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alysa111

  • 5 months ago

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

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  1. alysa111
    • 5 months ago
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    @kelliegirl33 can you help me

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

  3. alysa111
    • 5 months ago
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    @myininaya just plot -20,-21 ?

  4. myininaya
    • 5 months ago
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    Yep plot the point as a first step.

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

  6. alysa111
    • 5 months ago
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    ok

  7. alysa111
    • 5 months ago
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    |dw:1403474666905:dw|

  8. alysa111
    • 5 months ago
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    @myininaya

  9. myininaya
    • 5 months ago
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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.

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

  11. alysa111
    • 5 months ago
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    legnth of adjacent / legnth of hyp

  12. myininaya
    • 5 months ago
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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?

  13. alysa111
    • 5 months ago
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    yes

  14. myininaya
    • 5 months ago
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    Find the hyp of that triangle using Pythagorean Theorem.

  15. alysa111
    • 5 months ago
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    the adjacent side is -20

  16. myininaya
    • 5 months ago
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    You are correct to say that.

  17. myininaya
    • 5 months ago
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    Can you find the length of the hyp?

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

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

  20. alysa111
    • 5 months ago
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    29

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

  22. alysa111
    • 5 months ago
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    29 ?

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

  24. alysa111
    • 5 months ago
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    20/29

  25. myininaya
    • 5 months ago
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    don't forget the negative sign

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

  27. myininaya
    • 5 months ago
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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

  28. myininaya
    • 5 months ago
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    |dw:1403475467323:dw|

  29. alysa111
    • 5 months ago
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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?

  30. myininaya
    • 5 months ago
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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|

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

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

  33. alysa111
    • 5 months ago
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    in f(x) to 4 to 6 in g(x)

  34. myininaya
    • 5 months ago
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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)

  35. alysa111
    • 5 months ago
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    am i right

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

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

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

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

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

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

  42. alysa111
    • 5 months ago
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    its 2 to 8

  43. myininaya
    • 5 months ago
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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?

  44. alysa111
    • 5 months ago
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    yes

  45. myininaya
    • 5 months ago
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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?

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

  47. myininaya
    • 5 months ago
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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 ?

  48. alysa111
    • 5 months ago
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    -3

  49. myininaya
    • 5 months ago
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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)

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

  51. alysa111
    • 5 months ago
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    2 to 8

  52. myininaya
    • 5 months ago
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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

  53. alysa111
    • 5 months ago
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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

  54. myininaya
    • 5 months ago
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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)

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

  56. myininaya
    • 5 months ago
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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?

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

  58. alysa111
    • 5 months ago
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    Im not sure

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

  60. alysa111
    • 5 months ago
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    the same y-intercept

  61. myininaya
    • 5 months ago
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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?

  62. alysa111
    • 5 months ago
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    on the x axis

  63. myininaya
    • 5 months ago
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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.

  64. alysa111
    • 5 months ago
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    so 0-9 ? 0-4 ?

  65. myininaya
    • 5 months ago
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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.

  66. alysa111
    • 5 months ago
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    Yeah im confused now

  67. myininaya
    • 5 months ago
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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?

  68. alysa111
    • 5 months ago
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    functions has the greatest y-intercept

  69. alysa111
    • 5 months ago
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    1,1 ?

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

  71. myininaya
    • 5 months ago
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    |dw:1403477214481:dw|

  72. alysa111
    • 5 months ago
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    its 0,0 ?

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

  74. alysa111
    • 5 months ago
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    yes

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

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

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

  78. alysa111
    • 5 months ago
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    -5

  79. myininaya
    • 5 months ago
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    how did you get that?

  80. myininaya
    • 5 months ago
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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.

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

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

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

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

  85. myininaya
    • 5 months ago
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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?

  86. alysa111
    • 5 months ago
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    1.41

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

  88. alysa111
    • 5 months ago
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    1

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

  90. alysa111
    • 5 months ago
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    0

  91. alysa111
    • 5 months ago
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    same y

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

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

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

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

  96. myininaya
    • 5 months ago
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    you can use that

  97. alysa111
    • 5 months ago
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    |dw:1403477725033:dw|

  98. alysa111
    • 5 months ago
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    |dw:1403477767271:dw|

  99. myininaya
    • 5 months ago
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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)

  100. alysa111
    • 5 months ago
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    sin(15) = 0.650

  101. alysa111
    • 5 months ago
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    sin*17) = -0.96

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

  103. alysa111
    • 5 months ago
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    no

  104. myininaya
    • 5 months ago
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    why not?

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

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

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

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

  109. myininaya
    • 5 months ago
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    now solve for cos(theta)

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

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

  112. myininaya
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    and cos(theta) is the objective

  113. alysa111
    • 5 months ago
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    225/285

  114. alysa111
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    |dw:1403478332917:dw|

  115. myininaya
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    so what did you do to get that?

  116. alysa111
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    found cos of theta

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

  118. alysa111
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    yes

  119. myininaya
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    you might want to do that one more time

  120. myininaya
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    the denominator is right but your numerator is off

  121. alysa111
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    8 / 17

  122. myininaya
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    ok so cos could either be 8/17 or -8/17 when sin is 15/17

  123. alysa111
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    |dw:1403478613956:dw| my only option

  124. myininaya
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    Agreed

  125. alysa111
    • 5 months ago
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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).

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

  127. alysa111
    • 5 months ago
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    @myininaya

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