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:
5 milk shakes
Step-by-step explanation:
From the above question, we know that for the recipe
1/3 cup of milk = 1 milk shake
15/8 cup of milk = y milk shakes
We cross Multiply
15/8 cup × 1 = 1/3cup × y
y = 15/8 cup × 1 /1/3cup
= 15/8 ÷ 1/3
= 15/8 × 3
= 45/8
= 5.625 of 5 5/8 cups of milk
Therefore, the maximum number of
milkshakes the server can make following the recipe porpotians is 5 milk shakes.
Answer:
Step-by-step explanation:
17) HI ≅ UH ; GH ≅ TU ; GI ≅ TH
ΔHGI ≅ ΔUTH by Side Side Side congruent
∠G ≅ ∠T ; GI ≅ TH ; ∠GIH ≅ ∠THU
ΔHGI ≅ ΔUTH by Angel Side Angle congruent
19) IJ ≅ KD ; IK ≅ KC ; KJ ≅ CD
ΔIJK ≅ ΔKDC by Side Side Side congruent
∠J ≅ ∠D ; IJ ≅ KD ; ∠I ≅ ∠DKC
ΔIJK ≅ ΔKDC by Angle Side Angle congruent
Mario made the mistake on everything and he needs help
Answer:
The bottom question is wrong. You have to multiply 8 to get the amount earned. But the rest is correct.
Step-by-step explanation:
6.25 x 8 = 50
12.5 x 8 = 100
18.75 x 8 = 150
25 x 8 = 200
31.25 x 8 = 250
