I think the first one might be 18 second might be 10
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:
Option B. Amplitude =3 midline is y =2.
Step-by-step explanation:
In the graph attached we have to find the amplitude and midline of the periodic function.
Amplitude of the periodic function = (Distance between two extreme points on y asxis)/2
= (5-(-1))/2 = (5+1)/2 =6/2 =3.
Since amplitude of this function is 3 and by definition amplitude of any periodic function is the distance between the midline and the extreme point of wave on one side.
Therefore midline of the wave function is y=2 from which measurement of the amplitude is 3.
Answer:
33/40
Step-by-step explanation:
1/5x8= 8/40
5/8x5=25/40
add them
reduce answer:
33/40
