D dbag May 2019 11 0 moscow May 27, 2019 #1 Hello I have a folllowing question Function A(r)=2*pi*r^{2}+6*pi*r. I has a an inverse function r(A)=(-6*pi+-sqrt(4*pi(9*pi+2r))/4*pi Calculate at what radius does the area of the cylinder equal 200. do i just make the function r(A)=200? Thanks Last edited: May 27, 2019

Hello I have a folllowing question Function A(r)=2*pi*r^{2}+6*pi*r. I has a an inverse function r(A)=(-6*pi+-sqrt(4*pi(9*pi+2r))/4*pi Calculate at what radius does the area of the cylinder equal 200. do i just make the function r(A)=200? Thanks

I Idea Jun 2013 1,127 601 Lebanon May 27, 2019 #2 it should be \(\displaystyle r=\frac{-6 \pi + \sqrt{4\pi (9 \pi +2A)}}{4 \pi } \) now replace $A=200$ Last edited: May 27, 2019 Reactions: 2 people

it should be \(\displaystyle r=\frac{-6 \pi + \sqrt{4\pi (9 \pi +2A)}}{4 \pi } \) now replace $A=200$

T TKHunny Aug 2007 3,171 860 USA May 27, 2019 #3 dbag said: I has a an inverse function r(A)=(-6*pi+-sqrt(4*pi(9*pi+2r))/4*pi Click to expand... This is a very bad mistake. Understanding what a function is and isn't is SO fundamental. Please review the definition. Reactions: 1 person

dbag said: I has a an inverse function r(A)=(-6*pi+-sqrt(4*pi(9*pi+2r))/4*pi Click to expand... This is a very bad mistake. Understanding what a function is and isn't is SO fundamental. Please review the definition.