The circumference of the inner circle is 22 ft. The distance between the inner circle and the outer circle is 3 ft. By how many feet is the circumference of the outer circle greater than the circumference of the inner circle?

18.8 ft approximately

Step-by-step explanation:

The formula for the circumference of a circle is C = 2*pi*r, where pi is approximately 3.14 and r is the radius of the circle.

The circumference of the inner circle is 22 ft.  Thus, C = 22 ft = 2(3.14)(r).  Solving this for r, we get:

         22 ft

r = = 3.503 ft

         6.28

Since 3 ft separate the inner and outer circles, the radius of the outer circle is (3.5 ft+ 3 ft), or 6.5 ft, and so the circumference of the outer circle is

C = 2(3.14)(6.5 ft), or approximately 40.8 ft.

Thus, the circumference of the outer circle exceeds that of the inner circle by (40.8 ft - 22 ft), or 18.8 ft approximately.

The difference in the two circumferences is (157 ft - 22 ft), or 135 ft

dilation scale factor 2: starting with δabc, draw the dilation image of the triangle with a center at the origin and a scale factor of two. notice that every coordinate of the original triangle has been multiplied by the scale factor (x2). dilation with scale factor 2, multiply by 2.