Answer
For a speed range of $\left(40,50\right)$
Work Step by Step
$g=10+0.9v-0.01v^2$.
for $g\gt30$,
$10+0.9v-0.01v^2\gt30,$
$-0.01v^2+0.9v-20\gt0,$ multiplying both sides by 100.
$-v^2+90v-2000\gt0$
solving for the trinomial using quadratic formula for the quadratic equation of $ax^2+bx+c, x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
In this case, $-v^2+90v-2000\gt0, v=\frac{-90\pm\sqrt{(90)^2-4\times(-1)(-2000)}}{2(-1)}=\frac{-90\pm10}{-2}$
thus, $v=50$ or $v=40$
$-(v-50)(v-40)\gt0$.
$\begin{array}{lll}
intervals &testing cases& -(v-50)(v-40)\gt0 & -(v-50)(v-40)\gt0?\\
(-\infty, 40) & 0&(-)(-)(-)=(-)&F\\
(40,50)&45&(-)(-)(+)=(+)&T\\
(50,\infty)&60&(-)(+)(+)=(-)&F\\
\end{array}$
Therefore, for a speed range of $\left(40,50\right)$
