(1 point) A tank contains 2640 L of pure water. A solution that contains 0.09 kg of sugar per liter enters the tank at the rate 3 L/min. The solution is mixed and drains from the tank at the same rate. (a) How much sugar is in the tank at the beginning
2 answers:
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))
Step-by-step explanation:
In the attachment
You might be interested in
3 red and 3 yellow don't get ur question cause think u should re say it
4x^2-25= (2x)^2-5^2=(2x-5)(2x+5)
Used: a^2-b^2=(a+b)(a-b)
The answer to the problem is y=1/2x
Answer:
30/100=30%. 98-30%=68%, which will be your new grade. That is usually a D
Step-by-step explanation:
Answer:
2/3
Step-by-step explanation:
Because 7 doubled is 14, you just need to double 1/3 to get your answer. And 1/3 X 2 = 2/3