Answer
$$\frac{m^3}{3n}\frac{-9n^2}{m^4}=-\frac{9m^3n^2}{3m^4n}$$
Explanation
| $$ \begin{aligned}\frac{m^3}{3n}\frac{-9n^2}{m^4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-\frac{9m^3n^2}{3m^4n}\end{aligned} $$ | |
| ① | Multiply $ \dfrac{m^3}{3n} $ by $ \dfrac{-9n^2}{m^4} $ to get $ \dfrac{ -9m^3n^2 }{ 3m^4n } $. Step 1: Multiply numerators and denominators. Step 2: Simplify numerator and denominator. $$ \begin{aligned} \frac{m^3}{3n} \cdot \frac{-9n^2}{m^4} & \xlongequal{\text{Step 1}} \frac{ m^3 \cdot \left( -9n^2 \right) }{ 3n \cdot m^4 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ -9m^3n^2 }{ 3m^4n } \end{aligned} $$ |
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