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| 14 Sep 2012 12:21 AM |
i actually don't get this one
A circle has a radius of 9 inches. The radius is multiplied by 2/3 to form a second circle. How is the ratio of the areas related to the ratio of the radii?
thx |
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dc131
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| Joined: 16 Oct 2008 |
| Total Posts: 19334 |
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| 14 Sep 2012 01:02 AM |
i'll let r_0 be the old radius and r_1 be the new radius.
r_0 = r_0 r_1 = (2/3)r_0
the ratio of the areas is A_1 / A_0 = (pi*r_1^2) / (pi*r_0^2)
the pi's cancel and the ratio can be rewritten as
A_1 / A_0 = (r_1 / r_0)^2, replace r_1 with (2/3)r_0 to get
A_1 / A_0 = ((2/3)r_0 / r_0)^2, now the r_0's cancel
A_1 / A_0 = (2/3)^2... this means the ratio of the areas has nothing to do with what the original radius was.
A_1 / A_0 = 4/9 :) |
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| 14 Sep 2012 01:47 AM |
| Why ask us OTers? We might give you the wrong answers. |
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| 14 Sep 2012 10:36 PM |
No I gave the right answer
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