What is the inverse of y=100(1-t/(40))^2 ?

The inverse of y=100(1-t/(40))^2 is -4sqrt(t)+40,4sqrt(t)+40

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of y=100(1-t/(40))^2

inverse

y = 100 (1 - \frac{t}{40})^{2}

- 4 t + 40, 4 t + 40

Solution steps

y = 100 (1 - \frac{t}{40})^{2}

Replace

t

with

y

t = 100 (1 - \frac{y}{40})^{2}

Solve for

y, t = 100 (1 - \frac{y}{40})^{2}

- 4 t + 40, 4 t + 40

What is the inverse of y=100(1-t/(40))^2 ?
The inverse of y=100(1-t/(40))^2 is -4sqrt(t)+40,4sqrt(t)+40