Answer:
[tex]=\sqrt[10]{y}[/tex]
So option c is the correct answer
Step-by-step explanation:
[tex]\left(\frac{1}{\sqrt{y}}\right)^{\frac{-1}{5}}\\\mathrm{Apply\:exponent\:rule}:\quad \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c}\\\left(\frac{1}{\sqrt{y}}\right)^{\frac{-1}{5}}=\frac{1^{\frac{-1}{5}}}{\left(\sqrt{y}\right)^{\frac{-1}{5}}}\\=\frac{1^{\frac{-1}{5}}}{\left(\sqrt{y}\right)^{\frac{-1}{5}}}\\\mathrm{Apply\:rule}\:1^a=1\\1^{\frac{-1}{5}}=1\\=\frac{1}{\left(\sqrt{y}\right)^{\frac{-1}{5}}}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}[/tex]
[tex]=\frac{1}{\left(\sqrt{y}\right)^{-\frac{1}{5}}}\\\left(\sqrt{y}\right)^{-\frac{1}{5}}=\frac{1}{\sqrt[10]{y}}\\=\frac{1}{\frac{1}{\sqrt[10]{y}}}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{1}{\frac{b}{c}}=\frac{c}{b}\\=\frac{\sqrt[10]{y}}{1}\\\mathrm{Apply\:rule}\:\frac{a}{1}=a\\=\sqrt[10]{y}[/tex]
