Answer:
xy \sqrt{4y} /3[/tex]
Step-by-step explanation:
[tex]\sqrt[4]{16x^{11-7} }y^{8-2} } /\sqrt[4]{81} \\\= \sqrt[4]{4^{2} x^{4} }y^{6} } /\sqrt[4]{3^{4} } \\\\= ({4^{2} x^{4} }y^{6} })^{\frac{1}{4} } /( {3^{4})^{\frac{1}{4} } \\=({4^{\frac{1}{2} } x }y^{\frac{3}{2} } })/3\\=xy \sqrt{4y} /3[/tex]
