Simplify \(\left(\frac{8 \sqrt{n}}{m^{3}}\right)\left(\frac{4^{-1} m^{2}}{2 n^{-2}}\right)\)
A. \(128 n^{3} m^{-1}\)
B. \(8 n^{2} m^{1}\)
C. \(8 n^{4} m\)
D. \(8 n^{3} m\)
Correct Answer: B
Explanation
To simplify
\(\left(\frac{8 \sqrt{n}}{m}\right)^{2} \times\left(\frac{4^{-1} m^{2}}{2 n^{-2}}\right)=\left(\frac{8 \times n^{2}}{m^{2}}\right)^{2} \times \frac{4^{-1} m^{2}}{2 n^{-2}}\)xpanding the term on the L.H.S
\(\left.\frac{8^{2} \times n^{1}}{m^{2}}\right)^{2} \times \frac{4^{-1} m^{2}}{2 n^{-2}}\)
in another form. we have
\(64 m^{-2} \times \frac{1 \times m^{2} \times n^{2}}{4 \times 2}=\frac{64 m m^{2}}{8} \times m^{2} n^{2}\)
\(8 \mathrm{~mm}^{-2} \times \mathrm{m}^{2} \times \mathrm{n}^{2}=8 \times n^{1+2} \times \mathrm{m}^{-3+2}\)
\(=8 n m^{-1}\)