What is the inverse of h(t)=-4.9(t+3)^2+45.8 ?

The inverse of h(t)=-4.9(t+3)^2+45.8 is sqrt(-(10t-458)/(49))-3,-sqrt(-(10t-458)/(49))-3

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inverse of h(t)=-4.9(t+3)^2+45.8

inverse

h (t) = - 4.9 (t + 3)^{2} + 45.8

- \frac{10 t - 458}{49} - 3, - - \frac{10 t - 458}{49} - 3

Solution steps

y = - 4.9 (t + 3)^{2} + 45.8

Replace

t

with

y

t = - 4.9 (y + 3)^{2} + 45.8

Solve for

y, t = - 4.9 (y + 3)^{2} + 45.8

- \frac{10 t - 458}{49} - 3, - - \frac{10 t - 458}{49} - 3

What is the inverse of h(t)=-4.9(t+3)^2+45.8 ?
The inverse of h(t)=-4.9(t+3)^2+45.8 is sqrt(-(10t-458)/(49))-3,-sqrt(-(10t-458)/(49))-3