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2 years ago

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Mathematics

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dolphi86 [110]2 years ago

7 0

Answer:

113

Step-by-step explanation:

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A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in the solution. Water containing1 lb

devlian [24]

Answer:

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is $\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$ .

(b) The concentration (in lbs per gallon) when it is at the point of overflowing is $\frac{121}{125}\:\frac{lb}{gal}$ .

(c) The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$ .

Step-by-step explanation:

This is a mixing problem. In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank. The liquid entering the tank may or may not contain more of the substance dissolved in it. Liquid leaving the tank will of course contain the substance dissolved in it. If Q(t) gives the amount of the substance dissolved in the liquid in the tank at any time t we want to develop a differential equation that, when solved, will give us an expression for Q(t).

The main equation that we’ll be using to model this situation is:

Rate of change of Q(t) = Rate at which Q(t) enters the tank – Rate at which Q(t) exits the tank

where,

Rate at which Q(t) enters the tank = (flow rate of liquid entering) x (concentration of substance in liquid entering)

Rate at which Q(t) exits the tank = (flow rate of liquid exiting) x (concentration of substance in liquid exiting)

Let $C$ be the concentration of salt water solution in the tank (in $\frac{lb}{gal}$ ) and $t$ the time (in minutes).

Since the solution being pumped in has concentration $1 \:\frac{lb}{gal}$ and it is being pumped in at a rate of $3 \:\frac{gal}{min}$ , this tells us that the rate of the salt entering the tank is

$1 \:\frac{lb}{gal} \cdot 3 \:\frac{gal}{min}=3\:\frac{lb}{min}$

But this describes the amount of salt entering the system. We need the concentration. To get this, we need to divide the amount of salt entering the tank by the volume of water already in the tank.

$V(t)$ is the volume of brine in the tank at time t. To find it we know that at t = 0 there were 200 gallons, 3 gallons are added and 2 are drained, and the net increase is 1 gallons per second. So,

$V(t)=200+t$

Therefore,

The rate at which $C(t)$ enters the tank is

$\frac{3}{200+t}$

The rate of the amount of salt leaving the tank is

$C\:\frac{lb}{gal} \cdot 2 \:\frac{gal}{min}+C\:\frac{lb}{gal} \cdot 1\:\frac{gal}{min}=3C\:\frac{lb}{min}$

and the rate at which $C(t)$ exits the tank is

$\frac{3C}{200+t}$

Plugging this information in the main equation, our differential equation model is:

$\frac{dC}{dt} =\frac{3}{200+t}-\frac{3C}{200+t}$

Since we are told that the tank starts out with 200 gal of solution, containing 100 lb of salt, the initial concentration is

$\frac{100 \:lb}{200 \:gal} =0.5\frac{\:lb}{\:gal}$

Next, we solve the initial value problem

$\frac{dC}{dt} =\frac{3-3C}{200+t}, \quad C(0)=\frac{1}{2}$

$\frac{dC}{dt} =\frac{3-3C}{200+t}\\\\\frac{dC}{3-3C} =\frac{dt}{200+t} \\\\\int \frac{dC}{3-3C} =\int\frac{dt}{200+t} \\\\-\frac{1}{3}\ln \left|3-3C\right|=\ln \left|200+t\right|+D\\\\$

We solve for C(t)

$C(t)=1+D(200+t)^{-3}$

D is the constant of integration, to find it we use the initial condition $C(0)=\frac{1}{2}$

$C(0)=1+D(200+0)^{-3}\\\frac{1}{2} =1+D(200+0)^{-3}\\D=-4000000$

So the concentration of the solution in the tank at any time t (before the tank overflows) is

$C(t)=1-4000000(200+t)^{-3}$

(a) The amount of salt in the tank at any time prior to the instant when the solution begins to overflow is just the concentration of the solution times its volume

$(1-4000000(200+t)^{-3})(200+t)\\\left(1-\frac{4000000}{\left(200+t\right)^3}\right)\left(200+t\right)$

(b) Since the tank can hold 500 gallons, it will begin to overflow when the volume is exactly 500 gal. We noticed before that the volume of the solution at time t is $V(t)=200+t$ . Solving the equation

$200+t=500\\t=300$

tells us that the tank will begin to overflow at 300 minutes. Thus the concentration at that time is

$C(300)=1-4000000(200+300)^{-3}\\\\C(300)= \frac{121}{125}\:\frac{lb}{gal}$

(c) If the tank had infinite capacity the concentration would then converge to,

$\lim_{t \to \infty} C(t)= \lim_{t \to \infty} 1-4000000\left(200+t\right)^{-3}\\\\\lim _{t\to \infty \:}\left(1\right)-\lim _{t\to \infty \:}\left(4000000\left(200+t\right)^{-3}\right)\\\\1-0\\\\1$

The theoretical limiting concentration if the tank has infinite capacity is $1\:\frac{lb}{gal}$

4 0

4 years ago

What is the area of the grassy yard?

guajiro [1.7K]

384 because 24•16=384

5 0

3 years ago

Read 2 more answers

A corporation has 11 manufacturing plants. Of these, seven are domestic and four are outside the United States. Each year a perf

nikdorinn [45]

Answer:

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

Step-by-step explanation:

Total plants = 11

Domestic plants = 7

Outside the US plants = 4

Suppose X is the number of plants outside the US which are selected for the performance evaluation. We need to compute the probability that at least 1 out of the 4 plants selected are outside the United States i.e. P(X≥1). To compute this, we will use the binomial distribution formula:

P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ

where n = total no. of trials

x = no. of successful trials

p = probability of success

q = probability of failure

Here we have n=4, p=4/11 and q=7/11

P(X≥1) = 1 - P(X<1)

= 1 - P(X=0)

= 1 - ⁴C₀ * (4/11)⁰ * (7/11)⁴⁻⁰

= 1 - 0.16399

P(X≥1) = 0.836

The probability that a performance evaluation will include at least one plant outside the United States is 0.836.

7 0

3 years ago

Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials.

Fofino [41]

Answer:

Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and the profit on every wrap is $3. Sal made a profit of $1,470 from lunch specials last month. The equation 2x + 3y = 1,470 represents Sal's profits last month, where x is the number of sandwich lunch specials sold and y is the number of wrap lunch specials sold.

5. Suppose Sal's total profit on lunch specials for the next month is $1,593. The profit amounts are the same: $2 for each sandwich and $3 for each wrap. In a paragraph of at least three complete sentences, explain how the graphs of the functions for the two months are similar and how they are different.

6. Below is a graph that represents the total profits for a third month. Write the equation of the line that represents this graph. Show your work or explain how you determined the equations.

Please help.

Please Help

Oct 24, 2015

Step-by-step explanation:

pls brainliest

7 0

2 years ago

Does (2, -1) make this a true statement? Why or why not? 3x + 2y = 4

kvv77 [185]

Answer: Yes

Step-by-step explanation:

So a coordinate pair is always set up (x,y) so you plug the x term in the coordinate pair into the x in the equation and the y term in for the y. 3×2 + 2×-1. Multiply them together and you end up with 4.

3 0

3 years ago