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Mathematics 11 Online
OpenStudy (anonymous):

how do u make 'h' the subject of A=πr^2+2πrh ?

OpenStudy (ghazi):

you've to make h subject of what?

mathslover (mathslover):

It simply concludes that we have to solve for h

OpenStudy (anonymous):

Subtract \(\pi r^2\) both the sides first..

OpenStudy (anonymous):

Then divide by \( 2 \pi r\) you will get h as subject..

mathslover (mathslover):

\[\large{A=\pi r^2 +2\pi rh}\] \[\large{A-\pi r^2=\pi r^2-\pi r^2+2\pi rh}\] what do you get? @karenw12

OpenStudy (anonymous):

Subtract πr2 to balance out the equation

mathslover (mathslover):

right

mathslover (mathslover):

so what do you get?

mathslover (mathslover):

will it be : \[\large{A-\pi r^2=2\pi rh}\]?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

no?

OpenStudy (anonymous):

Just kidding @mathslover

mathslover (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 ?

OpenStudy (anonymous):

|dw:1345133600663:dw|

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