What is the answer to solvefor y,2y=x^{2/3}+4x^2 ?

The answer to solvefor y,2y=x^{2/3}+4x^2 is y=(x^{2/3}+4x^2)/2

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solvefor y,2y=x^{2/3}+4x^2

solve for

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

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

Solution steps

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

Divide both sides by

2

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

4 = b

\frac{x + 3}{7} = \frac{- 2}{3}

a, \frac{2}{( 7 + 3 5 )} = a - b \cdot 5

b, (b) (- \frac{5}{8})^{x - 9} = 1

11 n + 1 = 1

What is the answer to solvefor y,2y=x^{2/3}+4x^2 ?
The answer to solvefor y,2y=x^{2/3}+4x^2 is y=(x^{2/3}+4x^2)/2