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Mathematics 17 Online
OpenStudy (anonymous):

Solve for T, where 0≤T≤5 -4π^2cos2πt=0

OpenStudy (noelgreco):

Do you know how to calculate the period of the function?

OpenStudy (anonymous):

T = 2pi/k where k is a constant in this case u are having an equation which is simple harmonic motion.. k = 2 so T = pi

OpenStudy (noelgreco):

If the function is cos 2pi t, k is = 2pi, so the period is 1

OpenStudy (anonymous):

oo yea k was 2pi sorry ... sleepy me~ lolz guess i shud go sleep

OpenStudy (anonymous):

didnt see the pi lolx

OpenStudy (noelgreco):

Is mary even in on this?

OpenStudy (anonymous):

I know the answer to the problem (it's t= 1/4, 3/4, 5/4,7/4,9/4,11/4,13/4,15/4,17/4) I just don't understand how you get there

OpenStudy (phi):

if you plot cos( x) it looks like this |dw:1345925763378:dw| so you want your 2 pi t equal to those values

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