Answer
$$-\frac{20}{90}n^2=-\frac{2n^2}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}-\frac{20}{90}n^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-\frac{ 20 : \color{orangered}{ 10 } }{ 90 : \color{orangered}{ 10 }} \cdot n^2 \xlongequal{ } \\[1 em] & \xlongequal{ }-\frac{2}{9}n^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-\frac{2n^2}{9}\end{aligned} $$ | |
| ① | Divide both the top and bottom numbers by $ \color{orangered}{ 10 } $. |
| ② | Multiply $ \dfrac{2}{9} $ by $ n^2 $ to get $ \dfrac{ 2n^2 }{ 9 } $. Step 1: Write $ n^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{2}{9} \cdot n^2 & \xlongequal{\text{Step 1}} \frac{2}{9} \cdot \frac{n^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 2 \cdot n^2 }{ 9 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2n^2 }{ 9 } \end{aligned} $$ |
This page was created using
Simplify Rational Expressions