<span>1. $119 2. 64 beads 3. 4 candles 4</span>
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.
ABC = 408
C + 7 = B
A + 5 = B
B - 7 + B - 5 + B = 408
3B - 12 = 408
+ 12 = +12
3B = 420
/3 = /3
B = 120
A = 115
C = 113
The answer is A = 115, Abel has $ 115
<h2>39/6 = 6.5</h2><h2 /><h2>6 hours 30 minutes</h2>
Answer:
Copy all that then paste it to Google and there is a site that I found the answer on but I can't copy it it's to long.