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anonymous

  • one year ago

Could anyone help with this activity I have to do? It's trigonometry

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    I'd really appreciate it if someone could explain so I can figure it out . I'm terrible at trig . I'm taking an algebra 2 class and apparently trig is apart of algebra 2

  3. anonymous
    • one year ago
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    @jim_thompson5910 could you help me with this ? :/

  4. jim_thompson5910
    • one year ago
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    Your worksheet is a bit misleading. It says that B = 2pi/(period) but that's only true for sine, cosine, cosecant, secant. It doesn't apply to tangent and cotangent. That rule would be B = pi/(period). Luckily we won't be dealing with tangent or cotangent

  5. anonymous
    • one year ago
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    did I write that ?

  6. jim_thompson5910
    • one year ago
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    no it's printed on page 1

  7. anonymous
    • one year ago
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    apex is confusing and doesn't do the right information

  8. anonymous
    • one year ago
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    doesn't give *

  9. jim_thompson5910
    • one year ago
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    you wrote the correct max elevation what is the min elevation?

  10. anonymous
    • one year ago
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    it didn't give that . Idk

  11. jim_thompson5910
    • one year ago
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    |dw:1440386808612:dw|

  12. jim_thompson5910
    • one year ago
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    |dw:1440386828053:dw|

  13. jim_thompson5910
    • one year ago
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    since the ride starts 10 ft off the ground

  14. anonymous
    • one year ago
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    oh so it'd be 10 ? lol

  15. jim_thompson5910
    • one year ago
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    yeah

  16. jim_thompson5910
    • one year ago
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    and that would make the diameter 210 - 10 = 200 and not 210

  17. anonymous
    • one year ago
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    I have the diameter as 200

  18. jim_thompson5910
    • one year ago
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    oh nvm I misread that part

  19. jim_thompson5910
    • one year ago
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    don't forget the units. When it asks for the time to complete 1 revolution. Don't just write "90". Write "90 seconds". Same applies for the other values too

  20. jim_thompson5910
    • one year ago
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    other than that, page 1 looks great

  21. anonymous
    • one year ago
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    do the max and min have units ? oh and the diameter ?

  22. jim_thompson5910
    • one year ago
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    yeah, in this case, "feet"

  23. jim_thompson5910
    • one year ago
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    same for the diameter

  24. anonymous
    • one year ago
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    okay I just wanted to be sure

  25. jim_thompson5910
    • one year ago
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    ready for #2 ?

  26. anonymous
    • one year ago
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    yes

  27. jim_thompson5910
    • one year ago
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    I'm guessing you haven't started at all? Or do you have an idea where to start?

  28. anonymous
    • one year ago
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    The graph is where I got confused . I looked over it multiple times and I tried graphing but I just didn't understand

  29. jim_thompson5910
    • one year ago
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    At time t = 0 seconds, the height is h = 10 ft So one ordered pair is (0,10) and this is a min -------------------------------------- 45 seconds later at t = 45, the height moves to h = 210 one max point is (45,210) -------------------------------------- another 45 seconds pass by and we're now at t = 90 seconds we're back at the min height again and now at the point (90, 10)

  30. jim_thompson5910
    • one year ago
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    they want you to plot these mins and maxes for 2 full cycles

  31. anonymous
    • one year ago
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    I always forget how to graph lol . Just one second

  32. jim_thompson5910
    • one year ago
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    For #2, they don't want the full graph of the sinusoidal curve (ie they don't want sine or cosine plotted). They just want those discrete points.

  33. anonymous
    • one year ago
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    so don't plot all three of those points ?

  34. jim_thompson5910
    • one year ago
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    You do plot those points and any others that I didn't list until you've done 2 full cycles

  35. jim_thompson5910
    • one year ago
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    If you start at t = 0, 1 full cycle ends at t = 90 2 full cycles end at t = 180

  36. anonymous
    • one year ago
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    |dw:1440387875856:dw| so it'd look somewhat like that ?

  37. anonymous
    • one year ago
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    that's a horrible drawing I know lol . My mousepad sucks

  38. jim_thompson5910
    • one year ago
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    that's ok. You should get something like this (see attached). It looks like you did, so you are correct for #2

  39. anonymous
    • one year ago
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    do I connect the dots or leave them as is ?

  40. jim_thompson5910
    • one year ago
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    For #2, they just want the min/max points. That's all. So you leave them as is.

  41. anonymous
    • one year ago
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    Okay so that one is done

  42. jim_thompson5910
    • one year ago
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    what did you get for #3

  43. anonymous
    • one year ago
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    I'm reading it right now

  44. anonymous
    • one year ago
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    wouldn't the top of the ferris wheel be the max ?

  45. jim_thompson5910
    • one year ago
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    yes

  46. anonymous
    • one year ago
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    so it'd be 210 ?

  47. jim_thompson5910
    • one year ago
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    they want halfway to the top though

  48. jim_thompson5910
    • one year ago
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    |dw:1440388468748:dw|

  49. anonymous
    • one year ago
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    would you divide 210 by 2 ?

  50. jim_thompson5910
    • one year ago
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    you would find the midpoint of h = 10 and h = 210 add the values up and divide by 2: (10+210)/2 = ??

  51. jim_thompson5910
    • one year ago
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    `would you divide 210 by 2 ?` only if the starting height was 0

  52. anonymous
    • one year ago
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    so 110 ?

  53. jim_thompson5910
    • one year ago
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    yes, that's your midline |dw:1440388774233:dw|

  54. anonymous
    • one year ago
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    so their elevation would be 110 feet ?

  55. jim_thompson5910
    • one year ago
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    at the midway point, yes

  56. anonymous
    • one year ago
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    what do they mean by "use the diameter of the Ferris wheel to determine the correct vertical shift for your modeling function?"

  57. jim_thompson5910
    • one year ago
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    well I guess another way to do is to divide the diameter in half 200/2 = 100 the radius is 100. Add that radius to the starting height: 100+10 = 110

  58. jim_thompson5910
    • one year ago
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    I like the midpoint method better

  59. anonymous
    • one year ago
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    Midpoint takes out a step

  60. anonymous
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    so I'd graph it at 110 ?

  61. jim_thompson5910
    • one year ago
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    are you still on #3 ? or #4 now?

  62. anonymous
    • one year ago
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    on number 4

  63. jim_thompson5910
    • one year ago
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    yeah you'll have a dashed/dotted line that is horizontal and going through y = 110

  64. anonymous
    • one year ago
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    |dw:1440389236868:dw| so like this ?

  65. jim_thompson5910
    • one year ago
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    yeah

  66. jim_thompson5910
    • one year ago
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    which letter represents the vertical shift? A, B, C, or D?

  67. anonymous
    • one year ago
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    uh , how do you figure that out ?

  68. jim_thompson5910
    • one year ago
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    look back on page 1

  69. anonymous
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    D

  70. jim_thompson5910
    • one year ago
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    yes, D = 110

  71. jim_thompson5910
    • one year ago
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    for # 5, you can find the distance using two methods a) subtract the max and min height, then divide by 2 (max - min)/2 = (210-10)/2 = 200/2 = 100 OR b) Find the distance from the center (110) to either the peak or valley Center to peak: 210 - 110 = 100 Center to valley: 110 - 10 = 100 either way, the radius of this wheel is 100 ft. This is the amplitude of the sinusoidal curve the amplitude is |A|, so A = 100

  72. anonymous
    • one year ago
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    Yeah I just subtracted it and got 100

  73. anonymous
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    amplitude is distance ?

  74. jim_thompson5910
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    the distance from midline to the max or the distance from midline to the min

  75. jim_thompson5910
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    |dw:1440389822564:dw|

  76. jim_thompson5910
    • one year ago
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    or half the distance from the very top to the very bottom |dw:1440389869160:dw|

  77. anonymous
    • one year ago
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    so it just depends on exactly what they're asking ?

  78. jim_thompson5910
    • one year ago
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    what do you mean?

  79. anonymous
    • one year ago
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    or is all just the same ?

  80. jim_thompson5910
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    those are 2 different ways to get the same thing

  81. jim_thompson5910
    • one year ago
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    so it doesn't matter which path you take

  82. anonymous
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    ooh okay , I just wanted to make sure

  83. anonymous
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    for 6, do I keep it in terms of pi or simplify it completely ?

  84. jim_thompson5910
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    I'd leave it in terms of pi

  85. jim_thompson5910
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    reduce the fraction as much as possible though

  86. anonymous
    • one year ago
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    so it'd be \[B=\frac{ 2\pi }{ 90 } = \frac{ \pi }{ 45 }\]

  87. jim_thompson5910
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    yeah

  88. jim_thompson5910
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    which trig function will you use for #7 ? sine or cosine?

  89. anonymous
    • one year ago
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    sin

  90. jim_thompson5910
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    either works, but cosine is actually easier to work with if you graph the parent function y = cos(x), you'll notice it has the peak max point at (0,1). We want the peak max point to happen when t = 45 so that's why C = 45 is the phase shift. We shift everything over 45 units to the right

  91. jim_thompson5910
    • one year ago
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    A = 100 B = pi/45 C = 45 D = 110 \[\Large H(t) = A\cos\left(B(t-C)\right) + D\] \[\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110\] attached is the graph

  92. jim_thompson5910
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    you could use sine but the phase shift would be different. Everything else would stay the same

  93. anonymous
    • one year ago
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    Okay , it makes sense

  94. anonymous
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    in number 8 , what do they mean by horizontal shift ?

  95. jim_thompson5910
    • one year ago
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    the parent function y = cos(x) has a max when x = 0 but we want to force the max to be when x = 45 so that's why we do a horizontal shift and why C = 45

  96. jim_thompson5910
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    |dw:1440390964651:dw|

  97. jim_thompson5910
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    |dw:1440390975447:dw|

  98. jim_thompson5910
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    |dw:1440390992114:dw|

  99. anonymous
    • one year ago
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    So it is 45 . I'm used to shifts when it comes to 2 graphs

  100. jim_thompson5910
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    yeah it's 45 if we use cosine

  101. anonymous
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    Since with cosine x = 0

  102. jim_thompson5910
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    what do you mean?

  103. anonymous
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    idk . I'm probably confusing myself

  104. anonymous
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    so for number 9 , where are they getting at with the parent function being flipped ? I don't even know what the parent function is :/

  105. jim_thompson5910
    • one year ago
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    in this case, we used the parent function y = cos(x)

  106. jim_thompson5910
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    it's just the base function that you apply the transformations to

  107. jim_thompson5910
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    A = 100 which is positive, we aren't flipping the function over

  108. anonymous
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    if it were to be flipped , how would I know that was the case ?

  109. jim_thompson5910
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    well I started with y = cos(x) which has A = 1 as a positive number shifting everything over to the right 45 units doesn't flip things

  110. jim_thompson5910
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    I played around with the graph a bit, and I am just realizing that A could be negative, but that changes the phase shift.

  111. jim_thompson5910
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    So there are a few ways to get the h(t) function

  112. jim_thompson5910
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    if A = +100, then the phase shift is C = 45 if A = -100, then the phase shift is C = 0

  113. anonymous
    • one year ago
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    But since it's dealing with a ferris wheel , it couldn't have a negative value since you can't go in circles negative times right ?

  114. jim_thompson5910
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    t and h can't be negative, but A could be

  115. jim_thompson5910
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    the amplitude is |A| the absolute value ensures we have a positive amplitude

  116. jim_thompson5910
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    on your worksheet, draw vertical bars around the "A" where it says `A is the amplitude` on page 1

  117. anonymous
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    oh okay , so it doesn't really matter if it was negative cause it regardless it's going to be positive since it's absolute value ?

  118. jim_thompson5910
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    exactly

  119. jim_thompson5910
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    eg: y = -2*cos(x) has an amplitude of |-2| = 2

  120. anonymous
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    okay , I get it

  121. anonymous
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    in the equation on number 10, what's "Af" ?

  122. jim_thompson5910
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    I don't see #10 in any of the attachments

  123. anonymous
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    one moment

  124. anonymous
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    1 Attachment
  125. jim_thompson5910
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    oh, that's A*f

  126. jim_thompson5910
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    f is the trig function look back at page 1

  127. anonymous
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    so it's the amplitude times the trig function ? 100cosx ? or ...

  128. jim_thompson5910
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    yeah like how I wrote above when I wrote out H(t)

  129. anonymous
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    where did you write that ?

  130. jim_thompson5910
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    it's buried a bit when I wrote \[\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110\]

  131. anonymous
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    oooh okay , I remember

  132. anonymous
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    and t would b 90 , correct ?

  133. anonymous
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    be *

  134. jim_thompson5910
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    t is just a variable that represents the time in seconds

  135. jim_thompson5910
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    you plug in values of t to get corresponding values of h eg: plug in t = 0 to get h = 10 plug in t = 45 to get h = 210 etc etc

  136. anonymous
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    wait so how exactly is the equation set up ?

  137. jim_thompson5910
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    what do you mean?

  138. anonymous
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    I plug everything into that equation right ?

  139. jim_thompson5910
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    I'd just leave it as \[\Large H(t) = 100\cos\left(\frac{\pi}{45}(t-45)\right) + 110\]

  140. jim_thompson5910
    • one year ago
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    They want the function or equation that models the curve

  141. anonymous
    • one year ago
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    So no solving ?

  142. jim_thompson5910
    • one year ago
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    nope, they just want you to write the equation

  143. anonymous
    • one year ago
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    and for number 11 , i use that same equation, I'm just solving for t ?

  144. jim_thompson5910
    • one year ago
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    correct. replace h(t) with 100 and solve for t

  145. anonymous
    • one year ago
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    To start I'd subtract 110 over to 100 right ?

  146. jim_thompson5910
    • one year ago
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    yes

  147. anonymous
    • one year ago
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    what do I do from there ?

  148. jim_thompson5910
    • one year ago
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    you should have this \[\Large 100\cos\left(\frac{\pi}{45}(t-45)\right) = -10\]

  149. jim_thompson5910
    • one year ago
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    now divide both sides by 100

  150. anonymous
    • one year ago
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    so it'd be equal to -1/10 ?

  151. jim_thompson5910
    • one year ago
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    yeah or -0.1

  152. anonymous
    • one year ago
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    I feel like decimal is better

  153. jim_thompson5910
    • one year ago
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    Then apply the arccosine function to both sides \[\Large \Large \cos\left(\frac{\pi}{45}(t-45)\right) = -0.1\] \[\Large \Large \arccos\left(\cos\left(\frac{\pi}{45}(t-45)\right)\right) = \arccos(-0.1)\] \[\Large \Large \frac{\pi}{45}(t-45) = ??\]

  154. anonymous
    • one year ago
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    what's arccos ?

  155. jim_thompson5910
    • one year ago
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    arccos is \(\Large \cos^{-1}\), ie inverse cosine

  156. anonymous
    • one year ago
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    so csc ?

  157. jim_thompson5910
    • one year ago
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    no, that's a common misconception

  158. jim_thompson5910
    • one year ago
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    if you want to say 1/cos, you say secant

  159. jim_thompson5910
    • one year ago
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    you don't say \(\Large \cos^{-1}\) to mean secant \(\Large \cos^{-1} \ne \sec\)

  160. anonymous
    • one year ago
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    oh okay , so I haven't learned arccos

  161. jim_thompson5910
    • one year ago
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    your calculator should have a cos^(-1) or arccos button

  162. anonymous
    • one year ago
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    I have cos^-1

  163. jim_thompson5910
    • one year ago
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    ok use that

  164. anonymous
    • one year ago
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    so I find arccos-0.1

  165. jim_thompson5910
    • one year ago
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    yeah what is that numeric value?

  166. anonymous
    • one year ago
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    95.74 ?

  167. jim_thompson5910
    • one year ago
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    make sure you're in radian mode

  168. anonymous
    • one year ago
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    1.67

  169. jim_thompson5910
    • one year ago
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    yes or -1.67

  170. jim_thompson5910
    • one year ago
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    we will use the negative form to figure out when the height reaches 100 ft for the first time (the +1.67 is used to figure out when the height reaches 100 ft for the second time)

  171. jim_thompson5910
    • one year ago
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    so solve for t \[\Large \Large \frac{\pi}{45}(t-45) = -1.67\]

  172. anonymous
    • one year ago
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    do I distribute or multiply pi/45 by the recipricol on both sides ?

  173. jim_thompson5910
    • one year ago
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    I'd multiply by the reciprocal

  174. anonymous
    • one year ago
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    so it'd be \[t-45=\frac{ -75.15 }{ \pi }\]

  175. jim_thompson5910
    • one year ago
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    -75.15/pi = -23.9209879 approximately

  176. jim_thompson5910
    • one year ago
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    now just one more step to go

  177. anonymous
    • one year ago
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    adding the 45 over

  178. jim_thompson5910
    • one year ago
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    yes

  179. anonymous
    • one year ago
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    so it would equal 21.08 ?

  180. jim_thompson5910
    • one year ago
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    I'm getting 21.07901205 with a calculator and geogebra is telling me 21.06521 so it's pretty close

  181. jim_thompson5910
    • one year ago
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    so roughly 21 seconds

  182. anonymous
    • one year ago
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    I rounded the 21.07901205

  183. jim_thompson5910
    • one year ago
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    that works

  184. anonymous
    • one year ago
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    thank you so much !

  185. jim_thompson5910
    • one year ago
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    and desmos says the same thing more or less https://www.desmos.com/calculator/swknd4tau1

  186. jim_thompson5910
    • one year ago
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    you'll have to click on the first intersection point between the curve and the line

  187. anonymous
    • one year ago
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    You really helped a lot lol . I can finally release some stress . Thank you

  188. jim_thompson5910
    • one year ago
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    you're welcome

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