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.
Answer:
x=16.5
Step-by-step explanation:
-1/2 on both sides
17-0.5 or 1/2=16.5
Answer:
The answer to your question is x = 14.7
Step-by-step explanation:
Data
∠A = 20°
∠B = 46
a = 7
b = x
Process
To solve this problem use, the law of sines. This law states that the ratio of a side of a triangle to the sine of the opposite angle is the same for all three sides.
The law of sines for this problem is
x / sin 46 = 7 / sin 20
-Solve for x
x = 7 sin 46 / sin 20
-Simplification
x = 7 (0.719) / 0.342
x = 5.035/0.342
-Result
x = 14.7
Answer:
12
Step-by-step explanation:
5+7=12
5 have 7-14
7 have 15-22
The final answer is 12
