We have the following function:

So if we graph this function we will get the Figure below. Thus, let's study both the equation and the graph to get some conclusions. Therefore, we can assure these statements:
First. The function is defined only for

as shown in the Figure. This is also true because of

where

must be greater (or equal) than zero.
Second. The range of the function are the values of

.
Third. If

creases then

always creases, too.
Area of sector=fraction of sector times area of circle
fraction of sector=degrees/360
and area of circle=pir^2
so
hold a sec, we have the wiper on the outside, so we want to find the outside ring thing
do
area big sector-area small sector
or even better
(degrees/360) times (area big-areasmall)
big is 10
small is 4 (10-6=4)
areabig=10^2 times pi or 100pi
areasmall=4^2 times pi or 16pi
areabig-areasmall=100pi-16pi=84pi
then find the fraction
84pi times (150/360)=35pi square inches
First of all, the identity property of multiplication (which is what this is I'm assuming) is that the number 1 multiplied by any other number is that number itself. (An example would be 2 multiplied by 1, which would be two) So in this problem, this rule applies too, since 2/3 multiplied by 1 would be 2/3!
Hope this helped :)
Answer:
2⋅2⋅2⋅5=40 2 ⋅ 2 ⋅ 2 ⋅ 5 = 40