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how do u make 'h' the subject of A=πr^2+2πrh ?
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you've to make h subject of what?
It simply concludes that we have to solve for h
Subtract \(\pi r^2\) both the sides first..
Then divide by \( 2 \pi r\) you will get h as subject..
\[\large{A=\pi r^2 +2\pi rh}\] \[\large{A-\pi r^2=\pi r^2-\pi r^2+2\pi rh}\] what do you get? @karenw12
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Subtract πr2 to balance out the equation
right
so what do you get?
will it be : \[\large{A-\pi r^2=2\pi rh}\]?
yes
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no?
Just kidding @mathslover
\[\large{A-\pi r^2=\cancel{\pi r^2}\cancel{-\pi r^2}+2\pi rh}\] \[\large{\frac{A-\pi r^2}{2\pi r}=\frac{2\pi rh}{2\pi r}}\] what do you get now @karenw12 ?
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