Question
Answer
$$9\cdot\dfrac{y}{27}y^2=\dfrac{9y^3}{27}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}9 \cdot \frac{y}{27}y^2& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9y}{27}y^2 \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{9y^3}{27}\end{aligned} $$ | |
| ① | Multiply $9$ by $ \dfrac{y}{27} $ to get $ \dfrac{ 9y }{ 27 } $. Step 1: Write $ 9 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 9 \cdot \frac{y}{27} & \xlongequal{\text{Step 1}} \frac{9}{\color{red}{1}} \cdot \frac{y}{27} \xlongequal{\text{Step 2}} \frac{ 9 \cdot y }{ 1 \cdot 27 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 9y }{ 27 } \end{aligned} $$ |
| ② | Multiply $ \dfrac{9y}{27} $ by $ y^2 $ to get $ \dfrac{ 9y^3 }{ 27 } $. Step 1: Write $ y^2 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{9y}{27} \cdot y^2 & \xlongequal{\text{Step 1}} \frac{9y}{27} \cdot \frac{y^2}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 9y \cdot y^2 }{ 27 \cdot 1 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 9y^3 }{ 27 } \end{aligned} $$ |
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