course Mth 163 浾麇馾葂旷ㄡ|屋欤沈蔬暳湍K樸assignment #014
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20:18:07 `q001. Note that this assignment has 5 questions If a(n) = a(n-1) + 2^n, with a(0) = 3, then substitute in turn the values 1, 2, 3 and 4 into the equation to obtain the values a(1), a(2), a(3) and a(4).
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RESPONSE --> n=1 (1-1)+2^1 and a(0) is equal to 3 3+2 =5 n=2 (2-1) + 2^2= 5+4 =9 n=3 =17 n=4 =33 confidence assessment: 2
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20:18:16 If we substitute n = 1 into a(n) = a(n-1) + 2^n we get a(1) = a(1-1) + 2^1 or, since 1-1 = 0 and 2^1 = 2 a(1) = a(0) + 2. Since we are given a(0) = 3 we now have a(1) = 3 + 2 = 5. If we substitute n = 2 into a(n) = a(n-1) + 2^n we get a(2) = a(2-1) + 2^2 or, since 2-1 = 1 and 2^2 = 4 a(2) = a(1) + 4. Since we are given a(1) = 5 we now have a(2) = 5 + 4 = 9. If we substitute n = 3 into a(n) = a(n-1) + 2^n we get a(3) = a(3-1) + 2^3 or, since 3-1 = 2 and 2^3 = 8 a(3) = a(2) + 8. Since we are given a(2) = 9 we now have a(3) = 9 + 8 = 17. If we substitute n = 4 into a(n) = a(n-1) + 2^n we get a(4) = a(4-1) + 2^4 or, since 4-1 = 3 and 2^4 = 16 a(4) = a(3) + 16. Since we are given a(3) = 16 we now have a(4) = 17 + 16 = 33.
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RESPONSE --> self critique assessment: 3
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20:20:52 `q002. If a(n) = 2 * a(n-1) + n with a(0) = 3, then what are the values of a(1), a(2), a(3) and a(4)?
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RESPONSE --> you get the answers 7,16,35, and 74 by using the same kind of steps from the last problem confidence assessment: 3
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20:21:36 If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get a(1) = 2 * a(1-1) + 1 or since 1-1 = 0 a(1) = 2 * a(0) + 1. Since we know that a(0) = 3 we have a(1) = 2 * 3 + 1 = 7. If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get a(2) = 2 * a(2-1) + 2 or since 2-1 = 1 a(2) = 2 * a(1) + 2. Since we know that a(0) = 3 we have a(2) = 2 * 7 + 2 = 16. If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get a(3) = 2 * a(3-1) + 3 or since 3-1 = 2 a(3) = 2 * a(2) + 3. Since we know that a(0) = 3 we have a(3) = 2 * 16 + 3 = 35. If we substitute n = 1 into a(n) = 2 * a(n-1) + n we get a(4) = 2 * a(4-1) + 4 or since 4-1 = 3 a(4) = 2 * a(3) + 4. Since we know that a(0) = 3 we have a(4) = 2 * 35 + 4 = 74.
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RESPONSE --> self critique assessment: 3
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20:35:40 `q003. What are the average slopes of the graph of y = x^2 + x - 2 between the x = 1 and x= 3 points, between the x = 3 and x = 5 points, between the x = 5 and x = 7 points, and between the x = 7 and x = 9 points? What is the pattern of this sequence of slopes?
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RESPONSE --> 0, 10, 10, 28, 28, 54, 54, and 88 are the numbers we get when we substitiute the x's. and the pairs we get are (1,0), (3,10), (3,10), (5,28), (5,28), (7,54), (7,54), and (9,88). There are 2 points between the runs, so the rises are 10, 18, 26 and 34 and the slopes are: 6, 9, 13, and 17 confidence assessment: 3
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20:35:46 At x = 1, 3, 5 , 7 and 9 we find by substituting that y = 0, 10, 28, 54 and 88. The x = 1, 3, 5, 7 and 9 points are therefore (1,0), (3,10), (5,28), (7,54) and (9,88). The run from one point to the next is always 2. The rises are respectively 10, 18, 26 and 34. The slopes are therefore slope between x = 1 and x = 3: slope = rise / run = 10 / 2 = 6. slope between x = 3 and x = 5: slope = rise / run = 18 / 2 = 9. slope between x = 5 and x = 7: slope = rise / run = 26 / 2 = 13. slope between x = 7 and x = 9: slope = rise / run = 34 / 2 = 17.
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RESPONSE --> self critique assessment: 3
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20:49:38 `q004. If a solid stone sphere 4 inches in diameter weighs 3 pounds, then what would be the weight of a solid stone sphere 2 feet in diameter?
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RESPONSE --> well since a 4 inch diameter weighs 3 pounds that would mean every inch weighs .75 pounds so something 24 inches would be .75*24 which equals 18, but a sphere that is solid stone with a 2 ft diameter is gonna weigh a lot more than 18 pounds so I know I am probably not doing this right confidence assessment: 1
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20:50:10 The volume of a sphere is proportional to the cube of its diameters, and weight is directly proportional to volume so we have the proportionality w = k d^3, where w and d stand for weight and diameter and k is the proportionality constant. Substituting the known weight and diameter we get 3 = k * 4^3, where we understand that the weight is in pounds and the diameter in inches. This gives us 3 = 64 k so that k = 3 / 64. Our proportionality equation is now w = 3/64 * d^3. So when the diameter is 2 feet, we first recall that diameter must be in inches and say that d = 24, which we then substitute to obtain w = 3/64 * 24^3. A simple calculation gives us the final weight w = 748.
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RESPONSE --> yep totally did that wrong but now I see what I was supposed to do self critique assessment:
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20:56:27 `q005. Two boxes are each constructed of a single layer of cardboard. The first box is 12 inches by 18 inches by 24 inches and weighs 22 ounces; the second is 36 inches by 54 inches by 72 inches. Using proportionality determine the weight of the second box.
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RESPONSE --> 36/12, 54/18, and 72/24 are the ratios and they all equal three and we use w = k x^2 we get 9 and we multiply it by 22 and get 198 I think I might have actually done this right! confidence assessment: 2
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20:56:51 The ratios 36/12, 54/18 and 72/24 of the corresponding sides are all the same and all equal to 3, so the dimensions of the sides of the second box are 3 times those of the first. Since the thickness of the cardboard is the same on both boxes, only the dimensions of the rectangular sides change. The only thing that matters, therefore, is the surface area of the box. The proportionality is therefore of the form w = k x^2, where w is the weight of the box and x stands any linear dimension. It follows that w2 / w1 = (x2 / x1)^2. Since as we just saw x2 / x1 = 3, we see that w2 / w1 = 3^2 = 9. Since w1 = 22 oz, we write this as w2 / 22 oz = 9. Multiplying both sides by 22 oz we see that w2 = 22 oz * 9 = 198 oz.
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RESPONSE --> yes!!!!!!! I did it right self critique assessment: 2
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course Mth 163 蛝愋頭杳闸衚彧嚆艉翔晲敕assignment #014
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21:05:59 Query two examples and a picture ...explain the statement 'the rate of change of a quadratic function changes at a constant rate'
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RESPONSE --> the rate of change comes from intervals that change at a constant spans of time and the change is always the same amount confidence assessment: 2
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21:09:24 03-26-2009 21:09:24 ** We can calculate the rates of change of a quadratic function based on a series of consecutive intervals of constant length. We find that these rates change from interval to interval, and always by the same amount. Since the rates of change always change by the same amount, they are changing at a constant rate. **
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NOTES -------> ** We can calculate the rates of change of a quadratic function based on a series of consecutive intervals of constant length. We find that these rates change from interval to interval, and always by the same amount. Since the rates of change always change by the same amount, they are changing at a constant rate. **
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21:15:49 explain how to get the first few members of a sequence from its recurrence relation
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RESPONSE --> This is kind of hard to explain but you have to use the a(n) form, so the first number is n and you keep substituting using this method confidence assessment: 1
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21:19:01 03-26-2009 21:19:01 ** We let n be the first integer for which the value a(n) is not given, and we substitute this integer into the recurrence relation to evaluate a(n) for this 'new' integer, using values of a(n) for previous integers. If this is not possible then we have not been given enough information to evaluate the sequence. We then substitute the next integer and use values of a(n) for previous integers. We continue this process as long as necessary to get the results we need. **
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NOTES -------> We let n be the first integer for which the value a(n) is not given, and we substitute this integer into the recurrence relation to evaluate a(n) for this 'new' integer, using values of a(n) for previous integers. If this is not possible then we have not been given enough information to evaluate the sequence. We then substitute the next integer and use values of a(n) for previous integers. We continue this process as long as necessary to get the results we need. **
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