Any portion of the circumference of a circle is called a radius
Answer:
when you divide you get -11/12
if you have a calculator you could add the 5/8 and 3/4
then add the -2/3 and the -5/6
then divide the first answer by the second answer
Answer:
-1 is the answer , your welcome haha
This is one pathway to prove the identity.
Part 1

Part 2

Part 3

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.
We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity
in the second to last step. I broke the steps into three parts to hopefully make it more manageable.
Each folder must have 10 pages
!0 papers times 3 folders is 30
plus the 7 pages equals 37