Algebra 1: Common Core (15th Edition)

$9v^{\frac{5}{2}}$
$=(-3)(-3)\cdot(v^{2}\cdot v^{\frac{1}{2}})$ $=+9\cdot v^{2+\frac{1}{2}},\qquad$ ...because $a^{m}\cdot a^{n}=a^{m+n}$ $=9v^{\frac{4+1}{2}}$ $=9v^{\frac{5}{2}}$