course Mth 163 Ýsø钾“ó×¼õø†ƒå¹©öÚ¯®˜àassignment #011
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10:33:45 `q001. Note that this assignment has 11 questions How many squares one foot on a side would it take to construct a square two feet on a side?
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RESPONSE --> double the number of 1 foot squares confidence assessment: 3
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10:34:06 A common response is that it takes 2 one-ft. squares to make a 2-foot square. However, below thought shows that this isn't the case. If we put 2 one foot squares side by side we get a one-foot by two-foot rectangle, not a square. If we put a second such rectangle together with the first, so that we have 2 rows with 2 squares in a row, then we have a two-foot square. Thus we see that it takes 4 squares one foot on a side to make a square 2 ft. on a side.
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RESPONSE --> ok self critique assessment: 3
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10:34:25 `q002. How many cubes one foot on a side would it take to construct a cube two feet on a side?
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RESPONSE --> 4 confidence assessment: 3
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10:34:51 We could begin by constructing two rows with two cubes in a row, which would sit on a square two feet by two feet. However this would not give is a cube two feet on a side, because at this point the figure we have constructed is only one foot high. So we have to add a second layer, consisting of two more rows with two cubes a row. Thus we have 2 layers, each containing 2 rows with 2 cubes in a row. Each layer has 4 cubes, so our two layers contain 8 cubes.
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RESPONSE --> ok self critique assessment: 3
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10:35:36 `q003. How many squares one foot on a side would it take to construct a square three feet on a side?
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RESPONSE --> 9 confidence assessment: 3
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10:35:43 We would require three rows, each with 3 squares, for a total of 9 squares.
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RESPONSE --> ok self critique assessment: 3
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10:36:39 `q004. How many cubes one foot on a side would take to construct a cube three feet on a side?
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RESPONSE --> 18 confidence assessment: 3
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10:36:57 This would require three layers to make a cube three feet high. Each layer would have to contain 3 rows each with three cubes. Each layer would contain 9 cubes, so the three-layer construction would contain 27 cubes.
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RESPONSE --> oh ok i understand where the 27 comes from now self critique assessment: 3
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10:37:21 `q005. Suppose one of the Egyptian pyramids had been constructed of cubical stones. Suppose also that this pyramid had a weight of 100 million tons. If a larger pyramid was built as an exact replica, using cubical stones made of the same material but having twice the dimensions of those used in the original pyramid, then what would be the weight of the larger pyramid?
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RESPONSE --> 200 million confidence assessment: 3
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10:37:47 Each stone of the larger pyramid has double the dimensions of each stone of the smaller pyramid. Since it takes 8 smaller cubes to construct a cube with twice the dimensions, each stone of the larger pyramid is equivalent to eight stones of the smaller. Thus the larger pyramid has 8 times the weight of the smaller. Its weight is therefore 8 * 100 million tons = 800 million tons.
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RESPONSE --> oh ok self critique assessment: 3
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10:38:01 `q006. Suppose that we wished to paint the outsides of the two pyramids described in the preceding problem. How many times as much paint would it take to paint the larger pyramid?
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RESPONSE --> 8 confidence assessment: 3
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10:38:17 The outside of each pyramid consists of square faces of uniform cubes. Since the cubes of the second pyramid have twice the dimension of the first, their square faces have 4 times the area of the cubes that make up the first. There is therefore 4 times the area to paint, and the second cube would require 4 times the paint
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RESPONSE --> ok self critique assessment: 3
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10:38:37 `q007. Suppose that we know that y = k x^2 and that y = 12 when x = 2. What is the value of k?
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RESPONSE --> k=3 confidence assessment: 3
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10:38:49 To find the value of k we substitute y = 12 and x = 2 into the form y = k x^2. We obtain 12 = k * 2^2, which we simplify to give us 12 = 4 * k. The dividing both sides by 410 reversing the sides we easily obtain k = 3.
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RESPONSE --> ok self critique assessment: 3
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10:40:24 `q008. Substitute the value of k you obtained in the last problem into the form y = k x^2. What equation do you get relating x and y?
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RESPONSE --> y=0 and x=0 confidence assessment: 3
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10:40:56 We obtained k = 3. Substituting this into the form y = k x^2 we have the equation y = 3 x^2.
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RESPONSE --> oh i thought i was supposed to graph and find an x and y value
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10:43:08 `q009. Using the equation y = 3 x^2, determine the value of y if it is known that x = 5.
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RESPONSE --> y=75 confidence assessment: 3
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10:43:15 If x = 5, then the equation y = 3 x^2 give us y = 3 (5)^2 = 3 * 25 = 75.
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RESPONSE --> ok self critique assessment: 3
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10:45:24 `q010. If it is known that y = k x^3 and that when x = 4, y = 256, then what value of y will correspond to x = 9? To determine your answer, first determine the value of k and substitute this value into y = k x^3 to obtain an equation for y in terms of x. Then substitute the new value of x.
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RESPONSE --> k=4 which is y=4(9)^3 and you get y=2916 confidence assessment: 3
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10:45:34 To we first substitute x = 4, y = 256 into the form y = k x^3. We obtain the equation 256 = k * 4^3, or 256 = 64 k. Dividing both sides by 64 we obtain k = 256 / 64 = 4. Substituting k = 4 into the form y = k x^3, we obtain the equation y = 4 x^3. We wish to find the value of y when x = 9. We easily do so by substituting x equal space 9 into our new equation. Our result is y = 4 * 9^3 = 4 * 729 = 2916.
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RESPONSE --> ok self critique assessment: 3
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10:50:23 `q011. If it is known that y = k x^-2 and that when x = 5, y = 250, then what value of y will correspond to x = 12?
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RESPONSE --> k=6250 and then y=6250(12)^-2 and y=1.11*10^-6 confidence assessment: 3
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10:51:12 Substituting x = 5 and y = 250 into the form y = k x^-2 we obtain 250 = k * 5^-2. Since 5^-2 = 1 / 5^2 = 1/25, this becomes 250 = 1/25 * k, so that k = 250 * 25 = 6250. Thus our form y = k x^-2 becomes y = 6250 x^-2. When x = 12, we therefore have y = 6250 * 12^-2 = 6250 / 12^2 = 6250 / 144 = 42.6, approximately.
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RESPONSE --> oh ok I forgot about the neg exponent rule self critique assessment: 3
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course Mth 163 ƶqïêø‰ñ¼h²Ö÷àßgº~Ž€‡©ÓÇx}assignment #012
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10:59:10 `q001. Note that this assignment has 3 questions If we know that y = k x^2, then if (x2/x1) = 7, what is (y2/y1)?
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RESPONSE --> I dont really understand this confidence assessment: 3
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‰°ÒæЯóhžçÝãô]ŠSúxå{«·ÅþŠþ assignment #012 012. Precalculus I 02-25-2009
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11:06:47 `q001. Note that this assignment has 3 questions If we know that y = k x^2, then if (x2/x1) = 7, what is (y2/y1)?
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RESPONSE --> I kindof understand the question and I got the answer 49 but Im not positive confidence assessment: 2
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11:07:44 If y2 = k x2^2 and y1 = k x1^2, then y2 / y1 = (k x2^2) / ( k x1^2). Since k / k = 1 this is the same as y2 / y1 = x2^2 / x1^2, which is the same as y2 / y1 = (x2 / x1)^2. In words this tells us if y to is proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x2 to x1. Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^2 = 7^2 = 49.
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RESPONSE --> ok this is what I did but it didnt make this good of sense of to me but seeing it like this makes more sense self critique assessment: 3
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11:09:12 `q002. If we know that y = k x^3, then if (x2/x1) = 7, what is (y2/y1)?
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RESPONSE --> it would be the same as the last and it would be 7^3=343 confidence assessment: 3
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11:09:22 If y2 = k x2^3 and y1 = k x1^3, then y2 / y1 = (k x2^3) / ( k x1^3). Since k / k = 1 this is the same as y2 / y1 = x2^3 / x1^3, which is the same as y2 / y1 = (x2 / x1)^3. In words this tells us if y to is proportional to the cube of x, then the ratio of y2 to y1 is the same as the cube of the ratio of x2 to x1. Now if (x2 / x1) = 7, we see that y2 / y1 = (x2 / x1)^3 = 7^3 = 343.
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RESPONSE --> ok self critique assessment: 3
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11:09:56 `q003. If we know that y = k x^-2, then if (x2/x1) = 64, what is (y2/y1)?
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RESPONSE --> it would be 64^2=4096 confidence assessment: 3
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11:10:43 If y2 = k x2^-2 and y1 = k x1^-2, then y2 / y1 = (k x2^-2) / ( k x1^-2). Since k / k = 1 this is the same as y2 / y1 = x2^-2 / x1^-2, which is the same as y2 / y1 = (x2 / x1)^-2, which is the same as 1 / (x2 / x1)^2, which gives us (x1 / x2)^2. So if y = k x^-2, then (y2 / y1) = (x1 / x2)^2.( In words this tells us if y to is inversely proportional to the square of x, then the ratio of y2 to y1 is the same as the square of the ratio of x1 to x2 (note that this is a reciprocal ratio). Now if (x2 / x1) = 64, we see that y2 / y1 = (x1 / x2)^2 = (1/64)^2 = 1/ 4096.
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RESPONSE --> oh i see i forgot to do the recipricol self critique assessment: 3
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