this may be a stupid question and math isnt my forte so i am very terrible at it. however, do you mind explaining how did you get 24 pi over 6 on the last example?
So one of the way to solve these is to take out multiples of "2pi" which is a full revolution around the circle. So if the denominator is 6, that converts to 12pi/6, then 24pi/6, then 36pi/6, etc. Since the number given is 29pi/6, the most that can be removed is 24pi/6 leaving 5pi/6. Hope that helps. Good luck.
(11pie over 3) makes me go around three times with (2pie over three)left over but that is quadrant three and I passed it so how do I go about getting that reference point?
I assume you mean terminal point? So you have the logic correct. You move 3pi which takes you to the left side of the circle. You have 2pi/3 left which is a full quadrant plus a little bit more. So from the left side move a whole quadrant plus a little more which puts you in quadrant 4.
The reference number and terminal point are not the same point, however, when the denominator is 6, 4 or 3, no matter what the numerator is, the reference number will be pi over that number. For example, the reference number for 101pi/3 is pi/3. Hope that helps.
![Why you didn't learn tetration in school[Tetration]](http://i.ytimg.com/vi/Ea1_1aVfwl4/mqdefault.jpg)
May God bless you Mr. Buch. Thank you for all your help.
I wish I found this channel out earlier. Help out so much
George the legend. Thanks for making this easy to grasp brother.
My pleasure
Why is he such a good freaking teacher???
Thanks.
Amazing explanation and examples, thank you so much.
You're welcome. Good luck.
you are a great teacher thanks helped me alot
Thanks. Glad it helped. Good luck with your classes.
This saved my life
Glad they are helping!
this may be a stupid question and math isnt my forte so i am very terrible at it. however, do you mind explaining how did you get 24 pi over 6 on the last example?
So one of the way to solve these is to take out multiples of "2pi" which is a full revolution around the circle. So if the denominator is 6, that converts to 12pi/6, then 24pi/6, then 36pi/6, etc. Since the number given is 29pi/6, the most that can be removed is 24pi/6 leaving 5pi/6. Hope that helps. Good luck.
thank you go much it really helps me understand it better
Saved my life.
Thanks bro it's very helpful
Great 👍
(11pie over 3) makes me go around three times with (2pie over three)left over but that is quadrant three and I passed it so how do I go about getting that reference point?
I assume you mean terminal point? So you have the logic correct. You move 3pi which takes you to the left side of the circle. You have 2pi/3 left which is a full quadrant plus a little bit more. So from the left side move a whole quadrant plus a little more which puts you in quadrant 4.
So is the reference number and terminal point always the same point?
The reference number and terminal point are not the same point, however, when the denominator is 6, 4 or 3, no matter what the numerator is, the reference number will be pi over that number.
For example, the reference number for 101pi/3 is pi/3.
Hope that helps.
Thanks
How do you get the terminal points???
18sranchhod I believe I discuss terminal points in 5.1.2.3. If not, ask me again.
What of t= -7
You need to move 7 radians clockwise. This brings you all the way around the circle (2pi is about 6.28), and back into Q4.
George Buch thank you