An equation has as many solutions as the highest exponent.
For example: x^5+2x-3=0 would have a maximum of 5 solutions
6.20÷4 = 1.55
price per individual dounut =$1.55
1.55×12=18.6
12 donuts =$18.6
Answer: The area of the sector formed by central angle AOB is 5/8 th of the total area of the circle.
Step-by-step explanation:
Let r be the radius of the circle having the center O,
⇒ The area of the circle =
square unit.
And, the central angle AOB = 
( since,
= 180° )


Hence, the area of sector AOB

square unit.
Now,


⇒ Area of sector AOB = 5/8 × Area of the circle.
Hence, the area of the sector formed by central angle AOB is 5/8 th of the total area of the circle.
DF congruent to AC and EF is common side
25:1 describes price to uniforms because each one costs $25