The radius of the circle is increasing at a rate of 2 meters per minute and the sides of the square are increasing at a rate of 1 meter per minute. When the radius is 6 meters, and the sides are 24 meters, then how fast is the AREA outside the circle but inside the square changing?

Change in area=24-48

Step-by-step explanation:

Let s will be the side of square and r will be the radius of circle.

Then two given conditions are

1)dr/dt=2 m/s

2)ds/dt=1 m/s

Area enclosed=(Area of square)-(Area of circle)

Area of square=

Area of circle=

Area enclosed=

dA/dt=2r(dr/dt)-2s(ds/dt)

At s=24,and r=6

dA/dt=2()(6)(2)-2(24)(1)

Change in area=24-48

answer: x= -2y

step-by-step explanation:

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: