
It is prime because it has no common factors in 5a^2 and b.
The answer is
A<span>
.</span>
a.
has an average value on [5, 11] of

b. The mean value theorem guarantees the existence of
such that
. This happens for

See picture for answer and solution steps.
Answer:
C≈9.42m
Using any of this formulas
C=2πr
C=2πrd=2r
Solution:
C=πd=π·3≈9.42478m