A circle is inside a square. The radius of the circle is decreasing at a rate of 3 meters per hour and the sides of the square are decreasing at a rate of 5 meters per hour. When the radius is 4 meters, and the sides are 23 meters, then how fast is the AREA outside the circle but inside the square changing? The rate of change of the area enclosed between the circle and the square is square meters per hour.

Step-by-step explanation:

Let r = radius of circle

     s = length of a side of the square

dr/dt = -3m/hour

  ds/dt = 5m/hour

we know that

area of square= s^2

area of circle=πr^2

Let A be the area of outside

A = (area of square) - (area of circle)

          = s^2 - πr^2

dA/dt=?

when r = 4 and s = 23

 substituting we have

dA/dt = 2s(ds/dt) - 2πr(dr/dt)

             =2(23)(5) - 2π(4)(-3)

            = (230 + 24π) m^2/hour

            = (230 + 75.408) m^2/hour

            = 305.408m^2/hour

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.