course Mth 163 Kܘ痢H}[쌇ێassignment #014
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16:17:48 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 --> When finding the pattern of a sequence, there are three levels that we have been using, the first which is the y values or a(n) values, then the second level is the difference between the consecutive numbers, the value by which the first number is changed to arrive at the next number, the third level is the second difference. when something says ""the rate of change of a quadratic function changes at a constant rate"" it means that the values of the second difference are all the same number.
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16:18:04 ** 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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16:20:45 explain how to get the first few members of a sequence from its recurrence relation
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RESPONSE --> when given a recurrence relation of a(n+1)=a(n)+5n; a(1) = 2 for example, you take the numbers being substituted in for n, like 1,2,3,4 and substitue them in indiviudally, so when 1 is substituted in you get a(1+1) = ..., so the result of this equation will be the value of a(2). Sometime you have to substitue in a previous found value such as in a(1+1) = a(1) +5(1), you are given the value of a(1) as being 2, so you would substitute that in to find the value of a(2).
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16:21:01 ** 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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SzƒmƱᆷ Student Name: assignment #014
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14:53:32 `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 --> a(n)=a(n+1)+2^n; a(0) = 3 a(1) = a(1-1)+2^1 = 3+2 = 5 a(2) = a(2-1)+2^2 = 5+4 = 9 a(3) = a(3-1)+2^3 = 9+8 =17 a(4) = a(4-1)+2^4 =17+16=33
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14:53:39 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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14:58:26 `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 --> a(n) = 2a(n-1) + n; a(0) = 3 a(1) = 2a(1-1) +1 = 2*3+1=7 a(2) =2a(2-1)+2 = 2*7+2=16 a(3)=2a(3-1)+3=2*16+3=35 a(4)=2a(4-1)+4=2*35+4=74
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14:58:32 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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15:09:22 `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 --> y=x^2 + x -2 y=1^2 + 1 - 2 = 1 + 1 - 2 = 2 - 2 = 0 y=3^2 + 3 - 2 = 9 + 3 - 2 = 12 - 2 = 10 y=5^2 + 5 - 2 = 25 + 5 - 2 = 30 - 2 = 28 y=7^2 + 7 -2 = 49 + 7 - 2 = 56-2 = 54 y=9^2 + 9 - 2 = 81 + 9 - 2 = 90 - 2 = 88 average slopes; (10-0)/(3-1) = 10/2 = 5 (28-10)/(5-3) = 18/2 = 9 (54-28)/(7-5) = 26/2=13 (88-54)/(9-7) = 34/2 = 17 The difference between all of the slopes is 4 (9-5=4; 12-9 = 4; 17-13=4)
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15:10:01 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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15:20:15 `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 --> (4in, 3 lbs) 2ft*12in=24in. (24in, ? lbs) y=kx^3 would be used because we are working with a 3-dimensional object. x value ratio: 24/4 = 6 y value ratio: (6)^3 = 216 new y value: 246*3=648 lbs. a solid stone sphere 2 feet in diameter would weight 648 pounds.
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15:20:53 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 --> w = 3/64 * 24^3 = 648 instead of 748
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15:30:35 `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 --> I found the volume of the cubes because the volume would be proportional with the weight. first box volume = 12*18*24 = 5184 in^3 second box volume = 36*54*72 = 139968 in^3 (51874 in^3, 22oz.) and (139968 in^3, ? oz) x value ratio: 139968/5184 = 27 y value ratio: 27^3 = 19683 new y value: 19683*22=433026 (139968in^3, 433026oz)
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15:32:28 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 --> I used y=kx^3 correspoinding the volume and the weight instead of using y=kx^2 for surface area.
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