What is (4d^2t^5v^{-4})(-5dt^{-3}v^{-1}) ?

The solution to (4d^2t^5v^{-4})(-5dt^{-3}v^{-1}) is -(20d^3t^2)/(v^5)

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simplify (4d^2t^5v^{-4})(-5dt^{-3}v^{-1})

simplify

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

- \frac{20 d ^{3} t ^{2}}{v ^{5}}

Solution steps

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

Apply rule:

a (- b) = - ab

(4 d^{2} t^{5} v^{- 4}) (- 5 d t^{- 3} v^{- 1}) = - 4 d^{2} t^{5} v^{- 4} \cdot 5 d t^{- 3} v^{- 1}

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

Multiply the numbers:

- 4 \cdot 5 = - 20

= - 20 d^{2} t^{5} v^{- 4} d t^{- 3} v^{- 1}

Simplify

d^{2} d : d^{3}

= - 20 d^{3} t^{5} v^{- 4} t^{- 3} v^{- 1}

Simplify

t^{5} t^{- 3} : t^{2}

= - 20 d^{3} t^{2} v^{- 4} v^{- 1}

Simplify

v^{- 4} v^{- 1} : \frac{1}{v ^{5}}

= - 20 d^{3} t^{2} \frac{1}{v ^{5}}

Simplify

- 20 d^{3} t^{2} \frac{1}{v ^{5}} : - \frac{20 d ^{3} t ^{2}}{v ^{5}}

= - \frac{20 d ^{3} t ^{2}}{v ^{5}}

lo g_{5} (5) = x

\frac{y}{2} \geq 4

39 - 3 n = - 6 (4 n - 3)

9 b^{2} x - 16 y - 16 x + 9 b^{2} y

3 j^{2} + 5 j + 2

What is (4d^2t^5v^{-4})(-5dt^{-3}v^{-1}) ?
The solution to (4d^2t^5v^{-4})(-5dt^{-3}v^{-1}) is -(20d^3t^2)/(v^5)