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How do you find the amplitude and period of #y=5sin(2/3theta)#?

1 Answer

Jul 19, 2017

#A=5# and #T=3pi#

Explanation:

A function :

#y=Asinωθ# has amplitude equal to #A# and period equal to

#T=(2pi)/ω#

so the amplitude of the function #y=5sin(2/3θ)# is #A=5#

and the period is #T=(2pi)/(2/3)=3pi#

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