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cocoapeebles3

  • one year ago

Xavier is riding on a Ferris wheel at the local fair. His height can be modeled by the equation H(t) = 20cosine of the quantity 1 over 15 times t + 30, where H represents the height of the person above the ground in feet at t seconds. Part 1: How far above the ground is Xavier before the ride begins? Part 2: How long does the Ferris wheel take to make one complete revolution? Part 3: Assuming Xavier begins the ride at the top, how far from the ground is the edge of the Ferris wheel, when Xavier's height above the ground reaches a minimum?

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  1. cocoapeebles3
    • one year ago
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    -.- I just needed help figuring it out I have been stuck on it for hours and I thought openstudy would help but obviously not @IrishBoy123

  2. cocoapeebles3
    • one year ago
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    Okay for part 1 i think you just have to substitute in 0 for t. H(0)= 20cos(pi/15(o))+30 which is 50. But I'm not sure if that's right.

  3. IrishBoy123
    • one year ago
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    why not just draw the equation? i can guess sorta what you are trying to do, but that really defeats the point. loads of people here want to help and learn.

  4. IrishBoy123
    • one year ago
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    |dw:1439938025351:dw|

  5. IrishBoy123
    • one year ago
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    |dw:1439938102822:dw|

  6. IrishBoy123
    • one year ago
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    or post a url or do a screengrab or scribble it down and scan and link it

  7. IrishBoy123
    • one year ago
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    |dw:1439938283787:dw|

  8. IrishBoy123
    • one year ago
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    i am sorry to be such a pain but, once you get this bit sorted, you will get loads of help!

  9. cocoapeebles3
    • one year ago
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    |dw:1439938250052:dw|

  10. IrishBoy123
    • one year ago
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    an example of posting a screengrab https://gyazo.com/9bd2129de27eaf26d56f85c2ffee15eb

  11. IrishBoy123
    • one year ago
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    a url: http://openstudy.com/users/cocoapeebles3#/updates/55d3a6f1e4b0c5fe9806d2bc

  12. cocoapeebles3
    • one year ago
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    sorry I'm new to this

  13. IrishBoy123
    • one year ago
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    we all were once please stick with it your drawing is off the screen for me https://gyazo.com/944f3bbd967745366c90fa0c079f5f56

  14. IrishBoy123
    • one year ago
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    |dw:1439938633443:dw|

  15. cocoapeebles3
    • one year ago
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  16. cocoapeebles3
    • one year ago
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    which I think is equal to 50

  17. IrishBoy123
    • one year ago
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    and where is the \(\pi \) being brought in? it is not in the original post...

  18. IrishBoy123
    • one year ago
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    \(\pi\) https://gyazo.com/ca2e80c981ef716a06b30d2fe7207627

  19. IrishBoy123
    • one year ago
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    strongly suggest you close this and start a new thread with the same question i am pretty easy going but this is too much, even for me. having learned something about how to share info online, you will have better luck with a new thread, the audience will have less to guess. good luck, and wish you well!

  20. cocoapeebles3
    • one year ago
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    oh sorry the original equation is this\[H(t)= 20\cos(\pi/15\times t) + 30\]

  21. cocoapeebles3
    • one year ago
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    okay thanks

  22. IrishBoy123
    • one year ago
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    thanks for that @cocoapeebles3 if this is right, \(\large H(t)=20cos(\frac{π t}{15})+30\) then as cos(0) = 1 you have \(h(0) = 20+30=50\) the *period* of the wheel can be got from \(\omega \ t = \frac{π }{15}t= \frac{2 \pi}{T} t \) \(\frac{π }{15}t= \frac{2 \pi}{T} t \implies T = 30 \) finally, \(10 \le h(t) \le 50\) simply because \(-1 \le cos \le 1\)

  23. IrishBoy123
    • one year ago
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    there might be a load of jargon in there so please do persist if anything is unclear and you have to factor in errors on my part too cheers!

  24. cocoapeebles3
    • one year ago
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    I think I understand it all thank you so much!

  25. cocoapeebles3
    • one year ago
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    So for part 3 how would I word that?

  26. IrishBoy123
    • one year ago
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    for last part \(H = \pm 20 + 30\) because cosine goes from -1 to +1.

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