Question

please help me out with this question​

Answer 1

|| SOLUTION ||

$( {3}^{ - 1} + {6}^{ - 1} ) \div ( { \dfrac{3}{4} })^{ - 1}$

We can write the following terms like this ,

$\small {3}^{ - 1} = \dfrac{1}{3}$

$\small \: {6}^{ - 1} = \dfrac{1}{6}$

$\small \: ( { \dfrac{3}{4} })^{ - 1} = \dfrac{4}{3}$

The " - 1 " In the power raised refers the inverse of the given term , So we are taking the reciprocal of that number.

Now let us substitute simplified terms in the problem ,

$( \dfrac{1}{3} + \dfrac{1}{6} ) \div \dfrac{4}{3}$

$\dfrac{1}{3} + \dfrac{1}{6} = \dfrac{2 + 1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$

Therefore for 1/3 + 1/6 = 1/2

$\dfrac{1}{2} \div \dfrac{4}{3}$

$= \dfrac{1}{2} \times \dfrac{3}{4}$

$\therefore \: ( {3}^{ - 1} \: + {6}^{ - 1} ) \div ( { \frac{3}{4}) }^{ - 1} = \bf \: \frac{3}{8}$

Hence the simplified answer is 3/8 .