# How do you simplify the expression (16q^0r^-6)/(4q^-3r^-7) using the properties?

Apr 20, 2017

$\frac{16 {q}^{0} {r}^{-} 6}{4 {q}^{-} 3 {r}^{-} 7} = 4 {q}^{3} r$

#### Explanation:

$\frac{16 {q}^{0} {r}^{-} 6}{4 {q}^{-} 3 {r}^{-} 7}$

$= \frac{16 \cancel{\textcolor{red}{{q}^{0}}} \cancel{\textcolor{red}{1}} {r}^{-} 6}{4 {q}^{-} 3 {r}^{-} 7} \to$ Everything to the power of 0 is 1. Everything multiplied by 1 is itself/the same number.

$= \frac{16 \cancel{\textcolor{\mathmr{and} a n \ge}{{r}^{-} 6}} \textcolor{b l u e}{{q}^{3}} \textcolor{g r e e n}{{r}^{7}}}{4 \cancel{\textcolor{b l u e}{{q}^{-} 3}} \cancel{\textcolor{g r e e n}{{r}^{-} 7}} \textcolor{\mathmr{and} a n \ge}{{r}^{6}}} \to$ Negative exponents will bring the base and its POSITIVE exponent across the fraction bar.

$= \frac{\cancel{\textcolor{b r o w n}{16}} \textcolor{m a \ge n t a}{4 \cdot \cancel{4}} {q}^{3} {r}^{7}}{\cancel{\textcolor{m a \ge n t a}{4}} {r}^{6}} \to$ $16 = 4 \times 4$. You can now cancel a four in the numerator (above fraction bar) and the denominator (below fraction bar).

$= \frac{4 {q}^{3} \cancel{\textcolor{g o l d}{{r}^{7}}} \textcolor{g o l d}{r}}{\cancel{\textcolor{g o l d}{{r}^{6}}} \cancel{\textcolor{g o l d}{1}}} \to$ Cancel like exponents in the numerator and the denominator.

$= 4 {q}^{3} r \to$ Final answer!