Question

rewrite the literal equation to solve for h.
a = 2\pi rh + 2\pi r^2
\bigcirc\\ h = \frac{a + 2\pi r^2}{2\pi r}
\bigcirc\\ h = \frac{a - 2\pi r^2}{2\pi r}
\bigcirc\\ h = a - r
\bigcirc\\ h = a + r

Explanation:

Step1: Isolate the term with h

We start with the equation \( A = 2\pi rh + 2\pi r^2 \). First, we subtract \( 2\pi r^2 \) from both sides to get the term with \( h \) alone. So, \( A - 2\pi r^2=2\pi rh \).

Step2: Solve for h

Now, we divide both sides of the equation \( A - 2\pi r^2 = 2\pi rh \) by \( 2\pi r \) (assuming \( r
eq0 \) and \( \pi
eq0 \)) to solve for \( h \). This gives us \( h=\frac{A - 2\pi r^2}{2\pi r} \) (note: in the options, it's written as \( \frac{A - 2\pi r^2}{2\pi r} \) with \( \pi \) maybe typo as \( \pi \) but in the option it's \( 2\pi r \) as \( 2mr \) maybe a typo, but the structure is \( \frac{A - 2\pi r^2}{2\pi r} \) which matches the third option's numerator and denominator structure).

Answer:

\( h=\frac{A - 2\pi r^2}{2\pi r} \) (corresponding to the option \( h=\frac{A - 2\pi r^2}{2\pi r} \) (the third option in the list, written as \( h=\frac{A - 2\pi r^2}{2\pi r} \) with the denominator as \( 2\pi r \) (in the option it's \( 2mr \) maybe a typo for \( 2\pi r \))). So the correct option is the third one: \( h=\frac{A - 2\pi r^2}{2\pi r} \) (the option with \( h=\frac{A - 2\pi r^2}{2\pi r} \)).