Question

the cube of m varies inversely as the square root of n. which two equations model this relationship?\\(m^3 = \frac{k}{\sqrt{n}}\\)\\(\frac{m^3}{\sqrt{n}} = k\\)\\(m = \frac{k}{n}\\)\\(m^3 = kn^{\frac{1}{2}}\\)\\(m^3n^{\frac{1}{2}} = k\\)\\(mn = k\\)

Explanation:

Step1: Define inverse variation

If $a$ varies inversely as $b$, then $a = \frac{k}{b}$ where $k$ is a constant.
Here, $a = m^3$, $b = \sqrt{n} = n^{\frac{1}{2}}$.

Step2: Write base inverse equation

$m^3 = \frac{k}{\sqrt{n}}$

Step3: Rearrange to find equivalent form

Multiply both sides by $\sqrt{n}$:
$m^3 \sqrt{n} = k$
Or using exponent notation: $m^3 n^{\frac{1}{2}} = k$

Answer:

A. $m^3 = \frac{k}{\sqrt{n}}$
E. $m^3 n^{\frac{1}{2}} = k$