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Questions (1)
Water is poured into a container that has a small leak.The
mass 'm' of the water is given as a function of time 't' by,m=5t^0.8
5 answers
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Answers (3)
derivative V: V (average) = dx/dt = 128 - 16 x 2 t^2-1 = 128 - 32t substitute the time given t = 4: 128 - 32 x 4 = 0 t=0 t = 2: 128 - 32 x 2 = 64 t=64
only Time not the MAX Time get*
A) Time is ZERO T=0 because mass was in the max point before it started to leak. B) substitute the max time to the function: M= 5 x (0^0.8) - (3) x (0) + 20 Max MASS = 20g c*) to gite the RATE of the mass we take derivative of the Function M = (5) x (o.8)
