Answer
$x=138.065$
Work Step by Step
We are given $R(x)=10x^{2/3}$
and $C(x)=2x-9$
Let $R(x)=C(x)$
we will have $10x^{2/3}=2x-9 \rightarrow 10x^{2/3}-2x+9=0$
Let $f(x)=10x^{2/3}-2x+9$
$f'(x)=\frac{20}{3}x^{\frac{-1}{3}}-2$
Let $c_1=140$, using Newton's method to find
$c_2=138.07281$
$c_3=138.06477$
$c_4=138.06477$
Since $c_4=c_3=138.06477$, to the nearest thousandth, $x=138.065$
