course mth 163
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01:55:47 query the family of linear functions, Problem 2. Describe the graphs of y = A f(x) for A = -.3 and A = 1.3 and compare; explain the comparison.
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RESPONSE --> the a=-.3 passes thru the x axis at o as well as passing the y axis at 0 the graph for a=1.3 passes thru both axis at o and it is 1 point higher than the preceding graph
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01:55:53 ** For the basic linear function f(x) = x the A = -.3 graph is obtained by vertically stretching the y = x function by factor -.3, resulting in a straight line thru the origin with slope -.3, basic points (0,0) and (1, -.3), and the A = 1.3 graph is obtained by vertically stretching the y = x function by factor 1.3, resulting in is a straight line thru the origin with slope 1.3, basic points (0,0) and (1, 1.3). **
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RESPONSE -->
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01:57:37 describe the graphs of y = f(x) + c for c = .3 and c = -2.7 and compare; explain the comparison.
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RESPONSE --> the graph for -2.7 is very low on the axis crossing 2 points vertically as with the graph for .3 crosses at 0,0
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01:57:41 ** The graphs will have slopes identical to that of the original function, but their y intercepts will vary from -2.7 to .3. **
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01:58:13 query problem 4. linear function y = f(x) = -1.77 x - 3.87 What are your symbolic expressions, using x1 and x2, for the corresponding y coordinates y1 and y2.
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RESPONSE --> (x1,y1)/x2-y2
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02:00:09 ** y1 = f(x1) = -1.77 x1 - 3.87 y2 = f(x2) = -1.77 x2 - 3.87. `dy = y2 - y1 = -1.77 x2 - 3.87 - ( -1.77 x1 - 3.87) = -1.77 x2 + 1.77 x1 - 3.87 + 3.87 = -1.77 ( x2 = x1). Thus slope = `dy / `dx = -1.77 (x2 - x1) / (x2 - x1) = -1.77. This is the slope of the straight line, showing that these symbolic calculations are consistent. **
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RESPONSE --> ok
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02:09:32 query problem 5. graphs of families for y = mx + b. Describe your graph of the family: m = 2, b varies from -3 to 3 by step 1.
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RESPONSE --> my graph for some reason did not go straight as i thought that it would
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02:10:13 ** The graphs will all have slope 2 and will pass thru the y axis between y = -3 and y = 3. The family will consist of all such graphs. **
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RESPONSE -->
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02:17:01 ** This is of the form y = mx + b with b= 1. So the y intercept is (0, 1). The point 1 unit to the right is (1, 1.5). The x-intercept occurs when y = 0, which implies .5 x + 1 = 0 with solution x = -2, so the x-intercept is (-2, 0). **
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RESPONSE --> y intercept form is 0,1 x intercept is -2,0 unit to right is 1,1.5
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02:17:36 query problem 6. three basic points graph of y = .5 x + 1 What are your three basic points?
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RESPONSE --> y intercept is 0,1 x is -2,0 unit to right is 1,1.5
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02:17:45 ** The y intercept occurs when x = 0, at which point we have y = .5 * 0 + 1 = 1. So one basic point is (0, 1). The point 1 unit to the right of the y axis occurs at x = 1, where we get y = .5 * 1 + 1 = 1.5 to give us the second basic point (1, 1.5) }The third point, which is not really necessary, is the x intercept, which occurs when y = 0. This gives us the equation 0 = .5 x + 1, with solution x = -2. So the third basic point is (-2, 0). **
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02:31:30 query problem 7. simple pendulum force vs. displacement What are your two points and what line do you get from the two points?
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RESPONSE --> .77,2.9 1.22,5.2
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02:32:29 STUDENT RESPONSE: The two points are (1.1, .21) and (2.0, .54). These points give us the two simultaneous equations .21- m(1.1) + b .54= m(2.0) +b. If we solve for m and b we will get our y = mx + b form. INSTRUCTOR COMMENT: I believe those are data points. I doubt if the best-fit line goes exactly through two data points. In the future you should use points on your sketched line, not data points. However, we'll see how the rest of your solution goes based on these points. **
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RESPONSE -->
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02:32:51 what equation do you get from the slope and y-intercept?
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RESPONSE --> i dont think i got one
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02:34:26 STUDENT RESPONSE: b= .21 m=.19 INSTRUCTOR COMMENT: ** b would be the y intercept, which is not .21 since y = .21 when x = 1.1 and the slope is nonzero. If you solve the two equations above for m and b you obtain m = .367 and b = -.193. This gives you equation y = mx + b or y = .367 x - .193. **
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02:35:18 what is your linear regression model?
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RESPONSE --> ?
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02:35:31 ** Your linear regression model would be obtained using a graphing calculator or DERIVE. As a distance student you are not required to use these tools but you should be aware that they exist and you may need to use them in other courses. **
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RESPONSE -->
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02:37:01 What force would be required to hold the pendulum 47 centimeters from its equilibrium position? what equation did you solve to obtain this result?
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RESPONSE --> force=m*47+b the force function
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02:38:14 ** If your model is y = .367 x - .193, with y = force and x= number of cm from equilibrium, then we have x = 47 and we get force = y = .367 * 47 - .193 = 17 approx. The force would be 17 force units. **
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RESPONSE -->
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02:39:18 Why would it not make sense to ask what force would be necessary to hold the pendulum 80 meters from its equilibrium position? what equation did you solve to obtain this result?
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RESPONSE --> the data is not set up in meters and the measurements are unrealistic force=y=.367*80-.193
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02:39:36 STUDENT RESPONSE: I used the equation f= .10*47+.21 and got the answer 15.41 which would be to much force to push or pull INSTRUCTOR COMMENT: ** The problem is that you can't hold a pendulum further at a distance greater than its length from its equilibrium point--the string isn't long enough. **
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RESPONSE -->
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02:40:22 How far could you hold the pendulum from its equilibrium position using a string with a breaking strength of 25 pounds? what equation did you solve to obtain this result?
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RESPONSE --> ??????????
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02:40:50 ** Using the model y = .367 * x - .193 with y = force = 25 lbs we get the equation 25 = .367 x - .193, which we solve to obtain x = 69 (approx.). Note that this displacement is also unrealistic for this pendulum. **
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RESPONSE --> ok i am getting it finally
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02:42:53 What is the average rate of change associated with this model? Explain this average rate in common-sense terms.
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RESPONSE --> -1.90 i took m and divided it by b to obtain the average i got this my using points plotted in the preceeding graph
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02:43:18 ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium we can use any two (x, y) points to get the rate of change. In all cases we will get rate of change = change in y / change in x = .367. The change in y is the change in the force, while the change in x is the change in position. The rate of change therefore tells us how much the force changes per unit of change in position (e.g., the force increases by 15 pounds for every inch of displacement). **
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RESPONSE -->
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02:44:37 What is the average slope associated with this model? Explain this average slope in common-sense terms.
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RESPONSE --> the average slope would be the same as the preceeding average rate of change they are both represented by change in y divided by the change in x and the slope is constant ofr a straight line
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02:45:15 ** Using the model y = .367 * x - .193 with y = force and x = displacement from equilibrium the average slope is .367. **
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RESPONSE --> ok
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02:49:42 As you gradually pull the pendulum from a point 30 centimeters from its equilibrium position to a point 80 centimeters from its equilibrium position, what average force must you exert?
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RESPONSE --> if the pendulum is pulled by 30 cm to 80 cm the average equilibruim exerted would be .\0.3625
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02:50:44 ** if it was possible to pull the pendulum back this far and if the model applies you will get Force at 30 cm: y = .367 * 30 - .193 = 10.8 approx. and Force at 80 cm: y = .367 * 80 - .193 = 29 approx. so that ave force between 30 cm and 80 cm is therefore (10.8 + 29) / 2 = 20 approx.. **
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RESPONSE --> ok i took it a step too far
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02:53:55 query problem 8. flow range What is the linear function range(time)?
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RESPONSE --> ui
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02:56:45 ** STUDENT RESPONSE: I obtained model one by drawing a line through the data points and picking two points on the line and finding the slope between them. I then substituted this value for m and used one of my data points on my line for the x and y value and solved for b. the line I got was range(t) = -.95t + 112.38. y = -16/15x + 98 INSTRUCTOR COMMENT: This looks like a good model. According to the instructions it should however be expressed as range(time) = -16/15 * time + 98. **
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RESPONSE -->
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02:58:03 What is the significance of the average rate of change? Explain this average rate in common-sense terms.
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RESPONSE --> this should show the change in time and the rate at which the stream descends out of the uniform cylinder
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02:58:10 ** the average rate of change is change in range / change in clock time. The average rate of change indicates the average rate at which range in cm is changing with respect to clock time in sec, i.e., the average number of cm / sec at which the range changes. Thus the average rate tells us how fast, on the average, the range changes. **
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RESPONSE -->
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02:58:24 What is the average slope associated with this model? Explain this average slope in common-sense terms.
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RESPONSE --> -16/15
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02:59:00 ** it's the average rate at which the range of the flow changes--the average rate at which the position of the end of the stream changes. It's the speed with which the point where the stream reaches the ground moves across the ground. **
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03:04:49 query problem 9. If your total wealth at clock time t = 0 hours is $3956, and you earn $8/hour for the next 10 hours, then what is your total wealth function totalWealth( t ), where t is time in hours?
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RESPONSE -->
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03:06:05 ** Total wealth has to be expressed in terms of t. A graph of total wealth vs. t would have y intercept 3956, since that is the t = 0 value, and slope 8, since slope represents change in total wealth / change in t, i.e., the number of dollars per hour. A graph with y-intercept b and slope m has equation y = m t + b. Thus we have totalWealth(t) = 8 * t + 3956 . **
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RESPONSE -->
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03:07:02 At what clock time will your total wealth reach $4000? what equation did you solve to obtain this result?
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RESPONSE --> total wealth=8*t+4000 13 hours
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03:08:16 STUDENT RESPONSE: To find the clock time when my total wealth will reach 4000 I solved the equation totalWealth(t) = 4000. The value I got when I solved for t was t = 5.5 hours. 4.4 hours needed to reach 4000 4000 = 10x + 3956 INSTRUCTOR COMMENT: Almost right. You should solve 4000 = 8 x + 3956, obtaining 5.5 hours. This is equivalent to solving totalWealth(t) = 4000 = 8 t + 3956, which is the more meaningful form of the relationship. **
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RESPONSE -->
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03:08:44 What is the meaning of the slope of your graph?
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RESPONSE --> the average wage and the slope of the time it takes to reach the questioned amount of wealth
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03:09:08 GOOD STUDENT RESPONSE: The slope of the graph shows the steady rate at which money is earned on an hourly basis. It shows a steady increase in wealth.
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03:11:44 query problem 10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300. What linear function numberSold(price) describes this situation?
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RESPONSE --> i think tha t30=200*28+300
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03:11:48 query problem 10. Experience shows that when a certain widget is sold for $30, a certain store can expect to sell 200 widgets per week, while a selling price of $28 increases the number sold to 300. What linear function numberSold(price) describes this situation?
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RESPONSE -->
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03:13:01 If you make a graph of y = numberSold vs. x = price you have graph points (30, 200) and (28, 300). You need the equation of the straight line through these points. You plug these coordinates into the form y = m x + b and solve for m and b. Or you can use another method. Whatever method you use you get y = -50 x + 1700. Then to put this into the notation of the problem you write numberSold(price) instead of y and price instead of x. You end up with the equation numberSold(price) = -50 * price + 1700. **
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RESPONSE -->
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03:13:33 If the store must meet a quota by selling 220 units per week, what price should they set? what equation did you solve to obtain this result?
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RESPONSE --> number sold(price)=-50*price+220
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03:14:33 ** If the variables are y and x, you know y so you can solve for x. For the function numberSold(price) = -50 * price + 1700 you substitute 220 for numbersold(price) and solve for price. You get the equation 220 = -50 * price + 1700 which you can solve to get price = 30, approx. **
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03:14:55 If each widget costs the store $25, then how much total profit will be expected from selling prices of $28, $29 and $30? what equation did you solve to obtain this result?
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RESPONSE -->
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03:16:49 STUDENT RESPONSE: If each widget costs the store $25, then they should expect to earn a profit of 300 dollars from a selling price of $28, 250 dollars from a price of $29 and 200 dollars from a price of $30. To find this I solved the equations numberSold(28); numberSold(29), and numberSold(30). Solving for y after putting the price values in for p. They will sell 300, 250 and 200 widgets, respectively (found by solving the given equation). To get the total profit you have to multiply the number of widgets by the profit per widget. At $28 the profit per widgit is $3 and the total profit is $3 * 300 = $900; at $30 the profit per widgit is $5 and 200 are sold for profit $1000; at $29 what happens? **
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03:19:56 query problem 11. quadratic function depth(t) = .01 t^2 - 2t + 100 representing water depth vs. What is the equation of the straight line connecting the t = 20 point of the graph to the t = 60 point?
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RESPONSE --> 64=mx+b 16=mx+b
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03:21:38 ** The t = 20 point is (20,64) and the t = 60 point is (60, 16), so the slope is (-48 / 20) = -1.2. This can be plugged into the form y = m t + b to get y = -1.2 t + b. Then plugging in the x and y coordinates of either point you get b = 88. y = -1.2 t + 88 **
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03:27:09 query problem 13. quadratic depth function y = depth(t) = .01 t^2 - 2t + 100. What is `dy / `dt based on the two time values t = 30 sec and t = 40 sec.
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RESPONSE --> -0.7692
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03:27:58 ** For t = 30 we have y = 49 and for t = 40 we have y = 36. The slope between (30, 49) and (40,36) is (36 - 49) / (40 - 30) = -1.3. This tells you that the depth is changing at an average rate of -1.3 cm / sec. **
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03:30:54 what is `dy / `dt based on t = 30 sec and t = 31 sec.
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RESPONSE --> -1.39
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03:30:59 ** Based on t = 30 and t = 31 the value for `dy / `dt is -1.39, following the same steps as before **
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RESPONSE -->
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03:32:51 what is `dy / `dt based on t = 30 sec and t = 30.1 sec.
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RESPONSE --> -1.4
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03:33:09 ** STUDENT RESPONSE: The value for 'dy / `dt based on t = 30 sec and t = 30.1 sec is -1.4 INSTRUCTOR COMMENT: ** Right if you round off the answer. However the answer shouldn't be rounded off. Since you are looking at a progression of numbers (-1.3, -1.39, and this one) and the differences in these numbers get smaller and smaller, you have to use a precision that will always show you the difference. Exact values are feasible here and shoud be used. I believe that this one comes out to -1.399. **
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03:33:30 What do you think you would get for `dy / `dt if you continued this process?
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RESPONSE --> we would get the #2 eventually
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03:33:56 STUDENT RESPONSE: An even more and more accurate slope value. I don't think it would have to continue to decrease. INSTRUCTOR COMMENT **If you look at the sequence -1.3, -1.39, -1.399, ..., what do you think happens? It should be apparent that the limiting value is -1.4 **
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03:34:28 What does the linear function tell you?
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RESPONSE --> the slope as well as the decrease in depth in regards to the quadratic equation
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03:36:01 ** The function tells you that at any clock time t the rate of depth change is given by the function .02 t - 2. For t = 30, for example, this gives us .02 * 30 - 2 = -1.4, which is the rate we obtained from the sequence of calculations above. **
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03:36:51 query problem 14. linear function y = f(x) = .37 x + 8.09
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RESPONSE -->
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03:37:08 what are the first five terms of the basic sequence {f(n), n = 1, 2, 3, ...} for this function.
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RESPONSE --> 1,2,3,4,5
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03:37:33 ** The first five terms are 8.46, 8.83, 9.2, 9.57, and 9.94 **
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03:38:09 What is the pattern of these numbers?
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RESPONSE --> the increase gradually
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03:38:20 ** These numbers increase by .37 at each interval. **
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RESPONSE -->
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03:39:04 If you didn't know the equation for the function, how would you go about finding the 100th member of the sequence? How can you tell your method is valid?
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RESPONSE --> sub 100 in the equation. there is a checks and balance function
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03:39:46 ** You could find the 100th member by noting that you have 99 jumps between the first number and the 100 th, each jump being .37. Multiplying 99 times .37 and then adding the result to the 'starting value' (8.46). STUDENT RESPONSE: simply put 100 as the x in the formula .37x +8.09 INSTRUCTOR COMMENT: That's what you do if you have the equation. Given just the numbers you could find the 100th member by multiplying 99 times .37 and then adding the result to the first value 8.46. **
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03:39:53 for quadratic function y = g(x) = .01 x^2 - 2x + 100 what are the first five terms of the basic sequence {g(n), n = 1, 2, 3, ...}?
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RESPONSE -->
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03:40:20 ** We have g(1) = .01 * 1^2 - 2 * 1 + 100 = 98.01 g(2) = .01 * 2^2 - 2 * 2 + 100 = 96.04, etc. The first 5 terms are therefore {98.01, 96.04, 94.09, 92.16, 90.25}
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RESPONSE -->
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03:40:46 What is the pattern of these numbers?
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RESPONSE --> they decrease by 1.97 ate each interval
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03:41:05 ** The changes in these numbers are -1.97, -1.95, -1.93, -1.91. With each interval of x, the change in y is .02 greater than for the previous interval. **
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RESPONSE -->
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03:41:40 If you didn't know the equation for the function, how would you go about finding the next three members of the sequence?
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RESPONSE --> multiplying each needed # by .02
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03:41:58 ** According to the pattern established above, the next three changes are -1.89, -1.87, -1.85. This gives us g(6) = g(5) - 1.89, g(7) = g(6) - 1.87, g(7) = g(6) - 1.85. **
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RESPONSE -->
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03:42:09 How can you verify that your method is valid?
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RESPONSE --> plug your numbers in
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03:42:17 ** You can verify the result using the original formula; if you evaluate it at 5, 6 and 7 it should confirm your results. That's the best answer that can be given at this point. You should understand, though that even if you verified it for the first million terms, that wouldn't really prove it (who knows what might happen at the ten millionth term, or whatever). It turns out that to prove it would require calculus or the equivalent. **
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03:42:41 query problem 15. The difference equation a(n+1) = a(n) + .4, a(1) = 5 If you substitute n = 1 into a(n+1) = a(n) + .4, how do you determine a(2)?
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RESPONSE --> plug 2 in for a
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03:43:30 ** You get a(1+1) = a(1) + .4, or a(2) = a(1) + .4. Knowing a(1) = 5 you get a(2) = 5.4. **
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03:43:37 If you substitute n = 2 into a(n+1) = a(n) + .4 how do you determine a(3)?
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RESPONSE -->
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03:44:18 ** You have to do the substitution. You get a(2+1) = a(2) + .4, or since 2 + 1 = 3, a(3) = a(2) + .4 Then knowing a(2) = 5.4 you get a(3) = 5.4 + .4 = 5.8. **
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03:44:21 If you substitute n = 3 into a(n+1) = a(n) + .4, how do you determine a(4)?
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RESPONSE -->
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03:44:30 ** We get a(4) = a(3) +.4 = 5.8 + .4 = 6.2 **
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03:44:33 What is a(100)?
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03:44:45 ** a(100) would be equal to a(1) plus 99 jumps of .4, or 5 + 99*.4 = 44.6. **
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03:44:51 query problem 17. difference equation a(n+1) = a(n) + 2 n, a(1) = 4. What is the pattern of the sequence?
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03:45:06 ** You get a(2) = a(1) + 2 * 1 = 4 + 2 = 6, then a(3) = a(2) + 2 * 3 = 6 + 6 = 12 then a(4) = a(3) + 2 * 4 = 12 + 8 = 20; etc. The sequence is 6, 12, 20, 30, 42, ... . **
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03:45:29 What kind of function do you think a(n) is (e.g., linear, quadratic, exponential, etc.)?
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RESPONSE --> i think that a is a linear equation
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03:45:33 ** The differences of the sequence are 6, 8, 10, 12, . . .. The difference change by the same amount each time, which is a property of quadratic functions. **
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