Faster and higher I believe.
At -40.
-40 gives the same reading for Fahrenheit and Celsius scale.
Answer:
Therefore the surface area of the balloon is increased at 4 cm³/s.
Explanation:
The balloon is being filled with air at a rate of 10 cm³/s
It means the volume of the balloon is increased at a rate 10 cm³/s.
i.e 
Consider r be the radius of the balloon.
The volume of of a sphere is

Differentiate with respect to t



The surface of area of the balloon is(S) = 

Differentiate with respect to t


Putting the value of


Given that r = 5 cm
=4 cm³/s
Therefore the surface area of the balloon is increased at 4 cm³/s.
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