course Mth 163 YpJwכTΌassignment #012
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18:40:27 problem 1. box of length 30 centimeters capacity 50 liters . What is the proportionality for this situation, what is the proportionality constant and what is the specific equation that relates capacity y to length x?
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RESPONSE --> Since y = kx^3 and y = capacity and x = length then: 50 = k(30^3) 50 = 27000k 50/27000 = k = 1.85x10^-3 or .00185 This then gives us the equation y = .00185x^3 confidence assessment: 3
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18:40:47 ** The proportionality for volume is y = k x^3, where y is capacity in liters when x is length in cm. Since y = 50 when x = 30 we have 50 = k * 30^3 so that k = 50 / (30^3) = 50 / 27,000 = 1/540 = .0019 approx. Thus y = (1/540) * x^3. **
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RESPONSE --> ok self critique assessment: 3
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18:42:15 What is the storage capacity of a box of length 100 centimeters?
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RESPONSE --> y = .00185(100^3) y = .00185 * 1000000 y = 1850 confidence assessment: 3
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18:42:28 ** The proportionality is y = 1/540 * x^3 so if x = 100 we have y = 1/540 * 100^3 = 1900 approx. A 100 cm box geometrically similar to the first will therefore contain about 1900 liters. **
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RESPONSE --> ok self critique assessment: 3
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18:45:06 What length is required to obtain a storage capacity of 100 liters?
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RESPONSE --> 100 liters = .00185x^3 100/.00185 = x^3 54054.1 = x^3 'cubedrt54054.1 = x = 37.81 confidence assessment: 2
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18:50:02 ** If y = 100 then we have 100 = (1/540) * x^3 so that x^3 = 540 * 100 = 54,000. Thus x = (54,000)^(1/3) = 185 approx. The length of a box that will store 100 liters is thus about 185 cm. **
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RESPONSE --> Im a little confused about this one mainly because when get the cubed root of 54000 I get something around 38. Not sure where you are getting 185. self critique assessment: 2
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18:54:44 How long would a box have to be in order to store all the water in a swimming pool which contains 450 metric tons of water? A metric ton contains 1000 liters of water.
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RESPONSE --> Knowing that a metric ton is equal to 1000 liters and we need 450 metric tons then we multiply the two whcih equals 450000 liters. Now we substitute: 450000 = .00185x^3 450000/.00185 = x^3 243243243.2 = x^3 243243243.2^(1/3) = x = 624.23 confidence assessment: 1
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18:55:08 ** 450 metric tons is 450 * 1000 liters = 450,000 liters. Thus y = 450,000 so we have the equation 540,000 = (1/540) x^3 which we solve in a manner similar to the preceding question to obtain x = 624, so that the length of the box is 624 cm. **
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RESPONSE --> ok self critique assessment: 3
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19:01:10 problem 2. cleaning service scrub the surface of the Statute of width of finger .8 centimeter vs. 20-centimeter width actual model takes .74 hours. How long will it take to scrub the entire statue?
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RESPONSE --> .74 hrs = k(20^2) .74 hrs = 400k .74/400 = k = .00185 Now we substitue what we know: .8 = .00185x^2 .8 / .00185 = x^2 432.43 = x^2 'sqrt432.43 = x = 20.79 hours confidence assessment: 1
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19:02:15 ** y = k x^2 so .74 = k * .8^2. Solving for k we obtain k = 1.16 approx. so y = 1.16 x^2. The time to scrub the actual statue will be y = 1.16 x^2 with x = 20. We get y = 1.16 * 20^2 = 460 approx.. It should take 460 hrs to scrub the entire statue. **
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RESPONSE --> got that one confused, but I understand now. y = k x^2 so .74 = k * .8^2. Solving for k we obtain k = 1.16 approx. so y = 1.16 x^2. The time to scrub the actual statue will be y = 1.16 x^2 with x = 20. We get y = 1.16 * 20^2 = 460 approx.. It should take 460 hrs to scrub the entire statue. self critique assessment: 3
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19:06:18 problem 3. illumination 30 meters is 5 foot-candles. What is the proportionality for this situation, what is the value of the proportionality constant and what equation relates the illumination y to the distance x?
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RESPONSE --> 30 = k(5^-2) 30 = (1/25)k 750 = k The equation would look like this: y = 750x^-2 confidence assessment: 3
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19:07:16 ** The proportionality should be y = k x^-2, where y is illumination in ft candles and x the distance in meters. We get 5 = k * 30^-2, or 5 = k / 30^2 so that k = 5 * 30^2 = 4500. Thus y = 4500 x^-2. We get an illumination of 10 ft candles when y = 10. To find x we solve the equation 10 = 4500 / x^2. Multiplying both sides by x^2 we get 10 x^2 = 4500. Dividing both sides by 10 we have x^2 = 4500 / 10 = 450 and x = sqrt(450) = 21 approx.. For illumination 1000 ft candles we solve 1000 = 4500 / x^2, obtaining solution x = 2.1 approx.. We therefore conclude that the comfortable range is from about x = 2.1 meters to x = 21 meters. **
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RESPONSE --> Got the y and x confused. These types of problems make it very difficult to distinguish the x and y.
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19:13:47 problem 5. Does a 3-unit cube weigh more or less than 3 times a 1-unit cube? Why is this?
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RESPONSE --> A 3-unit cube would weigh more because a 3-unit cube is composed of 9 times the weight while the 1 -unit cube is only 3 times its weight. confidence assessment: 2
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19:15:32 ** A 3-unit cube is equivalent to 3 layers of 1-unit cubes, each layer consisting of three rows with 3 cubes in each row. Thus a 3-unit cube is equivalent to 27 1-unit cubes. If the weight of a 1-unit cube is 35 lbs then we have the following: Edge equiv. # of weight Length 1-unit cubes 1 1 35 2 4 4 * 35 = 140 3 9 9 * 35 = 315 4 16 16 * 35 = 560 5 25 25 * 35 = 875. Each weight is obtained by multiplying the equivalent number of 1-unit cubes by the 35-lb weight of such a cube. **
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RESPONSE --> A 3-unit cube is equivalent to 3 layers of 1-unit cubes, each layer consisting of three rows with 3 cubes in each row. Thus a 3-unit cube is equivalent to 27 1-unit cubes. If the weight of a 1-unit cube is 35 lbs then we have the following: Edge equiv. # of weight Length 1-unit cubes 1 1 35 2 4 4 * 35 = 140 3 9 9 * 35 = 315 4 16 16 * 35 = 560 5 25 25 * 35 = 875. Each weight is obtained by multiplying the equivalent number of 1-unit cubes by the 35-lb weight of such a cube. self critique assessment: 3
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19:17:38 ** To cover a 6-unit square requires 6 rows each containing 6 1-unit squares for a total of 36 one-unit squares. To cover a 7-unit square requires 7 rows each containing 7 1-unit squares for a total of 49 one-unit squares. To cover a 8-unit square requires 8 rows each containing 8 1-unit squares for a total of 64 one-unit squares. To cover a 9-unit square requires 9 rows each containing 9 1-unit squares for a total of 81 one-unit squares. To cover a 10-unit square requires 10 rows each containing 10 1-unit squares for a total of 100 one-unit squares. To cover an n-unit square requires n rows each containing n 1-unit squares for a total of n*n=n^2 one-unit squares. **
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RESPONSE --> To cover a 6-unit square requires 6 rows each containing 6 1-unit squares for a total of 36 one-unit squares. To cover a 7-unit square requires 7 rows each containing 7 1-unit squares for a total of 49 one-unit squares. To cover a 8-unit square requires 8 rows each containing 8 1-unit squares for a total of 64 one-unit squares. To cover a 9-unit square requires 9 rows each containing 9 1-unit squares for a total of 81 one-unit squares. To cover a 10-unit square requires 10 rows each containing 10 1-unit squares for a total of 100 one-unit squares. To cover an n-unit square requires n rows each containing n 1-unit squares for a total of n*n=n^2 one-unit squares. self critique assessment: 3
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19:22:18 ** right idea but you have the ratio upside down. The volume ratio of a 5-unit cube to a 3-unit cube is (5/3)^3 = 125 / 27 = 4.7 approx.. The edge ratio is 5/3 = 1.67 approx. VOlume ratio = edgeRatio^3 = 1.678^3 = 4.7 approx.. From this example we see how volume ratio = edgeRatio^3. If two cubes have edges 12.7 and 2.3 then their edge ratio is 12.7 / 2.3 = 5.5 approx.. The corresponding volume ratio would therefore be 5.5^3 = 160 approx.. If edges are x1 and x2 then edgeRatio = x2 / x1. This results in volume ratio volRatio = edgeRatioo^3 = (x2 / x1)^3. **
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RESPONSE --> The volume ratio of a 5-unit cube to a 3-unit cube is (5/3)^3 = 125 / 27 = 4.7 approx.. The edge ratio is 5/3 = 1.67 approx. Volume ratio = edgeRatio^3 = 1.678^3 = 4.7 approx.. From this example we see how volume ratio = edgeRatio^3. If two cubes have edges 12.7 and 2.3 then their edge ratio is 12.7 / 2.3 = 5.5 approx.. The corresponding volume ratio would therefore be 5.5^3 = 160 approx.. If edges are x1 and x2 then edgeRatio = x2 / x1. This results in volume ratio volRatio = edgeRatioo^3 = (x2 / x1)^3. self critique assessment: 3
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19:26:46 problem 9. Relating y and x ratios for a cubic proportionality. What is the y value corresponding to x = 3 and what is is the y value corresponding to x = 5?
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RESPONSE --> y2/y1 = (5/3)^3 (y2/y1) = 4.66 confidence assessment: 1
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19:27:58 ** If y = a x^3 then if x1 = 3 we have y1 = a * 3^3 and if x2 = 5 we have y2 = a * 5^3. This gives us ratio y2 / y1 = (a * 5^3) / (a * 3^3) = (a / a) * (5^3 / 3^3) = 1 * 125 / 27 = 125 / 27. In general if y1 = a * x1^3 and y2 = a * x2^3 we have } y2 / y1 = (a x2^3) / (a x1^3) = (a / a) * (x2^3 / x1^3) = (x2/x1)^3. This tells you that to get the ratio of y values you just cube the ratio of the x values. **
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RESPONSE --> ok self critique assessment: 3
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19:28:38 ** If y = a x^2 then y2 / y1 = (a x2^2) / (a x1^2) = (a / a) * (x2^2 / x1^2) = (x2/x1)^2. This tells you that to get the ratio of y values you just square the ratio of the x values. If y = f(x) = a x^p then y1 = f(x1) = a x1^p and y2 = f(x1) = a x2^p so that y2 / y1 = f(x2) / f(x1) = (a x2^p) / (a x1^p) = (a / a) ( x2^p / x1^p ) = x2^p / x1^p = (x2 / x1)^p. **
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RESPONSE --> If y = a x^2 then y2 / y1 = (a x2^2) / (a x1^2) = (a / a) * (x2^2 / x1^2) = (x2/x1)^2. This tells you that to get the ratio of y values you just square the ratio of the x values. If y = f(x) = a x^p then y1 = f(x1) = a x1^p and y2 = f(x1) = a x2^p so that y2 / y1 = f(x2) / f(x1) = (a x2^p) / (a x1^p) = (a / a) ( x2^p / x1^p ) = x2^p / x1^p = (x2 / x1)^p. self critique assessment: 3
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