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anonymous
 one year ago
*HELP*
anonymous
 one year ago
*HELP*

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@jim_thompson5910 please??

rishavraj
 one year ago
Best ResponseYou've already chosen the best response.3if speed at t1 is s1 and t2 is s2 ... then rate of change of speed = (s2  s1)/(t2  t1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is that like slope kinda?

rishavraj
 one year ago
Best ResponseYou've already chosen the best response.3value at pi = 2 and at (3 pi)/2 = 1 so rate of change \[\frac{ 2  (1) }{ \frac{ 3 \pi }{ 2 }  \pi }\]

rishavraj
 one year ago
Best ResponseYou've already chosen the best response.3oooh my bad.value at pi = 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how would the bottom work out? just cancel the pi?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0for the bottom, we need to multiply that part by \[\frac{2}{2} \] that way we can subtract that fraction because now they have the same denominators.

rishavraj
 one year ago
Best ResponseYou've already chosen the best response.3value at (3 pi)/2 = 2 dw:1435723454817:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you! can i get a second opinion on a question?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0depends on the question, but go ahead :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the amplitude, period, and phase shift of f(x) = −3 cos(4x + π) + 6? Amplitude = 3; period = pi over two; phase shift: x = negative pi over four Amplitude = −3; period = pi over two; phase shift: x = pi over four Amplitude = −3; period = negative pi over two; phase shift: x = negative pi over four Amplitude = 3; period = 2π; phase shift: x = negative pi over four Im stuck between a and b

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0uh oh. I forgot this already x.X!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@rishavraj @jim_thompson5910

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0sorry ^_^ it was too long ago... and it was my weak spot during my trigonometry class too.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its okay! thank you for helping with the other one though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you have anything? @rishavraj

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you have anything? @rishavraj

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435724341293:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435724350479:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435724371477:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435724392822:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435724440263:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it would be a! (:

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1dw:1435724524199:dw

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah looks like it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks! its that absolute values that confuse me alot

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1you mean the amplitude?
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