Ask a New Question

Question

npr=3024 find r
(2)dy/dx=secx-tanx/secx+tanx. plz help
10 years ago

Answers

Steve
If npr=3024, r=3024/np

y' = (secx-tanx)/(secx+tanx)
y' = (secx-tanx)^2/tan^2x
y' = csc^2x - 2cscx + 1

Those are straightforward integrals
10 years ago

Related Questions

find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivat... Find secx if sinx = -4/5 and 270 < x < 360. tan^2x+1=sec^2x (-4/5)^2+1=sec^2x 16/25+1=sec^2x 1... find x, tanxsinx= 3sinx-secx secx/secx-tanx=sec^2x+ secx+tanx (cosxcotx/secx+tanx)+(sinx/secx-tanx) (cotx)(cscx)+secx=(csc^2x)(secx) Please help to verify this trigonometric identity. cos2x=secx find x secx = 4 in Q2 find sin2x,cos2x,tan2x 24 of 3024 of 30 Items Question Which of the following correctly describes how heat mo... 24 of 3024 of 30 Items 23:36 Question What happened for the first time in the Presidential election...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use