What is ((-2a^3b^2c^0)/(3a^2b^3c^7))^{-2} ?

The solution to ((-2a^3b^2c^0)/(3a^2b^3c^7))^{-2} is (9b^2c^{14})/(4a^2)

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simplify ((-2a^3b^2c^0)/(3a^2b^3c^7))^{-2}

simplify

(\frac{- 2 a ^{3} b ^{2} c ^{0}}{3 a ^{2} b ^{3} c ^{7}})^{- 2}

\frac{9 b ^{2} c ^{14}}{4 a ^{2}}

Solution steps

(\frac{- 2 a ^{3} b ^{2} c ^{0}}{3 a ^{2} b ^{3} c ^{7}})^{- 2}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

(\frac{- 2 a ^{3} b ^{2} c ^{0}}{3 a ^{2} b ^{3} c ^{7}})^{- 2} = \frac{1}{( \frac{- 2 a ^{3} b ^{2} c ^{0}}{3 a ^{2} b ^{3} c ^{7}} ) ^{2}}

= \frac{1}{( \frac{- 2 a ^{3} b ^{2} c ^{0}}{3 a ^{2} b ^{3} c ^{7}} ) ^{2}}

Simplify

(\frac{- 2 a ^{3} b ^{2} c ^{0}}{3 a ^{2} b ^{3} c ^{7}})^{2} : \frac{4 a ^{2}}{9 b ^{2} c ^{14}}

= \frac{1}{\frac{4 a ^{2}}{9 b ^{2} c ^{14}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{9 b ^{2} c ^{14}}{4 a ^{2}}

5 x^{2} + 6 x - 7 = 0

\frac{( 4 x ^{- 3} y ^{5} ) ^{3}}{( 16 x ^{2} y ^{- 2} ) ^{- 4}}

x^{2} - 5 x + 15 = x (x - 3) - 14 + 5 (x - 2) + 3 (13 - 2 x)

x^{2} = 6 x + 91

\frac{1}{2} 12 - \frac{1}{3} 18 + \frac{3}{4} 48 + \frac{1}{6} 72

What is ((-2a^3b^2c^0)/(3a^2b^3c^7))^{-2} ?
The solution to ((-2a^3b^2c^0)/(3a^2b^3c^7))^{-2} is (9b^2c^{14})/(4a^2)