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.
Use this rule: <em>(x^a)^b = x^ab</em>
3(x + 2)^3/5 + 2 = 27
Subtract 3 from both sides
3(x + 2)^3/5 = 27 - 3
Simplify 27 - 3 to 24
3(x + 2)^3/5 = 24
Divide both sides by 3
(x + 2)^3/5 = 24/3
Simplify 24/3 to 8
(x + 2)^3/5 = 8
Take the cube root of both sides
x + 2 = 3/5√8
Invert and multiply
x + 2 = 8^5/3
Calculate
x + 2 = 2^5
Simplify 2^5 to 32
x + 2 = 32
Subtract 2 from both sides
x = 32 - 2
Simplify 32 - 3 to 30
<u>x = 30</u>
200 / 8 = 25
scale factor is 1:25 or 0.04
solve for "k", to find k or the "constant of variation"
then plug k's value back to
now.... what is "p" when q = 5? well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is
