How would i rearange A=2 pi r(r+h) to make h the subject?

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The equation A = 2πr(r + h) can be rearranged to isolate h as h = (A - 2πr²) / (2πr). This transformation allows for the calculation of h when the area A and radius r are known. The discussion emphasizes the importance of showing initial work when seeking help, as it encourages collaborative problem-solving and deeper understanding.

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How would i rearange A=2\pi r(r+h) to make h the subject?
 
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madmike159 said:
How would i rearange A=2\pir(r+h) to make h the subject?

do u mean to express it in terms of h??
then \ h=\frac{A-2^{pi}r^{2}}{2^{pi}r}
 
I think it means the same thing. Thanks, I'm really bad at rearranging
 
well, next time try to show some work of yours first. Because the people here won't just give u the answer unless u have given some thought to the problem.
 

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