Question

Help solve the above equation ​

Answer 1

Answer:

$\dfrac{1}{27}$

Step-by-step explanation:

$n^{-\frac{2}{3}}=9$

Rewrite:

$\dfrac{1}{\sqrt[3]{n^2}}=9\\\\9\sqrt[3]{n^2}=1$

Cube both sides:

$729n^2=1$

Divide both sides by 729:

$n^2=\dfrac{1}{729}$

Take the square root of both sides:

$n=\sqrt{\dfrac{1}{729}}=\dfrac{1}{27}$

Hope this helps!