Question

Find the product of $\frac{4}{45}{p}^2{q}^{5}and-7\frac{9}{13}p{q}^3$

Answer 1

Given that: Find the product of $\frac{4}{45}{p}^2{q}^{5}and-7\frac{9}{13}p{q}^3$.

Product of two monomials $=$ (Product of their numerical coefficients) $\times$ (Product of their literal coefficients)

$$\begin{gathered} (\frac{4}{45}{p}^2{q}^5)\times (-7\frac{9}{13}p{q}^3)=(\frac{4}{45}\times -7\frac{9}{13})(p^2{q}^5\times p{q}^3) \\ =(\frac{4}{45}\times -\frac{100}{13})(p^2{q}^5\times p{q}^3) \\ =(\frac{4}{9}\times -\frac{20}{13})(p^2{q}^5\times p{q}^3) \\ =-\frac{80}{117}\times p^3{q}^8 \end{gathered}$$

Therefore, we get $-\frac{80}{117}\times p^3{q}^8$