Question

The value of $\frac{(0.6{)}^0-(0.1{)}^{-1}}{\left(\frac{3}{2^3}{)}^{-1}\right(\frac{3}{2}{)}^3+(-\frac{1}{3}{)}^{-1}}$ is:

Answer 1

On solving the given expression,

$\frac{(0.6{)}^0-(0.1{)}^{-1}}{\left(\frac{3}{2^3}{)}^{-1}\right(\frac{3}{2}{)}^3+(-\frac{1}{3}{)}^{-1}}$

we get

$\frac{1-\frac{1}{(0.1{)}^1}}{\left(\frac{2^3}{3}{)}^1\right(\frac{3}{2}{)}^3+(\frac{-3}{1}{)}^1}$

$[\because x^0=1and(\frac{a}{b}{)}^{-n}=\left(\frac{b}{a}{)}^n\right]$ $\left(1Mark\right)$

$=\frac{1-10}{\frac{2^3}{3}\times \frac{3^3}{2^3}+(-3)}\left[\right(\frac{a}{b}{)}^n=\frac{a^n}{b^n}]$

$=\frac{-9}{3^2+(-3)}=\frac{-9}{9-3}=-\frac{9}{6}=-\frac{3}{2}$ $\left(2Marks\right)$