Question

If mn ≠ 0, then 1/n^2(m^5n^3/m^3)^2=
A. mn^4 B.m^4n^2 C.m^4n^3 D.m^4n^4 E.m^4n^5

Answer 1

\[\left(\begin{matrix}1 \\ n^2\end{matrix}\right)\left(\begin{matrix}m^5n^3 \\ m^3\end{matrix}\right)^2\]
When dividing by exponents, you subtract the exponent on the denominator from numerator of the same variable. So, m^5 - m^3 = m^2
Next step is to square it. (Remember PEMDAS?) When you square exponents you basically multiply the exponents. So, 2 * 2 = 4 and 2 * 3 = 6.
\[\left(\begin{matrix}1 \\ n^2\end{matrix}\right)\left(\begin{matrix}m^4n^6\\ 1\end{matrix}\right)\]
Notice that m^3 was subtracted from m^5, so the equation is no longer a fraction. But, in the above equation, I substituted 1 for the denominator so that I can multiply across.
\[\left(\begin{matrix}m^4n^6 \\ n^2\end{matrix}\right)\]
n^2 * 1 = n^2
Now, you can solve for the final answer.
\[\left( m^4n^4 \right)\]
You can only subtract exponential variables of the same kind, so n^6 - n^2 = n^4
You're final answer is D. m^4n^4.