assignmet 12 query

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00:37:12 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 --> 50liters=k(30^2) so k=1/2

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00:38:38 ** 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 --> i was thinking that for a box we would square the proportionality of the box, but i assume that a box is a cube?

Volume changes as the cube of the dimensions, and the question concerns the volume of the box.

Otherwise, good work.

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00:39:31 What is the storage capacity of a box of length 100 centimeters?

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RESPONSE --> y=.0019(100^3) so y=1900

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00:40:27 ** 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 got this one correct. i seem to be having trouble putting the equation together for myself, but once i have the equation the numbers come very easy as far as what to solve for.

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00:42:04 What length is required to obtain a storage capacity of 100 liters?

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RESPONSE --> 100=.0019(x^3)

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00:43:26 ** 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 have trouble with these problems because i have no idea where the 1/3 is coming from. i need you to clarify this for me.

To solve x^3 = c for x, you take the 1/3 power of both sides of the equation and you get

(x^3)^(1/3) = c^(1/3). By the laws of exponents (x^3)^(1/3) = x^(3 * 1/3) = x^1 = x, so x = c^(1/3).

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00:44:20 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 --> y=.0019(1000)^3=19000

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00:46:31 ** 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 understand how to get x. but i don't understand why and where we would get teh 1/540 and i do not understand why we would multiply 450*1000 if they are equal to each other.

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00:48:52 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 --> .8*20=16 so 16=.74(x^2) so x^2=.74-16 so x=15.26 it would take 15.26 hours to clean the statue

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00:50:49 ** 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 --> okay i did my equation backwards i went ahead and plugged an incorrect value into k instead of solving for k because i was under the impression that we had both values..

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00:52:18 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=5^2*k so k=1.2

If the equation is y = k x^2, you do get k = 1.2, so you're not doing badly.

However the proportionality here is inverse square, or y = k x^-2, same as y = k / x^2.

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00:54:59 ** 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 --> i am a little clue less on this problem. i understnad that i had a clerical error before but i don't understand where we got that x-10 and why would we take the square root?

If you know what x^2 is equal to, you take the square root of that quantity in order to get x. Also note that the negative of the square root also solves the equation.

x^2 = c is solved as (x^2)^(1/2) = c^(1/2) so x = c^(1/2),

but note that since the power is even, x could also be c^(-1/2).

Also note that the 1/2 power is the square root.

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00:57:24 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 --> it would weigh more because 1 unit cube by 3 times would only weigh 3 times the 35 pound 1 unti cubes when the 3 unit cubes will wiegh triple

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01:00:13 03-04-2006 01:00:13 ** 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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NOTES -------> ok so will which will weigh the most if we have this problem on a test will we need to do a table

Any correct line of reasoning, whether it uses a table or not, will get credit on the test.

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01:00:14 ** 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 -->

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01:02:32 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 --> i have no idea i am really confused would we need to use the form y-kx^2?

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01:03:56 ** 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 --> okay i was on the right track but i was scared of the answer, so for a 5 unit square we would need 25 one unit squares

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01:04:30 problem 8. Relating volume ratio to ratio of edges.

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RESPONSE --> wow this is hard part that i had lots of trouble with

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01:07:10 ** 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 --> okay i get it, but i have a question about a problem on this in the random problems

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01:08:10 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 --> i have no idea.

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01:10:34 ** 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

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01:10:53 problem 10. Generalizing to y = x^p.

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RESPONSE --> y=x^p

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01:12:31 ** 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 --> ok

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01:14:14 Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> this whole thing suprised because all in all I think that i actually understood. I do have a uquestion though about getting the fractions in the questions in the randomized problems. how would we know to put the p=1/3 or 1/2 do we have to do this or can we just work the equation our regular. I guess what i am asking is when should we use this?

Volumes are governed by y = k x^3; i.e., for volumes p = 3; this is proportionality to the cube. Areas are governed by y = k x^2; i.e., for volumes p = 2; this is proportionality to the square.

If the proportionality is an inverse square we have y = k / x^2, same as y = k x^-2. If the proportionality is an inverse cube we have y = k / x^3, same as y = k x^-3.

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01:14:17 21:40:33

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RESPONSE -->

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01:14:22 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 **

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Good work. Let me know if you have questions.