What is ((2x^2y^3)/(3xy^4))^2 ?

The solution to ((2x^2y^3)/(3xy^4))^2 is (4x^2)/(9y^2)

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simplify ((2x^2y^3)/(3xy^4))^2

simplify

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

\frac{4 x ^{2}}{9 y ^{2}}

Solution steps

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Simplify

(2 x^{2} y^{3})^{2} : 4 x^{4} y^{6}

= \frac{4 x ^{4} y ^{6}}{( 3 x y ^{4} ) ^{2}}

Simplify

(3 x y^{4})^{2} : 9 x^{2} y^{8}

= \frac{4 x ^{4} y ^{6}}{9 x ^{2} y ^{8}}

Apply exponent rule:

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

\frac{x ^{4}}{x ^{2}} = x^{4 - 2}

= \frac{4 y ^{6} x ^{4 - 2}}{9 y ^{8}}

Subtract the numbers:

4 - 2 = 2

= \frac{4 x ^{2} y ^{6}}{9 y ^{8}}

Apply exponent rule:

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

\frac{y ^{6}}{y ^{8}} = \frac{1}{y ^{8 - 6}}

= \frac{4 x ^{2}}{9 y ^{8 - 6}}

Subtract the numbers:

8 - 6 = 2

= \frac{4 x ^{2}}{9 y ^{2}}

3 < 15

x^{2} - 1 > 3

0.15 (\frac{1180}{2})

6 y^{2} (x - 8) + 5 (x - 8)

8 x + 7 = - 9

What is ((2x^2y^3)/(3xy^4))^2 ?
The solution to ((2x^2y^3)/(3xy^4))^2 is (4x^2)/(9y^2)