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:
103
Step-by-step explanation:
27) if a = 1/a
multiply both sides by a
a * a = 1/a * a
a^ 2 = 1
Answer is D) 1
28)
6 x 75 = 450
7 x 80 = 560
8 x 85 = 680
9 x 90 = 810
Total all students in class = 6 + 7 + 8 + 9 = 30 students
Total score of whole class:
450 + 560 + 680 + 810 = 2500
Average = 2500 /30 = 83.3 = 83 1/3
Answer is C) 83 1/3
0.006 is 1/10 of 0.06 .
Correct me if I am wrong.
Answer:
Whats the question?
Step-by-step explanation:
