Question

Which expression is equivalent to (3m^-1n^2)^4/(2m^-2n)^3?

Answer 1

We can rewrite the expression as

$\frac{ (3 m^{-1} n^{2} )^{4} }{(2 m^{-2} n)^{3} }$
$\frac{ ( \frac{3 n^{2} }{m} )^{4} }{( \frac{2n}{m^{2} } ) ^{3} }$

Using the indices law $(ab)^{2}= a^{2} b^{2}$

$\frac{3^{4} n^{8} } {m^{4} }$÷$\frac{2^{3} n^{3} } {m^{6} }$
$\frac{ 3^{4} n^{8} m^{6} } {2^{3} n^{3} m^{4} }$
$\frac{81 n^{5} m^{2} }{8}$

Answer 2

Answer:

Step-by-step explanation:

Given is an expression involving exponents of m and n.

We have to simplify and give the answer

Numerator is given as

$(3m^{-1} n^2)^4\\=\frac{3^4n^8}{m^4}$

Denominator =

$((2m^{-2}n)^{3} \\=\frac{8n^3}{m^6}$

Now simplify first constant term then m term and n term

i) 81/8 is the first term

ii) $\frac{\frac{1}{m^4} }{\frac{1}{m^6} } =m^2$

iii) $\frac{n^8}{n^3} =n^5$

Combining everything we get

answer as

$\frac{81m^2n^5}{8}$