What is the inverse of h(t)=160t-16t^2 ?

The inverse of h(t)=160t-16t^2 is -(-20+sqrt(-t+400))/4 ,(sqrt(-t+400)+20)/4

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inverse of h(t)=160t-16t^2

inverse

h (t) = 160 t - 16 t^{2}

- \frac{- 20 + - t + 400}{4}, \frac{- t + 400 + 20}{4}

Solution steps

y = 160 t - 16 t^{2}

Replace

t

with

y

t = 160 y - 16 y^{2}

Solve for

y, t = 160 y - 16 y^{2}

- \frac{- 20 + - t + 400}{4}, \frac{- t + 400 + 20}{4}

(0.2) (4.6)

f (x) = 10 x + 3

f (x) = \frac{4}{4 x + 1}

f (x) = \frac{2 x + 24}{x ^{2} + 4 x - 96}

f (x) = 3 x (x^{2} + 4 x + 5)

What is the inverse of h(t)=160t-16t^2 ?
The inverse of h(t)=160t-16t^2 is -(-20+sqrt(-t+400))/4 ,(sqrt(-t+400)+20)/4