What is the inverse of h(t)=-16t^2+30t+5 ?

The inverse of h(t)=-16t^2+30t+5 is -(-15+sqrt(-16t+305))/(16),(sqrt(-16t+305)+15)/(16)

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inverse of h(t)=-16t^2+30t+5

inverse

h (t) = - 16 t^{2} + 30 t + 5

- \frac{- 15 + - 16 t + 305}{16}, \frac{- 16 t + 305 + 15}{16}

Solution steps

y = - 16 t^{2} + 30 t + 5

Replace

t

with

y

t = - 16 y^{2} + 30 y + 5

Solve for

y, t = - 16 y^{2} + 30 y + 5

- \frac{- 15 + - 16 t + 305}{16}, \frac{- 16 t + 305 + 15}{16}

f (x) = 1.25 (38 - x)

y = 9 x

arccos (- 4)

f (x) = lo g_{10} (x - 5) - 4

f (x) = \frac{( 11 x + 9 )}{8 x + 12}

What is the inverse of h(t)=-16t^2+30t+5 ?
The inverse of h(t)=-16t^2+30t+5 is -(-15+sqrt(-16t+305))/(16),(sqrt(-16t+305)+15)/(16)