Question

Simplify the question and get 10 points

Answer 1

$\left\{\left(\dfrac{1}{3}\right)^{-2}-\left(\dfrac{1}{2}\right)^{-3}\right\}\div\bigg( \dfrac{1}{4}^{} \bigg)^{-2}$

SOLUTION:-

As we know that from exponent and laws,

$\bigg(\dfrac{a}{b} \bigg)^{-n} =\bigg(\dfrac{b}{a} \bigg)^{n}$

By using this formula we get

$\bigg(\dfrac{1}{3} \bigg)^{-2}=\bigg( \dfrac{3}{1} \bigg)^2$

$=(3)^2$

$=9$

So,

$\bigg(\dfrac{1}{3} \bigg)^{-2}= 9$

$\bigg(\dfrac{1}{2} \bigg)^{-3}=\bigg(\dfrac{2}{1} \bigg)^3$

$=(2)^3$

$=8$

So,

$\bigg(\dfrac{1}{2} \bigg)^{-3}=8$

$\bigg(\dfrac{1}{4} \bigg)^{-2}=\bigg(\dfrac{4}{1} \bigg)^2$

$=(4)^2$

$= 16$

So,

$\bigg(\dfrac{1}{4} \bigg)^{-2}=16$

$\left\{\left(\dfrac{1}{3}\right)^{-2}-\left(\dfrac{1}{2}\right)^{-3}\right\}\div\bigg(\dfrac{1}{4} \bigg)^{-2}$

$(9-8)\div 16$

$1\div 16$

$\dfrac{1}{16}$

Know more some exponent and laws :-

$a^m \times a^n = a^{m+n}$

$\bigg(\dfrac{a}{b} \bigg)^n=\dfrac{a^n}{b^n}$

$\dfrac{1}{a^n} =a^{-n}$

$\dfrac{a^m}{a^n} = a^{m-n}$

Bases are equal powers also should be equal

$a^0=1$

$1^n=1$

Answer 2

Answer:

{(1/3)-square - (1/2)-cube}/(1/4)-square

= ( 3 square - 2 cube)/4 square

= (9-8)/16

=1/16