Steps to solve:
3(x - 1) = 5x + 3 - 2x
~Distribute left side
3x - 3 = 5x + 3 - 2x
~Combine like terms
3x - 3 = 3x - 3
~Subtract 3x to both sides
-3 = -3
~Add 3 to both sides
0 = 0
All real numbers are solutions.
Best of Luck!
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.
Answer: I think the term is “Like Terms”
Answer:
the answer is f(x) = 6x + 1/6
Step-by-step explanation:
the g(x) ones y intercept is : 0,-6
the -6k(x) ones y intercept is : 0 , -6
the f ( x) ones y inetrcept is ; 0 , 1/6
the h (x) ones y intercept is : 0 , 6