Answer :
To find an equivalent expression for [tex]\(\left(\frac{1}{\sqrt{y}}\right)^{\frac{-1}{5}}\)[/tex], let's simplify it step-by-step.
1. Start with the expression:
[tex]\[ \left(\frac{1}{\sqrt{y}}\right)^{\frac{-1}{5}} \][/tex]
2. Rewrite the square root in the expression using exponents:
[tex]\[ \sqrt{y} = y^{\frac{1}{2}} \][/tex]
So,
[tex]\[ \frac{1}{\sqrt{y}} = y^{-\frac{1}{2}} \][/tex]
3. Substitute into the original expression:
[tex]\[ \left(y^{-\frac{1}{2}}\right)^{-\frac{1}{5}} \][/tex]
4. When raising an exponent to another exponent, multiply the exponents:
[tex]\[ y^{-\frac{1}{2} \cdot -\frac{1}{5}} \][/tex]
5. Calculate the product of the exponents:
[tex]\[ -\frac{1}{2} \times -\frac{1}{5} = \frac{1}{10} \][/tex]
6. The expression simplifies to:
[tex]\[ y^{\frac{1}{10}} \][/tex]
Hence, [tex]\(\left(\frac{1}{\sqrt{y}}\right)^{\frac{-1}{5}}\)[/tex] simplifies to [tex]\(y^{\frac{1}{10}}\)[/tex].
None of the provided options [tex]\(\sqrt[3]{y}\)[/tex], [tex]\(\frac{1}{\sqrt{v^3}}\)[/tex], [tex]\(\sqrt[4]{y^2}\)[/tex], or [tex]\(\frac{1}{197}\)[/tex] match this simplified form.
1. Start with the expression:
[tex]\[ \left(\frac{1}{\sqrt{y}}\right)^{\frac{-1}{5}} \][/tex]
2. Rewrite the square root in the expression using exponents:
[tex]\[ \sqrt{y} = y^{\frac{1}{2}} \][/tex]
So,
[tex]\[ \frac{1}{\sqrt{y}} = y^{-\frac{1}{2}} \][/tex]
3. Substitute into the original expression:
[tex]\[ \left(y^{-\frac{1}{2}}\right)^{-\frac{1}{5}} \][/tex]
4. When raising an exponent to another exponent, multiply the exponents:
[tex]\[ y^{-\frac{1}{2} \cdot -\frac{1}{5}} \][/tex]
5. Calculate the product of the exponents:
[tex]\[ -\frac{1}{2} \times -\frac{1}{5} = \frac{1}{10} \][/tex]
6. The expression simplifies to:
[tex]\[ y^{\frac{1}{10}} \][/tex]
Hence, [tex]\(\left(\frac{1}{\sqrt{y}}\right)^{\frac{-1}{5}}\)[/tex] simplifies to [tex]\(y^{\frac{1}{10}}\)[/tex].
None of the provided options [tex]\(\sqrt[3]{y}\)[/tex], [tex]\(\frac{1}{\sqrt{v^3}}\)[/tex], [tex]\(\sqrt[4]{y^2}\)[/tex], or [tex]\(\frac{1}{197}\)[/tex] match this simplified form.