Question

Multiply $\frac{-5}{8}{x}^2{y}^3$ by $\frac{-16}{15}{x}^{3}y$ and find the value for $x=-1$, $y=2$.

• A. $-\frac{32}{3}$
• B. $\frac{32}{4}$
• C. $-\frac{46}{5}$
• D. $\frac{67}{6}$

Answer 1

The correct option is A $-\frac{32}{3}$

$-\frac{5}{8}{x}^2{y}^3\times -\frac{16}{15}{x}^{3}y=\frac{2}{3}{x}^5{y}^4$

Put $x=-1$ and $y=2$

$-\frac{5}{8}{(-1)}^2{\left(2\right)}^3\times -\frac{16}{15}{(-1)}^3\left(2\right)=-\frac{32}{3}\frac{2}{3}{x}^5{y}^4=\frac{2}{3}{(-1)}^5{\left(2\right)}^4=-\frac{32}{3}$