Aspherical balloon is inflated with helium at the rate of 100pie ft^3/min/ a) how fast is the balloon's radius increasing at the interest the radius is 5 feet? b) how fast is the surface area increasing at that instant.

The problem at hand is on related rates, which could be solved using differential calculus

The volume of a sphere: V = (4/3)(pi)(r^3)

Differentiating with respect to time,

(dV/dt) = 4(pi)(r^2)(dr/dt)

Substituting the given,

(dV/dt) = 100 , r = 5

100 = 4(pi)(5^2)(dr/dt)

Solving for (dr/dt)

dr/dt= 0.318 ft/s

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.

d 25 pi/6

step-by-step explanation:

the diameter is 10 inches

the radius is 1/2 of the diameter or 5 inches

the area of the circle is

a = pi r^2

  = pi 5^2

  = 25 pi

the slice is 60 degrees of the entire circle.

60 degrees out of 360   60/360 = 1/6

the area of the slice is 1/6 of the circle

a slice = 1/6 a circle

a slice = 1/6 * 25 pi