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3 years ago

Is 7 more than, less than or equal to 24/4

Mathematics

1 answer:

vekshin13 years ago

5 0

Greater than since 24/4 is 6

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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

Abc = dfe find the value of x.

LiRa [457]

BC ≅ EF (CPCTC)

Set the two side lengths equal to each other.

38.4 = 5x - 10.6

Isolate the x. Add 10.6 to both sides

38.4 (+10.6) = 5x - 10.6 (+10.6)

5x = 49

Isolate the x, divide 5 from both sides

5x/5 = 49/5

x = 49/5

x = 9.8

9.8 should be your answer

hope this helps

8 0

3 years ago

Read 2 more answers

Help pleaseeeee :))) thanks!

stepladder [879]

Answer:

D. 8246

Step-by-step explanation:

3 0

3 years ago

Find the missing part. Use an improper fraction for your answer.

Pani-rosa [81]

Answer:

$x=\frac{15}{2}$ $z=\frac{35113}{2703}$ $a=\frac{15}{4}$ $y=\frac{35113}{5406}$ $b=\frac{45}{4}$

Step-by-step explanation:

To solve this, we will have to use multiple methods, including the pythagorean theorem and the 30 60 90 rule.

In the 30 60 90 rule, it is stated that the angle opposing the side with 30 degrees is represented by $s$ , the angle opposing the side with 60 is $s\sqrt{3}$ and the angle opposing 90 degree angle is represented by $2s$ . Looking at the whole triangle, we can see the 30, 60 and 90 degrees, and the angle with 90 has an opposing side of 15, which means we have to use 15=2s to find the other sides. When dividing 15 by 2, you get 7.5, which is s. Now we know that the angle opposing 30 is 15/2, which is x. Now that we have "s", we can also solve for z, represented by the 60 degrees. Multiply 7.5 and $\sqrt{3}$ to get z=35113/2703.

Now we can solve for the smaller triangles. Since we have already found x, we can use the 30 60 90 method in this too. The x (15/2) is now represented by 2s. Therefore, solve for your new s by doing this: 15/2 = 2s. You should get your 30 degree angle answer (a) with this, which is 15/4. Now that we have to use pythagorean theorem, which is $a^{2} + b^{2} = c^{2}$ to solve for the remaining side (y) of the small triangle. You should get 35113/5406.

Finally we have the last side left: b. Since we have so many sides now, all we have to use is the pythagorean theorem again. Use $b^2+y^2=z^2$ and solve for b. You should get 45/4.

8 0

3 years ago

What is the responsible estimate of 6207 divided 214

worty [1.4K]

Answer:

Step-by-step explanation:

We want to estimate 6207 / 214

Round 6207 to 6200

and 214 to 200

6200/200

62/2 = 31

We know that our estimate will be a little high since we are rounding the denominator down

5 0

3 years ago