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Mathematics 17 Online
OpenStudy (anonymous):

A spherical party balloon is being inflated with helium pumped in at a rate of 3 cubic feet per minute. How fast is the radius growing at the instant when the radius has reached 3 ft?

OpenStudy (anonymous):

V=3t=4/3*r^3*pi

OpenStudy (anonymous):

speed = flow rate / area => s = Q/A the area of a circle is A = pi*r^2 when r = 3 ft then A = pi*3^2 = 9*pi Q = 3 cubic feet per minute so finally: s = 3/(9*pi) = 1/(3*pi) feet/minute Thats how fast it is growing

OpenStudy (anonymous):

The problem has to do with volume and surface area The volume of a sphere of radius r is V = 4/3πr3. Round your answer to two decimal places.

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