What is the length of the indicated arc? (7/2 pi) (14 pi) (3.5 pi) (7 pi), or (none of the above)
Know the formula for arc length?
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If you think about it, the circumference of a circle is the measure of the arc length for a full 360 degrees. So \(2\pi r\) is the arc length for 360. You can then use that and a ratio to find what it is for 315.
I got 25.12
Well, you don't want to used 3.14 for pi. All the answers are exact. And let me see if that is right.
ok
Hmmm... That is not what I got. OK, do you know what the ratio or fraction you need to multiply by would be?
If you know how to convert from degrees to radians, that is another way to set it up. Ends up being the same thing in the end.
No I'm trying to figure it out but idk
OK. For any ratio like this, you are looking for a percentage of the whole. In general, that is the fraction: \(\dfrac{\rm portion}{\rm whole}\)
In this specific case, your potion is the degrees of the arc and the whole is the degrees of a circle.
so something divided by 360 degrees
Exactly! The degrees of the arc divided by 360.
315 divided by 360?
Yes. And you multiply that by the circumference, or \(2\pi r\). Now, because the answer has \(\pi\) in it, you can leave it out of the multiplying. So don't multiply by 3.14 because you know the "(7/2 pi) (14 pi) (3.5 pi) (7 pi) or none" takes that into account.
ok 315 divided by 360 is 0.875, then what if I can't multiply 3.14
Well, 2pi r without the pi is 2r. So multiply by 2r.
8
So 8 * .875
I get 70
70? Umm.... must have did something wrong.
7, I mean
=) And the pi is still there, we just did not multiply it, so \(7\pi\).
Thank you so much, This one problem helped with like four of my others and I made a 100 thank you, you're genius, I will now grant you your medal!
Glad you were able to learn how it worked. =)
Thanks Costello, anyways I got 3.75 for this ... IDK if it's right
3.75 for what?
oops... I accidentally wrote on the wrong question thread ...
But actually, you are the only person who I can understand this math from , so can you help me with one more problem
It's the latest one that I've put in the posts.
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