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alysa111 Group Title

For an angle Θ with the point (−20, −21) on its terminating side, what is the value of cosine?

  • 2 months ago
  • 2 months ago

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  1. alysa111 Group Title
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    @kelliegirl33 can you help me

    • 2 months ago
  2. myininaya Group Title
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    If it helps you graph the point. Can you do that for me first?

    • 2 months ago
  3. alysa111 Group Title
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    @myininaya just plot -20,-21 ?

    • 2 months ago
  4. myininaya Group Title
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    Yep plot the point as a first step.

    • 2 months ago
  5. myininaya Group Title
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    You can use the draw tool.

    • 2 months ago
  6. alysa111 Group Title
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    ok

    • 2 months ago
  7. alysa111 Group Title
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    |dw:1403474666905:dw|

    • 2 months ago
  8. alysa111 Group Title
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    @myininaya

    • 2 months ago
  9. myininaya Group Title
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    |dw:1403474755600:dw| Now I will simply make a right triangle using the center of the graph for my central angle.

    • 2 months ago
  10. myininaya Group Title
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    what is the ratio definition of cosine of an angle?

    • 2 months ago
  11. alysa111 Group Title
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    legnth of adjacent / legnth of hyp

    • 2 months ago
  12. myininaya Group Title
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    right you already have the adjacent side given in the picture the only thing you need now is the hyp of that triangle, right?

    • 2 months ago
  13. alysa111 Group Title
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    yes

    • 2 months ago
  14. myininaya Group Title
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    Find the hyp of that triangle using Pythagorean Theorem.

    • 2 months ago
  15. alysa111 Group Title
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    the adjacent side is -20

    • 2 months ago
  16. myininaya Group Title
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    You are correct to say that.

    • 2 months ago
  17. myininaya Group Title
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    Can you find the length of the hyp?

    • 2 months ago
  18. alysa111 Group Title
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    yes hold on wouldnt it be -20^2 + 21^2 = c^2

    • 2 months ago
  19. myininaya Group Title
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    20^2+21^2=c^2 and then take the square root of both sides to give you c.

    • 2 months ago
  20. alysa111 Group Title
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    29

    • 2 months ago
  21. myininaya Group Title
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    so that is the length of the hyp so what is cos(theta)=?

    • 2 months ago
  22. alysa111 Group Title
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    29 ?

    • 2 months ago
  23. myininaya Group Title
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    you said earlier cos(theta) is adj/hyp you never said cos(theta) is hyp

    • 2 months ago
  24. alysa111 Group Title
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    20/29

    • 2 months ago
  25. myininaya Group Title
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    don't forget the negative sign

    • 2 months ago
  26. alysa111 Group Title
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    thank you so much i have a few more cn yu help me

    • 2 months ago
  27. myininaya Group Title
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    if the angle was on the side of the graph like on the right hand side of the y-axis cos would have been indeed a positive number but since the angle was to the left of the y-axis cos will be negative

    • 2 months ago
  28. myininaya Group Title
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    |dw:1403475467323:dw|

    • 2 months ago
  29. alysa111 Group Title
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    How does changing the function from f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function?

    • 2 months ago
  30. myininaya Group Title
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    we could even talk about sine sin is positive when above the x-axis and sine is negative when below the x-axis |dw:1403475522177:dw|

    • 2 months ago
  31. myininaya Group Title
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    Well you are talking about shifting the graph up when you have that +5

    • 2 months ago
  32. alysa111 Group Title
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    so the function shifts up 5 units so the range changes from -1 to 1

    • 2 months ago
  33. alysa111 Group Title
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    in f(x) to 4 to 6 in g(x)

    • 2 months ago
  34. myininaya Group Title
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    For example, pretend we have y=sin(x) the range of sin(x) is from -1 to 1 (inclusive). but when we look at y=sin(x)+1 that means you are moving every point of y=sin(x) up one unit so the range of sin(x)+1 is from -1+1..1+1( inclusive) doing the addition gives 0..2 (inclusive)

    • 2 months ago
  35. alysa111 Group Title
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    am i right

    • 2 months ago
  36. myininaya Group Title
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    well first what is the range of y=3sin(x)?

    • 2 months ago
  37. alysa111 Group Title
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    f(x) = 3 sin 2x to g(x) = 3 sin 2x + 5 affect the range of the function

    • 2 months ago
  38. myininaya Group Title
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    yes the range of the graph of f would be affected if you need f+5

    • 2 months ago
  39. myininaya Group Title
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    did f+5 (not need)

    • 2 months ago
  40. myininaya Group Title
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    but the range of y=3sin(x) is not -1 to 1.

    • 2 months ago
  41. alysa111 Group Title
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    Which of the following functions has the greatest y-intercept?|dw:1403475803380:dw|

    • 2 months ago
  42. alysa111 Group Title
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    its 2 to 8

    • 2 months ago
  43. myininaya Group Title
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    Let y=sin(x). Let f(x)=3sin(x). Therefore you could say f(x)=3y where y=sin(x). If you have 3 times every y per x wouldn't the range be stretched more?

    • 2 months ago
  44. alysa111 Group Title
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    yes

    • 2 months ago
  45. myininaya Group Title
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    like the range for y=sin(x) is -1 to 1 (when we talk about range we are talking about the y values) the lowest y value for y=3sin(x) would be what then?

    • 2 months ago
  46. alysa111 Group Title
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    so f(x) has greater y-intercerpt

    • 2 months ago
  47. myininaya Group Title
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    the lowest y value for y=sin(x) is -1 so the lowest y value for f(x)=3sin(x) is ?

    • 2 months ago
  48. alysa111 Group Title
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    -3

    • 2 months ago
  49. myininaya Group Title
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    great so it shouldn't be hard for you to see the highest y-value for f(x)=3sin(x) would be 3(1)=3 so the range for y=3sin(x) is -3 to 3 (inclusive)

    • 2 months ago
  50. myininaya Group Title
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    now what is the range for y=3sin(x)+5

    • 2 months ago
  51. alysa111 Group Title
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    2 to 8

    • 2 months ago
  52. myininaya Group Title
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    omg you are great so now you can answer that one question about how the range is affected and actually state the ranges of both functions

    • 2 months ago
  53. alysa111 Group Title
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    now for Which of the following functions has the greatest y-intercept? wouldnt it be f(x) since the y is 9

    • 2 months ago
  54. myininaya Group Title
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    Also when answering the question it might be better to choose to answer in interval notation or inequality notation. 2 to 8 isn't really something people say to state an interval (not that I have ever seen anyways)

    • 2 months ago
  55. alysa111 Group Title
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    Okay , so it would be stated has 2 to 8 in g(x)

    • 2 months ago
  56. myininaya Group Title
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    well like i said you might want to choose to answer in interval notation (or inequality) y is an element of the interval [2,8] <--interval (i put brackets to mean to included endpoints inequality notation: \[2 \le y \le 8\] and to your next question.... |dw:1403476409004:dw| do you what the y-intercept is for a curve? It is the part of the curve that goes through the y-axis. When I say part, I'm talking about the specific point in which the curve does that. What do you know about all the points on the y-axis? What do they all have in common?

    • 2 months ago
  57. myininaya Group Title
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    I will give you a hint. All the x values of any point on the y-axis are _____?___

    • 2 months ago
  58. alysa111 Group Title
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    Im not sure

    • 2 months ago
  59. alysa111 Group Title
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    4-9 = -5 which is the same for g(x)

    • 2 months ago
  60. alysa111 Group Title
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    the same y-intercept

    • 2 months ago
  61. myininaya Group Title
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    Ok since you can't answer that, I will ask you a series of questions now? Where does (0,10) lie? Where does (0,5) lie? Where does (0,2) lie? Where does (0,0) lie? Where does (0,-1) lie? Where does (0,-6) lie? where does (0,-1000) lie?

    • 2 months ago
  62. alysa111 Group Title
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    on the x axis

    • 2 months ago
  63. myininaya Group Title
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    That is incorrect. They all lie on the y-axis. The answer to my question: All the x-values of any point on the y-axis is ____?__ All the x-values of any point on the y-axis is zero. So to find the y-intercept for a curve you just need to find y when x is zero. You just need to replace x with zero and solve for y.

    • 2 months ago
  64. alysa111 Group Title
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    so 0-9 ? 0-4 ?

    • 2 months ago
  65. myininaya Group Title
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    For the one function, it should be clear (0,0) is the y-intercept (that is the origin of the graph after all; is is all an x-intercept) The other function you did right even though it doesn't seem you know why you were doing it since you were not able to answer the question.

    • 2 months ago
  66. alysa111 Group Title
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    Yeah im confused now

    • 2 months ago
  67. myininaya Group Title
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    |dw:1403477105487:dw| This is the information given. You are asking to find the y-intercept for both graphs right? Can you tell me which point from f have the x is 0?

    • 2 months ago
  68. alysa111 Group Title
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    functions has the greatest y-intercept

    • 2 months ago
  69. alysa111 Group Title
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    1,1 ?

    • 2 months ago
  70. myininaya Group Title
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    I'm asking you to tell which of the points from f have the x value is 0?

    • 2 months ago
  71. myininaya Group Title
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    |dw:1403477214481:dw|

    • 2 months ago
  72. alysa111 Group Title
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    its 0,0 ?

    • 2 months ago
  73. myininaya Group Title
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    |dw:1403477221682:dw| Do you see any of these x's being 0?

    • 2 months ago
  74. alysa111 Group Title
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    yes

    • 2 months ago
  75. myininaya Group Title
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    yes the point (0,0) is a y-intercept because the x is 0

    • 2 months ago
  76. myininaya Group Title
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    now we need to look at g(x)=5cos(3x)-5

    • 2 months ago
  77. myininaya Group Title
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    what point on g gives us the x is 0?

    • 2 months ago
  78. alysa111 Group Title
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    -5

    • 2 months ago
  79. myininaya Group Title
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    how did you get that?

    • 2 months ago
  80. myininaya Group Title
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    Did you replace the x with 0 to see what y is? Because y is not -5 when x is 0.

    • 2 months ago
  81. myininaya Group Title
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    I only care about when x is 0 since you are asking about the y-intercept.

    • 2 months ago
  82. myininaya Group Title
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    Replace the x with 0 like so y-intercept of g is g(0)=5cos(3*0)-5

    • 2 months ago
  83. myininaya Group Title
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    evaluate 5cos(3*0)-5

    • 2 months ago
  84. alysa111 Group Title
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    5cos(0)-5 5cos

    • 2 months ago
  85. myininaya Group Title
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    you are right to say 5cos(3*0)-5 can be wrriten as 5cos(0)-5 but where do we go from there?

    • 2 months ago
  86. alysa111 Group Title
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    1.41

    • 2 months ago
  87. myininaya Group Title
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    :( 5cos(0)-5 what is cos(0)?

    • 2 months ago
  88. alysa111 Group Title
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    1

    • 2 months ago
  89. myininaya Group Title
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    5(1)-5 can you simplify from here now?

    • 2 months ago
  90. alysa111 Group Title
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    0

    • 2 months ago
  91. alysa111 Group Title
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    same y

    • 2 months ago
  92. myininaya Group Title
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    right so the y-intercept for g is (0,0) as well they have the same y-intercept

    • 2 months ago
  93. alysa111 Group Title
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    1f sin Θ = 15 over 17, use the Pythagorean Identity to find cos Θ.

    • 2 months ago
  94. myininaya Group Title
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    draw a right triangle using this information sin(theta)=15/17

    • 2 months ago
  95. alysa111 Group Title
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    \[\sin^2 \theta + \cos ^2\theta = 1\] will be what we use ?

    • 2 months ago
  96. myininaya Group Title
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    you can use that

    • 2 months ago
  97. alysa111 Group Title
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    |dw:1403477725033:dw|

    • 2 months ago
  98. alysa111 Group Title
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    |dw:1403477767271:dw|

    • 2 months ago
  99. myininaya Group Title
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    if you want to use that specific identity you can replace sin(theta) with 15/17 and solve for cos(theta)

    • 2 months ago
  100. alysa111 Group Title
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    sin(15) = 0.650

    • 2 months ago
  101. alysa111 Group Title
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    sin*17) = -0.96

    • 2 months ago
  102. myininaya Group Title
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    can you replace sin(theta) with 15/17 in the following equation \[\sin^2(\theta)+\cos^2(\theta)=1\]

    • 2 months ago
  103. alysa111 Group Title
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    no

    • 2 months ago
  104. myininaya Group Title
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    why not?

    • 2 months ago
  105. alysa111 Group Title
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    well wouldnt you put them in front of sin and cos

    • 2 months ago
  106. myininaya Group Title
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    just so you know \[\sin^2(\theta)\] and \[(\sin(\theta))^2\] have exactly the same meaning

    • 2 months ago
  107. myininaya Group Title
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    you are given sin(theta) is 15/17 so you replace the whole sin(theta) thing with 15/17

    • 2 months ago
  108. myininaya Group Title
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    \[\sin^2(\theta)+\cos^2(\theta)=1 \text{ means exactly that } (\sin(\theta))^2+(\cos(\theta))^2=1\] So we have \[(\sin(\theta))^2+(\cos(\theta))^2=1 \\ (\frac{15}{17})^2+(\cos(\theta))^2 =1 \text{ I did this because I was given that } \sin(\theta)=15/17\]

    • 2 months ago
  109. myininaya Group Title
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    now solve for cos(theta)

    • 2 months ago
  110. myininaya Group Title
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    if that confuses you solve the following equation for u \[(\frac{15}{17})^2+u^2=1 \]

    • 2 months ago
  111. myininaya Group Title
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    whatever value(s) you get for u is your answer since u is cos(theta)

    • 2 months ago
  112. myininaya Group Title
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    and cos(theta) is the objective

    • 2 months ago
  113. alysa111 Group Title
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    225/285

    • 2 months ago
  114. alysa111 Group Title
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    |dw:1403478332917:dw|

    • 2 months ago
  115. myininaya Group Title
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    so what did you do to get that?

    • 2 months ago
  116. alysa111 Group Title
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    found cos of theta

    • 2 months ago
  117. myininaya Group Title
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    you solve the equation (15/17)^2+u^2=1 for u (which is cos(theta) )?

    • 2 months ago
  118. alysa111 Group Title
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    yes

    • 2 months ago
  119. myininaya Group Title
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    you might want to do that one more time

    • 2 months ago
  120. myininaya Group Title
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    the denominator is right but your numerator is off

    • 2 months ago
  121. alysa111 Group Title
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    8 / 17

    • 2 months ago
  122. myininaya Group Title
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    ok so cos could either be 8/17 or -8/17 when sin is 15/17

    • 2 months ago
  123. alysa111 Group Title
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    |dw:1403478613956:dw| my only option

    • 2 months ago
  124. myininaya Group Title
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    Agreed

    • 2 months ago
  125. alysa111 Group Title
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    In Shadow Bay in July, high tide is at 1:30 pm. The water level is 4 feet at high tide and 0 feet at low tide. Assuming the next high tide is exactly 12 hours later and the height of the water can be modeled by a cosine curve, find an equation for Shadow Bay's water level in July as a function of time (t).

    • 2 months ago
  126. alysa111 Group Title
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    okay so i got f(t) = 4cos pie/2t +1

    • 2 months ago
  127. alysa111 Group Title
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    @myininaya

    • 2 months ago
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