What is y^{-5}(2x y/(\frac{x){3y^2}}) ?

The solution to y^{-5}(2x y/(\frac{x){3y^2}}) is 6/(y^2)

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simplify y^{-5}(2x y/(\frac{x){3y^2}})

simplify

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

\frac{6}{y ^{2}}

Solution steps

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

Apply exponent rule:

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

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

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

Simplify

\frac{y}{\frac{x}{3 y ^{2}}} : \frac{y ^{3} \cdot 3}{x}

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

Cross-cancel common factor:

x

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

Apply the fraction rule:

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

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

Multiply the numbers:

1 \cdot 2 \cdot 3 = 6

= \frac{6 y ^{3}}{y ^{5}}

Simplify

\frac{y ^{3}}{y ^{5}} : \frac{1}{y ^{2}}

= \frac{6}{y ^{2}}

\frac{c}{4} = 16

∣ 15 x ∣ \leq 0

2 x = 7 - x^{2}

x^{3} - 2 x^{2} - 5 x + 6

- 5 u = - 6

What is y^{-5}(2x y/(\frac{x){3y^2}}) ?
The solution to y^{-5}(2x y/(\frac{x){3y^2}}) is 6/(y^2)