How do you solve for #h# in the equation #S=2pirh + 2pir^2#?

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Answer 1 · 2017-02-11T21:23:32

See the entire solution process below:

First, subtract #color(red)(2pir^2)# from each side of the equation to isolate the #h# term:

#S - color(red)(2pir^2) = 2pirh + 2pir^2 - color(red)(2pir^2)#

#S - 2pir^2 = 2pirh + 0#

#S - 2pir^2 = 2pirh#

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

#(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#