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 $\frac{1}{\sqrt{y}} = y^{-\frac{1}{2}}$. Then the original expression $\left(\frac{1}{\sqrt{y}}\right)^{-\frac{1}{5}}$ becomes $\left(y^{-\frac{1}{2}}\right)^{-\frac{1}{5}}$.

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

The power - of - a - power rule states that $(a^{m})^{n}=a^{mn}$. For $\left(y^{-\frac{1}{2}}\right)^{-\frac{1}{5}}$, we have $a = y$, $m=-\frac{1}{2}$, and $n = -\frac{1}{5}$. So $mn=\left(-\frac{1}{2}\right)\times\left(-\frac{1}{5}\right)=\frac{1}{10}$. Then $\left(y^{-\frac{1}{2}}\right)^{-\frac{1}{5}}=y^{\frac{1}{10}}$.

Step3: Rewrite the fractional exponent as a radical

A fractional exponent $\frac{m}{n}$ can be written as a radical: $a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$. When $m = 1$ and $n = 10$, $y^{\frac{1}{10}}=\sqrt[10]{y}$.

Answer:

$\sqrt[10]{y}$