QUESTION IMAGE
Question
- $\frac{r^2}{2r^3}$
- $\frac{3n^4}{3n^3}$
- $\frac{3m^{-4}}{m^3}$
- $\frac{4x^0y^{-2}z^3}{4x}$
Problem 21: $\boldsymbol{\frac{r^2}{2r^3}}$
Step1: Use exponent rule $a^m / a^n = a^{m - n}$
$\frac{r^2}{2r^3} = \frac{1}{2} \cdot r^{2 - 3}$
Step2: Simplify the exponent
$r^{2 - 3} = r^{-1} = \frac{1}{r}$, so $\frac{1}{2} \cdot \frac{1}{r} = \frac{1}{2r}$
Step1: Cancel the common factor 3
$\frac{3n^4}{3n^3} = \frac{n^4}{n^3}$
Step2: Use exponent rule $a^m / a^n = a^{m - n}$
$n^{4 - 3} = n^1 = n$
Step1: Use exponent rule $a^m / a^n = a^{m - n}$
$3 \cdot m^{-4 - 3}$
Step2: Simplify the exponent
$m^{-7} = \frac{1}{m^7}$, so $3 \cdot \frac{1}{m^7} = \frac{3}{m^7}$
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$\frac{1}{2r}$