course mth 271 ڎɼTW͔assignment #001
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20:22:39 Section 0.1 solve x/2-x/3>5
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RESPONSE --> 3x/6-2x/6-30/6>0 x/6-5>0 x>30
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20:22:49 ** 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 you
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20:23:05 3x - 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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20:24:34 Section 0.1 solve 2x^2+1<9x-3
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RESPONSE --> (2x-1)(x-4)<0 0.5
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20:24:56 ** 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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̮Jw⫟ɤ assignment #002 ݺӯD Calculus I 06-12-2006
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12:54:01 Were you able to complete the DERIVE exercise?
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RESPONSE --> Most of it.
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12:54:11 ** If you weren't able to complete the exercise you should download the 30-day trial version of DERIVE and do so. Note that the instructor neither recommends nor discourages you from purchasing DERIVE. Except for the basic worksheet and other assignments that may occur in the first part of the coruse DERIVE is not required and references to DERIVE in the later part of the course will be optional. This ends Query 2.**
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ؕNy assignment #002 ݺӯD Calculus I 06-12-2006
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12:54:44 Were you able to complete the DERIVE exercise?
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12:54:46 ** If you weren't able to complete the exercise you should download the 30-day trial version of DERIVE and do so. Note that the instructor neither recommends nor discourages you from purchasing DERIVE. Except for the basic worksheet and other assignments that may occur in the first part of the coruse DERIVE is not required and references to DERIVE in the later part of the course will be optional. This ends Query 2.**
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]{ʘ assignment #002 ݺӯD Applied Calculus I 06-12-2006
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13:03:05 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 --> (0,95) (20,60) (40,41)
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13:03:11 ** Continue to the next question **
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CLa~wo assignment #002 ݺӯD Applied Calculus I 06-12-2006
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17:22:40 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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17:22:43 ** Continue to the next question **
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17:22:48 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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17:34:36 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> (7,81) (19,61) (31,48)
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17:34:38 ** Continue to the next question **
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À~St`芣 assignment #002 ݺӯD Applied Calculus I 06-12-2006
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17:40:35 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 --> (0,111) (16,9108563) (32,7578268)
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17:40:39 ** Continue to the next question **
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17:46:23 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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RESPONSE --> (7,111) (19,87) (31,77)
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17:46:25 ** Continue to the next question **
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17:47:20 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE --> (8,111) (24,82.93105) (48,64.02332)
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17:47:57 ** 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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17:50:45 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 64a + 8b +c =111 576a + 24b + c= 82.93105 2304a + 48b + c= 64.02332
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17:50:56 ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
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17:51:53 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 576a +24b + c=82.93105
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17:51:59 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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17:52:29 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE --> 2304a + 48b +c =64.02332
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17:52:33 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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18:00:58 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 --> eq'1 = eq3-eq1= 2304a+48b=64.02332 x 3
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18:01:02 ** 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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18:02:02 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 --> eq'2= eq3-eq2=1728+24b=-18.90773 x -5
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18:02:04 ** 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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18:03:25 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE --> b, 120 (40x3) and -120 (24x-5)
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18:03:27 ** 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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18:07:12 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 --> a= -0.024 y= -0.24t^2 + 2.5t +92.536
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18:07:20 ** 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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18:07:32 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE --> 92.536
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18:07:36 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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18:10:11 What is the resulting quadratic model?
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RESPONSE --> y=-0.024t^2 + 2.5t + 92.536
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18:10:15 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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18:22:56 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 --> (0,92.54) dev 18.46 (8,111) dev 0 (16,126) dev 35
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18:23:02 ** 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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18:23:16 What was your average deviation?
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RESPONSE --> 12
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18:23:20 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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18:25:10 Is there a pattern to your deviations?
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RESPONSE --> no, I must have my equation wrong. my deviation greatly increased with time. in fact my y increased instead of decreasing.
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18:25:17 ** 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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18:25:51 06-12-2006 18:25:51 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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NOTES -------> i think i understand. somewhere along the way i calculated wrong.
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18:25:52 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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18:25:57 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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18:26:12 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> no no
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18:26:24 ** 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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18:26:39 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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18:26:46 ** 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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18:26:49 What three points on your graph did you use as a basis for your model?
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18:26:52 ** 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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18:26:54 Give the first of your three equations.
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18:26:56 ** STUDENT SOLUTION CONTINUED: The point (5.3, 63.7) gives me the equation 28.09a + 5.3b + c = 63.7 **
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18:26:59 Give the second of your three equations.
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18:27:01 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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18:27:03 Give the third of your three equations.
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18:27:10 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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18:27:13 Give the first of the equations you got when you eliminated c.
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18:27:15 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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18:27:17 Give the second of the equations you got when you eliminated c.
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18:27:20 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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18:27:22 Explain how you solved for one of the variables.
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18:27:25 ** 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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18:27:28 What values did you get for a and b?
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18:27:30 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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18:27:32 What did you then get for c?
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18:27:34 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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18:27:36 What is your function model?
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18:27:38 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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18:27:41 What is your depth prediction for the given clock time (give clock time also)?
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18:27:44 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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18:27:47 What clock time corresponds to the given depth (give depth also)?
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18:27:51 ** 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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18:27:54 Completion of Model second problem: grade average Give your data in the form of grade vs. clock time ordered pairs.
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18:27:57 ** 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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}~}|LM assignment #002 ݺӯD Applied Calculus I 06-12-2006
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18:28:58 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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18:29:00 ** Continue to the next question **
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18:29:02 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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18:29:05 ** Continue to the next question **
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18:29:06 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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18:29:08 ** 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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18:29:10 What is the first equation you got when you substituted into the form of a quadratic?
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18:29:11 ** STUDENT SOLUTION CONTINUED: The equation that I got from the first data point (10,75) was 100a + 10b +c = 75.**
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18:29:13 What is the second equation you got when you substituted into the form of a quadratic?
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18:29:14 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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18:29:16 What is the third equation you got when you substituted into the form of a quadratic?
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18:29:17 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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18:29:19 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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18:29:20 ** 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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18:29:22 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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18:29:24 ** 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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18:29:26 Which variable did you eliminate from these two equations, and what was its value?
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18:29:28 ** 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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18:29:29 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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18:29:30 ** 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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18:29:32 What is the value of c obtained from substituting into one of the original equations?
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18:29:34 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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18:29:36 What is the resulting quadratic model?
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18:29:38 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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18:29:40 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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18:29:42 ** 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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18:29:44 What was your average deviation?
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18:29:46 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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18:29:48 Is there a pattern to your deviations?
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RESPONSE -->
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18:29:50 ** 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 -->
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18:29:52 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 -->
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18:29:54 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE -->
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18:29:56 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE -->
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18:29:58 ** 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 -->
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18:30:01 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 -->
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18:30:20 ** 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 -->
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18:30:27 What three points on your graph did you use as a basis for your model?
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RESPONSE -->
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18:30:34 ** 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 -->
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18:30:36 Give the first of your three equations.
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RESPONSE -->
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18:30:39 ** 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 -->
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18:30:41 Give the second of your three equations.
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18:30:43 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE -->
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18:30:45 Give the third of your three equations.
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RESPONSE -->
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18:30:48 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE -->
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18:30:51 Give the first of the equations you got when you eliminated c.
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18:30:56 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE -->
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18:30:58 Give the second of the equations you got when you eliminated c.
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RESPONSE -->
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18:31:00 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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18:31:02 Explain how you solved for one of the variables.
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RESPONSE -->
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18:31:05 ** 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 -->
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18:31:08 What values did you get for a and b?
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RESPONSE -->
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18:31:10 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE -->
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18:31:12 What did you then get for c?
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RESPONSE -->
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18:31:14 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE -->
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18:31:17 What is your function model?
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RESPONSE -->
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18:31:21 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE -->
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18:31:27 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE -->
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18:31:34 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE -->
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18:31:38 What clock time corresponds to the given depth (give depth also)?
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RESPONSE -->
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18:31:41 ** 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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18:31:47 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 -->
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ΫڻS~{zߣ assignment #002 ݺӯD Applied Calculus I 06-12-2006
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19:14:15 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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19:14:17 ** Continue to the next question **
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19:14:20 According to your graph what would be the temperatures at clock times 7, 19 and 31?
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19:14:22 ** Continue to the next question **
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19:14:24 What three points did you use as a basis for your quadratic model (express as ordered pairs)?
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RESPONSE -->
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19:14:26 ** 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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19:14:29 What is the first equation you got when you substituted into the form of a quadratic?
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RESPONSE -->
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19:14:31 ** 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 -->
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19:14:33 What is the second equation you got when you substituted into the form of a quadratic?
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RESPONSE -->
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19:14:37 ** STUDENT SOLUTION CONTINUED: The equation that I got from my second data point was 400a + 20b + c = 60 **
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RESPONSE -->
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19:14:41 What is the third equation you got when you substituted into the form of a quadratic?
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RESPONSE -->
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19:14:44 ** STUDENT SOLUTION CONTINUED: The equation that I got from my third data point was 3600a + 60b + c = 30. **
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RESPONSE -->
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19:14:47 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 -->
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19:14:49 ** 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 -->
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19:14:51 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 -->
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19:14:53 ** 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 -->
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19:14:55 Which variable did you eliminate from these two equations, and what was its value?
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RESPONSE -->
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19:14:58 ** 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 -->
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19:15:01 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 -->
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19:15:04 ** 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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19:15:07 What is the value of c obtained from substituting into one of the original equations?
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RESPONSE -->
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19:15:09 ** STUDENT SOLUTION CONTINUED: By substituting both a and b into the original equations, I found that c = 93 **
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RESPONSE -->
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19:15:12 What is the resulting quadratic model?
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RESPONSE -->
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19:15:14 ** STUDENT SOLUTION CONTINUED: Therefore, the quadratic model that I obtained was y = (.015) x^2 - (1.95)x + 93. **
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RESPONSE -->
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19:15:21 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 -->
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19:15:24 ** 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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19:15:27 What was your average deviation?
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RESPONSE -->
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19:15:29 ** STUDENT SOLUTION CONTINUED: My average deviation was .6 **
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19:15:32 Is there a pattern to your deviations?
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RESPONSE -->
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19:15:35 ** 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 -->
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19:15:41 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 -->
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19:15:43 ** STUDENT SOLUTION CONTINUED: Yes, I do completely understand the process after studying these outlines and explanations. **
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RESPONSE -->
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19:15:58 Have you memorized the steps of the modeling process, and are you gonna remember them forever? Convince me.
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RESPONSE --> no, hopefully
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19:16:01 ** 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 -->
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19:16:17 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 -->
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19:16:20 ** 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 -->
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19:17:37 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (10, 1.122194) (50,2.342635) (90,3.08151)
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19:17:40 ** 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 -->
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19:18:11 Give the first of your three equations.
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RESPONSE --> 100a2+10b+c=1.122194
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19:18:15 ** 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 -->
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19:18:58 Give the second of your three equations.
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RESPONSE --> 2500a^2 + 50b +c =2.342635
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19:19:00 ** STUDENT SOLUTION CONTINUED: The point (15.9, 46) gives me the equation 252.81a +15.9b + c = 46 **
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RESPONSE -->
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19:19:42 Give the third of your three equations.
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RESPONSE --> 8100a^2 + 90b +c =3.08151
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19:19:45 ** STUDENT SOLUTION CONTINUED: The point (26.5,32) gives me the equation 702.25a + 26.5b + c = 32. **
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RESPONSE -->
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19:20:31 Give the first of the equations you got when you eliminated c.
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RESPONSE --> eq 3- eq1 8000a^2 = 80b =1.959316
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19:20:33 ** STUDENT SOLUTION CONTINUED: Subtracting the second equation from the third gave me 449.44a + 10.6b = -14. **
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RESPONSE -->
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19:21:12 Give the second of the equations you got when you eliminated c.
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RESPONSE --> eq3-eq2 5600a^2 + 40b = .738875
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19:21:14 ** ** STUDENT SOLUTION CONTINUED: Subtracting the first equation from the third gave me 674.16a + 21.2b = -31.7. **
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RESPONSE -->
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19:21:52 Explain how you solved for one of the variables.
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RESPONSE --> multiplied the 2nd equation by -2 to eliminate b
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19:21:54 ** 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 -->
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19:23:47 What values did you get for a and b?
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RESPONSE --> a= .000148566 b= .024489243
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19:23:50 ** STUDENT SOLUTION CONTINUED: a = .0165, b = -2 **
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RESPONSE -->
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19:24:06 What did you then get for c?
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RESPONSE --> c= .877299365
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19:24:09 ** STUDENT SOLUTION CONTINUED: c = 73.4 **
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RESPONSE -->
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19:25:01 What is your function model?
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RESPONSE --> y= .000148566t + .024489243t + .877299365
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19:25:03 ** STUDENT SOLUTION CONTINUED: y = (.0165)x^2 + (-2)x + 73.4. **
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RESPONSE -->
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19:25:16 What is your depth prediction for the given clock time (give clock time also)?
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RESPONSE -->
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19:25:18 ** STUDENT SOLUTION CONTINUED: The given clock time was 46 seconds, and my depth prediction was 16.314 cm.**
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RESPONSE -->
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19:25:25 What clock time corresponds to the given depth (give depth also)?
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RESPONSE -->
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19:25:28 ** 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 -->
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19:25:54 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 -->
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19:26:02 ** 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 -->
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19:28:05 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (10,1.122194) (50,2.342635) (90,3.08151)
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19:28:08 ** STUDENT SOLUTION CONTINUED: (20, 2.118034) (50, 2.767767) (100, 3.5)**
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RESPONSE -->
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19:28:41 Give the first of your three equations.
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RESPONSE --> 100a^2 + 10b + c= 1.122194
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19:28:43 ** STUDENT SOLUTION CONTINUED: 400a + 20b + c = 2.118034**
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RESPONSE -->
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19:29:23 Give the second of your three equations.
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RESPONSE --> 2500a^2 + 50b + c= 2.342635
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19:29:25 ** STUDENT SOLUTION CONTINUED: 2500a + 50b + c = 2.767767 **
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RESPONSE -->
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19:29:51 Give the third of your three equations.
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RESPONSE --> 8100a^2 +90b +c=3.08151
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19:29:54 ** STUDENT SOLUTION CONTINUED: 10,000a + 100b + c = 3.5 **
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RESPONSE -->
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19:30:44 Give the first of the equations you got when you eliminated c.
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RESPONSE --> eq3-eq1 8000a^2 + 80b = 1.959316
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19:30:47 ** STUDENT SOLUTION CONTINUED: 7500a + 50b = .732233. **
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RESPONSE -->
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19:31:21 Give the second of the equations you got when you eliminated c.
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RESPONSE --> 5600a^2 + 40b = .738875
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19:31:23 ** STUDENT SOLUTION CONTINUED: Subracting the first equation from the third I go 9600a + 80b = 1.381966 **
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RESPONSE -->
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19:32:04 Explain how you solved for one of the variables.
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RESPONSE --> multiplied the 2nd equation by -2 to eliminate the b variable.
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19:32:07 ** 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 -->
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19:32:49 What values did you get for a and b?
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RESPONSE --> a= .000148566 b= .024489243
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19:32:54 ** STUDENT SOLUTION CONTINUED: a = -.0000876638 b = .01727 **
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RESPONSE -->
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19:33:12 What did you then get for c?
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RESPONSE --> c= .877299365
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19:33:14 ** STUDENT SOLUTION CONTINUED: c = 1.773. **
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RESPONSE -->
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19:34:26 What is your function model?
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RESPONSE --> y= .000148566t^2 + .024489243t + .877299365
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19:34:28 ** STUDENT ANSWER: y = (0) x^2 + (.01727)x + 1.773 **
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RESPONSE -->
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19:34:56 What is your percent-of-review prediction for the given range of grades (give grade range also)?
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RESPONSE --> I am not able to solve this.
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19:36:17 ** 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 -->
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19:39:29 What grade average corresponds to the given percent of review (give grade average also)?
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RESPONSE --> 1.073308391
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19:41:13 ** 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 --> I messed up by using .8 for 80% instead of 80.
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19:44:59 How well does your model fit the data (support your answer)?
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RESPONSE --> the largest deviation seems to be at 100%
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19:45:02 ** 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 -->
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19:59:32 illumination vs. distance Give your data in the form of illumination vs. distance ordered pairs.
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RESPONSE --> (3,115.5556) (6,28.88889) (9,12,83951)
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19:59:36 ** 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 -->
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20:00:34 What three points on your graph did you use as a basis for your model?
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RESPONSE --> (3,115.5556) (6,28.88889) (9,12.83951)
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20:00:37 ** STUDENT SOLUTION CONTINUED: (2, 264.4411) (4, 61.01488) (8, 16.27232) **
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RESPONSE -->
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20:01:09 Give the first of your three equations.
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RESPONSE --> 9a+3b+c=115.5556
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20:01:11 ** STUDENT SOLUTION CONTINUED: 4a + 2b + c = 264.4411**
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RESPONSE -->
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20:01:38 Give the second of your three equations.
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RESPONSE --> 36a+6b+c=28.88889
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20:01:39 ** STUDENT SOLUTION CONTINUED: 16a + 4b + c = 61.01488**
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RESPONSE -->
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20:02:02 Give the third of your three equations.
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RESPONSE --> 81a+9b+c=12.83951
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20:02:04 ** STUDENT SOLUTION CONTINUED: 64a + 8b + c = 16.27232**
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RESPONSE -->
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20:02:42 Give the first of the equations you got when you eliminated c.
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RESPONSE --> eq3-eq1 72a+6b=-102.7161
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20:02:45 ** STUDENT SOLUTION CONTINUED: 48a + 4b = -44.74256**
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RESPONSE -->
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20:03:25 Give the second of the equations you got when you eliminated c.
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RESPONSE --> eq3-eq2 45a+3b=-16.04938
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20:03:27 ** STUDENT SOLUTION CONTINUED: 60a + 6b = -248.16878**
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RESPONSE -->
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20:03:56 Explain how you solved for one of the variables.
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RESPONSE --> multiplied 2nd equation by -2
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20:03:58 ** 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 -->
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20:04:54 What values did you get for a and b?
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RESPONSE --> a= 3.92318556 b= -64.197
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20:04:57 ** STUDENT SOLUTION CONTINUED: a = 15.088, b = -192.24 **
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RESPONSE -->
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20:05:17 What did you then get for c?
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RESPONSE --> c= 272.83966
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20:05:20 ** STUDENT SOLUTION CONTINUED: c = 588.5691**
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RESPONSE -->
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20:06:07 What is your function model?
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RESPONSE --> y= 3.92318556t^2 -64.197t +272.83966
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20:06:09 ** STUDENT SOLUTION CONTINUED: y = (15.088) x^2 - (192.24)x + 588.5691 **
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RESPONSE -->
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20:06:54 What is your illumination prediction for the given distance (give distance also)?
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RESPONSE --> 1.6AU= 180.167815
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20:06:58 ** 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 -->
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20:07:08 What distances correspond to the given illumination range (give illumination range also)?
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RESPONSE -->
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20:07:09 ** 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 --> `sc2
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20:10:57 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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20:11:07 ** 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 -->
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20:14:21 0.2.22 (was 0.2.14 solve abs(3x+1) >=4
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RESPONSE --> 1
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20:14:32 ** 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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20:15:05 ** 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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20:17:40 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 --> -3
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20:17:49 ** 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 -->
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20:18:17 0.2.5 (was 0.2.23 describe [-2,2 ]
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RESPONSE -->
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20:18:28 ** 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 -->
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20:19:22 0.2.10 (was 0.2.28 describe [-7,-1]
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RESPONSE --> -7+-1= -4 abs value= 8
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20:19:28 ** 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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20:19:38 0.2.12 (was 0.2.30) describe (-infinity, 20) U (24, infinity)
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RESPONSE -->
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20:19:53 ** 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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20:20:07 0.2.42 (was 0.2 #36 collies, interval abs( (w-57.5)/7.5 ) < 1
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RESPONSE -->
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20:20:12 ** 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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20:22:06 0.2.40 (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 --> 31 1/8
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20:22:21 ** 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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