Monday
December 9, 2013

# Posts by Olivia

Total # Posts: 395

thanks alot!!!!!!!!!!!!!!

:)thanks!!! I've been trying to figure this out for a while now, can you explain this to me: what do I do when i get to something like this (cosx/sinx-1/(sinx))?

Sorry, it was a typo, it's supposed to be: sec/(secx-tanx)=sec^2x+secxtanx

secx/secx-tanx=sec^2x+ secx+tanx

precal
1/tanx-secx+ 1/tanx+secx=-2tanx so this is what I did: =tanx+secx+tanx-secx =(sinx/cosx)+ (1/cosx)+(sinx/cosx)-(1/cosx) =sinx/cosx+ sinx /cosx= -2tanxI but I know this can't be correct because what I did doesn't end as a negatvie:( please help

Pre Cal
the question is supposed to be: cos^2xtan^2xcos^2x=1 =cos^2x+ (sin^2x/cos^2x)(cos^2x/1) *the last 2 cos^2x will cancel out leaving you with: =cos^2x+sin^2x=1

biology
1. to prepare the uterus to support a pregnancy; instead of the body waiting until pregnancy occurs to start this process, each month the uterine lining gets ready to house a fertilized egg before an egg has even had the chance to become fertilized. That way, if an egg does ge...

precal
Thats ok, I think I kind of understand it.

precal
thanks

precal
cosx/secx+sinx/cscx=sec^2x-tan^2x

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