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.
1/5( 5/2 x 4/7 -11/7) + 3/28
Multiply the terms (5/2 x 4/7) in bracket then subtract it from 11/14. Result will be 10/7.
Then multiply 1/5 with 10/7. The result will be 2/7.
2/7 + 3/28 = 11/28
T
Answer:
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Step-by-step explanation:
Multiply through by 6 to eliminate the fraction
That's
Move - 18f³ to the right side of the equation
That's
Divide both sides by M to make B stand alone
That's
We have the final answer as
Hope this helps you
The triangle will be rotated with the rule of (x,y) → (-y,x). The image will be congruent to the pre-image, the side lengths will be the same and the angles too.
Answer:
1 mile = 0.2 hours
Step-by-step explanation:
Given that,
Eddie can travel 20 miles in 4 hours.
We need to find how many hours does it take Eddie to travel 1 mile.
20 mile = 4 hours
1 mile = (4/20) hours
= 0.2 hours
So, to travel 1 mile will take 0.2 hours.
