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.
Hypotenuse by Pythagoras theorem = 407.83.
Answer:
You multiply 5 by the exponents and 4^0 is just 1, so 4^45•1=4^45. Hope this helped!
Step-by-step explanation:
The first one would be A, the lower quartile is given, because the 5 number summary has been given. The second one would be 99, because 97 gives 89.6 and 98 gives 89.8. And the third one would be there are exactly 3 students with 4 pairs of shoes. Hope this helps! If you need anything else let me know.
Answer:
x=771
Step-by-step explanation:
−771 + x = 0
x − 771 = 0
x − 771 + 771 = 0 + 771
