Answer
$\dfrac{\sqrt{270y^{2}}}{5\sqrt{3y^{-4}}}=\dfrac{3}{5}y^{3}\sqrt{10}$
Work Step by Step
$\dfrac{\sqrt{270y^{2}}}{5\sqrt{3y^{-4}}}$
Rewrite this expression as $\dfrac{1}{5}\sqrt{\dfrac{270y^{2}}{3y^{-4}}}$ and evaluate the division inside the square root:
$\dfrac{\sqrt{270y^{2}}}{5\sqrt{3y^{-4}}}=\dfrac{1}{5}\sqrt{\dfrac{270y^{2}}{3y^{-4}}}=\dfrac{1}{5}\sqrt{90y^{2+4}}=\dfrac{1}{5}\sqrt{90y^{6}}=...$
Rewrite the expression inside the square root as $\sqrt{9\cdot10\cdot y^{6}}$ and simplify:
$...=\dfrac{1}{5}\sqrt{9\cdot10\cdot y^{6}}=\dfrac{3}{5}y^{3}\sqrt{10}$
