Answer:
Step-by-step explanation:
Volume of tank is 3000L.
Mass of salt is 15kg
Input rate of water is 30L/min
dV/dt=30L/min
Let y(t) be the amount of salt at any time
Then,
dy/dt = input rate - output rate.
The input rate is zero since only water is added and not salt solution
Now, output rate.
Concentrate on of the salt in the tank at any time (t) is given as
Since it holds initially holds 3000L of brine then the mass to volume rate is y(t)/3000
dy/dt= dV/dt × dM/dV
dy/dt=30×y/3000
dy/dt=y/100
Applying variable separation to solve the ODE
1/y dy=0.01dt
Integrate both side
∫ 1/y dy = ∫ 0.01dt
In(y)= 0.01t + A, .A is constant
Take exponential of both side
y=exp(0.01t+A)
y=exp(0.01t)exp(A)
exp(A) is another constant let say C
y(t)=Cexp(0.01t)
The initial condition given
At t=0 y=15kg
15=Cexp(0)
Therefore, C=15
Then, the solution becomes
y(t) = 15exp(0.01t)
At any time that is the mass.
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Let the actual distance be x mile.
1/4 inch = 1/2 mile
9/2 inch = x mile
(9/2)/(1/4)=x/(1/2)
18=2x
x=9
the actual distance is 9 mile.
If you were to make both x's equivalent, you'd have to multiply the first equation by 2 to get rid of the coefficient, 1/2. However, you'd have to multiply 2 onto -8 as well, therefore the equation would turn into x - 16 = 18. That is not equivalent to x - 8 = 18.
2.134 x 10^4 is equal to 21,340
