which expression is equivalent to $\\left(\\frac{1}{\\sqrt{y}}\\right)^{\\frac{-1}{5}}$?

Explanation:

Step1: Rewrite the square - root as a fractional exponent

We know that $\sqrt{y}=y^{\frac{1}{2}}$, so the original expression $\left(\frac{1}{\sqrt{y}}\right)^{-\frac{1}{5}}$ becomes $\left(\frac{1}{y^{\frac{1}{2}}}\right)^{-\frac{1}{5}}$. And $\frac{1}{y^{\frac{1}{2}}}=y^{-\frac{1}{2}}$, then the expression is $(y^{-\frac{1}{2}})^{-\frac{1}{5}}$.

Step2: Apply the power - of - a - power rule

The power - of - a - power rule is $(a^{m})^{n}=a^{mn}$. Here $a = y$, $m=-\frac{1}{2}$ and $n =-\frac{1}{5}$. So $(y^{-\frac{1}{2}})^{-\frac{1}{5}}=y^{(-\frac{1}{2})\times(-\frac{1}{5})}$.

Step3: Calculate the exponent

$(-\frac{1}{2})\times(-\frac{1}{5})=\frac{1}{10}$. So the expression simplifies to $y^{\frac{1}{10}}=\sqrt[10]{y}$.

Answer:

$\sqrt[10]{y}$