The radius r(t) of the base of a cone is increasing at a rate of 10 meters per second. The height h(t) of the cone is fixed at 6 meters. At a certain instant , the radius is 1 meter. What is the t 0 rate of change of the volume V(t) of the cone at that instant?

dV/dt = 40π

Step-by-step explanation:

We are told that The radius r(t) of the base of a cone is increasing at a rate of 10 meters per second. Thus;

dr/dt = 10 m

Height: h = 6 m

Volume of cone is given by the formula;

V = ⅓πr²h

dV/dr = ⅔πrh

We want to find the rate at which the volume is changing at radius of 1m.

Thus;

dV/dt = (dV/dr) × (dr/dt)

dV/dt = ⅔π(1 × 6) × (10)

dV/dt = 40π

(-1,-0.5)

step-by-step explanation:

a) y = -3x +2

b) y = 1/2x -6

these 2 can be combined into one equation.

-3x +2 = .5x -6

8 = 3.5x

x = 8 / 3.5 = 2.2857142857

y = -3 * (8 / 3.5) +2

y = (-24 / 3.5) + 2

y = -4.8571428571

step-by-step explanation: