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.
The inequality is t < 55
<em><u>Solution</u></em><em><u>:</u></em>
Given that, To qualify for the championship a runner must complete the race in less than 55 minutes
Let "t" represent the time in minutes of a runner who qualifies for the championship
Here it is given that the value of t is less than 55 minutes
Therefore, "t" must be less than 55, so that the runner qualifies the championship
<em><u>This is represented by inequality:</u></em>
The above inequality means, that time taken to complete the race must be less than 55 for a runner to qualify
Hence the required inequality is t < 55
The diagnonals of a parallelogram bisect each other ,
so
=》 BD = 3x × 2
=》BD = 6x
and there's given that
=》
=》
=》
=》
so, if x = 4 , then the given quadrilateral is a parallelogram.
Answer:
1/4
Step-by-step explanation:
