What is (8x^4y^{-3})(1/2 x^{-5}y^2) ?

The solution to (8x^4y^{-3})(1/2 x^{-5}y^2) is 4/(yx)

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simplify (8x^4y^{-3})(1/2 x^{-5}y^2)

simplify

(8 x^{4} y^{- 3}) (\frac{1}{2} x^{- 5} y^{2})

\frac{4}{y x}

Solution steps

(8 x^{4} y^{- 3}) (\frac{1}{2} x^{- 5} y^{2})

Simplify

8 x^{4} y^{- 3} : \frac{8 x ^{4}}{y ^{3}}

= \frac{8 x ^{4}}{y ^{3}} \cdot \frac{1}{2} x^{- 5} y^{2}

Apply exponent rule:

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

x^{- 5} = \frac{1}{x ^{5}}

= \frac{8 x ^{4}}{y ^{3}} \cdot \frac{1}{2} \cdot \frac{1}{x ^{5}} y^{2}

Apply the fraction rule:

a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{a \cdot b \cdot d}{c \cdot e}

= \frac{8 x ^{4} \cdot 1 \cdot 1 \cdot y ^{2}}{y ^{3} \cdot 2 x ^{5}}

Multiply the numbers:

8 \cdot 1 \cdot 1 = 8

= \frac{8 x ^{4} y ^{2}}{y ^{3} \cdot 2 x ^{5}}

Factor the number:

8 = 2 \cdot 4

= \frac{2 \cdot 4 x ^{4} y ^{2}}{y ^{3} \cdot 2 x ^{5}}

Cancel the common factor:

2

= \frac{4 x ^{4} y ^{2}}{y ^{3} x ^{5}}

Simplify

\frac{x ^{4}}{x ^{5}} : \frac{1}{x}

= \frac{4 y ^{2}}{y ^{3} x}

Simplify

\frac{y ^{2}}{y ^{3}} : \frac{1}{y}

= \frac{4}{y x}

\frac{6}{x - 4} \leq 2

3 x^{2} - 3 x - 5 = 0

88

\frac{1}{3 \cdot 0.00000005}

x (x + 2) = 5

What is (8x^4y^{-3})(1/2 x^{-5}y^2) ?
The solution to (8x^4y^{-3})(1/2 x^{-5}y^2) is 4/(yx)