Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of
(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min
and flows out at a rate of
(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min
Then the net flow rate is governed by the differential equation
Solve for S(t):
The left side is the derivative of a product:
![\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5Be%5E%7Bt%2F800%7DS%28t%29%5Cright%5D%3D%5Cdfrac8%7B100%7De%5E%7Bt%2F800%7D)
Integrate both sides:
There's no sugar in the water at the start, so (a) S(0) = 0, which gives
and so (b) the amount of sugar in the tank at time t is
As 
, the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.
Answer:
18 sq ft
Step-by-step explanation: all you have to do is add the sides. 5+5=10
4+4= 8
10+8=18
This makes the most sense with the answers givin
Answer:
4 years
Step-by-step explanation:
560 = 7000 × 2/100 × t
t = 4
Answer:
The answer for the first one is 1/f3
the second one is b-5 i think
third one is 8k^5/m4
last one is -9
hope this helped
Step-by-step explanation:
Answer:
90 : 1
Step-by-step explanation:
1 yard = 36 inches
5 yards = 36×5 = 180 inches
180 : 2
90 : 1
