Question

Which expression is equivalent to StartFraction (3 m Superscript negative 2 Baseline n) Superscript negative 3 Baseline Over 6 m n Superscript negative 2 Baseline EndFraction? Assume m not-equals 0, n not-equals 0.
StartFraction m Superscript 5 Baseline Over 162 n EndFraction
StartFraction 1 Over 2 m cubed n EndFraction
StartFraction 8 m Superscript 9 Baseline Over n Superscript 9 Baseline EndFraction
StartFraction 4 m Superscript 8 Baseline Over 3 n cubed EndFraction

Answer 1

GIVEN :

The expression is $\frac{(3m^{-3}n)^{-3}}{6mn^{-2}}$

TO FIND :

The equivalent expression to the given expression.

SOLUTION :

Given that the expression is $\frac{(3m^{-3}n)^{-3}}{6mn^{-2}}$

Assume $m\neq 0$, $n\neq 0$

To find the equivalent expression to the given expression:

Solving the given expression we get,

$\frac{(3m^{-2}n)^{-3}}{6mn^{-2}}$ for $m\neq 0$, $n\neq 0$

By using the property of power rule is given by:

$(abc)^m=a^mb^mc^m$

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

By using the property of power rule is given by:

$(a^m)^n=a^{mn}$

$=\frac{3^{-3}m^6n^{-3}}{6mn^{-2}}$

By using the property of Exponent is given by:

$a^{-m}=\frac{1}{a^m}$

$=\frac{m^6}{3^3.6mn^{-2}.n^3}$

By using the property of exponent is given by:

$a^m.a^n=a^{m+n}$

$=\frac{m^6.m^{-1}}{3^3.6n^{-2+3}}$

By using the property of exponent is given by:

$\frac{1}{a^m}=a^{-m}$

$=\frac{m^{6-1}}{162n^1}$

By using the property of exponent is given by:

$a^m.a^n=a^{m+n}$

$=\frac{m^{5}}{162n}$

∴ $\frac{(3m^{-2}n)^{-3}}{6mn^{-2}}=\frac{m^{5}}{162n}$ for $m\neq 0$, $n\neq 0$

∴  the equivalent expression to the given expression $\frac{(3m^{-2}n)^{-3}}{6mn^{-2}}$ is $\frac{m^{5}}{162n}$ for $m\neq 0$, $n\neq 0$

Option A) Start Fraction m Superscript 5 Baseline Over 162 n End Fraction is correct

Answer 2

Answer:

A

Step-by-step explanation: