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zaphod

  • 3 years ago

Please help...

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  1. zaphod
    • 3 years ago
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    http://screencast.com/t/N9H6p6CGYc

  2. zaphod
    • 3 years ago
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    @satellite73 @Shadowys @RadEn

  3. anonymous
    • 3 years ago
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    \(a\) 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

  4. anonymous
    • 3 years ago
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    that is, the range is of length 12 from -3 to 9, so the amplitude is half of that, therefore \(a=6\)

  5. zaphod
    • 3 years ago
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    IS THERE ANY OTHER METHOD TO SOLVE IT WITH EQUATIONS.

  6. anonymous
    • 3 years ago
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    from 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 \(b\)

  7. anonymous
    • 3 years ago
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    no there are no equations here, you have to visualize, since you are given a picture

  8. anonymous
    • 3 years ago
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    well 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

  9. zaphod
    • 3 years ago
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    and i know c, now can i substitute it in the main equation and find a?

  10. anonymous
    • 3 years ago
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    that is the entire point of this exercise, not to use equations, but to visualize the period, and amplitude from the picture

  11. anonymous
    • 3 years ago
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    we know \(c=3\) because this is the graph of sine lifted up 3 units

  12. zaphod
    • 3 years ago
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    can u explain why period of sin(bx) = 2pi/b

  13. anonymous
    • 3 years ago
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    it 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

  14. allamiro
    • 3 years ago
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    sub x = pi then x = 2pi when x = pi y =3 when x = 2pi y = 9 then solve the equations

  15. zaphod
    • 3 years ago
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    @satellite73 period is for one complete wave right, how come its 2 pi, it has to be pi

  16. allamiro
    • 3 years ago
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    when x = 0 y = 0 then solve the equations for a b and c

  17. zaphod
    • 3 years ago
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    cn u show the working @allamiro

  18. allamiro
    • 3 years ago
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    dsregard the 2 pi thing = 9 just x = pi and x = 0

  19. allamiro
    • 3 years ago
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    what 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

  20. zaphod
    • 3 years ago
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    yes?

  21. allamiro
    • 3 years ago
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    sorry again y = 3 when x = 0 I didnt focus at the graph

  22. allamiro
    • 3 years ago
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    so c = 3

  23. zaphod
    • 3 years ago
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    ok how do u find b now.

  24. allamiro
    • 3 years ago
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    yes 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

  25. zaphod
    • 3 years ago
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    b?

  26. allamiro
    • 3 years ago
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    y = 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

  27. allamiro
    • 3 years ago
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    y = 3 when x = 2pi * correctiion

  28. allamiro
    • 3 years ago
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    :)

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