course Math 271 I know I am behind, but it is taking me a lot longer than i thought and I am going to get caught up this week, I am not taking this lightly, it is really hard for me and I am determined that I am going to work on it until I get it. I work 32 hrs and have 3 other classes, and I am scheduling a lot more time for this class. ??~??????\?????Student Name:
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22:24:33 Note that there are four questions in this assignment. `q001. Sketch a graph similar to that you constructed for the stock values, this time for the depth of the water vs. clock time (depths 80, 40, 20 at clock times 10, 40, 90). Your first point, for example, will be (10, 80). Connect these points with straight lines and determine the slopes of the lines.
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RESPONSE --> the slope between the point of (10, 80) and the point of (40, 40) is -1.33 the slope between the point of (40, 40) and the point of (90, 200) is -.4
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22:24:43 The three points are (10, 80), (40, 40) and (90, 20). From the first point to the second the rise is from 80 to 40, or -40, and the run is from 10 to 40, or 30. So the slope is -40 / 30 = -1.33. From the second point to the third the rise is from 40 to 20, or -20, and the run is from 40 to 90, or 50, so the slope is -20 / 50 = -.4. Click on 'Next Picture' to see graph.
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RESPONSE --> ok
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22:30:08 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?
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RESPONSE --> the slope represents the average rate of change? each slope IS the average rate of change for that time period, but not for the entire time. right?
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22:30:31 The slopes and the rates of change are numerically equal. For example between the second and third points the rise of -20 represents the -20 cm change in depth and the run of 50 represents the 50 seconds required to make this change, so the slope represents the -20 cm / (50 sec) average rate of change over the second time interval. We therefore see that slope represents average rate of change.
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RESPONSE --> ok
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17:35:14 `q003. To what extent do you think your graph with three points and straight line segments between them accurately depicts the detailed behavior of the water over the 80-second period of observation? How do you think the actual behavior of the system differs from that of the graph? How do you think the graph of the actual behavior of the system would differ from that of the graph you made?
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RESPONSE --> This graph just gives 3 points and only gives and overall view of the behavior. More points on the graph inbetween these three would give us a more accurate depiction. I would think that the actual graph would be more rounded. The decrease in the depth would be decreasing at a decreasing rate. Graph of the actual behavior would be better and more accurate.
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17:35:24 The straight line segments would indicate a constant rate of change of depth. It is fairly clear that as depth decreases, the rate of change of depth will decrease, so that the rate of change of depth will not be constant. The graph will therefore never be straight, but will be a curve which is decreasing at a decreasing rate.
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RESPONSE --> ok
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17:44:29 `q004. From the given information, do you think you can accurately infer the detailed behavior of the water depth over the 80-second period? Do you think you can infer the detailed behavior better than you could the values of the stocks? Why or why not?
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RESPONSE --> I can understand the detailed behavior of the water depth over the 80 second period because of the knowledge that the water depth decreases at a decreasing rate. The behavior of the water is more predictable than the value of the stocks. While the water's rate of change can be determined in close aproximation, the value of the stocks does not necessarily act in a predictable way. The stock value does not change at a steady rate or even a decreasing or increasing rate. Because of the volitility of the stocks, stock value fluctuates, and a true graph most likely won't be a smooth steady one. Most likely a stock graph would be in a zig-zag pattern.
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17:44:54 It will turn out that three data points will be sufficient to infer the detailed behavior, provided the data are accurate. However you might or might not be aware of that at this point, so you could draw either conclusion. However it should be clear that the behavior of the water depth is much more predictable than the behavior of the stock market. We don't know on a given day whether the market will go up or down, but we do know that if we shoot a hole in the bottom of a full bucket the water level will decrease, and we expect that identical holes in identical buckets should result in the same depth vs. clock time behavior.
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RESPONSE --> ok
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q???o??ue?z??s???assignment #003 ?{?M??????????Applied Calculus I 09-27-2006
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17:51:15 0.3.24 (was 0.3.24 simplify z^-3 (3z^4)
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RESPONSE --> z^-3 (3z^4) ok, order of multiplication doesn't matter so 3 ( z^4-3) (add exponents) so this equals 3z
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17:51:20 ** z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. **
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RESPONSE --> ok
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17:58:17 0.3.30 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3
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RESPONSE --> (12 s^2 / (9s) ) ^ 3 I first reduced the fraction by dividing by 3s to get (4s / 3 ) ^ 3 simplify by raising the inside to the third power to get (64 s^3) / 27
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17:58:45 ** Starting with (12 s^2 / (9s) ) ^ 3 we simplify inside parentheses to get ( 4 s / 3) ^ 3, which is equal to 4^3 * s^3 / 3^3 = 64 s^3 / 27 It is possible to expand the cube without first simplifying inside, but the subsequent simplification is a little more messy and error-prone; however done correctly it gives the same result. It's best to simplify inside the parentheses first. **
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RESPONSE --> great
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18:45:59 0.3.38 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3)
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RESPONSE --> ( (3x^2 y^3)^4) ^ (1/3) First I raise the inside ()'s to the 4th power to get [(3^4) (x^8( (y^7)] ^(1/3) which = [(81) (x^8) (y^12)] ^(1/3) or [(3)(27)(x^6)(X^2)(y^12)] ^(1/3) which equals 3 x^2 y^3 (3x^2)^(1/3) or (3)(x^2)(y^3) (3x^2)^(1/3) next problem: (54 x^7) ^ (1/3) simplify [(2)(27)(x^6)(x)]^(1/3) simplify again 3x^2(2x)^(1/3)
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18:50:01 09-27-2006 18:50:01 ** To simplify (54 x^7)^(1/3) you have to find the maximum factor inside the parentheses which a perfect 3d power. First factor 54 into its prime factors: 54 = 2 * 27 = 2 * 3 * 3 * 3 = 2 * 3^3. Now we have (2 * 3^3 * x^7)^(1/3). 3^3 and x^6 are both perfect 3d Powers. So we factor 3^3 * x^6 out of the expression in parentheses to get ( (3^3 * x^6) * 2x ) ^(1/3). This is equal to (3^3 * x^6)^(1/3) * (2x)^(1/3). Simplifying the perfect cube we end up with 3 x^2 ( 2x ) ^ (1/3) For the second expression: The largest cube contained in 54 is 3^3 = 27 and the largest cube contained in x^7 is x^6. Thus you factor out what's left, which is 2x. Factoring 2x out of (54 x^7)^(1/3) gives you 2x ( 27 x^6) so your expression becomes [ 2x ( 27 x^6) ] ^(1/3) = (2x)^(1/3) * [ 27 x^6 ] ^(1/3) = (2x)^(1/3) * [ (27)^(1/3) (x^6)^(1/3) ] = (2x)^(1/3) * 3 x^2, which in more traditional order is 3 x^2 ( 2x)^(1/3). **
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NOTES -------> Apparently I got the one but I absolutely got lost on the other one - where did the Y value go? I should get the same anwer for both problems? Im really confused....
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18:53:36 0.3.58 (was 0.3.54 factor P(1+r) from expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ...
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RESPONSE --> if you factor out the P(1+r) from this expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ....... wouldn't you divide each by P(1+r) ? that would leave 1 for the first then P(1+r) for the next and P(1+r)^2 for the second and so on it would look like: 1 + P(1+r) + P(1+r)^2 + P(1+r)^3 + .......
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18:56:28 ** Few students get this one. If you didn't you've got a lot of company; if you did congratulations. It's important to understand how this problem illustrates the essence of factoring. It's important also because expressions of this form occur throughout calculus. Factor out P * (1 + r). Divide each term by P ( 1 + r), and your result is P (1 + r) * your quotient. Your quotient would be 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... . The factored form would therefore be P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... ]. You can verify that this is identical to the original expression if you multiply it back out. Analogy with different exponents and only three terms: A x^3 + A x^4 + A x^5 can be divided by A x^2 to give quotient x + x^2 + x^3, so the factored expression is A ( x + x^2 + x^3). **
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RESPONSE --> I forgot the p cancels out on each one as well had the basic idea though
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course Math 271 I know I am behind, but it is taking me a lot longer than i thought and I am going to get caught up this week, I am not taking this lightly, it is really hard for me and I am determined that I am going to work on it until I get it. I work 32 hrs and have 3 other classes, and I am scheduling a lot more time for this class. ??~??????\?????Student Name:
......!!!!!!!!...................................
22:24:33 Note that there are four questions in this assignment. `q001. Sketch a graph similar to that you constructed for the stock values, this time for the depth of the water vs. clock time (depths 80, 40, 20 at clock times 10, 40, 90). Your first point, for example, will be (10, 80). Connect these points with straight lines and determine the slopes of the lines.
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RESPONSE --> the slope between the point of (10, 80) and the point of (40, 40) is -1.33 the slope between the point of (40, 40) and the point of (90, 200) is -.4
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22:24:43 The three points are (10, 80), (40, 40) and (90, 20). From the first point to the second the rise is from 80 to 40, or -40, and the run is from 10 to 40, or 30. So the slope is -40 / 30 = -1.33. From the second point to the third the rise is from 40 to 20, or -20, and the run is from 40 to 90, or 50, so the slope is -20 / 50 = -.4. Click on 'Next Picture' to see graph.
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RESPONSE --> ok
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22:30:08 `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change?
......!!!!!!!!...................................
RESPONSE --> the slope represents the average rate of change? each slope IS the average rate of change for that time period, but not for the entire time. right?
.................................................
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22:30:31 The slopes and the rates of change are numerically equal. For example between the second and third points the rise of -20 represents the -20 cm change in depth and the run of 50 represents the 50 seconds required to make this change, so the slope represents the -20 cm / (50 sec) average rate of change over the second time interval. We therefore see that slope represents average rate of change.
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RESPONSE --> ok
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17:35:14 `q003. To what extent do you think your graph with three points and straight line segments between them accurately depicts the detailed behavior of the water over the 80-second period of observation? How do you think the actual behavior of the system differs from that of the graph? How do you think the graph of the actual behavior of the system would differ from that of the graph you made?
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RESPONSE --> This graph just gives 3 points and only gives and overall view of the behavior. More points on the graph inbetween these three would give us a more accurate depiction. I would think that the actual graph would be more rounded. The decrease in the depth would be decreasing at a decreasing rate. Graph of the actual behavior would be better and more accurate.
.................................................
......!!!!!!!!...................................
17:35:24 The straight line segments would indicate a constant rate of change of depth. It is fairly clear that as depth decreases, the rate of change of depth will decrease, so that the rate of change of depth will not be constant. The graph will therefore never be straight, but will be a curve which is decreasing at a decreasing rate.
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RESPONSE --> ok
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17:44:29 `q004. From the given information, do you think you can accurately infer the detailed behavior of the water depth over the 80-second period? Do you think you can infer the detailed behavior better than you could the values of the stocks? Why or why not?
......!!!!!!!!...................................
RESPONSE --> I can understand the detailed behavior of the water depth over the 80 second period because of the knowledge that the water depth decreases at a decreasing rate. The behavior of the water is more predictable than the value of the stocks. While the water's rate of change can be determined in close aproximation, the value of the stocks does not necessarily act in a predictable way. The stock value does not change at a steady rate or even a decreasing or increasing rate. Because of the volitility of the stocks, stock value fluctuates, and a true graph most likely won't be a smooth steady one. Most likely a stock graph would be in a zig-zag pattern.
.................................................
......!!!!!!!!...................................
17:44:54 It will turn out that three data points will be sufficient to infer the detailed behavior, provided the data are accurate. However you might or might not be aware of that at this point, so you could draw either conclusion. However it should be clear that the behavior of the water depth is much more predictable than the behavior of the stock market. We don't know on a given day whether the market will go up or down, but we do know that if we shoot a hole in the bottom of a full bucket the water level will decrease, and we expect that identical holes in identical buckets should result in the same depth vs. clock time behavior.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
q???o??ue?z??s???assignment #003 ?{?M??????????Applied Calculus I 09-27-2006
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17:51:15 0.3.24 (was 0.3.24 simplify z^-3 (3z^4)
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RESPONSE --> z^-3 (3z^4) ok, order of multiplication doesn't matter so 3 ( z^4-3) (add exponents) so this equals 3z
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17:51:20 ** z^-3 ( 3 z^4) = 3 * z^-3 * z^4 = 3 * z^(4-3) = 3 z. **
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RESPONSE --> ok
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17:58:17 0.3.30 (was 0.3.30 simplify(12 s^2 / (9s) ) ^ 3
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RESPONSE --> (12 s^2 / (9s) ) ^ 3 I first reduced the fraction by dividing by 3s to get (4s / 3 ) ^ 3 simplify by raising the inside to the third power to get (64 s^3) / 27
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17:58:45 ** Starting with (12 s^2 / (9s) ) ^ 3 we simplify inside parentheses to get ( 4 s / 3) ^ 3, which is equal to 4^3 * s^3 / 3^3 = 64 s^3 / 27 It is possible to expand the cube without first simplifying inside, but the subsequent simplification is a little more messy and error-prone; however done correctly it gives the same result. It's best to simplify inside the parentheses first. **
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RESPONSE --> great
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18:45:59 0.3.38 (was 0.3.38 simplify ( (3x^2 y^3)^4) ^ (1/3) and (54 x^7) ^ (1/3)
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RESPONSE --> ( (3x^2 y^3)^4) ^ (1/3) First I raise the inside ()'s to the 4th power to get [(3^4) (x^8( (y^7)] ^(1/3)
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18:50:01 09-27-2006 18:50:01 ** To simplify (54 x^7)^(1/3) you have to find the maximum factor inside the parentheses which a perfect 3d power. First factor 54 into its prime factors: 54 = 2 * 27 = 2 * 3 * 3 * 3 = 2 * 3^3. Now we have (2 * 3^3 * x^7)^(1/3). 3^3 and x^6 are both perfect 3d Powers. So we factor 3^3 * x^6 out of the expression in parentheses to get ( (3^3 * x^6) * 2x ) ^(1/3). This is equal to (3^3 * x^6)^(1/3) * (2x)^(1/3). Simplifying the perfect cube we end up with 3 x^2 ( 2x ) ^ (1/3) For the second expression: The largest cube contained in 54 is 3^3 = 27 and the largest cube contained in x^7 is x^6. Thus you factor out what's left, which is 2x. Factoring 2x out of (54 x^7)^(1/3) gives you 2x ( 27 x^6) so your expression becomes [ 2x ( 27 x^6) ] ^(1/3) = (2x)^(1/3) * [ 27 x^6 ] ^(1/3) = (2x)^(1/3) * [ (27)^(1/3) (x^6)^(1/3) ] = (2x)^(1/3) * 3 x^2, which in more traditional order is 3 x^2 ( 2x)^(1/3). **
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NOTES -------> Apparently I got the one but I absolutely got lost on the other one - where did the Y value go? I should get the same anwer for both problems? Im really confused....
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18:53:36 0.3.58 (was 0.3.54 factor P(1+r) from expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ...
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RESPONSE --> if you factor out the P(1+r) from this expression P(1+r) + P(1+r)^2 + P(1+r)^3 + ....... wouldn't you divide each by P(1+r) ? that would leave 1 for the first then P(1+r) for the next and P(1+r)^2 for the second and so on it would look like: 1 + P(1+r) + P(1+r)^2 + P(1+r)^3 + .......
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18:56:28 ** Few students get this one. If you didn't you've got a lot of company; if you did congratulations. It's important to understand how this problem illustrates the essence of factoring. It's important also because expressions of this form occur throughout calculus. Factor out P * (1 + r). Divide each term by P ( 1 + r), and your result is P (1 + r) * your quotient. Your quotient would be 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... . The factored form would therefore be P(1+r) [ 1 + (1+r) + (1+r)^2 + (1+r)^3 + ... ]. You can verify that this is identical to the original expression if you multiply it back out. Analogy with different exponents and only three terms: A x^3 + A x^4 + A x^5 can be divided by A x^2 to give quotient x + x^2 + x^3, so the factored expression is A ( x + x^2 + x^3). **
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RESPONSE --> I forgot the p cancels out on each one as well had the basic idea though
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