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anonymous
 one year ago
The function f(x)= cot(3xpi/2) has ___.?
A. period pi/3 and asymptote at x=pi/6
B. period 2pi/3 and asymptote at x=pi/6
C. period 2pi/3 and asymptote at x=0
D. period pi/3 and asymptote at x=0
anonymous
 one year ago
The function f(x)= cot(3xpi/2) has ___.? A. period pi/3 and asymptote at x=pi/6 B. period 2pi/3 and asymptote at x=pi/6 C. period 2pi/3 and asymptote at x=0 D. period pi/3 and asymptote at x=0

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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the period of the general cotangent function shown below is T = pi/B y = A*cot(Bx  C) + D

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1let me know if that's enough to get you started

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I still don't quiet understand

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1do you see how the Bx and 3x match up?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1so B = 3 and you plug that into T = pi/B to get pi/3 as the period

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh okay. And then how would I find the asymptote ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1what happens when you plug in x = 0 ?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yes, so because x = 0 produces a single number as an output, this means x = 0 is NOT a vertical asymptote

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1that leaves x = pi/6 to be an asymptote sure enough if you plug in x = pi/6, you should get NAN or UND or whatever error your calculator will spit out

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1it's A actually

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah D is out because f(0) is defined

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay i understand thank you so much:)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1glad to be of help
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