shorouq

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 -3t + 20 , with t>=0, t in seconds: A) at what time is the water mass greatest, B) what is the greatest mass? *In kilogram per...

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)...

only Time not the MAX Time get*

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