Aspherical snowball is melting in such a way that it maintains its shape. the snowball is decreasing in volume at a constant rate of 8 cubic centimeters per hour. at what rate, in centimeters per hour, is the radius of the snowball decreasing at the instant when the radius is 10 centimeters?

0.006366 cm/h

Step-by-step explanation:

First we need to know that the volume of a sphere is calculated with the following formula:

V = (4/3)*pi*r^3

Where V is the volume and r is its radius.

Then, to find the rate of change, we need to derivate the equation with respect to the radius:

dV/dr = (4/3)*pi*(3r^2) = 4*pi*r^2

Then, we can write dV/dr as being (dV/dt)*(dt/dr), where dt is the change of time (variable t)

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

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

The rate of change of the volume is 8 cm3/h when the radius is 10, so using these values, we can find the rate of change of the radius (dr/dt):

8 = 4*pi*10^2*(dr/dt)

8 = 400*pi*(dr/dt)

dr/dt = 8/(400*pi)

dr/dt = 0.006366 cm/h

v v cj van config k tbs eng r.c r.c fcb fb gb frfr

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92