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.
Lets write this out:-
2.4 + 0.8 =______ - 0.06 = _____+ 1.21=_____+1.78=_____- 5.14=___
So to solve d blanks we will do d following:-
2.4 + 0.8 = 3.2
Now lets write this out AGAIN.
2.4 + 0.8 = 3.2 - 0.06 = _____+ 1.21=_____+1.78=_____- 5.14=___
Now lets solve again:-
3.2 - 0.06 = 3.14
Now lets write this out AGAIN.
2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21=_____+1.78=_____- 5.14=___
Now lets solve again:-
3.14 + 1.21= 4.25
Now lets write this out AGAIN.
2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21 = 4.25 +1.78=_____- 5.14=___
Now lets solve again:-
4.25 +1.78= 6.03
Now lets write this out AGAIN.
2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21 = 4.25 +1.78= 6.03 - 5.14=___
Now lets solve again:-
6.03 - 5.14 = 0.89
Now lets write this out AGAIN.
2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21 = 4.25 +1.78= 6.03 - 5.14= 0.89
So, 2.4 + 0.8 = 3.2 - 0.06 = 3.14 + 1.21 = 4.25 +1.78= 6.03 - 5.14= 0.89
Hope I helped ya!! xD
Answer:
25pi. 36pi
10pi. 12pi
--------------------------
9pi. 49pi
6pi. 14pi
Answer:
YOU MUST HAVE ONC WONDERED WHAT IS CHICKEN PO0. WELL IM HERE TO TELL YOU. PLEASE STOP EATING IT. IT MUGHT BE BAD FOR YOU.
ALS OTHE ANSWEER IS A>6x+16=49>2
Step-by-step explanation: