How do you solve for $h$ $h$ in the equation $S = 2 π r h + 2 π r^{2}$ $S=2pirh + 2pir^2$ ?

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Answer 1 · 2017-02-11T21:23:32

See the entire solution process below:

First, subtract $2 π r^{2}$ $color(red)(2pir^2)$ from each side of the equation to isolate the $h$ $h$ term:

$S - 2 π r^{2} = 2 π r h + 2 π r^{2} - 2 π r^{2}$ $S - color(red)(2pir^2) = 2pirh + 2pir^2 - color(red)(2pir^2)$

$S - 2 π r^{2} = 2 π r h + 0$ $S - 2pir^2 = 2pirh + 0$

$S - 2 π r^{2} = 2 π r h$ $S - 2pir^2 = 2pirh$

Now, divide each side of the equation by $2 π r$ $color(red)(2pir)$ to solve for $h$ $h$ :

$\frac{S - 2 π r^{2}}{2 π r} = \frac{2 π r h}{2 π r}$ $(S - 2pir^2)/color(red)(2pir) = (2pirh)/color(red)(2pir)$

$(S - 2pir^2)/color(red)(2pir) = (color(red)(cancel(color(black)(2pir)))h)/cancel(color(red)(2pir))$

$(S - 2pir^2)/color(red)(2pir) = h$

$h = (S - 2pir^2)/color(red)(2pir)$

Or

$h = S/color(red)(2pir) - (2pir^2)/color(red)(2pir)$

$h = S/(2pir) - (color(red)(cancel(color(black)(2pi)))r^2)/cancel(color(red)(2pir))$

$h = S/(2pir) - r^2/r$

$h = S/(2pir) - r$

How do you solve for hhh in the equation S=2pirh + 2pir^2S=2πrh+2πr2S=2pirh + 2pir^2?