What is (-2c^{-4}d^{-1})(5c^2d^{-3}) ?

The solution to (-2c^{-4}d^{-1})(5c^2d^{-3}) is -(10)/(c^2d^4)

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simplify (-2c^{-4}d^{-1})(5c^2d^{-3})

simplify

(- 2 c^{- 4} d^{- 1}) (5 c^{2} d^{- 3})

- \frac{10}{c ^{2} d ^{4}}

Solution steps

(- 2 c^{- 4} d^{- 1}) (5 c^{2} d^{- 3})

Apply rule:

(- a) = - a

(- 2 c^{- 4} d^{- 1}) = - 2 c^{- 4} d^{- 1}

= - 2 c^{- 4} d^{- 1} \cdot 5 c^{2} d^{- 3}

Multiply the numbers:

- 2 \cdot 5 = - 10

= - 10 c^{- 4} d^{- 1} c^{2} d^{- 3}

Simplify

c^{- 4} c^{2} : \frac{1}{c ^{2}}

= - 10 \cdot \frac{1}{c ^{2}} d^{- 1} d^{- 3}

Simplify

d^{- 1} d^{- 3} : \frac{1}{d ^{4}}

= - 10 \cdot \frac{1}{c ^{2}} \cdot \frac{1}{d ^{4}}

Simplify

- 10 \cdot \frac{1}{c ^{2}} \cdot \frac{1}{d ^{4}} : - \frac{10}{c ^{2} d ^{4}}

= - \frac{10}{c ^{2} d ^{4}}

- 4 x \geq 0

\frac{π}{2} \cdot 4

(x^{\frac{1}{2}} y^{- \frac{2}{3}})^{- 6}

x^{2} + 2 x - 19 = 0

2 x^{3} - 5 x^{2} + 4 x - 10

What is (-2c^{-4}d^{-1})(5c^2d^{-3}) ?
The solution to (-2c^{-4}d^{-1})(5c^2d^{-3}) is -(10)/(c^2d^4)