
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RESPONSE --> the midpoint is the exact middle of a line. I dont know what the midpoint is. the midpoint between (3,8) and (7,12) is 5,10 and i got that by using the formula x1+x2/2 and y1+y2/2 confidence assessment: 3
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18:24:31 You are given two points. The points each have two coordinates. You have to average the x coordinates to get the x coordinate of the midpoint, then average the y coordinates to get the y coordinate of the midpoint. For example if the points are (3, 8) and (7, 12), the average of the x coordinates is (3 + 7) / 2 = 5 and the average of the y coordinates is (7 + 12) / 2 = 9.5 so the coordinates of the midpoint are (5, 9.5). **
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RESPONSE --> um, 7 was used twice instead of 8 thus giving the answer (5,10) self critique assessment: 3
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18:25:42 0.2.24 (was 0.2.14 solve abs(3x+1) >=4
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RESPONSE --> I got x=1 and the arrows should be pointing out towards the right from 1 with a closed circle confidence assessment: 2
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18:26:15 abs(a) >= b translates to a >= b OR a <= -b. In this case abs(3x+1) > 4 gives you 3x + 1 >= 4 OR 3x + 1 <= -4, which on solution for x gives x >= 1 OR x < = -5/3. **
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RESPONSE --> i got one but i did not get -5/3 self critique assessment: 2
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18:27:28 the given inequality is equivalent to the two inequalities 3x+1 >= 4 and 3x+1 =< -4. The solution to the first is x >= 1. The solution to the second is x <= -5/3. Thus the solution is x >= 1 OR x <= -5/3. COMMON ERROR: -5/3 > x > 1 INSTRUCTOR COMMENT: It isn't possible for -5/3 to be greater than a quantity and to have that same quantity > 1. Had the inequality read |3x+1|<4 you could have translated it to -4 < 3x+1 <4, but you can't reverse these inequalities without getting the contradiction pointed out here. **
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RESPONSE --> the solution should be x>=1 and x<= -5/3 self critique assessment: 3
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18:28:13 0.2.24 (was 0.2.16 solve abs(2x+1)<5. What inequality or inequalities did you get from the given inequality, and are these 'and' or 'or' inequalities? Give your solution.
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RESPONSE --> x<2 and confidence assessment: 1
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18:29:10 abs(a) < b means a < b AND -b < -a so from the given inequality abs(2x+1) < 5 you get -5 < 2x+1 AND 2x+1 < 5. These can be combined into the form -5 < 2x+1 < 5 and solved to get your subsequent result. Subtracting 1 from all expressions gives us -6 < 2x < 4, then dividing through by 2 we get -3 < x < 2. **
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RESPONSE --> i just needed to subtract 1 to get -6 and 4 leaving 2x and then divide both numbers to get -3 and 2(which i got) and i got the and self critique assessment: 2
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18:29:59 0.2.7 (was 0.2.23 describe [-2,2 ]
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RESPONSE --> it does not go past -2 and 2 so there is a line in between the two numbers with closed circles on both #s confidence assessment: 2
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18:30:39 The interval [-2, 2] is centered at the midpoint between x=-2 and x=2. You can calculate this midpoint as (-2 + 2) / 2 = 0. It is also clear from a graph of the interval that it is centered at x = 0 The center is at 0. The distance to each endpoint is 2. The interval is | x - center | < distance to endpoints. So the interval here is | x - 0 | < 2, or just | x | < 2. **
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RESPONSE --> The midpoint would be 0 and the distance to each endpoint is 2. self critique assessment: 2
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18:32:25 0.2.22 (was 0.2.28 describe [-7,-1]
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RESPONSE --> it goes from -7 to -1 and the midpoint is -4 confidence assessment: 2
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18:33:04 the interval is centered at -4 (midpt between -7 and -1). The distance from the center of the interval to -7 is 3, and the distance from the center of the interval to -1 is 3. This translates to the inequality | x - (-4) | < 3, which simplifies to give us | x + 4 | < 3. **
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RESPONSE --> the inequality can be the abs of x+4<3 self critique assessment: 2
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18:33:32 0.2.14 (was 0.2.30) describe (-infinity, 20) U (24, infinity)
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RESPONSE --> the arrow goes out to infinity on the right and left side. confidence assessment: 2
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18:34:22 22 is at the center of the interval. The endpoints are 2 units from the midpoint, and are not included. Everything that lies more than 2 units from 22 is in one of the intervals, and everything in either of the intervals lies at least 2 units from 22. So the inequality that describes this union of two intervals is | x - 22 | > 2. **
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RESPONSE --> 22 is the center. the inequality is abx-22>2 self critique assessment: 2
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18:35:08 0.2.44 (was 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1
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RESPONSE --> not sure on that one confidence assessment: 2
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18:36:05 The inequality is translated as -1<=(w-57.5)/7.5<=1. Multiplying through by 7.5 we get -7.5<=w-57.5<=7.5 Now add 57.5 to all expressions to get -7.5 + 57.5 <= x <= 7.5 + 57.5 or 50 < x < 65, which tells you that the dogs weigh between 50 and 65 pounds. **
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RESPONSE --> multiply both sides by 7.5 to get -7.5<=w-57.5<=7.5 then add 57.5 to both sides and get 50
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18:36:37 0.2.42 (was 0.2.38 stocks vary from 33 1/8 by no more than 2. What absolute value inequality or inequalities correspond(s) to this prediction?
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RESPONSE --> not sure confidence assessment: 0
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18:37:55 this statement says that the 'distance' between a stock price and 33 1/8 must not be more than 2, so this distance is <= 2 The distance between a price p and 33 1/8 is | p - 33 1/8 |. The desired inequality is therefore | p = 33 1/8 | < = 2. **
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RESPONSE --> it has to be <=2 because it cant be more than 2 and so the distance is absp-33 1/8 and gives us abs p=33 1/8<=2 self critique assessment: 2
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????????????assignment #002 002. `Query 2 Applied Calculus I 08-03-2007 18:27:28 the given inequality is equivalent to the two inequalities 3x+1 >= 4 and 3x+1 =< -4. The solution to the first is x >= 1. The solution to the second is x <= -5/3. Thus the solution is x >= 1 OR x <= -5/3. COMMON ERROR: -5/3 > x > 1 INSTRUCTOR COMMENT: It isn't possible for -5/3 to be greater than a quantity and to have that same quantity > 1. Had the inequality read |3x+1|<4 you could have translated it to -4 < 3x+1 <4, but you can't reverse these inequalities without getting the contradiction pointed out here. **
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RESPONSE --> the solution should be x>=1 and x<= -5/3 self critique assessment: 3
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18:28:13 0.2.24 (was 0.2.16 solve abs(2x+1)<5. What inequality or inequalities did you get from the given inequality, and are these 'and' or 'or' inequalities? Give your solution.
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RESPONSE --> x<2 and confidence assessment: 1
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18:29:10 abs(a) < b means a < b AND -b < -a so from the given inequality abs(2x+1) < 5 you get -5 < 2x+1 AND 2x+1 < 5. These can be combined into the form -5 < 2x+1 < 5 and solved to get your subsequent result. Subtracting 1 from all expressions gives us -6 < 2x < 4, then dividing through by 2 we get -3 < x < 2. **
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RESPONSE --> i just needed to subtract 1 to get -6 and 4 leaving 2x and then divide both numbers to get -3 and 2(which i got) and i got the and self critique assessment: 2
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18:29:59 0.2.7 (was 0.2.23 describe [-2,2 ]
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RESPONSE --> it does not go past -2 and 2 so there is a line in between the two numbers with closed circles on both #s confidence assessment: 2
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18:30:39 The interval [-2, 2] is centered at the midpoint between x=-2 and x=2. You can calculate this midpoint as (-2 + 2) / 2 = 0. It is also clear from a graph of the interval that it is centered at x = 0 The center is at 0. The distance to each endpoint is 2. The interval is | x - center | < distance to endpoints. So the interval here is | x - 0 | < 2, or just | x | < 2. **
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RESPONSE --> The midpoint would be 0 and the distance to each endpoint is 2. self critique assessment: 2
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18:32:25 0.2.22 (was 0.2.28 describe [-7,-1]
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RESPONSE --> it goes from -7 to -1 and the midpoint is -4 confidence assessment: 2
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18:33:04 the interval is centered at -4 (midpt between -7 and -1). The distance from the center of the interval to -7 is 3, and the distance from the center of the interval to -1 is 3. This translates to the inequality | x - (-4) | < 3, which simplifies to give us | x + 4 | < 3. **
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RESPONSE --> the inequality can be the abs of x+4<3 self critique assessment: 2
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18:33:32 0.2.14 (was 0.2.30) describe (-infinity, 20) U (24, infinity)
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RESPONSE --> the arrow goes out to infinity on the right and left side. confidence assessment: 2
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18:34:22 22 is at the center of the interval. The endpoints are 2 units from the midpoint, and are not included. Everything that lies more than 2 units from 22 is in one of the intervals, and everything in either of the intervals lies at least 2 units from 22. So the inequality that describes this union of two intervals is | x - 22 | > 2. **
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RESPONSE --> 22 is the center. the inequality is abx-22>2 self critique assessment: 2
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18:35:08 0.2.44 (was 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1
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RESPONSE --> not sure on that one confidence assessment: 2
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18:36:05 The inequality is translated as -1<=(w-57.5)/7.5<=1. Multiplying through by 7.5 we get -7.5<=w-57.5<=7.5 Now add 57.5 to all expressions to get -7.5 + 57.5 <= x <= 7.5 + 57.5 or 50 < x < 65, which tells you that the dogs weigh between 50 and 65 pounds. **
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RESPONSE --> multiply both sides by 7.5 to get -7.5<=w-57.5<=7.5 then add 57.5 to both sides and get 50
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18:36:37 0.2.42 (was 0.2.38 stocks vary from 33 1/8 by no more than 2. What absolute value inequality or inequalities correspond(s) to this prediction?
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RESPONSE --> not sure confidence assessment: 0
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18:37:55 this statement says that the 'distance' between a stock price and 33 1/8 must not be more than 2, so this distance is <= 2 The distance between a price p and 33 1/8 is | p - 33 1/8 |. The desired inequality is therefore | p = 33 1/8 | < = 2. **
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RESPONSE --> it has to be <=2 because it cant be more than 2 and so the distance is absp-33 1/8 and gives us abs p=33 1/8<=2 self critique assessment: 2
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????????????assignment #002 002. `Query 2 Applied Calculus I 08-03-2007 l??v???^~?b???assignment #003 003. `query 3 Applied Calculus I 08-03-2007
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18:43:39 0.3.22 (was 0.3.24 simplify z^-3 (3z^4)
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RESPONSE --> 3z^-3+4= 3z confidence assessment: 3
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18:43:46 z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. **
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RESPONSE --> that is what i got self critique assessment: 3
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18:44:56 0.3.28 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3
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RESPONSE --> 1728s^6/729s^3= 64s^6/27s^3= 64s^6-s^3/27= 64s^3/27 confidence assessment: 3
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18:45:12 Starting with (12 s^2 / (9s) ) ^ 3 we simplify inside parentheses to get ( 4 s / 3) ^ 3, which is equal to 4^3 * s^3 / 3^3 = 64 s^3 / 27 It is possible to expand the cube without first simplifying inside, but the subsequent simplification is a little more messy and error-prone; however done correctly it gives the same result. It's best to simplify inside the parentheses first. **
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RESPONSE --> that is what i got self critique assessment: 3
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18:46:30 0.3.34 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3)
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RESPONSE --> 3x^2y^3 and 3x^2* cubed 2x confidence assessment: 2
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18:47:36 To simplify (54 x^7)^(1/3) you have to find the maximum factor inside the parentheses which a perfect 3d power. First factor 54 into 18:47:36 `questionNumber 30000 To simplify (54 x^7)^(1/3) you have to find the maximum factor inside the parentheses which a perfect 3d power. First factor 54 into its prime factors: 54 = 2 * 27 = 2 * 3 * 3 * 3 = 2 * 3^3. Now we have (2 * 3^3 * x^7)^(1/3). 3^3 and x^6 are both perfect 3d Powers. So we factor 3^3 * x^6 out of the expression in parentheses to get ( (3^3 * x^6) * 2x ) ^(1/3). This is equal to (3^3 * x^6)^(1/3) * (2x)^(1/3). Simplifying the perfect cube we end up with 3 x^2 ( 2x ) ^ (1/3) For the second expression: The largest cube contained in 54 is 3^3 = 27 and the largest cube contained in x^7 is x^6. Thus you factor out what's left, which is 2x. Factoring 2x out of (54 x^7)^(1/3) gives you 2x ( 27 x^6) so your expression becomes [ 2x ( 27 x^6) ] ^(1/3) = (2x)^(1/3) * [ 27 x^6 ] ^(1/3) = (2x)^(1/3) * [ (27)^(1/3) (x^6)^(1/3) ] = (2x)^(1/3) * 3 x^2, which in more traditional order is 3 x^2 ( 2x)^(1/3). **
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RESPONSE --> the second one i didnt know what to call that but i see that i should have just done 2x^(1/3). and i am not sure of the other one. self critique assessment: 2
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18:48:17 `questionNumber 30000 0.3.62 (was 0.3.54 factor P(1+r) from expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ...
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RESPONSE --> i got A= P(1+r)(1+r)^n confidence assessment: 2
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18:50:47 `questionNumber 30000 Few students get this one. If you didn't you've got a lot of company; if you did congratulations. It's important to understand how this problem illustrates the essence of factoring. It's important also because expressions of this form occur throughout calculus. Factor out P * (1 + r). Divide each term by P ( 1 + r), and your result is P (1 + r) * your quotient. Your quotient would be 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... . The factored form would therefore be P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... ]. You can verify that this is identical to the original expression if you multiply it back out. Analogy with different exponents and only three terms: A x^3 + A x^4 + A x^5 can be divided by A x^2 to give quotient x + x^2 + x^3, so the factored expression is A ( x + x^2 + x^3). **
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RESPONSE --> P(1+r) is the quotient. which would be 1+(1+r)+(1+r)^2 and so on. it can be factored as P(1+r)[1+ (1+r)... and the factored expression of A is (x+x^2+x^3) self critique assessment: 2
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??|??C????????? assignment #003 003. `query 3 Applied Calculus I 08-03-2007 X?{?????L???????assignment #004 004. `query 4 Applied Calculus I 08-03-2007
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19:01:00 `questionNumber 40000 Query class notes #04 explain how we can prove that the rate-of-depth-change function for depth function y = a t^2 + b t + c is y' = 2 a t + b
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RESPONSE --> by using the power rule - by bringing down the exponent and gives us y'= 2at+b confidence assessment: 3
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19:01:31 `questionNumber 40000 You have to find the average rate of change between clock times t and t + `dt: ave rate of change = [ a (t+`dt)^2 + b (t+`dt) + c - ( a t^2 + b t + c ) ] / `dt = [ a t^2 + 2 a t `dt + a `dt^2 + b t + b `dt + c - ( a t^2 + b t + c ) ] / `dt = [ 2 a t `dt + a `dt^2 + b `dt ] / `dt = 2 a t + b t + a `dt. Now if `dt shrinks to a very small value the ave rate of change approaches y ' = 2 a t + b. **
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RESPONSE --> or we could use the average rate of change. self critique assessment: 2
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19:03:09 `questionNumber 40000 explain how we know that the depth function for rate-of-depth-change function y' = m t + b must be y = 1/2 m t^2 + b t + c, for some constant c, and explain the significance of the constant c.
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RESPONSE --> the constant c is 0 when it is a constant according to the constant rule. we get y=1/2mt^2 because in order to get mt we bring down the exponent and multiply 1/2 by 2 and get 1 thus mt and bt is one exponent and when it is brought down we get b confidence assessment: 3
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19:03:29 `questionNumber 40000 Student Solution: If y = a t^2 + b t + c we have y ' (t) = 2 a t + b, which is equivalent to the given function y ' (t)=mt+b . Since 2at+b=mt+b for all possible values of t the parameter b is the same in both equations, which means that the coefficients 2a and m must be equal also. So if 2a=m then a=m/2. The depth function must therefore be y(t)=(1/2)mt^2+bt+c. c is not specified by this analysis, so at this point c is regarded as an arbitrary constant. c depends only on when we start our clock and the position from which the depth is being measured. **
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RESPONSE --> i think that is what i said. self critique assessment: 2
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19:04:31 `questionNumber 40000 Explain why, given only the rate-of-depth-change function y' and a time interval, we can determine only the change in depth and not the actual depth at any time, whereas if we know the depth function y we can determine the rate-of-depth-change function y' for any time.
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RESPONSE --> because we are bringing down the exponents confidence assessment: 1
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19:05:51 `questionNumber 40000 Given the rate function y' we can find an approximate average rate over a given time interval by averaging initial and final rates. Unless the rate function is linear this estimate will not be equal to the average rate (there are rare exceptions for some functions over specific intervals, but even for these functions the statement holds over almost all intervals). Multiplying the average rate by the time interval we obtain the change in depth, but unless we know the original depth we have nothing to which to add the change in depth. So if all we know is the rate function, have no way to find the actual depth at any clock time. **
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RESPONSE --> we find the average rate by using the average initial and final rates. and unless its a linear function then the estimate won't be equal to the average rate. self critique assessment: 2
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19:06:32 `questionNumber 40000 In terms of the depth model explain the processes of differentiation and integration.
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RESPONSE --> we bring down the exponents and multiply them with any integer in front of the x or y or whatever letter. confidence assessment: 1
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19:07:36 `questionNumber 40000 Rate of depth change can be found from depth data. This is equivalent to differentiation. Given rate-of-change information it is possible to find depth changes but not actual depth. This is equivalent to integration. To find actual depths from rate of depth change would require knowledge of at least one actual depth at a known clock time. **
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RESPONSE --> rate of depth change can be found, which is differentiation and rate of change is integration but in order to find the actual depths and such we need one actual depth and a clock time self critique assessment: 2
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19:08:43 `questionNumber 40000 query modeling project #2 #5. $200 init investment at 10%. What are the growth rate and growth factor for this function? How long does it take the principle to double? At what time does the principle first reach $300?
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RESPONSE --> 2000 confidence assessment: 1
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19:10:12 `questionNumber 40000 At what time t is the principle equal to half its t = 20 value? What doubling time is associated with this result?
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RESPONSE --> I got the growth rate of .10 and the growth factor is 1.10 because we add 1 to .10. then we multiply 200*1.10^t and we just plug in numbers for t until we get to 300. confidence assessment: 2
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19:10:40 `questionNumber 40000 The t = 20 value is $200 * 1.1^20 = $1340, approx. Half the t = 20 value is therefore $1340/2 = $670 approx.. By trial and error or, if you know them, other means we find that the value $670 is reached at t = 12.7, approx.. For example one student found that half of the 20 value is 1345.5/2=672.75, which occurs. between t = 12 and t = 13 (at t = 12 you get $627.69 and at t = 13 you get 690.45). At 12.75 we get 674.20 so it would probably be about 12.72 This implies that the principle would double in the interval between t = 12.7 yr and t = 20 yr, which is a period of 20 yr - 12.7 yr = 7.3 yr. This is consistent with the doubling time we originally obtained, and reinforces the idea that for an exponential function doubling time is the same no matter when we start or end. **
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RESPONSE --> ok self critique assessment: 2
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19:11:07 `questionNumber 40000 query #8. Sketch principle vs. time for the first four years with rates 10%, 20%, 30%, 40%
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RESPONSE --> not sure confidence assessment: 1
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19:12:52 `questionNumber 40000 We find that for the first interest rate, 10%, we have amount = 1(1.10)^t. For the first 4 years the values at t=1,2,3,4 years are 1.10, 1.21, 1.33 and 1.46. By trial and error we find that it will take 7.27 years for the amount to double. for the second interest rate, 20%, we have amount = 1(1.20)^t. For the first 4 years the values at t=1,2,3,4 years are 1.20, 1.44, 1.73 and 2.07. By trial and error we find that it will take 3.80 years for the amount to double. Similar calculations tell us that for interest rate 30% we have $286 after 4 years and require 2.64 years to double, and for interest rate 40% we have $384 after 4 years and require 2.06 years to double. The final 4-year amount increases by more and more with each 10% increase in interest rate. The doubling time decreases, but by less and less with each 10% increase in interest rate. **
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RESPONSE --> ok, u just plug in 1,2,3,4, for t and get 1.10, 1.21. 1.33. 1.46 and then for 20% you had a .10 to the 1.20 and then do the same thing as before,etc... self critique assessment: 2
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19:13:16 `questionNumber 40000 query #11. equation for doubling time
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RESPONSE --> y=2(1.10)^t confidence assessment: 1
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19:14:08 `questionNumber 40000 the basic equation says that the amount at clock time t, which is P0 * (1+r)^t, is double the original amount P0. The resulting equation is therefore P0 * (1+r)^t = 2 P0. Note that this simplifies to (1 + r)^ t = 2, and that this result depends only on the interest rate, not on the initial amount P0. **
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RESPONSE --> it should be (1+r)^t=2 and it depends on the interest rate and not the P0 self critique assessment: 2
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19:14:33 `questionNumber 40000 Write the equation you would solve to determine the doubling time 'doublingTime, starting at t = 2, for a $5000 investment at 8%.
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RESPONSE --> 2=(5000+1.08)^t confidence assessment: 1
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19:16:19 `questionNumber 40000 dividing the equation $5000 * 1.08 ^ ( 2 + doubling time) = 2 * [$5000 * 1.08 ^2] by $5000 we get 1.08 ^ ( 2 + doubling time) = 2 * 1.08 ^2]. This can be written as 1.08^2 * 1.08^doublingtime = 2 * 1.08^2. Dividing both sides by 1.08^2 we obtain 1.08^doublingtime = 2. We can then use trial and error to find the doubling time that works. We get something like 9 years. **
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RESPONSE --> i got it somewhat right but i should have had 1.08^(2+doublingtime)*5000=2*[5000*1.08^2] and then simplify it to 1.08^2*1.08^doublingtime=2*1.08^2 self critique assessment: 2
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19:16:33 `questionNumber 40000 Desribe how on your graph how you obtained an estimate of the doubling time.
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RESPONSE --> by doubling the time confidence assessment: 1
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19:17:11 `questionNumber 40000 In this case you would find the double of the initial amount, $10000, on the vertical axis, then move straight over to the graph, then straight down to the horizontal axis. The interval from t = 0 to the clock time found on the horizontal axis is the doubling time. **
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RESPONSE --> double the initial amount and then move on the vertical axis then on the horizontal axis. self critique assessment: 2
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19:18:10 `questionNumber 40000 #17. At 10:00 a.m. a certain individual has 550 mg of penicillin in her bloodstream. Every hour, 11% of the penicillin present at the beginning of the hour is removed by the end of the hour. What is the function Q(t)?
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RESPONSE --> Q(t)= 550*1.11^t confidence assessment: 1
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19:18:53 `questionNumber 40000 Every hour 11% or .11 of the total is lost so the growth rate is -.11 and the growth factor is 1 + r = 1 + (-.11) = .89 and we have Q(t) = Q0 * (1 + r)^t = 550 mg (.89)^t or Q(t)=550(.89)^t **
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RESPONSE --> oh I should have subtracted the .11 from 1 instead of added because it was losing it not gaining it. self critique assessment: 3
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19:19:25 `questionNumber 40000 How much antibiotic is present at 3:00 p.m.?
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RESPONSE --> 387.73 confidence assessment: 1
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19:19:50 `questionNumber 40000 3:00 p.m. is 5 hours after the initial time so at that time there will be Q(5) = 550 mg * .89^5 = 307.123mg in the blood **
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RESPONSE --> i put 3 instead of 5 otherwise i would have gotten 307.123mg self critique assessment: 3
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19:20:01 `questionNumber 40000 Describe your graph and explain how it was used to estimate half-life.
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RESPONSE --> not sure confidence assessment: 1
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19:22:14 `questionNumber 40000 Starting from any point on the graph we first project horizontally and vertically to the coordinate axes to obtain the coordinates of that point. The vertical coordinate represents the quantity Q(t), so we find the point on the vertical axis which is half as high as the vertical coordinate of our starting point. We then draw a horizontal line directly over to the graph, and project this point down. The horizontal distance from the first point to the second will be the distance on the t axis between the two vertical projection lines. **
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RESPONSE --> the vertical coordinate is quantity Q(t) and it is half as high than the starting point, then we go over to the horizontal part of the graph and go downward. self critique assessment: 2
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19:22:23 `questionNumber 40000 What is the equation to find the half-life? What is the most simplified form of this equation?
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RESPONSE --> not sure confidence assessment: 0
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19:23:35 `questionNumber 40000 Q(doublingTime) = 1/2 Q(0)or 550 mg * .89^doublingTIme = .5 * 550 mg. Dividing thru by the 550 mg we have .89^doublingTime = .5. We can use trial and error to find an approximate value for doublingTIme (later we use logarithms to get the accurate solution). **
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RESPONSE --> Q(doublingtime)=1/2Q(0) so we just add doublinttime as the exponent on the side of 550mg*.89^doublingtime= .5*550mg which can be restated as .89^doublingtime=.5 self critique assessment: 2
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19:24:26 `questionNumber 40000 #19. For the function Q(t) = Q0 (1.1^ t), a value of t such that Q(t) lies between .05 Q0 and .1 Q0. For what values of t did Q(t) lie between .005 Q0 and .01 Q0?
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RESPONSE --> Q(t)=.05-.1(1.1^t) confidence assessment: 1
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19:25:53 `questionNumber 40000 Any value between about t = -24.2 and t = -31.4 will result in Q(t) between .05 Q0 and .1 Q0. Note that these values must be negative, since positive powers of 1.1 are all greater than 1, resulting in values of Q which are greater than Q0. Solving Q(t) = .05 Q0 we rewrite this as Q0 * 1.1^t = .05 Q0. Dividing both sides by Q0 we get 1.1^t = .05. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -31.4 approx. Solving Q(t) = .1 Q0 we rewrite this as Q0 * 1.1^t = .1 Q0. Dividing both sides by Q0 we get 1.1^t = .1. We can use trial and error (or if you know how to use them logarithms) to approximate the solution. We get t = -24.2 approx. (The solution for .005 Q0 is about -55.6, for .01 is about -48.3 For this solution any value between about t = -48.3 and t = -55.6 will work). **
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RESPONSE --> the equation is Q0*1.1^t=.05Q0 which can be rewritten as 1.1^t=.05 and plug in the numbers self critique assessment: 2
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19:26:25 `questionNumber 40000 explain why the negative t axis is a horizontal asymptote for this function.
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RESPONSE --> because it does not go down to 0 confidence assessment: 0
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19:27:03 `questionNumber 40000 The value of 1.1^t increases for increasing t; as t approaches infinity 1.1^t also approaches infinity. Since 1.1^-t = 1 / 1.1^t, we see that for increasingly large negative values of t the value of 1.1^t will be smaller and smaller, and would in fact approach zero. **
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RESPONSE --> the larger the negative values are the smaller it will get and would eventually get to 0 self critique assessment: 2
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19:27:25 `questionNumber 40000 #22. What value of b would we use to express various functions in the form y = A b^x? What is b for the function y = 12 ( e^(-.5x) )?
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RESPONSE --> b is e. confidence assessment: 1
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19:28:18 `questionNumber 40000 12 e^(-.5 x) = 12 (e^-.5)^x = 12 * .61^x, approx. So this function is of the form y = A b^x for b = .61 approx.. **
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RESPONSE --> 12e^(-.5x)=12(e^-.5)^x=12*.61^x. the b = .61 self critique assessment: 2
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19:28:28 `questionNumber 40000 what is b for the function y = .007 ( e^(.71 x) )?
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RESPONSE --> .71 confidence assessment: 1
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19:28:44 `questionNumber 40000 12 e^(.71 x) = 12 (e^.71)^x = 12 * 2.04^x, approx. So this function is of the form y = A b^x for b = 2.041 approx.. **
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RESPONSE --> i forgot to work it out but u would get 2.041 self critique assessment: 2
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19:42:17 `questionNumber 40000 what is b for the function y = -13 ( e^(3.9 x) )?
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RESPONSE --> 1.95 confidence assessment: 1
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19:42:40 `questionNumber 40000 12 e^(3.9 x) = 12 (e^3.9)^x = 12 * 49.4^x, approx. So this function is of the form y = A b^x for b = 49.4 approx.. **
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RESPONSE --> it should have been 49.4 self critique assessment: 2
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19:43:05 `questionNumber 40000 List these functions, each in the form y = A b^x.
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RESPONSE --> y=Ab^x confidence assessment: 1
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19:43:59 `questionNumber 40000 The functions are y=12(.6065^x) y=.007(2.03399^x) and y=-13(49.40244^x) **
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RESPONSE --> y=12(.6065^x) y=.007(2.03399^x and y=-13(49.40244^x) self critique assessment: 2
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19:44:40 `questionNumber 40000 0.4.44 (was 0.4.40 find all real zeros of x^2+5x+6
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RESPONSE --> (x+2)(x+3)=2,3 confidence assessment: 2
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19:44:59 `questionNumber 40000 We can factor this equation to get (x+3)(x+2)=0. (x+3)(x+2) is zero if x+3 = 0 or if x + 2 = 0. We solve these equations to get x = -3 and x = -2, which are our two solutions to the equation. COMMON ERROR AND COMMENT: Solutions are x = 2 and x = 3. INSTRUCTOR COMMENT: This error generally comes after factoring the equation into (x+2)(x+3) = 0, which is satisfied for x = -2 and for x = -3. The error could easily be spotted by substituting x = 2 or x = 3 into this equation; we can see quickly that neither gives us the correct solution. It is very important to get into the habit of checking solutions. **
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RESPONSE --> oh i forgot the minus signs in front of 2 and 3 self critique assessment: 3
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19:45:33 `questionNumber 40000 Explain how these zeros would appear on the graph of this function.
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RESPONSE --> they would appear on the x-axis confidence assessment: 3
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19:45:50 `questionNumber 40000 We've found the x values where y = 0. The graph will therefore go through the x axis at x = -2 and x = -3. **
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RESPONSE --> i think thats what i said self critique assessment: 3
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19:46:49 `questionNumber 40000 0.4.50 (was 0.4.46 x^4-625=0
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RESPONSE --> (x-5)(x+5)(x^2+5^2)= 5 and -5 confidence assessment: 2
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19:47:20 `questionNumber 40000 Common solution: x^4 - 625 = 0. Add 625 to both sides: x^4 = 625. Take fourth root of both sides, recalling that the fourth power is `blind' to the sign of the number: x = +-625^(1/4) = +-[ (625)^(1/2) ] ^(1/2) = +- 25^(1/2) = +-5. This is a good and appropriate solution. It's also important to understand how to use factoring, which applies to a broader range of equations, so be sure you understand the following: We factor the equation to get (x^2 + 25) ( x^2 - 25) = 0, then factor the second factor to get (x^2 + 25)(x - 5)(x + 5) = 0. Our solutions are therefore x^2 + 25 = 0, x - 5 = 0 and x + 5 = 0. The first has no solution and the solution to the second two are x = 5 and x = -5. The solutions to the equation are x = 5 and x = -5. **
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RESPONSE --> i believe that is what i got self critique assessment: 3
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19:49:41 `questionNumber 40000 0.4.70 (was 0.4.66 P = -200x^2 + 2000x -3800. Find the x interval for which profit is >1000
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RESPONSE --> 1000>-200x^2+2000x-3800 = 0>-200x^2+2000x-4800 = 0>-2000+ - sqrt160000/-400 = -2000+- 400/-400 = 4, 6 confidence assessment: 2
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19:50:12 `questionNumber 40000 You have to solve the inequality 1000<-200x^2+2000x-3800, which rearranges to 0<-200x^2+2000x-4800. This could be solved using the quadratic formula, but it's easier if we simplify it first, and it turns out that it's pretty easy to factor when we do: -200x^2+2000x-4800 = 0 divided on both sides by -200 gives x^2 - 10 x + 24 = 0, which factors into (x-6)(x-4)=0 and has solutions x=4 and x=6. So -200x^2+2000x-4800 changes signs at x = 4 and x = 6, and only at these values (gotta go thru 0 to change signs). At x = 0 the expression is negative. Therefore from x = -infinity to 4 the expression is negative, from x=4 to x=6 it is positive and from x=6 to infinity it's negative. Thus your answer would be the interval (4,6). COMMON ERROR: x = 4 and x = 6. INSTRUCTOR COMMENT: You need to use these values, which are the zeros of the function, to split the domain of the function into the intervals 4 < x < 6, (-infinity, 4) and (6, infinity). Each interval is tested to see whether the inequality holds over that interval. COMMON ERROR: Solution 4 > x > 6. INSTRUCTOR COMMENT: Look carefully at your inequalities. 4 > x > 6 means that 4 > x AND x > 6. There is no value of x that is both less than 4 and greater than 6. There are of course a lot of x values that are greater than 4 and less than 6. The inequality should have been written 4 < x < 6, which is equivalent to the interval (4, 6). **
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RESPONSE --> i do believe that is what i got (4,6) self critique assessment: 3
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?????Z?????h???? assignment #004 004. `query 4 Applied Calculus I 08-03-2007 f~??????????????assignment #005 005. `query 5 Applied Calculus I 08-03-2007
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19:57:30 `questionNumber 50000 explain why the slope of a depth vs. time trapezoid represents the average rate of change of the depth with respect to the time during the time interval represented
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RESPONSE --> not sure confidence assessment: 1
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19:58:02 `questionNumber 50000 the specific idea is that ave rate of depth change is [change in depth / change in time] ; rise represents change in depth and run represents change in time so slope = rise/run represents ave rate of depth change. **
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RESPONSE --> oh, slope= rise over run and it can be used to find the average rate of depth change self critique assessment: 3
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19:58:22 `questionNumber 50000 explain why the area of a rate vs. time trapezoid for a given time interval represents the change in the quantity corresponding to that time interval
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RESPONSE --> slope is rise over run confidence assessment: 2
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20:00:01 `questionNumber 50000 The average altitude represents the avg. velocity. The area of a trapezoid involves the altitude, which represents the avg. velocity, and the width, which represents the change in clock time. When you multiply ave altitude by width you are representing ave vel * change in clock time, which gives change in position. This reasoning isn't confined to velocities. For any rate vs. clock time graph, average altitude represents approximate average rate, which multiplied by the change in time (not by the time itself) gives you the change in quantity **
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RESPONSE --> the ave altitude can be ave velocity. when ave altitude is multiplied by the width will give you the change in position. ave altitude represents the app. ave rate and can be multiplied by the change in time which will give you the change in quantity self critique assessment: 2
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20:02:58 `questionNumber 50000 text problem 0.5 #8 add x/(2-x) + 2/(x-2)
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RESPONSE --> x(x-2)/ (2-x)(2-x) + 2(x-2)/(2-x)(2-x)= x^2-2x+4-2x/(2-x)(2-x)= x^2-4x=4/(2-x)(2-x)= (x-2)(x-2)/ (x-2)(2-x)= (x-2)/(2-x) confidence assessment: 3
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20:03:44 `questionNumber 50000 common denominator could be [ (2-x)(x-2) ]. In this case we have x / (2-x) + 2 / (x-2) = [ (x-2) / (x-2) ] * [ x / (2-x) ] + [ (2-x) / (2-x) ] * [ 2 / (x-2) ] = x(x-2) / [ (2-x)(x-2) ] + 2 (2-x) / [ (2-x)(x-2) ] = [x(x-2) + 2(2-x) ] / [ (2-x)(x-2) ] = [ x^2 - 2x + 4 - 2x ] / [ (2-x)(x-2) ] = (x^2-4x+4) / [ -x^2+4x-4 ] = (x-2)^2 / [-(x-2)^2] = -1. NOTE however that there is a SIMPLER SOLUTION: We can note that x-2 = -(2-x) so that the original problem is -x/(x-2) + 2 /(x-2) = (-x + 2) / (x-2) = -(x-2)/(x-2) = -1. **
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RESPONSE --> i should have factored the denominator and then i would have gotten -1 self critique assessment: 3
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20:04:30 `questionNumber 50000 text problem 0.5 #48 cost = 6 x + 900,000 / x, write as single fraction and determine cost to store 240 units
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RESPONSE --> 900,000+6x^2/x and for the 240 units i got 5,190 confidence assessment: 3
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20:04:40 `questionNumber 50000 express with common denominator x: [x / x] * 6x + 900,000 / x = 6x^2 / x + 900,000 / x = (6x^2 + 900,000) / x so cost = (6x^2+900,000)/x Evaluating at x = 240 we get cost = (6 * 240^2 + 900000) / 240 = 5190. **
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RESPONSE --> that is what i got self critique assessment: 3
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?????Q??y?????assignment #005 005. `query 5 Applied Calculus I 08-03-2007 20:10:51 `questionNumber 60000 Query class notes #06 If x is the height of a sandpile and y the volume, what proportionality governs geometrically similar sandpiles? Why should this be the proportionality?
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RESPONSE --> in order to get the volume you need the height confidence assessment: 1
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20:11:50 `questionNumber 60000 the proportionality is y = k x^3. Any proportionality of volumes is a y = k x^3 proportionality because volumes can be filled with tiny cubes; surface areas are y = k x^2 because surfaces can be covered with tiny squares. **
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RESPONSE --> y= kx^3 is the proportionality. the volume can be filled with the surface area because it can be covered in tiny squares like volume with tiny cubes. self critique assessment: 2
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20:12:19 `questionNumber 60000 If x is the radius of a spherical balloon and y the surface area, what proportionality governs the relationship between y and x? Why should this be the proportionality?
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RESPONSE --> y=pir^2 confidence assessment: 1
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20:13:27 `questionNumber 60000 Just as little cubes can be thought of as filling the volume to any desired level of accuracy, little squares can be thought of as covering any smooth surface. Cubes 'scale up' in three dimensions, squares in only two. So the proportionality is y = k x^2. Surfaces can be covered as nearly as we like with tiny squares (the more closely we want to cover a sphere the tinier the squares would have to be). The area of a square is proportional to the square of its linear dimensions. Radius is a linear dimension. Thus the proportionality for areas is y = k x^2. By contrast, for volumes or things that depend on volume, like mass or weight, we would use tiny cubes to fill the volume. Volume of a cube is proportional to the cube of linear dimensions. Thus the proportionality for a volume would be y = k x^3. **
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RESPONSE --> i got confused with the whole area of a circle... it should be y=kx^2 because the area is proportional to the linear dimensions, which the radius is. self critique assessment: 2
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20:14:20 `questionNumber 60000 Explain how you would use the concept of the differential to find the volume of a sandpile of height 5.01 given the volume of a geometrically similar sandpile of height 5, and given the value of k in the y = k x^3 proportionality between height and volume.
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RESPONSE --> 5.01=k5^3 = .04008 confidence assessment: 1
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20:15:36 `questionNumber 60000 The class notes showed you that the slope of the y = k x^3 graph is given by the rate-of-change function y' = 3 k x^2. Once you have evaluated k, using the given information, you can evaluate y' at x = 5. That gives you the slope of the line tangent to the curve, and also the rate at which y is changing with respect to x. When you multiply this rate by the change in x, you get the change in y. The differential is 3 k x^2 `dx and is approximately equal to the corresponding `dy. Since `dy / `dx = 3 k x^2, the differential looks like a simple algebraic rearrangement `dy = 3 k x^2 `dx, though what's involved isn't really simple algebra. The differential expresses the fact that near a point, provided the function has a continuous derivative, the approximate change in y can be found by multiplying the change in x by the derivative). That is, `dy = derivative * `dx (approx)., or `dy = slope at given point * `dx (approx), or `dy = 3 k x^2 `dx (approx). The idea is that the derivative is the rate of change of the function. We can use the rate of change and the change in x to find the change in y. The differential uses the fact that near x = 5 the change in y can be approximated using the rate of change at x = 5. Our proportionality is y = k x^3. Let y = f(x) = k x^3. Then y' = f'(x) = 3 k x^2. When x = 5 we have y' = f'(5) = 75 k, whatever k is. To estimate the change in y corresponding to the change .01 in x, we will multiply y ' by .01, getting a change of y ' `dx = 75 k * .01. } SPECIFIC EXAMPLE: We don't know what k is for this specific question. As a specific example suppose our information let us to the value k = .002, so that our proportionality is y = .002 x^3. Then the rate of change when x is 5 would be f'(5) = 3 k x^2 = 3 k * 5^2 = 75 k = .15 and the value of y would be y = f(5) = .002 * 5^3 = .25. This tells us that at x = 5 the function is changing at a rate of .15 units of y for each unit of x. Thus if x changes from 5 to 5.01 we expect that the change will be change in y = (dy/dx) * `dx = rate of change * change in x (approx) = .15 * .01 = .0015, so that when x = 5.01, y should be .0015 greater than it was when x was 5. Thus y = .25 + .0015 = .2515. This is the differential approximation. It doesn't take account of the fact that the rate changes slightly between x=5 and x = 5.01. But we don't expect it to change much over that short increment, so we expect that the approximation is pretty good. Now, if you evaluate f at x = 5.01 you get .251503. This is a little different than the .2515 approximation we got from the differential--the differential is off by .000003. That's not much, and we expected it wouldn't be much because the derivative doesn't change much over that short interval. But it does change a little, and that's the reason for the discrepancy. The differential works very well for decently behaved functions (ones with smooth curves for graphs) over sufficiently short intervals.**
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RESPONSE --> find k and then do the y' when x=5 and then multiply the rate by the change in x and u get the change of y. self critique assessment: 2
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20:16:32 `questionNumber 60000 What would be the rate of depth change for the depth function y = .02 t^2 - 3 t + 6 at t = 30? (instant response not required)
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RESPONSE --> -66 confidence assessment: 1
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20:17:18 `questionNumber 60000 You saw in the class notes and in the q_a_ that the rate of change for depth function y = a t^2 + b t + c is y ' = 2 a t + b. This is the function that should be evaluated to give you the rate. Evaluating the rate of depth change function y ' = .04 t - 3 for t = 30 we get y ' = .04 * 30 - 3 = 1.2 - 3 = -1.8. COMMON ERROR: y = .02(30)^2 - 2(30) + 6 =-36 would be the rate of depth change INSTRUCTOR COMMENT: This is the depth, not the rate of depth change. **
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RESPONSE --> i thought that the -2 was a -3 otherwise i would have gotten the -36 self critique assessment: 3
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20:17:44 `questionNumber 60000 modeling project 3 problem a single quarter-cup of sand makes a cube 1.5 inches on a side. How many quarter-cups would be required to make a cube with twice the scale, 3 inches on a side? Explain how you know this.
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RESPONSE --> not sure confidence assessment: 0
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20:19:22 `questionNumber 60000 You can think of stacking single cubes--to double the dimensions of a single cube you would need 2 layers, 2 rows of 2 in each layer. Thus it would take 8 cubes 1.5 inches on a side to make a cube 3 inches on a side. Since each 1.5 inch cube containts a quarter-cup, a 3 inch cube would contain 8 quarter-cups. COMMON ERROR: It would take 2 quarter-cups. INSTRUCTOR COMMENT: 2 quarter-cups would make two 1.5 inch cubes, which would not be a 3-inch cube but could make a rectangular solid with a square base 1.5 inches on a side and 3 inches high. **
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RESPONSE --> every side has 2 of everything thus giving us 8cubes. thus it would take a 8quarter cups because the 3 inch cube would contain that much self critique assessment: 2
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20:20:05 `questionNumber 60000 What value of the parameter a would model this situation? How many quarter-cups does this model predict for a cube three inches on a side? How does this compare with your previous answer?
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RESPONSE --> not sure confidence assessment: 0
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20:21:16 `questionNumber 60000 The proportionality would be y = a x^3, with y = 1 (representing one quarter-cup) when x = 1.5. So we have 1 = a * 1.5^3, so that a = 1 / 1.5^3 = .296 approx. So the model is y = .2963 x^3. Therefore if x = 3 we have y = .296 * 3^3 = 7.992, which is the same as 8 except for roundoff error. **
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RESPONSE --> y=ax^3 so plug in 1 for y and 1.5 for x and get .296 so then you get y=.296x^3 and if x is 3 then plug that into the equation and get 7.992 which is very close to 8 just rounded off self critique assessment: 2
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20:22:01 `questionNumber 60000 What would be the side measurement of a cube designed to hold 30 quarter-cups of sand? What equation did you solve to get this?
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RESPONSE --> it would be 80 confidence assessment: 1
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20:23:13 `questionNumber 60000 You are given the number of quarter-cups, which corresponds to y. Thus we have 30 = .296 x^3 so that x^3 = 30 / .296 = 101, approx, and x = 101^(1/3) = 4.7, approx..**
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RESPONSE --> i should have plugged in 30 for y and .296 for k and get 30=.296x^3 and then switch over the 30 and get x^3= 30/.296= 101 and then x=101^(1/3) which is 4.7 self critique assessment: 2
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20:24:07 `questionNumber 60000 query problem 2. Someone used 1/2 cup instead of 1/4 cup. The best-fit function was y = .002 x^3. What function would have been obtained using 1/4 cup?
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RESPONSE --> y=.002* .25^3 = 3.125*10^-5 confidence assessment: 0
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20:25:20 `questionNumber 60000 In this case, since it takes two quarter-cups to make a half-cup, the person would need twice as many quarter-cups to get the same volume y. He would have obtained half as many half-cups as the actual number of quarter-cups. To get the function for the number of quarter-cups he would therefore have to double the value of y, so the function would be y = .004 x^3. **
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RESPONSE --> since quarter cups are 2=a half cup then it needs to double so we get y=.004x^3 self critique assessment: 2
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20:25:42 `questionNumber 60000 query problem 4. number of swings vs. length data. Which function fits best?
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RESPONSE --> y=x^3 confidence assessment: 1
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20:26:45 `questionNumber 60000 If you try the different functions, then for each one you can find a value of a corresponding to every data point. For example if you use y = a x^-2 you can plug in every (x, y) pair and solve to see if your values of a are reasonably consistent. Try this for the data and you will find that y = a x^-2 does not give you consistent a values-every (x, y) pair you plug in will give you a very different value of a. The shape of the graph gives you a pretty good indication of which one to try, provided you know the shapes of the basic graphs. For this specific situation the graph of the # of swings vs. length decreases at a decreasing rate. The graphs of y = a x^.p for p = -.3, -.4, -.5, -.6 and -.7 all decrease at a decreasing rate. In this case you would find that the a x^-.5 function works nicely, giving a nearly constant value of a.
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RESPONSE --> it should be y=a*x^-2 but it isnt exactly correct so we do y=ax^p self critique assessment: 2
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20:27:06 `questionNumber 60000 problem 7. time per swing model. For your data what expression represents the number of swings per minute?
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RESPONSE --> y=ax^s confidence assessment: 1
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20:28:09 `questionNumber 60000 The model that best fits the data is a x^-.5, and with accurate data we find that a is close to 55. The model is pretty close to # per minute frequency = 55 x^-.5. As a specific example let's say we obtained counts of 53, 40, 33 and 26 cycles in a minute at lengths of 1, 2, 3 and 4 feet, then using y = a x^-.5 gives you a = y * x^.5. Evaluating a for y = 53 and x = 1 gives us a = 53 * 1^.5 = 53; for y = 40 and x = 2 we would get a = 40 * 2^.5 = 56; for y = 34 and x = 3 we get a = 33 * 3^.5 = 55; for y = 26 and x = 4 we get a = 26 * 4^.5 = 52. Since our value of a are reasonably constant the y = a x^.5 model works pretty well, with a value of a around 54. The value of a for accurate data turns out to be about 55.**
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RESPONSE --> it should be = 55x^-.5 and then it equals to #/min frequency the most accurate for a is 55 self critique assessment: 2
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20:28:30 `questionNumber 60000 If the time per swing in seconds is y, then what expression represents the number of swings per minute?
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RESPONSE --> not sure confidence assessment: 0
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20:29:47 `questionNumber 60000 To get the number of swings per minute you would divide 60 seconds by the number of seconds in a swing (e.g., if a swing takes 2 seconds you have 30 swings in a minute). So you would have f = 60 / y, where f is frequency in swings per minute. COMMON ERROR: y * 60 INSTRUCTOR COMMENT: That would give more swings per minute for a greater y. But greater y implies a longer time for a swing, which would imply fewer swings per minute. This is not consistent with your answer. **
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RESPONSE --> you have to divide by 60 seconds to gt it into minutes because there is 60 sec in a minute. f=frequency so f=60/y because y is swings per second. self critique assessment: 2
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20:30:13 `questionNumber 60000 If the time per swing is a x ^ .5, for the value determined previously for the parameter a, then what expression represents the number of swings per minute? How does this expression compare with the function you obtained for the number of swings per minute vs. length?
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RESPONSE --> f=60/y confidence assessment: 0
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20:31:11 `questionNumber 60000 Time per swing turns out to be a x^.5--this is what you would obtain if you did the experiment very accurately and correctly determined the power function. For x in feet a will be about 1.1. Since the number of swings per minute is 60/(time per swing), you have f = 60 / (a x^.5), where f is frequency in swings / minute. Simplifying this gives f = (60 / a) * x^.5. 60/a is just a constant, so the above expression is of form f = k * x^-.5, consistent with earlier statements. 60 / a = 60 / 1.1 = 55, approx., confirming our frequency model F = 55 x^-.5. **
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RESPONSE --> you plug in the ax^.5 for y and get f=60/(ax^.5) which gives us f=(60/a)*x^.5 self critique assessment: 2
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20:31:55 `questionNumber 60000 query problem 8. model of time per swing what are the pendulum lengths that would result in periods of .1 second and 100 seconds?
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RESPONSE --> not sure confidence assessment: 0
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20:34:08 `questionNumber 60000 You would use your own model here. This solution uses T = 1.1 x^.5. You can adapt the solution to your own model. According to the model T = 1.1 x^.5 , where T is period in seconds and x is length in feet, we have periods T = .1 and T = 100. So we solve for x: For T = .1 we get: .1 = 1.2 x^.5 which gives us x ^ .5 = .1 / 1.2 so that x^.5 = .083 and after squaring both sides we get x = .083^2 = .0069 approx., representing .0069 feet. We also solve for T = 100: 100 = 1.2 x^.5, obtaining x^.5 = 100 / 1.2 = 83, approx., so that x = 83^2 = 6900, approx., representing a pendulum 6900 ft (about 1.3 miles) long. **
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RESPONSE --> we would use T=1.1x^5. T= period in seconds and x is length (ft) so we have T=.1 and T=100. .1=1.2x^.5 and we get .0069ft and for T=100 we get 100=1.2x^.5 we get 6900ft self critique assessment: 2
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20:34:24 `questionNumber 60000 query problem 9. length ratio x2 / x1.
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RESPONSE --> x2/x1 confidence assessment: 0
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20:34:58 `questionNumber 60000 What expressions, in terms of x1 and x2, represent the frequencies (i.e., number of swings per minute) of the two pendulums?
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RESPONSE --> x1 equals the first pendulum and x2 is the 2nd pendulum confidence assessment: 1
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20:36:57 `questionNumber 60000 The solution is to be in terms of x1 and x2. If lengths are x2 and x1, you would substitute x2 and x1 for L in the frequency relationship f = 60 / (1.1 `sqrt(L)) to get 60 / (1.1 `sqrt(x1) ) and 60 / (1.1 `sqrt(x2)). Alternative form is f = 55 L^-.5. Substituting would give you 55 * x1^-.5 and 55 * x2^-.5. If you just had f = a L^-.5 (same as y = a x^-.5) you would get f1 = a x1^-.5 and f2 = a x2^-.5 **
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RESPONSE --> we would substitute x2 and x1 for L in f=60/1.1sqrt of L and get 60/1.1sqrt of x1 and 60/1.1sqrtx2 and if we use f=55L^-.5 we then get 55*x1^-.5 and 55*x2^-.5 and you could keep substituing L for x1 and x2 in equations self critique assessment: 2
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20:37:28 `questionNumber 60000 What expression, in terms of x1 and x2, represents the ratio of the frequencies of the two pendulums?
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RESPONSE --> x1=60/1.1^.5 and x2=60/1.1^.5 confidence assessment: 1
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20:38:40 `questionNumber 60000 We need to do this in terms of the symbols x1 and x2. If f = a x^-.5 then f1 = a x1^-.5 and f2 = a x2^-.5. With these expressions we would get f2 / f1 = a x2^-.5 / (a x1^-.5) = x2^-.5 / x1^-.5 = (x2 / x1)^-.5 = 1 / (x2 / x1)^.5 = (x1 / x2)^.5. Note that it doesn't matter what a is, since a quickly divides out of our quotient. For example if a = 55 we get f2 / f1 = 55 x2^-.5 / (55 x1^-.5) = x2^-.5 / x1^-.5 = (x2 / x1)^-.5 = 1 / (x2 / x1)^.5 = (x1 / x2)^.5. This is the same result we got when a was not specified. This shouldn't be surprising, since the parameter a divided out in the third step. **
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RESPONSE --> we would replace x and f with f1=ax1^-.5 and f2=ax2^-.5 and then if we did f2/f1 we would get (x1/c2)^.5 self critique assessment: 2
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20:39:23 `questionNumber 60000 query problem Challenge Problem for Calculus-Bound Students: how much would the frequency change between lengths of 2.4 and 2.6 feet
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RESPONSE --> there would be a .2 difference between the 2.4 and 2.6 confidence assessment: 1
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20:40:52 `questionNumber 60000 STUDENT SOLUTION: Note that we are using frequency in cycles / minute. I worked to get the frequency at 2.4 and 2.6 y = 55.6583(2.4^-.5) = 35.9273 and y = 55.6583(2.6^-.5)= 34.5178. subtracted to get -1.40949 difference between 2.4 and 2.6. This, along with the change in length of .2, gives average rate -1.409 cycles/min / (.2 ft) = -7.045 (cycles/min)/ft , based on the behavior between 2.4 ft and 2.6 ft. This average rate would predict a change of -7.045 (cycles/min)/ft * 1 ft = -7/045 cycles/min for the 1-foot increase between 2 ft and 3 ft. The change obtained by evaluating the model at 2 ft and 3 ft was -7.2221 cycles/min. The answers are different because the equation is not linear and the difference between 2.4 and 2.6 does not take into account the change in the rate of frequency change between 2 and 2.4 and 2.6 and 3 for 4.4 and 4.6 y = 55.6583(4.4^-.5) y = 55.6583(4.6^-.5) y = 26.5341 y = 25.6508 Dividing difference in y by change in x we get -2.9165 cycles/min / ft, compared to the actual change -2.938 obtained from the model. The answers between 4-5 and 2-3 are different because the equation is not linear and the frequency is changing at all points. **
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RESPONSE --> we would plug the numbers into the equation of y=55.6583(x^-.5) for x and get the answer for y self critique assessment: 2
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20:42:44 `questionNumber 60000 1.1.20 (was 1.1.18 are (0,4), (7,-6) and (-5,11) collinear?
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RESPONSE --> when i did the distance formula for (0,4) and (7,-6) i got sqrt of 149 and when i did the same thing with (7,-6) and (-5,11) i got sqrt 433 and when i did it with (0,4) and (-5,11) i got the sqrt 74 and i believe they are not collinear confidence assessment: 2
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20:43:28 `questionNumber 60000 The distance between (-5,11) and (7,-6) is approximately 20.81: d3 = sqr rt [(7+5)^2 + -(6 +11)^2] d3 = sqr rt 433 d3 = 20.81 Using the distance formula the distances between (-5,11) and (0,4) is 8.6 and the distance between (0,4) and (7, -6) is 12.2. 'Collinear' means 'lying along the same straight line'. If three points are collinear then the sum of the distances between the two closer pairs of points will equal the distance between the furthest two. Since 8.6 + 12.2 = 20.8 the points are on the same straight line. **
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RESPONSE --> I should have taken the sqrt of the 3 of them and then added the two smaller ones to see if the result equaled the larger number. self critique assessment: 3
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20:43:48 `questionNumber 60000 1.1.24 find x | dist (2,-1) to (x,2) is 5What value of x makes the distance 5?
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RESPONSE --> I got 7,-3 confidence assessment: 1
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20:46:06 `questionNumber 60000 The expression for the distance from (2, -1) to (x, 2) is = sqrt((x-2)^2 + (2+1)^2). This distance is to be 5, which gives us the equation5 = sqrt((x-2)^2 + (2+1)^2) Starting with the equation 5 = sqrt((x-2)^2 + (2+1)^2) we first square both sides to get 25 = (x-2)^2 + 9 or (x-2)^2 = 16. Solutions are found by taking the square root of both sides, keeping in mind that (x-2)^2 doesn't distinguish between positive and negative values of x - 2. We find that (x - 2) = +_ sqrt(16) = +- 4. (x-2) = 4 gives us the solution x = 6 and (x-2) = -4 gives us the solution x = -2. **
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RESPONSE --> I should have done the sqrt of ((x-2)^2+(2+1)^2) and make it equal to 5 and get it to (x-2)^2=16 or 25=(x-2)^2+9 and then do (x-2)=+_sqrt(16)= +-4 and this gives us x=6 and -2 self critique assessment: 2
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20:47:48 `questionNumber 60000 1.1.36 (was 1.1.34 percent increase in Dow What are the requested percent increases?
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RESPONSE --> 1801 and 1900 confidence assessment: 0
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20:49:05 `questionNumber 60000 Aug 99 to Nov 99 Change in value = 10600 - 10600 = 0 At a percent of the initial value we have 0/10600 = 0, or 0% increase May 2000 to September 2000: change in value = 10600-10800 = -200 As percent of initial value: -200/10600 = .019 or 1.9% approx.. **
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RESPONSE --> we have 10600 and we subtract from itself and get 0 thus 0% and then the value changed in may to sept so we subtract 10800 from 10600 and get a -200 and then divide that from 10600 and get .019 or 1.9% self critique assessment: 2
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