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zaphod
 2 years ago
Best ResponseYou've already chosen the best response.0@satellite73 @Shadowys @RadEn

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1\(a\) is the amplitude. since this starts and 3 and goes up do 9, then comes back down to 3 and then to 3, the amplitude is 6

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1that is, the range is of length 12 from 3 to 9, so the amplitude is half of that, therefore \(a=6\)

zaphod
 2 years ago
Best ResponseYou've already chosen the best response.0IS THERE ANY OTHER METHOD TO SOLVE IT WITH EQUATIONS.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1from your eyes you see that the period is \(\pi\) the period of \(\sin(bx)\) is \(\frac{2\pi}{b}\) so set \(\frac{2\pi}{b}=\pi\) and solve for \(b\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1no there are no equations here, you have to visualize, since you are given a picture

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1well there is an equation to find \(b\) . it is \[\frac{2\pi}{b}=\pi\] but you only know that the period is \(\pi\) from looking at the graph

zaphod
 2 years ago
Best ResponseYou've already chosen the best response.0and i know c, now can i substitute it in the main equation and find a?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1that is the entire point of this exercise, not to use equations, but to visualize the period, and amplitude from the picture

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1we know \(c=3\) because this is the graph of sine lifted up 3 units

zaphod
 2 years ago
Best ResponseYou've already chosen the best response.0can u explain why period of sin(bx) = 2pi/b

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1it is always the case that the period of \(\sin(bx)\) is \(\frac{2\pi}{b}\) we can think of it this way. since is periodic with period \(2\pi\) so it does everything on the interval \([0,2\pi)\) now if \(bx=0\) that means \(x=0\) and if \(bx=2\pi\) that means \(x=\frac{2\pi}{b}\) so that gives you the period

allamiro
 2 years ago
Best ResponseYou've already chosen the best response.0sub x = pi then x = 2pi when x = pi y =3 when x = 2pi y = 9 then solve the equations

zaphod
 2 years ago
Best ResponseYou've already chosen the best response.0@satellite73 period is for one complete wave right, how come its 2 pi, it has to be pi

allamiro
 2 years ago
Best ResponseYou've already chosen the best response.0when x = 0 y = 0 then solve the equations for a b and c

zaphod
 2 years ago
Best ResponseYou've already chosen the best response.0cn u show the working @allamiro

allamiro
 2 years ago
Best ResponseYou've already chosen the best response.0dsregard the 2 pi thing = 9 just x = pi and x = 0

allamiro
 2 years ago
Best ResponseYou've already chosen the best response.0what you mean show the work x = pi y = 3 3 = a sin ( b* pi ) + c x= 0 y = 0 0 = a sin ( b * 0 ) + c

allamiro
 2 years ago
Best ResponseYou've already chosen the best response.0sorry again y = 3 when x = 0 I didnt focus at the graph

allamiro
 2 years ago
Best ResponseYou've already chosen the best response.0yes so now lets say 9 = a sin ( bx) + 3 the highest value for sin when sinx b = 1 so bx = pi /2 so from there a =6

allamiro
 2 years ago
Best ResponseYou've already chosen the best response.0y = 3 when x = pi /2 3 = 6 sin ( b 2 pi ) + 3 6 sin ( 2 b pi) = 0 sin ( 2 b pi ) = sin ( 2b pi ) = sin ( pi ) b = 1/2

allamiro
 2 years ago
Best ResponseYou've already chosen the best response.0y = 3 when x = 2pi * correctiion
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