Question

$\left(\frac{-16}{7}{x}^4{y}^{3}z\right)\times \left(\frac{21}{96}{x}^2{yz}^3\right)$ is equal to

• A. $\frac{1}{6}{x}^4{y}^3{z}^4$
• B. $-\frac{1}{2}{x}^6{y}^4{z}^4$
• C. $\frac{1}{2}{x}^6{y}^4{z}^4$
• D. $\frac{5}{6}{x}^4{y}^4{z}^4$

Answer 1

Answer: (B)

$-\frac{1}{2}{x}^6{y}^4{z}^4$

Required product is
$=(-\frac{16}{7}{x}^4{y}^{3}z)\times \left(\frac{21}{96}{x}^2{yz}^3\right)$

$=(\frac{-16}{7}\times \frac{21}{96})\times (x^4\times x^2)\times (y^3\times y)\times (z\times z^3)$
$=-\frac{1}{2}{x}^6{y}^4{z}^4$