#$&* course Mth 163 2/27 012. `query 12
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Given Solution: OK ** The proportionality for volume is y = k x^3, where y is capacity in liters when x is length in cm. Since y = 50 when x = 30 we have 50 = k * 30^3 so that k = 50 / (30^3) = 50 / 27,000 = 1/540 = .0019 approx. Thus y = (1/540) * x^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat is the storage capacity of a box of length 100 centimeters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y= .0019(100^3)= .0019*1000000= 1900 The storage capacity of a 100 cm box is 1,900 liters. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: OK ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qWhat length is required of a geometrically similar box to obtain a storage capacity of 100 liters? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 100= 1/540(x^3) 100*540= 5400= 54000^1/3= 37.7976 The length required would be approximately 39 cm. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: OK ** If y = 100 then we have 100 = (1/540) * x^3 so that x^3 = 540 * 100 = 54,000. Thus x = (54,000)^(1/3) = 38 approx. The length of a box that will store 100 liters is thus about 38 cm. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qHow long would a geometrically similar box have to be in order to store all the water in a swimming pool which contains 450 metric tons of water? A metric ton contains 1000 liters of water. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 450*1000= 450,000 liters 450,000=(1/540)x^3 450000*540= 243000000^1/3= 624.025 The length would have to be approximately 624 cm. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: OK ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=k x^2 .74= k* .8^2 .74=.64k K= 1.15625 apprx. 1.16 Y= 1.16(x^2) Y= 1.16(20^2)= 400*1.16= 464 It would take approximately 464 hours. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: OK ** y = k x^2 so .74 = k * .8^2. Solving for k we obtain k = 1.16 approx. so y = 1.16 x^2. The time to scrub the actual statue will be y = 1.16 x^2 with x = 20. We get y = 1.16 * 20^2 = 460 approx.. It should take 460 hrs to scrub the entire statue. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 3. illumination 30 meters is 5 foot-candles. What is the proportionality for this situation, what is the value of the proportionality constant and what equation relates the illumination y to the distance x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=k x^-2 5= k (30^-2)= 5= .0011k = 4504.5045 K= approx. 4500 Y=4500(x^-2) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Ok ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 5. Does a 3-unit cube weigh more or less than 3 times a 1-unit cube? Why is this? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A 3 unit cube weighs more than a 1 unit cube, because a 3 unit cube consists of 3 layers of 9 cubes. With a 3 unit cube there will be a total of 27 cubes. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: OK ** A 3-unit cube is equivalent to 3 layers of 1-unit cubes, each layer consisting of three rows with 3 cubes in each row. Thus a 3-unit cube is equivalent to 27 1-unit cubes. If the weight of a 1-unit cube is 35 lbs then we have the following: Edge equiv. # of weight Length 1-unit cubes 1 1 35 2 8 8 * 35 = 360 3 27 27 * 35 = 945 4 64 64 * 35 = 2240 5 125 125 * 35 = 4375 Each weight is obtained by multiplying the equivalent number of 1-unit cubes by the 35-lb weight of such a cube. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 6. Give the numbers of 1-unit squares required to cover 6-, 7-, 8-, 9- and 10-unit square, and also an n-unit square. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6 unit squares require 6 rows, each having 6, 1 unit squares, for a total of 36 squares. 7 unit squares require 7 rows, each having 7, 1 unit squares, for a total of 49 squares. 8 unit squares require 8 rows, each having 8, 1 unit squares, for a total of 64 squares. 9 unit squares require 9 rows, each having 9, 1 unit squares, for a total of 81 squares. 10 unit squares require 10 rows, each having 10, 1 unit squares, for a total of 100 squares. n unit squares require n rows, each having n, 1 unit squares, for a total of n^2 squares. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: OK ** 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 (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 8. Relating volume ratio to ratio of edges. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Volume ratio is the same thing as an edge ratio, except a volume ratio is cubed. Volume ratio = edge ratio ^3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: OK ** right idea but you have the ratio upside down. The volume ratio of a 5-unit cube to a 3-unit cube is (5/3)^3 = 125 / 27 = 4.7 approx.. The edge ratio is 5/3 = 1.67 approx. VOlume ratio = edgeRatio^3 = 1.678^3 = 4.7 approx.. From this example we see how volume ratio = edgeRatio^3. If two cubes have edges 12.7 and 2.3 then their edge ratio is 12.7 / 2.3 = 5.5 approx.. The corresponding volume ratio would therefore be 5.5^3 = 160 approx.. If edges are x1 and x2 then edgeRatio = x2 / x1. This results in volume ratio volRatio = edgeRatioo^3 = (x2 / x1)^3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 9. Relating y and x ratios for a cubic proportionality. What is the y value corresponding to x = 3 and what is is the y value corresponding to x = 5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y= a x^3 Y= a 3^3= Y= 27a Y= a 5^3= Y= 125a confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: OK ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qproblem 10. Generalizing to y = x^p. Suppose that y = f(x) = a x^p. Let x1 and x2 represent two x values. What are the symbolic expressions, in terms of the symbols x1 and x2, for y1 = f(x1) and y2 = f(x2)? What then is the symbolic expression for y2 / y1? How does this expression tell you how to find the ratio of y values from the ratio of x values? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y1= f(x1) F(x1)= a x1^p Y2= f(x2) F(x2)= a x2^p Y2/y1= (a x2^p)/(a x1^p)= (x2^p/x1^p)= (x2/x1)^p To get the ratio of y values you have to square the ratio of the x values. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Ok ** 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. ** Add comments on any surprises or insights you experienced as a result of this assignment. this was a pretty easy assignment to comprehend, I did like the ratio stuff looks like it will come in handy ** this stuff is very important in most areas of study ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This assignment wasn’t too hard, I’m sure this information will come in handy to know as the class goes on. ------------------------------------------------ Self-critique Rating: