Question

Multiply - 4/3xy^3×6/7x^2y. verify result for x=2 y=1​

Answer 1

Answer:

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Answer 2

$\frac{4}{3} xy^{3} \times \frac{6}{7} {x}^{2} y \\ = (\frac{4}{3} \times \frac{6}{7} )(xy ^{3} \times x ^{2} y) \\ = \frac{8}{7} {x}^{3} y ^{4} \\ now \: put \: the \: value \: of \: x \: and \: y \: in \: \frac{4}{3} xy^{3} \times \frac{6}{7} {x}^{2} y \\= \frac{4}{3} (2)(1)^{3} \times \frac{6}{7} ( {2})^{2} (1)\\ \\ = \frac{8}{3} \times \frac{24}{7 } \\ = \frac{64}{7} \\ now \: put \: the \: value \: of \: x \: and \: y \: in \: \: \frac{8}{7} {x}^{3} y ^{4} \\ =\frac{8}{7} ( {2})^{3} (1)^{4} \\ = \frac{64}{7} \\ hence \: proved$