How are 10.5 radians and 5.25 rotations? By dividing 10.5π by π it becomes radians without pi... so would I then still divide by 3.14 since the multiplier is gone? I feel I am missing something fundamental here, im sorry.
Why did you divide by π instead of 2π? How did you know to divide by any...
Question: Where is an angle of 10.5π radians in standard position located?
A. On the positive vertical axis
B. On the negative vertical axis
C. In the second quadrant
D. In the fourth quadrant
I thought I had to divide 10.50 by 3.14, which I thought yielded 3.34π. Then I subtracted 3.34π...