g The radius of a spherical ball increases at a rate of 3 m/s. At what rate is the volume changing when the radius is equal to 2 meters

dV/dt = 150.79 m^3/s

Explanation:

In order to calculate the rate of change of the volume, you calculate the derivative, respect to the radius of the sphere, of the volume of the sphere, as follow:

                    (1)

r: radius of the sphere

You calculate the derivative of the equation (1):

(2)

where dr/dt = 3m/s

You replace the values of dr/dt and r=2m in the equation (2):

The rate of change of the sphere, when it has a radius of 2m, is 150.79m^3/s

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

where is the