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Mathematics 15 Online
OpenStudy (anonymous):

Can someone give me a real world example of a periodic function

Parth (parthkohli):

the clock.

OpenStudy (anonymous):

How do I put that into a function and graph it

Parth (parthkohli):

The time a clock shows \(n\) hours after 12 o'clock is \(12 + n \) modulo 12.

OpenStudy (anonymous):

I hate to be a pain but could you give me an example please

Parth (parthkohli):

\(n\) is your input and \(12 + n\) modulo 12 is the output. If you want the time shown on the clock \(4\) hours after \(12\), then you must calculate the remainder you get when you divide \(12 + 4\) by \(12\), which is \(4\).

Parth (parthkohli):

Be honest: am I being of any help here? :-)

OpenStudy (anonymous):

Yes, I am trying to see if I am doing it correctly. But I cant understand why we divide by 12

Parth (parthkohli):

It's a long concept. Have you heard of clock-12 arithmetic?

OpenStudy (anonymous):

no

Parth (parthkohli):

Well, it goes like this: If you wanna add two given times on the clock, you must first add them, then calculate the "extra" amount you got there after 12. So if you wanna add 6 to 7 o'clock, it won't be 13 o'clock. It'd be 1 o'clock instead because you are 1 "extra" after 12.

Parth (parthkohli):

OK, but you do understand that time on the clock keeps repeating right?

OpenStudy (anonymous):

Yes I understand that and I know that cos and sin do also, but I cant express it in terms of a function

Parth (parthkohli):

What if you want to know the time after \(x\) hours after \(12\) o'clock?

OpenStudy (anonymous):

Not sure what you mean

Parth (parthkohli):

After an hour, it is \(1\) o'clock. After two, it is \(2\) o'clock. You can make a table. y x 1 1 2 2 3 3 4 4 . . . . 1 13

OpenStudy (anonymous):

Okay, i think i know where I am confused. I do not add 24 hours, I stop after 12 hours and and start over again

OpenStudy (goformit100):

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OpenStudy (goformit100):

*Due

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