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.
They are already both positive numbers, so the absolute value of 2 is 2, and the absolute value of 1/3 is 1/3.
Hello!
(4y + 8) - (7y - 12) = 11 is 1(4y + 8) - 1(7y - 12) = 11
1(4y + 8) - 1(7y - 12) = 11 Given
4y + 8 - 7y + 12 = 11 Distribute the 1 and the -1
-3y + 20 = 11 Combine like terms
-3y = -9 Subtract 20 from both sides
y = 3 Divide both sides by -3
Answer:
y = 3
Here's the graph of your equation, instructions on how to plot is in the image. if there's a second line equation, you could use the method shown to plot that too, and the x and y values of the intersection point would be your answer.
you could also use algebra to simultaneously solve the two equations to obtain x and y as well, feel free to ask!
hope that helps :) sorry it's not a definite answer
9514 1404 393
Answer:
(d) x = 2
Step-by-step explanation:
Adding the two equations gives ...
(-8x +8y) +(3x -8y) = (8) +(-18)
-5x = -10 . . . . . sum of the two equations
x = 2 . . . . . . . . result of dividing by -5
