What is the value of $\frac{3 m^{2} - m n}{2 m n^{2}}$ $(3m^2 - mn)/(2mn^2)$ when $m = - 1 and n = 2$ $m = -1 and n=2$ ?

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What is the value of $\frac{3 m^{2} - m n}{2 m n^{2}}$ $(3m^2 - mn)/(2mn^2)$ when $m = - 1 and n = 2$ $m = -1 and n=2$ ?

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Answer 1 · 2016-10-20T21:34:42

$- 0.0625$ $-0.0625$

Explanation:

$\frac{3 m^{2} - m n}{2 m n^{2}}, m = - 1, n = 2 =$ $(3m^2-mn)/(2mn^2), m=-1, n=2=$

$\frac{3 {(- 1)}^{2} - (- 1) (2)}{2 (- 1) {(2)}^{2}} = \frac{3 - - 2}{2 (- 4)} = \frac{5}{- 8} = - 0.0625$ $(3(-1)^2-(-1)(2))/(2(-1)(2)^2)=(3--2)/(2(-4))=5/-8=-0.0625$

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What is the value of $\frac{3 m^{2} - m n}{2 m n^{2}}$ $(3m^2 - mn)/(2mn^2)$ when $m = - 1 and n = 2$ $m = -1 and n=2$ ?

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What is the value of $\frac{3 m^{2} - m n}{2 m n^{2}}$ $(3m^2 - mn)/(2mn^2)$ when $m = - 1 and n = 2$ $m = -1 and n=2$ ?