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itsmichelle29

  • one year ago

If sin θ = 1/4 and tan θ > 0, what is the value of cos θ?

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    draw a triangle, like this |dw:1439233810971:dw|

  3. misty1212
    • one year ago
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    there is a picture of an angle where the sine is \(\frac{1}{4}\) since the "opposite over adjacent" is \(\frac{1}{4}\)

  4. misty1212
    • one year ago
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    i meant "opposite over hypotenuse"

  5. misty1212
    • one year ago
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    what you need is the adjacent side, which you find via pythagoras let me know when you get it

  6. itsmichelle29
    • one year ago
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    7

  7. misty1212
    • one year ago
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    oh no not 7

  8. misty1212
    • one year ago
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    use pythagoras \[a^2+b^2=c^2\\ x^2+1^2=4^2\\ x^2+1=16\\ x^2=16-1=15\] so \[x=\sqrt{15}\]

  9. itsmichelle29
    • one year ago
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    okay is the answe root15 /4

  10. itsmichelle29
    • one year ago
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    @misty1212

  11. misty1212
    • one year ago
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    yes

  12. misty1212
    • one year ago
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    oh no!

  13. misty1212
    • one year ago
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    you want the tangent right? not the cosine

  14. itsmichelle29
    • one year ago
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    can u help me with like two more plz

  15. itsmichelle29
    • one year ago
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    the cos

  16. misty1212
    • one year ago
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    oh ok then yes

  17. misty1212
    • one year ago
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    and i will be happy to help with two more

  18. itsmichelle29
    • one year ago
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    thanks

  19. itsmichelle29
    • one year ago
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    What are the period and phase shift for f(x) = −4 tan(x − π)?

  20. itsmichelle29
    • one year ago
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    i think its pi and pi

  21. misty1212
    • one year ago
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    period of tangent is \(\pi\) so yeah

  22. misty1212
    • one year ago
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    phase shift is \(\pi\) as well you are right

  23. itsmichelle29
    • one year ago
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    thanks and

  24. itsmichelle29
    • one year ago
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    Use the graph below to answer the question that follows: cosine graph with points at 0, negative 1 and pi over 2, 3 and pi, negative 1 What are the amplitude, period, and midline of the function? Amplitude: 8; period: π; midline: y = 1 Amplitude: 8; period: 2π; midline: y = 5 Amplitude: 4; period: 2π; midline: y = 5 Amplitude: 2; period: π; midline: y = 1

  25. misty1212
    • one year ago
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    really need the graph for this one

  26. itsmichelle29
    • one year ago
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    1 Attachment
  27. itsmichelle29
    • one year ago
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    okay =)

  28. misty1212
    • one year ago
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    ok much better got a guess?

  29. misty1212
    • one year ago
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    only need to look at the amplitude to pick from your choices

  30. itsmichelle29
    • one year ago
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    Amplitude: 8; period: π; midline: y = 1 Amplitude: 8; period: 2π; midline: y = 5 Amplitude: 4; period: 2π; midline: y = 5 Amplitude: 2; period: π; midline: y = 1

  31. itsmichelle29
    • one year ago
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    well i thought it was -1 but thats not a chose

  32. misty1212
    • one year ago
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    from -1 to 3, that is from the lowest point to the highest, is a distance of 4 units the amplitude is half of that i.e. 2

  33. misty1212
    • one year ago
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    also you can see that the period is \(\pi\) since it goes up, and back down to -1 from 0 to \(\pi\)

  34. itsmichelle29
    • one year ago
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    Amplitude: 2; period: π; midline: y = 1 is that the answer

  35. misty1212
    • one year ago
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    yup

  36. itsmichelle29
    • one year ago
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    omg ur amazing thanks

  37. misty1212
    • one year ago
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    lol thanks\[\color\magenta\heartsuit\]

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