Divide the sum of $$\dfrac{-4}{9}$$ and $$\dfrac{1}{3}$$ by $$\dfrac{5}{6}$$.

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Hint: In this question we have to divide the sum $$\dfrac{-4}{9}$$ and $$\dfrac{1}{3}$$ by $$\dfrac{5}{6}$$, for this we have to know that how to add two fractions.

Complete step-by-step answer:
We are going to add $$\dfrac{-4}{9}$$ and $$\dfrac{1}{3}$$, and to add these fractions we need to have same denominator, so for this we are going to find the L.C.M(least common multiple) of denominator,
i.e L.C.M(9,3) = 9.

Now we are going to add these fractions,
$$\dfrac{-4}{9} +\dfrac{1}{3}$$
Since, L.C.M is 9 so we have make 9 in the denominator of both fractions
= $$\dfrac{-4}{9} +\dfrac{1\times 3}{3\times 3}$$
= $$\dfrac{-4}{9} +\dfrac{3}{9}$$
= $$\dfrac{\left( -4+3\right) }{9}$$ [ since denominator is same so we can add the numerators easily]
= $$\dfrac{\left( 3-4\right) }{9}$$
= $$\dfrac{-1}{9}$$

Now we have to divide the above result i.e $\dfrac{-1}{9}$ by $\dfrac{5}{6}$.
So we can write,
$$\dfrac{-1}{9} \div \dfrac{5}{6}$$
Since, as we know that $$\dfrac{a}{b} \div \dfrac{c}{d} =\dfrac{a}{b} \times \dfrac{d}{c}$$
So by this we can write, $$\dfrac{-1}{9} \div \dfrac{5}{6}$$ as
$\dfrac{-1}{9} \div \dfrac{5}{6} =\dfrac{-1}{9} \times \dfrac{6}{5}$$
= $$\dfrac{-6}{45}$
Dividing numerator and denominator by 3 we get,
= $\dfrac{-2}{15}$
So $$\dfrac{-2}{25}$$ is our required solution.

Note: In this type of question er generally forgot to take L.C.M of denominators while adding so keep it in mind and also the formula that we have used earlier i.e $$\dfrac{a}{b} \div \dfrac{c}{d} =\dfrac{a}{b} \times \dfrac{d}{c}$$.