#### Answer

$c=9$

#### Work Step by Step

RECALL:
There are two forms of perfect square trinomials:
(1) $m^2+2mn+n^2$, which is the square of $(m+n)^2$; and
(2) $m^2-2mn+n^2$, which is the square of $(m-n)^2$
The given trinomial has:
$m^2 = 100r^2=(10r)^2$, which means that $m=10r$
The middle term of the trinomial is negative therefore the perfect square trinomial will be in the same form as in (2) above.
Thus,
$-2mn=-60r$
Since $m=10r$, substitute $m$ with $10r$ to obtain:
$-2mn=-60r
\\-2(10r)(n) = -60r
\\-20r(n) = -60r$
Divide both sides by $-20r$ to obtain:
$\dfrac{-20r(n)}{-20r} = \dfrac{-60r}{-20r}
\\n=3$
The given trinomial is in the form $m^2-2mn+n^2$.
Square $n$ to obtain:
$n^2 = 3^2 = 9$
Therefore, the perfect square trinomial is $100r^2-60r+9$.
Hence, $c=9$