## Algebra 2 (1st Edition)

$\displaystyle \frac{7}{\sqrt{12}}=\frac{7\sqrt{3}}{6}$
$\displaystyle \frac{7}{\sqrt{12}}\qquad$ ...using the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ , rewrite $\sqrt{12}$ as $\sqrt{4\cdot 3}$ $=\displaystyle \frac{7}{\sqrt{4\cdot 3}}\qquad$ ...use the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $=\displaystyle \frac{7}{\sqrt{4}\cdot\sqrt{3}}\qquad$ ...simplify ($\sqrt{4}=2$) $=\displaystyle \frac{7}{2\sqrt{3}}\qquad$ ...rationalize the denominator by multyplying both the numerator and the denominator with $\sqrt{3}$. $=\displaystyle \frac{7\cdot\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}}\qquad$ ...simplify $=\displaystyle \frac{7\sqrt{3}}{6}$