Question

$\huge\underline\bold\red{QueStioN}$Divide the sum of 8/3 and 4/7 by the product of -3/7and 14/9.
$<marquee>Don't Spam here$​

Answer 1

Answer:

$the \: sum \: of \: \frac{8}{3} \: and \: \frac{4}{7} is$

$= \frac{8}{3} + \frac{4}{7}$

$= \frac{68}{21}$

$the \: product \: of \: \frac{ - 3}{7} \: and \: \frac{14}{9} \: is$

$= \frac{ - 3}{7} \times \frac{14}{9}$

$= \frac{ - 2}{3}$

Hence the answer is

$\frac{68}{21} \div \frac{ - 2}{3}$

$= \frac{68}{21} \times \frac{3}{ - 2}$

$= \frac{ - 34}{7}$

Answer 2

Answer:

$\frac{-34}{7} = -4 \frac{6}{7}$

Step-by-step explanation:

$\huge\star\bold\red{Given}$

sum of

$\frac{8}{3} \: \: and \: \: \frac{4}{7}$

product of

$\frac{ - 3}{7} \: \: and \: \: \frac{14}{9}$

$\huge\star\bold\blue{Answer}$

so,frist we have to find sum

sum =

$\frac{8}{3} + \frac{4}{7} \\ \frac{56 + 12}{21} \\ \frac{68}{21}$

secondly we have to find products

product =

$\frac{ - 3}{7} \times \frac{ 14}{9} \\ \frac{</strong><strong>-</strong><strong>2}{3}$

Now ,we have to Divide

$\frac{68}{21} \times \frac{3}{2} \\ \frac{-34}{7} = -4 \frac{6}{7}$

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Hope it helpful to you