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

The solution to ((8x^3y^-4))/((2x^5y^2)) is (16)/(x^2y^2)

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

simplify

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

\frac{16}{x ^{2} y ^{2}}

Solution steps

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

Multiply the numbers:

8 \cdot 4 = 32

= \frac{32 x ^{3} y ^{-}}{2 x ^{5} y ^{2}}

Divide the numbers:

\frac{32}{2} = 16

= \frac{16 x ^{3} y ^{-}}{x ^{5} y ^{2}}

Apply exponent rule:

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

\frac{x ^{3}}{x ^{5}} = \frac{1}{x ^{5 - 3}}

= \frac{16 y}{y ^{2} x ^{5 - 3}}

Subtract the numbers:

5 - 3 = 2

= \frac{16 y}{x ^{2} y ^{2}}

Apply exponent rule:

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

\frac{y ^{-}}{y ^{2}} = \frac{1}{y ^{2 -}}

= \frac{16}{x ^{2} y ^{2}}

\frac{y}{3 + y}

\frac{x ^{2} + 2 x - 80}{2 x ^{3} - 24 x ^{2} + 64 x}

\frac{3}{3} + 1

lo g_{10} (1.25)

(x + \frac{1}{3})^{2}

What is ((8x^3y^-4))/((2x^5y^2)) ?
The solution to ((8x^3y^-4))/((2x^5y^2)) is (16)/(x^2y^2)