Question

If $m=-2$ and $n=2$ ; find the value of :
(iii) $6m^{-3}+4n^2$

Answer 1

Given that : $m=-2$ and $n=2$
Considering $6m^{-3}+4n^2$
Substituting the values of $m$ and $n$ in the above expression , we have :
$6m^{-3}+4n^2=6(-2{)}^{-3}+4(2{)}^2$

$=6\times \frac{1}{-2}\times \frac{1}{-2}\times \frac{1}{-2}+4\times 2\times 2$

$=\frac{-3}{4}+16$

$=\frac{(-3+16\times 4)}{4}$

$=\frac{-3+64}{4}=\frac{61}{4}=15\frac{1}{4}$

Hence, the value of $6m^{-3}+4n^2$ is $15\frac{1}{4}$.