How do you get rid of the negative exponents and simplify $(\frac { 12m ^ { 2} n ^ { 3} p ^ { - 9} } { 20m ^ { - 7} n ^ { - 3} p ^ { - 6} } ) ^ { - 1}$ ?

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Answer 1 · 2018-05-16T00:04:42

$(5p^3)/(3m^9n^6)$

$x^-1=1/x$ . For example:

$2^2=4$
$2^1=2$
$2^0=1$
Note that for every one the exponent goes down, the result is divided by two. Following this pattern gives us:
$2^-1=1/2$

Applying this to the problem gives us

$((12m^2n^3p^-9)/(20m^-7n^-3p^-6))^-1=(20m^-7n^-3p^-6)/(12m^2n^3p^-9)$

Simplifying each individual exponent gives us

$(20p^(9-6))/(12m^(2+7)n^(3+3))=(20p^3)/(12m^9n^6)$

Because $a^x*a^y=a^(x+y)$

Finally, take out the common factors of the coefficients giving us:

$(20p^3)/(12m^9n^6)=(5p^3)/(3m^9n^6)$ , which is the final result

How do you get rid of the negative exponents and simplify (\frac { 12m ^ { 2} n ^ { 3} p ^ { - 9} } { 20m ^ { - 7} n ^ { - 3} p ^ { - 6} } ) ^ { - 1}(\frac { 12m ^ { 2} n ^ { 3} p ^ { - 9} } { 20m ^ { - 7} n ^ { - 3} p ^ { - 6} } ) ^ { - 1}?