Jill is trying to find the area of football field with the track around it at her high school. it is composed of a rectangular middle with a semicircle on each end. the rectangular field has length 130 feet and width 50 feet. what is the area of the whole thing?
a) 6500 + 625π ft2
b) 6500 + 2500π ft2
c) 6500 + 5000π ft2
d) 7125π ft2

Step-by-step explanation:

Refer the attached figure

Area of rectangle = Length * Breadth

So, area of rectangle ABCD : 130*50

                                              : 6500 square feet .

Now two semicircles are on the widths of the rectangular field .

So, diameter of the semicircles on the ends of the rectangular field = 50 feet

So, radius = diameter /2 = 50/2 = 25

Formula of area of circle =

Area of 2 semicircles =

                                   =

                                   =

So, the total area of the field = 6500 square feet + 625π square feet

                                               =

Hence  the area of the whole thing

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).

it is 10.6 according to my calculations