How do you simplify $\frac{3 b}{- 5 c^{- 1}}$ $(3b)/(-5c^-1)$ and write it using only positive exponents?

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Answer 1 · 2017-07-23T10:36:28

$\frac{3 b c}{- 5}$ $(3bc)/(-5)$ or $- \frac{3 b c}{5}$ $-(3bc)/5$

$x^{- a}$ $x^-a$ generally equates to $\frac{1}{x^{a}}$ $1/x^a$ , so:
$\frac{1}{c^{- 1}} = {(c^{- 1})}^{- 1}$ $1/c^-1=(c^-1)^-1$

$= c^{- 1 \cdot (- 1)}$ $=c^(-1\cdot(-1))$ applying exponential multiplication
$= c^{1}$ $=c^1$ simplified; also equals $c$ $c$

replacing into original:

$\frac{3 b}{- 5 c^{- 1}}$ $(3b)/(-5c^-1)$
$\frac{3 b c}{- 5}$ $(3bc)/(-5)$ or $- \frac{3 b c}{5}$ $-(3bc)/5$

How do you simplify (3b)/(-5c^-1)3b−5c−1(3b)/(-5c^-1) and write it using only positive exponents?