Answer
$\frac{18}{5}$
Work Step by Step
We are given $s(t)=5t^2-2t-7$
For $t = 2$ the instantaneous velocity is
$\lim\limits_{b \to 2}\frac{s(b)-s(2)}{b-2}$
$s(b)=5b^2-2b-7$
$s(6)=9$
The instantaneous rate of change is then
$\lim\limits_{b \to 2}\frac{5b^2-2b-7-9}{b-2}=\lim\limits_{b \to 2}\frac{(b-2)(b+\frac{8}{5})}{b-2}=\lim\limits_{b \to 2}(b+\frac{8}{5})=\frac{18}{5}$
