94. Juan is practicing archery using a simple circular target with a radius of 18 inches. The target's a bullseye has a circumference of 18pi inches. What is the area of the portion of the target that is not part of the bullseye?

763.02 in^2

Step-by-step explanation:

For a circle of radius R, we know that:

The area is:

A = pi*R^2

the circumference is:

C = 2*pi*R

where pi = 3.14

We know that the circumference of the bullseye is 18*pi, then the radius of this circle is:

2*pi*R = 18*pi

2*R = 18

R = 18/2 = 9

The radius of the bullseye is 9 inches.

Now, the area of the target that is not part of the bullseye is equal to the difference between the area of a circle of radius 18in (the total target area) and a circle of radius 9in (the bullseye area)

This is:

A = pi*(18in)^2 - pi*(9in)^2 = 3.14*( (18in)^2 - (9in)^2) = 763.02 in^2

its adding 5.5 each time

step-by-step explanation:

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870