Answer:
23
Step-by-step explanation:
Step 1:
3 - -5 (4)
Step 2:
3 - (-20)
Step 3:
3 + 20
Answer:
23
Hope This Helps :)
This graph will start at (.5, 0) as the vertex.
To find this, all you need to do if find the point where inside the absolute value sign is equal to 0. Since the graph can never have a negative value, this will be the lowest point.
2x - 1 = 0
2x = 1
x = .5
From there, you can plot each point by going up two and to the left one.
Examples of Points: (1.5, 2) and (2.5, 4)
You can also plot the other side of the graph by going up two and to the right one.
These points can be found using the slope, which is the number attached to x (2).
Example of Points (-.5, 2) and (-1.5, 4).
Answer:
the answer for apex is shift
Step-by-step explanation:
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.
