Answer
$36a^{\frac{1}{6}}$
Work Step by Step
Using the rule $a^{m}\times a^{n}=a^{m+n}$, we obtain:
$9a^{\frac{1}{2}}\times 4a^{-\frac{1}{3}}$
$=(9\times4)\times(a^{\frac{1}{2}}\times a^{-\frac{1}{3}})$
$=36(a^{\frac{1}{2}+(-\frac{1}{3})})$
$=36a^{\frac{1(3)-1(2)}{6}}$
$=36a^{\frac{3-2}{6}}$
$=36a^{\frac{1}{6}}$
