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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

y = - 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

\frac{x ^{\frac{1}{7}}}{5} + 6

f (x) = \frac{3}{3}

7 x^{2} + 9

f (x) = ln (x - 1) + 1

arccos (\frac{9}{14})

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