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3 years ago

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All of the edges of a moving box are of equal length. If the surface area is 864 square inches, what is the length; in inches of

each edge of the box?

1 answer:

andreyandreev [35.5K]3 years ago

6 0

Side makes 846 because side = area to the power of 2

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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 <em>Q(t)</em> 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 <em>Q(t)</em>.

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

Rate of change of <em>Q(t)</em> = Rate at which <em>Q(t)</em> enters the tank – Rate at which <em>Q(t)</em> exits the tank

where,

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

Rate at which <em>Q(t)</em> 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

3 years ago

a gas company in massachusetts charges $1.30 for 16.7 ft3 of natural gas. convert this rate to dollars per liter of gas

Katyanochek1 [597]

By using a change of units, we can see that the rate in dollars per liter is:

R = $0.0027 per liter.

<h3>How to convert the rate?</h3>

Here we have a rate in dollars per cubic foot, and we want to rewrite this in dollars per liter. So we just need to apply a change of units.

The original rate is in dollars per square foot, and it is given by:

R = $1.30 for 16.7ft^3

We know that the relation between cubic feet and liters is:

1ft^3 = 28.32 L

Then, if we apply a change of units, we can rewrite the volume as:

16.7ft^3 = 16.7*(28.32) L = 472.9 L

Then the rate is $1.30 per 472.9 L, taking the quotient we get:

R = ($1.30/472.9 L) = $0.0027 per liter.

If you want to learn more about changes of units:

brainly.com/question/9032119

#SPJ1

5 0

2 years ago

Use the correct function to find the missing angle y measurement to the

Doss [256]

Answer:

Cos;

26 degrees

Step-by-step explanation:

Recall: SOH CAH TOA

Reference angle = y°

Adjacent side = 18 in.

Hypotenuse = 20 in.

We would apply the trigonometric function CAH since we are dealing with the Adjacent side (A) and the Hypotenuse (H). Thus:

Cos y° = Adj/Hyp

Cos y° = 18/20

Cos y° = 0.9

y° = cos^{-1}(0.9)

y = 25.8419328° ≈ 26 degrees (nearest whole degree)

The answers are:

Cos and 26 degrees

4 0

3 years ago

Find the area A of the sector of a circle of radius 90 in formed by central angle 1/6 radian.

miv72 [106K]

Answer:

2121 $in^{2}$ Rounded to the nearest inch.

Step-by-step explanation:

A = $\pi r^{2}$ Is the formula for a circle, but we only have a portion of a circle so we will have to adjust the formula for a part of the circle.

A = $\frac{\frac{\pi }{6} }{2\pi }$ $\pi r^{2}$ The $\frac{\pi }{6}$ is the part of the circle and the whole circle is 2 $\pi$

$\frac{\frac{\pi }{6} }{2\pi }$ is the same as $\frac{\pi }{6}$ ÷ $\frac{2\pi }{1}$ Which is the same as $\frac{\pi }{6}$ x $\frac{1}{2\pi }$ Which is the same a $\frac{1}{12}$

A = $\frac{1}{12}$ $\pi$ $90^{2}$

A = $\frac{1}{12}$ (3.14)(90)(90) = 2119.5. When you use the $\pi$ on your calculator you get a slightly different number. This is due to rounding.

A = 2120.575041

5 0

2 years ago

what is the equation, in slope- intercept form, of the line that contains the points (-3,4) and (5,7)

gavmur [86]

Answer:

y = $\frac{3}{8}$ x + $\frac{41}{8}$

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

To calculate the slope use the gradient formula

m = ( y₂ - y₁ ) / ( x₂ - x₁ )

with (x₁, y₁ ) = (- 3, 4) and (x₂, y₂ ) = (5, 7)

m = $\frac{7-4}{5+3}$ = $\frac{3}{8}$ , hence

y = $\frac{3}{8}$ x + c ← is the partial equation

To find c substitute either of the 2 points into the partial equation

using (5, 7 ), then

7 = $\frac{15}{8}$ + c ⇒ c = 7 - $\frac{15}{8}$ = $\frac{41}{8}$

y = $\frac{3}{8}$ x + $\frac{41}{8}$ ← equation in slope-intercept form

6 0

3 years ago