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?

- cocoapeebles3

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- cocoapeebles3

-.- 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

- cocoapeebles3

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.

- IrishBoy123

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.

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## More answers

- IrishBoy123

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- IrishBoy123

|dw:1439938102822:dw|

- IrishBoy123

or post a url or do a screengrab or scribble it down and scan and link it

- IrishBoy123

|dw:1439938283787:dw|

- IrishBoy123

i am sorry to be such a pain but, once you get this bit sorted, you will get loads of help!

- cocoapeebles3

|dw:1439938250052:dw|

- IrishBoy123

an example of posting a screengrab
https://gyazo.com/9bd2129de27eaf26d56f85c2ffee15eb

- IrishBoy123

- cocoapeebles3

sorry I'm new to this

- IrishBoy123

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

- IrishBoy123

|dw:1439938633443:dw|

- cocoapeebles3

##### 1 Attachment

- cocoapeebles3

which I think is equal to 50

- IrishBoy123

and where is the \(\pi \) being brought in?
it is not in the original post...

- IrishBoy123

\(\pi\)
https://gyazo.com/ca2e80c981ef716a06b30d2fe7207627

- IrishBoy123

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!

- cocoapeebles3

oh sorry the original equation is this\[H(t)= 20\cos(\pi/15\times t) + 30\]

- cocoapeebles3

okay thanks

- IrishBoy123

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 \(-1 \le cos \le 1\)

- IrishBoy123

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!

- cocoapeebles3

I think I understand it all thank you so much!

- cocoapeebles3

So for part 3 how would I word that?

- IrishBoy123

for last part \(H = \pm 20 + 30\) because cosine goes from -1 to +1.

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