![andrecoral105](/avatars/10027.jpg)
andrecoral105
30.10.2020 •
Mathematics
Ok who wanna join a google meet? I just need to know age and gender
1. Why age?- So I know I'm not talking to an old man
2. Gender?- Because why not and I need a new ibf
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Ответ:
16 but im female
Ответ:
i want to but i need to know your age and gender
13 and female
Step-by-step explanation:
Ответ:
(a) Initial concentration of salt = 0.07 kg/L
(b)![\therefore Y'(t)= 0.28 -\frac{y}{125}](/tpl/images/0534/6441/723e2.png)
(c)![Y(t)=35 +35e^{-\frac{1}{125} t}](/tpl/images/0534/6441/705b6.png)
(d)Therefore the amount of salt after 5 hours is =38.18 kg
(e) The concentration of salt in the solution in the tank as time approaches infinity is = 0.035 kg/L
Step-by-step explanation:
Given that,
A tank contains 70 kg of salt and 1000 L of water.
(a)
=0.07 kg/L
(b)
Let Y(t) be the amount of salt at any instant time t.
Therefore
=(8×0.035) kg/min
=0.28 kg/min
Since the rate of water in and out are same , the amount of solution remain constant.
(c)
The above equation can be rewrite as
The coefficient of y is p(t)![=\frac{1}{125}](/tpl/images/0534/6441/02b10.png)
The integrating factor of the D.E is![=e^{\int p(t) dt=](/tpl/images/0534/6441/fbfd0.png)
![=e^{\frac{1}{125} t](/tpl/images/0534/6441/ae088.png)
Multiplying the integrating factor of both sides of D.E
Integrating both sides
At initial when t= 0, y =70
Therefore
⇒C= 70-35
⇒C=35
Therefore
(d)
When t= 5 hour = 300 min
To find the amount of salt after 5 hour , we need to put the value of t in the general solution of D.E
Therefore the amount of salt after 5 hours is =![Y(300)=35+35e^{-\frac{300}{125} }](/tpl/images/0534/6441/244d2.png)
= 38.18 Kg
(e)
When t=∞
= 35 Kg![[e^{-\infty}=0]](/tpl/images/0534/6441/caf96.png)
Since the amount of water is remain same i.e 1000 L
Therefore the concentration of salt is![=\frac{35}{1000}kg/L](/tpl/images/0534/6441/54c94.png)
=0.035 kg/L