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Suppose the formula A= 2(pie)(r)(h) + 2(pie)(r^2) is solved for r with the following result: r= A/ 2(pie)(h) + 2 (pie) (r) Is this an acceptable soloution?
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- Feb 4, 2004
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Note: The Greek letter is "pi"; "pie" is something one eats.
You have the equation:
. . . . .A = 2(pi)rh + 2(pi)r<sup>2</sup>
...the surface area of a right-circular cylinder with endcaps, having radius r and height h. You propose the following as being equivalent:
. . . . .r = A/(2(pi)h) + 2(pi)r
If you are trying to "solve for r", then you need to get r by itself on one side, with no r-containing terms on the other. Also, I don't see how "2(pi)r" would not placed outside the fraction as it is. It seems highly unlikely to me that this would be an "acceptable" solution.
I would suggest that you move the "A" over with everything else, you get a quadratic in r:
. . . . .(2pi)r<sup>2</sup> + (2pih)r - A = 0
Then apply the Quadratic Formula.
Eliz.
You have the equation:
. . . . .A = 2(pi)rh + 2(pi)r<sup>2</sup>
...the surface area of a right-circular cylinder with endcaps, having radius r and height h. You propose the following as being equivalent:
. . . . .r = A/(2(pi)h) + 2(pi)r
If you are trying to "solve for r", then you need to get r by itself on one side, with no r-containing terms on the other. Also, I don't see how "2(pi)r" would not placed outside the fraction as it is. It seems highly unlikely to me that this would be an "acceptable" solution.
I would suggest that you move the "A" over with everything else, you get a quadratic in r:
. . . . .(2pi)r<sup>2</sup> + (2pih)r - A = 0
Then apply the Quadratic Formula.
Eliz.