course Mth 163 assignment #012012. `query 12
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17:52:18 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 --> The prop. is y=kx^3 50=k*30^3 k=50/(30^3)=50/27000=1/540=.00185 That makes the equation y=1/540*x^3 confidence assessment: 2
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17:52:32 ** 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 --> When using the notes I understand. self critique assessment: 2
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17:53:56 What is the storage capacity of a box of length 100 centimeters?
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RESPONSE --> The storage capacity- proportionality=y=1/540*x^3, if x=100 then y=1/540*100^3=1900 confidence assessment: 2
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17:54:19 ** 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 --> I can figure out the answer when using the notes self critique assessment: 2
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17:55:43 What length is required to obtain a storage capacity of 100 liters?
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RESPONSE --> y=100 100=1/540*x^3 x^3=540*100=54000 x=(54000)^(1/3)=185=length of box confidence assessment: 2
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17:55:51 ** 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 --> I see self critique assessment: 2
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18:00:08 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 --> 450*100=450000=y 45000=(1/450)x^3 450*45*1/3 675 confidence assessment: 2
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18:00:32 ** 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 --> I totally got that wrong. self critique assessment: 2
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18:02:21 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 --> y=kx^2 .74=k*.8^2 K=1.16 y=1.16x^2 y=1.16*20^2=460 confidence assessment: 2
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18:02:33 ** 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 --> I used the notes self critique assessment: 2
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18:04:23 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 --> y=kx^-2 k=5*30^2=4500 1000=4500/x^2 x=2.1 confidence assessment: 2
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18:05:10 ** 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 --> As I wrote the equation out on paper, I forgot to type in part of the answer. I do understand self critique assessment: 2
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18:06:01 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 --> There are the same b/c a 3 unit cube is equivalent to 3 one unit cube. confidence assessment: 2
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18:06:09 ** 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 --> I see self critique assessment: 2
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18:07:13 problem 6. Give the numbers of 1-unit squares required to cover 6-, 7-, 8-, 9- and 10-unit square, and also an n-unit square.
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RESPONSE --> 6*6=36 7*7=49 8*8=64 9*9=81 10*10=100 confidence assessment: 2
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18:07:21 ** 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 --> I get it. self critique assessment: 2
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18:09:53 problem 8. Relating volume ratio to ratio of edges.
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RESPONSE --> The volume ratio = edge ratio^3 confidence assessment: 2
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18:10:28 ** 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 --> ok self critique assessment: 2
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