What is the inverse of f(A)=2pir^2+2pir ?

The inverse of f(A)=2pir^2+2pir is (-pi+sqrt(pi(2r+pi)))/(2pi),-(sqrt(pi(2r+pi))+pi)/(2pi)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(A)=2pir^2+2pir

inverse

f (A) = 2 π r^{2} + 2 π r

\frac{- π + π ( 2 r + π )}{2 π}, - \frac{π ( 2 r + π ) + π}{2 π}

Solution steps

y = 2 π r^{2} + 2 π r

Replace

r

with

y

r = 2 π y^{2} + 2 π y

Solve for

y, r = 2 π y^{2} + 2 π y

\frac{- π + π ( 2 r + π )}{2 π}, - \frac{π ( 2 r + π ) + π}{2 π}

f (x) = \frac{15}{2} lo g_{3} (x - 9) + 4

y = 2 lo g_{3} (x)

f (x) = \frac{( x + 3 )}{( x + 5 )}

f (x) = \frac{5 + x ^{2}}{4}

f (x) = 48 x^{2} - 36

What is the inverse of f(A)=2pir^2+2pir ?
The inverse of f(A)=2pir^2+2pir is (-pi+sqrt(pi(2r+pi)))/(2pi),-(sqrt(pi(2r+pi))+pi)/(2pi)