Answer:
t=0 Sugar = 0 Kg
t=1min Sugar=0.27 Kg
Step-by-step explanation:
Data
Tank = 2640 L (pure water)
Sol=0.09kg Sugar per liter
Vin = Vout = 3L/min
Sugar in the beginning = ?
if beginning is t = 0min Amount of sugar = 0, this is due to the fact that at the moment of entering the tank the content is only water , but if beginning is t= 1min then;
Answer: y(t)= 237.6(1-e^(-t/880))
In the attachment
19.5 because 6 minus 5.5 is equal to 0.5 and if you just add 0.5 to 19.5 it will be 19.5
20 I think I hope it is correct
A