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:
the question is incorrect. please write your points as coordinates or scan the diagram.
Answer:
f(5) = -2
Step-by-step explanation:
You look on the x-axis for 5 because f(5) means let x = 5. Then go down to the graph (the red zig-zaggy line) At the red line (you're still directly under the 5) look over to the y-axis to find the value. That's your answer. It is -2. The point (5, -2) is on the red graph.
f(5) = -2
Answer:
$120
Step-by-step explanation:
930-570=360 (Price difference in 3 years)
360 divided by 3= 120 (Price difference for each of the 3 years)
