What is the inverse of 100(1-x/(40))^2 ?

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

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inverse of 100(1-x/(40))^2

inverse

100 (1 - \frac{x}{40})^{2}

- 4 x + 40, 4 x + 40

Solution steps

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

Replace

x

with

y

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

Solve for

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

- 4 x + 40, 4 x + 40

f (x) = \frac{1}{x + 6}

f (x) = 5 x + 7

\frac{1}{2} sin (x + \frac{π}{4})

f (x) = lo g_{2} (1 - ∣ 3 - x ∣)

12 x + 4 y = 4

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