What is the answer to h=(-7536t+k)^{2/5} ?

The answer to h=(-7536t+k)^{2/5} is t=-(h^{5/2}-k)/(7536),t=-(-h^{5/2}-k)/(7536)

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h=(-7536t+k)^{2/5}

h = (- 7536 t + k)^{\frac{2}{5}}

t = - \frac{h ^{\frac{5}{2}} - k}{7536}, t = - \frac{- h ^{\frac{5}{2}} - k}{7536}

Solution steps

h = (- 7536 t + k)^{\frac{2}{5}}

Take both sides of the equation to the power of

5 : h^{5} = (- 7536 t + k)^{2}

h^{5} = (- 7536 t + k)^{2}

Solve

h^{5} = (- 7536 t + k)^{2} : t = - \frac{h ^{\frac{5}{2}} - k}{7536}, t = - \frac{- h ^{\frac{5}{2}} - k}{7536}

t = - \frac{h ^{\frac{5}{2}} - k}{7536}, t = - \frac{- h ^{\frac{5}{2}} - k}{7536}

x^{2} + 3 x + 3 x^{2} + 3 x = 10

4 + 4 x + 5 = x

u - 5 = 7

4 x + 2 = 10 x - 10

8 u + 3 = 6 u + 13

What is the answer to h=(-7536t+k)^{2/5} ?
The answer to h=(-7536t+k)^{2/5} is t=-(h^{5/2}-k)/(7536),t=-(-h^{5/2}-k)/(7536)