Answer:
1. Undefined
2. Negative
3. Zero
4. Positive
Step-by-step explanation:
1. Undefined, the line moves in both directions vertically. It's impossible to tell what the slope is.
2. Negative, the line is decreasing at a constant rate, the slope is negative.
3. Zero, the line doesn't increase or decrease, therefore the slope is 0. (No changes)
4. Positive, the line is increasing at a large rate, the slope is positive.
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Answer:
the second and last sentences.
Step-by-step explanation:
Its the most reasonable!
The answer is x=-2
As you could see, this is a factored function.
The un-factored version is x^2+4x-12
-b/2a=x
-4/2=-2= x of vertex
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.
